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Biology (044) · Class XII · Chapter-wise Board Questions

CBSE Board papers 2022 to 2026, plus CBSE Sample Question Papers. Chapters as per CBSE Curriculum 2026-27. 902 questions. Click Reveal answer under any question to see the official marking scheme.

Marks
Unit VI · Reproduction

1. Sexual Reproduction in Flowering Plants

Q1.2026 Main 57/1/1 · Q11 mark

Identify the plant in which emasculation is not required for artificial hybridization.

  1. Rice
  2. Wheat
  3. Pea
  4. Papaya
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(D) Papaya [1 mark]

Q2.2026 Main 57/2/1 · Q101 mark

Given below are stages of embryo sac formation from megaspore mother cell. Identify the correct option for the cell division marked a, b, c and d.

Figure

  1. a – Meiosis-I, b – No division, c – Mitosis, d – Meiosis-II
  2. a – Meiosis-I, b – Meiosis-II, c – No division, d – Mitosis
  3. a – Mitosis, b – Meiosis-I, c – Meiosis-II, d – No division
  4. a – No division, b – Mitosis, c – Meiosis-I, d – Meiosis-II
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(B) a – Meiosis I, b – Meiosis II, c – No division, d – Mitosis [1 mark]

Q3.2026 Main 57/2/1 · Q131 mark

Assertion (A): Apomixis leads to production of genetically identical offsprings.

Reason (R): Apomixis involves fertilization between two genetically similar plants.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
  3. Assertion (A) is true, Reason (R) is false.
  4. Assertion (A) is false, Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(C) Assertion (A) is true, Reason (R) is false. [1 mark]

Q4.2026 Main 57/3/1 · Q121 mark

Select the correct option for the pair of plants pollinated by water.

  1. Vallisneria and Hydrilla
  2. Zostera and Water lily
  3. Water lily and Parthenium
  4. Water hyacinth and Water lily
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(A) Vallisneria and Hydrilla [1 mark]

Q5.2026 Main 57/4/1 · Q11 mark

Choose the option that gives the correct number of pollen grains that will be formed after 325 microspore mother cells undergo microsporogenesis.

  1. 325
  2. 650
  3. 1300
  4. 975
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(C) 1300 [1 mark]

Q6.2026 Main 57/4/1 · Q131 mark

Assertion (A): Cleistogamous flowers have assured seed set.

Reason (R): Cleistogamous flowers are invariably autogamous and do not depend on any pollinating agents.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A) [1 mark]

Q7.2026 Main 57/5/1 · Q11 mark

Identify the parts P, Q, R, S in the figure given below and select the option in the correct order.

Figure

PQRS
(A)SeedThalamusMesocarpEndocarp
(B)ThalamusSeedEndocarpMesocarp
(C)SeedThalamusEndocarpMesocarp
(D)ThalamusSeedMesocarpEndocarp
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(B) P – Thalamus, Q – Seed, R – Endocarp, S – Mesocarp [1 mark]

Q8.2026 Supp. 57/7/1 · Q11 mark

The undifferentiated sheath that encloses the radicle and root cap in a monocot embryo, is known as :

  1. Coleoptile
  2. Coleorrhiza
  3. Epiblast
  4. Scutellum
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(B) Coleorrhiza [1 mark]

Q9.2026 Supp. 57/7/1 · Q131 mark

Assertion (A): Apomictic embryos are genetically identical to the parent plant.

Reason (R): Apomixis is the production of seeds without fertilisation.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]

Q10.2025 Main 57/1/1 · Q71 mark

The diagram given below shows labelling of four parts of a dicot embryo during its development as P, Q, R and S.

Figure

Choose the option that indicates correct labelling of ‘P’, ‘Q’, ‘R’ and ‘S’ of embryo in different stages of its development :

  1. P – Egg; Q – Suspensor; R – Radicle; S – Cotyledon
  2. P – Zygote; Q – Suspensor; R – Cotyledon; S – Plumule
  3. P – Egg; Q – Radicle; R – Suspensor; S – Cotyledon
  4. P – Zygote; Q – Suspensor; R – Cotyledon; S – Radicle
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Answer not available: no official marking scheme published.

Q11.2025 Main 57/1/1 · Q101 mark

How many pollen grains and ovules are likely to be formed in the anther and the ovary of an angiosperm bearing 50 microspore mother cells and 50 megaspore mother cells respectively ?

  1. 100, 25
  2. 200, 50
  3. 50, 50
  4. 200, 100
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Answer not available: no official marking scheme published.

Q12.2025 Main 57/2/1 · Q11 mark

Given below is a diagram of T.S. of a monocot seed with parts I, II & III labelled :

Figure

Choose the option where parts I, II and III are identified correctly.

  1. I – Pericarp; II – Endosperm; III – Scutellum
  2. I – Pericarp; II – Endosperm; III – Coleorhiza
  3. I – Scutellum; II – Pericarp; III – Coleorhiza
  4. I – Coleorhiza; II – Scutellum; III – Pericarp
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Answer not available: no official marking scheme published.

Q13.2025 Main 57/4/1 · Q61 mark

Select the statements that are true for a typical dicotyledonous embryo from the given options.

  1. It consists of an embryonal axis and scutellum.
  2. The portion of embryonal axis above the level of cotyledon is epicotyl.
  3. The portion of embryonal axis below the level of cotyledon is coleorhiza.
  4. The lower end of the embryo has radicle covered with a root cap.

Choose the correct answer :

  1. (i) and (ii)
  2. (i) and (iii)
  3. (iii) and (iv)
  4. (ii) and (iv)
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Answer not available: no official marking scheme published.

Q14.2025 Main 57/5/1 · Q61 mark

Select the statements that are true for pollination mechanism in flowering plants from the given options.

  1. In Vallisneria, the female flowers are pollinated by pollen grains inside the water.
  2. In Zostera, pollen grains are released on the surface of water.
  3. In most of the water-pollinated species, pollen grains are covered by a mucilaginous coating.
  4. Pollination by water is quite rare and limited to about 30 genera.

Choose the correct answer :

  1. (i) and (ii)
  2. (ii) and (iii)
  3. (iii) and (iv)
  4. (i) and (iv)
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Answer not available: no official marking scheme published.

Q15.2025 Main 57/6/1 · Q71 mark

Select the statements that are true for the seed of angiosperm from the given options :

  1. Non-albuminous seeds have no residual endosperm.
  2. Residual, persistent nucellus in wheat is known as perisperm.
  3. Integuments of ovules harden as tough protective seed coat.
  4. Metabolic activity of the embryo slows down in dormancy.

Choose the correct option :

  1. (i) and (ii)
  2. (ii) and (iii)
  3. (iii) and (iv)
  4. (i) and (iv)
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Answer not available: no official marking scheme published.

Q16.2025 Main 57/7/1 · Q41 mark

Given below are the events that are observed in an artificial hybridisation programme. Arrange them in the correct sequential order and select the correct option.

  1. Re-bagging
  2. Selection of parents
  3. Bagging
  4. Dusting the pollens on stigma
  5. Emasculation
  6. Collection of pollens from male parent

Options :

  1. (ii), (iii), (v), (vi), (iv), (i)
  2. (ii), (v), (iii), (vi), (iv), (i)
  3. (v), (ii), (iii), (vi), (i), (iv)
  4. (ii), (iii), (vi), (iv), (v), (i)
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Answer not available: no official marking scheme published.

Q17.2025 Main 57/7/1 · Q61 mark

Choose the option that correctly describes the gynoecium of Michelia :

  1. Multicarpellary, Apocarpous
  2. Bicarpellary, Apocarpous
  3. Multicarpellary, Syncarpous
  4. Bicarpellary, Syncarpous
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Answer not available: no official marking scheme published.

Q18.2025 Supp. 57/S/1 · Q21 mark

Identify the various parts, ‘a’, ‘b’ and ‘c’, in the given diagram of ovule :

Figure

  1. a – Embryo sac, b – Inner integuments, c – Outer integuments
  2. a – Inner integuments, b – Nucellus, c – Chalazal end
  3. a – Hilum, b – Funicle, c – Embryo sac
  4. a – Mycropylar end, b – Hilum, c – Inner integuments
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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(C) a – Hilum, b – Funicle, c – Embryo sac [1 mark]

Q19.2025 Supp. 57/S/1 · Q141 mark

Assertion (A): All the events starting from pollen landing on stigma till pollen tube entering the ovule are included in pollen-pistil interaction.

Reason (R): Pollen-pistil interaction ensures that only suitable pollens germinate on the stigma.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q20.2024 Main 57/1/1 · Q11 mark

In which of the following plants are both male and female flowers born on the same plant and the mode of pollination can be geitonogamy or xenogamy ?

  1. Papaya
  2. Date Palm
  3. Maize
  4. Spinach
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(C) Maize [1 mark]

Q21.2024 Main 57/2/1 · Q11 mark

An angiosperm embryo sac is located within the :

  1. Placenta
  2. Megasporangium
  3. Nucellus
  4. Ovary
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(C) Nucellus [1 mark]

Q22.2024 Main 57/3/1 · Q11 mark

The part of the ovule that develops into protective coats of a seed after fertilization in a typical flowering plant is :

  1. embryo sac
  2. nucellus
  3. integuments
  4. megaspore
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(C) integuments [1 mark]

Q23.2024 Main 57/3/1 · Q41 mark

In which one of the following floral plants are many embryos formed in the seeds without fertilisation of the egg cell ?

  1. Black pepper
  2. Mustard
  3. Groundnut
  4. Citrus
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(D) Citrus [1 mark]

Q24.2024 Main 57/3/1 · Q151 mark

Assertion (A): The zygote gives rise to heart-shaped embryo and subsequently proembryo in most angiosperms.

Reason (R): The zygote is present at the micropylar end of the embryo sac and develops into an embryo.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(D) Assertion (A) is false, but Reason (R) is true. [1 mark]

Q25.2024 Main 57/4/1 · Q11 mark

In a fertilized ovule of an angiosperm, the cells in which n, 2n and 3n conditions respectively occur are :

  1. antipodal, zygote and endosperm
  2. zygote, nucellus and endosperm
  3. endosperm, nucellus and zygote
  4. antipodals, synergids and integuments
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(A) antipodal, zygote and endosperm [1 mark]

Q26.2024 Main 57/4/1 · Q61 mark

Study the following diagram of Transverse Section of a young anther of an angiosperm :

Figure

Select the option where parts ‘A’, ‘B’ and ‘C’ are correctly identified.

  1. A – Connective, B – Endothecium, C – Pollen grain.
  2. A – Endothecium, B – Connective, C – Pollen grain.
  3. A – Pollen grain, B – Connective, C – Endothecium.
  4. A – Endothecium, B – Pollen grain, C – Connective.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(A) A – Connective, B – Endothecium, C – Pollen grain [1 mark]

Q27.2024 Main 57/5/1 · Q71 mark

A phenomenon where a male insect mistakenly identified the patterns of a orchid flower as the female insect partner, and tries to copulate and thereby pollinates the flower is said to be :

  1. Pseudocopulation
  2. Pseudopollination
  3. Pseudoparthenocarpy
  4. Pseudofertilisation
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(A) Pseudocopulation [1 mark]

Q28.2024 Main 57/5/1 · Q81 mark

Identify the correct labellings in the figure of a fertilised embryo sac of an angiosperm given below :

Figure

  1. A – zygote, B – degenerating synergids, C – degenerating antipodals, D – PEN
  2. A – degenerating synergids, B – zygote, C – PEN, D – degenerating antipodals
  3. A – degenerating antipodals, B – PEN, C – degenerating synergids, D – zygote
  4. A – degenerating synergids, B – zygote, C – degenerating antipodals, D – PEN
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(B) A – degenerating synergids, B – zygote, C – PEN, D – degenerating antipodals [1 mark]

Q29.2024 Supp. 57/S/1 · Q11 mark

Long ribbon-like pollen grains are seen in some :

  1. Aquatic plants
  2. Wind-pollinated grasses
  3. Gymnosperms
  4. Bird-pollinated flowers
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(A) Aquatic plants [1 mark]

Q30.2023 Main 57/1/1 · Q81 mark

Choose the option that gives the correct number of pollen grains that will be formed after 325 microspore mother cells undergo microsporogenesis.

  1. 325
  2. 650
  3. 1300
  4. 975
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(C) 1300 [1 mark]

Q31.2023 Main 57/1/1 · Q151 mark

Assertion (A): Endosperm is completely consumed during the development of embryo in ex-albuminous seeds.

Reason (R): Castor, pea and beans are all examples of ex-albuminous seeds.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(C) A is true, but R is false. [1 mark]

Q32.2023 Main 57/3/1 · Q61 mark

Which of the following seeds have remained alive for the longest period ?

  1. Phoenix dactylifera
  2. Striga asiatica
  3. Mangifera indica
  4. Yucca gigantea
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(A) Phoenix dactylifera [1 mark]

Q33.2023 Main 57/5/1 · Q121 mark

Select the option that shows the correctly identified ‘U’, ‘X’, ‘Y’ and ‘Z’ in a developing dicot embryo.

Figure

  1. X – Plumule (2n), Y – Suspensor (n), Z – Cotyledon (2n), U – Radicle (2n).
  2. X – Plumule (2n), Y – Suspensor (n), Z – Radicle (2n), U – Cotyledon (2n).
  3. X – Suspensor (2n), Y – Cotyledon (2n), Z – Radicle (2n), U – Plumule (2n).
  4. X – Cotyledon (2n), Y – Radicle (n), Z – Plumule (n), U – Suspensor (n).
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(C) X – Suspensor (2n), Y – Cotyledon (2n), Z – Radicle (2n), U – Plumule (2n). [1 mark]

Q34.2023 Supp. 57/C/1 · Q11 mark

Select the option that correctly matches with the labelling in the given diagram of T.S. of an apple which categorizes it as a false fruit.

Figure

  1. P – Thalamus
  2. Q – Seed
  3. R – Endocarp
  4. S – Mesocarp
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(A) P – Thalamus [1 mark]

Q35.2023 Supp. 57/C/1 · Q161 mark

Assertion (A): Banana is a parthenocarpic fruit.

Reason (R): It develops only from the ovary.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). [1 mark]

Q36.SQP 2025-26 · Q11 mark

The male gametes in plants are formed by:

  1. Mitotic division of nucleus of vegetative cell
  2. Meiotic division of nucleus of vegetative cell
  3. Mitotic division of nucleus of generative cell
  4. Meiotic division of nucleus of generative cell
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Official CBSE Sample Paper Marking Scheme · 2025-26

(C) Mitotic division of nucleus of generative cell [1 mark]

Q37.SQP 2026-27 · Q11 mark

A researcher observes that in a particular flowering plant, the pistil matures several days before the stamens in all flowers of the plant. What is the most significant consequence of this timing difference?

  1. It guarantees self-pollination
  2. It promotes autogamy
  3. It ensures cross-pollination
  4. It leads to geitonogamy
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Official CBSE Sample Paper Marking Scheme · 2026-27

(C) It ensures cross-pollination [1 mark]

Q38.SQP 2025-26 · Q21 mark

The primary endosperm nucleus is formed by fusion of which of the following?

  1. A male gamete and a female gamete
  2. A male gamete and two polar nuclei
  3. A female gamete and two synergids
  4. Two male gametes and an egg cell
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Official CBSE Sample Paper Marking Scheme · 2025-26

(B) A male gamete and two polar nuclei [1 mark]

Q39.SQP 2025-26 · Q131 mark

Assertion (A): The ability of the pistil to recognise the pollen is the result of a continuous dialogue between pollen and pistil.

Reason (R): This electrical dialogue allows only compatible pollen to germinate.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true and R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
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Official CBSE Sample Paper Marking Scheme · 2025-26

(C) A is true but R is false [1 mark]

Q40.2026 Main 57/1/1 · Q182 marks

(a) (i) How does a farmer use dormancy of seed to his advantage ?

(ii) Differentiate between pea seed and castor seed.

OR

(b) Identify the stage of follicle where primary oocyte undergoes first meiotic division.

Also mention the products of this division.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) (i) Dormancy is crucial for storage (dehydration) of seeds to be used to raise the crops in next season / seeds are used as source of food / for commercial purpose [1]

(ii)

Pea seedCastor seed
Non endospermic seedEndospermic seed
Non-albuminous seedAlbuminous seed
No endosperm is left in the seed as it is fully consumed during embryo development.It contains some amount of endosperm as it is not consumed completely during embryo development.

(Any one difference) [1]

OR

(b) • Tertiary follicle in the ovary [1]

• Secondary oocyte and 1st polar body. [½ + ½]

[Total: 2 marks]

Q41.2026 Main 57/4/1 · Q182 marks

(a) Write one way by which apomictic seeds can be developed.

(b) Why do farmers prefer apomictic seeds over hybrid seeds ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) In some species (Asteraceae and grasses) the diploid egg cell is formed without reduction division and develop into an embryo without fertilization such are apomictic seeds. / Some of the nucellar cells surrounding the embryo sac starts dividing and protrude into the embryo sac and develop into an embryo. [1]

(b) In Apomictic seeds there will be no segregation of the hybrid characters in the progeny plants so farmers can keep on using the hybrid seeds to raise new crop year after year / If hybrid seeds are made apomictic the farmer need not to buy hybrid seeds every year hence they are cost effective. [1]

[Total: 2 marks]

Q42.2026 Main 57/5/1 · Q172 marks

(a) Mention the functions of each of the following :

  1. Tassels of Corn Cob
  2. Scutellum

OR

(b) Farmers prefer apomictic seeds over hybrid seeds. Give any two reasons.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a) (i) Award marks if attempted. [1]

(ii) It provides nourishment to the embryo. [1]

OR

(b) Production of hybrid seed is very costly. If hybrid seeds are made into apomictic there is no segregation of characters of the hybrid progeny, farmers need not to buy hybrid seeds every year, use of apomictic seeds would be cost effective, any other correct reason (any two correct reasons) [1+1]

[Total: 2 marks]

Q43.2026 Supp. 57/7/1 · Q172 marks

Attempt either (a) or (b).

(a) How is it possible in Oxalis and Viola plants, to produce assured seed sets even in the absence of pollinators ?

OR

(b) Write the ploidy of the cells of the nucellus, MMC (Megaspore Mother Cell), the functional megaspore and female gametophyte.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) Oxalis and Viola produce cleistogamous flowers / flowers that do not open at all even at maturity / closed flowers [1]

These flowers are bisexual in which the anthers and stigma lie close to each other leading to fertilisation to produce assured seed sets even in absence of pollinators. [1]

OR

(b) Nucellus – 2n / diploid; MMC – 2n / diploid; functional megaspore – n / haploid; female gametophyte – n / haploid [½ × 4]

[Total: 2 marks]

Q44.2025 Main 57/2/1 · Q172 marks

Student to attempt either option (A) or (B) :

(A) (i) Write two crucial changes, the seed undergoes while reaching maturity that enable them to be in a viable state until the onset of favourable conditions.

(ii) Name the oldest viable seed excavated from Arctic Tundra as per the records.

OR

(B) (i) Pea flower produce assured seed sets. Give reason.

(ii) In case of Polyembryony, an embryo ‘P’ develops from a synergid and the embryo ‘Q’ develops from the nucellus. State the ploidy of embryo ‘P’ and ‘Q’.

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Answer not available: no official marking scheme published.

Q45.2025 Supp. 57/S/1 · Q172 marks

(a) How many meiotic divisions will a wheat plant undergo in order to produce 100 zygotes /100 grains ? Calculate and give explanation.

OR

(b) How many pollen grains and ovules are likely to be formed in the anther and the ovary of an angiosperm bearing 35 microspore mother cells and 35 megaspore mother cells respectively ?

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(a) To produce a zygote one megaspore mother cell will undergo one meiotic division to produce one ovule as it shows megasporic development therefore 100 meiotic divisions will take place to produce 100 zygotes, [½]

One microspore mother cell undergoes one meiotic division to produce four pollen grains therefore 25 meiotic divisions will take place to produce 100 zygotes or 100 pollen grains [½]

Total meiotic divisions 100 + 25 = 125 [1]

OR

(b) Microspore mother cells = 35; so, no. of pollen grains = 35 × 4 = 140 [1]

Megaspore mother cells = 35; so, no. of ovules = 35 (monosporic development) [1]

[Total: 2 marks]

Q46.2025 Supp. 57/S/1 · Q182 marks

Given below is a transverse sectional view of an apple. Label the following parts (i) and (ii) and write down their functions.

Figure

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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q47.2024 Main 57/1/1 · Q172 marks

Identify A, B, C and D in the table given below :

TermsPart of the plant it represents
Pericarp‘A’
‘B’Cotyledon in seed of grass family
Embryonal axis‘C’
‘D’Remains of nucellus in a seed
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1
  • A – Wall of fruit / Wall of ripened ovary
  • B – Scutellum
  • C – Radicle / Plumule / epicotyl / hypocotyl
  • D – Perisperm

[½ × 4 = 2 marks]

Q48.2024 Main 57/5/1 · Q172 marks
  1. “Farmers prefer apomictic seeds to hybrid seeds.” Justify giving two reasons.

OR

  1. Mention one advantage and one disadvantage of amniocentesis.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a) Any two: hybrid seeds show segregation of characters in the progeny; production of hybrid seeds is expensive / apomictic seeds would be cheaper if produced; hybrid seeds have to be produced every year; apomictic seeds bring homozygosity / apomictic seeds retain desirable characteristics of plants. [1 × 2]

OR

(b) Advantage – used to diagnose any chromosomal abnormality or genetic disorder such as Down syndrome, haemophilia, sickle cell anaemia in foetus (any one disease), determine survivability of foetus. (Any one point) [1]

Disadvantage – used to determine the sex of the foetus which may lead to female foeticide. [1]

[Total: 2 marks]

Q49.2023 Main 57/1/1 · Q182 marks
  1. From which end of the ovule, and how does the pollen tube gain its entry into the embryo sac of a Hibiscus flower ?
  2. State the fate of the male nuclei present in the pollen tube.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) From micropylar end, through the synergids (filiform apparatus)/filiform apparatus (within synergids) guides the entry of pollen tube. [½×2]

(b) One male nucleus fuses with two polar nuclei to form primary endosperm nucleus and termed triple fusion, other male nucleus fuses with egg cell nucleus to form zygote i.e. undergoes syngamy. [½×2]

[Total: 2 marks]

Q50.2023 Main 57/3/1 · Q172 marks
  1. Explain the process of the development of a male gametophyte in an angiosperm.
  2. Why is it called a male gametophyte ?
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) Cells of sporogenous tissue/Microspore mother cell/Pollen mother cell (PMC) in anther undergoes meiotic division, to form microspore tetrad which mature and dissociate to form pollen grains or male gametophyte. [½×2]

(b) Because its generative cell divides, to form two male gametes. [½×2]

[Total: 2 marks]

Q51.2023 Supp. 57/C/1 · Q172 marks

The male and female flowers of corn plant (maize) are well adapted for pollination by wind. Justify.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

Features of male flowers: Well exposed stamen, so that pollens are easily dispersed into wind current / light and non-sticky pollen grains, so that they are easily transported in wind current. [½ + ½]

Features of female flowers: Large often feathery stigma and style wave in the wind, to easily trap air borne pollen grains / numerous flowers packed into an inflorescence, to easily trap air borne pollen grains. [½ + ½]

(Or any other features with correct justification)

[Total: 2 marks]

Q52.SQP 2025-26 · Q172 marks

Attempt either option A or B.

A. During artificial hybridization, it is important to ensure that only desired pollen grains are used for pollination. How is it ensured?

OR

B. Why is manual transfer of desired pollen grains relatively easy in pea plants for hybridization, but in wheat, pollen grains are often utilized from pollen banks for hybridization purposes?

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Official CBSE Sample Paper Marking Scheme · 2025-26

A. It is achieved by emasculation and bagging techniques.

If the female parent bears bisexual flowers, removal of anthers from the flower bud before the anther dehisces using a pair of forceps is necessary. This step is referred to as emasculation. [1]

Emasculated flowers have to be covered with a bag of suitable size, generally made up of butter paper, to prevent contamination of its stigma with unwanted pollen. This process is called bagging. [1]

(1 × 2 = 2)

OR

B. In pea plants, manual emasculation is easier and pollen grains are viable for months as it belongs to Leguminosae. [1]

In wheat flowers (spikelets) pollen grain loses its viability half an hour after release. [1]

(1 × 2 = 2)

Q53.SQP 2026-27 · Q172 marks

Attempt either option A or B.

A. Geitonogamy can be compared with cross-pollination as well as self-pollination. Comment.

OR

B. Mohit wants to produce hybrid seeds in a plant in his field in which male and female flowers mature at the same time and are close together. Suggest one strategy to promote only desired cross pollination and justify your answer.

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Official CBSE Sample Paper Marking Scheme · 2026-27

A. Geitonogamy involves transfer of pollen grains from the anther to the stigma of another flower of the same plant. [1]

Although geitonogamy is functionally cross-pollination involving pollinating agents, genetically it is similar to autogamy since the pollen grains come from the same plant. [1]

OR

B. Mohit can emasculate (remove anthers) from selected flowers before pollen release and then manually pollinate them with pollen from the chosen male parent, followed by bagging. [1]

This prevents self-pollination and unwanted pollen and ensures that only the desired cross produces seeds. [1]

Q54.2026 Main 57/1/1 · Q223 marks

(a) Do all pollen grains remain viable for the same length of time ? Support your answer with two suitable examples.

(b) How are pollen grains stored in pollen banks ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) • No [½]

• In cereals or rice or wheat pollen grains lose viability within 30 minutes of their release [1]

• In some members of Rosaceae or Leguminoseae or Solanaceae – they maintain viability for months [1]

• or any other correct example

(b) They are stored using cryo-preservation techniques / in liquid nitrogen at –196°C [½]

[Total: 3 marks]

Q55.2026 Main 57/1/1 · Q253 marks

Draw a neat diagram of megasporangium of an angiosperm and label any six parts.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

Figure

(Label any six parts) [½ × 6]

[Total: 3 marks]

Q56.2026 Main 57/2/1 · Q243 marks

Draw a neat diagram of enlarged view of one microsporangium of a Young anther of an angiosperm. Label the different layers of the wall and write the functions of the outermost and innermost layer.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

Figure

(Award ½ mark for each correct labelling) [½ × 4]

  • Outermost layer: Epidermis helps in protection / in dehiscence of anther to release the pollens. [½]
  • Innermost layer: Tapetum provides nourishment to the developing pollen grains [½]

[Total: 3 marks]

Q57.2026 Main 57/3/1 · Q223 marks

Answer the following questions :

(a) Explain the different ways by which apomictic seeds can be developed. [2]

(b) Mention one advantage of apomictic seeds for farmers. [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a)

  • In some species (Asteraceae and grasses) the diploid egg cell is formed without reduction division and develop into an embryo without fertilization [1]
  • Some of the nucellar cells surrounding the embryo sac starts dividing and protrude into the embryo sac and develop into an embryo [1]

(b) In Apomictic seeds there will be no segregation of the hybrid characters in the progeny plants so farmers can keep on using the hybrid seeds to raise new crop year after year / If hybrid seeds are made apomictic the farmer need not to buy hybrid seeds every year hence they are cost effective [1]

[Total: 3 marks]

Q58.2026 Main 57/3/1 · Q263 marks

Draw a neat diagram of a maize grain showing the internal structure and label any five parts.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

Figure

(Diagram = ½, Any correct labelling = ½ × 5)

[Total: 3 marks]

Q59.2026 Main 57/4/1 · Q223 marks

Differentiate between wind-pollinated and insect-pollinated flowers. Give an example of each type.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(Any four differences) [½ × 4]

Wind pollinated flowersInsect Pollinated flowers
Flowers are dull or not colourful.Flowers are colourful.
Flowers do not produce nectar.Flowers produce nectar.
Flowers are not fragrant.Flowers are fragrant.
Flowers are generally of small size.Flowers are of large size.
Pollen grains are produced in large numbers.Pollen grains are not produced in large numbers.
Pollen grains are light weight and non sticky.Pollen grains are heavy and sticky.
Often have feathery stigma to easily trap air borne pollensStigma is often non feathery.

(Any other correct difference)

Example: (Any one) [½+½]

Wind pollinated flowersInsect Pollinated flowers
Maize, Cannabis, GrassesYucca, Sunflower

(Any other correct example)

[Total: 3 marks]

Q60.2026 Main 57/5/1 · Q223 marks

List any three outbreeding devices that flowering plants have developed for cross-pollination and explain how they help to encourage cross-pollination.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1
  • Self-incompatibility is the genetic mechanism which prevents self-pollination by inhibiting the germination of pollen from same flower or different flower of same plant on stigma.
  • Unisexual flowers have no chance for self-pollination
  • Receptivity of male and female parts: Pollen release and stigma receptivity are not synchronised.
  • Position of male and female parts: The anther and stigma are placed at different positions so that the pollen cannot come in contact with the stigma of the same flower.
  • Unisexual plants have no chance of self-pollination.
  • Any other correct outbreeding device.

(Any three) [1×3 = 3 marks]

Q61.2026 Supp. 57/7/1 · Q223 marks

(a) Draw a diagram of L.S. of an anatropous ovule of an angiosperm. Label the following parts :

  1. Nucellus
  2. Outer integument
  3. Antipodal cells
  4. Polar nuclei

(b) How does endosperm in angiosperms become triploid ?

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a)

Figure

(½ × 4)

(b) Endosperm is triploid because it is developed after triple fusion of one male gamete with two polar nuclei / develops by fusion of three haploid nuclei. [1]

[Total: 3 marks]

Q62.2025 Main 57/2/1 · Q223 marks

(a) A bilobed dithecous anther has 200 microspore mother cells per microsporangium. How many male gametophytes can be produced by this anther ?

(b) Write the composition of intine and exine layers of a pollen grains.

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Answer not available: no official marking scheme published.

Q63.2025 Main 57/4/1 · Q263 marks

Given below is a flower with its characteristic features specialised for the most common type of abiotic pollination.

Figure

Answer the following questions based on the above diagram :

  1. Name the mode of abiotic pollination that will be adopted by the given plant species in the above picture.
  2. State the need of exposed large feathery stigmas for the flower.
  3. What will be the two important adaptations in the pollen grains of the flowers pollinated by the above mode of pollination ?
  4. What could be the probable reason for the petals being small and non-green ?
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Answer not available: no official marking scheme published.

Q64.2025 Main 57/5/1 · Q283 marks

Many of the flowering plants producing hermaphrodite flowers have developed many devices to discourage self-pollination and to encourage cross-pollination. Given below is a picture of one such outbreeding device in a flowering plant. Study the picture and answer the questions that follow :

Figure

  1. Explain how the given type of pollination is advantageous to the plant.
  2. Can this flowering plant show geitonogamy ? Justify your answer.
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Answer not available: no official marking scheme published.

Q65.2025 Main 57/6/1 · Q223 marks

Flowering plants with hermaphrodite flowers have developed many reproductive strategies to ensure cross-pollination. Study the given outbreeding devices adopted by certain flowering plants and answer the questions that follow.

Stigma \ Pollen grainsPollen grains of Plant APollen grains of Plant BPollen grains of Plant C
Stigma of Plant A×✓✓
Stigma of Plant B✓×✓
Stigma of Plant C✓✓×

Note :

All plants belong to the same species.

× – No pollen tube growth/inhibition of pollen germination on stigma.

✓ – Pollen germination on stigma.

  1. Name and define the outbreeding device described in the above table.
  2. Explain what would have been the disadvantage to the plant in the absence of the given strategy.
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Answer not available: no official marking scheme published.

Q66.2024 Main 57/3/1 · Q223 marks

Draw a longitudinal section of pistil of a flower showing growth of the pollen tube. Label the part :

  1. through which the pollen tube moves down.
  2. the cell wherein the pollen tube releases its contents.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

Figure

(a) Style [or Stigma – any one]   (b) Synergid

(Note: 1 mark for correct diagram and 1 mark each for correct labelling) [1 + 1 + 1]

[Total: 3 marks]

Q67.2024 Main 57/4/1 · Q223 marks

Draw a well labelled diagram of sectional view of male gametophyte/microspore of an angiosperm and write the functions of any two parts labelled. (Any four labels).

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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

Figure

[Vegetative cell, generative cell, exine, intine, germ pore] (Any four labelling to be considered) [½ × 4]

  • Vegetative cell – contains abundant food reserve for pollen germination
  • Generative cell – divides to produce male gametes
  • Exine – protective layer
  • Intine – involved in the formation of pollen tube
  • Germ pore – pollen tube emerges from this point [½ × 2]

[Marks to be given to two correct labelling with their respective correct function only]

[Total: 3 marks]

Q68.2024 Main 57/5/1 · Q253 marks

If the cells in the leaves of a maize plant contain 10 chromosomes each, write the number of chromosomes in its endosperm and zygote. Name and explain the process by which an endosperm and a zygote are formed in maize.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1
  • Endosperm – 15 chromosomes [½]
  • Zygote – 10 chromosomes [½]
  • Endosperm is formed by the process of triple fusion, the fusion of two polar nuclei with one male gamete, [½ + ½]
  • Zygote is formed by the process of syngamy, the fusion of egg cell with a male gamete. [½ + ½]

[Total: 3 marks]

Q69.2023 Main 57/5/1 · Q253 marks

One of the major approaches of crop improvement programme is Artificial Hybridisation. Explain the steps involved in making sure that only the desired pollen grain pollinate the stigma of a bisexual flower by a plant breeder.

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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1
  • Emasculation: Removal of the anther from the flower bud before the anther dehisces.
  • Bagging: Covering the emasculated flower with a bag of suitable size to prevent contamination of its stigma with unwanted pollen.
  • Dusting of desired pollen on the stigma and re-bagging.

[1×3] (Half mark each if only name of process Emasculation and Bagging is mentioned)

[Total: 3 marks]

Q70.2023 Supp. 57/C/1 · Q223 marks

(a) Draw a neat diagram of a mature angiospermic embryo sac and label any four cellular components.

(b) Write the function of filiform apparatus.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a)

Figure

(½ mark each for any 4 correct labellings) [½ × 4]

(b) Guide pollen tube into synergids. [1]

[Total: 3 marks]

Q71.SQP 2026-27 · Q233 marks

In the post-fertilization embryo sac shown in figure, structure 'P' results from the fusion of one male gamete with the egg cell, and structure 'Q' results from the fusion of the other male gamete with the polar nuclei.

Figure

  1. Identify structure 'P' and structure 'Q' and state their ploidy.
  2. Evaluate the following statement: “The division of 'P' always precedes the division of 'Q'.” Justify your evaluation based on the nutritional requirements for seed development.
  3. Which characteristic feature of endosperm helps in its nutritive role?
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Official CBSE Sample Paper Marking Scheme · 2026-27

A. Structure P (Zygote) is diploid and Structure Q (Primary Endosperm Nucleus) is triploid. [1]

B. The statement is False. The Primary Endosperm Nucleus (PEN/'Q') divides much earlier than zygote ('P') to form the endosperm tissue for nutrition of embryo. [1]

C. The cells of endosperm tissue are filled with reserve food materials which act as the primary source of nutrition for the developing embryo. [1]

Q72.2025 Main 57/7/1 · Q304 marks

The following question is based on pollination. Study the figures carefully and answer the questions that follow.

Figure

  1. Give the scientific terms for the processes taking place in Figures A and B respectively.
  2. Mention two conditions necessary for the process occurring in Figure B.
  3. (i) State one advantage and one disadvantage of the process occurring in Figure B.

OR

  1. (ii) Name one plant where, in some flowers only, the process in Figure B takes place and give the reason responsible for it.
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Answer not available: no official marking scheme published.

Q73.SQP 2025-26 · Q294 marks

Given below is a set of information about some fruits and seeds.

FruitFruit and seed formation
PNucellar cells surrounding the embryo sac develop into embryos.
QOvary develops into the fruit by the application of growth hormones.
RThalamus contributes to fruit formation.
SOvary matures into a fruit after fertilisation.

On the basis of the information provided above, answer the following questions with justification for each answer.

A. How many embryo sacs will be present in each ovule of S before maturation and how many egg(s) will be present in each embryo sac when the embryo sac is developed from a single megaspore? (1)

B.

  1. Which of these fruits exhibits polyembryony? Will their embryos exhibit genetic variation? Justify.
  2. What will be ploidy of the embryonic cells in the above case? (2)

Attempt either subpart C or D.

C. Which of these fruits can be considered as parthenocarpic? Give a reason. (1)

OR

D. Which of the fruits P, Q, R or S is a true fruit with seeds? Give reason. (1)

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Official CBSE Sample Paper Marking Scheme · 2025-26

A. One embryo sac is present in each ovule and one egg is present in each embryo sac when the embryo sac is developed from a single megaspore. [1]

B.

(i) P exhibits polyembryony due to occurrence of more than one embryo in a seed. Embryos developed from nucellar cells by apomixis, a form of asexual reproduction, don't show genetic variation. [1]

(ii) Cells of embryos developed from diploid nucellar cells are diploid (2n) as these are formed by apomixis, a form of asexual reproduction. [1]

Student to attempt either subpart C or D.

C. Q, because this fruit is developed without fertilisation and will thus be seedless. [1]

OR

D. Fruit S is a true fruit with seeds. True fruits develop from the ripened ovary after fertilisation and fertilised ovules mature into seeds. [1]

Q74.2026 Main 57/2/1 · Q315 marks

(a) (i) Explain the role of stigma in pollen pistil interaction. [1]

(ii) Describe the post pollination events leading to double fertilization in angiosperms starting with 2-celled pollen grain. [4]

OR

(b) (i) Explain menstrual phase in the menstrual cycle of human female. [1]

(ii) Why is follicular phase also called proliferative phase ? [1]

(iii) Explain the events that occur in Graafian follicle at the time of ovulation and thereafter. [2]

(iv) Draw a Graafian follicle and label – Antrum and secondary oocyte. [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) (i) Stigma has the ability to recognise the right or compatible and wrong or incompatible type of pollen [1]

(ii) [1 × 4]

  • Compatible pollen grain germinates on the stigma to produce pollen tube that grows through the tissues of stigma and style and reaches to the ovary
  • Generative cell divides into two male gametes while moving through style and Pollen tube enters the ovule through the micropyle and then enters one of the synergids through the filiform apparatus
  • The pollen tube releases the two male gametes in the cytoplasm of Synergid and one male gamete moves towards the egg cell and fuses with its nucleus forming a zygote (syngamy)
  • The other male gamete moves towards the two polar nuclei present in the central cell and fuses with them to form Primary Endosperm Nucleus / PEN (triple fusion). (Since syngamy and triple fusion both occur in embryo sac so it is termed as Double Fertilization)

OR

(b) (i) During menstrual phase in the menstrual cycle of human female endometrium lining of the uterus and its blood vessels which forms liquid comes out through vagina and flow lasts for 3-5 days. [1]

(ii) As the regeneration of endometrium through proliferation takes place. [1]

(iii) Graafian follicle ruptures to release secondary oocyte, remaining part of graafian follicle transforms into corpus luteum [1+1]

(iv)

Figure

[½+½]

[Total: 5 marks]

Q75.2025 Main 57/1/1 · Q325 marks

Student to attempt either option-(A) or (B) :

(A) (a) Distinguish between the two cells enclosed in a mature male gametophyte of an angiosperm.

(b) Study the diagram given below showing the modes of pollination. Answer the questions that follow.

Figure

  1. The given diagram shows three methods of pollen transfer in plants. Examine them carefully and write the technical terms used for pollen transfer methods ‘1’, ‘2’ and ‘3’.
  2. How do the following plants achieve pollination successfully ?
    1. Water lily
    2. Vallisneria
  3. Write advantages of pollen transfer in method ‘3’.

OR

(B) Given below is the diagram of human ovum surrounded by a few sperms. Observe the diagram and answer the questions that follows :

Figure

  1. Compare the fate of sperms ‘P’, ‘Q’ and ‘R’ shown in the diagram.
  2. Write the role of Zona pellucida in this process.
  3. Analyse the changes occurring in the ovum after the entry of sperm.
  4. How acrosome and middle piece of a human sperm are able to play an important role in human fertilization ?
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Answer not available: no official marking scheme published.

Q76.2025 Main 57/2/1 · Q315 marks

Student to attempt either option (A) or (B).

(A) (i) Describe the process of megasporogenesis in an angiosperm.

(ii) Draw a diagram of a mature embryo sac of an angiosperm. Label its any four parts.

OR

(B) The reproductive cycle in the female primates is called menstrual cycle. The first menstruation begins at puberty.

Answer the following questions :

  1. Name the four phases of menstrual cycle in a proper sequence.
  2. How long does the menstrual phase last in a menstrual cycle ?
  3. When and why hormones estrogen and progesterone reach their peak levels respectively, in the menstrual cycle ?
  4. Give the significance of LH surge.
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Answer not available: no official marking scheme published.

Q77.2025 Main 57/4/1 · Q315 marks
    1. Explain how does double fertilisation take place in a flowering plant.
    2. Write the fate of the products of double fertilization in these plants.

OR

    1. Explain the structure of testicular lobules in human male reproductive system. Name the two types of cells present in the seminiferous tubules and state their role.
    2. Describe the role of hypothalamic hormone GnRH in spermatogenesis.
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Answer not available: no official marking scheme published.

Q78.2025 Main 57/5/1 · Q335 marks
    1. Explain the structure of a typical monocotyledonous embryo of a flowering plant.
    2. How are multiple embryos formed in a citrus fruit ? What is the mechanism known as ?

OR

    1. Name and explain the structural organisation of the male sex accessory ducts in the human male reproductive system.
    2. Describe the role of gonadotropin FSH in the regulation of spermatogenesis.
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Answer not available: no official marking scheme published.

Q79.2025 Main 57/6/1 · Q325 marks
    1. Explain the structure of a mature embryo sac of a typical flowering plant.
    2. How is triple fusion achieved in these plants ?

OR

  1. Describe the changes in the ovary and the uterus as induced by the changes in the level of pituitary and ovarian hormones during menstrual cycle in a human female.
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Answer not available: no official marking scheme published.

Q80.2025 Supp. 57/S/1 · Q335 marks

(a) (i) Draw a diagram of an enlarged view of transverse section (T.S.) of one microsporangium of an angiosperm and label the following parts :

  1. Tapetum
  2. Middle layers
  3. Endothecium
  4. Microspore mother cell

(ii) Write the function of Tapetum.

(iii) Explain the following giving reasons :

  1. Pollen grains are well-preserved as fossils.
  2. Pollen tablets are in use by people these days.

OR

(b) (i) Where does spermatogenesis occur in human testes ? Describe the process of spermatogenesis up to the formation of spermatozoa.

(ii) Trace the path of movement of spermatozoa from the testes upto the ejaculatory duct.

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(a) (i)

Figure

(½ × 4)

(ii) It nourishes the developing pollen grains [1]

(iii) (I) Presence of sporopollenin in exine leads to preservation of pollen as fossil as it is most resistant organic material and can withstand high temperature and strong acid and alkali [1]

(II) Pollen tablets are rich in nutrients [1]

OR

(b) (i) • Spermatogenesis occurs in seminiferous tubules of testes [½]

• Spermatogonia (at puberty) → (mitosis) Primary spermatocytes → (1st meiosis) Secondary spermatocytes → (2nd meiosis) Spermatids → (differentiation) Spermatozoa [½ × 5]

(ii) Seminiferous tubule → rete testis → vasa efferentia → epididymis → vas deferens → ejaculatory duct [½ × 4]

[Total: 5 marks]

Q81.2024 Main 57/1/1 · Q315 marks

(a) (i) Explain the development of male gametophyte in an angiosperm.

(ii) Draw a labelled diagram of a three-celled male gametophyte.

OR

(b) (i) Draw a diagrammatic sectional view of ovary of human female and label the following :

  1. Blood vessels
  2. Primary follicle
  3. Tertiary follicle
  4. Ovum

(ii) At which stage of life are primary follicles formed in a human female ?

(iii) Explain the events (both hormonal and structural) that occur at the time of ovulation till the onset of the next menstrual cycle.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a)

(i) Each microspore mother cell in sporogenous tissue undergoes meiosis, forming microspores tetrad by the process called microsporogenesis; microspores form cluster of four cells called microspore tetrad; as the anther matures and dehydrates, each microspore undergoes asymmetric division to produce a vegetative cell and a generative cell. [½ × 5]

(ii)

Figure

[½ × 5]

OR

(b)

(i)

Figure

[½ × 4]

(ii) At embryonic stage [½]

(iii) Both LH and FSH attain a peak level in mid of menstrual cycle; rapid secretion of LH leading to maximum level is called LH surge, which induces rupture of Graafian follicle and cause ovulation; ovulatory phase is followed by luteal phase during which remaining parts of Graafian follicle transform as corpus luteum; it produces large amount of progesterone needed for maintaining endometrium; in absence of fertilization corpus luteum degenerates which causes disintegration of the endometrium leading to menstruation. [½ × 5]

[Total: 5 marks]

Q82.2024 Main 57/2/1 · Q315 marks

(a) (i) Explain the process of double fertilization in an angiosperm starting from the germination of pollen grains on the stigma, mentioning the ploidy of the end products formed at the end. State the role of synergids during the course of the process.

(ii) Why does the development of endosperm precede that of the embryo ?

OR

(b) (i) Mention the site where fertilisation of the ovum occurs in a human female. Explain the process of fertilization and mention how polyspermy is prevented.

(ii) Name the embryonic stage that gets implanted in the uterus. Explain the process of implantation in a human female.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a)

(i)

  • Pollen tube passes down the style, generative cell divides mitotically into two male gametes and male gametes are discharged into the embryo sac through the micropyle. [½ + ½]
  • One male gamete + Egg/ovum → Zygote (2n) / zygote is diploid; this process is known as syngamy. [½ + ½]
  • Other male nucleus (n) + secondary nucleus (2n) → PEN / Primary Endosperm Nucleus (3n) / PEN is triploid; this is known as triple fusion. (Ploidy of zygote and PEN) [½ + ½]
  • Since two types of fusion (syngamy and triple fusion) occur in an embryo sac, it is called double fertilization. [½]
  • Filiform apparatus of synergids guide entry of the pollen tube (at the micropylar end). [½]

(ii) To provide assured nutrition to the developing embryo. [1]

OR

(b)

(i)

  • Ampullary-isthmic junction / ampulla of fallopian tube [½]
  • A sperm comes in contact with zona pellucida (layer of ovum); the secretion of the acrosome of the sperm helps the sperm to enter into the cytoplasm of the ovum; this induces completion of meiosis II to form haploid ovum (ootid); haploid nucleus of the sperm and of the ovum fuse together to form the diploid zygote. [½ × 4]
  • On contact of sperm with zona pellucida induces changes in the membrane of the ovum that blocks the entry of the additional sperms. [½]

(ii)

  • Blastocyst [½]
  • Trophoblast layer of the blastocyst gets attached to the endometrium, inner cell mass gets differentiated into an embryo, the uterine cells divide rapidly and blastocyst gets embedded in the endometrium of the uterus. [½ × 3]

[Total: 5 marks]

Q83.2024 Main 57/3/1 · Q315 marks
    1. Explain any four devices that flowering plants have developed to encourage cross-pollination.
    2. Why do plants discourage self-pollination ? State any one reason.

OR

  1. Explain the ovarian and uterine events taking place along with the role of pituitary and ovarian hormones, during menstrual cycle in a normal human female under the following phases :
    1. Follicular phase/proliferative phase
    2. Luteal phase/secretory phase
    3. Menstrual phase
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) (i)

  • Pollen release and stigma receptivity are not synchronized / either the pollen is released before the stigma becomes receptive or stigma becomes receptive much before the release of pollen.
  • Anther and the stigma are placed at different positions so that pollen cannot come in contact with the stigma of the same flower.
  • Self-incompatibility / genetic mechanism / prevention of self-pollen from fertilizing the ovules by inhibiting pollen germination or pollen tube growth in the pistil.
  • Production of unisexual flower. [1 × 4]

(ii) To prevent inbreeding depression / to have more variations for better adaptation. [1]

OR

(b) Menstrual cycle in a normal human female

PhaseUterine eventOvarian eventMarks
(i)Proliferative phase / Follicular phaseEndometrium of the uterus regenerates through proliferation under the influence of estrogen from follicles.Primary follicles in the ovary grow to become Graafian follicles under FSH. (Secretion of estrogens by the growing follicles).[1 + 1]
(ii)Luteal phase / Secretory phaseMaintenance of the endometrium (thickness) in the presence of progesterone.Ruptured Graafian follicle transforms into corpus luteum. (secretes large amounts of progesterone).[1 + 1]
(iii)Menstrual phaseBreakdown of the endometrium lining of the uterus in the absence of progesterone from corpus luteum.Primary follicles start maturing and developing in the ovary under the effect of FSH.[½ + ½]

[Total: 5 marks]

Q84.2024 Supp. 57/S/1 · Q315 marks

(a) (i) Draw a labelled diagram of a fertilised embryo sac of an angiosperm. (label any four parts)

(ii) Explain double fertilisation in angiospermic plant.

OR

(b) (i) Draw a labelled diagram of a cross-section of human seminiferous tubule. (label any four parts)

(ii) Explain the hormonal regulation of spermatogenesis.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) (i)

Figure

(Any four correct labelling) [½ × 4]

(ii) One of the male gametes fuses with the egg, called syngamy, results in formation of diploid zygote. [½ + ½]

The other male gamete fuses with polar nuclei, called triple fusion, results in formation of triploid primary endosperm nucleus (3n). [½ + ½]

Since two types of fusions, syngamy and triple fusion, take place in angiosperm it is called double fertilization. [1]

OR

(b) (i)

Figure

[½ × 4]

(ii) Hypothalamus → GnRH → Anterior pituitary [½]; anterior pituitary secretes LH and FSH [½ + ½]

LH → acts on Leydig cells → synthesis and secretion of androgens → stimulate spermatogenesis; FSH → acts on Sertoli cells → secretion of some factors → help in spermatogenesis [½ + ½ + ½]

[Total: 5 marks]

Q85.2023 Main 57/1/1 · Q335 marks
  1. Study the diagram given below showing the modes of pollination. Answer the questions that follow.

    Figure

    1. The given diagram shows three methods of pollen transfer in plants. What are the technical terms used for pollen transfer methods ‘1’, ‘2’ and ‘3’ ?
    2. How do the following plants achieve pollination successfully ?
      1. Water lily
      2. Vallisneria
    3. Flowering plants have developed many devices to avoid inbreeding depression. Explain one hereditary and one physiological device which helps plants to achieve this target.

OR

  1. Observe the diagram given below showing the menstrual cycle of a normal human female and answer the questions that follow :

    Figure

    Numbers indicate the days of the menstrual cycle

    1. What are the suitable technical terms used for the following ?
      1. Days 1 – 7
      2. Days 8 – 12
      3. Days 16 – 28
      4. Days 13 – 15
    2. Explain the role of ovarian and pituitary hormones during the following time periods :
      1. Days 8 – 12
      2. Days 13 – 15
      3. Days 16 – 28
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) (i) (1) Autogamy; (2) Geitonogamy; (3) Xenogamy [½×3]

(ii) (1) Water lily: pollinated by insects/wind. [½]

(2) Vallisneria: Female flowers on long stalks reach water surface; male flowers or pollen released on water and carried by water current to female flowers to achieve pollination. [1]

(iii) Genetic: Self-incompatibility/prevents self-pollen (same flower or other flowers of same plant) from fertilizing the ovules by inhibiting pollen germination, pollen tube growth in pistil. [½×2]

Physiological: Pollen release and stigma receptivity are not synchronized, either pollen matures earlier and stigma later or pollen matures later than stigma. [½×2]

OR

(b) (i) (1) Menstrual period; (2) Follicular phase/proliferative phase; (3) Luteal phase/secretory phase; (4) Ovulatory phase [½×4]

(ii)

DaysOvarian hormonesPituitary hormones
18–12Follicular growth/proliferation of endometrial cells.Stimulates follicular development/secretion of estrogen by growing follicles
213–15Maturation of ovarian follicles/formation of graafian follicles/thickening of endometrium.Rupture of graafian follicle to release ovum.
316–28Maintenance of endometriumSecretion of progesterone from corpus luteum.

[½×6]

[Total: 5 marks]

Q86.2023 Main 57/5/1 · Q315 marks
    1. Explain the monosporic development of embryo sac in the ovule of an angiosperm.
    2. Draw a diagram of the mature embryo sac of an angiospermic ovule and label any four parts in it.

OR

    1. Explain the formation of placenta after the implantation in a human female.
    2. Draw a diagram showing human foetus within the uterus and label any four parts in it.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) (i) Megaspore mother cell (2n) — (Meiosis) → Tetrad (n) → Degeneration of 3 cells (one functional megaspore left) → 3 Mitosis without cell wall formation/Free nuclear division → 8 nuclear stage → (Cell wall formation) → Embryo sac (7 celled 8 nucleated stage) [½×6]

(ii) Mature embryo sac: labels Chalazal end, Antipodals, Polar nuclei, Central cell, Egg, Synergids, Micropylar end. (Any four parts labelling) [½×4]

Figure

OR

(b) (i) After implantation finger like projections appear on the trophoblast called chorionic villi which are surrounded by the uterine tissue and maternal blood, these tissues become interdigitated with each other, and jointly form a structural and functional unit between developing embryo and maternal body called placenta. [1×3]

(ii) (Diagram of human foetus within uterus; any four parts labelling) [½×4]

Figure

[Total: 5 marks]

Unit VI · Reproduction

2. Human Reproduction

Q1.2026 Main 57/1/1 · Q31 mark

Select the option that shows correct statements :

  1. Corpus luteum secretes progesterone.
  2. Only FSH attains a peak level in middle of cycle.
  3. LH surge induces rupture of graafian follicle.
  4. Luteal phase of menstrual cycle is also called proliferative phase.

Options :

  1. Only (ii) is correct.
  2. (i) and (iii) are correct.
  3. Only (iv) is correct.
  4. (ii) and (iv) are correct.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(B) (i) and (iii) are correct [1 mark]

Q2.2026 Main 57/1/1 · Q121 mark

Which connective tissue connects ovary to pelvic wall and uterus ?

  1. Tendons
  2. Ligaments
  3. Cartilage
  4. Bone
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(B) Ligaments [1 mark]

Q3.2026 Main 57/3/1 · Q11 mark

Foetal ejection reflex in human female triggers the release of which hormone ?

  1. Oxytocin from foetal pituitary gland
  2. hCG from placenta
  3. Progesterone from corpus luteum
  4. Oxytocin from maternal pituitary gland
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(D) Oxytocin from maternal pituitary gland [1 mark]

Q4.2026 Main 57/4/1 · Q21 mark

Select the phase of menstrual cycle in which the remaining parts of Graafian follicle transform into the corpus luteum.

  1. Secretory phase
  2. Proliferative phase
  3. Menstrual phase
  4. Ovulatory phase
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(A) Secretory phase [1 mark]

Q5.2026 Main 57/5/1 · Q21 mark

The foetal ejection reflex in human triggers the release of _________ hormone from __________.

Fill in the blanks by choosing right option.

  1. oxytocin, foetal pituitary
  2. oxytocin, maternal pituitary
  3. human chorionic gonadotropin, placenta
  4. progesterone, corpus luteum
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(B) oxytocin, maternal pituitary [1 mark]

Q6.2026 Main 57/5/1 · Q51 mark

In a human female, menstrual cycle of 28 days is represented by the diagram given below.

Figure

Select the correct statement related with this diagram.

  1. Primary follicle of ovary matures into Graafian follicle in phase II.
  2. The hormone secreted in large amounts in phase III is also responsible for maintaining pregnancy in human females.
  3. Corpus luteum secretes progesterone in phase I, however it degenerates completely in phase III.
  4. Both (A) and (B).
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(D) Both (A) and (B) [1 mark]

Q7.2026 Main 57/5/1 · Q131 mark

Assertion (A): The embryo with 8 to 16 blastomeres is called a morula.

Reason (R): The morula continues to divide and transform into trophoblast.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(C) Assertion (A) is true but Reason (R) is false. [1 mark]

Q8.2026 Supp. 57/7/1 · Q21 mark

The path of the sperm transport is shown below. Identify P and Q and select the correct option.

Figure

  1. P – Vasa efferentia, Q – Ureter
  2. P – Ejaculatory duct, Q – Vas deferens
  3. P – Vasa efferentia, Q – Vas deferens
  4. P – Vas deferens, Q – Vasa efferentia
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(C) P – Vasa efferentia, Q – Vas deferens [1 mark]

Q9.2025 Main 57/1/1 · Q61 mark

Match the following items of Column-I with that of Column-II :

Column-IColumn-II
(a) Trophoblast(i) Embedding of blastocyst in the endometrium
(b) Implantation(ii) Group of cells that would differentiate as embryo
(c) Inner cell mass(iii) Embryo with 8-16 blastomeres
(d) Morula(iv) Outer layer of blastocyst

Choose the option that matches Column-I with Column-II correctly.

  1. (a)–(iv), (b)–(i), (c)–(ii), (d)–(iii)
  2. (a)–(i), (b)–(ii), (c)–(iii), (d)–(iv)
  3. (a)–(ii), (b)–(i), (c)–(iv), (d)–(iii)
  4. (a)–(ii), (b)–(iv), (c)–(iii), (d)–(i)
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Answer not available: no official marking scheme published.

Q10.2025 Main 57/2/1 · Q21 mark

The number of autosomes present in a human secondary spermatocyte

  1. 44
  2. 22
  3. 23
  4. 46
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Answer not available: no official marking scheme published.

Q11.2025 Main 57/2/1 · Q131 mark

Assertion (A): Corpus luteum secretes the hormone, progesterone.

Reason (R): Hormone Progesterone is essential for maintenance of the endometrium.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Answer not available: no official marking scheme published.

Q12.2025 Main 57/4/1 · Q21 mark

Given below are a few statements with reference to the accessory ducts of the human male reproductive system :

  1. The seminiferous tubules of the testes open into rete testis then into the vas deferens.
  2. The vasa efferentia leave the testes and open into the epididymis.
  3. The epididymis leads to vas deferens that ascends into the abdomen.
  4. The vas deferens receives a duct from the prostrate gland and opens into the urethra as ejaculatory duct.
  5. The urethra originates from the urinary bladder and extends through the penis to its external opening, urethral meatus.

Choose the option with all true statements from the given options :

  1. (i), (ii), (iv)
  2. (ii), (iii), (v)
  3. (ii), (iv), (v)
  4. (i), (iii), (iv)
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Answer not available: no official marking scheme published.

Q13.2025 Main 57/5/1 · Q21 mark

Given below are few statements with reference to the uterus in the female reproductive system :

  1. The myometrium exhibits strong contractions during the delivery of the baby.
  2. The uterus opens into the cervix through a narrow opening called vagina.
  3. The cavity of the cervix and the vagina forms the birth canal.
  4. The outermost layer of uterus is a thin membranous perimetrium.
  5. The uterus is supported by tendons attached to the pelvic wall.

Choose the option with all true statements from the given options :

  1. (i), (ii) and (iv)
  2. (ii), (iii) and (v)
  3. (ii), (iv) and (v)
  4. (i), (iii) and (iv)
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Answer not available: no official marking scheme published.

Q14.2025 Main 57/6/1 · Q51 mark

Given below are a few statements with respect to spermatogenesis in a human male.

  1. Sperms are released from the seminiferous tubules by the process of spermiation.
  2. Spermiogenesis involves the maturation of spermatids into sperms.
  3. Spermatogonia produce spermatids by the process of spermiogenesis.
  4. Meiosis II in secondary spermatocytes results in the formation of four equal haploid spermatids.
  5. Primary spermatocyte completes the first meiotic division forming two equal, diploid cells called secondary spermatocytes.

Choose the option with all true statements from the given options :

  1. (i), (ii) and (iv)
  2. (ii), (iii) and (v)
  3. (ii), (iv) and (v)
  4. (i), (iii) and (iv)
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Answer not available: no official marking scheme published.

Q15.2025 Supp. 57/S/1 · Q31 mark

Which of the following statements are true in respect of chorionic villi in humans ?

  1. It appears after implantation of human embryo in the uterus.
  2. It becomes interdigitated with cervical tissue of female reproductive tract.
  3. It increases the surface area for exchange of materials.
  4. It develops from the inner cell mass of blastocyst.
  1. (i) and (ii)
  2. (ii) and (iii)
  3. (i) and (iv)
  4. (i) and (iii)
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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q16.2024 Main 57/1/1 · Q21 mark

Which one of the following hormones is secreted by the human placenta that helps in the maintenance of pregnancy ?

  1. Relaxin
  2. Human Chorionic Gonadotropin
  3. Oxytocin
  4. Human Placental Lactogen
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(B) Human Chorionic Gonadotropin [1 mark]

Q17.2024 Main 57/2/1 · Q31 mark

In humans, the secondary oocyte completes meiotic division when :

  1. it gets implanted in the uterine endometrium.
  2. it is released from the matured Graafian follicle.
  3. it is penetrated by the sperm cell.
  4. acrosomal enzymes break down the zona pellucida.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(C) it is penetrated by the sperm cell. [1 mark]

Q18.2024 Main 57/4/1 · Q21 mark

Select the option that gives the correct identification of ovum, morula and blastocyst in a human female reproduction system as shown in the following diagram :

Figure

  1. Ovum – B, Morula – D, Blastocyst – F
  2. Ovum – A, Morula – B, Blastocyst – G
  3. Ovum – A, Morula – E, Blastocyst – G
  4. Ovum – B, Morula – D, Blastocyst – G
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(C) Ovum – A, Morula – E, Blastocyst – G [1 mark]

Q19.2024 Supp. 57/S/1 · Q21 mark

Estrogen is secreted by :

  1. Corpus luteum
  2. Membrane granulosa of Graafian follicle
  3. Pituitary gland
  4. Germinal epithelium
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(B) Membrane granulosa of Graafian follicle [1 mark]

Q20.2023 Main 57/1/1 · Q101 mark

In which one of the following options does the endocrine gland correctly match with its hormonal secretion and its function ?

  1. Sertoli cells – Testosterone – Development of secondary sexual characteristics
  2. Placenta – Estrogen – Initiates secretion of milk
  3. Leydig cells – Androgen – Initiates the production of sperms
  4. Ovary – FSH – Stimulates follicular development
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(C) Leydig cells – Androgen – Initiate the production of sperms. [1 mark]

Q21.2023 Main 57/3/1 · Q51 mark

Which of the following options correctly matches the name of the hormone to its site of production in the human body ?

Name of the HormoneSite of Production
P.Oxytocini.Placenta
Q.Relaxinii.Corpus Luteum
R.hCGiii.Pituitary Gland
S.Progesteroneiv.Ovaries
  1. P-i, Q-iii, R-ii, S-iv
  2. P-i, Q-iv, R-iii, S-ii
  3. P-iii, Q-iv, R-i, S-ii
  4. P-ii, Q-iii, R-i, S-iv
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(C) P-iii, Q-iv, R-i, S-ii [1 mark]

Q22.SQP 2025-26 · Q31 mark

During the menstrual cycle of a human female, formation of graafian follicle is stimulated by secretion of which of the following gonadotropin hormones?

  1. Estrogen and progesterone
  2. FSH and Estrogen
  3. FSH and LH
  4. Progesterone and LH
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Official CBSE Sample Paper Marking Scheme · 2025-26

(C) FSH and LH [1 mark]

Q23.SQP 2026-27 · Q31 mark

During the early stage of human embryonic development, the blastomeres are arranged into an outer layer and an inner group of cells known as the Inner Cell Mass (ICM). What is the fundamental significance of the outer layer, the trophoblast?

  1. It is responsible for forming the primary germ layers (ectoderm, mesoderm, endoderm).
  2. It provides essential nutrients to developing embryo and helps in its implantation.
  3. It secretes colostrum which contains antibodies.
  4. It gives rise to all the organs of the future embryo (organogenesis).
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Official CBSE Sample Paper Marking Scheme · 2026-27

(B) It provides essential nutrients to developing embryo and helps in its implantation. [1 mark]

Q24.SQP 2026-27 · Q131 mark

Assertion (A): Spermatogenesis is an ongoing process throughout the reproductive life of a human male, while oogenesis is a discontinuous process.

Reason (R): Oogenesis begins during the embryonic stage, gets arrested at prophase-I, resumes after puberty, and is arrested again at metaphase-II.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
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Official CBSE Sample Paper Marking Scheme · 2026-27

(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). [1 mark]

Q25.2026 Main 57/2/1 · Q212 marks

(a) (i) Name the organ that forms a connection between foetus and the uterus of a pregnant human female. [½]

(ii) How does it support both nutrition and hormonal regulation ? [1½]

OR

(b) Give reasons for :

  1. All copulations do not lead to pregnancy.
  2. Meiosis I in spermatogenesis and oogenesis are markedly different.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) (i) Placenta [½]

(ii) Nutrition: It facilitates the supply of nutrients to embryo. [½]

Hormonal regulation: Acts as an endocrine gland and produces hCG / Human Chorionic Gonadotropin, hPL / Human Placental Lactogen, estrogen, progestogens. (Any two hormones) [½+½]

OR

(b) (i) Fertilisation can only occur if the ovum and sperms are transported simultaneously to the ampullary region / use of contraceptive devices / low sperm count or poor sperm quality / absence of ovulation / lactational amenorrhea / coitus interruptus / periodic abstinence (Any other correct reason) [1]

(ii) (Any one correct difference) [1]

Meiosis I in SpermatogenesisMeiosis I in Oogenesis
Starts at pubertyStarts during embryonic stage
Results in production of two equal haploid secondary spermatocyteResults in production of unequal haploid secondary oocyte and first polar body

[Total: 2 marks]

Q26.2025 Main 57/6/1 · Q192 marks
    1. Explain why the milk produced by the mother during the initial days of lactation is considered to be very essential for the newborn infant.
    2. What is the term used for the milk produced during the initial days of lactation ?

OR

  1. Many children in the metro cities are suffering from a very common exaggerated response of the immune system to certain weak antigens in air.
    1. What is the term used for the above mentioned disease ?
    2. Name the main type of antibody produced by the immune system in response to this disease.
    3. Which two main inflammation-causing chemicals are produced by the mast cells in such an immune response ?
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Answer not available: no official marking scheme published.

Q27.2025 Main 57/7/1 · Q182 marks
  1. “A fully developed foetus initiates its delivery from the mother’s womb.” Explain.

OR

  1. Give reasons for the following :
    1. Why can a woman generally not conceive a child after 50 years of age ?
    2. Polar bodies are formed during oogenesis and not during spermatogenesis.
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Answer not available: no official marking scheme published.

Q28.2024 Main 57/4/1 · Q172 marks

Study the graph given below that represents the changes in the thickening of the uterine wall in women ‘X’ and women ‘Y’ over a period of one month :

Figure

What does the graph with respect to woman ‘X’ and woman ‘Y’ indicate ? Give suitable reason.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1
  • X – In woman X thickness of uterine wall increases after mid of menstrual cycle; reason – due to fertilization of egg / pregnancy / conceived [½ + ½]
  • Y – In woman Y thickness of uterine wall decreases after mid of menstrual cycle; reason – egg has not been fertilized / leading to the breakdown of lining of the uterus / menstrual flow / bleeding [½ + ½]

[Total: 2 marks]

Q29.2024 Supp. 57/S/1 · Q172 marks

State the cellular nature and functions of myometrium and endometrium.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

Myometrium

Smooth muscular layer of uterus [½]

It exhibits strong contraction of the uterus during delivery of the baby (parturition) [½]

Endometrium

Glandular [½]

Undergoes cyclic changes during menstruation / implantation of the developing embryo or blastocyst / other events of pregnancy [½]

[Total: 2 marks]

Q30.2023 Main 57/1/1 · Q172 marks

The diagram given below shows a developmental stage of human embryo. Answer the following questions with reference to it :

Figure

  1. Identify and name the human embryonic stage shown.
  2. Mention its exact location in the normal pregnancy of a woman.
  3. Write one function of each of the two parts labelled ‘X’ and ‘Y’.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) Blastocyst [½]

(b) Uterine wall/endometrium/innermost layer of uterine wall. [½]

(c) (Outer layer/trophoblast) ‘X’ – helps in implantation in uterus/attachment to endometrium. [½]

(Inner cell mass) ‘Y’ – gets differentiated into embryo. [½]

[Total: 2 marks]

Q31.2023 Main 57/5/1 · Q172 marks

The graph given below shows the number of primordial follicles per ovary in women at different ages. Study the graph and answer the questions that follow.

Figure

  1. What is the average age of the women at the onset of menopause ?
  2. At what age are maximum primordial follicles present in the ovary, according to the given graph ?
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) Accept all values within 47±2 (years of age) [1]

(b) Accept all values from 6–8 years of age [1]

[Total: 2 marks]

Q32.2023 Supp. 57/C/1 · Q212 marks

Differentiate between Follicular phase and Luteal phase of the Menstrual cycle in human females on the basis of the following criteria :

  1. Days of their occurrence in the cycle
  2. Stage of the follicle
  3. Hormones influencing the phases
  4. State of endometrium
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(½ mark to be awarded for every correct difference)

Follicular phaseLuteal phase
(a) Days of their occurrence in the cycle6th–13th / 6th–14th day15th–28th / 15th–29th day
(b) Stage of the follicleDevelopment of primary follicle into Graafian follicleTransformation of Graafian follicle into corpus luteum
(c) Hormones influencing the phasesLH / FSH / EstrogenProgesterone
(d) State of endometriumRegeneration of endometrium through proliferation.Endometrium further proliferates and thickens.

[½ × 4 = 2 marks]

Q33.2025 Main 57/4/1 · Q223 marks

Explain the process of formation of placenta in a human female after the implantation of the blastocyst in the endometrium of the uterus.

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Answer not available: no official marking scheme published.

Q34.2025 Main 57/5/1 · Q223 marks

Explain the neuroendocrine mechanism involved in the process of parturition in a human female leading to the expulsion of the baby out of the uterus through the birth canal.

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Answer not available: no official marking scheme published.

Q35.2025 Main 57/6/1 · Q263 marks
  1. Name and explain the role of inner and middle walls of the human female uterus.
  2. Write the location and function of fimbriae in human female.
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Answer not available: no official marking scheme published.

Q36.2024 Supp. 57/S/1 · Q223 marks

Given below is the diagram showing a part of the reproductive system of a human female.

Figure

(a) Identify ‘Y’ and write its function.

(b) Identify ‘Z’ and write the events that may occur here.

(c) Mention the stage at which the process of oogenesis temporarily gets arrested and name the cells thus formed in the ‘X’ of a baby girl at the time of her birth.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) ‘Y’ – Fimbriae; it helps in collection of the ovum after ovulation. [½ + ½]

(b) ‘Z’ – Isthmus; zygote undergoes cleavage to form morula. [½ + ½]

(c) Prophase I of meiotic division. Primary oocyte. [½ + ½]

[Total: 3 marks]

Q37.2023 Main 57/1/1 · Q243 marks
  1. Write the characteristics of ‘stem cells’.
  2. From where can one obtain ‘stem cells’ in humans ?
  3. State any two applications of ‘stem cells’ in curing human diseases.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) They have the ability of self-renewal/to divide, and differentiate into any kind of cell/tissue/organ. [½×2]

(b) Inner cell mass of blastocyst/umbilical cord/bone marrow (or any other correct source) [1]

(c) (Any two) Diabetes treatment via forming islets of Langerhans, restoration of vision by injecting stem cells, to treat rheumatoid arthritis, reduces pancreatic cancer, to treat genetic disorder like cystic fibrosis, spinal cord injuries, heart disease, any other correct application [½×2]

[Total: 3 marks]

Q38.2023 Main 57/3/1 · Q243 marks

Given below is a diagrammatic sectional view of a seminiferous tubule. State the developmental process of :

Figure

    1. ‘b’ from ‘a’.
    2. ‘e’ from ‘d’.
    3. ‘d’ from ‘b’.
  1. Identify ‘a’, ‘b’ and ‘c’.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) (i) ‘b’ (secondary spermatocytes) formed after first meiotic division/Meiosis-I in ‘a’ (primary spermatocytes) [½]

(ii) ‘e’ (sperms) formed via a process called spermiogenesis/differentiation from ‘d’ (spermatids) [½]

(iii) ‘d’ (Spermatids) formed after second meiotic division/Meiosis-II from ‘b’ (secondary spermatocyte). [½]

(b) a – Primary spermatocyte [½]

b – Secondary spermatocyte [½]

c – Sertoli cell [½]

[Total: 3 marks]

Q39.SQP 2025-26 · Q233 marks

Given below is a figure showing transport of ovum, fertilisation and passage of growing embryo through fallopian tube in a human female. Answer the questions that follow:

  1. What will be the ploidy of cells shown in (a) and (c) stage in the figure given below?
  2. What will happen if component L as shown in the figure (g) given below does not attach properly to the endometrium?
  3. In a pregnant mother (case X), during early pregnancy, the fertilised egg splits into two embryos at stage C shown in the figure given below, resulting in the formation of twins. Will the genome of cells of these two embryos exhibit variation? Justify your answer.

Figure

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Official CBSE Sample Paper Marking Scheme · 2025-26

(i) (a) ovum is haploid (n) and (c) – blastomeres are diploid (2n).

(ii) If the trophoblast (L) does not attach to the endometrium properly, it can lead to implantation failure, pregnancy loss, and other pregnancy complications.

(iii) In case X, the cells of these embryos will have identical genome as they have developed from the same zygote.

(1 × 3 = 3)

Q40.2025 Main 57/1/1 · Q294 marks

Study the graphs given below for Case 1 and Case 2 showing different levels of certain hormones and answer the question that follows :

Figure

(a) Which hormone is responsible for the peak observed in Case 1 and Case 2 ? Write one function of that hormone.

(b) Write changes that take place in the ovary and uterus during follicular phase.

Student to attempt either sub-part (c) or (d) :

(c) Name the hormone Q of Case 2. Write one function of hormone Q.

OR

(d) Which structure in the ovary will remain functional in Case 2 ? How is it formed ?

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Answer not available: no official marking scheme published.

Q41.2024 Main 57/2/1 · Q294 marks

Read the following passage and answer the questions that follow.

Spermatogenesis is an important primary sex characteristic in humans and all other vertebrates. The process is coordinated and controlled under the influence of hormones. It starts with the onset of puberty in humans and thereafter continues. The primordial cells within the embryonic testis which differentiate into spermatogonia are the precursors of the sperms. These are located at the outer walls of the seminiferous tubules where the process of spermatogenesis proceeds.

(a) State the site of action of FSH in the testes and describe its action thereafter.

OR

(a) Describe the role of LH in the process of spermatogenesis.

(b) Name the cells and their products which undergo :

  1. Mitosis and Differentiation
  2. Meiosis I and Meiosis II

during the process of spermatogenesis.

(c) Name the accessory ducts that the sperms travel through from seminiferous tubules to reach the epididymis.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a) Sertoli cells in seminiferous tubule; induces release of some factors which induce spermiogenesis. [1 + 1]

OR

(a) LH acts on Leydig cells, and stimulates the synthesis and secretion of androgens for spermatogenesis. [1 + 1]

(b) (i) Spermatogonia → (mitosis / differentiation) → Primary spermatocyte [½]

(ii) Primary spermatocyte → (meiosis I) → Secondary spermatocyte → (meiosis II) → Spermatid [½]

(c) Rete testis, vasa efferentia [½ + ½]

[Total: 4 marks]

Q42.2024 Main 57/5/1 · Q294 marks

In a human female, the reproductive phase starts on the onset of puberty and ceases around middle age of the female. Study the graph given below regarding menstrual cycle and answer the questions that follow :

Figure

  1. Name the hormones and their source organ, which are responsible for menstrual cycle at puberty.
  2. For successful pregnancy, at what phase of the menstrual cycle an early embryo (upto 3 blastomeres) should be Implanted in the Uterus (IUT) of a human female who has opted for Assisted Reproductive Technology (ART) ? Support your answer with a reason.
  3. Name the hormone and its source organ responsible for the events occurring during proliferative phase of menstrual cycle. Explain the event.

OR

  1. In a normal human female, why does menstruation only occurs if the released ovum is not fertilised ? Explain.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a)

HormoneSource Organ
FSHPituitary gland
LHPituitary gland
Estrogenovary / graafian follicle
Progesteroneovary / corpus luteum

(any two hormones with their relevant source organ) [½ + ½]

(b) Luteal phase or Secretory phase of the menstrual cycle, as endometrium is grown and suitable for implantation of an early embryo or blastocyst. [½ + ½]

(c) Estrogen, ovary / graafian (mature) follicles [½ + ½]; endometrium of uterus regenerates through proliferation. [1]

OR

(c) If ovum is not fertilized corpus luteum degenerates, progesterone level falls, disintegration of endometrium (and its blood vessels), leading to menstrual flow. [½ × 4]

[Total: 4 marks]

Q43.SQP 2026-27 · Q294 marks

Madhuri, a 17-year-old girl, has been experiencing irregular periods for the past one year. Sometimes, her cycles are as short as 20 days and sometimes as long as 45 days. She occasionally has very heavy bleeding that lasts over a week, causing fatigue and weakness. A doctor reviews her menstrual diary and explains that these changes can be linked to hormonal fluctuations in the hypothalamic-pituitary-ovarian (HPO) axis. He shows Madhuri the lack of a clear ovulation peak and low progesterone levels in her recent cycles.

A. Madhuri experiences heavy bleeding lasting over a week causing fatigue. What possible complication could arise from this condition? (1)

B. Examine how the absence of a distinct LH (luteinizing hormone) surge in the menstrual cycle might explain both the irregularity and heaviness of Madhuri's periods. (2)

Attempt either subpart C or D.

C. Madhuri's cycles vary between 20 and 45 days. What might this irregularity indicate about her ovarian function? (1)

OR

D. If a doctor suspects hypothalamic-pituitary dysfunction in Madhuri, which hormone secretion pattern would you expect to be disrupted? (1)

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Official CBSE Sample Paper Marking Scheme · 2026-27

A. Anaemia due to excessive blood loss. [1]

B. Without a distinct LH surge, ovulation may not occur, so no corpus luteum forms to secrete progesterone. As a result, the endometrial lining grows excessively under unopposed estrogen and is then unpredictably or heavily shed, causing irregular and heavy periods. [2]

Attempt either subpart C or D.

C. It suggests lack of ovulation due to hormonal imbalance affecting ovarian function. [1]

OR

D. GnRH secretion from hypothalamus and subsequent LH/FSH secretion from pituitary would be disrupted. [1]

Q44.2026 Main 57/1/1 · Q315 marks

(a) Trace the events from fertilization till implantation of blastocyst in human female. Also mention the site where fertilization takes place.

OR

(b) (i) Case studies of some couples are given below. They were not able to have kids even though the parents were not taking any precautions. They also do not wish to adopt a child or take the help of donors. Study these cases and suggest an appropriate Assisted Reproductive Technology that could help them.

Case I – Female – Normal Reports
Male – Normal sperms but no connection between epididymis and vas deferens.

Case II – Female – Normal eggs but Fallopian tube blocked.
Male – Normal sperms. Male reports are normal.

Case III – Female – Normal Reports.
Male – Low sperm count with less motility.

(ii) Copper releasing intrauterine devices are very effective and popular contraceptive devices. Name any two copper releasing IUDs. Write two reasons that make them effective contraceptives.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) • After the entry of sperm into female gamete, it undergoes meiotic division II, to produce haploid ovum i.e. ootid, then the nucleus of sperm and ootid fuse together to form zygote, zygote moves through the isthmus of oviduct towards uterus and it undergoes mitotic division or cleavage to form blastomeres, the embryo with 8 to 16 blastomere is called morula and it further divides and transforms into blastocyst, blastocyst have outer layer trophoblast and inner cell mass and the trophoblast layer of blastocyst gets attached to endometrium, uterine wall divide rapidly and covers blastocyst and thus it gets embedded in the endometrium of uterus, this is called implantation. [½ × 8]

• The site of fertilization is ampullary region of fallopian tube / isthmus-ampullary junction of fallopian tube. [1]

OR

(b) (i) Case I – In vitro fertilization followed by zygote intra fallopian transfer or ZIFT / in vitro fertilization followed by intra uterine transfer or IUT / intra uterine insemination or IUI / artificial insemination or AI. [1]

Case II – In vitro fertilization followed by IUT [1]

Case III – Intra Cytoplasmic Sperm Injection or ICSI / artificial insemination or AI / intra uterine insemination or IUI. [1]

(ii) • The copper releasing IUDs are CuT, Cu7, Multiload 375 (any two) [½ + ½]

• Copper releasing IUDs increase phagocytosis of sperms within the uterus; Cu ions released suppress sperm motility and the fertilising capacity of sperms. [½ + ½]

[Total: 5 marks]

Q45.2026 Main 57/3/1 · Q335 marks

(a) Oogenesis is a discontinuous process that begins before birth and is completed after puberty.

(i) Trace the development of a gamete mother cell till its release from the ovary during ovulation. [4]

(ii) Name the two pituitary hormones that play an important role in the process. [1]

OR

(b) (i) Draw a labelled two-celled structure of male gametophyte of an angiosperm. [1]

(ii) Name the three layers that surround the cytoplasm of a male gametophyte starting from innermost to outermost layer. [1½]

(iii) Which organic material makes the outermost layer ? Mention its advantage. [1½]

(iv) Why is the outermost layer of male gametophyte not continuous ? [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) (i)

  • Oogonia (gamete mother cells) 2n multiply by mitosis during foetal ovary and differentiate into primary oocytes, each primary oocyte gets surrounded by follicle cells to form a primary follicle. [½+½]
  • The primary oocyte remains arrested in prophase-I till puberty, after puberty oocyte inside the tertiary follicle completes meiosis-I to form a haploid secondary oocyte and a first polar body. [½+½]
  • Tertiary follicle changes into Graafian follicle. [1]
  • The secondary oocyte starts meiosis-II and then stops, Graafian follicle ruptures to release secondary oocyte (ovum) from the ovary (ovulation). [½+½]

(Diagram given in the scheme:)

Figure

(ii) Luteinizing hormone (LH) and Follicle stimulating hormone (FSH) [½+½]

OR

(b) (i)

Figure

[½+½]

(ii) Three layers: Plasma membrane [½], Intine [½], Exine [½]

(iii) Sporopollenin. [1] It is very resistant / cannot be degraded by enzymes / acids / alkali / temperature. [½]

(iv) Outermost layer is disrupted at some places to form the germ pores through which pollen germinates to form pollen tube. [1]

[Total: 5 marks]

Q46.2026 Main 57/4/1 · Q315 marks

(a) Study the flow chart given below. Name the hormones involved at each stage and explain their functions.

Hypothalamus
↓
Pituitary
↓
Ovary
↓
Pregnancy

OR

(b) (i) Draw a diagram of L.S. of maize grain and label its any six parts.

(ii) Differentiate between perisperm and endosperm giving one example of each.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a)

  • Gonadotropin releasing hormone / GnRH – It stimulates the pituitary gland to release follicle stimulating hormone or FSH and Luteinising hormone or LH both act on ovary. [½+½]
  • FSH / Follicle stimulating hormone – stimulate follicular development and secretion of estrogen by the follicle cells. [½+½]
  • LH / Luteinising hormone – stimulates ovulation / formation of corpus luteum from the ruptured graafian follicle. [½+½]
  • Progesterone – Maintenance of endometrium necessary for implantation and other events of pregnancy. [½+½]
  • Estrogen – responsible for the maturation of primary follicle into graafian follicle / regeneration of endometrium lining. [½+½]

OR

(b) (i)

Figure

(½ mark for any 6 correct labelling) [½ × 6]

(ii) (any one difference) [1]

PerispermEndosperm
In seeds residual persistent nucellus is called perisperm.It is the product formed from the primary endosperm cell in the ovule.
The cells are diploid (2n).Cells are triploid (3n).

Examples: [½+½]

PerispermEndosperm
Eg. Black pepper / beet / any other exampleeg. Castor / Wheat / Maize / Barley / Sunflower / Coconut / any other example

[Total: 5 marks]

Q47.2026 Main 57/5/1 · Q315 marks

(a) Briefly explain the events of fertilisation and implantation in an adult human female.

OR

(b) (i) Arrange the following hormones in sequence of their secretion in a pregnant woman : hCG; LH; FSH; Relaxin. [1]

(ii) Mention the source and the functions of the above mentioned hormones. [4]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a)

Fertilisation: Fertilisation occurs in the ampulla / ampullary-isthmic junction of the fallopian tube, when sperm comes in contact with the ovum, it induces changes in the zona pellucida of the ovum that block the entry of additional sperms and thus ensures only one sperm, the secretions of the acrosome help the sperm to enter the cytoplasm of the ovum by dissolving the zona pellucida and plasma membrane, after the entry of sperm into the cytoplasm of ovum induces completion of meiosis II (of secondary oocyte), fusion occurs between the nucleus of ootid (ovum) and that of sperm to form a diploid zygote [½ × 6]

Implantation:

  • The trophoblast layer gets attached to the endometrium.
  • The uterine cells divide rapidly and cover the blastocyst as a result the blastocyst becomes embedded in the endometrium of uterus [1×2]

OR

(b) (i) Hormones in sequence of their secretion in a pregnant woman: FSH; LH; hCG; Relaxin. [1]

(ii) [½+½ each]

  • FSH: Source: Anterior Pituitary. Function: Stimulates follicular development
  • LH: Source: Anterior Pituitary. Function: It causes rupture of Graafian follicle and release of ovum into fallopian tube / Ovulation
  • hCG: Source: Placenta. Function: It supports foetal growth, metabolic changes in mother and maintenance of pregnancy.
  • Relaxin: Source: Ovary. Function: It facilitates parturition.

[Total: 5 marks]

Q48.2026 Supp. 57/7/1 · Q315 marks

Attempt either (a) or (b).

(a) Draw a flow chart to show hormonal control of the human male reproductive system.

OR

(b) Explain the changes that the primary oocyte undergoes while in different follicular stages before ovulation.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a)

Figure

[½ × 10]

OR

(b) • Each primary oocyte is surrounded by a layer of granulosa cells and become the primary follicle. [1]

• When a primary follicle becomes surrounded by more layers of granulosa cells and a new theca called secondary follicle [1]

• The secondary follicle transforms into a tertiary follicle with the development of a fluid filled cavity (antrum) around the primary oocyte. [1]

• At this stage primary oocyte completes meiosis-I and forms a large haploid secondary oocyte and tiny first polar body. [1]

• The tertiary follicle grows further and changes into the mature follicle or Graafian follicle. [1]

[Total: 5 marks]

Q49.2025 Main 57/7/1 · Q325 marks
  1. Study the given diagram showing the sectional view of a seminiferous tubule.

    Figure

    Answer the following questions :

    1. Label A, B, C and D in the figure.
    2. What will be the number of chromosomes in secondary spermatocyte and spermatid respectively ?
    3. Explain the terms – Spermiogenesis and Spermiation.

OR

  1. Study the figures given below showing initial stages in the formation of female gametophyte and answer the questions that follow.

    Figure

    1. Identify (P) and (Q).
      1. What kind of division does cell (P) undergo to form (Q) ?
      2. How many (Q) cells form the embryo sac ? What is the name given to such kind of development ?
      3. How many free nuclear mitotic divisions will the functional megaspore undergo to form the embryo sac ?
      4. Describe the structure of a mature female gametophyte.
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Answer not available: no official marking scheme published.

Q50.2024 Main 57/4/1 · Q315 marks
  1. The given diagram shows the sectional view of a seminiferous tubule of Human testis :

Figure

  1. Name and describe the process depicted in the diagram which results in the development of spermatozoa.
  2. Identify the cell where you are seeing a cluster of spermatozoa attached in the diagram. Write the function of the cell.

OR

  1. Figure

    Observe the picture of Commelina plant bearing two types of flowers given above.

    1. Identify the two types of flowers labelled ‘A’ and ‘B’ in the picture.
    2. Compare the two types of flowers with reference to :
      1. Characteristic feature
      2. modes of pollination
    3. List any two ‘out breeding devices’ in flowering plants. Explain why do plants develop such devices.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) (i) Spermatogenesis [½]

Figure

Spermatogonia → (mitosis, differentiation) → Primary spermatocytes → (1st meiotic division) → Secondary spermatocytes → (2nd meiotic division) → Spermatids → (differentiation / spermiogenesis) → Spermatozoa. (Any 7 labelling to be considered, ½ mark each) / Statement in lieu of the above pointers should be considered. [½ × 7]

(ii) Sertoli cells; provide nourishment to the developing sperm. [½ + ½]

OR

(b) (i) A – chasmogamous flower; B – cleistogamous flower [½ + ½]

(ii)

‘A’‘B’
(1)open flower / blossomclosed flower / do not blossom
(2)Autogamy as well as geitonogamy and xenogamy / cross pollinationonly autogamy / self-pollination

[1 + 1]

(iii) Any two:

  • In some species, pollen release and stigma receptivity are not synchronized. Either the pollen is released before the stigma becomes receptive or stigma becomes receptive much before the release of pollen.
  • The anther and stigma are placed at different positions so that the pollen cannot come in contact with the stigma of the same flower.
  • Self-incompatibility. This is a genetic mechanism and prevents self-pollen from fertilizing the ovules.
  • Production of unisexual flowers. [½ + ½]

To prevent inbreeding depression / to discourage self-pollination and to encourage cross pollination / to develop genetic variation. [1]

[Total: 5 marks]

Q51.2024 Main 57/5/1 · Q325 marks
    1. Draw a diagram of a human sperm. Label any four parts and write their functions.
    2. In a human female, probability of an ovum to get fertilized by more than one sperm is impossible. Give reason.

OR

    1. With the help of labelled diagram only, show the different stages of embryo development in a dicot plant.
    2. Endosperm development precedes embryo development. Justify.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a) (i)

Figure

  • Head: Contains chromosomal / genetic / nuclear material
  • Acrosome: Filled with enzymes that help in fertilisation of the ovum / its secretion helps the sperm enter into the cytoplasm of the ovum
  • Middle piece: Possesses mitochondria (which produce energy)
  • Mitochondria: Produce energy (for the movement of the tail / for swimming)
  • Tail: Provides motility / helps in swimming
  • Plasma membrane: Envelope of the sperm / protection (while passing through acidic medium of reproductive tract)
  • Nucleus: Contains genetic / chromosomal material / DNA

(any correct four labellings with their respective function) [½ × 8 = 4]

(ii) When a sperm comes in contact with Zona Pellucida (layer of the ovum), it induces changes in the membrane which blocks the entry of any additional sperm. [½ + ½]

OR

(b) (i)

Figure

(Eight labellings) [½ × 8 = 4]

(ii) Providing nutrition to the developing embryo [1]

[Total: 5 marks]

Q52.2023 Main 57/3/1 · Q335 marks
  1. Given below is a diagrammatic representation of a human ovum.

    Figure

    1. Identify the parts ‘a’, ‘b’ and ‘c’.
    2. This ovum is released from the ovary with incomplete meiotic division. When, where and how is the meiotic division completed ?
    3. How does an ovum ensure the entry of a single sperm during fertilisation ?

OR

    1. Double fertilisation is an event unique to all flowering plants. Explain the process.
    2. Give a reason for the following :
      1. A seed of an orange has many embryos.
      2. Cashew is a false fruit but Guava is a true fruit.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) (i) a – Cells of corona radiata. b – Zona pellucida/Perivitelline space. c – Haploid nucleus [½×3]

(ii) • Once the sperm enters the cytoplasm of the ovum [½]

• The whole process is completed within the fallopian tube. [½]

• Entry of sperm in the cytoplasm of the ovum induces the completion of the 2nd meiotic division of the secondary oocyte, it is unequal division and results in formation of a second polar body and a haploid ovum (ootid). [½×2]

(iii) During fertilisation as the sperm comes in contact with the zona pellucida layer of the ovum, it induces changes in the membrane, that block the entry of any additional sperms. [½×3]

OR

(b) (i) • Double fertilisation is the occurrence of two types of fusion syngamy and triple fusion in an embryo sac of the angiosperm. [1]

• Syngamy – Fusion of one of the male gamete and the egg cell resulting in formation of a zygote (diploid). [1]

• Triple fusion – Fusion of another male gamete with two haploid polar nuclei to produce a (triploid) primary endosperm nucleus. [1]

(ii) (1) Some of the nucellar cells surrounding the embryo start dividing and protrude into the embryo sac to form embryos. [1]

(2) In case of Cashew thalamus also contribute in fruit formation along with ovary/development of fruit after fertilisation from the part other than ovary, Guava fruit develops only from the ovary after fertilisation. [½×2]

[Total: 5 marks]

Q53.2023 Supp. 57/C/1 · Q315 marks

(a) (i) Draw a labelled diagram of a matured human ovum ready for fertilization. Explain the events thereafter, leading to zygote formation.

(ii) It is said that the sex of the baby is decided as soon as the process of fertilization begins. Justify the statement.

OR

(b) (i) Write any four steps that you would recommend to improve the human reproductive health standards in India.

(ii) Refer to the following diagrams ‘A’ and ‘B’ showing sterilization in humans and answer the questions.

Figure

  1. Identify the labellings in the diagrams ‘A’ and ‘B’ where the procedure is carried out.
  2. Write the technical terms, and the procedures followed in the above-mentioned techniques.
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a)

(i)

Figure

(½ mark each for any three correct labelling) [½ × 3]

During fertilisation a sperm comes in contact with the zona pellucida layer of the ovum and induces changes in the membrane that block the entry of additional sperms. The secretions of the acrosome help the sperm enter into the cytoplasm of the ovum through the zona pellucida and the plasma membrane. This induces the completion of the meiotic division of the secondary oocyte, and results in the formation of a second polar body and a haploid ovum (ootid). The haploid nucleus of the sperm and that of the ovum fuse together to form a diploid zygote. [½ × 5]

(ii) If sperm carrying X chromosome fertilizes the egg, the child born will be a female baby. If sperm carrying Y chromosome fertilizes the egg, the child born will be a male baby. [½ + ½]

OR

(b)

(i)

  • Educating them about safe and hygienic sexual practices and sexually transmitted diseases.
  • Educating people in marriageable age group about the birth control measures, pre-natal and post-natal care of the mother and significance of breastfeeding.
  • Sex education should be provided to the school going children so as to discourage myths and misconceptions about sex related aspects.
  • A legal ban on the checking of the gender of the foetus.
  • Create awareness about sex-abuse and drawbacks of population explosion.
  • Proper infrastructural and professional facilities to attain reproductive health standards. (Any other correct recommendation)

(Any four points) [½ × 4]

(ii)

(1) ‘A’ – X, ‘B’ – P [½ + ½]

(2)

  • ‘A’ – Vasectomy; small part of vas deferens is removed or tied up [½ + ½]
  • ‘B’ – Tubectomy; small part of fallopian tube is removed or tied up [½ + ½]

[Total: 5 marks]

Unit VI · Reproduction

3. Reproductive Health

Q1.2026 Main 57/2/1 · Q61 mark

How many of the following features are related to Nirodh – a popular brand of condoms ?

  1. Prevention of sexually transmitted infections.
  2. Phagocytosis of sperms.
  3. Male contraceptive device.
  4. Acts as chemical barrier.
  5. Physiological changes in cervical mucus.
  1. (i), (ii), (iv)
  2. (i) and (iii)
  3. (i), (ii), (v)
  4. (ii) and (iv)
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(B) (i) and (iii) [1 mark]

Q2.2026 Main 57/4/1 · Q31 mark

A female undergoing IVF treatment has blocked fallopian tube. The technique by which the embryo with more than 8 blastomeres will be transferred into the female for further development is :

  1. ZIFT
  2. GIFT
  3. IUT
  4. AI
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(C) IUT [1 mark]

Q3.2026 Supp. 57/7/1 · Q141 mark

Assertion (A): Determining the sex of an unborn child followed by MTP is an illegal practice.

Reason (R): Amniocentesis is a practice to test the presence of genetic disorders.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). [1 mark]

Q4.2025 Main 57/1/1 · Q131 mark

Assertion (A): Saheli is the World’s first non-steroidal oral contraceptive pill.

Reason (R): It has been developed by National Institute of Immunology, New Delhi.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Answer not available: no official marking scheme published.

Q5.2025 Main 57/4/1 · Q151 mark

Assertion (A): ‘Saheli’, an oral contraceptive inhibits ovulation and increases phagocytosis of sperms.

Reason (R): It is a non-steroidal preparation.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q6.2025 Main 57/5/1 · Q161 mark

Assertion (A): Cu-T, Cu-7 and LNG-20 are the most widely used copper-releasing IUDs.

Reason (R): Cu-ions in IUDs effectively suppress sperm motility and the fertilising capacity of sperms.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q7.2025 Main 57/6/1 · Q141 mark

Assertion (A): Male contraceptive ‘Nirodh’ works on the principle of avoiding chances of ovum and sperm meeting.

Reason (R): It is made of thin rubber/latex sheath and is used to cover the penis before coitus.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q8.2025 Supp. 57/S/1 · Q101 mark

Given below is Column A with a list of certain Assisted Reproductive Technologies (ART), and Column B, with the procedures followed during ART :

Column A (ART)Column B (Procedures)
a.GIFTi.Transfer of ovum from a donor into the fallopian tube of another female
b.ICSIii.Transfer of semen from the donor into the vagina of female
c.ZIFTiii.Injecting sperms directly into the ovum
d.IUIiv.Transfer of early embryos into the fallopian tube

Select the correct match :

  1. a-i, b-ii, c-iii, d-iv
  2. a-iv, b-i, c-ii, d-iii
  3. a-iv, b-iii, c-i, d-ii
  4. a-i, b-iii, c-iv, d-ii
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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(D) a– i, b – iii, c – iv, d – ii [1 mark]

Q9.2024 Main 57/1/1 · Q31 mark

The periodic abstinence by a couple for family planning should be from :

  1. Day 5 to 10 of menstrual cycle
  2. Day 13 to 15 of menstrual cycle
  3. Day 10 to 17 of menstrual cycle
  4. Day 16 to 20 of menstrual cycle
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(C) Day 10 to 17 of menstrual cycle [1 mark]

Q10.2024 Main 57/1/1 · Q131 mark

Assertion (A): The laws of our country permit legal adoption and it is as yet, one of the best methods for childless couples looking for parenthood.

Reason (R): Emotional, religious and social factors are no deterrents to the legal adoption of orphaned and destitute children in India.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(C) Assertion (A) is true, but Reason (R) is false. [1 mark]

Q11.2024 Main 57/4/1 · Q31 mark

Study the table given below :

Contraceptive / Contraceptive MethodMode of Action
A. The pillI. Prevent sperm reaching cervix
B. CondomII. Prevent implantation
C. VasectomyIII. Inhibits ovulation
D. Copper-TIV. Semen contains no sperm

Select the option where contraceptive/contraceptive method are correctly matched with their mode of action.

  1. A – III, B – II, C – I, D – IV
  2. A – II, B – III, C – I, D – IV
  3. A – III, B – I, C – IV, D – II
  4. A – IV, B – III, C – II, D – I
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(C) A – III, B – I, C – IV, D – II [1 mark]

Q12.2024 Main 57/5/1 · Q31 mark

Observe the schematic representation of assisted reproductive technology given below :

Figure

Identify the most appropriate technique depicted in the above diagram.

  1. IUT
  2. IUI
  3. ICSI
  4. ZIFT
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(C) ICSI [1 mark]

Q13.2024 Supp. 57/S/1 · Q31 mark

Amniocentesis is a technique that is used to :

  1. determine any disease of the heart
  2. determine any genetic disorder of the foetus
  3. determine any disorder of the brain
  4. detect any abnormality in the bone formation
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(B) Determine any genetic disorder of the foetus [1 mark]

Q14.2024 Supp. 57/S/1 · Q131 mark

Assertion (A): Periodic abstinence is a method in which couples avoid coitus from day 10 to 17 of menstrual cycle.

Reason (R): Periodic abstinence has limited effectiveness because menstrual cycles are not always regular.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). [1 mark]

Q15.2023 Main 57/1/1 · Q91 mark

Given below are two columns. In Column I the names of four contraceptive devices are given and in Column II the modes of action of the contraceptives are given. Select the option where the contraceptive devices are correctly matched with their respective modes of action.

Column I (Contraceptive devices)Column II (Modes of action)
P.Lippes loopi.Inhibition of ovulation
Q.Multiload 375ii.Phagocytosis of sperms in uterus
R.Subcutaneous Norplantiii.Causes thickening of cervical mucous
S.Saheliiv.Makes cervix hostile to sperms
  1. P-ii, Q-iv, R-iii, S-i
  2. P-i, Q-ii, R-iii, S-iv
  3. P-iii, Q-i, R-iv, S-ii
  4. P-iv, Q-iii, R-ii, S-i
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(A) P– (ii), Q – (iv), R – (iii), S – (i) [1 mark]

Q16.2023 Main 57/3/1 · Q41 mark

A human male decides to adopt a surgical method for contraception. Identify the point in the diagram where a cut would be made and tied.

Figure

  1. Point S
  2. Point R
  3. Point Q
  4. Point P
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(C) Point Q [1 mark]

Q17.2023 Main 57/5/1 · Q21 mark

Given below are Column A with a list of certain Assisted Reproductive Technologies (ART) and in Column B the procedures followed during ART :

S. No.Names of ARTS. No.Procedures
(A)GIFT(i)Transfer of ovum from a donor into the fallopian tube of another female.
(B)ICSI(ii)Transfer of semen from the donor into the vagina of the female.
(C)ZIFT(iii)Injecting sperms directly into the ovum.
(D)IUI(iv)Transfer of early embryos into the fallopian tube.

Choose the option where ART correctly matches with the procedure.

  1. (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
  2. (A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)
  3. (A)-(iv), (B)-(iii), (C)-(i), (D)-(ii)
  4. (A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(D) (A) – (i), (B) – (iii), (C) – (iv), (D) – (ii). [1 mark]

Q18.2023 Main 57/5/1 · Q131 mark

Assertion (A): Determining the sex of an unborn child followed by MTP is an illegal practice.

Reason (R): Amniocentesis is a practice to test the presence of genetic disorders also.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). [1 mark]

Q19.2023 Supp. 57/C/1 · Q21 mark

Which one of the following techniques is employed in test tube baby programme ?

  1. Intra Cytoplasmic Sperm Injection
  2. Intra Uterine Insemination
  3. Gamete Intra Fallopian Transfer
  4. Zygote Intra Fallopian Transfer
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(A) Intra Cytoplasmic Sperm Injection

OR

(D) Zygote Intra Fallopian Transfer [1 mark]

Q20.2023 Supp. 57/C/1 · Q141 mark

Assertion (A): Cervical caps and vaults are barrier methods of contraception used by human females.

Reason (R): They prevent conception by phagocytosis of sperms.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(C) (A) is true, but (R) is false. [1 mark]

Q21.SQP 2026-27 · Q21 mark

A male patient undergoes a medical procedure where a small section of the vas deferens is cut and tied at both ends. Which of the following effects is the intended and direct result of this procedure?

  1. Blockage of sperm transport into the ejaculatory duct and urethra.
  2. Inhibition of testosterone production by the testes.
  3. Suppression of the formation of seminal plasma by accessory glands.
  4. Prevention of spermatogenesis in the seminiferous tubules.
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Official CBSE Sample Paper Marking Scheme · 2026-27

(A) Blockage of sperm transport into the ejaculatory duct and urethra [1 mark]

Q22.2026 Main 57/4/1 · Q172 marks

How do ‘implants’ act as an effective method of contraception in human females ? Mention one advantage of implants over contraceptive pills.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1
  • Implants work by inhibiting ovulation / inhibiting implantation / changing the quality of cervical mucus to retard entry of sperms. [1]
  • The effective periods are much longer than those of pills. (Any other correct advantage) [1]

[Total: 2 marks]

Q23.2026 Main 57/5/1 · Q182 marks

Define the term ‘amniocentesis’. Why has the government imposed a statutory ban in spite of its importance in the medical field ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1
  • Amniocentesis is test or procedure where some foetal cells in the amniotic fluid is analysed for certain genetic disorders. [1]
  • Sex determination is done with the help of amniocentesis which may be misused for female foeticide / to legally check increasing menace of female foeticide. [1]

[Total: 2 marks]

Q24.2026 Supp. 57/7/1 · Q182 marks

How can childless couples get help by the following assisted reproductive technologies ?

(a) GIFT

(b) ICSI

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) GIFT (Gamete Intra Fallopian Transfer) – Transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one but can provide suitable environment for fertilization. [1]

(b) ICSI (Intra Cytoplasmic Sperm Injection) – It is a specialised procedure in which the sperm is directly injected into the cytoplasm of the ovum to form zygote in the laboratory condition. [1]

[Total: 2 marks]

Q25.2025 Main 57/4/1 · Q192 marks
  1. Explain what is meant by the term MTP. What was the main reason to legalize MTP by the Government of India ?

OR

  1. Name any two STIs which might occur in a human female. State its two early symptoms.
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Answer not available: no official marking scheme published.

Q26.2025 Main 57/5/1 · Q192 marks
  1. Explain what is meant by the term amniocentesis. How is this technique misused in India ?

OR

  1. Name any two VDs which might occur in a human female. State any two complications in a female if it is left untreated.
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Answer not available: no official marking scheme published.

Q27.2025 Main 57/6/1 · Q212 marks
  1. Why are restrictions imposed on MTP in India ? Up to how many weeks or trimesters, is MTP considered relatively safe for a female, if necessary to perform, by a medical practitioner ?

OR

  1. Expand PID. Name any two common viral infections transmitted through sexual contact in human females.
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Answer not available: no official marking scheme published.

Q28.2024 Main 57/1/1 · Q192 marks

Write the basic steps followed in the Assisted Reproductive Technologies (ART) programme to help childless couples. Why is it also known as test tube baby programme ?

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1
  • Ova from wife or female donor and sperm from husband or male donor are collected, induced to form a zygote under simulated conditions in the laboratory (in vitro) (outside body); zygote or embryo are transferred into the female body for development. [½ × 3]
  • Test tube baby programme – because initial process is carried out in the laboratory / in vitro [½]

[Total: 2 marks]

Q29.2024 Main 57/2/1 · Q172 marks

(a) Name any two copper releasing intra-uterine devices. State two reasons that make them effective contraceptives.

OR

(b) Name any two outbreeding devices that flowering plants have developed and explain how they help in encouraging cross-pollination.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a) Copper releasing IUDs – CuT, Cu-7, Multiload-375 (any two) [½ + ½]

Cu+ released from IUDs suppress sperm motility, reduces fertilising capacity of sperms, increase phagocytosis of sperms (any two) [½ + ½]

OR

(b)

  • Unisexuality – production of unisexual flowers, prevent self pollination
  • Release and receptivity of stigma are not synchronized; either pollen is released before the stigma becomes receptive or stigma becomes receptive much before the release of pollen
  • Anther and stigma are placed at different positions, so that the pollen cannot come in contact with stigma
  • Self-incompatibility / genetic incompatibility – genetic mechanism that prevents self-pollen from fertilizing the ovule by inhibiting pollen germination or pollen tube growth in the pistil.

(Any two devices with explanation) [½ × 4 = 2 marks]

Q30.2024 Main 57/3/1 · Q172 marks

Oral contraceptives are widely accepted for controlling the increasing rate of population. Name the two important components of oral contraceptives. Why is ‘Saheli’ considered a preferred contraceptive by women ?

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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1
  • Progesterone, Progestogen – Estrogen combination [½ × 2]
  • ‘Saheli’ is a non-steroidal preparation, ‘once a week’ pill with very few side effects, high contraceptive value, easily available, cheaper in cost. (Any one) [1]

[Total: 2 marks]

Q31.2024 Main 57/4/1 · Q182 marks
  1. Intensely lactating mothers generally do not conceive. Why ?
  2. Our government has intentionally imposed strict conditions for MTP. Why ?
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) No ovulation therefore menstrual cycle does not occur during the period / due to lactational amenorrhea. [1]

(b) As majority of MTP’s are performed illegally by quacks which are not safe / to prevent female foeticide. [1]

[Total: 2 marks]

Q32.2024 Supp. 57/S/1 · Q182 marks

How do ‘implants’ act as an effective method of contraception in human females ? Mention their one advantage over contraceptive pills.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

Progestogen alone or in combinations with estrogen inhibit ovulation and implantation / alter the quality of cervical mucus to prevent or retard the entry of sperm. [1]

Their effective periods are much longer than contraceptive pills. [1]

[Total: 2 marks]

Q33.2026 Main 57/2/1 · Q253 marks

Study the diagram given below :

Figure

(a) Name and explain the procedure shown in the diagram. Mention one of its advantage and one disadvantage. [2]

(b) How is ‘Saheli’ different from other oral pills in terms of its composition and advantage ? [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a)

  • Amniocentesis [½]
  • Some of the amniotic fluid of the developing foetus is taken to analyse the fetal cells and dissolved substances. [½]
  • Advantage: It tests the presence of certain genetic disorders / down syndrome / haemophilia / sickle cell anaemia / determine the survivability of the foetus [½]
  • Disadvantage: to determine the sex of unborn child and if found female may lead to female foeticide. [½]

(b) Saheli is different from other pills in term of:

  • Composition: It is non-steroidal. [½]
  • Advantage: Once a week pill / low side effects / high contraceptive value [½]

[Total: 3 marks]

Q34.2026 Main 57/3/1 · Q253 marks

In vitro fertilization (IVF) is a popular method these days that is helping childless couples to bear a child.

(a) Write the different steps that are carried out in this technique. [2]

(b) Would you consider gamete intra fallopian transfer as a type of IVF ? Justify your answer. [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) Steps in IVF

  • Ova from wife/donor female and sperms from husband/donor male, are collected and induced to form a zygote under simulated conditions, in a laboratory. [½×3]
  • The zygote or early embryos with up to 8 blastomeres could then be transferred in the fallopian tube (ZIFT) / embryos with more than 8 blastomeres could be transferred in the uterus (IUT) [½]

(b) No [½]

During IVF fertilization takes place in the lab (outside the female body) / In this case fertilization takes place in female’s body. [½]

[Total: 3 marks]

Q35.2025 Main 57/1/1 · Q263 marks

Shyam and Radha are expecting their first child with Radha being in her second month of pregnancy with no complications. Shyam’s family has a history of cystic fibrosis while Radha’s family has a history of Down’s syndrome, leading to a concern that the baby may have one of these conditions.

(a) Suggest and explain a way of testing if their baby is at risk for any genetic disorders.

(b) In case of presence of one or both of the abnormalities and posing a risk to the mother’s health, mention one possible option for them to consider. Is that option safe for Radha at the current gestational age ? Justify.

(c) Under what conditions is the process mentioned in (b) illegal ?

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Answer not available: no official marking scheme published.

Q36.2025 Main 57/2/1 · Q233 marks

(a) List two reasons that make copper releasing IUDs as effective contraceptives.

(b) Explain how the intake of oral contraceptive pills prevent pregnancy in humans.

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Answer not available: no official marking scheme published.

Q37.2025 Main 57/7/1 · Q223 marks

A village health worker was taking a session with women. She tells them that one has to be careful while using oral pills as a method of birth control. Wrong usage can actually promote conception.

  1. Analyse the above statement and compare the merits and demerits of using oral pills and surgical methods of birth control.
  2. Village women are confused as to how a thin metallic copper loop can provide protection against pregnancy. Explain the mode of action of IUDs. Give two points.
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Answer not available: no official marking scheme published.

Q38.2025 Supp. 57/S/1 · Q223 marks

A woman has certain queries as listed below, before starting with contraceptive pills. Answer them.

(a) What do contraceptive pills contain ? Why are they considered as contraceptives ?

(b) What schedule should be followed for taking these pills ?

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(a) • The contraceptive pills contain hormones like progestogen or combination of progestogen–estrogen [1]

• Pills bring about the inhibition of ovulation and implantation / they act by altering the quality of cervical mucus to prevent or retard the entry of sperms / they prevent pregnancy [1]

(b) These pills have to be taken regularly for 21 days starting within first five days of menstrual cycle and to be repeated after a gap of 7 days / Saheli pill is to be taken once a week [1]

[Total: 3 marks]

Q39.2024 Main 57/1/1 · Q223 marks

Explain the mode of action of contraceptive pills taken by human females. Mention the schedule to be followed for effective outcome.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1
  • Pills contain progestogens or progestogen–estrogen combination. [½]
  • They inhibit ovulation and implantation as well as alter the quality of cervical mucus to prevent or retard the entry of sperms. [½ × 3]
  • Pills have to be taken daily for a period of 21 days starting within first five days of menstrual cycle. [½]
  • After a gap of 7 days it has to be repeated in the same pattern till the female desires to prevent conception. [½]

[Total: 3 marks]

Q40.2024 Main 57/2/1 · Q223 marks

(a) Why is “in vitro fertilization (IVF)” so named ? State its importance.

(b) Distinguish between GIFT and ZIFT.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a)

  • IVF – Fertilisation outside the human body in almost similar conditions as that of the body. [1]
  • Helps infertile couples to enjoy parenthood [1]

(b)

GIFTZIFT
Transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce one but can provide suitable environment for fertilization.Transfer of zygote or early embryos (with upto 8 blastomeres) into the fallopian tube.

[1]

[Total: 3 marks]

Q41.2024 Main 57/3/1 · Q233 marks

Explain the IUI and IUT methods of assisted reproductive technologies.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1
  • IUI (Intra uterine insemination) [½]
  • The semen collected either from the husband or a healthy donor is artificially introduced either into the vagina or into the uterus of the female. [1]
  • IUT (Intra uterine transfer) [½]
  • The embryo with more than 8 blastomeres is introduced into the uterus of the female. [1]

[Total: 3 marks]

Q42.2023 Main 57/1/1 · Q283 marks
    1. Expand the abbreviations given below, used for different modes of assisted reproductive technologies :
      1. ZIFT
      2. ICSI
      3. IUT
      4. GIFT
    2. Which one of them cannot be considered as a procedure of IVF ? Give reasons in support of your answer.

OR

  1. Differentiate between the following :
    1. Perisperm and Pericarp
    2. Syncarpous pistil and Apocarpous pistil
    3. Plumule and Radicle
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) (i) (1) ZIFT: Zygote intrafallopian transfer.

(2) ICSI: Intracytoplasmic sperm injection.

(3) IUT: Intra uterine transfer.

(4) GIFT: Gamete intrafallopian transfer. [½×4]

(ii) • GIFT [½]

• GIFT allows the eggs to fertilize and develop in the fallopian tube/IVF places a directly fertilized egg (zygote) into the uterus/in vivo fertilisation is involved in GIFT. [½]

OR

(b) (i)

PerispermPericarp
Persistent nucellus in some seedsThe wall of ovary develops into wall of fruit.

[1]

(ii)

SyncarpousApocarpous
fused pistils.free pistils.

[1]

(iii)

PlumuleRadicle
Future stem/terminal part of epicotyl/shoot tip of embryonal axisFuture root/terminal part of hypocotyl/root tip of embryonal axis

[1]

[Total: 3 marks]

Q43.2023 Main 57/3/1 · Q233 marks

Expand and explain the following techniques used in the ‘Test Tube Baby’ programme :

  1. GIFT
  2. ZIFT
  3. IUI
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) Gamete Intra Fallopian Transfer, transfer of an ovum collected from a donor into the fallopian tube of another female who cannot produce an ovum but can provide suitable environment for fertilization and further development/It has no role in test tube baby program. [½×2]

(b) Zygote Intra Fallopian Transfer, zygote or early embryo up to 8 blastomeres transferred into fallopian tube. [½×2]

(c) Intra Uterine Insemination, semen collected either from husband or a healthy donor is artificially introduced either into vagina or uterus of the female/It has no role in test tube baby program. [½×2]

[Total: 3 marks]

Q44.2023 Main 57/5/1 · Q223 marks

Name and explain a surgical contraceptive method that can be adopted by the male partner of a couple.

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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

• Vasectomy [1]

• A small part of the vas deferens is cut/removed, tied-up to block sperm transport [1×2]

[Total: 3 marks]

Q45.SQP 2025-26 · Q223 marks

Suggest a suitable contraceptive device for the following cases with justification.

  1. Mohini does not want to take the risk of conception and sexually transmitted infections (STIs).
  2. Lalita has two children and does not want any more children.
  3. Geeta wants a contraceptive that she can take till she wants to avoid conception and can resume back to her fertile life without the intervention of the doctor. Also, it should have a lower failure rate.
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Official CBSE Sample Paper Marking Scheme · 2025-26

(i) Condoms; these act by blocking the entry of microbes and sperms in the cervix.

(ii) Surgical methods such as Tubectomy by her or vasectomy by her husband as these are highly effective but irreversible methods.

(iii) Oral pills containing progestogens or progestogen – estrogen combination are very effective and can be repeated as long as the female desires to prevent conception.

(1 × 3 = 3)

Q46.SQP 2026-27 · Q223 marks

A couple, A and B, visit a fertility clinic. Examination reveals that female A has a blockage in her fallopian tubes, but her ovaries can produce viable ova. Male B produces sperm with a very low sperm count, making natural fertilization unlikely.

Which two different Assisted Reproductive Technologies (ARTs) would a doctor most likely recommend for this couple to increase their chances of conceiving? Justify your selection by briefly stating the principle behind each chosen technique concerning the couple's specific issues.

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Official CBSE Sample Paper Marking Scheme · 2026-27

ART 1: Intracytoplasmic Sperm Injection (ICSI)

Justification: ICSI is required to address Male B's low sperm count. It is an advanced procedure to form the embryo in vitro, where a sperm is directly injected into the ovum. [1.5]

ART 2: In vitro fertilisation followed by Intrauterine Transfer (IUT)

Justification: IVF followed by IUT is required to address Female A's blocked fallopian tubes. Since the natural path for the egg is blocked, the embryo (formed via IVF/ICSI) must be transferred directly to the uterus. [1.5]

Unit VII · Genetics and Evolution

4. Principles of Inheritance and Variation

Q1.2026 Main 57/1/1 · Q81 mark

In the following figure, two ways of pairing of two homologous pairs of chromosomes are shown. Which of the following phenomena is expressed ?

Figure

  1. Linkage of genes
  2. Independent assortment of genes
  3. Multiple alleles
  4. Incomplete dominance
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(B) Independent assortment of genes [1 mark]

Q2.2026 Main 57/2/1 · Q31 mark

In a dihybrid cross, 2400 individuals are produced in F2 generation. Approximately how many will be phenotypically similar to parents ?

  1. 2000
  2. 1500
  3. 580
  4. 450
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(B) 1500 [1 mark]

Q3.2026 Main 57/3/1 · Q101 mark

The smallest part of DNA molecule that can be changed by point mutation is :

  1. Oligonucleotide
  2. Codon
  3. Gene
  4. Nucleotide
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(D) Nucleotide [1 mark]

Q4.2026 Main 57/4/1 · Q41 mark

Select the correct statement. All genes located on the same chromosome :

  1. form different groups depending upon their relative distance.
  2. form one linkage group.
  3. will not form linkage groups.
  4. form interactive groups that affect the phenotype.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(B) form one linkage group [1 mark]

Q5.2026 Main 57/4/1 · Q141 mark

Assertion (A): The human blood group ‘AB’ exemplifies co-dominance.

Reason (R): The gene for human blood group trait exists in three allelic forms IA, IB and i.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion(A) [1 mark]

Q6.2026 Main 57/5/1 · Q41 mark

In Pisum sativum, the flower colour may be violet (V) or white (v). What proportion of the offspring in a cross of ‘VV’ × ‘vv’ would be expected to be violet ?

  1. 25%
  2. 50%
  3. 75%
  4. 100%
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(D) 100% [1 mark]

Q7.2026 Supp. 57/7/1 · Q31 mark

Morgan hybridised Drosophila white-eyed and yellow-bodied female with red-eyed and brown-bodied male (wild type) and intercrossed their F1 progeny. He observed that the two genes :

  1. Did not segregate independent of each other.
  2. Showed very low percentage of parental gene combinations.
  3. Segregated independently of each other.
  4. Showed very high percentage of recombinations.
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(A) Did not segregate independent of each other [1 mark]

Q8.2025 Main 57/1/1 · Q11 mark

A man whose father was colour-blind marries a woman who had a colour-blind mother and normal father. What percentage of male children of this couple will be colour-blind ?

  1. 25%
  2. 0%
  3. 50%
  4. 75%
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Answer not available: no official marking scheme published.

Q9.2025 Main 57/1/1 · Q161 mark

Assertion (A): When the two genes in a dihybrid cross are situated on the same chromosome, the proportion of parental gene combinations is much higher than non-parental type.

Reason (R): Higher parental gene combinations can be attributed to crossing over between two genes.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Answer not available: no official marking scheme published.

Q10.2025 Main 57/2/1 · Q31 mark

A child with blood group A has father with blood group B and the mother with blood group AB. Choose the option that gives the correct genotypes of father, mother and the child :

  1. Father: IAi; Mother: IBi; Child: IAi
  2. Father: IAIB; Mother: IAi; Child: IAIA
  3. Father: IBi; Mother: IAIB; Child: IAi
  4. Father: IBIB; Mother: IAIB; Child: IAIA
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Answer not available: no official marking scheme published.

Q11.2025 Main 57/2/1 · Q41 mark

In a pedigree chart the symbol

Figure

represents :

  1. unrelated mating
  2. affected individuals
  3. mating between relatives (consanguineous mating)
  4. Non-identical twins
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Answer not available: no official marking scheme published.

Q12.2025 Main 57/4/1 · Q41 mark

In the given pedigree chart, a cross between a normal couple resulted in a son who was haemophilic and a normal daughter. In course of time, when the daughter was married to a normal man, to their surprise the grandson was also haemophilic.

Figure

Choose the option that indicates the correct inheritance of trait in the above pedigree chart :

  1. Autosome linked dominant trait
  2. Sex-linked dominant trait
  3. Autosomal recessive trait
  4. Sex-linked recessive trait
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Answer not available: no official marking scheme published.

Q13.2025 Main 57/4/1 · Q121 mark

In a pea plant (Pisum sativum) inflated pod shape is dominant over constricted pod shape. The expected ratio of phenotypes of the offspring in a cross between both the parents with heterozygous inflated pod shape will be :

  1. 1 : 0
  2. 1 : 1
  3. 2 : 1
  4. 3 : 1
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Answer not available: no official marking scheme published.

Q14.2025 Main 57/4/1 · Q161 mark

Assertion (A): ABO blood grouping in humans is an example of multiple allelism.

Reason (R): More than two genes in a population govern the same character in ABO blood grouping in humans.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q15.2025 Main 57/5/1 · Q41 mark

Which of the following is not an example of aneuploidy ?

  1. Turner’s syndrome
  2. Down’s syndrome
  3. Phenylketonuria
  4. Klinefelter’s syndrome
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Answer not available: no official marking scheme published.

Q16.2025 Main 57/5/1 · Q91 mark

During a monohybrid cross involving a tall pea plant with a dwarf pea plant, the offspring populations were tall and dwarf in equal ratio. Find out the genotype of parent pea plants.

  1. Tt × tt
  2. TT × Tt
  3. tt × tt
  4. Tt × Tt
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Answer not available: no official marking scheme published.

Q17.2025 Main 57/5/1 · Q151 mark

Assertion (A): Gene pairs present on the same chromosome may be tightly linked or loosely linked.

Reason (R): Frequency of recombination between gene pairs on different chromosomes as a measure of the distance between genes can be used for ‘mapping’ their position on the chromosomes.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q18.2025 Main 57/6/1 · Q61 mark

Study the pedigree chart of a family sharing the inheritance of sickle cell anemia.

Figure

The trait traced in the above pedigree chart is :

  1. Dominant X-linked
  2. Autosomal dominant
  3. Recessive X-linked
  4. Autosomal recessive
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Answer not available: no official marking scheme published.

Q19.2025 Main 57/6/1 · Q81 mark

Which of the following do not follow the law of independent assortment ?

  1. Genes on non-homologous chromosomes and absence of linkage
  2. Two or more genes on homologous chromosomes
  3. Linked genes located on the same chromosomes
  4. Two or more distant genes present on the same chromosome
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Answer not available: no official marking scheme published.

Q20.2025 Main 57/6/1 · Q131 mark

Assertion (A): In dihybrid crosses involving sex-linked genes in Drosophila generation of non-parental gene combinations are observed.

Reason (R): Two genes present on different chromosomes show linkage and recombination in Drosophila.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q21.2025 Main 57/7/1 · Q51 mark

Which of the following conditions correctly describes the manner of determining the sex in the given options ?

  1. Homozygous sex chromosomes (ZZ) determine female sex in birds.
  2. XO type of chromosomes determine male sex in grasshoppers.
  3. XO condition in humans determines female sex.
  4. Homozygous sex chromosomes (XX) produce male in Drosophila.
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Answer not available: no official marking scheme published.

Q22.2025 Main 57/7/1 · Q151 mark

Assertion (A): A male individual always inherits haemophilia from his mother.

Reason (R): The gene for haemophilia lies on X-chromosome.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q23.2025 Supp. 57/S/1 · Q41 mark

A man with a certain disease marries a normal woman. They have four children (three daughters and one son). All the daughters suffer from their father’s disease but the son is not affected. Which of the following mode of inheritance do you suggest for this disease ?

  1. Autosome Linked Recessive
  2. Sex-linked Dominant
  3. Autosome Linked Dominant
  4. Sex-linked Recessive
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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q24.2025 Supp. 57/S/1 · Q61 mark

What would be the genotype of parents if the F1 offspring have the phenotypes in 1 : 1 proportion ?

  1. Aa × Aa
  2. AA × AA
  3. Aa × AA
  4. Aa × aa
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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(D) Aa × aa [1 mark]

Q25.2025 Supp. 57/S/1 · Q81 mark

Which of the following animals exhibit male heterogamety ?

  1. Fruit fly
  2. Fowl
  3. Human
  4. Honey bee
  1. (i) and (iii)
  2. (ii) and (iv)
  3. (ii) and (iii)
  4. (i) and (iv)
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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q26.2024 Main 57/1/1 · Q41 mark

Select the incorrect match from the following :

Human KaryotypeCharacters
(A)45 + XXBroad palm with characteristic palm crease
(B)44 + XXYOverall feminine development
(C)44 + XOSterile females as ovaries are rudimentary
(D)44 + XYNormal male
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(B) 44 + XXY – Overall feminine development [1 mark]

Q27.2024 Main 57/1/1 · Q141 mark

Assertion (A): Linked genes do not show dihybrid F2 ratio 9 : 3 : 3 : 1.

Reason (R): Linked genes do not undergo independent assortment.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). [1 mark]

Q28.2024 Main 57/2/1 · Q21 mark

Match the items in Column I with those in Column II and select the correctly matched option from those given below :

Column I
Cross
Column II
Phenotypic Ratio
1.Mendelian monohybrid(i)1 : 2 : 1 (F2)
2.Mendelian dihybrid(ii)1 : 1
3.Incomplete dominance(iii)3 : 1 (F2)
4.Test cross (monohybrid)(iv)9 : 3 : 3 : 1 (F2)

Options :

  1. 1 (ii), 2 (iv), 3 (i), 4 (iii)
  2. 1 (iii), 2 (i), 3 (iv), 4 (ii)
  3. 1 (iii), 2 (iv), 3 (i), 4 (ii)
  4. 1 (ii), 2 (i), 3 (iv), 4 (iii)
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(C) 1-(iii), 2-(iv), 3-(i), 4-(ii) [1 mark]

Q29.2024 Main 57/2/1 · Q81 mark

A woman with normal vision has a colour blind father. She marries a man with normal vision. The percentage chance of their progeny being colour blind is :

  1. 25%
  2. 50%
  3. 75%
  4. 100%
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(A) 25% [1 mark]

Q30.2024 Main 57/3/1 · Q51 mark

A Snapdragon plant bearing pink colour flowers is crossed with a Snapdragon plant bearing white colour flowers. The expected phenotypic percentage of the offspring is :

  1. 50% Red : 50% White
  2. 25% Red : 50% Pink : 25% White
  3. 50% Pink : 50% White
  4. 25% Pink : 50% Red : 25% White
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(C) 50% pink : 50% white [1 mark]

Q31.2024 Main 57/3/1 · Q61 mark

In which of the given chromosomal disorders does the individual have tall stature with feminised character ?

  1. Klinefelter’s syndrome
  2. Down’s syndrome
  3. Turner’s syndrome
  4. Edwards’ syndrome
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(A) Klinefelter’s syndrome [1 mark]

Q32.2024 Main 57/4/1 · Q41 mark

Identify the category of genetic disorder depicted in the pedigree chart given below :

Figure

  1. X-Linked recessive
  2. X-Linked dominant
  3. Autosomal recessive
  4. Autosomal dominant
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(C) Autosomal recessive [1 mark]

Q33.2024 Main 57/4/1 · Q71 mark

Turner’s syndrome in humans occurs due to

  1. Aneuploidy
  2. Euploidy
  3. Polyploidy
  4. Autosomal abnormality
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(A) Aneuploidy [1 mark]

Q34.2024 Main 57/5/1 · Q11 mark

A single gene that controls the expression of more than one trait is said to show

  1. Multiple allelism
  2. Polygenic inheritance
  3. Incomplete dominance
  4. Pleiotropism
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(D) Pleiotropism [1 mark]

Q35.2024 Main 57/5/1 · Q21 mark

A person with trisomy of 21st chromosome shows

  1. Furrowed tongue
  2. Characteristic palm crease
  3. Rudimentary ovaries
  4. Gynaecomastia

Select the correct option, from the choices given below :

  1. (ii) and (iv)
  2. (i), (ii) and (iv)
  3. (ii) and (iii)
  4. (i) and (ii)
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(D) (i) and (ii) [1 mark]

Q36.2024 Main 57/5/1 · Q91 mark

Study the pedigree chart of a family showing the inheritance pattern of a certain disorder. Select the option that correctly identifies the nature of the trait depicted in the pedigree chart.

Figure

  1. Dominant X – linked
  2. Recessive X – linked
  3. Autosomal dominant
  4. Autosomal recessive
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(D) Autosomal recessive [1 mark]

Q37.2024 Main 57/5/1 · Q111 mark

If both the parents are carriers for thalassaemia, the chances of an afflicted child to be born to them is :

  1. 25%
  2. 50%
  3. 75%
  4. 100%
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(A) 25 % [1 mark]

Q38.2024 Main 57/5/1 · Q151 mark

Assertion (A): In birds the sex of the offspring is determined by males.

Reason (R): Males are homogametic while females are heterogametic.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(D) (A) is false, but (R) is true. [1 mark]

Q39.2024 Supp. 57/S/1 · Q41 mark

In humans, non-disjunction of the 21st pair of chromosomes leads to :

  1. Acquired Immune Deficiency Syndrome
  2. Klinefelter’s Syndrome
  3. Turner’s Syndrome
  4. Down’s Syndrome
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(D) Down’s Syndrome [1 mark]

Q40.2023 Main 57/1/1 · Q11 mark

Which one of the following processes results in the production of recombinants in future generations ?

  1. Mutation
  2. Independent assortment during meiosis I
  3. Independent assortment during meiosis II
  4. Crossing over of bivalents
  1. (iv) only
  2. (ii) and (iv)
  3. (i), (ii) and (iii)
  4. (i), (ii), (iii) and (iv)
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(B) (ii) and (iv) [1 mark]

Q41.2023 Main 57/1/1 · Q161 mark

Assertion (A): Birds like pigeon have heterogametic females whereas the males are homogametic.

Reason (R): In pigeons, females have Z and W sex chromosomes whereas males have ZZ sex chromosomes.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(A) A and R are true and R is correct explanation of A. [1 mark]

Q42.2023 Main 57/3/1 · Q11 mark

Choose the set of sex determining chromosomes that indicates the correct sex of the respective organism.

  1. Homozygous sex chromosomes (ZZ) determine female sex in birds
  2. XO type of sex chromosomes determine male sex in grasshoppers
  3. XXY condition in humans, as found in Turner Syndrome, determines male sex
  4. Homozygous sex chromosomes (XX) produce male sex in Drosophila
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(B) XO type of sex chromosome determines male sex in grasshopper. [1 mark]

Q43.2023 Main 57/3/1 · Q131 mark

Assertion (A): In humans the genotype with all the dominant alleles (AABBCC) will have the darkest skin colour.

Reason (R): In a polygenic trait, phenotype reflects the contribution of each allele.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(A) Both A and R are true and R is correct explanation of A. [1 mark]

Q44.2023 Main 57/5/1 · Q161 mark

Assertion (A): In Thalassemia an abnormal myoglobin chain is synthesized due to a gene defect.

Reason (R): α-Thalassemia is controlled by genes HBA1 and HBA2 on chromosome 16.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(D) (A) is false, but (R) is true. [1 mark]

Q45.2023 Supp. 57/C/1 · Q31 mark

When round yellow seed bearing heterozygous pea plant is selfed, the frequency of occurrence of RrYY genotype among the F2 offsprings is :

  1. 9/16
  2. 3/16
  3. 1/16
  4. 2/16
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(D) 2/16 [1 mark]

Q46.2023 Supp. 57/C/1 · Q71 mark

In a pedigree analysis, the symbol shown below represents :

Figure

  1. Unrelated mating
  2. Affected individuals
  3. Consanguineous mating
  4. Non-identical twins
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(C) Consanguineous mating [1 mark]

Q47.SQP 2025-26 · Q71 mark

Albinism is known to be due to an autosomal recessive mutation. The first child of a couple with normal skin pigmentation was an albino. What is the probability that their second child will also be an albino?

  1. 100%
  2. 25%
  3. 50%
  4. 75%
Reveal answerHide answer
Official CBSE Sample Paper Marking Scheme · 2025-26

(B) 25% [1 mark]

Q48.SQP 2026-27 · Q81 mark

Study the given pedigree and identify the disease represented

Figure

  1. Myotonic Dystrophy
  2. Sickle cell anaemia
  3. Haemophilia
  4. Colour blindness
Reveal answerHide answer
Official CBSE Sample Paper Marking Scheme · 2026-27

(B) Sickle cell anaemia [1 mark]

Q49.SQP 2026-27 · Q141 mark

Assertion (A): There is no variation in the percentage recombination due to loosely or tightly linked genes.

Reason (R): Linked genes result in occurrence of higher proportion of parental gene combinations.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
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Official CBSE Sample Paper Marking Scheme · 2026-27

(D) A is false but R is true. [1 mark]

Q50.2026 Main 57/1/1 · Q202 marks

(a) Analyse the following two cases of sex determination in different organisms :

Case I : Males have (XO) sex chromosomes and females have two copies of same sex chromosome (XX).

Case II : Females have two different sex chromosomes (ZW) and males have two copies of same sex chromosome (ZZ).

Identify the type of heterogamety in each case, giving one example of each.

OR

(b) (i) Differentiate between a DNA and RNA nucleotide.

(ii) Name the two types of bonds present within a single nucleotide.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) Case I – Male heterogamety [½]; organism – many insects / grasshopper / any correct example [½]

Case II – Female heterogamety [½]; organism – birds / hen / any correct example [½]

OR

(b) (i)

DNA nucleotideRNA nucleotide
Has deoxyribose sugarHas ribose sugar
Types of nitrogen bases are A, G, C, TTypes of nitrogen bases are A, G, C, U

(Any one difference) [1]

(ii) Bonds in a nucleotide: N-glycosidic linkage (between sugar and base) [½]; phosphoester linkage (between sugar and phosphate) [½]

[Total: 2 marks]

Q51.2026 Main 57/2/1 · Q172 marks

(a) The diagram below shows the sequence of amino acids in part of haemoglobin molecule.

Figure

(i) If the base T* is substituted with A, in ‘CTT*’ of DNA strand how would it affect the haemoglobin chain ? [1]

(ii) Name the condition and the effects associated with the above substitution. [1]

OR

(b) DNA is considered a more stable genetic material than RNA. Justify this statement with suitable reasons.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a)

(i) Substitution of Glutamic acid by Valine at the sixth position of beta globin chain of haemoglobin molecule / Polymerisation of haemoglobin molecule under low oxygen tension [1]

(ii) Sickle cell anaemia [½]

Change in shape of the RBC from biconcave to elongated sickle like structure / Oxygen carrying capacity of haemoglobin reduces. [½]

OR

(b) (Any Two)

  • DNA is double stranded and the two strands being complimentary make it more stable than RNA
  • The presence of thymine at the place of uracil confers additional stability to DNA.
  • 2’-OH group in RNA is reactive group and makes it labile and easily degradable therefore DNA chemically is less reactive
  • DNA being more stable mutate at slower rate than RNA

[1+1 = 2 marks]

Q52.2026 Main 57/3/1 · Q192 marks

What did Henking observe in his experiments on insects that led to the discovery of sex chromosome ? How did his observation contribute to our understanding of sex determination ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

Henking traced a specific nuclear structure all through spermatogenesis in a few insects, and observed that 50 per cent of the sperm received a special structure after spermatogenesis, whereas the other 50 per cent sperm did not receive it. [½+½]

This observation later helped in sex determination as eggs fertilised by a sperm having X chromosome developed into females. [½]

Eggs fertilised by sperms not having X chromosome developed into males. [½]

[Total: 2 marks]

Q53.2025 Main 57/1/1 · Q192 marks

Study the cross given below :

Figure

Identify the abnormalities ‘1’ and ‘2’ in the offsprings of a cross done between a couple and distinguish between them.

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Answer not available: no official marking scheme published.

Q54.2025 Main 57/1/1 · Q202 marks

Study the pedigree chart given below, showing the inheritance pattern of blood group in a family :

Figure

Answer the following questions :

(a) Give the possible genotypes of individual 1 and 2.

(b) Which antigen or antigens will be present on the plasma membranes of the R.B.Cs of individuals ‘5’ and ‘8’ ?

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Answer not available: no official marking scheme published.

Q55.2025 Main 57/2/1 · Q182 marks

Study the given pedigree chart in which neither of the parents shows the trait but the trait is present in both male and female children.

Figure

Answer the following questions :

(a) Write about the trait, also explain the inheritance of such trait in the progeny on the basis of given pedigree chart.

(b) Give one example of such trait in human beings.

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Answer not available: no official marking scheme published.

Q56.2025 Main 57/7/1 · Q212 marks

A karyotype of a human suffering from a certain disorder is given below :

Figure

  1. Identify the disorder.
  2. Write the symptoms of the disorder.
  3. Give reason for such a disorder.
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Answer not available: no official marking scheme published.

Q57.2025 Supp. 57/S/1 · Q192 marks

(a) A couple believed that hospital authorities have exchanged babies and they have been given another baby by them. The wife is blood group ‘A’, her husband is blood group ‘B’ and the child is blood group ‘O’.

  1. Could the baby be theirs ?
  2. Justify your answer with the help of a cross.

OR

(b) Observe the diagram given below and answer the following questions.

Figure

  1. Name the molecule ‘X’ synthesized by ‘i’ gene.
  2. Which one of the structural genes codes for β-galactosidase ?
  3. When will the transcription of this gene stop ?
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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(a) (i) Yes [½]

(ii) Since the blood group of husband and wife is B and A respectively, so their genotype can be IBi and IAi respectively [½ + ½]

Parents IBi and IAi

Figure

½ mark for blood group O of baby [½]

OR

(b) (i) The molecule ‘X’ is repressor [½]

(ii) z gene codes for β-galactosidase [½]

(iii) In the absence of lactose or inducer [1]

[Total: 2 marks]

Q58.2024 Main 57/2/1 · Q182 marks

Although Haemophilia and sickle cell anemia are two blood related Mendelian disorders, yet, they differ in their pattern of inheritance. State any two differences.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1
HaemophiliaSickle cell anaemia
Sex linked / X-linked recessive disorderAutosomal recessive disorder
More males than females are affected.Affects both males and females equally.
Carrier / unaffected female transmits the disease to some of her male offspring.When both the parents are carriers, the disease is transmitted to the offspring.

(Any two corresponding differences) [1 + 1 = 2 marks]

Q59.2023 Main 57/5/1 · Q192 marks

By using Punnett square depict the genotypes and phenotypes of test crosses (where green pod colour (G) is dominant over yellow pod colour (g)) in Garden pea with unknown genotype.

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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

Case I – Homozygous dominant

GG × gg (Green Pods × Yellow Pods)

GG
gGgGg
gGgGg

Phenotype: All green coloured pod [½+½]

Case II – Heterozygous dominant

Gg × gg (Green Pods × Yellow Pods)

Gg
gGggg
gGggg

Phenotype: 50% green & 50% yellow colour pod [½+½]

[Total: 2 marks]

Q60.2026 Main 57/1/1 · Q233 marks

(a) What is pedigree analysis ? Mention its importance in human genetics. (2 pts.)

(b) Analyse the following pedigree and write the

(i) Pattern of inheritance.

(ii) Give one example of disease showing such an inheritance pattern.

Figure

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) • Pedigree analysis – It is analysis of traits or inheritance of traits in several generations of a family. [1]

• Importance : To trace the inheritance of a trait, abnormality or a disease. [½ + ½]

(b) (i) Autosomal recessive [½]

(ii) Sickle cell anaemia / Thalassemia / Phenylketonuria / any other correct example. [½]

[Total: 3 marks]

Q61.2026 Main 57/2/1 · Q233 marks

(a) Give the scientific name of the organism used by Morgan to study sex-linked traits. [1]

(b) Write any four reasons why this organism made it an ideal experimental model. [2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) Drosophila melanogaster [1]

(b) (Any four)

  • They could be grown on simple synthetic medium in the laboratory.
  • Complete their life cycle in about two weeks.
  • Single mating produces large number of progeny flies.
  • Clear differentiation of sexes – male and female flies are easily distinguishable.
  • Has many types of hereditary variations that can be seen under low power microscopes. / (Any other correct reason)

[½ × 4 = 2 marks; Total: 3 marks]

Q62.2026 Main 57/3/1 · Q283 marks

(a) Describe the experiment conducted by T.H. Morgan on Drosophila melanogaster involving eye colour and body colour. [1]

(b) How did the results deviate from Mendelian inheritance pattern ? [1]

(c) Explain the two genetic terms used by Morgan for his observations. [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) Morgan hybridised yellow bodied white eyed females with brown bodied red eyed males and inter-crossed their progeny. [1]

(b) He observed that the two genes did not segregate independently of each other as in Mendelian inheritance pattern. The ratio deviated from 9:3:3:1 ratio (expected when genes segregate independently) [1]

(c) Linkage – He gave this term to describe the physical association of the two genes present on the same chromosome. [½]

Recombination – Generations of nonparental gene combinations. [½]

[Total: 3 marks]

Q63.2026 Supp. 57/7/1 · Q233 marks

Define test cross. How does test cross help to determine the genotype of an individual ? Explain with the help of an example.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

• It is a cross to know the genotype of an unknown dominant trait by crossing it with homozygous recessive trait. [1]

• e.g. :

Figure

(Award 1 mark to each cross) [1 + 1]

[Total: 3 marks]

Q64.2025 Main 57/2/1 · Q243 marks

Using a Punnett square workout the distribution of an autosomal phenotypic feature in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

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Answer not available: no official marking scheme published.

Q65.2025 Main 57/4/1 · Q233 marks

Gregor Mendel conducted hybridisation experiments in garden pea for seven years and proposed the law of inheritance.

  1. Why was he successful in his hybridisation experiments ? Give two reasons.
  2. State the law of independent assortment as proposed by Mendel after his dihybrid crosses.
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Answer not available: no official marking scheme published.

Q66.2025 Main 57/5/1 · Q233 marks

During a medical investigation, an infant was found to possess an extra copy of chromosome no. 21. Identify the disorder the child is suffering from. Describe the symptoms the child is likely to develop later in life.

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Answer not available: no official marking scheme published.

Q67.2025 Main 57/5/1 · Q263 marks

Figure

Study the diagram above and answer the following questions :

  1. How many alleles are involved in blood grouping ?
  2. A person having ‘AB’ blood group has both dominant alleles. What is this inheritance type called ?
  3. A man with ‘A’ blood group marries a woman with ‘B’ blood group. Can they have a child with ‘O’ blood group ? Explain with the help of a cross.
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Answer not available: no official marking scheme published.

Q68.2025 Main 57/6/1 · Q243 marks

Answer the following questions with respect to the sex determining mechanism observed in honey bee.

  1. Name the type of sex determination system observed in honey bee.
  2. Fill in the blanks (i), (ii) and (iii) in the given question.
    ParentMale
    Type of cell division involved during gamete formation in males(i) ____________
    Number of chromosomes in the gametes(ii) ____________
    Number of chromosomes in the diploid cell of the progeny(iii) ____________
  3. What will be the sex and chromosome number of the progeny formed from the unfertilised eggs of honey bee ?
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Answer not available: no official marking scheme published.

Q69.2024 Main 57/1/1 · Q263 marks

In a family, the father, the daughter and the son are colour blind, whereas the mother has normal vision. Do you think the son and the daughter have inherited the disease from their father ? Work out a cross to justify your answer.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

No, [½]

Son inherited disease from the mother and daughter inherited disease from both mother and father. [½]

Mother: XCX (Normal, carrier) [½] × Father: XCY (Colour blind) [½]

Figure

[1]

[Total: 3 marks]

Q70.2024 Main 57/2/1 · Q233 marks

(a) (i) Write the karyotype and the genetic disorder of an individual who has developed from a zygote formed from an ‘XX’ egg fertilised by a ‘Y’ sperm.

(ii) Mention any two symptoms of this genetic disorder.

(iii) Write the possible reason that leads to the formation of this ‘XX’ egg.

OR

(b) In case of any dispute, a very small sample of tissue or even a drop of blood can help us to determine the paternity of a child. Provide a scientific explanation to substantiate the statement.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a)

(i) Karyotype – 44 + XXY / 47-XXY / AA+XXY; Genetic disorder – Klinefelter’s syndrome [½ + ½]

(ii) Gynaecomastia / development of breast, sterile individuals, overall masculine development (any two symptoms) [½ + ½]

(iii) Failure of segregation of chromatids or chromosomes / non-disjunction of chromatids or chromosomes during cell division or gametogenesis. [1]

OR

(b) DNA fingerprinting is the basis of paternity testing, high degree of polymorphism in DNA, the polymorphisms are inheritable. [1 × 3]

//

Isolation of DNA and amplification by PCR, digestion of DNA, separation of DNA fragments by electrophoresis, blotting and hybridisation, detection of hybridised fragments by autoradiography, matching the DNA bands of father and child. [½ × 6]

[Total: 3 marks]

Q71.2024 Main 57/3/1 · Q243 marks

Three crosses were carried out in pea plants with respect to flower colour violet/white (V/v) and flower position axial/terminal (A/a). Study in the table the crosses ‘a’, ‘b’ and ‘c’ where parental phenotypes and their F1 progeny phenotypes are given.

Parental plants (Phenotypes)F1 Progeny (Phenotypes)
(a) Violet, axial × white, axial6/16 white, axial
2/16 white, terminal
6/16 violet, axial
2/16 violet, axial
(b) Violet, axial × white, terminal1/4 violet, axial
1/4 violet, terminal
1/4 white, axial
1/4 white, terminal
(c) Violet, axial × violet, axial3/4 violet, axial
1/4 white, axial

Find the genotypes of each of the parental pairs of crosses ‘a’, ‘b’ and ‘c’.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) 1 mark is to be awarded to all the students if attempted; parental genotype is not possible to find from the given information. [1]

(b) Violet, axial × White, terminal – parents are heterozygous dominant × homozygous recessive: VvAa × vvaa [1]

(c) Violet, axial × Violet, axial – heterozygous for flower colour and homozygous dominant for flower position: VvAA × VvAA [1]

[Total: 3 marks]

Q72.2024 Main 57/4/1 · Q233 marks
  1. A man with blood group ‘A’ marries a woman with blood group ‘AB’. The first child born to them has blood group ‘B’. Work out a cross to find the genotype of the father. Give the possible blood groups and their genotypes of the children that could be born to this couple. (Use a Punnet square).
  2. State the basis of ‘ABO’ blood grouping in humans.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a)

  • Father genotype – IAi [½]
  • Parents – IAi × IAIB [½]

Figure

  • Possible genotypes – IAIA, IAi, IBi, IAIB [½]
  • Possible blood groups – A, AB, B [½]

(b) It is controlled by gene ‘I’ having 3 alleles (IA, IB, i). The blood group is determined by the type of sugar polymer present on the plasma membrane of the RBC. [½ + ½]

[Total: 3 marks]

Q73.2024 Supp. 57/S/1 · Q233 marks

Given below is the pedigree chart up to 3 generations (Generation I, Generation II, Generation III) of a family with respect to their blood groups.

Figure

(a) Mention the blood group along with the genotype of the offspring numbered ‘1’ in Generation II.

(b) Write the possible blood groups along with their genotypes of the offsprings numbered ‘2’ and ‘3’ in Generation III.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) Offspring 1: Blood group A – genotype IAi; Blood group B – genotype IBi [½ + ½]

(b) Offspring 2: Blood group A – genotype IAi; Blood group O – genotype ii [½ + ½]

Offspring 3: Blood group A – genotype IAi, IAIA [½]; Blood group AB – genotype IAIB [½]

(Award half mark if blood group along with its possible genotypes is correct)

[Total: 3 marks]

Q74.2023 Main 57/1/1 · Q233 marks

Study the given pedigree chart and answer the questions that follow.

Figure

  1. Is the trait given in the chart dominant or recessive ? Give reason in support of your answer.
  2. Is this trait autosomal or sex-linked ? Give reason in support of your answer.
  3. Write the possible genotypes of the children numbers ‘1’ and ‘3’ of the second generation.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) Recessive trait, both the parents in generation I do not express the trait yet it appears in the progeny. [½×2]

(b) Autosomal trait, both male and females have equal chances of getting the trait. [½×2]

(c) Child ‘1’: Aa/AA, Child ‘3’: Aa [½×2]

//

(a) Recessive trait, both the parents in generation I do not express the trait, yet it appears in the progeny. [½×2]

(b) Sex linked trait, comparatively more male are getting affected. [½×2]

(c) Child ‘1’: XY, Child ‘3’: X′X (carrier) [½×2]

[Total: 3 marks]

Q75.2023 Supp. 57/C/1 · Q233 marks

(a) The male fruit fly and female fowl are heterogametic, whereas female fruit fly and male fowl are homogametic. Explain.

(b) Male honey bees do not have a father and hence no sons, but can have a grandfather and grandson. Justify.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a) Male fruit fly has XY and female fowl has ZW sex chromosomes, they produce two different types of gametes in terms of sex chromosomes, while female fruit fly has XX and male fowl has ZZ sex chromosomes and they produce same type of gametes in terms of sex chromosomes. [½ × 3]

(b) Females are diploid and produces egg by meiosis, males are haploid produces sperm by mitosis. Unfertilized eggs develop into male by parthenogenesis (so do not have a father and will not have sons) and fertilized eggs develop into female. [½ × 3]

(Consider given points if flow chart is drawn)

[Total: 3 marks]

Q76.SQP 2025-26 · Q243 marks

In guinea pigs, black coat colour (G) dominates over white (g) and brown eyes (B) dominate over blue (b). The alleles for coat colour and eye colour are not linked. What will be the probability of the offspring having blue eyes and a white coat if both parents are heterozygous for eye and coat colour? Find the probability using a Punnett square.

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Official CBSE Sample Paper Marking Scheme · 2025-26

BbGg male is crossed with BbGg female which are both heterozygous of both the characters for eye and coat colour. [1]

Punnett square [1]

Figure

1/16 or 6.25 % - blue eyes and white coat [1]

Q77.SQP 2026-27 · Q243 marks

In mice, the traits for running ability and hair colour assort independently. Running (R) is dominant over walking in circles (r). Black hair (B) is dominant over brown hair (b). A heterozygous running, brown-haired female mouse was artificially inseminated using sperm from a heterozygous running, heterozygous black male.

  1. Write the genotypes of both parents and draw a Punnett square showing all possible offspring combinations. (1)
  2. Using the Punnett square, state the phenotypic ratio for: Running–Black, Running–Brown, Walking–Black, Walking–Brown. (1)
  3. If 64 offspring were produced, calculate how many would be expected to show the Running–Black phenotype. (1)
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Official CBSE Sample Paper Marking Scheme · 2026-27

A. Female mice: Heterozygous for running (Rr) and brown hair (bb), so its genotype is Rrbb. Male mice: Heterozygous for running (Rr) and heterozygous for black hair (Bb), so its genotype is RrBb.

Possible gametes

Female (Rrbb): Rb, Rb, rb, rb    Male (RrBb): RB, Rb, rB, rb [1]

Punnett square

RBRbrBrb
RbRRBbRRbbRrBbRrbb
RbRRBbRRbbRrBbRrbb
rbRrBbRrbbrrBbrrbb
rbRrBbRrbbrrBbrrbb

B. Probable phenotypes of the progeny

Running black (RRBb × 2, RrBb × 4): 6/16
Running brown (RRbb × 2, Rrbb × 4): 6/16
Walking black (rrBb × 2): 2/16
Walking brown (rrbb × 2): 2/16
Ratio will be 6:6:2:2 OR 3:3:1:1 [1]

C. 24 will show running black phenotype. [1]

Q78.SQP 2026-27 · Q253 marks

How can a single gene defect caused by the mutation result in Phenylketonuria (PKU)? Explain the molecular mechanism underlying pleiotropy.

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Official CBSE Sample Paper Marking Scheme · 2026-27

The disease is caused by a single gene mutation resulting in an inborn error of metabolism where the affected individual lacks an enzyme that converts the amino acid phenylalanine into tyrosine. [1]

As a result of this phenylalanine is accumulated and converted into phenyl pyruvic acid and other derivatives. Accumulation of these in the brain results in mental retardation. [1]

In pleiotropy a single gene can exhibit multiple phenotypic expressions by affecting metabolic pathways in different ways. [1]

Q79.2026 Main 57/1/1 · Q304 marks

A student performed some crosses in plants and represented the result in the form of bar graphs as shown below. Each graph displays the phenotypic proportion of the progeny. Study the graphs and answer the questions :

Figure

(a) What can you infer about the genotype of parents in crosses I and II ?

OR

(a) Which genetic cross is represented by these crosses ?

(b) Looking at bar graph of cross III, identify the type of cross performed and its importance in genetics.

(c) What conclusion can you draw from the results of bar graphs of crosses I and II ? Name the genetic principle illustrated.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) Cross I – Genotypes TT and tt / TT and Tt / TT and TT [½]

Cross II – Genotypes Tt and Tt. [½]

OR

(a) Monohybrid cross [1]

(b) • Test cross. [1]

• It is done to find the genotype of an organism showing dominant phenotype. [1]

(c) • It shows that tall trait is dominant over dwarf. [½]

• Genetic principle – Law of Dominance. [½]

[Total: 4 marks]

Q80.2026 Supp. 57/7/1 · Q294 marks

Read the following passage and answer the questions that follow :

Haemophilia is a sex-linked recessive disease. Study the pedigree analysis given below showing the inheritance of the disease in a family and answer the questions that follow :

Figure

(a) Analyse the given pedigree chart to explain why Haemophilia is considered a sex-linked recessive disorder.

(b) (i) The possibility of a human female becoming a haemophilic is extremely rare. Explain.

(ii) Why do the sons of a haemophilic father never suffer from this disease ?

(c) (i) Write the possible genotype of the individuals 2 and 5.

OR

(c) (ii) Why is haemophilia categorised as a Mendelian disorder ?

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) Since males have single copy of X chromosome and those who receive defective or mutated copy from mother get affected. [1]

(b) (i) The possibility of a female becoming haemophilic is extremely rare because mother of such a female has to be at least carrier (XhX) and father should be haemophilic (XhY) (unviable in later stage of life) [1]

(ii) Haemophilia is an X-linked disorder and father passes only the Y chromosome to the son. [1]

(c) (i) Individual 2 – XXh [½]; Individual 5 – XhY [½]

OR

(c) (ii) Haemophilia is categorised as Mendelian disorder because this occurs due to alteration or mutation in a single gene. [1]

[Total: 4 marks]

Q81.2025 Supp. 57/S/1 · Q294 marks

Read the following passage and answer the questions that follow.

The chromosome number is fixed for all normal organisms leading to species specification, whereas any abnormality in the chromosome number of an organism results into abnormal individuals. Sometimes an error may occur during meiosis of cell cycle, where the sister chromatids fail to segregate, called non-disjunction, leading to the production of abnormal gametes with altered chromosome number. On fertilization, such gametes develop into abnormal individuals.

(a) State what is aneuploidy.

(b) A normal human sperm (22 + Y) fertilizes an ovum with karyotype (22 + XX). Name the disorder of the offspring thus produced and write any two symptoms of the disorder.

(c) (i) What is meant by trisomy ? Mention the chromosome number that shows trisomy in Down’s syndrome.

OR

(c) (ii) Write any two symptoms of Down’s syndrome.

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(a) Failure of segregation of chromatids during cell division cycle results in gain or loss of chromosome(s) called aneuploidy [1]

(b) • Klinefelter syndrome [1]

• Symptoms – Overall masculine development, however the feminine development (development of breast / gynecomastia) is expressed, they are sterile, tall stature with feminized character (any two symptoms) [½ + ½]

(c) (i) • Presence of additional copy of a chromosome [½]

• Chromosome no. 21 [½]

OR

(c) (ii) Symptoms – Short statured with small round head, furrowed tongue, partially open mouth, broad palm with characteristic crease, physical / psychomotor / mental development is retarded, flat back of head, broad flat face, many loops on finger tips, congenital heart disease, big and wrinkled tongue (any two symptoms) [½ × 2]

[Total: 4 marks]

Q82.2023 Main 57/3/1 · Q294 marks

The following pedigree chart shows the inheritance of a genetic disorder up to three generations of a family. Observe the chart and answer the questions that follow.

Figure

  1. Is the disease sex-linked or autosomal as per the chart ? Give reasons in support of your answer.
  2. Is it a recessive or a dominant disorder ?
  3. Write the genotypes of the individuals ‘C’, ‘D’ and ‘H’.
    1. If the female ‘D’ marries a normal man, what will be the probability of their daughter being a sufferer of this disease ?

    OR

    1. If the mother ‘B’ is a carrier of the disease, what will be the probability of their daughter being a sufferer of this disease ?
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(i) • Sex linked disorder [½]

• More males are affected in the family as males have only one X chromosome which if affected expresses [½]

(ii) Recessive disorder [1]

(iii) C – XXc; D – XXc; H – XXc [½×2]

‘c’ is affected allele, accept other symbols used for the same. (If any two genotypes are correct then award 1 mark)

(iv) (a) Probability 0% [1]

OR

(iv) (b) Probability 50% [1]

[Total: 4 marks]

Q83.2023 Main 57/5/1 · Q304 marks

The chromosome number is fixed for all normal organisms leading to species specification whereas any abnormality in the chromosome number of an organism results into abnormal individuals. For example, in humans 46 is the fixed number of chromosomes both in male and female. In male it is ‘44 + XY’ and in female it is ‘44 + XX’. Thus the human male is heterogametic, in other words produces two different types of gametes one with ‘22 + X’ chromosomes and the other with ‘22 + Y’ chromosomes respectively. Human female, on the other hand, is homogametic i.e. produces only one type of gamete with ‘22 + X’ chromosomes only.

Sometimes an error may occur during meiosis of cell cycle, where the sister chromatids fail to segregate called nondisjunction, leading to the production of abnormal gametes with altered chromosome number. On fertilisation such gametes develop into abnormal individuals.

  1. State what is aneuploidy.
  2. If during spermatogenesis, the chromatids of sex chromosomes fail to segregate during meiosis, write only the different types of gametes with altered chromosome number that could possibly be produced.
  3. A normal human sperm (22 + Y) fertilises an ovum with karyotype ‘22 + XX’. Name the disorder the offspring thus produced would suffer from and write any two symptoms of the disorder.

OR

  1. Name a best known and most common autosomal aneuploid abnormality in human and write any two symptoms.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) Gain or loss of chromosome due to failure of segregation of chromatids during cell division cycle is known as aneuploidy. [1]

(b) 22 + XY, 22 + 0 [½×2]

(c) Klinefelter’s syndrome [1]

Gynaecomastia/feminine development, sterile individual, tall stature, overall masculine development (Any two) [½×2]

OR

(c) Down’s Syndrome: [1]

Short statured with small round head, furrowed tongue, partially open mouth, broad palm with palm crease, physical/psychomotor/mental retardation, flat back of head, loops on finger tips, congenital heart disease, big and wrinkled tongue, broad flat face (Any two) [½×2]

[Total: 4 marks]

Q84.2026 Main 57/4/1 · Q325 marks

(a) Write the types and locations of the genes causing thalassemia in humans. State the cause and symptoms of the disease. How is sickle-cell anemia different from this disease ?

OR

(b) (i) Enlist any four prime goals of the Human Genome Project (HGP).

(ii) List the two methodologies which were involved in the human genome project. Mention how they were used.

(iii) Expand ‘YAC’ and mention what it was used for.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) Types and location of gene:

α-Thalassemia is under the control of two genes HBA1 and HBA2, both located on chromosome-16 [½+½]

β-Thalassemia is the control of a single gene HBB, present on chromosome-11 [½+½]

Causes: It is due to mutation / deletion of one or more of the four alleles of the genes controlling the disease. [1]

Symptoms: Anaemia / extreme tiredness / pale skin / breathlessness / any other correct symptom [1]

Difference (Any one) [1]

ThalassemiaSickle cell anaemia
It is quantitative where less amount of haemoglobin is synthesised.It is qualitative where defective haemoglobin is synthesised.
Both α and β chain are affectedOnly β chain is affected

OR

(b) (i) Major goals of HGP: (any four) [½ × 4]

  • Identify all the genes (20,000-25,000 in human DNA)
  • Determine the sequence of the three billion base pairs present in human DNA.
  • Store the information in database.
  • Improve the tools for data analysis.
  • Transfer the technologies to other sectors (like industries).
  • Address the ethical legal and social issues that may arise from this project.

(ii) Methodologies:

  • Expressed Sequence Tags (ESTs) – It focuses on identifying all the genes that are expressed as RNAs. [½+½]
  • Sequence Annotation – it aims at sequencing the whole set of genome that includes all coding sequences and non coding sequences and then assigning functions to different regions in the sequence. [½+½]

(iii) Yeast artificial chromosomes; It is used as vector. [½+½]

[Total: 5 marks]

Q85.2026 Main 57/5/1 · Q325 marks

(a) Work out separate monohybrid crosses up to F2 generation between two pea plants and two Antirrhinum plants, both having contrasting traits with respect to the colour of the flower. Comment on the patterns of inheritance in the crosses carried out in such two cases.

OR

(b) Answer the following questions : [2+3 = 5]

(i) How does a chromosomal disorder differ from a Mendelian disorder ? Write one example for each.

(ii) Name the phenomenon that leads to situations like ‘XO’ abnormality in humans. Also name this genetic disorder. How are individuals with an XO chromosomal abnormality affected ? Write its symptoms as well as karyotype.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a) In Pea plants: [2½ for the cross (½ each: gametes, F1, F2 ...)]

Figure

Phenotypic ratio: violet : white = 3 : 1 [½]

Conclusion: This is the case of dominance [½]

In Antirrhinum plant [2½]

Figure

Phenotypic ratio: Red : Pink : white = 1 : 2 : 1 [½]

Conclusion: This is a case of incomplete dominance [½]

OR

(b) (i) [1]

Mendelian disorderChromosomal disorder
Occurs due to mutation / alteration in single gene / follow mendelian inheritance patternOccurs due to addition or deletion of one or more chromosomes / Does not follow mendelian inheritance pattern

Example of mendelian disorder: Haemophilia / Cystic Fibrosis / Sickle Cell Anaemia / Colour Blindness / Phenylketonuria / Thalassemia / any other correct example. [½]

Example of chromosomal disorder: Turner’s syndrome / Down’s syndrome / Klinefelter’s syndrome / any other correct example. [½]

(ii) Aneuploidy / Monosomy [½]

Turner’s syndrome [½]

Symptoms (any three) [½ × 3]

  • lack of secondary sexual characters.
  • ovaries are rudimentary
  • sterile
  • short stature
  • underdeveloped feminine characters

Karyotype – 44+XO (45) [½]

[Total: 5 marks]

Q86.2025 Main 57/1/1 · Q335 marks

Student to attempt either option-(A) or (B) :

(A) (i) Perform a cross between two sickle cell carriers. What ratio is obtained between carrier, disease free and diseased individuals in F1 progeny ? Name the nitrogenous base substituted, in the haemoglobin molecule in this disease.

(ii) Explain the difference in inheritance pattern of flower colour in garden pea plant and snap-dragon plant with the help of monohybrid crosses.

OR

(B) Explain with the help of well-labelled diagrams how lac operon operates in E. coli :

  1. In presence of an inducer.
  2. In absence of an inducer.
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Answer not available: no official marking scheme published.

Q87.2024 Main 57/2/1 · Q325 marks

(a) (i) Compare the pattern of inheritance of flower colour in garden pea plant (violet/white) with snapdragon plant (red/white) on the basis of the following :

  1. F1 phenotypic expression;
  2. expected phenotypic and genotypic expression of F2 generation;
  3. the conclusion you reached at the end of the comparison made.

(ii) List any two characteristics of pattern of inheritance of human blood group ABO.

OR

(b) (i) Draw a schematic, self-explanatory labelled diagram of lac operon in ‘switched on’ condition.

(ii) Why is regulation of lac operon referred to as negative regulation ?

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a) (i)

Pea plant (Violet/White)Snapdragon plant (Red/White)
(1)F1 – All violet (100%) [½]F1 – All pink flowers (100%) [½]
(2)F2 phenotype – 3 : 1 Violet : White [1]
Genotype – 1 : 2 : 1, VV : Vv : vv
F2 phenotype – 1 : 2 : 1 Red : Pink : White [1]
Genotype – 1 : 2 : 1, RR : Rr : rr
(3)Conclusion: Inheritance of flower colours in pea plant shows violet colour gene is completely dominant over white colour gene (recessive) / shows dominance [½]Whereas in snapdragon red colour gene shows incomplete dominance over white colour gene in heterozygous state / shows incomplete dominance [½]

(Note: one mark each for correct phenotype and genotype)

(ii) Any two:

  • It shows multiple allelism with alleles IA, IB, i
  • IA, IB genes show co-dominance in blood group AB,
  • IA & IB show complete dominance over i.

[½ × 2]

OR

(b) (i)

Figure

[½ × 8]

(ii) Repressor binds to operator to inhibit the gene expression [1]

[Total: 5 marks]

Q88.2024 Main 57/3/1 · Q325 marks
  1. “The influence of both the alleles in a heterozygous state is clearly expressed in codominance.” Explain with the help of inheritance of ABO blood group in humans.

OR

  1. “A group of genes are regulated and expressed together as a unit in lac operon.”
    1. Explain the mechanism of switching ‘on’ of the structural genes of lac operon.
    2. “Regulation of ‘lac operon’ is referred to be negatively regulated.” Justify giving a reason.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a)

  • The plasma membrane of the red blood cell has sugar polymers that protrude from its surface and the kind of sugar is controlled by the gene I present in blood.
  • The gene (I) has three alleles IA, IB and i
  • The alleles IA and IB produce a slightly different form of the sugar.
  • IA and IB are completely dominant over i
  • When IA and IB are present as in AB blood group they both express their own type of sugars and show codominance. [1 × 5]

(Note: Marks to be awarded if the above points are represented in the form of a table or cross)

OR

(b) (i) Switching ‘on’ of the lac operon

  • The lac operon consists of one regulatory gene and three structural genes (z, y and a)
  • The i gene codes for the repressor of the lac operon
  • The Z gene codes for beta-galactosidase (β-gal) (β-galactosidase is responsible for hydrolysis of (disaccharide) lactose into galactose and glucose)
  • y gene codes for permease (which increases the permeability of the cell to β-galactosides/lactose)
  • The repressor protein binds to the operator region and prevents RNA polymerase from transcribing the operon
  • Lactose is the inducer and regulates switching on and off of the operon
  • In the presence of inducer, lactose or allolactose, the repressor is inactivated by interaction with the inducer
  • This allows RNA polymerase to access the promoter and transcription proceeds [½ × 8]

OR (diagram, each labelled point [½ × 8]):

Figure

(ii) Repressor binds to the operator to inhibit gene expression therefore it is referred to be negatively regulated. [1]

[Total: 5 marks]

Q89.2023 Supp. 57/C/1 · Q325 marks

(a) Work out a monohybrid cross up to F2 generation (1) between two pea plants (2) between two Antirrhinum plants, both having contrasting traits with respect to colour of the flower. Comment on their pattern of inheritance.

OR

(b) (i) Expand hnRNA. Name the type of cells and the enzyme that transcribes it.

(ii) Is it a precursor ? Explain the process that it undergoes to become a functional molecule. Name this molecule.

(iii) Differentiate between mRNA and tRNA with respect to their roles in the cells.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a)

In case of pea plant

Parents: VV × vv (Violet × White) [½]

F1: Vv (Violet) [½]

Selfing

F2 (Punnett square): gametes V, v (male/female): VV (Violet), Vv (Violet), Vv (Violet), vv (White) [½]

Phenotype: Violet : White = 3 : 1 [½]

It shows complete dominance. [½]

In case of Snapdragon / Dogflower / Antirrhinum

Parents: RR × rr (Red × White) [½]

F1: Rr (Pink) [½]

Selfing

F2 (Punnett square): gametes R, r (male/female): RR (Red), Rr (Pink), Rr (Pink), rr (White) [½]

Phenotype: Red : Pink : White = 1 : 2 : 1 [½]

One allele is incompletely dominant over the other allele / Incomplete dominance. [½]

OR

(b)

(i) Heterogenous nuclear RNA; Eukaryotic cell; RNA polymerase-II [½ × 3]

(ii) Yes [½]

hnRNA is subjected to a process called splicing where the introns are removed and exons are joined in a defined order; in capping an unusual nucleotide (methyl guanosine triphosphate) is added to the 5′-end of hnRNA; in tailing adenylate residues (200–300) are added at 3′-end. [½ × 3]

mRNA [½]

(iii)

mRNAtRNA
Provides template for protein synthesisActs as adaptor molecule / brings amino acids and reads genetic code

[1]

[Total: 5 marks]

Unit VII · Genetics and Evolution

5. Molecular Basis of Inheritance

Q1.2026 Main 57/2/1 · Q11 mark

To separate repetitive DNA from bulk genomic DNA, which of the following method is used ?

  1. Elution
  2. Southern Blotting
  3. Polymerase Chain Reaction
  4. Density Gradient Centrifugation
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(D) Density Gradient Centrifugation [1 mark]

Q2.2026 Main 57/2/1 · Q71 mark

A bacterium having radioactive thymidine in its DNA is allowed to multiply in a medium having non-radioactive thymidine for two generations. What percentage of bacteria will have radioactive thymidine in its DNA ?

  1. 100
  2. 50
  3. 75
  4. 25
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(B) 50 [1 mark]

Q3.2026 Main 57/3/1 · Q51 mark

Four different transcription units are shown below. Choose the option with the correct image.

Options (A)–(D) are shown in the figure.

Figure

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(C) [1 mark]

Q4.2026 Main 57/3/1 · Q161 mark

Assertion (A): Repetitive sequences are stretches of DNA sequences that are thought to have no direct coding functions.

Reason (R): They shed light on chromosome structure, dynamics and evolution.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A) [1 mark]

Q5.2026 Main 57/4/1 · Q51 mark

Choose the correct statements :

  1. Polycistronic mRNA is generally found in eukaryotes.
  2. The RNA polymerase II transcribes precursor of mRNA, the hnRNA.
  3. The process of translation of mRNA begins when the mRNA encounters the large subunit of ribosomes.
  4. Stop codons do not have any tRNAs.

Options :

  1. (i) and (ii)
  2. (ii) and (iii)
  3. (ii) and (iv)
  4. (i), (ii) and (iv)
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(C) (ii) and (iv) [1 mark]

Q6.2026 Main 57/5/1 · Q31 mark

A DNA molecule is 160 base pairs long. It has 30% Guanine. How many Adenine bases are present in this DNA molecule ?

  1. 48
  2. 64
  3. 96
  4. 192
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(B) 64 [1 mark]

Q7.2026 Main 57/5/1 · Q141 mark

Assertion (A): Repetitive sequences make up a very large portion of human genome.

Reason (R): Repetitive sequences do not have direct coding functions in the genome.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(B) Both Assertion (A) & Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A) [1 mark]

Q8.2026 Supp. 57/7/1 · Q41 mark

Select the option that has only termination codons.

  1. UAA, UGA, UAG
  2. UAA, UUU, UGG
  3. UAA, UAG, UAC
  4. UUU, UCC, UGG
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(A) UAA, UGA, UAG [1 mark]

Q9.2025 Main 57/1/1 · Q91 mark

The sequence of nitrogenous bases in a segment of a coding strand of DNA is 5′ – AATGCTAGGCAC – 3′. Choose the option that shows the correct sequence of nitrogenous bases in the mRNA transcribed by the DNA.

  1. 5′ – UUACGAACCGAG – 3′
  2. 5′ – AAUGCUAGGCAC – 3′
  3. 5′ – UUACGUACCGUG – 3′
  4. 5′ – AACGUAGGCAGC – 3′
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Answer not available: no official marking scheme published.

Q10.2025 Main 57/1/1 · Q141 mark

Assertion (A): One of the property of genetic code is degeneracy.

Reason (R): Some amino acids can be coded by more than one codon.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Answer not available: no official marking scheme published.

Q11.2025 Main 57/2/1 · Q61 mark

The phosphoester linkage in the nucleotides is between

  1. phosphate group and OH of 3′C of a nucleoside.
  2. phosphate group and OH of 5′C of a nucleoside.
  3. phosphate group and H of 3′C of a nucleoside.
  4. phosphate group and H of 5′C of a nucleoside.
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Answer not available: no official marking scheme published.

Q12.2025 Main 57/2/1 · Q71 mark

Given below is a heterogeneous RNA formed during Eukaryotic transcription :

Figure

How many introns and exons respectively are present in the hnRNA ?

  1. 7, 7
  2. 8, 7
  3. 8, 8
  4. 7, 8
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Answer not available: no official marking scheme published.

Q13.2025 Main 57/2/1 · Q81 mark

Study the items of Column-I and those of Column-II :

Column-IColumn-II
(a) RNA polymerase I(i) 18s rRNA
(b) RNA polymerase II(ii) SnRNAs
(c) RNA polymerase III(iii) hnRNA

Choose the option that correctly matches the items of Column-I with those of Column-II :

  1. (a)–(i), (b)–(ii), (c)–(iii)
  2. (a)–(iii), (b)–(ii), (c)–(i)
  3. (a)–(ii), (b)–(iii), (c)–(i)
  4. (a)–(i), (b)–(iii), (c)–(ii)
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Answer not available: no official marking scheme published.

Q14.2025 Main 57/2/1 · Q101 mark

If Meselson and Stahl’s experiment is continued for 80 minutes (till III generation) then what would be the ratio of DNA containing N15/N15 : N15/N14 : N14/N14 in the medium ?

  1. 1 : 1 : 0
  2. 0 : 1 : 3
  3. 0 : 1 : 8
  4. 1 : 4 : 0
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Answer not available: no official marking scheme published.

Q15.2025 Main 57/4/1 · Q11 mark

In its extended ‘beads-on-string’ form of chromatin, the ‘beads’ in the string represent :

  1. Linker DNA
  2. Histone proteins
  3. Nucleosomes
  4. NHC proteins
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Answer not available: no official marking scheme published.

Q16.2025 Main 57/4/1 · Q31 mark

The substrate used during DNA replication by the enzyme DNA-dependent DNA polymerase is :

  1. Deoxyribonucleotide triphosphate
  2. Deoxyribonucleoside triphosphate
  3. Ribonucleotide triphosphate
  4. Ribonucleoside triphosphate
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Answer not available: no official marking scheme published.

Q17.2025 Main 57/4/1 · Q81 mark

Use the given information to select the amino acid attached to the 3′ end of tRNA during the process of translation, if the coding strand of the structural gene being transcribed has the nucleotide sequence TAC.

Codons for the amino acids :
AUCIsoleucine
AUGMethionine
UACTyrosine
GUAValine

Options :

  1. Isoleucine
  2. Methionine
  3. Tyrosine
  4. Valine
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Answer not available: no official marking scheme published.

Q18.2025 Main 57/4/1 · Q101 mark

The correct depiction of the experiment performed by Matthew Meselson and Franklin Stahl to prove that DNA replicates semi-conservatively on separation of DNA by centrifugation after 40 minutes is :

Options (A)–(D) are shown in the figure.

Figure

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Answer not available: no official marking scheme published.

Q19.2025 Main 57/5/1 · Q11 mark

The histone core in a nucleosome of chromatin thread is a/an :

  1. pentamer
  2. hexamer
  3. heptomer
  4. octamer
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Answer not available: no official marking scheme published.

Q20.2025 Main 57/5/1 · Q31 mark

During the process of transcription, after binding to a promoter, RNA polymerase catalyses and makes the bases in the template strand of DNA available for base pairing, with the bases of :

  1. Deoxyribonucleotide triphosphate
  2. Deoxyribonucleoside triphosphate
  3. Ribonucleotide triphosphate
  4. Ribonucleoside triphosphate
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Answer not available: no official marking scheme published.

Q21.2025 Main 57/5/1 · Q101 mark

What would happen if a gene encoding a polypeptide of 50 amino acids, 25th Codon (UAU) is mutated to “UAA” ?

  1. A polypeptide of 49 amino acids will be formed.
  2. A polypeptide of 25 amino acids will be formed.
  3. A polypeptide of 24 amino acids will be formed.
  4. A polypeptide of 50 amino acids will be formed.
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Answer not available: no official marking scheme published.

Q22.2025 Main 57/5/1 · Q121 mark

The correct depiction of the centrifugation step of the experiment conducted by Alfred Hershey and Martha Chase on using radioactive labelled phages to prove that DNA is the genetic material is :

Options (A)–(D) are shown in the figure.

Figure

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Answer not available: no official marking scheme published.

Q23.2025 Main 57/6/1 · Q11 mark

The process of splicing in eukaryotes represents the dominance of the :

  1. DNA world
  2. RNA world
  3. Protein world
  4. Lipid world
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Answer not available: no official marking scheme published.

Q24.2025 Main 57/6/1 · Q21 mark

Regulation of lac operon by repressor is referred to as :

  1. Inducible regulation
  2. Repressible regulation
  3. Negative regulation
  4. Positive regulation
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Answer not available: no official marking scheme published.

Q25.2025 Main 57/6/1 · Q31 mark

SNPs in Human Genome Project refers to :

  1. Polymorphism in repetitive sequences.
  2. Single-base DNA differences.
  3. Single changes in nucleotide of mRNA.
  4. Loss or gain of a gene function.
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Answer not available: no official marking scheme published.

Q26.2025 Main 57/7/1 · Q11 mark

What are minisatellites ?

  1. 10 – 40 bp sized small sequences within the genes.
  2. Short coding repetitive sequences region on the eukaryotic genome.
  3. Short non-coding repetitive sequences forming a large portion of eukaryotic genome.
  4. Regions of coding strand of DNA.
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Answer not available: no official marking scheme published.

Q27.2025 Main 57/7/1 · Q141 mark

Assertion (A): Replication of DNA takes place in S phase of the cell cycle.

Reason (R): DNA replication and cell division cycle should be highly coordinated.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q28.2025 Supp. 57/S/1 · Q91 mark

What would happen if in a gene encoding a polypeptide of 51 amino acids, codon present at 32 position (UAU) is mutated to UGA ?

Select the option that indicates the polypeptide with correct number of amino acids formed after mutation.

  1. A polypeptide of 50 amino acids will be formed.
  2. A polypeptide of 32 amino acids will be formed.
  3. Two polypeptides of 20 and 31 amino acids will be formed.
  4. A polypeptide of 31 amino acids will be formed.
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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q29.2025 Supp. 57/S/1 · Q151 mark

Assertion (A): DNA chemically is less reactive and structurally more stable when compared to RNA.

Reason (R): DNA has 2– OH group present at every nucleotide, which makes it more stable than RNA.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q30.2024 Main 57/1/1 · Q51 mark

You know that there are twenty different types of naturally occurring amino acids and four different types of bases in the DNA. A combination of 3 such bases code for a specific amino acid. If instead there are 96 different amino acids and 12 different bases in the DNA, then the minimum number of combination of bases required to form a codon is :

  1. 6
  2. 8
  3. 2
  4. 4
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(C) 2 [1 mark]

Q31.2024 Main 57/1/1 · Q61 mark

The type of bond represented by the dotted line ‘– – – – –’ in a schematic polynucleotide chain is :

Figure

  1. Hydrogen bond
  2. Peptide bond
  3. N-glycosidic linkage
  4. Phosphodiester bond
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(C) N-glycosidic linkage [1 mark]

Q32.2024 Main 57/2/1 · Q61 mark

In the double helical structure of DNA molecule, the strands are :

  1. identical and complementary
  2. identical and non-complementary
  3. anti-parallel and complementary
  4. anti-parallel and non-complementary
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(C) anti-parallel and complementary [1 mark]

Q33.2024 Main 57/2/1 · Q71 mark

In a ‘transcription unit’, the ‘terminator’ is located towards the :

  1. 3′ end of the template strand
  2. 5′ end of the template strand
  3. 5′ end of the coding strand
  4. 3′ end of the coding strand
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(D) 3′ end of the coding strand. [1 mark]

Q34.2024 Main 57/2/1 · Q131 mark

Assertion (A): RNA is unstable and can mutate at a faster rate.

Reason (R): The presence of 2′ – OH group in every nucleotide of RNA makes it labile and easily degradable.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]

Q35.2024 Main 57/3/1 · Q21 mark

A DNA fragment has 2000 nucleotides, out of which 140 are Adenine. How many bases does this DNA segment possess that have triple hydrogen bonds between them ?

  1. 280
  2. 860
  3. 1720
  4. 1860
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(C) 1720 [1 mark]

Q36.2024 Main 57/3/1 · Q81 mark

In an experiment, E. coli is grown in a medium containing 14NH4Cl. (14N is the light isotope of Nitrogen) followed by growing it for six generations in a medium having heavy isotope of nitrogen (15N). After six generations, their DNA was extracted and subjected to CsCl density gradient centrifugation. Identify the correct density (Light/Hybrid/Heavy) and ratio of the bands of DNA in CsCl density gradient centrifugation.

  1. Hybrid : Heavy, 1 : 16
  2. Light : Heavy, 1 : 31
  3. Hybrid : Heavy, 1 : 31
  4. Light : Heavy, 1 : 05
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(C) Hybrid : Heavy, 1 : 31 [1 mark]

Q37.2024 Main 57/3/1 · Q131 mark

Assertion (A): Primary transcripts in eukaryotes are subjected to splicing to remove the introns.

Reason (R): Primary transcripts contain both exons and introns and the introns are non-functional in eukaryotes.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]

Q38.2024 Main 57/4/1 · Q51 mark

Which was the last of the 24 human chromosome to be completely sequenced ?

  1. Chromosome – 1
  2. Chromosome – 11
  3. Chromosome – 21
  4. Chromosome – X
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(A) Chromosome – 1 [1 mark]

Q39.2024 Main 57/4/1 · Q81 mark

Which of the options has correct identification of ‘P’, ‘Q’ and ‘R’ in the illustration of ‘Central Dogma’ given below ?

Figure

  1. P – Replication, Q – rRNA, R – Transcription
  2. P – Translation, Q – mRNA, R – Transcription
  3. P – Replication, Q – mRNA, R – Translation
  4. P – Transcription, Q – mRNA, R – Translation
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(D) P – Transcription, Q – mRNA, R – Translation [1 mark]

Q40.2024 Main 57/5/1 · Q101 mark

Match the following genes of the lac operon listed in column ‘A’ with their respective products listed in column ‘B’ :

A
Gene
B
Products
a. ‘i’ gene(i) β-galactosidase
b. ‘z’ gene(ii) lac permease
c. ‘a’ gene(iii) repressor
d. ‘y’ gene(iv) transacetylase

Select the correct option :

  1. a – (i), b – (iii), c – (ii), d – (iv)
  2. a – (iii), b – (i), c – (ii), d – (iv)
  3. a – (iii), b – (i), c – (iv), d – (ii)
  4. a – (iii), b – (iv), c – (i), d – (ii)
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(C) (iii), (i), (iv), (ii) [1 mark]

Q41.2024 Main 57/5/1 · Q121 mark

If the sequence of nitrogen bases of the coding strand in a transcription unit is 5′ – ATGAATG – 3′, the sequence of bases in its RNA transcript would be

  1. 5′ – AUGAAUG – 3′
  2. 5′ – UACUUAC – 3′
  3. 5′ – CAUUCAU – 3′
  4. 5′ – GUAAGUA – 3′
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(A) 5′ – AUGAAUG – 3′ [1 mark]

Q42.2024 Supp. 57/S/1 · Q51 mark

Which one of the following codons has dual function ?

  1. AUG
  2. AUC
  3. ACU
  4. ACA
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(A) AUG [1 mark]

Q43.2023 Main 57/1/1 · Q21 mark

Identify the region ‘X’, the factor ‘Y’ and the enzyme ‘Z’ involved in the process of transcription in prokaryote as shown in the schematic representation given below.

Figure

  1. Region ‘X’: Terminator; Factor ‘Y’: Sigma (σ); Enzyme ‘Z’: RNA polymerase
  2. Region ‘X’: Promoter; Factor ‘Y’: Rho (ρ); Enzyme ‘Z’: RNA polymerase
  3. Region ‘X’: Promoter; Factor ‘Y’: Sigma (σ); Enzyme ‘Z’: RNA polymerase
  4. Region ‘X’: Promoter; Factor ‘Y’: Sigma (σ); Enzyme ‘Z’: DNA polymerase
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(C) X = Promoter, Y = Sigma factor, Z = RNA polymerase. [1 mark]

Q44.2023 Main 57/1/1 · Q61 mark

Study the DNA profiles obtained as a result of DNA fingerprinting of a child ‘X’ and three individuals 1, 2 and 3. Which one of the following options shows the possible parents of the child ‘X’ ?

Figure

  1. 1 and 2
  2. 2 and 3
  3. 1 and 3
  4. Only individual 3
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(C) 1 and 3 [1 mark]

Q45.2023 Main 57/3/1 · Q21 mark

Given below is a list of steps Meselson and Stahl carried out in their experiment to prove that DNA replication is semi-conservative. Select the option that gives the correct sequence of steps followed by them.

  1. Bacteria transferred to a N14 medium and sampled every 20 minutes.
  2. All bacteria contain hybrid DNA (N14 DNA and N15 DNA).
  3. Bacteria grown in N15 medium for many generations.
  4. All bacteria contain N15 DNA.
  5. Bacteria contain either all N14 DNA or all hybrid DNA.
  1. (ii) → (iv) → (iii) → (i) → (v)
  2. (i) → (ii) → (v) → (iv) → (iii)
  3. (iii) → (iv) → (i) → (ii) → (v)
  4. (iv) → (iii) → (ii) → (v) → (i)
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(C) (iii) → (iv) → (i) → (ii) → (v) [1 mark]

Q46.2023 Main 57/3/1 · Q121 mark

DNA profiles of the child and three individuals 1, 2 and 3 who claim to be the parents of the child are given below. Select the option that shows the correct actual parent/parents of the child.

Figure

  1. Individual 1 and 3
  2. Individual 1 and 2
  3. Individual 2 and 3
  4. Individual 1 is the only parent of the child amongst 1, 2 and 3
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(A) Individual 1 and 3 [1 mark]

Q47.2023 Main 57/5/1 · Q41 mark

Given below is a sequence of bases in mRNA of a bacterial cell. Identify the amino acid that would be incorporated at codon position 3 and codon position 5 during the process of its translation.

3′ AUCAGGUUUGUGAUGGUACGA 5′

  1. Phenylalanine, Methionine
  2. Cysteine, Glycine
  3. Alanine, Proline
  4. Serine, Valine
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Answer not available: the official marking scheme gives no answer key for this question and awards the mark to everyone who attempts it.

Q48.2023 Supp. 57/C/1 · Q51 mark

Automated DNA sequences work on the principle of a method developed by :

  1. Alec Jeffreys
  2. Francis Crick
  3. Friedrich Miescher
  4. Frederick Sanger
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(D) Frederick Sanger [1 mark]

Q49.SQP 2025-26 · Q41 mark

The experimental proof on the thermal stability of genetic material was first provided by experiments of

  1. Hershey and Chase
  2. Meselson and Stahl
  3. Frederick Griffith
  4. Jacob and Monod
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Official CBSE Sample Paper Marking Scheme · 2025-26

(C) Frederick Griffith [1 mark]

Q50.SQP 2025-26 · Q51 mark

Short stretches of DNA used to identify complementary sequences in a sample are called

  1. Probes
  2. Markers
  3. Primers
  4. Minisatellites
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Official CBSE Sample Paper Marking Scheme · 2025-26

(A) Probes [1 mark]

Q51.SQP 2026-27 · Q51 mark

The initiation step during the process of transcription in bacteria is shown below. Identify A, B, C and D by selecting the option:

Figure

  1. A-RNA polymerase, B-Promoter, C-Sigma factor, D-DNA helix
  2. A-Promoter, B-Sigma factor, C-RNA polymerase, D-DNA helix
  3. A-Promoter, B-RNA polymerase, C-DNA helix, D-Sigma factor
  4. A-Promoter, B-RNA polymerase, C-Sigma factor, D-DNA helix
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Official CBSE Sample Paper Marking Scheme · 2026-27

(D) A - Promoter, B - RNA polymerase, C - Sigma factor, D - DNA helix [1 mark]

Q52.SQP 2026-27 · Q61 mark

Given below is the illustration of the different steps of experiments conducted by MacLeod, McCarty and Avery to find the chemical nature of the 'transforming principle' as DNA. Select the option that incorrectly depicts the step of the experiment.

Options (A)–(D) are shown in the figure.

Figure

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Official CBSE Sample Paper Marking Scheme · 2026-27

(A) No transformation [1 mark]

Q53.SQP 2026-27 · Q71 mark

A scientist was culturing E.coli in a 15NH4Cl medium initially, then he shifted E.coli to a 14NH4Cl medium. The DNA was extracted from E.coli after 60 minutes of transfer to 14NH4Cl Medium and performed density gradient centrifugation to check the isotope distribution in newly replicated DNA. What will be the densities of DNA molecules formed?

  1. 25% hybrid and 75% light
  2. 50% hybrid and 50% light
  3. 75% hybrid and 25% light
  4. 85% hybrid and 15% light
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Official CBSE Sample Paper Marking Scheme · 2026-27

(A) 25% hybrid and 75% light [1 mark]

Q54.SQP 2025-26 · Q101 mark

Which mRNA will be translated to a polypeptide chain containing 8 amino acids?

  1. AUGUUAAUAGACGAGUAGCGACGAUGU
  2. AUGAGACGGACUGCAUUCCCAACCUGA
  3. AUGCCCAACCGUUAUUCAUGCUAG
  4. AUGUCGACAGUCUAAAACAGCGGG
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Official CBSE Sample Paper Marking Scheme · 2025-26

(B) AUGAGACGGACUGCAUUCCCAACCUGA [1 mark]

Q55.SQP 2026-27 · Q101 mark

A stretch of an euchromatin has 200 nucleosomes, how many base pairs (bp) will be there in the stretch and what would be the length of the typical euchromatin?

  1. 20,000 bp and 13,000×10-9 m
  2. 10,000 bp and 10,000×10-9 m
  3. 40,000 bp and 13,600×10-9 m
  4. 40,000 bp and 13,900×10-9 m
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Official CBSE Sample Paper Marking Scheme · 2026-27

(C) 40,000 bp and 13,600×10-9 m [1 mark]

Q56.2026 Supp. 57/7/1 · Q192 marks

Attempt either (a) or (b).

(a) Name the scientist who proposed the central dogma in molecular biology. Give its schematic representation.

OR

(b) Write four criteria to be fulfilled by a molecule to act as a genetic material.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) • Francis Crick [½]

Figure

(Award ½ mark for representing each process) [1½]

OR

(b) A molecule that can act as a genetic material must fulfil the following criteria: [½ × 4]

  • It should be able to generate its replica (Replication).
  • It should be stable chemically and structurally.
  • It should provide the scope for slow changes (mutation) that are required for evolution.
  • It should be able to express itself in the form of Mendelian Characters.

[Total: 2 marks]

Q57.2025 Main 57/4/1 · Q182 marks

Assume that the given mRNA (start site is not depicted) is theoretically translated in two reading frames.

  1. Translation starting from the first nucleotide (Reading frame 1)
  2. Translation starting from the second nucleotide (Reading frame 2)

Frame 1

5′ – CUCGCUUGCCGAUCAAGGGUUA – 3′

Frame 2

5′ – GUGGCACUCAGUCCUUAAUGGCG – 3′

Answer the following question :

How many amino acids will be specified in case (a) and case (b) on translation ? Justify your answer.

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Answer not available: no official marking scheme published.

Q58.2025 Main 57/5/1 · Q182 marks

Given below are the diagrammatic representations of the replicating fork of DNA in E. coli. Study the diagrams and answer the questions that follow.

Figure

  1. Which one of the three diagrams (i), (ii) or (iii) is the correct representation of the replicating fork of DNA replication ? Explain your answer.
  2. Name the enzyme used in E. coli to join the newly synthesised fragments of DNA.
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Answer not available: no official marking scheme published.

Q59.2025 Main 57/6/1 · Q202 marks

Study the given molecular structure of double-stranded polynucleotide chain of DNA and answer the questions that follow.

Figure

  1. How many phosphodiester bonds are present in the given double-stranded polynucleotide chain ?
  2. How many base pairs are there in each helical turn of double helix structure of DNA ? Also write the distance between a base pair in a helix.
  3. In addition to H-bonds, what confers additional stability to the helical structure of DNA ?
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Answer not available: no official marking scheme published.

Q60.2024 Main 57/3/1 · Q192 marks

Consider the given data of a hypothetical small portion of mRNA that codes for a functional polypeptide chain and answer the questions that follow :

mRNA   5′ – UCAUUACCACGAUUCUUUAAAAGA – 3′

  1. How many amino acids will be formed from the given codons, if substitution of ‘U’ by ‘C’ takes place at the 5th codon ? Explain your answer.
  2. Write the number of amino acids that would be in the polypeptide synthesised by a similar mRNA as above, where in the fourth codon instead of ‘C’ there is ‘U’. Justify your answer.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) 8 amino acids; genetic code is read in triplets and there is no change in the number of triplets / no change in reading frame. [½ × 2]

(b) 3 amino acids; the 4th codon now reads as UGA – a stop codon. [½ × 2]

[Total: 2 marks]

Q61.2024 Main 57/5/1 · Q212 marks

If the base adenine constitutes 31% of an isolated DNA fragment, then write what will be the expected percentage of the base cytosine in it. Explain how did you arrive at the answer given.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1
  • Amount of A = T and G = C and A + G = C + T / ratio between adenine and thymine and ratio between guanine and cytosine are constant and equals one / A/T = G/C = 1 [½]
  • If A = 31% then T = 31% [½]
  • A + T = 62%; C + G = 100 – 62 = 38% [½]
  • C = 38/2 = 19% [½]

[Total: 2 marks]

Q62.2024 Supp. 57/S/1 · Q192 marks

(a) Given below is a representation of the “Central dogma” :

Figure

Identify ‘A’, ‘B’ and ‘C’ in the above representation.

(b) What does the Central dogma state in molecular biology ? Write an example where it is not applicable.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) A – Replication; B – Transcription; C – Translation [1] (Award 1 mark, if any two are correct)

(b) Central dogma states the flow of genetic information in a cell (DNA → RNA → Protein) [½]

Viruses in which flow of information is in reverse direction, that is from RNA to DNA / HIV / Retrovirus. [½]

[Total: 2 marks]

Q63.2023 Supp. 57/C/1 · Q182 marks

List the different criteria (4 points) that a molecule must fulfil to act as a genetic material.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

A molecule that can act as a genetic material must fulfil the following criteria:

  • It should be able to generate its replica (Replication).
  • It should be stable chemically and structurally.
  • It should provide the scope for slow changes (mutation) that are required for evolution.
  • It should be able to express itself in the form of ‘Mendelian characters’.

[½ × 4 = 2 marks]

Q64.SQP 2025-26 · Q182 marks

How is the rate of initiation of RNA polymerase at a given promoter in a transcriptional unit of prokaryotes regulated?

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Official CBSE Sample Paper Marking Scheme · 2025-26

In a transcription unit, the activity of RNA polymerase at a given promoter is regulated by accessory proteins that have an ability to recognise start sites. [1]

These regulatory proteins can act both positively (activators) and negatively (repressors) with the operator which is adjacent to the promoter in an operon unit. [1]

Q65.2026 Main 57/2/1 · Q273 marks

(a) What is a genetic code ? [1]

(b) Why did scientists feel the need to propose a genetic code ? [1]

(c) Genetic code is said to be ‘Universal’. Why ? [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) Genetic Code: It is the sequence of three nitrogenous bases (codon) on mRNA that directs the sequence of amino acids during synthesis of proteins. [1]

(b) To provide complementarity between nucleotides and amino acids / to support the notion that change in nucleic acids was responsible for change in amino acids in proteins / to establish the correlation between genetic information and protein synthesis / to constitute a combination of four bases to code twenty amino acids. [1]

(c) As a codon from bacteria to human code for the same amino acid. [1]

[Total: 3 marks]

Q66.2026 Main 57/3/1 · Q233 marks

(a) What is meant by translation in protein synthesis ? [1]

(b) Explain charging of tRNA (aminoacylation of tRNA) and mention its importance in the process of translation. [2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) It is the process of polymerisation of amino acids to form a polypeptide as per the sequence coded in mRNA [1]

(b) Amino acids are activated in the presence of ATP and then linked to their corresponding cognate tRNA. [1]

Importance – The formation of peptide bond is favoured energetically. [1]

[Total: 3 marks]

Q67.2026 Main 57/4/1 · Q233 marks

(a) Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical coding strand given below :

Figure

(b) Write the RNA strand transcribed from the above transcription unit along with its polarity.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) [½ × 4]

Figure

(b) The RNA transcribed is 5′ A U G C A U G C A U A G 3′ [1]

// (alternative orientation)

(a) [½ × 4]

Figure

(b) The RNA transcribed as 5′ G A U A C G U A C G U A 3′ [1]

[Total: 3 marks]

Q68.2026 Main 57/5/1 · Q233 marks

Draw a labelled diagram of a ‘replicating fork’ showing the polarity. Why does DNA replication occur within such ‘fork’ ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

Figure

[2]

Replication of DNA occurs in small replication fork because DNA being a long molecule separation of the two strands along its entire length requires a very high amount of energy. [1]

[Total: 3 marks]

Q69.2025 Main 57/1/1 · Q253 marks

Study the diagram given below and answer the questions that follows :

Figure

(a) Identify the structure shown in the above figure.

(b) Identify the labels P and Q.

(c) Write the nature of histone proteins.

(d) Distinguish between Euchromatin and Heterochromatin.

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Answer not available: no official marking scheme published.

Q70.2025 Main 57/4/1 · Q243 marks

Study the given below single strand of deoxyribonucleic acid depicted in the form of a “stick” diagram with 5′ – 3′ end directionality, sugars as vertical lines and bases as single letter abbreviations and answer the questions that follow.

Figure

  1. Name the covalent bonds depicted as (a) and (b) in the form of slanting lines in the diagram.
  2. How many purines are present in the given “stick” diagram ?
  3. Draw the chemical structure of the given polynucleotide chain of DNA.
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Answer not available: no official marking scheme published.

Q71.2025 Main 57/6/1 · Q253 marks

Explain how the addition of lactose in the medium regulates the switching on of the lac operon in bacteria.

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Answer not available: no official marking scheme published.

Q72.2025 Main 57/7/1 · Q243 marks

A segment of DNA, ‘TTG AGG GGG ATG’ was translated into an oligopeptide with the amino acids, Lysine – Serine – Proline – Tyrosine.

  1. Write the codons in correct sequence for the four amino acids.
  2. If first adenine in DNA is substituted by guanine, what will be the sequence of amino acids in the new oligopeptide ?
  3. Write the anticodons for these amino acids.
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Answer not available: no official marking scheme published.

Q73.2025 Supp. 57/S/1 · Q243 marks

Draw a labelled schematic sketch of replication fork of DNA. Explain the role of any two enzymes involved in DNA replication.

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

Figure

(Any two correct labelling) [½ + ½]

Award marks for continuous and discontinuous synthesis only when polarity is correct.

• DNA dependent DNA polymerase – causes polymerization of deoxyribonucleotides [½ + ½]

• Ligase – joins fragments of short discontinuous strand during DNA replication [½ + ½]

(Any other enzyme with correct role)

[Total: 3 marks]

Q74.2024 Main 57/1/1 · Q233 marks

(a) Name and write two characteristics of the type of DNA that forms the basis of DNA fingerprinting technique.

(b) Mention any two applications of this technique.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a) Satellite DNA / Repetitive DNA / VNTR [1]

Do not code for any protein, form a large portion of human genome, show high degree of polymorphism (any two points) [½ + ½]

(b) Useful in forensic applications, helps in determining population and genetic diversities, forms the basis of paternity testing, to study evolution, to trace path of hereditary diseases. (any two applications) [½ + ½]

[Total: 3 marks]

Q75.2024 Main 57/2/1 · Q243 marks

(a) Explain the process by which amino acid gets attached to the tRNA molecule during translation process.

(b) How does the translation process get terminated ?

(c) Expand ‘UTR’. Where are they located ?

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a) Amino acids are activated in the presence of ATP, and linked to their cognate tRNA – a process called charging of tRNA or aminoacylation of tRNA. [½ + ½]

(b) At the end of translation a release factor binds to the stop codon terminating translation / when ribosome moves to the stop codon (UAA/UAG/UGA) release factor binds to the stop codon terminating translation. [1]

(c) Untranslated regions (UTR); UTRs are present at both 5′-end (before start codon / AUG) and at 3′-end (after stop codon) of mRNA. [½ + ½]

[Total: 3 marks]

Q76.2024 Main 57/5/1 · Q263 marks
  1. Why does DNA replication occur within a replication fork and not in its entire length simultaneously ?
  2. “DNA replication is continuous and discontinuous on the two strands within the replication fork.” Explain with the help of a schematic representation.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a) Energetically very expensive process or there is requirement of high energy in the process of replication. [1]

(b) DNA-dependent DNA polymerases catalyse polymerisation according to template strand only in one direction that is 5′ → 3′. On one strand with polarity 3′ → 5′, the replication is continuous, while on other strand with polarity 5′ → 3′ the replication is discontinuous. [½ × 4]

Figure

(Labels: polarity, template DNA / parental strands, continuous synthesis, discontinuous synthesis – ½ mark each.) No marks will be awarded if polarity is not correctly shown.

[Total: 3 marks]

Q77.2024 Main 57/5/1 · Q273 marks

Explain the processing of heterogeneous nuclear RNA (hnRNA) into a fully functional mRNA in eukaryotes. Where does this processing occur in the cell ?

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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1
  • hnRNA undergoes capping at 5′-end (methyl guanosine triphosphate or mGppp), and tailing at 3′-end (with polyA tail or adenylate residues). Further splicing is carried out, where the non-coding sequences or introns are removed, and coding sequences or exons are joined together in a defined manner. [½ × 5] (or by the diagram below)

Figure

  • Site of processing of hnRNA – nucleus. [½]

[Total: 3 marks]

Q78.2024 Supp. 57/S/1 · Q243 marks

(a) Name the enzyme responsible for the transcription of tRNA.

(b) Draw a labelled diagram of an initiator tRNA charged with its respective amino acid.

(c) Explain the role of the initiator tRNA in initiation of protein synthesis.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) RNA polymerase III / RNA polymerase [1]

(b) [½ + ½]

Figure

(½ mark for correct anticodon and ½ mark for correct amino acid)

(c) It recognizes the start codon (AUG) and binds to mRNA. [1]

[Total: 3 marks]

Q79.2023 Main 57/3/1 · Q223 marks
    1. How many types of RNA polymerases are there in a eukaryote cell ? Mention which one of them transcribes hnRNA.
    2. Write the changes that hnRNA undergoes before it leaves the nucleus as mRNA.

OR

  1. The length of DNA in any cell is far greater than the dimension of its nucleus. Explain how this enormous DNA is packaged in a eukaryotic cell.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) (i) • 3 types [½]

• RNA Polymerase II [1]

(ii) Splicing/Introns are removed and exons are joined in a definite order, undergoes capping/at 5′ end where unusual nucleotide (methyl guanosine triphosphate) is added, tailing/at 3′ end where (200–300) adenylate residues are added. [½×3]

OR

(b) A set of positively charged proteins called histones, due to presence of lysine and arginine (basic amino acids), holds the negatively charged DNA around it in a coiled manner, histones are organised to form a unit of eight molecules (histone octamer), a typical nucleosome contains 200 bp of DNA helix, Nucleosomes constitute repeating units of a structure in nucleus called chromatin thread (like bodies as “beads on string” structure in a nucleus). [½×6]

[Total: 3 marks]

Q80.2023 Main 57/5/1 · Q233 marks

Human Genome Project (HGP) was a mega project launched in the year 1990 with some important goals.

  1. Enlist any four prime goals of HGP.
  2. Name any one common non-human animal model organism which has also been sequenced thereafter.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) (i) Identify all the approximately 20,000 – 25,000 genes in human DNA

(ii) Determine the sequences of the 3 billion chemical base pairs that make up human DNA.

(iii) Store this information in databases.

(iv) Improve tools for data analysis.

(v) Transfer related technologies to other sectors, such as industries.

(vi) Address the ethical/legal/social issues (ELSI) that may arise from the project.

(Any four) [½×4]

(b) Caenorhabditis elegans/Drosophila (or any other correct example) [1]

[Total: 3 marks]

Q81.2023 Supp. 57/C/1 · Q253 marks

(a) Mention the difference between the ‘R’ strain and ‘S’ strain in Streptococcus pneumoniae.

(b) Write the steps followed by Griffith during the course of his experiment and the conclusion he arrived at, at the end of his experiment.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a)

‘R’ Strain‘S’ Strain
Non-virulentVirulent
No polysaccharide coatHave polysaccharide coat

(½ mark for any one difference) [½]

(b)

S strain → Inject into mice → Mice die

R strain → Inject into mice → Mice live

S strain (heat killed) → Inject into mice → Mice live

S strain (heat killed) + R strain → Inject into mice → Mice die [½ × 4]

Conclusion: He concluded that the R strain bacteria had somehow been transformed by the heat-killed S strain bacteria / Some ‘transforming principle’ transferred from heat killed S strain and transform R strain into S strain. [½]

[Total: 3 marks]

Q82.2026 Main 57/4/1 · Q294 marks

Read the following passage and answer the questions that follow :

Gene expresses itself in a cell system as a protein/enzyme. How an expression of gene occurs in a cell system, when it needs to occur and how the gene expression is regulated in prokaryotic cell system was studied by the combined efforts of Francois Jacob, a geneticist and Jacques Monod, a biochemist. For their work on lactose metabolism in E. coli and introducing the concept of ‘lac operon’, they were awarded the Nobel Prize in 1965.

(a) It is said that ‘lac operon’ has to be operational at a very low level in the bacterial cell all the time. Justify. [1]

(b) (i) How are the structural genes activated in the lac operon in E. coli ? [1]

OR

(b) (ii) Why is the regulation of operon called negative regulation ? [1]

(c) Why is lac operon said to be a transcriptionally regulated system ? [2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) It is necessary for the entry of lactose into the cell. [1]

(b) (i) In the presence of lactose the repressor is inactivated by the interaction with the inducer that allow RNA polymerase access to promoter and transcription proceeds. [1]

OR

(b) (ii) It is called negative regulation because the operon is switched off in the absence of inducer and transcription is prevented. [1]

(c) In the absence of the inducer the repressor binds to the operator and prevents transcription of the structural gene. [1]

When inducer or lactose is present it prevents the repressor from binding to the operator hence regulation is at the transcriptional level. [1]

[Total: 4 marks]

Q83.2026 Main 57/5/1 · Q294 marks

Read the following passage and answer the questions that follow :

The process of copying genetic information from template strand of DNA into RNA is called transcription. It is mediated by RNA polymerase. Transcription takes place in the nucleus of eukaryotic cells. In transcription, only a segment of DNA and only one of the strands is copied into RNA.

(a) Why is the strand of DNA with 3′ → 5′ polarity transcribed and not the other strand of 5′ → 3′ polarity ? [1]

(b) (i) Why is hnRNA required to undergo splicing ? [1]

OR

(b) (ii) Mention the two additional processes which hnRNA needs to undergo after splicing to become functional. [1]

(c) Why is only one strand of the DNA transcribed ? Give two reasons. [1+1 = 2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a) Because RNA polymerase carries out polymerisation in 5′ → 3′ direction only. [1]

(b) (i) Since hnRNA contains both exons or coding sequence and introns or non coding sequence, it undergoes splicing to remove the introns or non-coding sequence / for the removal of introns. [1]

OR

(b) (ii) Capping / addition of mGppp (Methyl Guanosine Triphosphate) at 5′ of hnRNA. [½]

Tailing / Adenylate residue (200-300) are added at the 3′ end in a template independent manner as poly A tail. [½]

(c) If a segment of DNA produces two polypeptides the genetic information machinery becomes complicated / two different polypeptides will be formed and one of which may not be useful for the cell. [1]

The two complementary RNA molecules produced simultaneously will form a double-stranded RNA that will not be translated into polypeptides. [1]

[Total: 4 marks]

Q84.2025 Main 57/2/1 · Q304 marks

The process of copying the genetic information from one strand of DNA into RNA is termed as transcription. The principle of complementarity of bases governs the process of transcription, also except that uracil comes in place of thymine.

Study the complete transcription unit given below and answer the following questions :

Figure

(a) Name the main enzyme involved in the process of transcription.

(b) Identify coding strand and template strand of DNA in the transcription unit.

Attempt either sub-part (c) or (d) :

(c) Identify (C) and (D) in the diagram, mention their significance in the process of transcription.

OR

(d) Describe the location of (C) and (D) in the transcription unit.

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Answer not available: no official marking scheme published.

Q85.2024 Main 57/4/1 · Q304 marks

Read the following passage :

Generally, in eukaryotic cells the average length of a transcription unit along a DNA molecule is about 8,000 nucleotides, so the RNA product of the transcription is also that long. But it only takes about 1200 nucleotides from the above RNA product to translate average sized polypeptide of 400 Amino acids.

  1. Name this RNA product transcribed from the DNA that subsequently translates into a polypeptide of 400 amino acids. Mention the enzyme responsible for transcribing this type of RNA from the DNA.
  2. Name and explain the process the RNA molecule transcribed from 8000 nucleotide long DNA undergoes to be able to translate a polypeptide of 400 amino acids.
  3. Write the number of RNA polymerases involved in the transcription of DNA in a prokaryote and eukaryotes.

OR

  1. Mention the difference in the site of transcription in a prokaryote and eukaryote cell.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) hnRNA / heterogeneous nuclear RNA [½]; RNA polymerase II [½]

(b) hnRNA undergoes capping at 5′ end (methyl guanosine triphosphate / mGppp), and tailing at 3′ end (with poly A tail or adenylate residue), further splicing is carried out, where non-coding sequences or introns are removed and coding sequences or exons are joined together / diagrammatic representation with given markers can also be considered. [½ × 4]

(c) In prokaryotes – 1 [½]; in eukaryotes – 3 [½]

OR

(c) In prokaryotes the transcription takes place in the cytoplasm / cytosol whereas in eukaryotes transcription occurs in the nucleus. [1]

[Total: 4 marks]

Q86.2023 Main 57/1/1 · Q294 marks

Gene expresses itself in a cell system as a protein/enzyme. How does an expression of gene occur in a cell system and when does it need to occur, and how the gene expression is regulated in a prokaryote cell system was studied by the combined efforts of Jacque Monod, the biochemist and Francois Jacob, the geneticist. For their work on lactose metabolism in E. coli and introducing the concept of “lac operon” they were awarded the Nobel Prize in 1965.

  1. Why is lac operon said to be a transcriptionally regulated system ?
  2. It is said that “the lac operon has to be operational at a very low level in the bacterial cell all the time.” Justify.
  3. Why is the regulator gene in lac operon marked as ‘i’ gene ?
  4. Draw a schematic diagram of lac operon in absence of inducer in the culture medium of the bacteria.

OR

  1. Draw a schematic diagram of lac operon in the presence of inducer in the culture medium of the bacteria.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) In presence of lactose repressor protein does not bind to the operator region (O) and allows RNA polymerase to transcribe the operon. [1]

//

In absence of lactose repressor protein binds to the operator region (O) and prevents RNA polymerase from transcribing the operon. [1]

(b) Presence of Permease enzyme coded by gene ‘y’ is required that allows lactose to enter the cell for switching on the operon/so that lactose enter inside the cell. [½]

(c) ‘i’ stands for ‘inhibitor’/this gene transcribes repressor protein which binds to the ‘operator’ site and switch off the operon. [½]

(d) (In absence of inducer)

Figure

[½×4]

OR

(d) (In presence of inducer)

Figure

[½×4]

[Total: 4 marks]

Q87.2026 Main 57/1/1 · Q325 marks

(a) (i) Watson and Crick’s discovery of double helical structure was based on which two findings. Also mention the name of the scientists associated with these findings.

(ii) Write down the salient features of double helix structure of DNA (any three points).

OR

(b) Given below is a stretch of DNA showing the coding strand of structural gene of transcription unit.

5′ - ATG ACC GTA TTT TCT GTA GTG CCC GTA CTT CAG GCA TTA 3′

(i) Write the corresponding template strand and m-RNA strand that will be transcribed along with its polarity.

(ii) If GUA of transcribed mRNA is an intron, then depict the sequence involved in formation of mRNA / mature / processed hnRNA strand :

(1) In a bacterium

(2) In humans

(iii) How many amino acids the resulting polypeptide will have after the process of translation in humans ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) (i) • Maurice Wilkins and Rosalind Franklin – X-ray diffraction studies of DNA. [½ + ½]

• Erwin Chargaff – For a double stranded DNA the ratios between Adenine and Thymine and Guanine and Cytosine are constant and equal to one. [½ + ½]

(ii) The salient features of double helix structure of DNA are as follows (any three points) [1 × 3]

  • It is made of two polynucleotide chains, in which backbone is constituted by sugar-phosphate and the bases project inside.
  • The two chains have antiparallel polarity.
  • The bases in two strands are paired through hydrogen bonds forming base pairs (bp). Adenine forms double hydrogen bond with thymine and Guanine forms triple hydrogen bond with cytosine.
  • The two chains are coiled in a right handed fashion. The pitch of helix is 3.4 nm and 10 bp in each turn. Distance between base pair in a helix is approximately 0.34 nm.
  • The plane of one base pair stacks over the other in double helix.

OR

(b) (i) Template strand
3′ – TAC TGG CAT AAA AGA CAT CAC GGG CAT GAA GTC CGT AAT – 5′ [1]

m-RNA
5′ – AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA – 3′ [1]

(ii) (1) In a bacterium
5′ – AUG ACC GUA UUU UCU GUA GUG CCC GUA CUU CAG GCA UUA – 3′ [1]

(2) In humans
5′ – mGppp AUG ACC UUU UCU GUG CCC CUU CAG GCA UUA – poly A tail 3′ [1]

(iii) It will have 10 amino acids. [1]

[Total: 5 marks]

Q88.2026 Main 57/2/1 · Q325 marks

(a) Observe the segment of mRNA given below and answers the questions given below :

Figure

(i) Explain and illustrate the steps involved to make fully processed hn RNA. [3]

(ii) Gene encoding RNA polymerase I and RNA polymerase III have been affected by mutation. Explain its impact on the synthesis of polypeptide giving reasons. [2]

OR

(b) In a particular cattle, the coat colour red and white are controlled by single pair of alleles. A calf that receives allele for red coat from mother and allele for white coat from father shows equal number of red and white hairs on its coat.

(i) Is this an example of Co-dominance or incomplete dominance. Give a reason for your answer. [1]

(ii) With the help of a genetic cross upto F2 generation, explain what will be the consequent phenotype of calf when

  1. Red is dominant over white, [2]
  2. Red is incompletely dominant. [2]
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) (i)

  • Splicing – Introns or non-coding sequences are removed and axon [exon] or coding sequences are joined. [1]

Figure

  • Capping – An unusual nucleotide Methyl guanosine triphosphate is added at 5' end of hn RNA [1]

Figure

  • Tailing – About 200 – 300 adenylate residues are added at 3' end of hnRNA in a template independent manner [1]

Figure

(ii) Polypeptide will not be synthesized as translation will not take place because RNA I transcribes rRNAs which plays catalytic and structural role in protein synthesis, RNA polymerase III helps in transcription of tRNA which is the adaptor molecule and transfers amino acids to the site of protein synthesis. [1+1]

OR

(b) (i) Co-dominance, as two different alleles for a genetic trait are able to express / both characters are expressed equally. [½+½]

(ii) (1) Red is dominant over white

If pure breeding red coat cattle are represented as ‘RR’ and pure breeding white coat cattle as ‘rr’. Then

RR (Red coat) × rr (White coat) [½]

Gametes: R, R and r, r

F1 generation: Rr (All with red coat) [½]

Rr × Rr (gametes R, r and R, r)

F2 generation [½]

Rr
RRR
Red coat
Rr
Red coat
rRr
Red coat
rr
White coat

F2 phenotypic ratio: 3 Red Coat : 1 White Coat [½]

(2) Red is incompletely dominant

RR (Red coat) × rr (White coat) [½]

Gametes: R, R and r, r

F1 generation: Rr (All with Pink coat) [½]

Rr × Rr (gametes R, r and R, r)

F2 generation [½]

Rr
RRR
Red coat
Rr
Pink coat
rRr
Pink coat
rr
White coat

F2 phenotypic ratio: 1 Red : 2 Pink : 1 White [½]

[Total: 5 marks]

Q89.2026 Main 57/3/1 · Q315 marks

(a) (i) Describe the series of experiments conducted by Frederick Griffith. Comment on the significance of the result obtained. [3]

(ii) State the contribution of Avery, MacLeod and McCarty. [2]

OR

(b) The following questions are based on haemophilia.

(i) Work out the crosses between :

  1. Normal female and Haemophilic male
  2. Carrier female and Normal male
  3. Carrier female and Haemophilic male

(ii) Write the conclusions you draw from these crosses. Comment on the type of inheritance of the disease.

(Use : X – Normal, Xh – Haemophilic) [1+1+1+2 = 5]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) (i) Griffith worked on Streptococcus pneumoniae bacteria having two strains: S strain (Smooth) with mucous (polysaccharide) coat was virulent / pathogenic; R strain (Rough) has no coat and was non virulent / non pathogenic [½]

Griffith took a Mice and injected these bacteria in a series of experiments

S strain → Inject into Mice → Mice Died [½]

R strain → Inject into Mice → Mice live [½]

He killed S strain bacteria with heat: S strain (heat killed) → Inject into Mice → Mice live [½]

S strain (heat killed) + R strain (live) → Inject into mice → Mice die. He recovered living S bacteria from the body of dead mice. [½]

Conclusion: Griffith concluded that some “transforming principle” transferred from the heat-killed S strain, had enabled the R strain to synthesise a smooth polysaccharide coat and become virulent due to the transfer of the genetic material. [½]

(ii) MacLeod, McCarty & Avery: They worked to determine the biochemical nature of “transforming principle” in Griffith’s experiment.

They purified biochemical (DNA, RNA, protein etc) from heat killed S cells to see which could transform live R cells into S cells. [½]

They discovered that DNA alone from S cells transformed R cells. [½]

They also discovered that protein digesting Proteases and RNA digesting RNases did not affect transformation. [½]

But when DNA was digested by DNase there was no transformation. So DNA is hereditary material. [½]

OR

(b) (i) (I) Normal female × Haemophilic male [½+½]

Figure

(II) Carrier female × Normal male [½+½]

Figure

(III) Carrier female × Haemophilic male [½+½]

Figure

(ii) Conclusions

  • Haemophilia is transmitted from unaffected carrier female to some of the male progeny / The possibility of a female being haemophilic is extremely rare because mother of such a female has to be a carrier and father haemophilic which is not possible in the later stage of life. [1]
  • Sex linked recessive disorder. [1]

[Total: 5 marks]

Q90.2026 Supp. 57/7/1 · Q325 marks

Attempt either (a) or (b).

(a) (i) How and why is charging of tRNA essential in the process of translation ?

(ii) State the function of ribosome as a catalyst in translation in bacteria.

(iii) Explain the process of binding of ribosomal units to mRNA during protein synthesis.

(iv) Where are untranslated regions located in mRNA ? Mention their role.

OR

(b) (i) State the arrangement of different genes, that in bacteria is referred to as ‘operon’.

(ii) Draw a schematic labelled illustration of lac operon in a switched on state.

(iii) Describe the role of lactose in lac operon.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) (i) • The amino acids are activated in the presence of ATP and linked to their cognate tRNA [1]

• It is necessary for translation as the formation of peptide bond between two such charged tRNAs would be favoured energetically when they are brought close enough. [1]

(ii) It catalyses the process of peptide bond formation [1]

(iii) When small subunit of ribosome binds to the mRNA at the start codon (AUG) that is recognised by the initiator tRNA the process of protein synthesis from mRNA begins, and the two sites on the large subunit of ribosome bind to the amino acids which are close to each other for the formation of peptide bond. [½ + ½]

(iv) • At 5′ end before the start codon and at the 3′ end after the stop codon [½]

• They are required for efficient translation process. [½]

OR

(b) (i) A regulatory gene, a promoter, an operator, structural genes [1]

(ii)

Figure

[½ × 6]

(iii) Lactose is the inducer / it inactivates repressor. [1]

[Total: 5 marks]

Q91.2025 Main 57/7/1 · Q315 marks
  1. Write the features a molecule should have to act as a genetic material. In the light of the above features, evaluate and justify the suitability of the molecule that is preferred as an ideal genetic material.

OR

  1. Differentiate between the following :
    1. Polygenic Inheritance and Pleiotropy
    2. Dominance, Codominance and Incomplete dominance
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Answer not available: no official marking scheme published.

Q92.2024 Main 57/1/1 · Q325 marks

(a) Stability, as one of the properties of genetic material, was very evident in one of the very early experiments in genetics. Name the scientist and describe his experiment. State the conclusion he arrived at.

OR

(b) A tall pea plant bearing violet flowers with unknown genotype is given. Find the genotype by working out different crosses only by selfing the plants. Write the genotypic and phenotypic ratios of each cross shown by you.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a)

Frederick Griffith [½]

Took two strains of Streptococcus pneumoniae bacteria and inject them into mice

R strain – Rough and non-virulent [½]

S strain – Smooth and virulent (with mucous coat) [½]

S strain → Inject into mice → Mice die [½]

R strain → Inject into mice → Mice live [½]

S strain (heat-killed) → Inject into mice → Mice live [½]

S strain (heat-killed) + R strain (live) → Inject into mice → Mice die [½]

Conclusion:

R-strain bacteria had been transformed by heat killed S-strain. [½]

‘Transforming principle’ transferred from heat killed S-strain and enabled R-strain to synthesise a smooth polysaccharide coat. This must be due to the transfer of the genetic material. [½]

Heat which killed bacteria did not destroy some of the properties of genetic material which shows stability of genetic material. [½]

OR

(b)

Tall pea plant with violet flowers can have 4 types of genotypes: TTVV, TtVV, TtVv, TTVv [½ × 4]

Case I – TTVV × TTVV

Genotype: TTVV [½]; Phenotype: All will be tall and violet. [½]

Case II – TtVV × TtVV

Gametes: TV, tV

GametesTVtV
TVTTVV Tall VioletTtVV Tall Violet
tVTtVV Tall VioletttVV Dwarf Violet

Phenotypic ratio: Tall Violet : Dwarf Violet = 3 : 1 [½]

Genotypic ratio: TTVV : TtVV : ttVV = 1 : 2 : 1 [½]

Case III – TtVv × TtVv

Gametes: TV, Tv, tV, tv

GametesTVTvtVtv
TVTTVV
Tall violet
TTVv
Tall Violet
TtVV
Tall Violet
TtVv
Tall Violet
TvTTVv
Tall Violet
TTvv
Tall white
TtVv
Tall Violet
Ttvv
Tall white
tVTtVV
Tall violet
TtVv
Tall Violet
ttVV
dwarf violet
ttVv
dwarf Violet
tvTtVv
Tall violet
Ttvv
Tall white
ttVv
dwarf violet
ttvv
dwarf white

Phenotypic ratio: Tall Violet : Dwarf violet : Tall white : dwarf white = 9 : 3 : 3 : 1 [½]

Genotypic ratio: TTVV : TTVv : TtVV : TtVv : TTvv : ttVV : ttVv : Ttvv : ttvv = 1 : 2 : 2 : 4 : 1 : 1 : 2 : 2 : 1 [½]

Case IV – TTVv × TTVv

Gametes: TV, Tv

GametesTVTv
TVTTVV
Tall Violet
TTVv
Tall Violet
TvTTVv
Tall Violet
TTvv
Tall White

Phenotypic ratio: Tall violet : Tall white = 3 : 1 [½]

Genotypic ratio: TTVV : TTVv : TTvv = 1 : 2 : 1 [½]

(Any three self cross shown and explained with genotypic and phenotypic ratio)

[Total: 5 marks]

Q93.2024 Main 57/4/1 · Q325 marks
  1. Study the schematic diagram given below and answer the questions that follow :

Figure

  1. Identify the polarity from ‘X’ to ‘X̄’ in the mRNA segment shown. Mention how many more amino acids can be added to the polypeptide that is being translated and why.
  2. Write the initiating codon for translation, its anticodon and the amino acid it codes for.
  3. Explain the charging of an adaptor molecule. Why this molecule needs to be charged ?

OR

  1. Why is sickle-cell anaemia, a human blood disorder so named ?
  2. Explain the genetic basis that results in the expression of this disorder.
  3. Work out a cross to explain how normal parents may have a sickle-cell anaemic child.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) (i) X to X̄ is 5′ → 3′; no more amino acids will be added as the last codon UAA is a stop codon. [½ × 3]

(ii) AUG; anticodon – UAC; amino acid – methionine [½ × 3]

(iii) The amino acids are activated in the presence of ATP, and linked to their cognate tRNA or the adapter molecule. [½ + ½] Amino acids are activated so peptide bonds can be formed using this energy. [1]

OR

(b) (i) The RBC in such patients takes up a sickle shape instead of biconcave. [1]

(ii) The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val), at the sixth position of the beta globin chain of the haemoglobin molecule / The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene, from GAG to GUG. [1 + 1]

(iii)

Figure

[½ × 4]

[Total: 5 marks]

Q94.2024 Supp. 57/S/1 · Q325 marks

(a) (i) Absence of lactose in the culture medium affects the expression of lac operon in E. coli. Why and how ? Explain.

(ii) Write any two ways in which the gene expression is regulated in eukaryotes.

OR

(b) (i) By taking two suitable traits in a pea plant, work out a dihybrid cross up to F2 generation.

(ii) State the Mendel’s law derived from such a cross only.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) (i) Lactose acts as an inducer molecule; in its absence the repressor protein binds to the operator region. [1 + 1]

RNA polymerase is prevented from transcribing the operon. [1]

(ii) By preventing formation of primary transcript, by preventing splicing, by preventing transport of mRNA from nucleus to the cytoplasm, by preventing formation of protein from mRNA. (Any two) [1 + 1]

OR

(b) (i) [½ + ½ + ½ + ½ + 1 + 1]

Figure

(ii) The law states that ‘when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters’. [1]

[Total: 5 marks]

Q95.2023 Main 57/1/1 · Q315 marks
  1. Meselson and Stahl carried out an experiment to prove the nature of DNA replication. Recall the experiment and answer the following questions.
    1. Which two types of nitrogen were used by them in their experiment and why ?
    2. Why did they take samples of E. coli at definite time intervals for their observation ?
    3. State the role of caesium chloride density gradient in their experiment.
    4. Write the conclusions they arrived at.

OR

    1. A true breeding tall pea plant with round seeds is crossed with a recessive dwarf pea plant having wrinkled seeds. Work out the cross up to F2 generation giving the phenotypic ratios of F1 and F2 generation respectively.
    2. State the Mendelian principle that can be derived only with the help of such a cross.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) (i) Normal nitrogen/14N, heavy nitrogen/15N [½+½]

to produce two types of DNA/light and heavy DNA respectively. [1]

(ii) E. coli has generation time of 20 minutes so the samples taken at intervals of 20 minutes, to understand the mode of replication when E. coli with 15N DNA was cultured in medium 14N (normal) nitrogen. [1]

(iii) To distinguish or separate heavy DNA from light DNA on the basis of density. [1]

(iv) Mode of DNA replication is semiconservative. [1]

OR

(b) (i) Parents: TTRR (Tall round seeds) × ttrr (Dwarf wrinkled seeds) [½]

Gametes: TR and tr [½]

F1 generation: TtRr (All tall round) [½]

Selfing: TtRr × TtRr [½]

(Male and Female gametes along with Punnett square are to be awarded two marks) [2]

Figure

F2 gen.: Tall Round : Dwarf Round : Tall wrinkled : Dwarf wrinkled = 9 : 3 : 3 : 1 [½]

(ii) ‘When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters’ (law of independent assortment). [½]

[Total: 5 marks]

Q96.2023 Main 57/3/1 · Q325 marks
  1. Protein synthesis requires the services of all three types of RNAs, namely t-RNA, m-RNA and r-RNA. Explain the role of each of them during the process of protein synthesis in prokaryotes.

OR

  1. A homozygous tall pea plant with green seeds is crossed with a homozygous dwarf pea plant with yellow seeds.
    1. Write the possible phenotype and genotype of F1 generation.
    2. State the laws of Mendel that are proved true by the F1 generation.
    3. Mention the F2 phenotypic ratio along with their possible phenotypes.
    4. Write the genotypes of the male and female gametes produced by F1 progeny.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) tRNA – Act as adaptor molecule, with amino acid binding site and anticodon loop, brings specific amino acid to the amino acid binding site on the ribosome, initiator t-RNA starts the process of translation. [½×4]

mRNA – Act as a template for protein synthesis, carries information in the form of codon, It has the initiator codon/start codon/AUG to initiate the process, it has the stop codon/UAA/UAG/UGA for termination of the protein synthesis (Any two) [½×2]

rRNA – rRNA organizes itself into ribosomes, there are two sites in the large subunit in which one site is for binding the incoming t-RNA with the corresponding amino acid second site for peptide bond formation, when the small subunit of rRNA (ribosomes) encounters the mRNA the process of translation begins, ribosome also acts as a catalyst for the formation of peptide bond. [½×4]

OR

(b) (i) Phenotype – Tall plant with Yellow seeds [½]

Genotype – TtYy [½]

(ii) • Law of Dominance: Out of two contrasting traits only one trait will appear in F1 generation and is called dominant trait while the one which remain unexpressed is called recessive trait. [1]

• Law of segregation: In a hybrid union trait simply remain together and segregate at the time of gamete formation. [1]

(iii) Phenotypic ratio of F2: Tall yellow : Tall green : Dwarf yellow : Dwarf green = 9 : 3 : 3 : 1 [½+½]

(iv) Male gametes of F1: TY, Ty, tY, ty [½]

Female gametes of F1: TY, Ty, tY, ty [½]

[Total: 5 marks]

Q97.2023 Main 57/5/1 · Q325 marks
  1. Name and describe the steps involved in the technique widely used in forensics that serves as the basis of paternity testing in case of disputes.

OR

  1. It is sometimes observed that the F1 progeny has a phenotype that does not resemble either of the two parents and has intermediate phenotype. Explain by taking a suitable example and working out the cross upto F2 progeny.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) DNA fingerprinting [1]

Steps

i) Isolation of DNA [½]

ii) Digestion of DNA by restriction endonuclease enzyme [½]

iii) Separation of DNA fragments, by electrophoresis [½×2]

iv) Transferring of separated DNA fragments to synthetic membrane such as nitrocellulose or nylon. [½]

v) Hybridization, using labelled VNTR probe. [½×2]

vi) Detection of hybridized DNA fragments by autoradiography [½]

OR

(b) Incomplete Dominance [½]

Snapdragon flower/Dog flower/Antirrhinum [½]

RR × rr (Red × White) [½]

F1: Rr (Pink) [½]

Selfing: Rr × Rr

Gametes: R, r [½]

Rr
RRR (Red)Rr (Pink)
rRr (Pink)rr (White)

[½]

Phenotype – Red : Pink : White = 1 : 2 : 1 [½+½]

Genotype – RR : Rr : rr = 1 : 2 : 1 [½+½]

[Total: 5 marks]

Q98.SQP 2025-26 · Q315 marks

A. Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical template strand given below.

Figure

B. How is transcription a more complex process in eukaryotic cells? Explain the additional processes that a precursor mRNA has to undergo in these organisms.

OR

A. Explain the process of aminoacylation of tRNA. Mention its role in translation.

B. How do ribosomes in the cells act as factories for protein synthesis?

C. Given below is a strand of mRNA undergoing the process of translation, what will be the sequence of Amino acids that will be translated? Name the triplet codons that should be added to bring to the end of translation at X.

Figure

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Official CBSE Sample Paper Marking Scheme · 2025-26

A.

Figure

[2]

B. [3]

  • Transcription in eukaryotes involves one of three types of polymerases, depending on the gene being transcribed. RNA polymerase II transcribes all of the protein-coding genes, whereas RNA polymerase I transcribes rRNA genes, and RNA polymerase III transcribes rRNA, tRNA, and small nuclear RNA genes. [1]
  • The primary transcripts contains the coding region, exon, and non-coding region, intron, hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA by the process called splicing. [1]
  • The hnRNA undergoes two additional processes called capping and tailing. In capping, an unusual nucleotide, methyl guanosine triphosphate, is added to the 5′-end of hnRNA. In tailing, adenylate residues (about 200–300) are added at 3′-end in a template independent manner. [1]

OR

A. Aminoacylation is the process by which amino acids become activated by binding with its aminoacyl tRNA synthetase in the presence of ATP. If two charged tRNAs come close during translation process the formation of peptide bond between them is energetically favourable. [1]

B. The cellular factory responsible for synthesising proteins is the ribosome. In its inactive state it exists as two subunits: a large subunit and a small subunit. When the small subunit encounters an mRNA the process of translation of the mRNA to protein begins. There are two sites in the large subunit for subsequent amino acids to bind to and thus be close enough to each other for the formation of a peptide bond. The ribosome also acts as a catalyst 23S rRNA in bacteria is the enzyme-ribozyme for the formation of peptide bonds. [2]

C. ARGININE, THREONINE, HISTIDINE, LEUCINE, PROLINE, GLYCINE, PROLINE, PROLINE, PROLINE, LEUCINE, SERINE [1]

Stop codon UGA/UAG/UAA [1]

Q99.SQP 2026-27 · Q315 marks

Attempt either option A or B.

A. A short stretch of DNA strand that codes for a polypeptide is shown below:

3′—TACAAAAAGAAGAAAAAGAAGATT—5′

Due to exposure to harmful UV radiation two different types of errors occurred during DNA replication affecting their sequences of bases in the DNA in two different cells.

  1. What is the term given to a segment of DNA that codes for a polypeptide?
  2. Which type of mutation could have occurred in each of the following due to the error during replication?
    a) 3′—TACAAAAAGAAGAAAAAGAAAATT—5′
    b) 3′—TACAAAAAGAAGAAAAAGAGATT—5′
  3. If DNA strand (a) codes for Methionine and sequences of Phenylalanine, mention the sequence of amino acids that will constitute the polypeptide chain after translation. How many amino acids will be translated from this strand?
  4. If all the codons of the stretch code for Phenylalanine excluding the initiator codon which feature does it depict?

OR

B.

  1. If GUA is an intron how many amino acids will be translated from the mature or processed RNA. Pick out the untranslated region from the following messenger RNA and mention their location and sequence. What is its significance?
    5′ACGUCGAUGACCGUAGCGUUUGUAUCUUUAGUAGUGGUAUUAGUAGGCUAAAAA3′
  2. If AAA is the anticodon of phenylalanine, draw the tRNA adaptor molecule, Mention the position of amino acid Phenylalanine in the sequence of polypeptide counting the triplet codons.
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Official CBSE Sample Paper Marking Scheme · 2026-27

A.

I. The term given to a segment of DNA that codes for a polypeptide is Cistron. [0.5]

II.

(a) 3′—TACAAAAAGAAGAAAAAGAAAATT—5′ — Point Mutation. [1]

(b) 3′—TACAAAAAGAAGAAAAAGAAATT—5′ — Deletion / Frameshift mutation. [1]

III. Methionine - Phenylalanine - Phenylalanine - Phenylalanine - Phenylalanine - Phenylalanine - Phenylalanine; 7 amino acids. [2]

IV. Degenerate. [0.5]

OR

B.

I. AUGACCGUAGCGUUUGUAUCUUUAGUAGUGGUAUUAGUAGGC

Nine amino acids will be translated from the mature or processed RNA. [1]

5′ACGUCG ------------------ UAAAAA3′ Location of the untranslated region from the above messenger RNA will be at the 5′ end and 3′ end and their sequences are 5′ACGUCG and 3′ end UAAAAA. [2]

They are required for efficient translation process. [0.5]

II.

Figure

[1]

4th triplet codon codes for phenylalanine. [0.5]

Unit VII · Genetics and Evolution

6. Evolution

Q1.2026 Main 57/1/1 · Q101 mark

The idea of use and disuse of organs for evolution of organism was proposed by

  1. Charles Darwin
  2. Thomas Malthus
  3. Hugo De Vries
  4. Lamarck
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(D) Lamarck [1 mark]

Q2.2026 Main 57/1/1 · Q131 mark

Assertion (A): The population of melanized moths increased in industrial areas after Industrial Revolution.

Reason (R): In Industrial environment lichen covered trees were replaced by soot-covered trees offering better camouflage to dark coloured moths.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A). [1 mark]

Q3.2026 Main 57/2/1 · Q51 mark

What conclusion can be drawn if the forelimbs of an aquatic animal and a terrestrial animal have similar anatomical structure ?

  1. Live in same habitat
  2. Share common ancestors
  3. Forelimbs perform same functions
  4. Forelimbs are analogous
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(B) Share common ancestors [1 mark]

Q4.2026 Main 57/3/1 · Q61 mark

Appearance of antibiotic-resistant bacteria is an example of evolution due to :

  1. Adaptive radiation
  2. Divergent evolution
  3. Artificial selection
  4. Anthropogenic action
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(D) Anthropogenic action [1 mark]

Q5.2026 Main 57/3/1 · Q71 mark

Identify the statements that correctly describe Darwin’s Theory of Evolution.

  1. Overproduction of organisms leads to competition.
  2. Variation is inherited and causes evolution.
  3. Acquired characters are inherited.
  4. Survival depends on favourable traits.
  5. New species arise due to accumulation of favourable variations.

Choose the correct option.

  1. (i), (ii), (iv) and (v) are correct
  2. (ii), (iii) and (iv) are correct
  3. All are correct
  4. Only (i) and (iv) are correct
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(A) (i), (ii), (iv) and (v) are correct [1 mark]

Q6.2026 Main 57/4/1 · Q61 mark

Name the extinct ancestor of humans who ate only fruits and hunted with stone weapons.

  1. Ramapithecus
  2. Australopithecus
  3. Dryopithecus
  4. Homo erectus
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(B) Australopithecus [1 mark]

Q7.2026 Main 57/4/1 · Q91 mark

The given diagram shows the skulls of two different mammals.

Figure

Which of the following options accurately describes the differences between these skulls ?

  1. Skull B is the ancestor of skull A.
  2. Skull A has more brain capacity than skull B.
  3. Skull A is of a human and skull B is of an adult chimpanzee.
  4. Skull A is of an adult chimpanzee and skull B is of a human.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(D) Skull A is of an adult chimpanzee and skull B is of human [1 mark]

Q8.2026 Main 57/5/1 · Q61 mark

Arrange the following in the correct sequence of their evolution and select the correct option :

  1. Seaweed
  2. Invertebrates
  3. Jawless fish

Options :

  1. (i), (ii), (iii)
  2. (i), (iii), (ii)
  3. (ii), (iii), (i)
  4. (ii), (i), (iii)
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(C) (ii), (iii), (i) [1 mark]

Q9.2026 Main 57/5/1 · Q111 mark

Select the odd option in the context of convergent evolution.

  1. Eyes of Octopus and Mammals
  2. Flippers of Penguins and Dolphins
  3. Flying Squirrel and Flying Phalanger
  4. Thorns of Bougainvillea and Tendrils of Cucurbita
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(D) Thorns of Bougainvillea and Tendrils of Cucurbita. [1 mark]

Q10.2026 Supp. 57/7/1 · Q51 mark

Choose the correct option that refers to adaptive radiation.

  1. Evolution of different species from a common ancestor to other geographical areas
  2. Migration of members of different species to a common geographical area
  3. Power of adaptation in an individual to a variety of environments
  4. Evolution by anthropogenic action
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(A) Evolution of different species from a common ancestor to other geographical areas [1 mark]

Q11.2025 Main 57/1/1 · Q111 mark

Evolution of modern man involves the following man-like primates. Choose the correct series of human evolution.

  1. Dryopithecus → Homo erectus → Australopithecines → Homo sapiens
  2. Australopithecines → Homo erectus → Neanderthal → Homo sapiens
  3. Australopithecines → Ramapithecus → Dryopithecus → Homo sapiens
  4. Homo erectus → Australopithecines → Homo sapiens → Neanderthal
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Answer not available: no official marking scheme published.

Q12.2025 Main 57/2/1 · Q51 mark

Which one of the following options shows the correct evolutionary order of the plants mentioned below ?

  1. Ferns
  2. Ginkgo
  3. Zosterophyllum
  4. Gnetales

Choose the correct option.

  1. (i), (iii), (ii), (iv)
  2. (iii), (i), (ii), (iv)
  3. (i), (ii), (iii), (iv)
  4. (iv), (ii), (i), (iii)
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Answer not available: no official marking scheme published.

Q13.2025 Main 57/2/1 · Q141 mark

Assertion (A): The number of white winged moths decreased after industrialisation in England.

Reason (R): Effects of industrialisation were more marked in rural areas of England.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Answer not available: no official marking scheme published.

Q14.2025 Main 57/4/1 · Q71 mark

About 15 mya during human evolution, the primates which used to walk like gorillas and chimpanzees were :

  1. Australopithecine and Neanderthal
  2. Dryopithecus and Ramapithecus
  3. Homo erectus and Homo sapiens
  4. Homo habilis and Homo erectus
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Answer not available: no official marking scheme published.

Q15.2025 Main 57/5/1 · Q71 mark

Which of the following combinations is a correct example of convergent evolution in Australian marsupials and Placental mammals ?

Australian MarsupialsPlacental Mammals
(A)LemurSpotted cuscus
(B)Tasmanian tiger catAnteater
(C)BobcatLemur
(D)NumbatAnteater
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Answer not available: no official marking scheme published.

Q16.2025 Main 57/6/1 · Q41 mark

If a natural population with 50 individuals is in Hardy-Weinberg equilibrium for a gene with two alleles A and a, with the gene frequency of allele A of 0·6, the genotype frequency of Aa will be :

  1. 0·16
  2. 0·36
  3. 0·24
  4. 0·48
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Answer not available: no official marking scheme published.

Q17.2025 Main 57/6/1 · Q121 mark

In his observations of small black birds in the Galapagos Islands, Darwin found that all the finches arose from the original ancestor :

  1. Insect-eating finches
  2. Seed-eating finches
  3. Cactus-eating finches
  4. Fruit-eating finches
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Answer not available: no official marking scheme published.

Q18.2025 Main 57/7/1 · Q71 mark

The given graph shows the range of variation among population members, for a trait determined by multiple genes. If this population is subjected to disruptive selection for several generations, which of the following distributions is most likely to result ?

Figure

Choose the correct option :

Options (A)–(D) are shown in the figure.

Figure

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Answer not available: no official marking scheme published.

Q19.2025 Supp. 57/S/1 · Q51 mark

The chronological order of evolution of modern man from man-like primates is given below.

Choose the option that indicates the evolution of man in correct order from early to recent times :

  1. Australopithecines → Ramapithecus → Homo habilis → Homo erectus → Homo sapiens
  2. Ramapithecus → Australopithecines → Homo habilis → Homo erectus → Homo sapiens
  3. Ramapithecus → Homo habilis → Australopithecines → Homo erectus → Homo sapiens
  4. Australopithecines → Homo habilis → Homo sapiens → Ramapithecus → Homo erectus
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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q20.2024 Main 57/2/1 · Q41 mark

Which one of the following statements is not true ?

  1. Flippers of whales and dolphins are homologous organs.
  2. Homologous organs have similar anatomical structure, but perform different functions.
  3. Homology indicates common ancestry.
  4. Homologous structures are a result of convergent evolution.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(D) Homologous structures are a result of convergent evolution. [1 mark]

Q21.2024 Main 57/2/1 · Q51 mark

A population is in genetic equilibrium/Hardy-Weinberg equilibrium for a gene with 2 alleles (dominant allele is ‘A’ and recessive allele ‘a’). If the frequency of allele ‘A’ is 0·6, then the frequency of genotype ‘Aa’ is :

  1. 0·21
  2. 0·42
  3. 0·48
  4. 0·32
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(C) 0·48 [1 mark]

Q22.2024 Main 57/3/1 · Q31 mark

During the 1850s in the pre-industrialisation era in England, the expected effect of natural selection on the number of dark-winged moths as compared to white-winged moths was :

  1. more in number
  2. less in number
  3. both were equal in number
  4. both were less in number
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(B) less in number [1 mark]

Q23.2024 Main 57/3/1 · Q71 mark

S.L. Miller in 1953, to support the theory of chemical evolution, created conditions in the closed flask that included :

  1. CH3, O2, NH3, H2O vapour at 1800°C
  2. CH4, H2, NH3, H2O vapour at 800°C
  3. CH4, CO2, H2, H2O vapour at 1800°C
  4. CH4, NH4, SO2, H2O vapour at 800°C
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(B) CH4, H2, NH3, H2O vapour at 800°C [1 mark]

Q24.2024 Main 57/4/1 · Q91 mark

Who proposed the mutation theory in favour of organic evolution ?

  1. Weisman
  2. Louis Pasteur
  3. Darwin
  4. Hugo de Vries
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(D) Hugo de Vries [1 mark]

Q25.2024 Supp. 57/S/1 · Q61 mark

Which one of the following options gives the correct temperature condition and the mixture of the gaseous components that were used by S.L. Miller in 1953 to prove abiogenesis of life ?

  1. CH4, H2, NO2 and water vapour at 1800 °C
  2. CH4, H2, NH3 and water vapour at 1800 °C
  3. CO2, H2, NH3 and water vapour at 800 °C
  4. CH4, H2, NH3 and water vapour at 800 °C
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(D) CH4, H2, NH3 and water vapour at 800 °C [1 mark]

Q26.2024 Supp. 57/S/1 · Q91 mark

The most primitive ancestor of humans is :

  1. Homo habilis
  2. Australopithecus
  3. Ramapithecus
  4. Homo neanderthalensis
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(C) Ramapithecus [1 mark]

Q27.2023 Main 57/1/1 · Q71 mark

Select the option that gives the correct description of the process of Natural Selection with respect to the length of the neck of giraffe.

Figure

  1. Stabilising selection as giraffes with longer neck lengths are selected further.
  2. Disruptive selection as giraffes with smaller and longer neck lengths are selected.
  3. Directional selection as giraffes with longer neck lengths are selected.
  4. Stabilising selection as giraffes with medium neck lengths are selected.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(C) Directional selection as giraffes with longer neck lengths are selected [1 mark]

//

(D) Stabilizing selection as giraffe with medium neck lengths are selected. [1 mark]

(The scheme accepts either (C) or (D).)

Q28.2023 Main 57/3/1 · Q31 mark

Identify the option that gives the correct type of evolution exhibited by the two animals shown, living in the same habitat in Australia.

Figure

  1. Convergent Evolution
  2. Disruptive Selection
  3. Divergent Evolution
  4. Homologous Ancestry
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(A) convergent evolution [1 mark]

Q29.2023 Main 57/3/1 · Q141 mark

Assertion (A): The number of white winged moths decreased drastically after industrialisation in England.

Reason (R): Effects of industrialisation were more marked in rural areas of England.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(C) A is true but R is false. [1 mark]

Q30.2023 Main 57/5/1 · Q11 mark

At which stage during evolution did human use hides to protect their bodies and buried their dead ?

  1. Homo habilis
  2. Neanderthal man
  3. Java man
  4. Homo erectus
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(B) Neanderthal man. [1 mark]

Q31.2023 Supp. 57/C/1 · Q131 mark

Assertion (A): Most “evolutionary trees” place information about pattern of relationship on horizontal axis and time on vertical axis.

Reason (R): An “evolutionary tree” depicts pattern of relationship among parents and offsprings.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(C) (A) is true, but (R) is false. [1 mark]

Q32.SQP 2026-27 · Q41 mark

Select the option that gives the correct description of the process of natural selection with respect to the length of the neck of the giraffe.

Figure

  1. Stabilising selection as giraffes with longer neck lengths are selected further.
  2. Disruptive selection as giraffes with smaller and longer neck lengths are selected.
  3. Directional selection as giraffes with longer neck lengths are selected.
  4. Stabilising selection as giraffes with medium neck lengths are selected.
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Official CBSE Sample Paper Marking Scheme · 2026-27

(C) Directional selection as giraffes with longer neck lengths are selected. [1 mark]

Q33.SQP 2025-26 · Q61 mark

Select the incorrect statement among the following.

  1. p2+2pq+q2 = 1. This is binomial expansion of (p+q)2.
  2. When frequency measured differs from expected values, the difference (direction) indicates the extent of evolutionary change.
  3. Hardy-Weinberg principle says that phenotype frequencies in a population are stable and are constant from generation to generation.
  4. The gene pool (total genes and their alleles in a population) remains constant. This is called genetic equilibrium. Sum total of all the allelic frequencies is 1.
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Official CBSE Sample Paper Marking Scheme · 2025-26

(C) Hardy-Weinberg principle says that phenotype frequencies in a population are stable and are constant from generation to generation. [1 mark]

Q34.SQP 2025-26 · Q81 mark

“In Cricket species, the sound produced by rubbing the wings or legs together play a crucial role in attracting mates, any change in the morphology of Cricket legs could potentially affect their ability to produce sound”.

A mutant Cricket had thicker hind legs. What would you expect for this cricket species?

  1. The leg mutation will not lead to speciation if they diversify into new habitats.
  2. The leg mutation will have little effect on other external features, and therefore have little effect on speciation.
  3. The leg mutation will have no effect on behavior, and thus have little effect on speciation.
  4. The leg mutation might lead to reproductive isolation and speciation due to an effect on the mating call.
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Official CBSE Sample Paper Marking Scheme · 2025-26

(D) The leg mutation might lead to reproductive isolation and speciation due to an effect on the mating call. [1 mark]

Q35.2026 Main 57/3/1 · Q182 marks

(a) (i) Mention any two characteristics of Neanderthal man that lived near East and Central Asia.

(ii) Identify the following hominids with respect to human evolution :

  1. Brain capacity 900 cc, probably ate meat.
  2. First human-like with brain capacity 650 – 800 cc.

[1 + ½ + ½ = 2]

OR

(b) The allele for brown eyes (B) is dominant over blue eyes (b). In a population, 36% of individuals have blue eyes.

Assuming Hardy-Weinberg equilibrium, calculate :

  1. Frequency of allele B
  2. Percentage of heterozygous individuals
  3. Percentage of individuals that are homozygous dominant

(Working to be shown) [½ + ½ + ½ + ½ = 2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) (i) (Any two) [½+½]

  • Brain size was 1400 cc
  • They used hides to protect their bodies
  • They buried their dead bodies

(ii) (I) Homo erectus [½]

(II) Homo habilis [½]

OR

(b) Given: Frequency of blue-eyed individuals (bb) = 36%

As per Hardy-Weinberg equilibrium, if q2 = 0.36, then q = 0.6 [½]

According to Hardy-Weinberg equilibrium (p + q) = 1. Therefore p = 1 – 0.6 = 0.4

(i) Frequency of allele B (p) = 0.4 [½]

(ii) Heterozygous individuals (2pq): 2pq = 2 × 0.4 × 0.6 = 0.48 = 48% [½]

(iii) Homozygous dominant (p2): p2 = 0.4 × 0.4 = 0.16 = 16% [½]

[Total: 2 marks]

Q36.2025 Main 57/7/1 · Q192 marks

A few stages and their respective time period in the evolutionary history of human beings are mentioned in the flowchart given below :

15 mya–Primates walking like gorillas and chimpanzees existed. (a)
↓
3 – 4 mya–Man-like primates walked in Eastern Africa. Fossils of their bones were discovered. (b)
↓
2 mya–This ancestor lived in the East African grasslands and ate fruits. (c)
↓
1·5 mya–This hominid had a brain size of 900 cc and probably ate meat. (d)

Based on the above information, answer the following questions :

  1. Name one primate about 15 mya.
  2. Name one place where fossils of primates were discovered in Eastern Africa.
  3. Name the ancestor that lived in the East African grasslands about 2 mya.
  4. Name the hominid that was found around 1·5 mya.
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Answer not available: no official marking scheme published.

Q37.2024 Main 57/1/1 · Q182 marks

Observe the picture given below. Name the naturalist and write the explanation given by him that evolution of life forms had occurred on the basis of this example.

Figure

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

Name of the naturalist – Lamarck [½]

Evolution of life forms driven by use and disuse of organs. [½]

He said giraffes in an attempt to forage leaves on tall trees had to adapt by elongation of their necks. [½]

They passed on this acquired character of elongated neck to their succeeding generations and slowly over the years all the future generations had long necks. [½]

[Total: 2 marks]

Q38.SQP 2026-27 · Q182 marks

The movement of landmasses over geological time changed the distribution and evolution of animals. Explain this idea using two examples from different regions of the world.

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Official CBSE Sample Paper Marking Scheme · 2026-27

There were in South America mammals resembling horse, hippopotamus, bear and rabbit. Due to continental drift, when South America joined North America, these animals were overridden by North American fauna. [1]

Due to continental drift, a number of Australian marsupials, each different from the other, evolved from an ancestral stock, but all within the Australian island continent. These pouched mammals of Australia survived because of lack of competition from any other mammals. [1]

Q39.2026 Main 57/1/1 · Q243 marks

(a) How is Hardy-Weinberg expression (p2 + 2pq + q2) = 1 derived ?

(b) List any two factors that disturb the genetic equilibrium.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) According to Hardy-Weinberg principle – sum total of all the allelic frequencies in a population = 1. Let there be two alleles A and a in the population. Let their frequencies be ‘p’ and ‘q’ respectively. p + q = 1 [½]

The frequency of AA will be p2 because the probability that an allele A with frequency ‘p’ would appear on both the chromosomes of a diploid individual will be product of probabilities i.e. p2. [½]

Similarly frequency of ‘aa’ will be q2, [½]

and Aa is 2pq [½]

Hence p2 + 2pq + q2 = 1

(b) Gene migration or gene flow, genetic drift, mutation, genetic recombination, natural selection (any two factors) [½ + ½]

[Total: 3 marks]

Q40.2026 Main 57/2/1 · Q263 marks

S.L. Miller conducted an experiment in 1953 to simulate the conditions of early earth in a laboratory setup. Answer the following questions based on the experiment :

(a) What was the aim of his experiment ? [1]

(b) Name the gases used and the source of energy. [2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) To prove that formation of life was preceded by chemical evolution / organic molecules can form spontaneously from inorganic molecules [1]

(b) Gases used: CH4, NH3, H2, Water vapours. [1]

Source of energy – Electric discharge [1]

[Total: 3 marks]

Q41.2026 Main 57/4/1 · Q243 marks

Explain the three different ways in which natural selection operates. Explain with the help of graphs.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1
  • Stabilising – more individuals acquire the mean character value i.e. variation is much reduced. [½]

Figure

[½]

  • Directional – more individuals acquire value other than the mean character value. [½]

Figure

[½]

  • Disruptive – more individuals acquire peripheral character value at both ends of the distribution curve; two peaks are formed. [½]

Figure

[½]

// (alternative: combined diagram)

Figure

[½ × 6]

[Total: 3 marks]

Q42.2026 Main 57/5/1 · Q243 marks

Study the given chart showing evolution of plants. Answer the following questions :

Figure

  1. Identify the plant which acts as an immediate ancestor of both ferns and conifers.
  2. Name the nearest ancestors of flowering plants.
  3. Name the most primitive group of plants.
  4. Psilophyton provides common ancestry to which classes ?
  5. Name the common ancestor of psilophyton and seed ferns.
  6. Name the common ancestor of mosses and tracheophytes.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a) Psilophyton [½]

(b) Seed ferns [½]

(c) Chlorophyte ancestors [½]

(d) Ferns, Conifers, Seed ferns [½]

(e) Tracheophyte ancestors [½]

(f) Chlorophyte ancestors [½]

[Total: 3 marks]

Q43.2026 Supp. 57/7/1 · Q243 marks

Explain the Hardy-Weinberg principle with the help of an algebric equation.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

Hardy-Weinberg’s principle states that allele frequencies in a population are stable and remain constant from generation to generation / the gene pool remain constant and this is called genetic equilibrium. (Any deviation in frequency of alleles represents evolutionary change) [1]

p – represents the frequency of allele A / p2 represents the frequency of allelic combination AA [½]

q – represents the frequency of allele a / q2 represents the frequency of allelic combination aa [½]

2pq represents the frequency of allelic combination Aa [½]

Hence p2 + 2pq + q2 = 1 [½]

[Total: 3 marks]

Q44.2025 Main 57/1/1 · Q283 marks

Compare and contrast convergent and divergent evolution.

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Answer not available: no official marking scheme published.

Q45.2025 Main 57/2/1 · Q253 marks

How does the process of Natural Selection affect Hardy-Weinberg equilibrium ? Explain with the help of graphs.

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Answer not available: no official marking scheme published.

Q46.2025 Main 57/7/1 · Q233 marks
  1. “The process of evolution of different species in a given geographical area starts from a point and literally radiates to other geographical areas.” Explain it with an example.
  2. Cite an example where more than one adaptive radiation has occurred in an isolated geographical area. Name the type of evolution your example depicts.
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Answer not available: no official marking scheme published.

Q47.2024 Main 57/1/1 · Q243 marks

(a) Explain the significance of the experiment carried out by S.L. Miller. Name the scientists whose hypothesis prompted him to carry out this experiment.

(b) How does meteorite analysis favour this hypothesis ?

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a)

Miller experimentally showed formation of amino acids [½]

and this proved theory of chemical evolution of life / formation of organic molecules from inorganic molecules. [1]

Oparin, Haldane [½ + ½]

(b) Analysis of meteorite content also revealed similar compounds indicating that similar processes are occurring in space. [½]

[Total: 3 marks]

Q48.2024 Main 57/3/1 · Q253 marks

A population of snakes lived in a desert with brown sand. Study the drawings given below showing the change in the population from ‘one’ to ‘two’ over time and answer the question that follows. Brown snakes and Grey snakes are represented by alleles A/a (Dominant/recessive).

Figure

  1. If the frequency of the recessive trait is 9% in population-one, work out the frequency of homozygous dominant and heterozygous dominant snakes.
  2. Name the mechanism of evolution that must have operated so that population-two evolved from population-one.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) Hardy–Weinberg Equilibrium: p2 + 2pq + q2 = 1; AA + 2Aa + aa = 1

  • Since frequency of grey snakes in the population = 9% = 9/100 = 0.09; q2 = 0.09; q = 0.3
  • Since p + q = 1, p = 1 – q = 1 – 0.3; p = 0.7 [½ + ½]
  • The frequency of homozygous dominant (AA) is equal to p2 = 0.49, or % frequency of homozygous dominant = 49% [½]
  • The frequency of heterozygous dominant (Aa) is equal to 2pq = 2 × 0.7 × 0.3 = 0.42, or % frequency of Aa = 42% [½]

(b) Natural selection [1]

[Total: 3 marks]

Q49.2024 Main 57/4/1 · Q243 marks
  1. Whose skulls ‘A’, ‘B’, and ‘C’ are shown below ? Which of the two are more similar to each other ?

Figure

  1. Name the (i) ape like (ii) man like primates that existed 1.5 million years ago.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) A – Modern human being; B – Baby chimpanzee; C – Adult chimpanzee [½ × 3]

  • Skull of baby chimpanzee is more like adult human skull (A and B) [½]

(b) (i) Dryopithecus   (ii) Ramapithecus [½ + ½]

[Total: 3 marks]

Q50.2024 Supp. 57/S/1 · Q253 marks

Differentiate between the explanations given by Darwin and de Vries respectively on the mechanism of evolution. Write any three differences.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1
Darwin’s Evolutionde Vries Evolution
Evolution was gradual (stepwise)Evolution occurred in a single step (saltation)
Variations and natural selection occur through a number of generations and are responsible for speciationSingle step mutation causes speciation
Variations are small and directionalMutations are random and directionless
Theory is based on natural selectionTheory is based on mutation

(Any three points) [1 × 3]

[Total: 3 marks]

Q51.2023 Main 57/1/1 · Q273 marks
  1. Name the two primate ancestors of the present day humans, who existed approximately about 15 million years ago.
  2. According to geological records, when and where did Australopithecines live ?
  3. Give two differences between Homo habilis and Homo erectus.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) Dryopithecus, Ramapithecus [½+½]

(b) Time period: 2 million years ago [½]

Place: East African grasslands [½]

(c)

Homo habilisHomo erectus
Brain capacity between 650 – 800 cc [½]Brain capacity 900 cc [½]
probably did not eat meat. [½]probably ate meat. [½]

[Total: 3 marks]

Q52.2023 Main 57/3/1 · Q253 marks
  1. Darwin’s theory of Natural Selection is widely accepted but some limitations have been identified by modern biologists. Mention the limitations identified.
  2. Name and state the most accepted theory of evolution in modern times.
  3. Mention any two ways the limitations identified in Darwin’s theory of evolution are explained in modern biology.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) Darwin’s theory could not explain how the variations arose. [1]

(b) • Synthetic theory of evolution [½]

• Origin of species is based on the interaction of genetic variation and natural selection. [½]

(c) Mutation, recombinants formed during meiosis/crossing over/hybridization/sexual reproduction [½×2]

[Total: 3 marks]

Q53.2023 Main 57/5/1 · Q243 marks

Industrial melanism in England after 1850 is an excellent example of Natural selection. Explain how ?

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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

Before industrialization thick growth of almost white lichens covered tree trunks and walls, from the mixed population of white and black winged moths white winged moth survived and increased in number due to camouflaging, dark coloured moth were picked out by predators and thus decreased in their numbers.

After industrialization tree trunks became dark due to industrial smoke and soot/absence of lichen, dark coloured moth survived and increased in number due to camouflaging, white winged moths were picked out by predators and thus decreased in their numbers. [½×6]

[Total: 3 marks]

Q54.2023 Supp. 57/C/1 · Q243 marks

Explain with the help of a diagrammatic representation, how natural selection operates on different traits in a population.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

Figure

(½ mark for each figure with correct labelling) [½ × 6 = 3 marks]

Q55.SQP 2025-26 · Q253 marks

Shown below are certain aquatic vertebrates, where natural selection has favoured them to develop certain characteristics which enable them to live in water.

Figure

  1. Name and explain the phenomenon exhibited by the above animals.
  2. Which one of the above is the most primitive one? What is its significance?
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Official CBSE Sample Paper Marking Scheme · 2025-26

A. These animals exhibit convergent evolution, structures that are not anatomically similar are evolved to perform similar functions adapted to the same habitat, the fins of Salmon and Shark, flippers of Dolphins, Seals and whales. One can say that it is the similar habitat that has resulted in selection of similar adaptive features in different groups of organisms but toward the same function: hence, analogous structures are a result of convergent evolution. [2]

B. Ichthyosaurs. Some of the land reptiles went back into water to evolve into fish like reptiles probably 200 mya. [1]

Q56.2025 Main 57/4/1 · Q294 marks

Read the following passage and answer the questions that follow.

The most convincing evidence to trace evolutionary relationships between humans and different groups of animals come from the basic similarities seen at the molecular level. Study the table given below that depicts the number of amino acid differences between the haemoglobin polypeptide of few animals with that of humans and answer the questions that follow.

S.No.AnimalNo. of amino acid differences in haemoglobin with that of humans
1.Macaque08
2.Dog32
3.Bird45
4.Frog67
5.Lamprey125
  1. To which category of evolution (Divergent or Convergent) do the following evolutionary relationships belong to :
    1. Humans and Macaque
    2. Humans and Frog
  2. What do the biochemical similarities in haemoglobin suggest about the evolutionary relationship between humans, frog and lamprey ?
  3. (i) Which one of the two – lampreys’ or macaques’ evolution is more closely related to humans and why ?

OR

  1. (ii) Which one of the two – frogs’ or dogs’ evolution is more closely related to humans and why ?
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Answer not available: no official marking scheme published.

Q57.2025 Main 57/5/1 · Q294 marks

Highly conserved proteins such as Haemoglobin and Cytochrome-C provide the best biochemical evidences to trace evolutionary relationships between different groups. Cytochrome-C is formed of 104 amino acids. Cytochrome-C is the respiratory pigment present in all eukaryotic cells. It has evolved at a constant rate during evolution. In chimpanzees and humans, Cytochrome-C genes are identical. The given data shows the evolution of the Cytochrome-C gene in different mammals from kangaroos, cows, rodents to humans :

GroupsNucleotide substitution in the gene of Cytochrome-CMillions of years ago
Human/Kangaroo100125 mya
Human/Cow75120 mya
Human/Rodent6075 mya
  1. Select the correct option for the time of separation of two groups and the number of nucleotide substitutions in the gene of Cytochrome-C :
    OptionsTime of separation of two groups during evolutionNumber of nucleotide substitutions
    (i)LesserGreater
    (ii)GreaterLesser
    (iii)GreaterGreater
  2. What do you infer about the type of evolution (convergent or divergent) for the given pair of groups and why ?
    1. Human and Kangaroo
    2. Human and Rodent
  3. (i) Define convergent evolution.

OR

  1. (ii) Define divergent evolution.
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Answer not available: no official marking scheme published.

Q58.2025 Main 57/6/1 · Q294 marks

Read the following passage and answer the questions that follow.

According to evolutionary theory, every evolutionary change involves the substitution of a new gene for the old one and the new allele arises from the old one. Continuous accumulation of changes in the DNA coding for proteins leads to evolutionary differences. The chemical composition of DNA is basically the same in all living beings, except for differences in the sequence of nitrogenous bases. Given below are percentage relative similarities between human DNA and DNA of other vertebrates :

S.No.VertebratesPercentage similarities
1.Chimpanzee100
2.Gibbon94
3.Rhesus Monkey88
4.Lemur47
5.Treeshrew28
6.Mouse21
7.Hedgehog19
8.Chicken10
  1. What is the term used for the substitution of a new gene for the old one and the new allele arising from the old one during evolutionary process ?
  2. Which one of the following holds true for the data provided in the above table ?
    1. Greater the evolutionary distance, greater are the differences in the nitrogenous bases.
    2. Lesser the evolutionary distance, greater are the differences in the nitrogenous bases.
    3. Greater the evolutionary distance, lesser are the differences in the nitrogenous bases.
    4. Lesser the evolutionary distance, lesser are the differences in the nitrogenous bases.
  3. (i) To which category of evolution (divergent or convergent) does the following relationship belong to ? Justify your answer.
    Human and Rhesus Monkey

OR

  1. (ii) Differentiate between Convergent and Divergent evolution.
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Answer not available: no official marking scheme published.

Q59.2025 Supp. 57/S/1 · Q325 marks

(a) (i)

Figure

Natural selection operates in different ways in nature.

  1. In England after industrialisation, the population of dark winged moths were more favoured than white winged moths. Explain.
  2. Identify the type of natural selection depicted in the graph above.

(ii) State Hugo de Vries’ theory of mutation. How is it different from Darwin’s theory of natural selection ?

OR

(b) (i) (I) Why did Hershey and Chase use 35S and 32P in their experiments ? Explain.

(II) Write the conclusion they reached at the end of the experiment.

(ii) Draw a clover leaf-like structure of tRNA showing the following :

  1. Tyrosine attached to the amino acid site.
  2. Anticodon for this amino acid in its correct site (codon for tyrosine is UAC).
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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(a) (i) (I) After industrialisation tree trunk became dark due to deposition of soot and smoke, so in that background dark winged moths could survive better than white, so population of dark winged moths increased. [1 + 1]

(II) Directional natural selection [1]

(ii) Hugo de Vries – single step large mutation or saltation leads to speciation. [1]

Hugo de VriesDarwin
1. Single step large mutation causes evolution.1. Minor variations cause evolution.
2. Mutations are random and directionless.2. Variations are small and directional.
3. Evolution is sudden3. Evolution is gradual

(Any one difference) [1]

OR

(b) (i) (I) Sulphur is the component of protein and not of DNA, so when they used radioactive sulphur (35S) it was incorporated into protein and no radioactivity was found in DNA; phosphorus is a component of DNA and not of proteins, so when they used radioactive phosphorus (32P) no radioactivity was found in protein, it was only incorporated in DNA; the bacteria infected with bacteriophage labelled with 35S did not show any radioactivity, while bacteria infected with bacteriophage labelled with 32P showed radioactivity. [½ × 4]

(II) They concluded that DNA is the genetic material, not the protein. [1]

(ii)

Figure

(I) Tyr at the amino acid site [1]; (II) anticodon at its correct site [1]

[Total: 5 marks]

Q60.2024 Main 57/5/1 · Q335 marks

Figure

  1. Natural selection operates in different ways in nature.
    1. Identify the type of natural selection depicted in the graph above.
    2. In England after industrialisation, the population of dark winged moths were more favoured than white winged moth. Explain.
    3. Anthropogenic action can enhance the rate of evolution. Explain with the help of an example.

OR

    1. Why did Hershey and Chase use ‘35S’ and ‘32P’ in their experiment ? Explain.
    2. State the importance of (1) blending and (2) centrifugation in their experiment.
    3. Write the conclusion they arrived at the end of their experiment.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a) (i) Directional selection [1]

(ii) During post-industrialisation period due to smoke and soot tree trunks became dark, lichens were not able to grow on the tree trunks, white moths were not able to camouflage and were eaten by the predators and their population size decreased, while dark moth population size increased due to their natural selection or camouflage with the background. [½ × 4]

(iii) Industrialisation leads to the natural selection of melanised moth, causing dominance of dark moth over white moth / excess use of herbicides or pesticides or antibiotics or drugs, leads to selection of resistant varieties or organisms or cells or any correct example in lesser time scale. (Any one explanation) [1 + 1]

OR

(b) (i) 35S is a component of protein and 32P is a component of DNA which help in identification (and confirmation) of protein and genetic material or DNA / to detect and confirm whether DNA or protein passed from virus to bacteria. [½ + ½]

(ii) (1) Blending removes viral coat from bacteria or E. coli by agitating [1]

(2) Centrifugation: separation of virus coat from bacteria or E. coli by spinning in a centrifuge [1]

(iii) DNA is the genetic material [2]

[Total: 5 marks]

Unit VIII · Biology and Human Welfare

7. Human Health and Disease

Q1.2026 Main 57/1/1 · Q21 mark

Which of the following is not a component of immune system ?

  1. Lymphocytes
  2. Plasma Proteins
  3. Red Blood Cells
  4. Bone Marrow
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(C) Red Blood Cells [1 mark]

Q2.2026 Main 57/2/1 · Q81 mark

What is the role of histamine released by mast cells ?

  1. Lowers blood pressure
  2. Causes allergic symptoms
  3. Stimulates antibody formation
  4. Destroys pathogens
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(B) Causes allergic symptoms [1 mark]

Q3.2026 Main 57/3/1 · Q81 mark

Which of the following is not a symptom of Ascariasis ?

  1. Intestinal blockage
  2. Muscular pain and fever
  3. Anaemia
  4. Skin ulcers
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(D) Skin ulcers [1 mark]

Q4.2026 Main 57/3/1 · Q111 mark

Some statements regarding drugs and alcohol are given below :

  1. Adolescents are more vulnerable to peer pressure.
  2. Drug addiction leads to increased tolerance level of the receptors present in our body.
  3. Alcohol affects the central nervous system.
  4. Excessive use of drugs helps to increase energy.
  5. Anabolic steroids decrease aggressiveness.

Choose the option with correct statement(s).

  1. Statements (i), (iv) and (v) are correct
  2. Statements (i), (ii) and (iii) are correct
  3. Only statements (i) and (v) are correct
  4. Only statement (iii) is correct
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(B) Statements (i), (ii) and (iii) are correct [1 mark]

Q5.2026 Main 57/4/1 · Q71 mark

Nicotine affects blood pressure and heart rate due to release of certain adrenal hormones. Following are some of the adrenal hormones :

  1. Adrenaline
  2. Cortisol
  3. Noradrenaline
  4. Aldosterone

Select the correct option :

  1. (i) and (ii)
  2. (i) and (iv)
  3. (ii) and (iii)
  4. (i) and (iii)
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(D) (i) and (iii) [1 mark]

Q6.2026 Supp. 57/7/1 · Q81 mark

Tetanus antitoxin, when injected into the human body, immediately provides which kind of immunity ? Choose the correct option.

  1. Innate immunity
  2. Passive immunity
  3. Autoimmunity
  4. Active immunity
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(B) Passive immunity [1 mark]

Q7.2026 Supp. 57/7/1 · Q151 mark

Assertion (A): Mucosa-associated lymphoid tissues are specialised immune barriers located on the skin.

Reason (R): MALT constitutes about 50 percent of the lymphoid tissue in a human body.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(D) Assertion (A) is false, but Reason (R) is true. [1 mark]

Q8.2025 Main 57/1/1 · Q51 mark

Transplantation of tissues/organs to some patients often fails due to rejection of such tissues/organs by the body of the patient. Which type of immune response is responsible for such rejections ?

  1. Autoimmune response
  2. Humoral immune response
  3. Physiological immune response
  4. Cell mediated immune response
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Answer not available: no official marking scheme published.

Q9.2025 Main 57/2/1 · Q121 mark

The decrease in the T-Lymphocytes count in human blood will finally result in

  1. decrease in antigens
  2. decrease in antibodies
  3. increase in antibodies
  4. increase in antigens
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Answer not available: no official marking scheme published.

Q10.2025 Main 57/2/1 · Q151 mark

Assertion (A): Streptococcus pneumoniae and Haemophilus influenzae are responsible for causing infectious disease in human beings.

Reason (R): A healthy person acquires the infection by inhaling the aerosols released by an infected person.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Answer not available: no official marking scheme published.

Q11.2025 Main 57/4/1 · Q51 mark

In which of the following human diseases does the body’s self-defence mechanism attack self-cells ?

  1. Thalassemia
  2. Phenylketonuria
  3. Filariasis
  4. Rheumatoid arthritis
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Answer not available: no official marking scheme published.

Q12.2025 Main 57/4/1 · Q91 mark

The technique for the early detection of a disease based on the principle of antigen-antibody interaction is :

  1. RNAi
  2. EST
  3. PCR
  4. ELISA
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Answer not available: no official marking scheme published.

Q13.2025 Main 57/4/1 · Q141 mark

Assertion (A): A person infected with malaria suffers from chill and high fever, recurring every three or four days.

Reason (R): The parasite attacks the RBC resulting in their rupture and release of haemozoin.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q14.2025 Main 57/5/1 · Q51 mark

Colostrum secreted by the mother’s mammary glands in a human female during the initial days of lactation is rich in antibody :

  1. IgE
  2. IgD
  3. IgA
  4. IgG
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Answer not available: no official marking scheme published.

Q15.2025 Main 57/6/1 · Q91 mark

A characteristic property that distinguishes a malignant tumor from a benign tumor is :

  1. Metamorphosis
  2. Metastasis
  3. Metabolism
  4. Metagenesis
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Answer not available: no official marking scheme published.

Q16.2025 Main 57/6/1 · Q161 mark

Assertion (A): In humans, filariasis is characteried by inflammation in the lower limbs.

Reason (R): Filarial worm usually lives in the lymphatic vessels of the lower limbs.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q17.2025 Main 57/7/1 · Q31 mark

The following information is about drugs and tobacco. Select the correct statement from the options given below.

  1. Cocaine is given to patients after surgery as it stimulates recovery.
  2. Chewing tobacco lowers blood pressure and heart rate.
  3. Barbiturates when given to criminals makes them tell the truth.
  4. Morphine is often given to persons who have undergone surgery as a painkiller.
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Answer not available: no official marking scheme published.

Q18.2025 Main 57/7/1 · Q81 mark

Which one of the following immune system components does not correctly match with its respective role ?

  1. Interferons – Secreted by virus-infected cells and protect non-infected cells from further viral infection.
  2. Macrophages – Mucus-secreting cells that trap microbes entering into the body.
  3. B-Lymphocytes – Produce antibodies in response to pathogens into blood to fight with them.
  4. IgA – Present in colostrum in early days of lactation to protect infants from diseases.
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Answer not available: no official marking scheme published.

Q19.2025 Supp. 57/S/1 · Q11 mark

Diseases like pneumonia and common cold are caused by pathogens. Choose the option that correctly indicates the difference between Pneumonia and Common cold :

  1. Pneumonia is a communicable disease, whereas Common cold is a non-communicable disease.
  2. Pneumonia is caused by a virus, while the Common cold is caused by bacteria.
  3. Pneumonia pathogen infects alveoli of the lungs, whereas Common cold affects nose and respiratory passage but not the lungs.
  4. Pneumonia cannot be prevented by vaccine, whereas Common cold can be prevented by vaccine.
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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(C) Pneumonia pathogen infects alveoli of the lungs, whereas common cold affects nose and respiratory passage but not lungs [1 mark]

Q20.2024 Main 57/1/1 · Q71 mark

In which of the following conditions/diseases is there a substantial increase in the activity of mast cells observed in the human body ?

  1. Typhoid
  2. Allergy
  3. Ascariasis
  4. AIDS
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(B) Allergy [1 mark]

Q21.2024 Main 57/2/1 · Q91 mark

The vector for dengue fever is :

  1. Female Aedes mosquito
  2. Female Anopheles mosquito
  3. Male Aedes mosquito
  4. Female Culex mosquito
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(A) Female Aedes mosquito [1 mark]

Q22.2024 Main 57/2/1 · Q141 mark

Assertion (A): Virus-infected cells produce interferons.

Reason (R): Interferons can cause inflammation of virus-infected cells.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(C) Assertion (A) is true, but Reason (R) is false [1 mark]

Q23.2024 Main 57/3/1 · Q91 mark

Which disease is the patient suffering from who is showing symptoms such as sustained high fever (39°C to 40°C), stomach pain, constipation, headache, loss of appetite and weakness ?

  1. Pneumonia
  2. Malaria
  3. Typhoid
  4. Amoebiasis
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(C) Typhoid [1 mark]

Q24.2024 Main 57/3/1 · Q141 mark

Assertion (A): The chronic use of alcohol by a person leads to cirrhosis.

Reason (R): Alcohol addiction at times becomes the cause of mental and financial distress to the entire family of the addicted person.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). [1 mark]

Q25.2024 Main 57/4/1 · Q121 mark

ELISA technique is based on the principle of

  1. DNA replication
  2. antigen-antibody interaction
  3. pathogen – antigen interaction
  4. antigen – protein interaction
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(B) Antigen-antibody interaction [1 mark]

Q26.2024 Main 57/5/1 · Q41 mark

Interferons are proteins secreted by

  1. RBC
  2. WBC
  3. Bacteria infected cell
  4. Virus infected cell
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(D) Virus infected cell [1 mark]

Q27.2024 Main 57/5/1 · Q131 mark

Assertion (A): AIDS is a syndrome caused by HIV.

Reason (R): HIV is a virus that damages the immune system with DNA as its genetic material.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(C) (A) is true, but (R) is false. [1 mark]

Q28.2024 Supp. 57/S/1 · Q71 mark

The mosquito-borne disease in humans causing chronic inflammation of the lymphatic vessels is :

  1. Elephantiasis
  2. Ascariasis
  3. Ringworm
  4. Amoebiasis
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(A) Elephantiasis [1 mark]

Q29.2024 Supp. 57/S/1 · Q141 mark

Assertion (A): Streptococcus pneumoniae and Haemophilus influenzae are responsible for causing infectious diseases in human beings.

Reason (R): A healthy person acquires the infection by inhaling the droplets/aerosols released by an infected person.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). [1 mark]

Q30.2023 Main 57/3/1 · Q71 mark

Select the options which is/are incorrect statement(s) with respect to T-lymphocytes in the human body.

  1. They are a type of white blood cells.
  2. They are produced in bone marrow.
  3. They remain active at all times in the body.
  4. They mature in the bone marrow.
  1. (i) and (iv) only
  2. (iii) only
  3. (iv) only
  4. (iii) and (iv) only
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(B) (iii) only [1 mark]

//

(D) (iii) and (iv) only [1 mark]

(The scheme accepts either (B) or (D).)

Q31.2023 Main 57/3/1 · Q151 mark

Assertion (A): More and more children in metro cities of India suffer from allergies and asthma due to sensitivity to the environment.

Reason (R): Modern day lifestyle and a protected environment in early life has resulted in lowering the immunity.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(A) Both A and R are true and R is correct explanation of A. [1 mark]

Q32.2023 Main 57/5/1 · Q31 mark

The decrease in the T-lymphocytes count in human blood will result in :

  1. Decrease in antigens
  2. Decrease in antibodies
  3. Increase in antibodies
  4. Increase in antigens
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(D) Increase in antigens. [1 mark]

Q33.2023 Main 57/5/1 · Q61 mark

Select the pathogen mismatched with the symptoms of disease caused by it from the list given below :

  1. Entamoeba histolytica : Constipation, abdominal pain.
  2. Epidermophyton : Dry scaly lesions on nail.
  3. Wuchereria bancrofti : Chronic inflammation of lymphatic vessels of lower limb.
  4. Haemophilus influenzae : Blockage of the intestinal passage.
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(D) Haemophilus influenzae: Blockage of the intestinal passage. [1 mark]

Q34.2023 Supp. 57/C/1 · Q61 mark

Choose the correct option in which the diagram of the flowering plant shown below is correctly matched with the drug obtained from it :

Figure

  1. Datura – Hallucinogen
  2. Cannabis – Stimulant
  3. Datura – Depressant
  4. Poppy – Depressant
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(A) Dhatura – Hallucinogen [1 mark]

Q35.2023 Supp. 57/C/1 · Q151 mark

Assertion (A): Cocaine produces a sense of euphoria and energy.

Reason (R): It interferes with the transport of neurotransmitter dopamine.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(A) Both (A) and (R) are true, and (R) is the correct explanation of (A). [1 mark]

Q36.SQP 2025-26 · Q91 mark

Plasmodium is a pathogen that causes malaria. Identify the correct sequence of transmission of the pathogen.

I
Stage of pathogen as it is transferred by vector bite
II
First site in the host body where pathogens infect and proliferates
III
Second site in the host body where the pathogen infects and manifests clinical symptoms
IV
Stage of pathogen as it is transferred to a new vector
ASporozoitesErythrocyte infectionLiver infectionGametocytes
BGametocytesErythrocyte infectionLiver infectionSporozoites
CGametocytesLiver infectionErythrocyte infectionSporozoites
DSporozoitesLiver infectionErythrocyte infectionGametocytes
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Official CBSE Sample Paper Marking Scheme · 2025-26

(D) Sporozoites, Liver infection, Erythrocyte infection, Gametocytes [1 mark]

Q37.SQP 2026-27 · Q91 mark

A patient with fever, chest pain, cough and breathing difficulty undergoes a chest X-ray. The film shows patchy opacities and reduced clarity of lung fields. Which observation best supports pneumonia?

  1. Enlarged heart size
  2. Inflammation or fluid buildup in the lungs
  3. Fractured ribs
  4. Air pockets in the lungs
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Official CBSE Sample Paper Marking Scheme · 2026-27

(B) Inflammation or fluid buildup in the lungs [1 mark]

Q38.SQP 2025-26 · Q151 mark

Assertion (A): Excess dose of coke or crack produces a sense of euphoria, increased energy and causes hallucinations.

Reason (R): It interferes with the transport of dopamine.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true and R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
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Official CBSE Sample Paper Marking Scheme · 2025-26

(A) Both A and R are true and R is the correct explanation of A [1 mark]

Q39.2026 Main 57/1/1 · Q172 marks

(a) Define Humoral immune response.

(b) Name any two types of antibodies found to give humoral immune response in humans.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) An immune response where the B-lymphocytes produce an army of proteins in response to pathogens into our blood to fight with them. [1]

(b) Antibodies – IgA, IgE, IgM, IgG (any two) [½ + ½]

[Total: 2 marks]

Q40.2026 Main 57/3/1 · Q172 marks

(a) A person complains of persistent itching in the groin region and scalp. The person also shows dry and scaly lesions on skin, nails, etc.

  1. Identify the disease.
  2. Name its two causative agents.
  3. Mention its mode of transmission.

[½ + 1 + ½ = 2]

OR

(b) A girl develops sneezing, watery eyes and difficulty in breathing every time she visits a flower garden.

  1. Identify the condition.
  2. Name the immune component responsible for this condition.
  3. Name the chemical released.
  4. State one medicine used to treat such a condition.

[½ + ½ + ½ + ½ = 2]

Reveal answerHide answer
Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) (i) Ringworm [½]

(ii) Epidermophyton, Trichophyton, Microsporum (Any two) [½+½]

(iii) From soil / using towels / clothes / comb of infected person (Any one) [½]

OR

(b) (i) Allergy [½]

(ii) IgE antibodies [½]

(iii) Histamine, Serotonin (Any one) [½]

(iv) Anti histamine, adrenaline, Steroids (Any one) [½]

[Total: 2 marks]

Q41.2026 Main 57/5/1 · Q192 marks

Name the lymphoid organ that acts as a large reservoir of erythrocytes. Explain its role as a lymphoid organ.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1
  • Spleen. [1]
  • Acts as a filter of the blood by trapping blood-borne microorganisms. [1]

[Total: 2 marks]

Q42.2026 Supp. 57/7/1 · Q212 marks

Attempt either (a) or (b).

(a) (i) Name the source plant of heroin drug. How is it obtained from the plant ?

(ii) Write the effects of heroin on the human body.

OR

(b) What is ‘withdrawal syndrome’ ? List any two symptoms it is characterised by.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) (i) • Papaver somniferum [½]

• Morphine is extracted from the latex of poppy plant and by acetylation of morphine heroin is formed. [½]

(ii) It is a depressant / it slows down the body functions. [1]

OR

(b) • Withdrawal syndrome refers to the characteristic and unpleasant manifestation of the body of an addict when regular dose of drugs / alcohol is abruptly discontinued. [1]

• Symptoms : Anxiety, Shakiness, Excess sweating, Nausea (any two) [½ × 2]

[Total: 2 marks]

Q43.2025 Main 57/1/1 · Q172 marks

Student to attempt either option-(A) or (B) :

(A) How are morphine and heroin related ? Mention their effect on the human body.

OR

(B) (i) Name an alcoholic drink which is produced by the help of microbes :

  1. With distillation
  2. Without distillation

(ii) Explain how cyanobacteria can be used as bio-fertilizer.

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Answer not available: no official marking scheme published.

Q44.2025 Main 57/2/1 · Q192 marks

Student to attempt either option (A) or (B).

(A) Describe any two situations where a medical doctor would recommend injection of a pre-formed antibodies (antitoxins) into the body of a patient.

OR

(B) The symptoms of malaria do not appear immediately after the entry of sporozoites into the human body when bitten by female Anopheles mosquito. Explain why it happens.

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Answer not available: no official marking scheme published.

Q45.2025 Main 57/4/1 · Q172 marks
  1. Explain how the immunity of a person is affected if there is atrophy (degeneration) of the thymus gland at an early stage of life.

OR

    1. What are interferons ? Explain their role in providing immunity to a person.
    2. Which category of innate immunity defence barrier can interferons be classified into ?
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Answer not available: no official marking scheme published.

Q46.2025 Main 57/5/1 · Q172 marks
  1. Give an account of the generalised structure of an antibody molecule produced by B-lymphocytes in response to the pathogen.

OR

  1. Other than public awareness and counselling, enlist four measures taken up by NACO, WHO and other NGOs to prevent the spread of HIV infection in the society.
Reveal answerHide answer

Answer not available: no official marking scheme published.

Q47.2025 Main 57/7/1 · Q202 marks
  1. How does a human body respond when vaccine is introduced into it ? It is said that vaccinations are a must for a healthy society. Justify.

OR

  1. Humans have innate immunity for protection against pathogens that may enter the gut along with food. What are the two barriers in our body that protect it from such pathogens ?
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Answer not available: no official marking scheme published.

Q48.2024 Main 57/1/1 · Q202 marks

(a) A farmer while working on his farm was bitten by a poisonous snake. He was rushed to a nearby health centre where the doctor gave him an injection to save his life.

  1. What did the doctor inject and why ?
  2. Name the kind of immunity provided by this injection.

OR

(b) Why do organic farmers not recommend complete eradication of insect pests ? Explain giving reason.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a)

(i) Readymade or preformed antibodies or antitoxins against the snake venom [½]

A quick immune response is required in this case. [1]

(ii) Passive immunity [½]

OR

(b) The organic farmers hold the view that complete eradication of pests is not only impossible but also undesirable because without them beneficial predatory and parasitic insects which depend upon them as food or hosts will not survive / One of the key beliefs of organic farmers is that biodiversity furthers health. Greater biodiversity leads to more sustainable ecosystem. [2]

[Total: 2 marks]

Q49.2024 Main 57/2/1 · Q192 marks

Identify A, B, C and D in the following table :

Scientific name of the plantDrugEffect on the human body/human system
(a)Papaver somniferumADepressant/slows down body function
(b)Cannabis sativaCannabinoidsB
(c)Erythroxylum cocaCD
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1
  • A – Heroin / smack / diacetylmorphine / opioid
  • B – Cardiovascular system
  • C – Cocaine / coca alkaloid / coke / crack
  • D – Stimulates central nervous system / produces a sense of euphoria and increased energy / hallucination

[½ × 4 = 2 marks]

Q50.2024 Main 57/3/1 · Q182 marks

What is a vaccine ? Write the basis on which it acts when administered in the body.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1
  • Vaccine is a preparation of antigenic proteins of pathogen or inactivated/weakened pathogen. [1]
  • It is based on the property of memory of immune system, the antibodies produced in the body in response to these antigens would neutralize the pathogenic agents and body will show amnestic or quick response. [½ × 2]

[Total: 2 marks]

Q51.2023 Supp. 57/C/1 · Q192 marks

(a) Name the lymphoid organ in humans that produces different types of cells.

(b) Mention the name of the blood cells that migrate from this lymphoid organ to another lymphoid organ, and state the changes it undergoes so as to provide immunity.

Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a) Bone marrow [1]

(b) T-Lymphocytes; Maturation / Proliferation / become antigen sensitive / become effector cells. [½ + ½]

[Total: 2 marks]

Q52.2022 Main 57/1/1 · Q12 marks

Mention the parts of human body that get affected by Pneumonia and common cold infections. Write the causative agents of the two diseases.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

Parts affected: Common cold – nose / respiratory passage [½]; Pneumonia – (alveoli of) lungs / (air sacs of) lungs [½]

Causative agents: Pneumonia – Streptococcus pneumoniae / Haemophilus influenzae [½]; Common cold – Rhino virus [½]

[Total: 2 marks]

Q53.2022 Main 57/1/1 · Q42 marks

What is the importance of female Anopheles mosquitoes in the life of a malarial Parasite Plasmodium?

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

Female Anopheles mosquito takes up gametocytes with blood meal, Fertilization and development take place in the mosquito's gut, Mature infective stage (sporozoites) escape from the gut and migrate to mosquito salivary gland, Female Anopheles bites the humans transmitting the plasmodium to its second host to complete its life cycle. [½ × 4]

OR (alternative acceptable answer, as shown by '//' in the scheme)

Completes sexual stages of life cycle in female Anopheles mosquito / female Anopheles mosquito acts as primary host, acts as vector to transfer parasites from infected person to healthy person to complete its life cycle. [1 + 1]

OR (alternative acceptable answer, as shown by '//' in the scheme: labelled diagram)

Figure

[½ × 4]

[Total: 2 marks]

Q54.2022 Main 57/2/1 · Q12 marks

Though filariasis is not fatal, but the disease in humans is responsible for considerable suffering, gross deformities and disability. Write the scientific name of any two helminth worms causing the disease and state two chronic manifestations of filariasis.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

Helminth worms: Wuchereria bancrofti and Wuchereria malayi [½ + ½]

Chronic manifestations (any two): inflammation of the organs in which they live for many years; inflammation of lymphatic vessels of the lower limbs; deformities in genital organs [½ + ½]

[Total: 2 marks]

Q55.2022 Main 57/2/1 · Q32 marks

(a) A person suffering from AIDS dies of opportunistic infections (ARC) i.e., infections that could have been otherwise overcome.

  1. State one reason as to why an 'HIV' patient dies of 'opportunistic infections'.
  2. Give the scientific name of one bacterium and one parasite which mainly attack a person suffering from AIDS.
  3. Write the full form of the name of the widely used diagnostic test for AIDS.

OR

(b) In spite of the statutory warning on cigarette packets against 'smoking' and its injurious effects on health, smoking is still prevalent in the society. Enumerate four important health hazards to a smoker.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

(a)

  1. Decrease in the number of helper T-lymphocytes / weakened immunity [½]
  2. Bacterium – Mycobacterium [½]; Parasite – Toxoplasma [½]
  3. Enzyme Linked Immuno-sorbent Assay / Polymerase Chain Reaction [½]

OR

(b) Increased CO content in blood, reduced concentration of haem bound oxygen / oxygen deficiency in the body, increased incidence of cancers (of lung / urinary bladder / throat), bronchitis, emphysema, coronary heart disease, gastric ulcer, raised blood pressure, increased heart rate (any four points) [½ × 4]

[Total: 2 marks]

Q56.2022 Main 57/3/1 · Q12 marks

A boy developed some allergic reactions when he straight entered into his air conditioned room after a game of football outside his house. Write any two symptoms that could be noticed in such condition. How does our body combat such conditions?

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

Sneezing, watery eyes, running nose, difficulty in breathing and any other symptom (any two) [½ + ½]

Body produces IgE antibodies / give exaggerated response to the antigens / release chemicals like histamine and serotonin from mast cells. [1]

[Total: 2 marks]

Q57.2022 Main 57/3/1 · Q22 marks

(a) (i) Write the Scientific name of the plant from where natural cannabinoids are obtained.

(ii) Mention the parts of the plant that are used for extracting the drug.

(iii) How does the drug affect human body?

OR

(b) Epithelial lining of our intestine is considered as secondary lymphoid organ. Justify the statement.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

(a) (i) Cannabis sativa [½]

(ii) Flower tops or inflorescences, leaves and resin (any two) [½ + ½]

(iii) Influences cardio vascular system [½]

OR

(b) Provide site for interaction of lymphocytes with the antigen [1], lymphocytes proliferate to become effector cells. [1]

[Total: 2 marks]

Q58.2022 Main 57/4/1 · Q12 marks

Explain, giving two reasons, how immune response by “vaccine” is different from that by “antitoxin” in humans.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1
Immune response by vaccineImmune response by antitoxin
Immune response is generated by antibodies produced in response to weakened antigen introduced in the bodyImmune response is generated by pre-formed antibodies or antitoxin is injected
Response takes sometimeQuick immune response
Memory cells are generated / long lastingOne time response
Vaccine provides active immunityAntitoxin provides passive immunity

[1+1] (Any two differences)

[Total: 2 marks]

Q59.2022 Main 57/5/1 · Q12 marks

Ringworm is one of the most common infectious fungal disease in humans. Name any two genera of fungi which cause ringworm and state any of its two symptoms.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

● Trichophyton, Epidermophyton, Microsporum [½×2] (Any two)

● Symptoms: Appearance of dry lesions, scaly lesions on different parts of the body (as skin/nails/scalp), intense itching [½×2] (Any two symptoms)

[Total: 2 marks]

Q60.2022 Main 57/5/1 · Q32 marks

(a) State the mode of action of cocain on human body. Write the scientific name of the source plant it is obtained from.

OR

(b) Enumerate four most common warning signs of drug and alcohol abuse amongst the youth.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

(a) ● Interferes with the transport of neurotransmitter (dopamine) / stimulating action on central nervous system (CNS) / hallucinations / sense of euphoria / increased energy [1]

● Erythroxylum coca [1]

OR

(b) Drop in academic performance, unexplained absence from school/college, lack of interest in personal hygiene, withdrawal, isolation, depression, fatigue, aggressive and rebellious behaviour, deteriorating relationships with family and friends, loss of interest in hobbies, change in sleeping and eating habits, fluctuations in weight, appetite, stealing/mental stress/financial stress or any other relevant point [½×4] (Any four points)

[Total: 2 marks]

Q61.2022 Supp. 57/6/1 · Q12 marks

A major cause of death among infants is because of Acute Respiratory Infection (ARI) of the lower respiratory tract, which further affects the alveoli. Name the associated disease, a causative bacteria and give any two symptoms of the disease.

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

● Pneumonia [½]

● Streptococcus pneumoniae / Haemophilus influenzae [½]

● Symptoms – fever, chills, cough, headache, in severe cases lips and fingers nails may turn grey to bluish in colour. [½+½] (any two)

[Total: 2 marks]

Q62.2022 Supp. 57/6/1 · Q32 marks

(a) In our country NACO and other Non-Government Organisations (NGOs) are doing a lot to educate people about AIDS. Expand ‘NACO’. Enlist any three ways by which transmission of HIV-AIDS infection occurs.

OR

(b) Write the scientific name of the plant and its part from which opioids are extracted. How does it affect our body functions?

Reveal answerHide answer
Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

(a) ● NACO – National AIDS Control Organisation [½]

● Transmission of HIV – sexual contact with infected person, by transfusion of contaminated blood and blood products, by sharing infected needles as in the case of intravenous drug abusers, from infected mother to her child through placenta [½×3] (any three)

(½ mark to be deducted if infected / contaminated not mentioned)

OR

(b) ● Papaver somniferum [½]

● fruit / latex of poppy plant / inflorescence [½]

● acts as depressant / slows down body functions by binding to the opioid receptors present in the central nervous system and gastrointestinal tract. [1]

[Total: 2 marks]

Q63.SQP 2025-26 · Q192 marks

The table below shows a hypothetical blood report of a patient.

Test descriptionObserved valueUnitReference range
Leucocytes
Total leucocyte count1100Per Microliter4400-11000
Neutrophils31%55-70
Lymphocytes25%20-40
Basophils0.5%0.5 - 1
Eosinophils02%1-4
Monocytes0%1-8
  1. Looking at the values suggest which defense mechanism/ immunity is affected and state how this defense mechanism provides immunity.
  2. Name the barrier with least count and enumerate its role in providing immunity.
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Official CBSE Sample Paper Marking Scheme · 2025-26

A. Innate immunity/ non-specific type of defense/immunity present at the time of birth is affected. It provides different types of barriers to the entry of the foreign agents into our body/destroys microbes/ prevents microbial growth. [1]

B. Cellular barrier-monocytes is with least count and it can phagocytose and destroy microbes. [1]

Q64.SQP 2026-27 · Q192 marks

The following is a realistic example dataset and not a real world patient dataset:

Sample IDTumor size (mm3)Glucose level mM/litreOxygen level (mmHg)Neighboring cell's viability/ vitality (%)
11502.01560
2504.53085
32001.31050
41003.02070
52500.9740
  1. Interpret the relationship between tumor size and glucose level.
  2. Evaluate the data to determine how tumor progression may affect the viability/vitality of neighboring cells. Provide a brief explanation.
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Official CBSE Sample Paper Marking Scheme · 2026-27

A. As tumor size increases, glucose levels decrease, suggesting that larger tumors consume more glucose due to higher metabolic demand, possibly indicating a more aggressive growth pattern. [1]

B. As tumor size increases, the viability of neighboring cells decreases suggesting that growing tumors create a hostile microenvironment, due to nutrient competition and less oxygen is available leading to stress or death of surrounding normal cells. / As these cells actively divide and grow they also starve the normal cells by competing for vital nutrients. [1]

Q65.2026 Main 57/1/1 · Q263 marks

(a) Mention the scientific name of the source plant and the part from which opioids are extracted.

(b) How are morphine and heroin related ? Mention the effect each one of them has on human body.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) Source plant – Papaver somniferum [½]

Part – latex of the plant [½]

(b) • Heroin is formed by acetylation of morphine. [1]

• Morphine → It is effective sedative / used as pain killer / useful in patient who have undergone surgery [½]

• Heroin is a depressant / slows down body function [½]

[Total: 3 marks]

Q66.2026 Main 57/2/1 · Q283 marks

(a) A person inhales dust and pathogens but does not always fall sick. Explain how the body protects against such infections. [1]

(b) Name and describe any two barriers involved in this protection. [2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) Due to immune response shown by the body / due to ability of the body to fight against foreign agents (pathogens) [1]

(b) (Any Two) [½ × 4]

  • Physical barrier: Skin on our body prevents entry of pathogens / Mucus coating (of respiratory, gastrointestinal, urinogenital tract) help in trapping microbes entering our body.
  • Physiological barrier: Acid from stomach / saliva in the mouth / tears from eyes – all prevent microbial growth.
  • Cellular barrier: Neutrophils and monocytes in the blood as well as macrophages in tissues can phagocytose and destroy microbes.
  • Cytokine barrier: Virus infected cell secrete proteins interferons that protect non-infected cells from further viral infection

[Total: 3 marks]

Q67.2026 Main 57/4/1 · Q253 marks

(a) How do some viruses cause cancer in humans ?

(b) How do benign tumours turn malignant ? How do malignant tumours harm the human body ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) The cancer causing viruses called oncogenic viruses have genes called viral oncogenes which transform normal cell into cancerous cells. [1]

(b) When the cells of a benign tumour acquire the property of metastasis under certain conditions. [1]

These cells damage the surrounding normal cells of the body / They compete with the normal cells and starve them of vital nutrients. [1]

[Total: 3 marks]

Q68.2026 Supp. 57/7/1 · Q253 marks

(a) How does the human body respond to an allergen ? Which type of antibody is produced during allergic reactions ?

(b) Name the types of drugs that help in reducing the symptoms of allergic reactions.

(c) Name the chemicals released in the body during an allergic reaction.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) The immune system shows an exaggerated response to allergen. [½] It produces antibodies IgE. [½]

(b) Drugs like adrenalin, antihistamine, steroids can reduce the symptoms of allergic reactions (any 2 drugs) [½ + ½]

(c) Serotonin and histamine. [½ + ½]

[Total: 3 marks]

Q69.2025 Main 57/1/1 · Q233 marks

Study a part of life cycle of Plasmodium given below :

Figure

Answer the following questions :

(a) Name the infective stage of Plasmodium that is stored in the female Anopheles mosquito.

(b) Where does fertilization and development of parasite take place ?

(c) Identify labels P and Q in the given diagram.

(d) Asexual and sexual phase of the life cycle of the Plasmodium takes place in two different hosts. Write their names.

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Answer not available: no official marking scheme published.

Q70.2025 Main 57/2/1 · Q263 marks

Samples of blood and urine of a sportsperson are collected before any sports event for drug tests.

(a) Why there is a need to conduct such tests ?

(b) Name the drugs the authorities usually look for.

(c) Write the generic names of two plants from which these drugs are obtained.

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Answer not available: no official marking scheme published.

Q71.2025 Main 57/7/1 · Q253 marks

A person is suffering from high grade fever. Which symptoms will help to identify if he/she is suffering from Typhoid, Pneumonia or Malaria ?

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Answer not available: no official marking scheme published.

Q72.2025 Supp. 57/S/1 · Q233 marks

The graph given below indicates the administration of the first (L) and second dose (M) of a vaccine. The corresponding response of the body is indicated by X and Y. Interpret the graph and explain the reason for such a response shown by the body.

Figure

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

On administration of first dose (L) of a vaccine the body response is of low intensity (X) as the immune system is coming in contact with weak antigens first time (Primary response), but on administration of second dose (M) with same antigen body response becomes intensified as secondary response (Y) [1 + 1]

Because of formation of memory cells on first encounter with the antigen the secondary response is faster and stronger [1]

[Total: 3 marks]

Q73.2025 Supp. 57/S/1 · Q263 marks

(a) Why do sportspersons often fall victim to coccaine addiction ? Name the plant from which this drug is derived. Also write down the mode of action of this drug.

(b) Name one drug and its harmful effect obtained from latex of Poppy plant.

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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q74.2024 Main 57/1/1 · Q253 marks

Identify A, B, C, D, E and F in the table given below :

Name of Human DiseaseCausative OrganismSymptoms
PneumoniaStreptococcus‘A’
Typhoid‘B’High fever, weakness, headache, stomach pain
Common ColdRhino virus‘C’
Ringworm‘D’Dry scaly lesions on body parts, redness, itching
AscariasisAscaris‘E’
‘F’Entamoeba histolyticaConstipation, cramps, stools with mucous and blood clots
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1
  • A – Fever / chills / cough / headache / greyish blue lips and nails / severe problem in respiration
  • B – Salmonella typhi
  • C – Nasal congestion / discharge / sore throat / cough / hoarseness / tiredness
  • D – Microsporum / Trichophyton / Epidermophyton
  • E – Internal bleeding / fever / muscular pain / anaemia / blockage of intestinal passage
  • F – Amoebiasis / Amoebic dysentery

[½ × 6 = 3 marks]

Q75.2024 Main 57/2/1 · Q253 marks

(a) Differentiate between humoral immune response and cell-mediated immune response.

(b) Draw a schematic diagram of an antibody molecule and label any four parts.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a)

Humoral immune responseCell-mediated immune response
(i)Mediated by B-lymphocytesMediated by T-lymphocytes
(ii)Antibodies are produced by B-lymphocytes in the blood.T-cells do not secrete antibodies but help B-cells to produce them.
(iii)This is not responsible for graft rejection.This is responsible for the graft rejection.

(Any two corresponding differences) [½ + ½]

(b)

Figure

(Any correct four labels) [½ × 4]

[Total: 3 marks]

Q76.2024 Main 57/4/1 · Q253 marks
    1. Name the group of drugs whose skeletal molecule is shown below :

    Figure

    1. How are such drugs consumed ?
    2. Name the human body organ affected by the consumption of these drugs.

OR

  1. Draw a schematic diagram of an antibody molecule and label any 4 parts. Mention their chemical nature. Name the cells which produce them.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) (i) Cannabinoids [1]

(ii) Inhalation, oral ingestion [½ + ½]

(iii) Brain / Heart (cardiovascular system) [1]

OR

(b)

Figure

[Four labelling – antigen binding site, light chain, heavy chain, C-terminal, N-terminal, disulphide bridge – ½ mark each] [½ × 4]

  • They are chemically proteins [½]
  • B-lymphocytes / B-cells [½]

[Total: 3 marks]

Q77.2023 Main 57/1/1 · Q253 marks
  1. Differentiate between malignant and benign tumours.
  2. Name and explain the most feared property of a malignant tumour.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a)

S. No.Malignant tumorBenign tumor
1Cells grow very rapidly and invade and damage the surrounding normal tissue.Comparatively slow growth and remain confined to their original location and do not spread to other parts of the body
2Show metastasisDo not show metastasis

(Any one difference) [1]

(b) • Metastasis [1]

• Cells from these tumors slough off and reach distant sites through blood, wherever they get lodged in the body they start a new tumor there. [½×2]

[Total: 3 marks]

Q78.2023 Main 57/3/1 · Q263 marks

Figure

  1. Name the category of drugs represented by the chemical structure given above.
  2. If the methyl group is substituted by acetyl group we get a bitter crystalline compound. Name the compound.
  3. Name the natural source of these compounds.
  4. State the harmful effects of this class of drugs on the human body.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) Opioids/Morphine [½]

(b) Diacetylmorphine/Smack [½]

(c) Papaver somniferum/Poppy plant [1]

(d) Slows down body function, act as depressant [½×2]

[Total: 3 marks]

Q79.2023 Main 57/5/1 · Q263 marks
  1. “Plasmodium protozoan needs both a mosquito and a human host for its continuity.” Explain.

OR

  1. We all must work towards maintaining good health because ‘health is wealth’. Enlist any six ways of achieving good health.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) Sporozoites (infectious form of Plasmodium) enter in the blood of human through female Anopheles mosquito’s bite, sporozoites multiply asexually (asexual phase) in liver, Red Blood Cells (RBCs), and form gametocytes in RBCs of human host which are taken by female Anopheles mosquito with blood meal, fertilization and further development take place in the mosquito’s gut leading to formation of sporozoites (sexual phase), that are stored in its salivary glands hence needs both host for its continuity. [½×6]

//

Figure

[½×6]

OR

(b) Balanced diet, personal hygiene, regular exercise/yoga, vaccination, proper disposal of waste, control of vectors, maintenance of hygienic food and water, any other relevant point (Any six) [½×6]

[Total: 3 marks]

Q80.2022 Main 57/2/1 · Q73 marks

When Plasmodium sp. (malarial parasite) enters the human body, it reproduces asexually and increases its number. During this course, the affected person suffers from malaria. After a number of asexual cycles, the parasite, in order to complete its life cycle, enters the sexual phase of its life. Trace the sexual stages in the life cycle of Plasmodium in a female Anopheles mosquito, when it bites an infected person, up to the formation of mature infective stages of the parasite.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

Sexual stage (gametocytes) develop in RBC, gametocytes are taken by female Anopheles mosquito with blood meal, fertilisation and development take place in the mosquito's gut, mature infective stages (sporozoites) escape from gut, migrate to mosquito's salivary glands, when the mosquito bites another human sporozoites are injected with bite [½ × 6]

// (alternative: shown as a diagram, ½ × 6)

Figure

[Total: 3 marks]

Q81.2022 Main 57/2/1 · Q83 marks

(a) Name and state the type of immunity that acts as a 'biological shield' to the developing foetus against many infections in the mother's womb.

(b) A large proportion of infants remain free from potent infections at least three months after birth, or even longer if breast fed. Explain, giving suitable reasons.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

(a) Passive immunity [½]; when readymade antibodies are directly given to protect the body against foreign agents [½]

(b) The yellowish fluid colostrum secreted by mothers during initial days of lactation [1], abundant antibodies (IgA) to protect the infant. [1]

[Total: 3 marks]

Q82.2022 Main 57/3/1 · Q73 marks

(a) Write the complete name of the diagnostic test for AIDS. Explain the principle it works on.

(b) Name the type of genetic material present in AIDS causing pathogen.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

(a) Enzyme-Linked Immunosorbent Assay [1]; it is based on the principle of antigen-antibody interaction / HIV infection can be detected by the presence of antigen or by detecting the antibody synthesised against the pathogen. [1]

//

Polymerase chain reaction [1]; early detection and amplification of genetic material of pathogen / virus. [1]

(b) RNA [1]

[Total: 3 marks]

Q83.2022 Main 57/3/1 · Q83 marks

A patient complains of suffering from constipation, stomach ache, stool with blood clots and excess mucous. The physician diagnosed it as amoebiasis, after stool test.

(a) Write the scientific name of the microbe identified in the stool sample.

(b) How do you think, the patient must have contracted it?

(c) Write your suggestions to the patient to avoid infection in future.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

(a) Entamoeba histolytica [1]

(b) Through contaminated water / contaminated food [1]

(c) Consuming clean drinking water, consuming non-contaminated food (vegetables, fruits), taking care of personal hygiene, periodic cleaning and disinfecting of water reservoirs (pools, tanks) (any two; or any other correct answer) [½ + ½]

[Total: 3 marks]

Q84.2022 Main 57/4/1 · Q113 marks

(a) Explain the roles of (i) primary, and (ii) secondary lymphoid organs that are responsible for developing defence to combat the action of pathogens/foreign antigens which enter our body.

(b) Doctors generally advise not to undergo surgery of tonsils. Why?

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

(a) (i) Origin and/or proliferation and maturation of lymphocytes / the immature lymphocytes differentiate into antigen sensitive lymphocytes / provide micro environment for development and maturation of T-lymphocytes. [1]

(ii) Traps the microorganisms (antigens) and activate the lymphocytes that then proliferate to become effector cells (cause immune response) / provides site for interaction of lymphocytes with antigens [1]

(b) Tonsils are secondary lymphoid organs / trap the antigen (pathogen) which then activate the mature lymphoid organs to become effector cells / provides site for interaction of lymphocytes with antigens [1]

[Total: 3 marks]

Q85.2022 Main 57/5/1 · Q73 marks

‘An HIV patient normally doesn’t die of ‘AIDS’, but death is caused due to many other infections.’ Do you agree with the statement? Give explanatory reasons in support of your answer.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

● Yes [1]

● HIV attacks macrophages and helper T-lymphocytes (T-helper cells), leads to decrease in T-helper cells, patient becomes immuno-deficient / reduced immunity, becomes prone to infection by pathogens (Mycobacterium/viruses/fungi and even parasites like Toxoplasma). [½×4]

[Total: 3 marks]

Q86.2022 Main 57/5/1 · Q83 marks

(a) Why do doctors have to carry many tests for selecting a person to be a suitable donor for someone who is going for an organ transplant and not take the organ from just anybody? Explain giving reason.

(b) Name the drug a patient who has undergone a successful organ transplant, has to take all his/her life.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

(a) ● To avoid graft rejection [1]

● Grafts would be rejected because of mismatch of blood groups or tissues / the body is able to differentiate between ‘self’ and ‘non-self’ leading to rejection / the cell-mediated response of T-lymphocytes will lead to graft or organ be rejected. [1]

(b) Immuno-suppresants / Immuno-suppressive agents / Cyclosporin A [1]

[Total: 3 marks]

Q87.2022 Supp. 57/6/1 · Q73 marks

Name any two techniques which can be used to generate a three-dimensional image to detect cancers of the internal organs and explain any one of the techniques.

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

● CT (Computed Tomography), MRI (Magnetic Resonance Imaging) [1+1]

● Computed Tomography – uses X-rays to generate a three dimensional image of the internals of an object. / MRI – uses strong magnetic fields / non-ionising radiations to accurately detect the cancer in internal organs. [1] (explain any one technique)

[Total: 3 marks]

Q88.2022 Supp. 57/6/1 · Q83 marks

Given below is a diagrammatic representation of immune system of the human body:

Figure

(a) Identify ‘X’ and ‘Y’ in the given diagram.

(b) Explain two major functions of the organs that you have identified.

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

(a) ‘X’ – Lymph nodes [½]   ‘Y’ – Thymus [½]

(b) ● Lymph nodes (secondary lymphoid organ) – to trap the microorganisms or other antigens present in lymph or tissue fluid. [1]

● Thymus (primary lymphoid organ) – provide microenvironment for the development and maturation of T-lymphocytes. [1]

[Total: 3 marks]

Q89.SQP 2026-27 · Q263 marks

Drug-use disorders often begin with occasional consumption but may progress into a compulsive state.

Analyse how addiction differs from dependence in both origin and consequence, and justify why long-term drug intake forces the user to increase dose over time.

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Official CBSE Sample Paper Marking Scheme · 2026-27

Addiction – Psychological attachment to effects of drugs/alcohol (½) → driven by desire for euphoria/well-being even when not needed (½). [1]

Dependence – Physical tendency of the body → manifests withdrawal symptoms (anxiety, shakiness, nausea, sweating) when drug is stopped. [1]

Reason for increased intake – Repeated drug use increases tolerance level of body receptors → receptors respond only to higher doses for the same effect. [1]

Q90.SQP 2025-26 · Q273 marks

Expand ELISA. On what principle is the ELISA test based? List two ways by which an infection can be detected by this test.

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Official CBSE Sample Paper Marking Scheme · 2025-26

ELISA – Enzyme Linked Immunosorbent Assay. [1]

ELISA is based on antigen-antibody interaction. [1]

The ways to detect the presence of infection or disease by ELISA are as follows: The presence of antigens (proteins, glycoproteins, etc.) is detected. / Antibodies produced against the pathogens are detected. [1]

Q91.2026 Main 57/3/1 · Q294 marks

Read the following passage and answer the question that follow :

The data below shows the concentration of nicotine smoked by a smoker taking 10 puffs/minute.

Figure

(a) (i) With reference to the above graph, explain the concentration of nicotine in the blood at 10 minutes. [1]

OR

(ii) How will this affect the concentration of carbon monoxide and haem-bound oxygen at 10 minutes ? [1]

(b) How does cigarette smoking result in high blood pressure and increase in heart rate ? [1]

(c) To which class of compounds does nicotine belong ? Name one other drug from the same class. [2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) (i) Maximum / 45 mg per cm³ at 10 minutes. [1]

OR

(ii) Concentration of Carbon monoxide increases and concentration of haem-bound oxygen decreases [½+½]

(b) Nicotine stimulates adrenal gland which releases adrenaline / noradrenaline in blood resulting in increase of blood pressure and heart beat. [1]

(c) Nicotine is an Alkaloid [1]

Cocaine / coke / crack / morphine / cannabinoids. [1]

[Total: 4 marks]

Q92.2026 Main 57/4/1 · Q304 marks

Read the following passage and answer the questions that follow :

Immunity in our body is of two types (i) Innate immunity, and (ii) Acquired immunity. Innate immunity is a non-specific defence mechanism, whereas Acquired immunity is pathogen specific. Acquired immunity is characterised by memory. Antibodies are specific to antigens and there are four types of antibodies produced in our body — IgA, IgE, IgG and IgM. It shows primary response where it encounters the pathogen for the first time and secondary response during the subsequent encounters with the same antigen/pathogen. These immune responses are carried out with the help of two special types of lymphocytes present in the blood.

(a) Name the two types of specialised cells which carry out the primary and secondary responses. [1]

(b) (i) Name any two barriers present in our body that provide innate immunity. [1]

OR

(b) (ii) The organ transplants (grafts) are often rejected if not taken from suitable compatible persons. Mention the characteristic of our immune system that is responsible for the rejection. [1]

(c) How is active immunity different from passive immunity ? Write any two differences. [2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) B-lymphocytes / B cells, and T-lymphocytes / T cells [½+½]

(b) (i) Physical barrier, physiological barriers, Cellular barriers, Cytokine barriers (any two) [½+½]

OR

(b) (ii) Our immune system has the ability to distinguish between ‘self’ and ‘non-self’ cells or tissues. [1]

(c) (Any two correct differences) [1+1]

Active ImmunityPassive Immunity
When antibodies are produced in our body in response to an antigen then it is called active immunity.When preformed antibodies are injected into the body for protection then it is called passive immunity.
It is slow and takes time to give its full effective response.It gives quick response.
It lasts for longer period.It lasts for shorter period.
Memory cells are formed.Memory cells are not formed.

[Total: 4 marks]

Q93.2025 Main 57/2/1 · Q294 marks

Immunity in our body is of two types : (i) Innate immunity and (ii) acquired immunity. Innate immunity is a non-specific defence mechanism, whereas acquired immunity is pathogen-specific; it is called specific immunity too. Acquired immunity is characterised by memory. Antibodies are specific to antigens and there are different types of antibodies produced in our body : they are IgA, IgE, IgG and IgM. It shows primary response when it encounters the pathogen for the first time and secondary response during the subsequent encounters with the same Antigen/Pathogen.

(a) Name the two types of specialised cells which carry out the primary and secondary immune response.

(b) Why is the antibody-mediated immunity also called as humoral immune response ?

Attempt either sub-part (c) or (d) :

(c) The organ transplants are often rejected if not taken from suitable compatible persons.

  1. Mention the characteristic of our immune system that is responsible for the graft rejection.
  2. Name the type of immune response and the cell involved in it.

OR

(d) How is active immunity different from passive immunity ?

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Answer not available: no official marking scheme published.

Q94.2025 Main 57/4/1 · Q304 marks

Read the following passage and answer the questions that follow.

Deaths related to the use of drugs were estimated at about 5,00,000 in 2019, 17·5 percent more than in 2009. Liver diseases attributed to Hepatitis B are a major cause of drug-related deaths, according to UNODC, accounting for more than half of the total number of deaths attributed to the use of drugs. Drug overdoses account for a quarter of drug-related deaths.

Opioids contribute to account for the most severe drug-related harm, including fatal overdoses, when used non-medically. At the global level, two-third of direct drug-related deaths are due to opioids, and in some sub-regions the proportion can be as high as three-quarters of such deaths.

  1. Why are people taking opioids more prone to liver diseases attributed to Hepatitis B ?
  2. What is meant by direct drug-related disease ?
  3. (i) What is the scientific name of the plant from which the opioids are derived and from which part of the plant is it extracted ?

OR

  1. (ii) State two common warning signs of drug abuse among the youth.
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Answer not available: no official marking scheme published.

Q95.2025 Main 57/5/1 · Q304 marks

In 2021, 5·3 percent of 15 to 16-year-olds worldwide (13·5 million individuals) had used Cannabis in the past year according to UNODC. The adolescent brain is still developing and drug use can have long-term negative effects. Early drug use initiation can lead to faster development of dependence than in adults and other problems in adulthood. Parts of the Amazon Basin are at the intersection of multiple forms of organised crimes that are accelerating devastation, with severe implications for the security, health and well-being of the population across the region. The direct impact of coca cultivation on deforestation is minimal, but indirectly it acts as a catalyst for “Narco-deforestation”. The laundering of drug trafficking profits into land speculation etc. is posing a growing danger to the world’s largest rainforest.

  1. Which age group or period of growth people are more vulnerable to drug abuse ?
  2. Explain the negative impact of coca cultivation on the world’s largest rainforest.
  3. (i) From which part of the plant are cannabinoids mainly obtained ? Mention any one negative effect of this drug on adolescents.

OR

  1. (ii) State the scientific name of the plant from which coca alkaloids are derived and state one negative impact of use of excessive dosage of cocaine.
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Answer not available: no official marking scheme published.

Q96.2025 Main 57/6/1 · Q304 marks

Read the following passage and answer the questions that follow.

Prevention is the frontline response to drug use. Effective interventions address the underlying conditions contributing to drug use, such as a lack of connection to family or community, instability, insecurity, trauma, mental health issues, etc. When addressed, these factors can effectively prevent the initiation of drug use and the progression to drug use disorders. Study the few key figures of drug use given below and answer the questions that follow.

Figure

  1. What do you infer from the figures in Table No. 1 about the people with drug use disorders, 2022 (in million) ? State any two of your observations.
  2. How are Hepatitis C and HIV related to drug use disorders by people, as shown in Table No. 2 ? State the correlation between the two.
  3. (i) Give the scientific name of (p) shown in Table No. 1.

OR

  1. (ii) Give the scientific name of (q) shown in Table No. 1.
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Answer not available: no official marking scheme published.

Q97.2024 Main 57/2/1 · Q304 marks

Read the following passage and answer the questions that follow.

In 1981, the health workers of United States of America had become aware of the increased frequency of Kaposi’s sarcoma, cancer of the skin and blood vessels and another disease pneumocystis pneumonia, a respiratory infection caused by a protozoan. Both these diseases were very rare in the general population, but occurred frequently in more severely “immunosuppressed” individuals. This led to the recognition of the immune system disorder that was named Acquired Immune Deficiency Syndrome (AIDS).

In 1983, virologists working in the USA and France had identified a causative agent for ‘AIDS’, now known as Human Immunodeficiency Virus (HIV). ‘HIV’ follows a set path to attack the human body to cause the disease.

(a) Name the group of cells the HIV attacks after gaining entry into the human body and write the various events that occur within this cell.

(b) Write the expanded form of the diagnostic test used for detecting AIDS. Write the possible treatment available for the disease at present.

(c) Mention any two steps suggested by WHO for preventing the spread of this disease.

OR

(c) “A patient suffering from AIDS does not die of this disease but from some other infection.” Justify the statement.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a) Macrophages; virus replication (RNA genome) [½ + ½]

(b) Enzyme-linked immuno-sorbent assay (ELISA) / Polymerase Chain Reaction (PCR) [½]

Treatment available – Antiviral drugs that are only partially effective as they only prolong the life of the patient. [½]

(c) Making blood HIV safe in blood banks, use of only disposable needles and syringes in hospitals, free distribution of condoms, controlling drug abuse, advocating safe sex, regular check-ups for HIV susceptible population. (any two) [1 + 1]

OR

(c) This is because of drastic reduction of helper T-lymphocytes that are responsible to fight infections, [1]

person become immune-deficient, [½]

unable to protect oneself from other bacterial or viral or fungal or parasitic infection [½]

[Total: 4 marks]

Q98.2024 Main 57/5/1 · Q304 marks

Read the following passage and answer the questions that follow :

“Mosquitoes are drastically affecting the human health in almost all the developing tropical countries. Different species of mosquitoes cause very fatal diseases so much so that many humans loose their life and if they survive, are unable to put in productive hours to sustain their life. With the result the health index of the country goes down.”

  1. Name the form in which Plasmodium gains entry into (i) human body (ii) the female Anopheles body.
  2. Why do the symptoms of malaria not appear in a person immediately after being bitten by an infected female Anopheles ? Give one reason. Explain when and how do the symptoms of the disease would appear.

OR

  1. Explain the events which occur within a female Anopheles mosquito after it has sucked blood from a malaria patient.
  1. Name a species of mosquito other than female Anopheles and the disease, for which it carries the pathogen.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a) (i) Sporozoites, (ii) gametocytes [½ + ½]

(b)

  • The sporozoites after entering the body need to undergo asexual reproduction in liver and RBC [1]
  • RBC burst, released haemozoin which is responsible for chill and high fever recurring every 3-4 days. [½ + ½]

OR

(b) Gametocytes (male and female) enter female mosquito body via blood meal, fertilization in gut / stomach, sporozoites escape from the gut, and migrate into salivary glands (of mosquito). [½ × 4]

(c) Aedes – dengue / chikungunya, or Culex – filariasis or elephantiasis (any other correct example with disease can be considered) [½ + ½]

[Total: 4 marks]

Q99.2024 Supp. 57/S/1 · Q304 marks

Read the following passage and answer the questions that follow.

A senior student, sneezing very badly, with watery eyes and having difficulty in breathing was brought to the school medical room by his fellow friends. The medical room nurse enquired as to what had happened. One of the students said “after the football match we all were sweating profusely, one of our friends sprayed deodorant on us. Soon after the deodorant was sprayed on the senior boy, the symptoms appeared.”

(a) What are the symptoms seen in the student indicative of ? Elaborate.

(b) State how the cause of such responses can be confirmed.

(c) (i) Name the cells in our body and the chemicals produced by them that are responsible for such reactions. Name any two drugs used to reduce these symptoms.

OR

(c) (ii) ‘In recent times there is a rise in such reactions amongst urban human population.’ Do you agree ? Give three reasons in support of your answer.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) Allergy, the exaggerated response of the immune system to certain antigens present in the environment. [½ + ½]

(b) By analysing the production of IgE type of antibodies / by injecting very small doses of possible allergens and studying the reaction. [1]

(c) (i) Mast cells [½]; histamine, serotonin [½]

Drugs are – anti-histamine, adrenalin, steroids (any two drugs) [½ + ½]

OR

(c) (ii) Yes, lowering of immunity due to modern lifestyle, more sensitivity to allergens, protected environment provided in early life. [½ × 4]

[Total: 4 marks]

Q100.2023 Main 57/3/1 · Q304 marks

The diagram shows the life cycle of a pathogenic protozoan.

Figure

  1. Name the parasitic stage that is being transferred from host ‘X’ to host ‘Y’.
  2. Write the changes the parasite undergoes in the liver.
  3. Write the changes the parasite undergoes when it enters the RBC.
    1. Trace the changes the parasite undergoes when the host ‘X’ takes its blood meal from infected host ‘Y’.

    OR

    1. At which stage during the life cycle of the pathogen does the host ‘Y’ experience the symptoms of the disease ? Name the disease and the toxic substance responsible for these symptoms.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(i) Sporozoites. [1]

(ii) Undergoes asexual reproduction [1]

(iii) The number increase asexually, parasites ultimately change into gametocytes/undergoes gametogenesis [½×2]

(iv) (a) Fertilisation and development takes place in host ‘X’, ultimately forming the infective stage sporozoites. [½×2]

OR

(iv) (b) • Bursting of RBCs

• Malaria

• Haemozoin

(Award 1 mark if any two are correct) [½×2]

[Total: 4 marks]

Q101.2023 Main 57/5/1 · Q294 marks

When a microorganism invades a host, a definite sequence of events usually occur leading to infection and disease, causing suffering to the host. This process is called pathogenesis. Once a microorganism overcomes the defense system of the host, development of the disease follows a certain sequence of events as shown in the graph. Study the graph given below for the sequence of events leading to appearance of a disease and answer the questions that follow :

Figure

  1. In which period, according to the graph there are maximum chances of a person transmitting a disease/infection and why ?
  2. Study the graph and write what is an incubation period. Name a sexually transmitted disease that can be easily transmitted during this period. Name the specific type of lymphocytes that are attacked by the pathogen of this disease.

OR

  1. Draw a schematic labelled diagram of an antibody.
  1. In which period, the number of immune cells forming antibodies will be the highest in a person suffering from pneumonia ? Name the immune cells that produce antibodies.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) • Incubation period [½]

• Infected person may not take precautions during this period as there are no symptoms of the diseases. [½]

//

• Period of illness [½]

• Number of microorganisms is high and can be transmitted via droplet infection. [½]

(b) • Time period between infection and appearance of its symptoms. [1]

• AIDS/any other correct example [½]

• T lymphocytes [½]

OR

(b) (Diagram of an antibody)

Figure

[½×4]

(c) • Period of illness [½]

• B lymphocytes/B cells [½]

[Total: 4 marks]

Q102.SQP 2025-26 · Q304 marks

The graph below shows the Antibody concentration in young calves. Study the graph and answer the questions that follow:

Figure

A. What do you think is the difference between passive and active immunity in this case? (1)

B. What kind of immunity will be observed when a vaccine is administered to the calf and why? (2)

Attempt either subpart C or D.

C. What kind of trend does passive immunity show and why? (1)

OR

D. Why does active immunity start after 14 days as per the above graph? (1)

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Official CBSE Sample Paper Marking Scheme · 2025-26

A. Passive immunity – Ready made antibodies from colostrum / less effective / transient / no memory cells involved.

Active immunity – Made by host's immune system/ Lag phase/ memory based/ largely effective. [1]

B. Active immunity. As the vaccine contains a weakened or inactivated form of the pathogen (microbe), which is harmless but still recognized by the immune system as foreign. The immune system responds by activating lymphocytes, which produce specific antibodies to fight the pathogen. This process also leads to the creation of memory cells, which provide long-term protection by “remembering” to fight this pathogen if the person is exposed again. [2]

Attempt either subpart C or D.

C. This passive immunity peaks at day 1 and declines as the calf ages as the antibodies are used up to develop immunity against diseases/no memory. [1]

OR

D. Active immunity is slow, and it takes time to give its full effective response. [1]

Q103.2026 Main 57/1/1 · Q335 marks

(a) (i) Name the protozoan species that is responsible for causing the most serious and even fatal malarial disease.

(ii) Name the host in which the parasite completes its sexual stages and explain the changes taking place.

(iii) How does the parasite damage the human body after entering the blood stream ?

(iv) Suggest two effective preventive measures to control the spread of this disease in endemic regions.

OR

(b) (i) What are bio-fertilizers ?

(ii) Name the different types of microorganisms used as bio-fertilizers in organic farming and explain how each contributes to soil fertility.

(iii) Write two advantages of using bio-fertilizers over chemical fertilizers.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) (i) Plasmodium falciparum [½]

(ii) • Female Anopheles mosquito [½]

• Mosquito picks up gametocytes from the infected person. Fertilization and development takes place in mosquito gut. Sporozoites from gut migrate to salivary glands of mosquito. [½ × 3]

(iii) After entering the blood stream they enter liver, and then into red blood cells and reproduce asexually, RBCs rupture and a toxic substance haemozoin is released causing chills and high fever. [½ × 3]

(iv) • Avoiding stagnation of water in and around residential areas. • Regular cleaning of household coolers. • Use of mosquito nets. • Introduce fishes like Gambusia in ponds – they feed on mosquito larvae. • Spraying insecticides in ditches / doors and windows to have wire mesh. • Any other appropriate preventive measures. (Any two) [½ + ½]

OR

(b) (i) Bio-fertilizers – Organisms that enrich the nutrient quality of soil. e.g. bacteria, fungi and cyanobacteria. [1]

(ii) • Bacteria / Rhizobium / Azotobacter / Azospirillum [½]

Rhizobium forms symbiotic association with roots of leguminous plants and fixes atmospheric nitrogen into organic forms to be used by the plant; Azotobacter or Azospirillum fix atmospheric nitrogen while free living in the soil. [½] (Award one mark for any one organism with its correct contribution)

• Fungi / genus Glomus. [½]

Fungi break down organic matter to release nutrients; Glomus forms mycorrhiza with roots of the plants. This association helps to absorb phosphorus from soil and pass to the plant; plants show resistance to root-borne pathogens; plants show tolerance to salinity and drought; there is increase in plant growth. (Any one) [½]

• Cyanobacteria / Anabaena / Nostoc / Oscillatoria [½]

Fix atmospheric nitrogen, serve as important biofertilizers in paddy fields (Any one) [½]

(iii) • Bio-fertilizers do not pollute our environment (soil, ground water). • They conserve the beneficial soil microbes. • Do not cause contamination and harm human health. (Any two) [½ + ½]

[Total: 5 marks]

Q104.2026 Main 57/2/1 · Q335 marks

(a) A 30 year old man was admitted to a hospital with recurrent infections, weight loss, prolonged fever. His blood report showed a decline in count of T-Lymphocytes thus affecting immune system.

(i) Identify the disease and its causative agent. [2]

(ii) How did the pathogen affect the immune system once it entered the body ? Explain. [2]

(iii) Mention two major modes of its transmission. [1]

OR

(b) (i) Why are microbes considered effective biocontrol agents in sustainable agriculture ? [1]

(ii) Explain the role of the following as biocontrol agents naming the pests or diseases they help to control :

  1. A Bacterium [1½]
  2. A Fungus [1]
  3. Virus [1½]
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) (i) Disease: Acquired Immuno Deficiency Syndrome / AIDS [1]

Causative Agent: Human Immuno deficiency Virus / HIV [1]

(ii) [½ × 4]

  • After getting the entry into the body of the person virus enters macrophages
  • RNA genome of the virus replicates to form viral DNA with the help of enzyme reverse transcriptase
  • The viral DNA gets incorporated into the host cell’s DNA and directs the infected cells to produce virus particles.
  • Simultaneously HIV enters into helper T-lymphocytes (TH) and produce progeny viruses which attack other T-lymphocytes leading to weakened immune system.

// (alternatively, the following diagram)

Figure

(iii) (Any Two) [½+½]

  • By Sexual contact with infected person.
  • Transfusion of contaminated blood and blood products.
  • Sharing of infected needles by intravenous drug abusers.
  • Infected mother to her child through placenta.

OR

(b) (i) They reduce our dependence on toxic chemicals and pesticides / They create a system where insects called pests are not eradicated but kept at manageable level by complex system of checks and balances / No negative impact on plants mammals birds fish or non-target insects (Any other correct reason) [1]

(ii) (1) A Bacterium: Bacillus thuringiensis, controls butterfly caterpillars, larvae eat the spores and toxin is released in the gut of larvae thus killing it [½+½+½]

(2) A Fungus: Trichoderma are free living fungi common in root ecosystem, effective biocontrol agents against several plant pathogens. [1]

(3) Virus: Baculoviruses under genus Nucleopolyhedrovirus, shows species specific, narrow spectrum insecticidal application. [½ × 3]

[Total: 5 marks]

Q105.2026 Main 57/5/1 · Q335 marks

(a) (i) Draw the structure of an antibody molecule and label any four of its parts. [2]

(ii) Differentiate between active and passive immunity. Write any three differences. [3]

OR

(b) (i) (I) Write the scientific names of the two species of filarial worms causing filariasis.

(II) How do they affect the body of infected persons ?

(III) How does the disease spread ?

(ii) Mention the source and the role of the following in providing defence against infection in the human body :

(I) Histamine

(II) Interferons [1+1+1+2 = 5]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a) (i) (Any four labellings) [½ × 4]

Figure

(ii) (any three correct difference) [1×3]

Active ImmunityPassive Immunity
When antibodies are produced in our body in response to an antigen then it is called active immunityWhen preformed antibodies (readymade) are injected into the body for protection then it is called passive immunity.
It is slow and take time to give its full effective responseIt gives quick response
It lasts for longer periodIt lasts for shorter period
Memory cells are formedMemory cells are not formed

OR

(b) (i) (I) Wuchereria bancrofti, and Wuchereria malayi [½+½]

(II) Chronic inflammation of the organs / swelling in lower limbs / gross deformities in genital organs. [1]

(III) Transmitted by the bite of female mosquito vector. [1]

(ii) (I) Mast cells, cause allergy [½+½]

(II) Virus infected cells, Protect the non-infected cells of our body from viral infections [½+½]

[Total: 5 marks]

Q106.2025 Main 57/1/1 · Q315 marks

Student to attempt either option-(A) or (B) :

(A) Answer the following questions :

  1. State what do you understand by “MALT” ? Where it is located inside our body ?
  2. Explain cytokine barriers.
  3. Name the diagnostic test for AIDS. On what principle does it work ?
  4. Bone marrow and thymus play an important role in human immune system. Explain how are they able to achieve this.

OR

(B) (i) Study the following table & fill ‘H’, ‘I’, ‘J’, ‘K’, ‘L’ and ‘M’ in following table with suitable words :

Chemical / Bioactive MoleculeMicro-organismCategoryUse
(a)Butyric acidHIImportant applications in food, chemical & pharma industry
(b)JMonoscus purpureusKInhibit cholesterol biosynthesis pathway
(c)Cyclosporin ALFungusM

(ii) Why are baculoviruses used as biological control agents ?

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Answer not available: no official marking scheme published.

Q107.2025 Supp. 57/S/1 · Q315 marks

(a) (i) Identify the disease in which the body cells of the patient lose the property of contact inhibition. State its possible causes (any three). Also name any two methods to detect and to diagnose this disease.

(ii) Your classmate complains of headache and cough. The doctor confirms that he is suffering from pneumonia and not common cold, on the basis of certain symptoms. List these symptoms and write any two precautions to be followed to prevent the spread of this disease.

OR

(b) Study the diagram of a typical biogas plant and answer the following questions :

Figure

  1. Identify ‘x’ and ‘y’.
  2. Write three components of biogas. Name one microorganism which is involved in biogas production.
  3. Name two institutes which developed the technology of biogas production in India.
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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q108.2024 Main 57/1/1 · Q335 marks

(a) (i) Name and explain the property present in normal cells but is lost in cancer cells.

(ii) All normal human cells have genes that may become cancerous under certain conditions. Name them and explain how.

(iii) State the role of the following techniques in detection and diagnosis of cancer :

  1. Biopsy and Histopathology
  2. Magnetic Resonance Imaging

OR

(b) Large quantities of sewage are generated every day in cities as well as in towns and are treated in Sewage Treatment Plants (STPs) to make them less polluting. Given below is the flow diagram of stages of STP.

Study the flow diagram and answer the questions that follow :

Primary effluent passed into large aeration tanks
↓
Effluent passed into settling tanks to form ‘sediment’

(i)

  1. Why is primary effluent passed into large aeration tanks ?
  2. What is the ‘sediment’ formed, referred to as ? Mention its significance.
  3. Explain the final step in the settling tank before the treated effluent is released into water bodies.

(ii) Name any two organisms commonly used as biofertilisers, belonging to different kingdoms. Write how each one acts as a biofertiliser.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a)

(i) Contact inhibition is present in normal cells but not in cancer cells. [½] When normal cells come in contact with other cells it inhibits their uncontrolled growth. [1]

(ii) Cellular oncogenes / Proto-oncogenes, when activated under certain conditions could lead to oncogenic transformation of the cells. [½ + ½]

(iii)

(1) Biopsy and histopathology – A piece of suspected tissue cut into thin sections is stained, and examined under microscope by pathologist for increased cell counts. [½ + ½]

(2) MRI – detects cancer of internal organs, [½] uses strong magnetic fields and non-ionising radiations to detect pathological and physiological changes in living tissue. [1]

OR

(b)

(i)

(1) In aeration tanks there is growth of aerobic microbes and fungi (flocs) that consume major part of organic matter in effluent thus reducing BOD. [1]

(2) Activated sludge [½]; used as inoculum in aeration tanks. [½]

(3) Bacterial flocs are allowed to sediment (activated sludge). [1]

(ii) Rhizobium (Bacteria) – lives symbiotically in nodules of roots of leguminous plants and fixes atmospheric nitrogen into organic form and provides nitrogen to the plant. [½ + ½]

Glomus (fungi) – lives in symbiotic association with roots of higher plants and absorbs phosphorus from the soil and passes it to plants. [½ + ½]

Cyanobacteria (Anabaena, Nostoc, Oscillatoria) – add organic matter to the soil and increase fertility (paddy fields). [½ + ½]

(Any two can be explained)

[Total: 5 marks]

Q109.2024 Main 57/3/1 · Q335 marks
    1. Describe the life cycle of Plasmodium from the time it enters the human body till a female Anopheles mosquito bites an infected person.
    2. Mention the two events of Plasmodium life cycle that occur within the female Anopheles body.

OR

    1. Write two differences between malignant tumor and benign tumor.
    2. Explain any three diagnostic techniques for the detection of cancer.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) (i) Life cycle of the Plasmodium from the period it enters the human body till a female Anopheles mosquito bites an infected person.

  • Plasmodium enters the human body as sporozoites (through the bite of infected Anopheles mosquito)
  • The Plasmodium reproduces asexually in the liver cells initially;
  • And then on releasing it in blood to attack the Red Blood cells (RBCs)
  • Parasite (Plasmodium) reproduces asexually in Red Blood cells (RBCs)
  • Resulting in rupture/bursting of the RBCs
  • Several sexual stages (gametocytes) develop in RBCs
  • Female Anopheles mosquito takes up gametocytes with blood meal. [½ × 6]

(or as in the diagram, each labelled stage [½ × 6])

Figure

(ii) Any two: fertilization of male and female gametocytes; production of sporozoites; migration of sporozoites into salivary glands of mosquito. [1 + 1]

OR

(b) (i)

Malignant tumorBenign tumor
1) Mass of proliferating cells called neoplastic or tumor cells grows rapidly.1) Tumor cells grow rapidly.
2) It spreads to other parts of the body or causes metastasis.2) It is confined to their original location, or causes no metastasis.
3) Invading and damaging the normal tissues.3) Causes little damage in the tissues.

(Any two corresponding differentiation points) [1 + 1]

(ii)

  • Biopsy and/histopathological studies – a piece of suspected tissue or blood or bone marrow cut into thin section is stained and examined by pathologist.
  • Radiography – use of X-rays.
  • CT / Computed tomography – to generate 3-dimensional image of the internal organs by using X-rays.
  • MRI – uses strong magnetic field and non-ionising radiations to accurately detect pathological and physiological changes in the living tissue.
  • Antibodies against specific antigens – used for detection of certain cancers.
  • Technique of Molecular Biology – applied to detect genes in an individual with inherited susceptibility to certain cancers.

(Any three techniques with explanation) [½ + ½] × 3

[Total: 5 marks]

Q110.2024 Main 57/4/1 · Q335 marks
  1. Describe the life cycle of HIV from the time of its entry into the human body till full blown AIDS sets in.

OR

  1. Write the symptoms of malaria in human and explain what causes these symptoms.
  2. Describe the different steps in the sexual mode of reproduction in the life cycle of a malarial parasite from the time of its initiation till where it is completed and ready to start a fresh cycle.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) HIV enters human body and enters macrophage cells, where viral RNA is introduced in the cell; viral DNA is produced by reverse transcription; viral DNA incorporated in host genome; new viral RNA is produced by the infected cell; new viruses are produced, which enter T-helper cells, replicate and produce progeny; progeny viruses attack other T-helper cells, this is repeated leading to progressive decline in the number of T-helper cells; person becomes immunodeficient and AIDS develops. [½ × 10]

(or diagram, each labelled step ½ mark)

Figure

OR

(b) (i)

  • Chills and high fever occur in a cyclic order / every 2-3 days [½ + ½]
  • fever is due to the toxic haemozoin [½]
  • released at the time of RBC rupture [1]

(ii) [½ × 5]

Figure

[Total: 5 marks]

Q111.2023 Main 57/1/1 · Q325 marks
  1. A time-bound vaccination programme is followed for the children in our country from their birth up to ten years of age. A graph plotted below shows the effect of the vaccination followed by infection by the same pathogen, and the antibody concentration in the blood of the child.

    Figure

    1. Explain why the administration of a vaccine causes an increase in the antibody concentration.
    2. If the child is infected with the same pathogen almost four months later, the antibody concentration in his/her blood increases very fast. Explain why.
    3. A table given below gives information about different types of immunity and how they are attained. Identify ‘P’, ‘Q’, ‘R’, ‘S’ and ‘T’ in the table.
      Type of immunityProduction of antibodiesPresence of memory cellsMode attained
      (1)Natural, activeYes‘P’‘Q’
      (2)Natural, passiveNo‘R’Across the placenta during pregnancy/breast feeding
      (3)Acquired, active‘S’YesGetting a vaccine during breast feeding
      (4)Acquired, passive‘T’NoGetting an injection of antibodies

OR

    1. What is the chemical name of ‘smack’ ? Why is the consumption of smack considered as an abuse ?
    2. Name the source plant and one effect of the following drugs on the human body :
      1. Marijuana
      2. Cocaine
      3. Morphine
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) (i) The vaccine contains the antigen, which stimulates or activates immune cells to produce antibodies (by B lymphocytes)/which generates primary response or humoral immune response. [½+1]

(ii) Memory cells generate, anamnestic response/secondary response [½+½]

(iii) P = Yes; Q = Catching an infection/getting infected; R = No; S = Yes; T = No [½×5]

OR

(b) (i) Diacetylmorphine [1]

as it is highly addictive, and being a depressant it slows down body functions. [½+½]

(ii) (1) Cannabis sativa, affects the cardiovascular system of the body. [½+½]

(2) Erythroxylum coca/coca plant, interferes with the transport of neurotransmitter dopamine/produces sense of euphoria/increased energy. [½+½]

(3) Papaver somniferum, acts as depressant/slows down body function/reduces pain/sedative [½+½]

[Total: 5 marks]

Q112.2023 Supp. 57/C/1 · Q335 marks

(a) Explain the following with reference to ‘Cancer’ :

  1. Contact Inhibition
  2. All normal cells have proto-oncogenes
  3. Name of carcinogens and their role
  4. Difference between benign and malignant tumours

OR

(b) (i) Why is it necessary to treat sewage water before it can be discharged into rivers and streams ? Explain different steps carried out in the treatment of sewage water and the production of fuel during this process.

(ii) Mention the initiative taken by the Ministry of Environment and Forests to save the major rivers of our country from pollution.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a)

(i) Cancer cells appear to have lost the property of contact inhibition as a result they continue to divide to form mass of cells (tumor). [1]

(ii) When proto-oncogenes are activated under certain conditions it could lead to oncogenic transformation of the cells. [1]

(iii)

  • X-rays / UV rays / Nicotine / Caffeine / Tobacco smoke / Oncogenic viruses [½]
  • Damage DNA which causes neoplastic transformation. [½]

(iv)

Benign tumoursMalignant tumours
Remain confined to their original locationNot remain confined to their original location / Show property of metastasis
Cause little damageDamage surrounding tissue and starve normal cells by competing for vital nutrients

(1 mark for each correct difference) [1 + 1]

OR

(b)

(i)

  • To make it less polluting [1]
  • Steps for sewage treatment:
  • Primary treatment – physical removal of floating debris through sequential filtration and sedimentation
  • Secondary treatment / biological treatment – primary effluent is passed to large aeration tank where it is constantly agitated and air is pumped into it
  • This allows vigorous growth of aerobic microbes into floc which significantly reduces organic matter or BOD
  • Flocs are allowed to settle in settling tank; this sediment is called activated sludge
  • Major part of sludge is pumped into anaerobic sludge digestor to produce biogas [½ × 6]

(ii) Ganga Action Plan, Yamuna Action Plan; to build a large number of sewage treatment plants so that only treated sewage may be discharged into the rivers. (Any two points to be considered) [½ + ½]

[Total: 5 marks]

Unit VIII · Biology and Human Welfare

8. Microbes in Human Welfare

Q1.2026 Main 57/1/1 · Q151 mark

Assertion (A): Flocs are masses of bacteria associated with fungal filaments in secondary treatment of sewage.

Reason (R): Flocs help in digestion of solid waste by anaerobic respiration.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(C) Assertion (A) is true, but Reason (R) is false. [1 mark]

Q2.2026 Main 57/2/1 · Q141 mark

Assertion (A): Microbes are used for commercial production of alcohol and beverages.

Reason (R): Yeast converts glucose into ethanol and carbon dioxide during fermentation.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
  3. Assertion (A) is true, Reason (R) is false.
  4. Assertion (A) is false, Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A). [1 mark]

Q3.2026 Main 57/3/1 · Q131 mark

Assertion (A): Activated sludge contains large population of bacteria.

Reason (R): These bacteria help in anaerobic digestion of human waste in biogas plant.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(C) Assertion (A) is true, but Reason (R) is false [1 mark]

Q4.2026 Main 57/4/1 · Q81 mark

Microbes used for biocontrol of pest butterfly caterpillars is :

  1. Saccharomyces cerevisiae
  2. Bacillus thuringiensis
  3. Streptococcus sp.
  4. Trichoderma
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(B) Bacillus thuringiensis [1 mark]

Q5.2026 Main 57/4/1 · Q151 mark

Assertion (A): Nucleopolyhedrovirus are useful as they are good biocontrol agents.

Reason (R): They are species-specific, narrow spectrum bioinsecticides.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A) [1 mark]

Q6.2026 Main 57/5/1 · Q71 mark

Monascus purpureus is a yeast used commercially in the production of which one of the following ?

  1. Ethanol
  2. Streptokinase
  3. Citric acid
  4. Statins
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(D) Statins [1 mark]

Q7.2026 Main 57/5/1 · Q151 mark

Assertion (A): Trichoderma species are free living fungi that are very common in the root ecosystems.

Reason (R): They are effective bio-control agents of several plant pathogens.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(B) Both Assertion (A) & Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A) [1 mark]

Q8.2026 Supp. 57/7/1 · Q71 mark

Identify the microbe that is responsible for producing large-holes in Swiss cheese.

  1. Propionibacterium sharmanii
  2. Saccharomyces cerevisiae
  3. Penicillium notatum
  4. Acetobacter aceti
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(A) Propionibacterium sharmanii [1 mark]

Q9.2025 Main 57/1/1 · Q31 mark

Match the items in Column-A with that of Column-B :

Column-AColumn-B
(i) Lady bird beetle(a) Methanobacterium
(ii) Mycorrhiza(b) Trichoderma
(iii) Biological control(c) Aphids
(iv) Biogas(d) Glomus

Choose the option that matches the items of Column A with that of B correctly :

  1. (i)–(b), (ii)–(d), (iii)–(c), (iv)–(a)
  2. (i)–(c), (ii)–(d), (iii)–(b), (iv)–(a)
  3. (i)–(d), (ii)–(a), (iii)–(b), (iv)–(c)
  4. (i)–(c), (ii)–(b), (iii)–(a), (iv)–(d)
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Answer not available: no official marking scheme published.

Q10.2025 Main 57/2/1 · Q91 mark

For commercial and industrial production of citric acid, which one of the following microbes is used ?

  1. Aspergillus niger
  2. Lactobacillus sp.
  3. Clostridium butylicum
  4. Saccharomyces cerevisiae
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Answer not available: no official marking scheme published.

Q11.2025 Main 57/4/1 · Q111 mark

Bioactive molecule Cyclosporin A used for human welfare is derived from :

  1. Propionibacterium sharmanii
  2. Monascus purpureus
  3. Trichoderma polysporum
  4. Aspergillus niger
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Answer not available: no official marking scheme published.

Q12.2025 Main 57/5/1 · Q111 mark

Large scale industrial production of Butyric acid for human welfare is done using the microbe :

  1. Aspergillus sp.
  2. Streptococcus sp.
  3. Clostridium sp.
  4. Trichoderma sp.
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Answer not available: no official marking scheme published.

Q13.2025 Main 57/5/1 · Q141 mark

Assertion (A): Biogas plants are more often built in rural areas.

Reason (R): The excreta or ‘gobar’ of cattle is rich in Methanobacterium.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q14.2025 Main 57/6/1 · Q111 mark

Bottled fruit juices are clearer as compared to those made at home, as they are clarified by the use of :

  1. Lipases and pectinases
  2. Pectinases and proteases
  3. Proteases and cellulases
  4. Nucleases and lipases
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Answer not available: no official marking scheme published.

Q15.2025 Main 57/7/1 · Q131 mark

Assertion (A): Swiss cheese is characterized by large holes due to CO2 production.

Reason (R): It is ripened by growing a specific fungi.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer

Answer not available: no official marking scheme published.

Q16.2024 Main 57/1/1 · Q81 mark

Lactobacillus that sets milk into curd is categorised as :

  1. Cyanobacteria
  2. Archaebacteria
  3. Chemosynthetic bacteria
  4. Heterotrophic bacteria
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(D) Heterotrophic bacteria [1 mark]

Q17.2024 Main 57/2/1 · Q101 mark

Which one of the following pairs is not correctly matched ?

  1. Clostridium butylicum – Butyric acid
  2. Trichoderma polysporum – Cyclosporin A
  3. Monascus purpureus – Citric Acid
  4. Streptococcus – Streptokinase
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(C) Monascus purpureus – Citric Acid [1 mark]

Q18.2024 Main 57/4/1 · Q101 mark

Study the following list of bioactive substances and their action :

Bioactive SubstanceRole
A. StatinI. Removal of oil stains
B. Cyclosporin AII. Removal of clots from blood vessels
C. StreptokinaseIII. Lowering of blood cholesterol
D. LipaseIV. Immuno-suppressive agent

Select the option in which the bioactive substances are correctly matched with their action.

  1. A – II, B – III, C – I, D – IV
  2. A – III, B – IV, C – II, D – I
  3. A – IV, B – I, C – II, D – III
  4. A – IV, B – II, C – I, D – III
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(B) A – III, B – IV, C – II, D – I [1 mark]

Q19.2024 Main 57/5/1 · Q51 mark

During biological treatment of sewage, the masses of bacteria held together by fungal filaments to form mesh like structures are called

  1. primary sludge
  2. flocs
  3. activated sludge
  4. anaerobic sludge
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(B) Flocs [1 mark]

Q20.2024 Supp. 57/S/1 · Q81 mark

In plants, which one of the following helps in the absorption of phosphorus from soil ?

  1. Glomus
  2. Rhizobium
  3. Frankia
  4. Anabaena
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(A) Glomus [1 mark]

Q21.2023 Main 57/1/1 · Q41 mark

The graph plotted below is based on the data collected by biology students with respect to the levels of oxygen at the specific points in the river flowing outside their city. Which point in the graph indicates the entry of untreated sewage in the river ?

Figure

  1. Point (i)
  2. Point (ii)
  3. Point (iii)
  4. Point (iv)
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(A) Point (i) [1 mark]

Q22.2023 Main 57/3/1 · Q101 mark

Identify the fungus that ripens the famous ‘Roquefort’ cheese :

  1. Saccharomyces cerevisiae
  2. Propionibacterium sharmanii
  3. Monascus purpureus
  4. Penicillium notatum
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Answer not available: the official marking scheme gives no answer key for this question and awards the mark to everyone who attempts it.

Q23.2023 Main 57/5/1 · Q101 mark

Given below are the list of the commercially important products and their source organisms. Select the option that gives the correct matches.

S. No.List A: Bioactive ProductsS. No.List B: Microbes (Source Organism)
(A)Cyclosporin A(i)Streptococcus
(B)Statins(ii)Tricoderma polysporum
(C)Streptokinase(iii)Penicillium notatum
(D)Penicillin(iv)Monascus purpureus
  1. (A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
  2. (A)-(iii), (B)-(iv), (C)-(ii), (D)-(i)
  3. (A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)
  4. (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(D) A – (ii), B – (iv), (C) – (i), (D) – (iii). [1 mark]

Q24.SQP 2025-26 · Q121 mark

Integrated Pest Management involves

  1. Using pesticides/insecticides judiciously
  2. Using biocontrol agents
  3. Engaging in organic farming
  1. Only I
  2. Only II
  3. Both I and II
  4. Only III
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Official CBSE Sample Paper Marking Scheme · 2025-26

(C) Both I and II [1 mark]

Q25.SQP 2026-27 · Q121 mark

A rice farmer's field, which typically thrives in waterlogged conditions and requires soil rich in nutrients like phosphorus, is experiencing a severe drought this year. The crop is also susceptible to root infestation by pathogens. Which of the following techniques is best to address all these issues?

  1. Adding nitrogen-fixing bacteria to the soil
  2. Applying chemical fertilizers
  3. Adding spores of Glomus to the soil
  4. Flooding the field artificially to mimic normal conditions
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Official CBSE Sample Paper Marking Scheme · 2026-27

(C) Adding spores of Glomus to the soil [1 mark]

Q26.SQP 2026-27 · Q151 mark

Assertion (A): The primary goal of sewage treatment is to eliminate all organic matter from the effluent.

Reason (R): High BOD indicates a low level of organic matter in the water, which can lead to increased oxygen consumption by microorganisms, potentially harming aquatic life.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
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Official CBSE Sample Paper Marking Scheme · 2026-27

(C) A is true but R is false. [1 mark]

Q27.2026 Main 57/2/1 · Q192 marks

After significant reduction of BOD during secondary treatment of sewage, what further processes are carried out ? Mention any one important outcome of these processes.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1
  • The effluent from aeration tank is passed into settling tanks where flocs settle down forming activated sludge [½]
  • A small part of activated sludge is pumped back into the aeration tank to serve as inoculum and major remaining part is pumped into large tanks called anaerobic sludge digesters [½]
  • Anaerobic bacteria digest the sludge and produce a mixture of gases as methane, hydrogen sulphide and carbon dioxide or biogas [½]
  • Outcome: Biogas produced can be used as a source of energy / Effluent from secondary treatment plant is generally released into natural water bodies like rivers and streams. [½]

[Total: 2 marks]

Q28.2026 Main 57/4/1 · Q192 marks

A laboratory assistant collected three water samples, namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant. But the laboratory attendant did not note which was which. The samples were labelled A, B and C and subjected to a BOD test. The BOD values of the three samples A, B and C were recorded as 30 mg/L, 8 mg/L and 400 mg/L respectively.

(a) Which sample of the water is most polluted ?

(b) Can you assign the correct label to each, assuming the river water is relatively clean ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) Sample C is most polluted (Highest BOD) [½]

(b) Sample A – River water [½]

Sample B – Secondary effluent (Least BOD) [½]

Sample C – Untreated sewage water (Highest BOD) [½]

[Total: 2 marks]

Q29.2024 Main 57/5/1 · Q202 marks

Explain how are plants benefitted by their association with “Glomus species”.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

Absorbs phosphorus from soil and passes it to plants, resistant to root-borne diseases, tolerance to salinity and drought, overall increases in plants growth and development. (Any two points) [1 × 2]

[Total: 2 marks]

Q30.2023 Main 57/3/1 · Q182 marks
  1. Write the first step the primary effluent undergoes when it enters the secondary treatment plant and state the purpose.
  2. What is the level of B.O.D indicative of in the secondary treatment plant ? Mention its significance.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) Primary effluent passes to aeration tank/it is constantly agitated mechanically and air is pumped into it, this allows vigorous growth of flocs (masses of bacteria associated with fungal filaments to form mesh like structure)/to consume organic matter/to reduce BOD level of primary effluents [½×2]

(b) It is a measure of organic matter/polluting potential in the water, greater BOD more is polluting potential. [½×2]

[Total: 2 marks]

Q31.2023 Main 57/5/1 · Q202 marks
    1. Give an example of a genus of virus used as narrow spectrum insecticidal biocontrol agent.
    2. How does its use serve as an aid in overall integrated pest management programme ?

OR

  1. Why a malignant tumour considered to be more damaging than a benign tumour ? Explain.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) (i) Nucleopolyhedrovirus [1]

(ii) Species specific, no negative impact on other plants/mammals/birds/fish/non target insects, beneficial insects are conserved, useful for ecologically sensitive area. (Any two) [½×2]

OR

(b) Malignant tumor grows very rapidly, and starves the normal cells by competing for vital nutrients, invades and damages surrounding normal tissue, shows metastasis/spread to other body parts or tissues. (Any two) [1×2]

[Total: 2 marks]

Q32.2022 Main 57/1/1 · Q22 marks

State the impact of constant mechanical agitation and pumping of air in the aeration tank on the sewage during the biological treatment.

OR

(a) Cattle excreta is important source for producing a domestic fuel. Name the fuel and write its main components.

(b) Write the biological process that is responsible for the production of this fuel.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

Allows vigorous growth of microbes, formation of flocs to form mesh like structures, microbes consume major part of organic waste in the sewage, reduces BOD of the effluent / reduces polluting potential [½ × 4]

OR

(a) Biogas / Gobar gas [½]; Methane, CO2 / H2 / H2S [½ + ½]

Note: any one component along with methane to be awarded one mark.

(b) Anaerobic digestion of cellulose [½]

[Total: 2 marks]

Q33.2022 Main 57/1/1 · Q32 marks

(a) Give an example of viral biocontrol agent.

(b) Why are they considered to be desirable when an ecologically sensitive area is being treated?

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

(a) Baculoviruses / Nucleopolyhedrovirus [1]

(b) Species specific, narrow spectrum insecticidal application, no negative impact on plants / mammals / birds / fish / non-target insects, desirable for integrated pest management (any two) [½ + ½]

[Total: 2 marks]

Q34.2022 Main 57/2/1 · Q22 marks

"Curd is easier to digest by humans than milk." Justify, giving two suitable reasons.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

LAB produce acids that coagulate milk [1]

Partially digests milk proteins [1]

[Total: 2 marks]

Q35.2022 Main 57/2/1 · Q42 marks

Study the given diagrams of Biogas plants – 'A' and 'B' and answer the questions that follow:

Figure

Which one of the two biogas plants can be used for generating gas fuel and electricity and why? Give suitable reasons in support of your answer.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

Biogas plant A [1]

Methanogens present in the cow dung (grow anaerobically on cellulosic material), produce large amounts of methane along with carbon dioxide and hydrogen. [½ + ½]

[Total: 2 marks]

Q36.2022 Main 57/3/1 · Q32 marks

Farmers are often suggested to use the following organisms in their crop land so as to improve the soil fertility.

(i) Rhizobium (ii) Anabaena

Explain.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

(i) Forms symbiotic association with roots of leguminous plants and helps in nitrogen fixation [1]

(ii) Helps in nitrogen fixation in aquatic or terrestrial environment and serves as biofertilizer / increases organic matter of soil. [1]

[Total: 2 marks]

Q37.2022 Main 57/3/1 · Q42 marks

Organic farmer use Trichoderma and Baculovirus as biological control agents. Explain.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

Trichoderma – The fungus controls plant pathogens. [1]

Baculovirus – Specific to insects and arthropods, narrow spectrum insecticidal applications, used in IPM, used in ecologically sensitive area, no negative impact on plants / mammals / birds / fishes / non-target insects (any two) [½ + ½]

[Total: 2 marks]

Q38.2022 Main 57/4/1 · Q22 marks

(a) Mention the common bacterium found in the anaerobic sludge during sewage treatment and also in the rumen of cattle. How is this bacterium commercially useful?

OR

(b) Name the effective biocontrol agents of several plant pathogens belonging to group of viruses. Also write about the ways they support the environment.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

(a) ● Methanobacterium / Methanogen [1]

● It helps in production of biogas / gobar gas [1]

OR

(b) ● Baculoviruses / Nucleopolyhedrovirus [1]

● Species specific, narrow spectrum insecticidal application, no negative impact on plants / vertebrates / non-target insects, help to conserve beneficial insects, useful in IPM, useful in ecologically sensitive area [½×2] (Any two)

[Total: 2 marks]

Q39.2022 Main 57/4/1 · Q32 marks

Some of the microbes used as biofertilizers are prokaryotes. Name the taxonomic group they come under. With the help of an example, mention how they act as biofertilizers.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

● Bacteria / Cyanobacteria / Blue Green Algae [1]

● Rhizobium / Azospirillum / Azotobacter / Nostoc / Anabaena / Oscillatoria / or any other correct example [½]

● Fixes atmospheric nitrogen into organic form / Enrich nitrogen content of soil / Increases soil fertility [½]

[Total: 2 marks]

Q40.2022 Main 57/4/1 · Q42 marks

Different species belonging to genus Trichoderma are useful to humans as well as to plants. Justify their roles by giving one instance of each.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

Trichoderma act as biocontrol agent for several plant pathogens, Trichoderma produces immunosuppressive agent / cyclosporin A / immunosuppressant useful for humans [1+1]

[Total: 2 marks]

Q41.2022 Main 57/5/1 · Q22 marks

Can we use slurry of human excreta instead of cowdung slurry to produce biogas in a typical biogas plant? Support your answer giving reasons.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

● No [1]

● Cowdung has methanogens or Methanobacterium / In human excreta methanogens or Methanobacteria are absent, [½]

Which grow anaerobically on cellulose material or gobar and produce methane, CO2, H2 / Methanogens or Methanobacterium are essential for bio gas production [½]

[Total: 2 marks]

Q42.2022 Main 57/5/1 · Q42 marks

Study the given diagram of Sewage Treatment Plant (S.T.P.) and answer the questions that follow:

Figure

(a) Which one of the two ‘S.T.P.’ (A) or (B) will be more effective in treating the human excreta in the municipal waste?

(b) Write the steps followed in carrying the treatment of the sewage in step (Z), once the BOD of sewage is reduced significantly till it is passed on to the “anaerobic sludge digesters”.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

(a) S.T.P (B) / B [½]

(b) ● The effluent is passed into the settling tanks [½]

● The bacterial ‘flocs’ are allowed to settle (sediment is called activated sludge) [½]

● a small part of the ‘activated sludge’ is pumped back into the aeration tanks as inoculum (for secondary/biological treatment) [½]

[Total: 2 marks]

Q43.2022 Supp. 57/6/1 · Q22 marks

A particular cyanobacteria is spread by the farmers in their fields while growing paddy. Name the cyanobacteria used and give two advantages of it.

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

● Anabaena / Nostoc / Oscillatoria (or any other correct example) [1]

● fix atmospheric nitrogen, acts as biofertiliser, add organic matter to the soil, increases soil fertility, reduces dependence on chemical fertilisers, replenish the soil nutrients [½+½] (any two)

[Total: 2 marks]

Q44.2022 Supp. 57/6/1 · Q42 marks

Study the given diagrams of Sewage Treatment Plants (STP) and answer the questions that follow:

Figure

(a) Which of the two Sewage Treatment Plants, (A) or (B), will be more effective in treating human excreta in the municipal waste?

(b) How is the primary effluent treated in the aeration tanks till there is significant reduction in the BOD of the effluent?

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

(a) (A) [½]

(b) – In the aeration tanks the effluent is constantly agitated mechanically and air is pumped into it. [½]

– Vigorous growth of aerobic microbes into flocs (masses of bacteria associated with fungal filaments to form mesh like structures) takes place. [½]

– While growing these microbes consume the major part of the organic matter in the effluent thus decreasing / reducing BOD. [½]

[Total: 2 marks]

Q45.2026 Main 57/3/1 · Q273 marks

Cow dung and water are mixed and fed into a biogas plant to allow digestion of biowastes. The person performing this process says that there is no need to provide an inoculum.

(a) Do you agree with him ? Justify your answer. [1]

(b) What happens to the biowaste inside the digester ? [1]

(c) Name the useful by-products obtained from this process and mention how they are used. [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) Yes, there is no need of inoculum as the cow dung contains a lot of methanogens. [½+½]

(b) Inside the digester the biowaste is digested by anaerobic bacteria (methanogens) present in the cow dung. [1]

(c) Useful By-products: Biogas, used for cooking and lighting / Slurry, used as fertiliser / manure [½+½]

(Anyone with correct correlation – ½ mark for name of byproduct and ½ mark for its use)

[Total: 3 marks]

Q46.2026 Main 57/5/1 · Q253 marks

Mention a product of human welfare obtained with the help of each one of the following :

  1. Saccharomyces cerevisiae
  2. Propionibacterium shermanii
  3. Aspergillus niger
  4. Trichoderma polysporum
  5. Acetobacter aceti
  6. Streptococcus
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a) Beverages / Ethanol / Bakery products / Bread [½]

(b) Swiss Cheese [½]

(c) Citric acid [½]

(d) Cyclosporin-A / Immunosuppressive agent [½]

(e) Acetic acid [½]

(f) Streptokinase / Clot buster enzyme [½]

[Total: 3 marks]

Q47.2025 Main 57/4/1 · Q253 marks

Explain the biological treatment of primary effluent when passed into the large aeration tanks in a sewage treatment plant (STP).

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Answer not available: no official marking scheme published.

Q48.2025 Main 57/5/1 · Q243 marks
  1. Write the full form of BOD.
  2. Define BOD. Explain how it is a measure of the organic matter present in the water body.
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Answer not available: no official marking scheme published.

Q49.2025 Main 57/6/1 · Q273 marks
  1. What do you mean by activated sludge in an STP ?
  2. Explain the biological treatment of the major part of the sludge transferred from the large aeration tank into the anaerobic sludge digesters before its final release into the natural water bodies.
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Answer not available: no official marking scheme published.

Q50.2024 Main 57/3/1 · Q263 marks
    1. List two major reasons for using cow-dung in a biogas plant instead of using domestic sewage.
    2. Mention one use of the unspent slurry of the biogas plant.

OR

  1. Name the bioactive molecule and its microbial source generally used by physicians to treat the patients for :
    1. Myocardial infarction
    2. High blood cholesterol level
    3. Organ transplantation
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) (i)

  • Cow dung has methanogens or Methanobacterium
  • Cow dung is rich in cellulosic material
  • Bacteria grows anaerobically on cellulosic material to produce large amount of methane. (Any two points) [1 × 2]

(ii) Used as manure / fertilizer (organic) [1]

OR

(b)

Bioactive moleculeMicrobial source
(i)Myocardial infarctionStreptokinaseStreptococcus
(ii)High blood cholesterol levelStatinsMonascus purpureus
(iii)Organ transplantationCyclosporin ATrichoderma polysporum

[½ × 2] for each of (i), (ii), (iii)

[Total: 3 marks]

Q51.2024 Main 57/4/1 · Q263 marks

Explain the role of the following during the sewage treatment :

  1. flocs
  2. anaerobic sludge digester
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) These consume the major part of organic matter in the effluent, and significantly reduce the BOD. [1 + 1]

(b) Anaerobic sludge digesters – anaerobic digestion of sludge takes place, producing ‘Biogas’ in the process. [½ + ½]

[Total: 3 marks]

Q52.2024 Main 57/5/1 · Q223 marks

Identify a, b, c, d, e and f in the table given below :

Sl. No.OrganismBioactive MoleculeUse
1.Monascus purpureusab
2.cdAntibiotic
3.eCyclosporin Af
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

1. (a) Statin   (b) Blood-cholesterol lowering agent

2. (c) Penicillium notatum   (d) Penicillin

(In (c) and (d) any other correct organism and its bioactive molecule can be considered)

3. (e) Trichoderma polysporum   (f) Immunosuppressant

[½ × 6]

[Total: 3 marks]

Q53.2023 Main 57/1/1 · Q263 marks

Treatment of wastewater is done in a sewage treatment plant to make it less polluting. Explain the following with reference to this treatment process :

  1. Primary sludge
  2. Activated sludge
  3. Anaerobic sludge digesters
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) Primary sludge: All the solids that settle down, during the primary treatment of sewage water. [½×2]

(b) Activated sludge: Produced during the secondary treatment or biological treatment of sewage, primary effluent + aerobic microbes flocs (bacteria and fungus) – get converted to a sediment whose BOD has reduced significantly. [½×2]

(c) Anaerobic sludge digesters: Large tanks where activated sludge is treated with anaerobic bacteria which digest the bacteria and fungi, and produce a mixture of CH4, H2S and CO2/Biogas [½×2]

[Total: 3 marks]

Q54.2023 Supp. 57/C/1 · Q273 marks

Name the antibiotic that was used to treat wounded soldiers in World War II. Explain its chance discovery.

Name the scientists who were awarded the Nobel Prize for this discovery.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1
  • Penicillin [½]
  • Alexander Fleming while working on Staphylococci bacteria, once observed a mould growing in one of his unwashed culture plates around which Staphylococci could not grow. He found out that it was due to a chemical produced by the mould and he named it Penicillin after the mould Penicillium notatum. [½ × 3]
  • Alexander Fleming, Ernest Chain, Howard Florey. (½ mark each for any two correct names) [½ + ½]

[Total: 3 marks]

Q55.2022 Main 57/1/1 · Q103 marks

Enumerate the main sources of bio-fertilisers giving one example of each.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

Bacteria – Rhizobium / Azospirillum / Azotobacter [½ + ½]

Fungi – Mycorrhiza / Glomus [½ + ½]

Cyanobacteria – Anabaena / Nostoc / Oscillatoria [½ + ½]

[Total: 3 marks]

Q56.2022 Main 57/1/1 · Q113 marks

Explain giving reason the action plan followed by organic farmers that support their key belief "biodiversity furthers health of crop lands".

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

Do not eradicate pests but keep them at manageable level by various checks and balances, pests may be food for otherwise beneficiary predators and parasites who would suffer and not survive, more varieties in a landscape more sustainable it is (or any other correct reason) [1 × 3]

[Total: 3 marks]

Q57.2022 Main 57/4/1 · Q93 marks

Bacillus thuringiensis plays an important role in Integrated Pest Management strategy. Explain how. Name any two crops that are protected efficiently from pests.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

● Spores of Bacillus thuringiensis available as sachets are mixed with water and sprayed on vulnerable plants, insect will die if they eat the plant. [1+1]

● Examples: Brassica, fruit trees [½+½]

//

● Scientists have introduced Bt toxin gene into plants, insect will die if they eat the plant. [1+1]

● Examples: Bt rice, Bt cotton, Bt tomato, Bt potato, Bt soyabean, or any other examples [½+½]

[Total: 3 marks]

Q58.SQP 2025-26 · Q263 marks

The Biological Oxygen Demand (BOD) of a primary effluent during sewage treatment is reduced. Enlist the process of how this is achieved.

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Official CBSE Sample Paper Marking Scheme · 2025-26

‘Flocs’ and ‘activated sludge’ in sewage treatment help to reduce the BOD.

(i) Flocs: These are masses of bacteria held together by slime and fungal filaments to form mesh-like structures. These are used during the secondary sewage treatment in the aeration tank to increase the rate of decomposition. The microbes digest a lot of organic matter, converting it into microbial biomass and releasing a lot of minerals. As a result, the BOD of sewage is reduced. As the BOD of waste is reduced to 1% of raw sewage, it is passed into the settling tank. In these tanks, flocs are allowed to undergo sedimentation.

(ii) Activated sludge: The sediment of the settling tank is called activated sludge. A part of it is used as an inoculum in aeration tanks. The remaining part is passed into a large tank called an anaerobic sludge digester. In these tanks, anaerobic microbes are present that digest the organic mass as well as aerobic microbes of activated sludge. The remaining sludge is used as manure or compost.

Q59.2026 Supp. 57/7/1 · Q304 marks

Read the following passage and answer the questions that follow :

Sewage is generated in large quantities in the urban areas. It is treated in sewage treatment-plants (STPs) to make it less polluted before discharging into any natural water body. Observe the flow chart given below and answer the questions that follow :

Figure

(a) Name the two products obtained after primary treatment of sewage.

(b) What happens to the activated sludge when it is passed into an anaerobic sludge digester ? Explain.

(c) (i) Name the two groups of microbes present in the primary effluent after it is passed into large aeration tanks.

OR

(c) (ii) Define BOD. What happens to the BOD when organic matter present in the wastewater increases ?

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) Primary sludge, primary effluent. [½ + ½]

(b) • The anaerobic bacteria digest the other bacteria and fungi or flocs in the sludge. [1]

• A mixture of gases (methane, hydrogen sulphide & CO2) called biogas is produced which can be used as source of energy. [1]

(c) (i) Aerobic bacteria and fungi. [½ + ½]

OR

(c) (ii) • BOD refers to the amount of oxygen that is required by bacteria to oxidise organic matter in one litre of water. [½]

• BOD will increase. [½]

[Total: 4 marks]

Q60.2025 Main 57/7/1 · Q294 marks

Study the figure given below and answer the questions that follow.

Figure

  1. Identify the figure and state its importance.
  2. Why is air diffused into the aerator tank ? Explain.
  3. (i) What changes take place in the settling tank ?

OR

  1. (ii) How is BOD related to organic matter present in the water ?
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Answer not available: no official marking scheme published.

Q61.SQP 2026-27 · Q304 marks

A tribal farmer is cultivating Brassica oleracea (cabbage) in a forest region rich in pollinators and rare butterflies. Recently, his crops have been heavily damaged by cabbage looper caterpillars. To manage the pest outbreak, he applies Nucleopolyhedrovirus (NPV). After a few weeks, the pest population decreases significantly without affecting other insects or animals in the area.

A. Justify with reasons why the use of Nucleopolyhedrovirus (NPV) was a suitable choice for controlling cabbage looper caterpillars. (1)

B. Suggest another IPM strategy suitable for controlling caterpillar pests, and briefly describe its mode of action. (2)

Attempt either subpart C or D.

C. Name the group of pathogens to which Nucleopolyhedrovirus belong. (1)

OR

D. Justify why the use of NPV is considered a biocontrol method. (1)

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Official CBSE Sample Paper Marking Scheme · 2026-27

A. NPV is species-specific and affects only target insect pests, narrow spectrum insecticide; causes no harm to plants, pollinators, mammals, birds, fish, or non-target insects. [1]

B. He could have sprayed the spores of Bacillus thuringiensis, in the gut of the larvae, the bacterial toxin is released and the larvae get killed. [2]

C. Baculovirus [1]

OR

D. Because it is a biological method for controlling plant diseases and pests. [1]

Q62.2024 Supp. 57/S/1 · Q335 marks

(a) Describe the process of wastewater (sewage) treatment under the following heads :

(i) Primary treatment

(ii) Secondary treatment

OR

(b) (i) Explain the sequence of events occurring in a biogas plant.

(ii) Write the components of biogas.

(iii) Write any three advantages of using cow dung as the raw material.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) (i) During primary treatment the floating debris is removed by sequential filtration, then the grit (soil and small pebbles) is removed by sedimentation; all solids that settle form the primary sludge, and the supernatant forms the effluent. [½ × 4]

(ii) Secondary treatment – it is biological treatment in which the primary effluent is passed into large aeration tanks where it is constantly agitated mechanically and air is pumped into it; this allows vigorous growth of useful aerobic microbes into flocs (masses of bacteria associated with fungal filaments to form mesh-like structures); while growing, these microbes consume the major part of the organic matter in the effluent, which significantly reduces the BOD (biochemical oxygen demand) of the effluent; once the BOD of sewage or waste water is reduced significantly the effluent is then passed into a settling tank where the bacterial ‘flocs’ are allowed to sediment, called activated sludge, and effluent is released in natural water bodies like river; a small part of the activated sludge is pumped back into the aeration tank to serve as the inoculum, and the remaining major part of the sludge is pumped into large tanks called anaerobic sludge digesters to produce biogas. [½ × 6]

OR

(b) (i) The biogas plant consists of a concrete tank (10–15 feet deep) in which bio-wastes are collected and a slurry of dung is fed; certain bacteria (methanogens) found in anaerobic sludge help in breakdown of cellulose and production of biogas; outlet pipe of biogas plant is connected to supply of biogas to nearby houses. [½ × 4]

(ii) Methane, CO2 and H2 [½ × 3]

(iii) Cow dung is available in large quantities in rural areas, rich in methanogen bacteria, cheap, used as manure (sludge), rich in cellulosic material, used for generation of biogas, eco-friendly. (Any three) [½ × 3]

[Total: 5 marks]

Unit IX · Biotechnology and its Applications

9. Biotechnology: Principles and Processes

Q1.2026 Main 57/1/1 · Q91 mark

Match Column-I with Column-II and choose the correct option :

Column-IColumn-II
a.Biolistic guni.Bacterial cell
b.Chitinaseii.Tumour inducing
c.Tiiii.Animal cell
d.Ca++iv.Fungal cell
v.Plant cell

Options :

  1. a – i, b – ii, c – iii, d – iv
  2. a – ii, b – v, c – i, d – iii
  3. a – v, b – iv, c – ii, d – i
  4. a – v, b – i, c – iv, d – ii
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(C) a – v, b – iv, c – ii, d – i [1 mark]

Q2.2026 Main 57/1/1 · Q111 mark

Which of the following statements about plasmids is incorrect ?

  1. Plasmids have the ability to replicate within the bacterial cell.
  2. Their replication is controlled by chromosomal DNA.
  3. They are autonomously replicating circular extra-chromosomal DNA.
  4. They often carry antibiotic resistant genes.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(B) Their replication is controlled by chromosomal DNA [1 mark]

Q3.2026 Main 57/2/1 · Q91 mark

Which of the following is used to visualize DNA bands under UV light after gel electrophoresis ?

  1. Acetocarmine
  2. Safranine
  3. Ethidium Bromide
  4. Potassium Iodide
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(C) Ethidium Bromide [1 mark]

Q4.2026 Main 57/2/1 · Q161 mark

Assertion (A): DNA Ligase is used to join DNA fragments.

Reason (R): It catalyses the formation of glycosidic bonds between nucleotides.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
  3. Assertion (A) is true, Reason (R) is false.
  4. Assertion (A) is false, Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(C) Assertion (A) is true, Reason (R) is false. [1 mark]

Q5.2026 Main 57/3/1 · Q31 mark

Match Column I with Column II and select the correct option :

Column IColumn II
a. Primersi. β-galactosidase
b. Insertional inactivationii. To separate and purify products
c. Bioreactoriii. Chemically synthesized oligonucleotides
d. Downstream processingiv. Large scale production of specific product of activity of microbes

Options :

  1. a-iii, b-i, c-iv, d-ii
  2. a-iii, b-i, c-ii, d-iv
  3. a-i, b-iii, c-iv, d-ii
  4. a-i, b-iv, c-ii, d-iii
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(A) a-iii, b-i, c-iv, d-ii [1 mark]

Q6.2026 Main 57/3/1 · Q151 mark

Assertion (A): DNA cannot pass through cell membrane of a bacterial cell.

Reason (R): DNA is a hydrophilic molecule.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of the Assertion (A) [1 mark]

Q7.2026 Main 57/5/1 · Q81 mark

Given below is the restriction site of a restriction endonuclease Pst I and the cleavage sites on a DNA molecule.

Choose the option that gives the correct resultant fragments.

Options (A)–(D) are shown in the figure.

Figure

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(D) 5′ C-T-G-C-A   G 3′ / 3′ G   A-C-G-T-C 5′ [1 mark]

Q8.2026 Main 57/5/1 · Q161 mark

Assertion (A): The use of chitinase enzyme is necessary for the separation of DNA from yeast cells but not in the case of Spirogyra.

Reason (R): Fungal cell wall is made up of fungal chitin.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A) [1 mark]

Q9.2026 Supp. 57/7/1 · Q61 mark

P and Q are required in the extension step of PCR to extend the primer sequences from the 3′ end of each primer. Choose the option which correctly fills the blanks P and Q.

  1. P – Ribonucleoside triphosphate; Q – DNA polymerase
  2. P – Deoxyribonucleotide triphosphate; Q – Thermolabile Taq-polymerase
  3. P – Deoxynucleotides; Q – Thermostable DNA-polymerase
  4. P – Deoxyribonucleoside monophosphate; Q – Thermostable DNA-polymerase
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(C) P – Deoxynucleotide(s), Q – Thermostable DNA polymerase [1 mark]

Q10.2026 Supp. 57/7/1 · Q161 mark

Assertion (A): Vector DNA and foreign DNA are cut by the same restriction enzyme.

Reason (R): Vector DNA and foreign DNA, cut with same enzyme, produces complimentary sticky ends.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]

Q11.2025 Main 57/1/1 · Q81 mark

Amplification of gene of interest by using DNA polymerase may go upto

  1. 0.1 million times
  2. 1 million times
  3. 1 billion times
  4. 1 trillion times
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Answer not available: no official marking scheme published.

Q12.2025 Main 57/1/1 · Q151 mark

Assertion (A): A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions.

Reason (R): The most commonly used bioreactors are of stirring type.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Answer not available: no official marking scheme published.

Q13.2025 Main 57/2/1 · Q111 mark

Select the correct statement from the following biotechnological procedures :

  1. The polymerase enzyme joins the gene of interest and the vector DNA.
  2. Gel electrophoresis is used for amplification of a DNA segment.
  3. PCR is used for isolation and separation of gene of interest.
  4. Plasmid DNA acts as vector to transfer the piece of DNA attached to it.
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Answer not available: no official marking scheme published.

Q14.2025 Main 57/2/1 · Q161 mark

Assertion (A): Restriction endonuclease recognises palindromic sequence in DNA and cuts them.

Reason (R): Palindromic sequence has two unique recognition sites PstI and PvuI recognised by restriction endonuclease.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Answer not available: no official marking scheme published.

Q15.2025 Main 57/5/1 · Q81 mark

Isolation of DNA from a fungal cell can be achieved by using :

  1. Cellulase
  2. Chitinase
  3. Lysozyme
  4. Protease
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Answer not available: no official marking scheme published.

Q16.2025 Main 57/6/1 · Q101 mark

The cloning site present in the tetracycline resistance gene of E. coli cloning vector pBR322 is :

  1. EcoR I
  2. Pvu II
  3. Sal I
  4. Pst I
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Answer not available: no official marking scheme published.

Q17.2025 Main 57/7/1 · Q21 mark

Identify the incorrect statement regarding PCR.

  1. Two sets of primers are required during polymerisation.
  2. The process of replication is repeated multiple times to produce one billion copies.
  3. Thermostable DNA polymerase is used for extension of primers.
  4. Annealing is required to separate both the strands of template DNA.
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Answer not available: no official marking scheme published.

Q18.2025 Main 57/7/1 · Q161 mark

Assertion (A): A piece of DNA inserted into an alien organism generally does not replicate, if not inserted into a chromosome.

Reason (R): Chromosomes have specific sequences called ‘ori’ region, where DNA replication is initiated.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q19.2025 Supp. 57/S/1 · Q71 mark

Figure

Identify the material that helps in the precipitation of purified DNA during the process of isolation of DNA as shown in the given diagram.

  1. Protease
  2. Lysozyme
  3. Cellulase
  4. Chilled ethanol
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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(D) Chilled Ethanol [1 mark]

Q20.2025 Supp. 57/S/1 · Q111 mark

What is the process of binding a primer to the denatured strand called ?

  1. Denaturation
  2. Annealing
  3. Renaturation
  4. Polymerisation
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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(B) Annealing [1 mark]

Q21.2025 Supp. 57/S/1 · Q131 mark

Assertion (A): In bioreactors, raw materials are biologically converted into desired products.

Reason (R): In bioreactors, optimum growth conditions for microbes such as temperature, pH, substrate, salts, etc. are available.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q22.2024 Main 57/1/1 · Q101 mark

Restriction Endonuclease Hind II always cuts DNA molecules at a particular point by recognising a specific sequence of :

  1. Six base pairs
  2. Four base pairs
  3. Seven base pairs
  4. Three base pairs
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(A) Six base pairs [1 mark]

Q23.2024 Main 57/1/1 · Q151 mark

Assertion (A): Agrobacterium tumefaciens is a pathogen of several monocot plants.

Reason (R): It is able to deliver a piece of DNA known as T-DNA to transform normal plant cells into a tumor.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(D) Assertion (A) is false, but Reason (R) is true [1 mark]

Q24.2024 Main 57/2/1 · Q111 mark

Which one of the following is not a feature of plasmids ?

  1. Circular
  2. Self-replicating
  3. Single stranded
  4. Extra-chromosomal
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(C) Single stranded [1 mark]

Q25.2024 Main 57/2/1 · Q151 mark

Assertion (A): Specific enzymes are used to degrade the cell wall in organisms to isolate the DNA from the cell.

Reason (R): Fungal cell wall is degraded by the enzyme cellulase.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(C) Assertion (A) is true, but Reason (R) is false [1 mark]

Q26.2024 Main 57/3/1 · Q101 mark

Which native plasmid did Stanley Cohen and Herbert Boyer use for the construction of the first recombinant DNA ?

  1. Salmonella typhimurium
  2. Streptococcus pneumoniae
  3. Escherichia coli
  4. Haemophilus influenzae
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(A) Salmonella typhimurium [1 mark]

Q27.2024 Main 57/3/1 · Q111 mark

Which one of the following represents the correct annealing of primers to the DNA to be amplified in the PCR ?

Options (A)–(D) are shown in the figure.

Figure

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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(B) [1 mark]

Figure

Q28.2024 Main 57/3/1 · Q161 mark

Assertion (A): The stirrer facilitates the even mixing of oxygen availability in a bioreactor.

Reason (R): Stirred-tank bioreactors generally have a flat base.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(C) Assertion (A) is true, but Reason (R) is false. [1 mark]

Q29.2024 Main 57/4/1 · Q111 mark

The ‘molecular scissors’ fall in the category of :

  1. Cleaving enzyme
  2. Endonuclease
  3. Exonuclease
  4. Restriction enzymes
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(D) Restriction enzymes [1 mark]

Q30.2024 Main 57/4/1 · Q141 mark

Assertion (A): Plasmids are autonomously replicating circular extra-chromosomal DNA.

Reason (R): Plasmids are usually present in Eukaryotic cells.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(C) Assertion (A) is true, but Reason (R) is false. [1 mark]

Q31.2024 Main 57/5/1 · Q61 mark

Which one of the following statements is correct in the context of observing DNA separation by agarose gel electrophoresis ?

  1. DNA can be seen in visible light.
  2. DNA can be seen without staining in visible light.
  3. Ethidium bromide stained DNA can be seen in visible light
  4. Ethidium bromide stained DNA can be seen under UV light.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(D) Ethidium bromide stained DNA can be seen under UV light [1 mark]

Q32.2024 Supp. 57/S/1 · Q101 mark

The sequence that controls the copy number of linked DNA in the vector is termed :

  1. Selectable marker
  2. Ori site
  3. Palindromic sequence
  4. Recognition site
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(B) Ori site [1 mark]

Q33.2023 Main 57/1/1 · Q51 mark

Given below are two columns. In Column I is the list of four enzymes and in Column II is the list of functions of the given enzymes. Which one of the following options shows the enzymes matched with their respective functions correctly ?

Column I (Enzyme)Column II (Function)
P.DNA Ligasei.Removes nucleotides from ends of DNA
Q.Restriction exonucleaseii.Extends primer on a DNA template
R.Taq polymeraseiii.Joins the DNA fragments
S.Restriction endonucleaseiv.Cuts DNA at a specific position
  1. P-i, Q-ii, R-iv, S-iii
  2. P-iv, Q-iii, R-ii, S-i
  3. P-i, Q-iv, R-iii, S-ii
  4. P-iii, Q-i, R-ii, S-iv
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(D) P - (iii), Q - (i), R - (ii), S - (iv) [1 mark]

Q34.2023 Main 57/1/1 · Q111 mark

The organism used in construction of the first artificial recombinant DNA by Cohen and Boyer in 1972 was :

  1. E. coli
  2. Salmonella typhimurium
  3. Agrobacterium tumefaciens
  4. Bacillus thuringiensis
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(B) Salmonella typhimurium. [1 mark]

Q35.2023 Main 57/3/1 · Q111 mark

The given schematic illustration shows three steps ‘P’, ‘Q’ and ‘R’ of the polymerase chain reaction.

Figure

Which of the following statements are correct with reference to the illustration given above ?

  1. Step ‘P’ is showing denaturation at low temperature.
  2. Step ‘Q’ is a denaturation of DNA strand at high temperature, followed by annealing.
  3. Step ‘R’ is an extension of DNA in presence of thermostable DNA polymerase.
  4. Step ‘Q’ is extension with two sets of primers.
  1. (i) and (iii) only
  2. (ii) and (iii) only
  3. (ii) only
  4. (i) only
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(B) (ii) and (iii) only [1 mark]

Q36.2023 Main 57/5/1 · Q81 mark

Given below is the restriction site of a restriction endonuclease Pst-I and the cleavage sites on a DNA molecule.

Figure

Choose the option that gives the correct resultant fragments by the action of the enzyme Pst-I.

Options (A)–(D) are shown in the figure.

Figure

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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(D) 5′ C-T-G-C-A    G 3′ / 3′ G    A-C-G-T-C 5′ (the option shown as (d) in the figure) [1 mark]

Q37.2023 Main 57/5/1 · Q141 mark

Assertion (A): Synthetic oligonucleotide polymers are used during Annealing in a PCR.

Reason (R): The primers bind to the double stranded DNA at their complementary regions.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(C) (A) is true, but (R) is false. [1 mark]

Q38.2023 Supp. 57/C/1 · Q81 mark

The vector for ‘T-DNA’ is :

  1. Escherichia coli
  2. Agrobacterium tumefaciens
  3. Thermus aquaticus
  4. Bacillus thuringiensis
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(B) Agrobacterium tumefaciens [1 mark]

Q39.2023 Supp. 57/C/1 · Q91 mark

Plasmids are suitable vectors for gene cloning because :

  1. They are small circular DNA molecules which can integrate with the host chromosomal DNA.
  2. They are small circular DNA molecules with their own origin of replication site.
  3. They can shuttle between prokaryotic and eukaryotic cells.
  4. They carry antibiotic resistance genes.
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(B) They are small circular DNA molecules with their own origin of replication site. [1 mark]

Q40.SQP 2025-26 · Q111 mark

In order to isolate genetic material of a bacterium, the cell must be treated with

  1. Lysozyme, ribonuclease, protease, chilled ethanol
  2. Cellulase, ribonuclease, protease, chilled ethanol
  3. Chitinase, ribonuclease, chilled ethanol, water
  4. Ribonuclease, protease, chilled ethanol, water
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Official CBSE Sample Paper Marking Scheme · 2025-26

(A) Lysozyme, ribonuclease, protease, chilled ethanol [1 mark]

Q41.SQP 2026-27 · Q111 mark

A biotechnology company grew Streptomyces in a fermenter to produce an antibiotic. It was observed that Day 7 onwards, the actual yield drops sharply, even though nutrient level, temperature, and aeration remain unchanged. Also, Microscopy revealed new colonies of bacteria which were not present in the original culture. So the fermenter was stopped. What measures should be taken to prevent the fall in yield in future fermentations?

  1. Increasing temperature only at the end of fermentation to kill contaminants.
  2. Adding antibiotics to the culture so unwanted microbes cannot grow.
  3. Ensuring sterile handling, sterilised equipment, and aseptic transfer before and during fermentation.
  4. Using a genetically modified strain designed to grow faster than contaminants.
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Official CBSE Sample Paper Marking Scheme · 2026-27

(C) Ensuring sterile handling, sterilised equipment, and aseptic transfer before and during fermentation. [1 mark]

Q42.2026 Main 57/1/1 · Q192 marks

Study the sequence of bases in a DNA molecule that is shown below and answer the questions.

5′ – GAATTC – 3′
3′ – CTTAAG – 5′

(a) What are such sequences called ? Name the enzyme used that recognises such nucleotide sequences.

(b) How do these enzymes function ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) • Palindromic sequences [½]

• Restriction endonuclease / EcoRI. [½]

(b) Restriction endonucleases cut the strands of DNA a little away from the centre of the palindrome sites to form sticky end on each strand / Restriction endonucleases cut at specific recognition site between two same bases in opposite strands to produce sticky ends / EcoRI cuts the DNA between bases G and A in both the strand in the sequence 5′-GAATTC-3′ / 3′-CTTAAG-5′ [1]

[Total: 2 marks]

Q43.2026 Main 57/2/1 · Q182 marks

How do scientists use microinjection and gene gun in biotechnology ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1
  • Microinjection: Recombinant DNA is directly injected into the nucleus of an animal cell. [1]
  • Gene gun: Plant cells are bombarded with high velocity micro particles of gold or tungsten coated with DNA. [1]

[Total: 2 marks]

Q44.2026 Main 57/3/1 · Q202 marks

(a) (i) Given below is the schematic representation of the process of electrophoresis. Identify the alphabets representing the :

  1. Anode end, and
  2. Lightest/Smallest DNA in matrix.

Figure

(ii) What is Agarose gel and why is it used in this process ? [1+1 = 2]

OR

(b) After performing gel electrophoresis, a student observes that DNA fragments are not visible.

  1. Suggest a reason for this and explain how fragments can be made visible.
  2. Also explain how a specific DNA fragment can be collected from the gel for further use.

[1+1 = 2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) (i) (I) Anode end is at S [½]

(II) Lightest/Smallest DNA is shown as R [½]

(ii) Agarose gel is made from a natural polymer extracted from sea weeds. [½]

Agarose gel provides sieving effect due to which DNA fragments separate according to their size. [½]

OR

(b) (i) DNA fragments might not be stained or exposed to UV rays. [½]

They can be seen only after staining with ethidium bromide followed by exposure to UV radiation when DNA bands appear orange in colour. [½]

(ii) The desired DNA fragment is cut from the gel and extracted by elution. [1]

[Total: 2 marks]

Q45.2026 Main 57/4/1 · Q202 marks

Write the role of ‘ori’ and ‘restriction’ site in a cloning vector pBR322.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1
  • ori – Definite regions in DNA where replication originates / It also decides the copy number of the linked DNA (alien DNA). [1]
  • Restriction Site – Site or sequence on vector DNA where the alien DNA is linked / If the alien DNA is ligated at a restriction site present in the antibiotic resistance gene the plasmid will lose the antibiotic resistance due to insertional inactivation hence antibiotic genes act as selectable marker. [1]

[Total: 2 marks]

Q46.2026 Main 57/5/1 · Q202 marks

(a) Suggest how a virus-free healthy plant can be obtained from a diseased sugarcane plant.

OR

(b) In order to force bacteria to take up the recombinant DNA, they must be made competent. Explain how it can be achieved.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a) The meristem is removed from a diseased plant, and grown in-vitro / by tissue culture method plant produced is virus free. [1+1]

OR

(b) [½ × 4]

  • Treat the bacterial cells with a specific concentration of a divalent cation such as Ca++.
  • Incubating bacteria (host) with recombinant DNA on ice.
  • Place them at 42°C / give heat shock at 42°C
  • Put them back on ice this enables the bacteria to take up the recombinant DNA.

[Total: 2 marks]

Q47.2025 Main 57/2/1 · Q202 marks

Observe the given sequence of nitrogenous bases on a DNA fragment and answer the following questions :

5′ – C A G A A T T C T T A – 3′
3′ – G T C T T A A G A A T – 5′

(a) Name the restriction enzyme which can recognise the DNA sequence.

(b) Write the sequence after restriction enzyme cut the palindrome.

(c) Why are the ends generated after digestion called as ‘Sticky Ends’ ?

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Answer not available: no official marking scheme published.

Q48.2025 Main 57/4/1 · Q202 marks

The basic scheme of the essential steps involved in the process of recombinant DNA technology is summarized below in the form of a flow diagram. Study the given flow diagram and answer the questions that follow.

Step-1Vector DNA (cut using Restriction Enzyme EcoR I) + Alien DNA (cut using Restriction Enzyme EcoR I)
↓ Enzyme A
Step-2Recombinant DNA molecule
↓
Step-3Transfer of recombinant DNA molecule in E. coli (Host)
↓
Step-4Replication of recombinant DNA molecule in E. coli
  1. Name the enzyme used in Step-1 to join the cut plasmid and alien DNA.
  2. State the technical term used for Step-3.
  3. Justify the use of same Restriction Enzyme EcoR I to cut both the vector DNA and the alien DNA.
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Answer not available: no official marking scheme published.

Q49.2025 Main 57/5/1 · Q202 marks

The basic scheme of the essential steps involved in the process of recombinant DNA technology is summarised below in the form of a flow diagram. Study the given flow diagram and answer the questions that follow :

Step-1Plasmid DNA (cut using Restriction Enzyme EcoR I) + Foreign DNA (cut using Restriction Enzyme EcoR I)
↓ DNA ligase
Step-2Recombinant DNA molecule
↓
Step-3Transfer of recombinant DNA molecule to the host cell
↓
Step-4Replication of recombinant DNA molecule in the host cell
  1. Name the specific enzyme that might have been used to make the multiple copies of foreign DNA before undergoing Step-1 of the process.
  2. How does the use of restriction enzyme EcoR I in Step-1 facilitate the action of DNA ligase to form the recombinant DNA molecule ? Explain.
  3. Name the most commonly used host in the above process.
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Answer not available: no official marking scheme published.

Q50.2025 Main 57/6/1 · Q182 marks

The basic scheme of the essential steps involved in the process of recombinant DNA technology is summarised below in the form of a flow diagram. Study the given flow diagram and answer the questions that follow.

Step 1Vector DNA (Plasmid) (cut using Restriction Enzyme) + Alien DNA (cut using Restriction Enzyme)
↓
Step 2Recombinant DNA molecule
↓
Step 3Transfer of recombinant DNA molecule in E. coli (Host)
↓
Step 4Replication of the recombinant DNA molecule in E. coli to form multiple copies of the alien gene
  1. What is the technical term used for Step 4 in the above process ?
  2. Which of the given two combinations of restriction enzyme should be used in Step 1 ? Justify your answer.
    1. EcoR I to cut the plasmid and Hind III to cut the alien DNA.
    2. EcoR I to cut both the plasmid and alien DNA.
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Answer not available: no official marking scheme published.

Q51.2025 Supp. 57/S/1 · Q202 marks

Given below is the diagram representing the observation made for separating DNA fragments by gel electrophoresis. Observe the illustration and answer the questions that follow :

Figure

(a) DNA fragments move in the direction A→B. Explain.

(b) Name the matrix used and its role in gel electrophoresis.

(c) Mention how the separated fragments can be visualized for further technical use.

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Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q52.2024 Main 57/2/1 · Q202 marks

Write the role of ‘ori’ and restriction site in the cloning vector pBR322.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1
  • Ori – site where replication starts / Responsible for controlling the copy number of the linked DNA. [1]
  • Restriction site – Site of ligation of alien DNA or foreign DNA or desirable DNA to the vector. [1]

[Total: 2 marks]

Q53.2024 Main 57/3/1 · Q202 marks

With reference to the set-ups (A, B and C) given below, of the electrophoretic separation of a mixture of DNA fragments of varied lengths, answer the questions that follow :

Figure

  1. In which one of the two Set-ups, A or B, would you see the DNA fragments separated and why ? Justify your answer.
  2. In Set-up C, which one of the two, I / II, are the bands of longer fragments of DNA ? Justify your answer.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) Set-up A [½]

  • DNA fragments being negatively charged move towards the anode on applying the electric field [½]

(b) I [½]

  • Smaller fragment will move faster as compared to longer fragments of DNA / longer fragments of DNA will move slower as compared to smaller fragments of DNA. [½]

[Total: 2 marks]

Q54.2024 Main 57/5/1 · Q182 marks

Figure

  1. Name the restriction enzyme that recognises the given specific sequence of bases. What are such sequence of bases referred to as ?
  2. What are the arrows in the given figure indicating ? Write the result obtained thereafter.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a) Enzyme – EcoRI; palindrome / palindromic nucleotide sequences. [½ + ½]

(b) Indicates the site at which EcoRI makes a cut in the two strands of DNA / restriction sites or recognition sites of EcoRI [½ + ½]; thereafter gives rise to “sticky ends.” [½ + ½]

Figure

[Total: 2 marks]

Q55.2024 Supp. 57/S/1 · Q202 marks

Why is making host cells ‘competent’ essential for rDNA technology ? Mention any two ways by which this can be achieved.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

To take up the DNA, as the DNA molecule is hydrophilic (it cannot pass through the cell membrane). [½ + ½]

By treating host cells with a specific concentration of a divalent cation (like Ca2+ ion) and then incubating the cells with rDNA on ice, followed by placing them briefly at 42 °C (heat shock) and then putting them back on ice. [1]

[Total: 2 marks]

Q56.2023 Main 57/1/1 · Q192 marks
    1. Identify and name the structures ‘A’ and ‘B’ marked in the image given below :

    Figure

    1. State their importance in various biotechnology experiments.

OR

  1. Explain the process by which a bacterial cell can be made ‘competent’ to take up foreign DNA from its surroundings, using divalent cations and temperature treatment.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) (i) ‘A’: Circular DNA/Plasmid [½]

‘B’: Bacteriophage [½]

(ii) (Plasmid) – Can carry foreign gene into the host cell/acts as cloning vector/has selectable marker/independent of the control of chromosomal DNA/high copy number. [½]

(Bacteriophage) – Cloning vector have the ability to replicate in bacterial cells/independent of the control of chromosomal DNA/high copy number per cell. [½]

OR

(b) Treating bacteria with specific concentration of calcium (ions) which increases the efficiency with which DNA enters the bacteria through pores in its cell wall, recombinant DNA can then be forced into such cells by incubating the cells with recombinant DNA on ice, followed by placing them briefly at 42°C (heat shock), then putting them back on ice. [½×4]

[Total: 2 marks]

Q57.2023 Main 57/3/1 · Q212 marks
  1. ‘Insertional inactivation’ is a method to detect recombinant DNA. Explain the method.

OR

  1. Explain how recombinant DNA technology is used to detect a disease even before any clinical symptom appears.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) Recombinant DNA/Desired DNA is inserted into the coding sequence of an enzyme β-galactosidase, this results into inactivation of the gene for the synthesis of this enzymes, presence of chromogenic substrate gives blue coloured colonies if the plasmid in bacteria does not have the insert (Non recombinants), but presence of insert (Recombinants) leads to the growth of bacterial colonies with no colour. [½×4]

OR

(b) A single stranded DNA or RNA tagged with a radioactive molecule (probe), is allowed to hybridize with its complementary DNA in a clone of cells, followed by detection using autoradiography, the clone having the mutated gene will hence not appear on the photographic film. [½×4]

//

Polymerase Chain Reaction/PCR is used to detect a disease even before any clinical symptoms appears, involves denaturation, annealing, to amplify DNA of the pathogen using pathogen specific primers [½×4]

[Total: 2 marks]

Q58.2023 Main 57/5/1 · Q212 marks
  1. Write the scientific name of the source organism of the themostable DNA polymerase used in PCR.
  2. State the advantage of using Thermostable DNA polymerase.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) Thermus aquaticus [1]

(b) Thermostable DNA polymerase remains active during the high temperature induced denaturation of double stranded DNA/It is not required to be added every time after denaturation in every cycle. [1]

[Total: 2 marks]

Q59.SQP 2025-26 · Q202 marks

Attempt either option A or B.

A. A cheese maker claims to be a biotechnologist. How will you support the same?

OR

B. Why does the 'insertional inactivation' method to detect recombinant DNA is preferred to 'antibiotic resistance' procedure?

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Official CBSE Sample Paper Marking Scheme · 2025-26

A. Anyone who can use/modify any living organism or their products using technology is considered to be a biotechnologist. [1]

Thus, cheese maker can be a biotechnologist as he uses microbes like bacteria, fungi to make cheese for commercial purpose [1]

OR

B. Selection of recombinants due to inactivation of antibiotics is a cumbersome procedure because it requires simultaneous plating on two plates having different antibiotics. [1]

In insertional inactivation method, insertion of recombinant DNA within the coding sequence of the enzyme beta galactosidase results into inactivation of this enzyme and the colonies do not produce any colour. The presence of a chromogenic substrate gives blue coloured colonies in absence of an insert. [1]

Q60.2026 Main 57/1/1 · Q283 marks

Observe the given picture carefully. A mixture of DNA with fragments ranging from 100 base pairs to 1800 base pairs were separated by electrophoresis on agarose gel with the following arrangement :

Figure

(a) What result will be obtained in staining with ethidium bromide ? Explain with reasons.

(b) The above setup was modified as shown below and a band with 100 base pairs was obtained at X.

Figure

What changes were made to the previous design to get a band at ‘X’. Why did the band appear at ‘X’ ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) • No bands will be seen in agarose gel. [½]

• DNA fragments being negatively charged will not move towards negative end or cathode / DNA being negatively charged will remain positioned at anode or positive end. [½]

(b) • Position of positive terminal or anode and negative terminal or cathode has been interchanged. [1]

• Band at X being the smallest fragment of DNA move farthest towards the anode. [1]

[Total: 3 marks]

Q61.2025 Main 57/1/1 · Q223 marks

(a) Write the palindromic nucleotide sequence for following sequence of DNA segment :

5′ – GAATTC – 3′

(b) Name the restriction endonuclease that recognizes this sequence.

(c) How are sticky ends produced ? Mention their role.

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Answer not available: no official marking scheme published.

Q62.2025 Main 57/1/1 · Q273 marks

Explain the basis on which gel electrophoresis technique works. Write any two ways the products obtained through this technique can be utilised.

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Answer not available: no official marking scheme published.

Q63.2025 Supp. 57/S/1 · Q253 marks

A vector is genetically engineered with three features which facilitate gene-cloning within the host cell. List the three features and explain each one of them.

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

Features of cloning vectors –

• Origin of Replication (Ori): [½] This is the sequence of DNA from where replication starts / any piece of foreign DNA linked to it is made to replicate within host cell / it also decides the copy number of the linked DNA. [½]

• Selectable marker: [½] helps in identifying and eliminating non-transformants and selectively permitting the growth of the transformants / help in selection of recombinants [½]

• Cloning sites: [½] The vector should have a few, preferably single, recognition site(s) to link the foreign DNA / the ligation of alien DNA is carried out at a restriction site present in one of the two antibiotic resistance genes [½]

(Any other feature with correct explanation)

[Total: 3 marks]

Q64.2024 Main 57/4/1 · Q273 marks

Study the steps shown below, that are carried during a specific technique :

Figure

  1. Identify the steps ‘A’ and ‘D’ in the diagram.
  2. What does ‘B’ represent ?
  3. Write what is ‘C’ ? Name its source organism.
  4. Mention the use of this technique in molecular diagnostics.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) A – Denaturation, D – Extension [½ + ½]

(b) B – Primers [½]

(c) C – Taq polymerase; Source – Thermus aquaticus [½ + ½]

(d) Use – Detection of AIDS / Cancer / Genetic disorder / Amplification of mutated gene / Early diagnosis of disease (Any one) [½]

[Total: 3 marks]

Q65.2023 Main 57/1/1 · Q223 marks

With the help of a schematic diagram only, show in three steps, the formation of recombinant DNA by the action of restriction endonuclease – EcoRI and DNA ligase.

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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

Figure

[½×6]

[Total: 3 marks]

Q66.2023 Supp. 57/C/1 · Q283 marks

A restriction enzyme digests a certain DNA into fragments. The fragments are subjected to a technique, the result obtained is in the illustration given below. Observe and answer the questions that follow.

Figure

(a) Name the technique and its purpose.

(b) Explain the procedure carried out under the following steps :

  1. Matrix used and its role
  2. Staining and extraction of the DNA
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a) Gel electrophoresis, to separate DNA fragments. [½ + ½]

(b)

(i) Agarose gel, to separate DNA fragments according to their size through sieving effect. [½ + ½]

(ii) Stained with Ethidium bromide followed by exposure to UV light which can be seen as orange coloured DNA band; separated DNA bands are cut out from the gel and extracted from gel piece. [½ + ½]

[Total: 3 marks]

Q67.2022 Main 57/1/1 · Q93 marks

Name two naturally occurring sources, one that transfers pathogenic genes into a plant cells and the other into an animal cell respectively, for their benefit. Write how have these naturally occurring sources been used for the benefit of human race by the biotechnologists.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

Agrobacterium tumefaciens – (plant pathogen) [½]; the disarmed Ti plasmid is used as a vector to introduce the gene of interest in the variety of plants. [1]

Retrovirus – (animals) [½]; disarmed retroviruses are used to transfer genes of interest into mammalian host cells. [1]

[Total: 3 marks]

Q68.2022 Main 57/1/1 · Q123 marks
  1. State the role of a selectable marker in r-DNA technology.
  2. Name one such selectable marker which is considered to be useful for E. coli.
  3. Give one reason why is it considered to be a useful marker.

OR

What are plasmids? How are they different from cloning vectors? Give one example each for a viral and a bacterial cloning vector.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1
  1. Helps in identifying non-transformants from transformants / recombinants from non recombinants [1]
  2. Genes encoding resistance to antibiotics such as ampicillin / tetracycline / kanamycin / chloramphenicol / ampR / tetR [1]
  3. (The normal E. coli cells do not carry resistance against any of these antibiotics.) It helps to identify and select transformants / identification of recombinants. [1]

OR

  • Plasmids are extra chromosomal self-replicating (double stranded) circular DNA molecules (generally found in bacterial cell) [1]
  • Plasmid is circular extra chromosomal DNA of bacterial cells whereas cloning vector is a vehicle that carries foreign DNA into another cell. [1]
  • Bacteriophage, pBR322 [½ + ½]

[Total: 3 marks]

Q69.2022 Main 57/2/1 · Q93 marks

With the advent of sophisticated techniques of genetic engineering, we can now readily purify and isolate DNA. Name and explain the different steps involved in the separation and isolation of DNA fragments once cut by restriction endonucleases.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

Gel electrophoresis [½]

Negatively charged DNA fragments (produced by restriction endonuclease) move towards anode through agarose gel, smaller fragments move further, separated fragments are stained with ethidium bromide, followed by exposure to UV radiation, extraction of DNA bands by elution. [½ × 5]

[Total: 3 marks]

Q70.2022 Main 57/3/1 · Q123 marks

(a) Given below is the stepwise schematic representation of the process of electrophoresis. Identify the 'alphabets' representing (i) Anode end (ii) smallest/lightest DNA strand in the matrix (iii) Agarose gel

Figure

(b) What is elution? State the importance of elution in this process.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

(a) (i) S; (ii) R; (iii) T [½ × 3]

(b) Process of cutting of separated bands of DNA, and extracting from the agarose gel. [½ + ½] Purified DNA is used in rDNA technology / genetic experiments. [½]

[Total: 3 marks]

Q71.2022 Main 57/4/1 · Q123 marks

Causative agents of HIV-AIDS and COVID-19 belong to the same group of viruses. To diagnose and amplify the genetic material for further study of COVID-19 virus, ‘RT-PCR’ test is carried out.

(a) What does ‘RT-PCR’ stand for?

(b) Explain the various steps of PCR technique.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

(a) Real Time-Polymerase Chain Reaction / Reverse Transcriptase-Polymerase Chain Reaction [1]

Note: One mark is to be awarded for attempting this question

(b) ● Denaturation by heating / DNA strands are separated by heating, Annealing of two primers to complementary region of DNA / Joining of primer to complementary region of DNA, Extension of primers, using thermostable DNA polymerase or Taq polymerase [½×4]

(The process is repeated many times / amplification)

//

(b) (Figure in the marking scheme, ½ mark each for denaturation, annealing, extension and repetition/amplification)

Figure

[Total: 3 marks]

Q72.2022 Main 57/5/1 · Q93 marks

A cell free method of amplifying DNA first developed in the mid 1980’s revolutionized the field of biotechnology. Name the method and explain the basic steps of the technique involved.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

● Polymerase Chain Reaction / PCR [1]

● Denaturation by heating / DNA strands are separated by heating. [½]

Annealing of two primers to complementary region of DNA / Joining of primer to complementary region of DNA [½]

Extension of primers, using thermostable DNA polymerase or Taq polymerase. [½+½]

(The process is repeated many times / amplification)

//

● Polymerase Chain Reaction / PCR [1]

Figure

(½ mark each for denaturation, annealing, extension and repetition as shown in the figure)

[Total: 3 marks]

Q73.2022 Supp. 57/6/1 · Q93 marks

(a) Simple stirred-tank bioreactors are used to produce large quantities of recombinant proteins, stirring the contents and mixing it with oxygen. Write any four other advantages of using stirred tank.

(b) After downstream processing, the product of the biosynthetic stage cannot be marketed directly. Why? Give two reasons.

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

(a) Advantages: large volumes (100 – 1000 litres) of culture can be processed, foam control system can control foam formation, temperature can be controlled by temperature control system, pH control system, optimum growth conditions can be maintained, substrate / salt / vitamins can be periodically added, small volumes of cultures can be withdrawn periodically through sampling ports [½×4] (any four)

(b) Product has to be formulated with suitable preservatives, clinical trials are required in case of drugs, strict quality control testing of each product is required [½×2] (any two)

[Total: 3 marks]

Q74.SQP 2026-27 · Q273 marks

Pathogens are often considered harmful, yet they have been adapted as powerful tools in genetic engineering. Explain how pathogens have been modified into useful vectors. Describe the role of a similarly modified plant vector and animal vector.

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Official CBSE Sample Paper Marking Scheme · 2026-27

I. Certain pathogens naturally have the ability to transfer their DNA into host cells. Scientists have modified these pathogens by removing their disease-causing genes, while retaining their ability to deliver DNA, making them useful cloning vectors. [1]

II. In animals, retroviruses have been engineered to remove cancer-causing genes, while keeping their capacity to integrate DNA into the host genome, making them effective vectors for gene delivery. [1]

III. In transgenic plants, nematode-specific genes were introduced via Ti plasmid of Agrobacterium tumefaciens to produce dsRNA, which triggered RNA interference (RNAi) and silenced nematode mRNA, protecting the plant from the parasite. [1]

Q75.2026 Main 57/2/1 · Q294 marks

Gene manipulation is a fast emerging science. It started with the development of recombinant DNA molecules. This technology which mostly involves cutting and pasting of desired DNA fragments is based on two most important discoveries in bacteria – presence of plasmid and restriction endonucleases. The science of r-DNA technology took birth when Cohen and Boyer (1973) were able to produce a piece of gene containing foreign DNA introduced into plasmid of E.coli.

(a) Name one core technique of biotechnology that alters the genetic material. [1]

(b) What is a plasmid ? Mention its importance in biotechnology. [2]

(c) Which gene was isolated by Cohen and Boyer and from which bacteria ? [1]

OR

(c) How is the action of exonuclease different from that of endonuclease ? [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) Genetic Engineering / Recombinant DNA Technology [1]

(b) Plasmid is an autonomously replicating circular extra chromosomal DNA. [1]

It is used as a vector to transfer foreign genes into host cells. [1]

(c) Antibiotic resistant gene, Salmonella typhimurium [½+½]

OR

(c) Exonuclease: Remove the nucleotides from the ends of DNA strand. [½]

Endonuclease: Make cuts at specific positions within DNA strand. [½]

[Total: 4 marks]

Q76.2026 Main 57/5/1 · Q304 marks

Read the following passage and answer the questions that follow :

We know that plasmids and bacteriophages are the most commonly used vectors in biotechnology experiments. If we can link an alien piece of DNA to the plasmid DNA, the alien DNA can be multiplied equal to the copy number of the plasmid. Engineered vectors are used these days. Study the diagram of the E. coli cloning vector pBR322 and answer the questions that follow :

Figure

(a) Why are plasmids and bacteriophages used as cloning vectors ? [1]

(b) (i) Identify : [½ + ½ = 1]

  1. The gene in the cloning vector that controls the copy number of the vector.
  2. The restriction site – C in the ‘rop’ gene.

OR

(b) (ii) Identify and name two selectable markers shown in the diagram. [1]

(c) Name the two restriction sites each in the two genes you have identified as selectable markers. [2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a) Plasmids and bacteriophages have the ability to replicate / They multiply within the bacterial cells independently of the chromosomal DNA / an alien piece of DNA can be ligated to them / They can be multiplied equal to their specific copy numbers. [1]

(b) (i) (I) ori [½]

(II) Pvu II [½]

OR

(b) (ii) G : ampR / ampicillin-resistance gene, H : tetR / tetracycline-resistance gene. [½+½]

(c) Pst I, Pvu I, BamH I, Sal I [½ × 4]

[Total: 4 marks]

Q77.2024 Main 57/1/1 · Q304 marks

Study the diagram given below that shows the steps involved in the procedure of selecting transformed bacteria and answer the questions that follow :

Figure

(a) Identify the colony that has got transformed. Justify your answer.

(b) What are the sites in a plasmid called where ampicillin and tetracycline resistance genes are inserted ? State their role in genetic engineering.

(c) Name two enzymes playing an important role in genetic engineering.

OR

(c) State the role of β-galactosidase in insertional inactivation.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a) Colony 4 is transformed with plasmid containing recombinant DNA, as they will not show resistance towards tetracycline. [½ + ½]

(b) Award 2 marks to each student. [2]

(c) Restriction endonuclease / ligase / Taq DNA Polymerase [1]

OR

(c) Insertional inactivation of gene encoding for β-galactosidase will lead to colourless bacterial colonies (recombinant). [1]

[Total: 4 marks]

Q78.2024 Main 57/3/1 · Q294 marks

Read the passage given below and answer the questions that follow.

In recombinant DNA technology, restriction enzymes are used as they recognize and cut DNA within a specific recognition sequence. BamH I is one such restriction enzyme which binds at the recognition sequence 5′ G-G-A-T-C-C 3′ and cleaves this sequence between G and G on each strand, whereas Alu I binds at the recognition sequence 5′ A-G-C-T 3′ and cleaves these sequences between G and C on each strand.

  1. If Alu I is used to cut the given DNA strand, how many DNA fragments would be formed ? Write the sequence of each fragment formed with its polarity.

5′  C – C – G – T – A – G – C – T – A – T – C – A – G – C – T – G – G  3′
     |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |
3′  G – G – C – A – T – C – G – A – T – A – G – T – C – G – A – C – C  5′

  1. Which one of the two restriction enzymes BamH I or Alu I will preferably be used on the same given DNA strand to make a recombinant DNA molecule and why ?
  2. After binding to the two strands of the double helix DNA, where specifically does the restriction enzyme act to cut the two strands of DNA ? Write the specific term used for the specific nucleotide sequences of DNA recognised by a restriction endonuclease.

OR

  1. Write the specific sequence of DNA segment recognised by the restriction endonuclease EcoRI.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a)

  • 3 / Three fragments [½]
  • Fragments with polarity: [½]

Figure

(b) Alu I; Alu I site is present in the given sequence and BamH I site is not given. [1]

(c)

  • Sugar phosphate backbone
  • Palindrome sequence / recognition site / restriction site

[2]

OR

(c) 5′ G-A-A-T-T-C 3′
     3′ C-T-T-A-A-G 5′

(Note: 1 mark for polarity and 1 mark for correct sequence) [2]

[Total: 4 marks]

Q79.2024 Supp. 57/S/1 · Q294 marks

Read the following passage and answer the questions that follow.

Restriction endonuclease was isolated for the first time by W. Arber in 1962, in bacteria. DNA, the genetic material can be manipulated by addition or substitution of the desired gene by using a specific restriction endonuclease resulting in rDNA. This is one of the major steps in biotechnology.

(a) Name the first isolated restriction endonuclease. Why are restriction endonucleases so called ?

(b) Write the palindrome recognised by EcoRI.

(c) (i) How does restriction endonuclease function ? Explain.

OR

(c) (ii) Write the convention for naming a restriction endonuclease with the help of an example.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) Hind II [½]

It restricts the growth of bacteriophage in E. coli. [½]

(b) 5′-GAATTC-3′
3′-CTTAAG-5′ [1]

(c) (i) It inspects the length of a DNA sequence and binds to the specific recognition palindromic sequence, and cuts each of the two strands at specific points in the sugar–phosphate backbones. [1 + 1]

OR

(c) (ii) In EcoRI (comes from Escherichia coli RY13): E represents genus Escherichia, co represents species coli, R represents RY13 strain, I represents the order in which the enzyme was isolated from that strain of bacteria. [½ × 4]

[Total: 4 marks]

Q80.2023 Supp. 57/C/1 · Q294 marks

An illustration given below was on the display board in the class. Study the illustration and answer the questions that follow.

Figure

(i) (a) Name the particular technique shown.

OR

(b) Name the enzymes needed to isolate the DNA from bacterial and fungal cells respectively.

(ii) Taking the help of the illustration, explain the steps used by the scientists at Eli Lilly Company to produce a drug that proved to be highly beneficial to diabetic patients.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(i)

(a) Recombinant DNA technology [1]

OR

(b) Bacterial – Lysozyme, Fungal – Chitinase [½ + ½]

(ii) Scientists at Eli Lilly company prepared two DNA sequences, corresponding to A and B peptide chains of human insulin, and introduced them in plasmids of E. coli to produce insulin chains. Chains A and B were produced separately, extracted, and combined by creating disulfide bonds to form human insulin. [½ × 6 = 3]

[Total: 4 marks]

Q81.2026 Main 57/3/1 · Q325 marks

(a) (i) Explain any two basic principles/core techniques on which biotechnology is based. [2]

(ii) Describe any three key tools used in Recombinant DNA technology. [3]

OR

(b) “Early and accurate diagnosis of diseases is vital in medical technology.”

(i) Name the conventional methods of diagnosis and their disadvantages. [2]

(ii) Which three diagnostic techniques have been developed through Biotechnology ? Explain how each one helps in detecting diseases. [3]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) (i)

  • Genetic Engineering [½]: Technique to alter the chemistry of genetic material (DNA and RNA) to introduce these into host organisms and thus change the phenotype of the host organism. [½]
  • Bioprocess Engineering [½]: Maintenance of sterile ambience in chemical engineering process to enable growth of only the desired microbe/eukaryotic cell in large quantities for the manufacture of biotechnological/biological products like antibiotics, vaccines, enzymes etc. [½]

(ii)

  • Restriction Enzymes (Molecular Scissors) [½]: Cut DNA at specific point on sugar phosphate backbone / They cut DNA at specific recognition sequences, called palindromes. They create sticky ends [½]
  • Cloning vectors [½]: These are DNA molecules used to carry foreign genes into a host cell. [½]
  • Competent host [½]: These are living organisms in which recombinant DNA is introduced [½]

OR

(b) (i) Urine analysis / stool analysis / sputum analysis / Serum Analysis [1]

Disadvantages: Early detection not possible / Detect the disease when it has spread / shows symptoms [1] (Any other method with its disadvantages)

(ii) Three molecular diagnostic techniques:

  • Polymerase Chain Reaction (PCR) [½]: Amplifies specific DNA sequences from very small samples. [½]
  • ELISA (Enzyme Linked Immuno Sorbent Assay) [½]: Based on antigen-antibody interaction, infection by pathogen can be detected by the presence of antigen / by detecting the antibodies synthesised against the pathogen. [½]
  • Probe (single stranded DNA or RNA tagged with radioactive molecule) [½]: it binds to its complementary DNA in a clone of cells followed by detection using autoradiography. Clone having mutated gene will not appear on photographic film as it does not have complementarity. [½]

[Total: 5 marks]

Q82.2026 Main 57/4/1 · Q335 marks

(a) (i) EcoR I has played a very important role in rDNA technology.

  1. Explain the convention for naming EcoR I. [2]
  2. Write the recognition site of this endonuclease restriction enzyme. [1]

(ii) Name the overhanging stretches of DNA produced by these restriction enzymes. Describe their role in the formation of rDNA. [2]

OR

(b) (i) Insertional inactivation is a method to detect the recombinant DNA. Explain this method. [3]

(ii) Name the commonly used vector for transferring genes in plants from bacteria. [1]

(iii) How is this able to transform a normal plant cell into a tumour ? [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) (i) (I) Naming of restriction enzymes (EcoRI):

  • First letter ‘E’ of the name comes from genus (Escherichia) [½]
  • The second two letters ‘co’ come from the species (coli) [½]
  • The letter ‘R’ derived from the name of the strain [½]
  • Roman number (I) indicate the order in which the enzymes were isolated from the strain of the bacteria. [½]

(II) The restriction enzyme EcoRI recognised the palindromic sequence

5′ – GAATTC – 3′

3′ – CTTAAG – 5′ [1]

(ii) These overhanging stretches are called ‘Sticky ends’ [1]

They form hydrogen bond with their complementary cut counterpart that facilitates the action of enzyme DNA ligase. [1]

OR

(b) (i)

  • When the alien DNA is inserted within the coding sequence of the enzyme β-galactosidase it results in inactivation of the enzyme which is referred as insertional inactivation. [1]
  • The presence of chromogenic substrate gives blue coloured colonies if the plasmid in the bacteria does not have insert or is non recombinant. [1]
  • Colonies which do not produce any colour are identified as recombinants as their gene for enzyme is inactivated due to an insert. [1]

(ii) Agrobacterium tumefaciens. [1]

(iii) When this Agrobacterium infects a plant cell it delivers ‘T’-DNA or Ti tumor-inducing plasmid into the plant cell and transforms it into a tumor cell. [1]

[Total: 5 marks]

Q83.2026 Supp. 57/7/1 · Q335 marks

Attempt either (a) or (b).

(a) (I) (i) State the role of selectable markers in rDNA technology.

(ii) Name two such selectable markers which are considered to be useful for E. coli cloning vector.

(iii) State two significances of ‘Ori’ in E. coli cloning vector.

(II) How have biotechnologists effectively used Agrobacterium tumefaciens in plants and retroviruses in animals ? Explain.

OR

(b) Name and describe the technique that helps in separation and isolation of DNA fragments after cutting of DNA by restriction endonucleases.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) (I) (i) A selectable marker helps in identifying and eliminating the non-transformants and selectively permitting the growth of transformants / a selectable marker helps in differentiating between recombinants and non-recombinants. [1]

(ii) Ampicillin resistance / (ampR) gene, Tetracycline resistance / (tetR) gene (any other correct example of gene) [½ + ½]

(iii) • ‘Ori’ is a sequence from where replication process starts. [½]

• It is also responsible for controlling the copy number of the linked DNA. [½]

(II) • Tumor inducing plasmid or Ti plasmid of Agrobacterium tumefaciens has been manipulated in such a way that it is no more pathogenic but still able to deliver genes of interest ligated to it into plant cell. [1]

• Retroviruses are disarmed and are used to deliver gene of interest into animal cells. [1]

OR

(b) • Gel electrophoresis [1]

• The DNA fragments (provided in the sample) being negatively charged can be separated by forcing them to move towards the anode under an electric field through a medium/matrix made up of agarose; the DNA fragments separate out according to their size, i.e. smaller the fragment size the farther it will move; separated DNA fragments are visualised as orange bands after staining with ethidium bromide followed by exposure to UV radiation; the separated bands of DNA are cut out and extracted from the gel-piece called elution. [1 × 4]

[Total: 5 marks]

Q84.2025 Main 57/2/1 · Q325 marks

Student to attempt either option (A) or (B).

(A) (i) Explain how is a bacterial cell made ‘competent’ to take up recombinant DNA from the medium.

(ii) Explain the steps of amplification of gene of interest using PCR technique.

OR

(B) (i) What are transgenic animals ?

(ii) Why are these animals being produced ? Explain any four reasons.

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Answer not available: no official marking scheme published.

Q85.2024 Main 57/2/1 · Q335 marks

(a) (i) Why should a cell be made competent to take up an alien DNA ? How can a bacterial cell be made competent using calcium ions ? Explain.

(ii) (1) State the importance of gel electrophoresis in biotechnology.

(2) Explain the principle on which this technique works.

(3) Mention why ethidium bromide is used in this technique.

OR

(b) Bt cotton, the genetically modified crop, has greatly helped the cotton farmers to increase their crop yield.

(i) How was Bt cotton plant made resistant to bollworm ? Explain.

(ii) Describe the mechanism that leads to the death of bollworms feeding on Bt cotton plants.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a) (i)

  • DNA is a hydrophilic molecule and cannot pass through the cell membrane. [1]
  • A bacterial cell is made competent by treating the bacterial cell with a specific concentration of a divalent cation such as calcium, which increases efficiency with which the DNA enters through pores in its cell wall / this creates certain transient pores in its cell and increases the efficiency of the cell to take up DNA. [½ + ½]

(ii)

(1) Separation of DNA fragments. [1]

(2) DNA fragments are negatively charged molecules, they can be separated according to their size by forcing them to move toward the anode under an electric field through agarose gel. [½ + ½]

(3) To stain the DNA to visualize by exposure to UV radiation. [1]

OR

(b) (i) Specific Bt toxin gene cryIAc / cryIIAb, isolated from Bacillus thuringiensis bacteria and incorporated into the cotton plant to provide resistance to bollworm. [1 + 1]

(ii) Bacillus thuringiensis forms toxic insecticidal protein or Bt toxin protein during a particular growth phase; Bt toxin protein exist as inactive pro-toxin; on ingestion by the bollworm inactive toxin is converted into active form due to alkaline pH of the gut; activated toxin binds to the surface of the mid-gut epithelial cells, create pores and causes cell swelling, lysis and death of the insect. [½ × 6]

[Total: 5 marks]

Q86.2024 Main 57/5/1 · Q315 marks
    1. Draw a schematic diagram of the cloning vector pBR 322 and label (1) Bam HI site (2) gene for ampicillin resistance (3) ‘ori’ (4) ‘rop’ gene.
    2. State the role of ‘rop’ gene.
    3. A cloning vector does not have a selectable marker. How will it affect the process of cloning ?
    4. Why is insertional inactivation preferred over the use of selectable markers in cloning vectors ?

OR

    1. Name the nematode (scientific name) that infects the roots of tobacco plant and reduces its yield.
    2. Name the vector that is used to introduce nematode-specific genes into the host plant (tobacco).
    3. How do sense and anti-sense RNAs function ?
    4. Why could parasite not survive in a transgenic tobacco plant ?
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a) (i)

Figure

[½ × 4]

(ii) ‘rop’ codes for proteins involved in the replication of the plasmid. [1]

(iii) It will not be possible to differentiate transformants from the non-transformants or to differentiate recombinants from non-recombinants. [1]

(iv) (1 mark to be awarded if attempted) [1]

OR

(b) (i) Meloidogyne incognita [1]

(ii) Agrobacterium tumefaciens / Ti plasmid [1]

(iii) Both sense and antisense RNA are complementary to each other, form a double stranded RNA (dsRNA). [1 + 1]

(iv) The double stranded RNA binds to a specific mRNA / initiates RNAi, prevents translation of mRNA / silencing of specific mRNA of parasite or nematode. [½ + ½]

[Total: 5 marks]

Q87.2023 Main 57/3/1 · Q315 marks
  1. Answer the following questions with respect to recombinant DNA technology :
    1. Why is plasmid considered to be an important tool in rDNA technology ? From where can plasmids be isolated ? (Any two sources)
    2. Explain the role of ‘ori’ and selectable marker in a cloning vector.
    3. “r-DNA technology cannot proceed without restriction endonuclease.” Justify.

OR

  1. Answer the following questions based on Bt-crops :
    1. Why do farmers prefer to grow Bt cotton crop than genetically unmodified cotton crops ?
    2. Name any two insects that are killed by Bt toxin.
    3. Explain the mechanism by which Bt toxin kills the insects but not the bacterium which possesses the toxin.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) (i) • Can act as vector/can self-replicate to form multiple copies/have selectable markers/small in size will facilitate insertion/presence of ‘Ori’ [1]

• E. coli, Agrobacterium tumefaciens, Salmonella typhi, Bacteria, any other correct example (any two) [½+½]

(ii) • ‘Ori’ – this is a sequence from where replication starts/any piece of DNA when linked to this sequence can be made to replicate within the host cells/controls copy number of linked DNA. [½×2]

• Selectable marker helps in identifying and eliminating non-transformants, and selectively permitting the growth of transformants during recombinant DNA technology. [½×2]

(iii) Restriction endonuclease identifies a specific palindromic sequence of DNA and cut the DNA at the specific sites in both the host as well in desired/foreign DNA, thereby creates “sticky ends” facilitating ligation to form a recombinant DNA.

OR

(b) (i) Bt cotton crop is pest resistant/insect resistant/with increase productivity [1]

(ii) Cotton bollworms/corn borer/tobacco budworm/army worm/coleopterans (beetles)/dipterans (flies, mosquitoes) (Any two) [½×2]

(iii) Bt toxin protein exists as an inactive protoxin in the bacterium, but once the insect ingests this toxin it is converted into an active form, due to the alkaline pH of the gut, which solubilizes inactive crystals of toxic insecticide; the activated toxin binds to the surface of midgut epithelial cells of the insect, creates pores, that cause cell swelling and lysis and eventually the death of the insect. [½×6]

[Total: 5 marks]

Q88.2023 Main 57/5/1 · Q335 marks
  1. Bioreactors are the containment vehicles of any biotechnology-based production process. For large scale production and for economic reasons the final success of biotechnological process depends on the efficiency of the bioreactor.

    Answer the following questions w.r.t. the given paragraph :

    1. List the operational guidelines that must be adhered to so as to achieve optimisation of the bioreactor system. Enlist any four.
    2. Mention the phase of the growth we refer to in the statement “Optimisation of growth and metabolic activity of the cells”.
    3. Is the biological product formed in the bioreactor suitable for the intended use immediate ? Give reason in support of your answer.

OR

  1. ‘EcoRI’ has played very significant role in r-DNA technology.
      1. Explain the convention for naming EcoRI.
      2. Write the recognition site and the cleavage sites of this restriction endonuclease.
    1. What are the protruding and hanging stretches of DNA produced by these restriction enzymes called ? Describe their role in formation of r-DNA.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) (i) Optimum growth conditions: Temperature, pH, Substrate, Salts, Vitamins, Oxygen (Any four) [½×4]

(ii) Log phase/Exponential phase [1]

(iii) • No [1]

• it needs separation and purification/down-streaming process/quality control testing/needs to be formulated with suitable preservatives/clinical trials. [1]

OR

(b) (i) (I) E – Genus; co – Species; R – Strain; I – Order of isolation of enzyme [½×4]

(II) Recognition site – 5′GAATTC 3′ / 3′CTTAAG 5′ [½]

Cleavage site – Between G and A from both sides [½]

//

5′GAATTC 3′ / 3′CTTAAG 5′ [1]

(Note: half mark to be awarded for correct recognition site and half mark for cleavage site)

(ii) • Sticky ends [1]

• Sticky ends form hydrogen bonds with their complimentary cut counterparts, this stickiness facilitates the action of the enzyme DNA ligase. [½×2]

[Total: 5 marks]

Q89.2022 Main 57/2/1 · Q135 marks

(a) Development of recombinant DNA technology has opened gates to many breakthroughs in the fields of medicine and agriculture. This has enabled scientists to isolate, sequence and manipulate individual genes obtained from diverse living or dead cells. Given below is a diagram showing the basic steps involved in genetically modifying an organism. Study the given diagram and answer the questions that follow:

Figure

  1. Are two different types of restriction endonucleases used, one to cut the vector DNA and another to cut the desired DNA to be cloned? Support your answer, giving reason.
  2. Which enzyme is used at step (X) to integrate the foreign DNA with the vector DNA?
  3. What is the term used for step (Y) showing multiple copies of the foreign DNA being formed in transformed E. coli?
  4. Draw a diagram of E. coli cloning vector pBR322 to show the following:
    1. Any one restriction endonuclease site in tetracycline resistance gene
    2. Any one restriction endonuclease site in ampicillin resistance gene
    3. 'ori' site
  5. What does "rop" code for in plasmid pBR322?

OR

(b) RNA interference (RNAi) holds great potential as a therapeutic agent for the treatment of human diseases and as a biocontrol agent for controlling pests in the agricultural fields. An experiment was conducted to study the use of 'RNAi' for the potential treatment of disorders of cholesterol metabolism. Some people possess genetic mutations with elevated levels of 'ApoB' gene which predisposes them to coronary artery diseases.

Lowering the amount of 'ApoB' can reduce the number of lipoproteins and lower the blood cholesterol.

Tracy Zimmerman and her colleagues used RNAi in 2006 to reduce the level of 'ApoB' in non-human primates Cynomolgus monkeys. (i) One group of monkeys were given RNAi treatment (Small interfering RNAs, SiRNAs) (doses 1 mg/kg SiRNAs), (ii) Second group of monkeys were given RNAi treatment (doses 2·5 mg/kg SiRNAs), and (iii) Third group of monkeys were injected saline as control.

The results of the experiments are illustrated in the graph given below:

Figure

  1. What do you interpret from the bars (X and Z) obtained after 264 hours of treatment of monkeys with saline and 2·5 mg/kg SiRNAs treatment?
  2. Name the category of organisms in the living world, where RNA interference (RNAi) takes place. Why?
  3. What is the basic principle involved in RNA interference (RNAi) in silencing the preferred genes?
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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

(a)

  1. No [½]; Cut with same restriction endonuclease to obtain complementary sticky ends / to obtain DNA fragments with same kind of sticky ends (which can be joined together end to end) [½]
  2. DNA ligase [1]
  3. Cloning / Host Cloning / Gene Cloning [1]
  4. Diagram of pBR322 (shown in the scheme) [½ + ½]

    Figure

    I. BamHI or SalI; II. PvuI or PstI; III. Ori (any two correct labellings)

  5. Codes for the proteins involved in replication of the plasmid [1]

OR

(b)

  1. X – cholesterol level increases [1]; Z – cholesterol level decreases [1]
  2. In Eukaryotes [1]; method of cellular defense [1]
  3. Silencing of specific mRNA due to binding of a complementary ds RNA [1]

[Total: 5 marks]

Q90.2022 Main 57/3/1 · Q135 marks

(a) Read the paragraph given below and answer the questions that follow:

Enzyme Taq polymerase, is extracted from a eubacterial microorganism Thermus aquaticus from Yellowstone National Park in Montana, USA and isolated by Chien et al. (1976). Taq polymerase successfully replaced the DNA polymerase from E. coli that was being used in PCR earlier and this shift revolutionised the PCR technique.

  1. Taq polymerase after its discovery replaced E. coli DNA polymerase in PCR technique. Explain giving reasons why was the need felt for the change?
  2. What is a primer and its importance in PCR?
  3. Write the importance of PCR as a diagnostic tool.

OR

(b) Read the following paragraph and answer the questions that follow:

Biotechnology revolves around the "gene of interest", with an objective to open various avenues for human welfare in health, medicine, pharma, agriculture etc. using different techniques, tools and processes. One of the breakthroughs of biotechnology in medicine is the gene therapy.

  1. Name the human disease for which the gene therapy was used for the first time.
  2. Explain the steps of gene therapy carried to cure the disease using the lymphocytes of the patient. Why is this therapy not a permanent cure of the disease?
  3. Write the possible permanent cure of the disease by the gene therapy that is in progress.
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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

(a)

  1. The E. coli DNA polymerase cannot carry out PCR at high temperature (as it becomes inactive) [1]; whereas Taq polymerase being thermostable remains active even at high temperatures. [1]
  2. Primers are small chemically synthesised oligonucleotides that are complementary to the regions of genomic DNA strand [1]; primers help in extension of complementary DNA strand [1]
  3. Early detection of diseases like cancer / AIDS / genetic disorder, by amplification of desired genes (when very low concentration of bacteria or virus before setting of the disease symptoms). [½ + ½]

OR

(b)

  1. Adenosine deaminase (ADA) deficiency. [1]
  2. Lymphocytes from the patient's blood are grown in a culture medium outside the body [½]; using retroviral vector, a functional ADA cDNA is introduced into these lymphocytes [½ + ½]; modified lymphocytes injected back to the patient body [½]. Lymphocytes are mortal / not immortal, the patient requires repeated / periodic infusion. [½ + ½]
  3. Gene is obtained from bone marrow cells producing ADA is introduced into cells at early embryonic stage. [1]

[Total: 5 marks]

Q91.2022 Main 57/5/1 · Q135 marks

(a) Cloning of genes, play a very significant role in genetic engineering, helping the transfer of desirable foreign genes into different hosts. The scientists, to make this process easier and effective are creating engineered vectors in such a way that they help easy linking of foreign DNA and selection of recombinants from non recombinants. ‘pBR322’ is one such engineered vectors developed by scientists. A diagram of an engineered vector pBR322 is given below:

Figure

  1. Name the host for this cloning vector.
  2. Identify ‘Rop’ and ‘Ori’ in the diagram from ‘U’, ‘V’, ‘W’, ‘X’, ‘Y’ and ‘Z’. Write their functions.
  3. Draw the fragments that will be formed by the action of ‘Z’ (marked in the diagram) on the specific site of the DNA segment given below:
    5′ --- GTACGAATTCCTGA---3′
    3′---CATGCTTAAGGACT---5′

OR

(b) RNA interference (RNAi) holds great potential as a therapeutic agent for the treatment of human diseases and as biocontrol agents in controlling pests in the field of agriculture. An experiment was carried to study the use of ‘RNAi’ for the potential treatment of disorders of cholesterol metabolism. Some people possess genetic mutations with elevated levels of ApoB gene which predisposes them to coronary artery diseases.

Lowering the amount of ApoB can reduce the number of lipoproteins and lower the blood cholesterol.

Tracy Zimmerman and her colleagues used RNAi in 2006 to reduce the level of ApoB in non human primates Cynomolgus monkeys.

One group of monkeys were given RNAi treatment (small interfering RNAs, SiRNAs) (doses 1 mg/kg SiRNAs), second group of monkeys were given RNAi treatment (doses 2.5 mg/kg SiRNAs) and third group of monkeys were injected with saline.

The results of the study are depicted in the graph below:

Figure

  1. How does the treatment with 2.5 mg/kg brings an effect on cholesterol metabolism when compared from 24 hours and 144 hours.
  2. Write any two natural sources from where dsRNA molecule could be obtained for silencing the specific mRNA.
  3. How is RNAi used in controlling the infection on the roots of tobacco plants by the nematode Meloidogyne incognitia.
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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

(a) (i) E. coli / Escherichia coli [1]

(ii) rop — ‘W’, — code for the proteins involved in the replication of the plasmid. [½+½]

ori — ‘U’, — this is a sequence from where replication starts / control the copy number of the linked DNA [½+½]

(iii) 5′---GTACG 3′    5′ AATTCCTGA---3′
3′---CATGCTTAA 5′    3′ GGACT---5′ [1+1]

OR

(b) (i) There is a considerable decrease in the level of serum cholesterol after 144 hrs. as compared to 24 hrs. [1]

(ii) ● Infection by RNA viruses / Retrovirus [½] ● mobile genetic elements / transposons / jumping genes [½]

(iii) Using Agrobacterium vectors, nematode-specific genes are introduced into the host plant, introduced DNA forms both sense and anti-sense RNA in the host cell, these two RNAs being complementary to each other form a dsRNA (double-stranded RNA), that initiates RNAi thus silencing the specific mRNA of the nematode, nematode is unable to survive in the transgenic plant [½×6]

[Total: 5 marks]

Q92.2022 Supp. 57/6/1 · Q135 marks

(a) Restriction endonucleases are a class of bacterial enzymes that recognize a specific short sequence of nucleotides within a double-stranded DNA molecule. The natural purpose of these enzymes is to protect bacteria from pathogens, notably bacteriophages. There are different classes of restriction enzymes, but type-II restriction enzymes are widely used in manipulating DNA as they recognize short sequence nucleotides that are typically palindromes.

  1. Name a specific restriction endonuclease and write the palindromic nucleotide sequence in the DNA recognised by this enzyme. Also, indicate the site at which it makes the cut.
  2. A piece of DNA is cut by a restriction enzyme. The fragments are then separated by gel electrophoresis and stained by ethidium bromide. Write the principle on which gel electrophoresis works.

OR

(b) RNA interference (RNAi) holds great potential as a therapeutic agent for the treatment of human diseases and as a bicontrol agent for controlling pests in the field of agriculture.

The graph given below illustrates the use of ‘RNAi’ for the potential treatment of disorders of cholesterol metabolism. Some people possess genetic mutations with elevated levels of ‘ApoB’ gene which predisposes them to coronary artery diseases.

Lowering the amount of ‘ApoB’ can reduce the number of lipoproteins and lower the blood cholesterol.

Tracy Zimmerman and her colleagues used RNAi in 2006 to reduce the level of ‘ApoB’ in non-human primates Cynomolgus monkeys. One group of monkeys were given RNAi treatment (Small interfering RNAs, SiRNAs) (doses 1 mg/kg SiRNAs), a second group of monkeys were given RNAi treatment (doses 2·5 mg/kg SiRNAs) and a third group of monkeys were injected with saline as control.

The results of the experiments are illustrated in the graph given below:

Figure

  1. Write your interpretation from the bars X and Z obtained after 264 hours of treatment of monkeys with saline and 2·5 mg/kg SiRNAs treatment respectively, in comparison to the bars obtained after 24 hours of treatment with saline and 2·5 mg/kg SiRNAs.
  2. A tobacco plant made transgenic using RNA interference is protected from the parasite Meloidogyne incognitia. How is the transgenic tobacco plant able to prevent itself from infestation by the nematode? Explain briefly.
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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

(a) (i) ● EcoRI [1]

● 5′ – GAATTC – 3′
3′ – CTTAAG – 5′ [1]

● EcoRI cuts the DNA between bases G and A from 5′ end of both DNA strands. /
5′ G↓AATTC 3′
3′ CTTAA↑G 5′ [1] (or any other correct example with relevant answer)

(ii) – DNA molecule being negatively charged moves towards the anode / positive electrode through a medium of agrose gel under an electric field. [1]

– DNA fragments separate according to their size / molecular weight (smaller the fragment size, the farther it moves) [1]

OR

(b) (i) ● When monkeys are treated with saline solution, serum cholesterol level increases from 24 hours to 264 hours. [1]

● When monkeys are treated with 2.5mg/kg SiRNAs, level of serum cholesterol decreases from 24 hours to 264 hours. [1]

(ii) using Agrobacterium vectors, nematode specific genes are introduced into the host plant, introduced DNA forms both sense and anti-sense RNA in the host cell, these two RNAs being complementary to each other, form a double stranded RNA, that initiates RNAi and thus silencing the specific mRNA of the nematode, nematode is unable to survive in the transgenic plant. [½×6]

[Total: 5 marks]

Q93.SQP 2025-26 · Q325 marks

‘Some plant and animal pathogens serve as one of the tools of recombinant DNA (rDNA) technology.’

A. Name one animal and one plant pathogen and discuss the pathogenic nature of both. State how they serve as a tool in rDNA technology.

B. What are the enzymes needed for rDNA technology?

C. A farmer owns a cotton farm land which is getting infested with coleopteran pests. He is not willing to use the microbes to protect his farm.

  1. Name an alternate method to introduce the gene of interest the pathogen would have otherwise delivered and discuss how the alternate method would deliver the gene.
  2. State how this gene would control the pest.

OR

BamH1 is a restriction enzyme which recognizes the sequence 5′ - GGATCC - 3′. The restriction activity of this enzyme is between G and G.

A. Construct the palindrome for the above sequence.

B. Draw a labeled diagram to show the formation of recombinant DNA (rDNA) using BamH1.

C. PBR322 is a plasmid that has a restriction site for this enzyme at the tetracycline resistant gene. If BamH1 were to be used, how will it impact the response of the transformant with rDNA to antibiotics - ampicillin and tetracycline. Justify.

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Official CBSE Sample Paper Marking Scheme · 2025-26

A. Retrovirus in animals and Agrobacterium tumefaciens can transform normal cells into cancerous cells / Agrobacterium tumefaciens is responsible for causing crown gall disease / it can transfer its T-DNA to transform normal plant cell into tumorous cells [1]. They are used as cloning vectors to deliver desirable genes into animal/plant cells. [1]

B. restriction enzyme and ligase. [1]

C.

(i) Biolistic/ gene guns can be used. The plant cells are bombarded with high velocity micro particles of gold or tungsten coated with DNA. [1]

(ii) the cry gene will create Bt toxin inside the pest which will get activated in the alkaline gut of the pest and cause the gut epithelial lining to disintegrate. [1]

OR

A. 5′ - GGATCC - 3′
3′ - CCTAGG - 5′ [1]

B. drawing and labelling [2]

Figure

C. If ampicillin is added, the bacteria will show resistance as the gene is intact and will survive; if tetracycline is added, the bacteria will die as it will show insertional inactivation / gene is not functional owing to insertion of the gene of interest in the tetracycline region of selectable marker. [2]

Q94.SQP 2026-27 · Q325 marks

Attempt either option A or B.

A. A scientist is screening bacterial transformants engineered with DNA fragments of the β-globin gene. A radioactive DNA probe complementary to the normal β-globin sequence is used. After autoradiography: Plates A and C show strong and weak radioactive signals respectively. Plate B shows no radioactivity.

Answer the following:

  1. Why does plate B fail to show radioactivity, while A and C do? (2)
  2. How can this experiment help in distinguishing a normal individual, a carrier, and a patient of sickle cell anaemia? (2)
  3. Name one other commonly used technique (apart from autoradiography) to differentiate recombinants from non-recombinants. (1)

OR

B. A student placed DNA from three different sources: bacteria, plant tissue, and fungal cells in a test tube A, B and C respectively. During the process of isolating and purifying DNA, the student added cellulase to all test tubes.

  1. Predict and explain in which trial(s) DNA isolation would fail, and why? Also suggest how it could be corrected. (2)
  2. After rectifying the mistake made in ‘a’, outline the steps to obtain visible spooled DNA from the sample. (2)
  3. Suggest one application of the spooled DNA in biotechnology. (1)
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Official CBSE Sample Paper Marking Scheme · 2026-27

A.

I. Plate B fails to show radioactivity because the probe could not hybridise with its DNA sequence, indicating the presence of the mutated β-globin gene (sickle mutation). Plates A and C show signals since their DNA has the normal complementary sequence. [2]

II.

  • Normal individual: Probe binds fully → strong radioactivity observed
  • Carrier (heterozygote): Both normal and mutated sequences present → partial hybridisation (mixed pattern).
  • Patient (homozygous mutant): No hybridisation with the probe → no radioactivity seen.

[2]

III. Insertional inactivation [1]

OR

B.

I.

  • Test tube A (bacteria): DNA isolation will fail because bacteria require lysozyme, not cellulase, to digest the peptidoglycan cell wall.
  • Test tube B (plant): DNA isolation will succeed because plants require cellulase to digest cellulose walls.
  • Test tube C (fungus): DNA isolation will fail because fungi require chitinase to digest chitin walls.
  • Correction: Use lysozyme for bacteria, cellulase for plants, and chitinase for fungi.

[2]

II. The further steps are

  • Break open cells → DNA released with RNA, proteins, polysaccharides, and lipids.
  • Remove contaminants: RNase digests RNA; Protease digests proteins
  • Remove polysaccharides and lipids by specific treatments.
  • Add chilled ethanol → DNA precipitates as white threads, spooled on a glass rod.

[2]

III. Spooled DNA can be used in genetic engineering, PCR amplification, DNA fingerprinting, or sequencing. [1]

Unit IX · Biotechnology and its Applications

10. Biotechnology and its Applications

Q1.2026 Main 57/1/1 · Q51 mark

Pomato was produced by fusing protoplasts of

  1. Tomato and Potato
  2. Pomegranate and Tomato
  3. Pomegranate and Potato
  4. Pomegranate, Potato and Tomato
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(A) Tomato and Potato [1 mark]

Q2.2026 Main 57/1/1 · Q141 mark

Assertion (A): The milk produced by transgenic cow ‘Rosie’ was nutritionally more balanced product for human babies than natural cow milk.

Reason (R): It was human protein enriched milk containing human alpha lactaglobulin.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(C) Assertion (A) is true, but Reason (R) is false. [1 mark]

Q3.2026 Main 57/2/1 · Q111 mark

Genetically modified tobacco plants were produced using Agrobacterium vectors. Which property of Agrobacterium helps in carrying nematode specific genes into the host plant ?

  1. Infectious for monocot plant ; has Ti plasmid
  2. Infectious for dicot plant ; has Ti plasmid
  3. Infectious for plants and animals ; has Ti plasmids
  4. Infectious for monocot and dicot plants ; has cry gene
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(B) Infectious for dicot plant; has Ti plasmid. [1 mark]

Q4.2026 Main 57/3/1 · Q41 mark

Golden rice is a promising transgenic crop. When released for cultivation, it will help in :

  1. Producing petrol-like fuel
  2. Reduction of Vitamin A deficiency in humans
  3. Pest resistance
  4. Herbicide tolerance
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(B) Reduction of Vitamin A deficiency in humans [1 mark]

Q5.2026 Main 57/4/1 · Q101 mark

In RNAi, genes are silenced using which of the following biomolecules ?

  1. ssDNA
  2. dsDNA
  3. dsRNA
  4. ssRNA
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(C) dsRNA [1 mark]

Q6.2026 Main 57/4/1 · Q121 mark

Protein encoded by gene Cry-I Ab controls the infestation of which of the following ?

  1. Cotton bollworms
  2. Mosquitoes
  3. Corn borer
  4. Agrobacterium
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(C) Corn borer [1 mark]

Q7.2026 Main 57/5/1 · Q101 mark

Transgenic animals have their genetic material manipulated. Select the correct option in reference to transgenic animals.

  1. Foreign DNA and RNA are present in their cells.
  2. Foreign RNA is present in all of their cells.
  3. Foreign DNA is present in all of its cells.
  4. Foreign RNA is present in some of its cells.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(C) Foreign DNA is present in all of its cells. [1 mark]

Q8.2026 Supp. 57/7/1 · Q91 mark

Cry proteins coded by the gene Cry IAb controls which one of the following pests ?

  1. Cotton bollworm
  2. Corn borer
  3. Tobacco budworm
  4. Mosquito
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(B) Corn borer [1 mark]

Q9.2026 Supp. 57/7/1 · Q121 mark

Genetically modified tobacco plants were produced using Agrobacterium vectors. Which properties of Agrobacterium help in carrying nematode-specific genes into the host plant ?

  1. Infectious for monocot plants; has Ti plasmid
  2. Infectious for dicot plants; has Ti plasmid
  3. Infectious for plants and animals; has Ti plasmid
  4. Infectious for monocots and dicots; has cry genes
Reveal answerHide answer
Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(B) Infectious for dicot plant; has Ti plasmid [1 mark]

Q10.2025 Main 57/1/1 · Q21 mark

GEAC stands for

  1. Genome Engineering Action Committee
  2. Ground Environment Action Committee
  3. Genetic and Environment Approval Committee
  4. Genetic Engineering Approval Committee
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Answer not available: no official marking scheme published.

Q11.2025 Main 57/1/1 · Q121 mark

RNA interference (RNAi) helps in making tobacco-plant resistant to a nematode (Meloidegyne incognitia).

Choose the correct option that shows how RNAi is achieved :

  1. Preventing the process of translation of mRNA of the nematode.
  2. Preventing the process of replication of DNA of the nematode.
  3. Preventing the process of transcription of DNA of the plant.
  4. Preventing the process of replication of DNA of the plant.
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Answer not available: no official marking scheme published.

Q12.2025 Main 57/4/1 · Q131 mark

Assertion (A): The meristems are grown ‘in vitro’ to obtain virus-free plants from an infected plant.

Reason (R): If the plant is infected with a virus, the roots and the stems are free of virus.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q13.2025 Main 57/5/1 · Q131 mark

Assertion (A): To generate only a part of the plant from a cell is totipotency.

Reason (R): Suitable special nutrient media and sterile conditions are required in ‘in vitro’ conditions for the division of cells in explants.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q14.2025 Main 57/6/1 · Q151 mark

Assertion (A): Isolated single cells can be fused to produce somatic hybrids.

Reason (R): Cells selected for somatic hybridisation have desirable characters.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Answer not available: no official marking scheme published.

Q15.2025 Main 57/7/1 · Q91 mark

Which one of the following is not the product of transgenic experiments ?

  1. Pest-resistant crop variety
  2. High nutritional value in grains
  3. Drought-resistant crops
  4. Production of insulin by rDNA technique
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Answer not available: no official marking scheme published.

Q16.2025 Main 57/7/1 · Q121 mark

Out of the following, select the correct match :

  1. Transgenic cow milk – Human beta-lactalbumin protein
  2. ELISA – Antigen antibody interaction
  3. Corn Borer – Cry II Ab gene
  4. Cotton plant – Meloidegyne Incognitia
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Answer not available: no official marking scheme published.

Q17.2025 Supp. 57/S/1 · Q121 mark

Golden rice is a promising transgenic crop. When released for cultivation, it will help in :

  1. Producing a petrol-like fuel from rice
  2. Enrichment of Vitamin A
  3. Pest resistance
  4. Herbicide tolerance
Reveal answerHide answer

Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q18.2024 Main 57/1/1 · Q91 mark

Which one of the following transgenic animals is being used to test the safety of the polio vaccine ?

  1. Sheep
  2. Goat
  3. Pig
  4. Mice
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(D) Mice [1 mark]

Q19.2024 Main 57/1/1 · Q111 mark

The improved trait that is found in the genetically modified transgenic crop Golden rice is :

  1. High lysine content
  2. Insect resistant
  3. High protein content
  4. High vitamin-A content
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(D) High vitamin – A content [1 mark]

Q20.2024 Main 57/1/1 · Q161 mark

Assertion (A): Indian Government has set up an organisation known as GEAC to decide the validity of GM research.

Reason (R): Genetic modification of organisms has no effect when such organisms are introduced in the ecosystem.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(C) Assertion (A) is true, but reason (R) is false. [1 mark]

Q21.2024 Main 57/4/1 · Q151 mark

Assertion (A): Patents are granted by government to an inventor.

Reason (R): Patents prevents others from commercial use of an invention.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]

Q22.2024 Main 57/5/1 · Q141 mark

Assertion (A): In molecular diagnosis, single stranded DNA or RNA tagged with radioactive molecule is called a probe.

Reason (R): A probe always searches and hybridises with its complementary DNA in a clone of cells.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(A) Both (A) and (R) are true and (R) is the correct explanation of (A). [1 mark]

Q23.2024 Supp. 57/S/1 · Q111 mark

Crystals of Bt toxin produced by some bacteria, do not kill the bacteria producing them because :

  1. bacteria are resistant to the toxin
  2. toxin is immature
  3. toxin is inactive
  4. bacteria encloses ‘toxin’ in a special capsule
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(C) toxin is inactive [1 mark]

Q24.2024 Supp. 57/S/1 · Q151 mark

Assertion (A): Biotechnology produces transgenic micro-organisms that act as microfactories for proteins.

Reason (R): To produce proteins for human use like insulin, transgenic microorganisms can be developed.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). [1 mark]

Q25.2023 Main 57/1/1 · Q121 mark

Who among the following challenged the patent right granted to the University of Mississippi Medical Centre for ‘use of turmeric in wound healing’ ?

  1. Mr. Ajay Phadke
  2. Ms. Vandana Shiva
  3. Dr. Venugopalan
  4. Dr. R.A. Mashelkar
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(D) Dr. R.A. Mashelkar. [1 mark]

Q26.2023 Main 57/1/1 · Q131 mark

Assertion (A): A patient of ADA deficiency undergoing treatment for gene therapy requires periodic infusion of genetically engineered lymphocytes.

Reason (R): Lymphocytes are immortal.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(C) A is true, but R is false. [1 mark]

Q27.2023 Supp. 57/C/1 · Q121 mark

Nematode resistant tobacco plants have been developed by introduction of the DNA that produces :

  1. Toxin protein
  2. A particular hormone
  3. Sense and Antisense RNA
  4. Only Sense RNA
Reveal answerHide answer
Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(C) Sense and Antisense RNA [1 mark]

Q28.SQP 2025-26 · Q161 mark

Assertion (A): Rosie was the first transgenic cow to make more nutritionally balanced milk for consumption by human babies.

Reason (R): The milk of Rosie cow contained human beta-lactalbumin which made the milk rich in protein.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true and R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Reveal answerHide answer
Official CBSE Sample Paper Marking Scheme · 2025-26

(C) A is true but R is false [1 mark]

Q29.SQP 2026-27 · Q161 mark

Assertion (A): Tissue culture is used when traditional breeding technique is insufficient to keep pace with the demand of crop improvement.

Reason (R): Totipotency is the ability of a single plant cell or explant to regenerate into a whole plant under sterile conditions in nutrient media.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true but R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
Reveal answerHide answer
Official CBSE Sample Paper Marking Scheme · 2026-27

(B) Both A and R are true but R is not the correct explanation of A. [1 mark]

Q30.2026 Main 57/3/1 · Q212 marks

GEAC is a regulatory body under the Ministry of Environment, Forest and Climate Change in India.

(a) Write the full form of GEAC.

(b) Mention two of its important functions related to biotechnology.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) Genetic Engineering Approval Committee [1]

(b) To make decisions regarding validity of GM Research. [½]

Safety of introducing GM organisms for public services. [½]

[Total: 2 marks]

Q31.2025 Main 57/1/1 · Q212 marks

Student to attempt either option-(A) or (B) :

(A) A patient with ADA deficiency requires periodic infusion of genetically engineered lymphocytes. Explain why such periodic infusion is required and also suggest a permanent cure for such ADA deficiency.

OR

(B) Describe in brief any two techniques that can be utilised to transfer recombinant DNA into the host cell directly without using any vector.

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Answer not available: no official marking scheme published.

Q32.2025 Main 57/7/1 · Q172 marks
  1. Name the Indian crop variety for which, in 1997, an American company got patent rights through the US Patent and Trademark Office. This is considered a case of Biopiracy. Justify.

OR

  1. State two purposes for which the Indian Government has set up GEAC (Genetic Engineering Approval Committee).
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Answer not available: no official marking scheme published.

Q33.2024 Main 57/4/1 · Q192 marks
  1. Name the source from which insulin was extracted in earlier times. Why is this insulin no more in use by the diabetic patients ?
  2. Why does the insulin synthesised in the human body undergo processing whereas the insulin produced by Eli Lilly company does not need to undergo any processing ? Explain.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) Insulin was earlier extracted from pancreas of slaughtered cattle or pigs (animals). This insulin from an animal source may develop allergy or other types of reactions in human body to foreign proteins. [½ + ½]

(b) In human, insulin is synthesized as a prohormone. It contains an extra stretch called C-peptide which needs to be removed before it becomes functional; this ‘C’ peptide is absent in Eli Lilly company insulin. [½ + ½]

[Total: 2 marks]

Q34.2023 Supp. 57/C/1 · Q202 marks

(a) How have transgenic animals proved to be beneficial in the study of :

  1. Normal physiology and development ?
  2. Chemical safety testing ?

OR

(b) Mention four applications of plants whose genes have been altered by manipulation.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a)

(i) Transgenic animals are designed to allow the study of gene regulation and effects of gene on the normal functions of the body / The result obtained by introducing gene from other species that alter the formation of any growth like factors (such as insulin like growth factor) is used to know its biological role. [1]

(ii) Transgenic animals are made more sensitive to toxic substances and when exposed to these substances the effect is studied to get results in lesser time. [1]

OR

(b)

  • Made crops more tolerant to abiotic stresses (cold, drought, salt, heat).
  • Reduced reliance on chemical pesticides (pest-resistant crops).
  • Helped to reduce post-harvest losses.
  • Increased efficiency of mineral usage by plants (this prevents early exhaustion of fertility of soil).
  • Enhanced nutritional value of food, e.g., golden rice, i.e., Vitamin ‘A’ enriched rice.
  • In addition to these uses, GM has been used to create tailor-made plants to supply alternative resources to industries, in the form of starches, fuels and pharmaceuticals.

(Any four applications) [½ × 4 = 2 marks]

Q35.SQP 2026-27 · Q202 marks

Attempt either option A or B.

A. The table below shows hypothetical data comparing α-lactalbumin concentration in three types of milk, including that of Rosie, a transgenic cow designed to express human milk proteins.

Source of Milkα-lactalbumin Content approximately (g/L)
Normal cow's milk0.1
Rosie cow's milk2.4
Human milk3.5
  1. Assuming all other nutritional components are identical, which milk would provide the lowest nutritional contribution from α-lactalbumin?
  2. If further genetic modification increases α-lactalbumin levels in Rosie to 3.5 g/L, predict one ethical concern that might arise from such a development.

OR

B. A farmer replaces the GM crop with a conventional variety. Predict any two challenges she/he may face. Support your answer with two reasons.

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Official CBSE Sample Paper Marking Scheme · 2026-27

A.

I. Normal cow's milk (~0.1 g/L) is least suitable. Reason: Very low α-lactalbumin compared to human milk. [1]

II.

  • (i) Going beyond morality genetical modification of organisms can have long term unpredictable results.
  • (ii) patenting for resource
  • (iii) loss of genetic diversity of natural cows
  • (iv) economic exploitation

[Any one, 1]

OR

B.

  • (i) Increased pesticide use due to lack of pest resistance.
  • (ii) Lower yield under abiotic stress conditions.
  • (iii) Faster depletion of soil nutrients due to inefficient mineral usage.

[Any two, 1+1]

Q36.2026 Main 57/1/1 · Q273 marks

A farmer grew two varieties of corn crop in field A and B. He grew normal corn crops in field A and GM corn crops in field B.

He observed that corn borers attacked only in field A. To control it, spores of Bt were sprayed on field A.

(a) Name the gene in the spores responsible for control of pest.

(b) What effect will the spores of Bt have on insect pest ?

(c) How has corn plants of field B developed resistance against this pest ? Explain.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) Cry I Ab [½]

(b) The spores of Bt were mixed with water and sprayed in field A where these are eaten by the insect larvae, in the gut of the insect toxins are released, and insect get killed. [½ × 3]

(c) The genetically modified plants (GM) express Bt gene / produce Bt toxin which when ingested will kill the pest. [1]

[Total: 3 marks]

Q37.2026 Main 57/2/1 · Q223 marks

There are limitations of traditional breeding techniques that led to promotion of micro-propagation ?

(a) Mention two advantages of micropropagation. [2]

(b) Give two examples where it is commercially adopted. [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) (Any two) Propagation of large number of plants in short duration, new plants are genetically identical to the original plants from which they have grown, production of disease free plants. [1+1]

(b) Tomato, Banana, Apple (Any other two correct examples) [½+½]

[Total: 3 marks]

Q38.2026 Main 57/4/1 · Q263 marks

(a) How is insulin different from proinsulin ?

(b) Explain all the stages of how human insulin is produced through rDNA technology.

[1+2 = 3]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) Proinsulin contains an extra stretch called the C-peptide in addition to chain A and B / C-peptide is not present in the (mature) insulin. [1]

(b) Steps to produce insulin through rDNA technology. [½ × 4]

  • Eli Lilly an American Company prepared two DNA sequences corresponding to A and B chain of human insulin.
  • These chains A and B were then introduced in plasmid E. coli to produce multiple copies.
  • These chains were extracted separately.
  • Chain A and B were combined by creating disulphide bonds to form human insulin.

[Total: 3 marks]

Q39.2026 Main 57/5/1 · Q263 marks

Transgenic animals are produced by man. Explain any three ways in which such animals can be beneficial to humans.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1
  • Normal physiology and development: to study how genes are regulated / how genes affect normal body function and its development / study of complex factors involved in growth such as insulin-like growth factor.
  • Study of disease: to study how genes contribute to the development of a disease / model animals help us to study many diseases such as cancer or cystic fibrosis or rheumatoid arthritis or Alzheimer’s / help us to investigate new treatment methods.
  • Biological products: To produce specific biological products / medicines / Medicines required to treat certain human diseases / produce human protein α-1-antitrypsin (which treat emphysema) / treatment of phenylketonuria / medicine required for the treatment of cystic fibrosis / produce human protein (alpha lactalbumin) rich milk by Rosie cow (it is nutritionally more balanced for human babies)
  • Vaccine safety: transgenic mice being developed for use in testing safety of vaccines before they are used on humans / Transgenic mice are being used to test the safety of the polio vaccine / after testing on transgenic animals is successful and found reliable they could replace use of monkeys to test the safety of batches of vaccine.
  • Chemical safety testing: Transgenic animals are made to carry genes which make them more sensitive to toxic substances than non-transgenic animals. They are then exposed to the toxic substances and the effects studied.

(Any three with correct explanation) [1×3 = 3 marks]

Q40.2026 Supp. 57/7/1 · Q263 marks

Explain any three different approaches used in the treatment of a person suffering from Adenosine Deaminase deficiency.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

The three different approaches include –

  • Enzyme replacement therapy – The functional enzyme adenosine deaminase is injected into the patient periodically. [1]
  • Bone marrow transplantation – Bone marrow from a normal individual is transplanted in the patient which can produce the functional enzyme adenosine deaminase. [1]
  • Gene therapy – A functional ADA-cDNA is introduced into the cultured lymphocytes through a retroviral vector which are subsequently returned to the patient. [1]

[Total: 3 marks]

Q41.2025 Main 57/2/1 · Q273 marks

(a) The insulin synthesised in our body is different from that synthesised by Eli Lilly company using recombinant DNA technology. Differentiate between them.

(b) Why the insulin extracted from an animal source is not in use these days ?

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Answer not available: no official marking scheme published.

Q42.2025 Main 57/4/1 · Q283 marks

Enlist one advantage and two disadvantages of green revolution.

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Answer not available: no official marking scheme published.

Q43.2025 Main 57/5/1 · Q253 marks

Enlist three advantages of genetically modified plants.

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Answer not available: no official marking scheme published.

Q44.2025 Main 57/6/1 · Q283 marks

Explain the beneficial role of the following, produced as a result of the processes of biotechnology, to mankind :

  1. Cow named Rosie
  2. α-1-antitrypsin
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Answer not available: no official marking scheme published.

Q45.2025 Main 57/7/1 · Q263 marks
  1. Why are transgenic animals so called ?
  2. With the help of an example each, explain the role of transgenic animals in the following :
    1. Vaccine safety
    2. Biological products
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Answer not available: no official marking scheme published.

Q46.2025 Main 57/7/1 · Q273 marks
  1. Mention any two advantages of micropropagation techniques.
  2. Write in brief how the process is carried out in the laboratory.
  3. Name any two important food plants grown commercially by this method.
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Answer not available: no official marking scheme published.

Q47.2025 Supp. 57/S/1 · Q283 marks

Biotechnology has helped farmers to get pest-resistant cotton crops. Explain the technique adopted along with its mode of action.

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

Protoxin or inactive protein present in Bt-cotton plant, when ingested by Lepidopteran, gets converted into an active form of toxin due to alkaline pH of its gut; the activated toxin binds to the surface of midgut epithelial cells and creates pores, that causes cell swelling and lysis and eventually causes death of the insect. [½ × 4]

Q48.2024 Main 57/1/1 · Q273 marks

(a) (i) What are transgenic animals ?

(ii) Name the first transgenic cow and state its importance.

OR

(b) (i) Explain the convention for naming EcoRI.

(ii) With the help of an illustration only, show the action of EcoRI on a DNA Polynucleotide.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a)

(i) Transgenic animals are those which have their DNA manipulated to possess and express a foreign gene. [1]

(ii) First transgenic cow – Rosie [1]

Produced human protein enriched milk (2.4 g/litre) / cow milk containing human alpha-lactalbumin protein is nutritionally more balanced product for human babies than natural cow milk. [1]

OR

(b)

(i) In EcoRI (comes from Escherichia coli RY13):

  • E represents Genus Escherichia
  • co represents species coli
  • R represents RY 13 strain
  • I represents order in which the enzyme were isolated from that strain of bacteria

[½ × 4]

(ii)

Figure

[½ + ½]

[Total: 3 marks]

Q49.2024 Main 57/2/1 · Q263 marks

The picture given below shows :

(a) Roots of a typical control tobacco crop plant (infected).

(b) Transgenic tobacco plant showing healthy roots even after deliberate infection by nematode.

Figure

Explain how this transformation was achieved in the tobacco plant.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

Introduction of nematode specific gene using Agrobacterium vector, production of both sense and anti-sense RNA in host cell, production of dsRNA, RNAi / RNA interference is initiated, silencing of specific mRNA of the nematode, parasite could not survive in the transgenic host and transgenic plant is protected from the parasite. [½ × 6 = 3 marks]

Q50.2024 Main 57/3/1 · Q273 marks
  1. Give the scientific name of the bacteria widely used in biotechnology to create a GM cotton crop resistant to bollworm attacks.
  2. Explain how GM cotton crop is able to resist insect attacks.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) Bacillus thuringiensis [1]

(b) GM cotton crop contains Bt toxin protein which exists as inactive protoxins, but once bollworm ingest the inactive toxin it is converted into an active form due to the alkaline pH of the gut of the worm, which solubilizes the crystals of the prototoxin; activated toxin binds to the surface of the midgut epithelial cells of the worm and creates pores that cause swelling and lysis and death of the insect caterpillar. [½ × 4]

[Total: 3 marks]

Q51.2024 Main 57/4/1 · Q283 marks

Explain the role of transgenic animals in :

  1. Production of Biological products
  2. Studying diseases
  3. Chemical safety testing
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) Production of biological products – Human protein α-1 antitrypsin used to treat emphysema produced by transgenic organism / Transgenic cow ‘Rosie’ produced human protein-enriched milk (α-lactalbumin). [1]

(b) Studying diseases – Transgenic models of human diseases like cystic fibrosis, cancer etc. are designed to increase our understanding of how genes contribute to development of disease. [1]

(c) Chemical safety testing – Transgenic animals are made that carry genes which make them more sensitive to toxic substances. They are then exposed to toxic substance and the effect studied. [1]

[Total: 3 marks]

Q52.2024 Main 57/5/1 · Q243 marks
  1. What are transgenic animals ?
  2. Name the transgenic animal having the largest number amongst all the existing transgenic animals.
  3. State any 3 reasons for which these types of animals are being produced.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a) Animals that have had their DNA manipulated to possess and express an extra or foreign gene or desirable foreign gene. [1]

(b) Mice [½]

(c) Any three reasons: [½ × 3]

  • To study the effect of gene on normal physiology and development.
  • To study how gene contribute the development of disease
  • To obtain useful biological products
  • To use in testing the safety of vaccines
  • For chemical safety testing / used for testing toxicity of drugs.

[Total: 3 marks]

Q53.2024 Supp. 57/S/1 · Q263 marks

(a) What is gene therapy ?

(b) Describe the procedure of such a therapy that could be a permanent cure for a genetic disease. Name the genetic disease.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) Gene therapy is a collection of methods that allows correction of a gene defect that has been diagnosed in a child / embryo. [1]

(b) A normal gene is isolated from bone marrow cells and introduced into the cells at an early embryonic stage. [½ + ½]

ADA deficiency (adenosine deaminase deficiency) / SCID (severe combined immunodeficiency disease) / any other correct genetic disease. [1]

[Total: 3 marks]

Q54.2023 Main 57/3/1 · Q273 marks
  1. Write the scientific name of the nematode that infests the tobacco plants and the part that it infests.
  2. How is Agrobacterium used to protect tobacco plant from this attack ?
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) • Meloidogyne incognita [½]

• Roots [½]

(b) By using Agrobacterium vector, Nematode specific genes were introduced into host plant, introduction of DNA produced both sense and anti-sense RNA in the host cells; these two RNAs being complementary to each other form a double stranded RNA (dsRNA), that initiated RNAi and thus silenced the specific mRNA of the nematode hence parasite could not survive in transgenic host. [½×4]

[Total: 3 marks]

Q55.2023 Main 57/5/1 · Q273 marks

On spraying Bacillus thuringiensis on an infected cotton crop field the pests are killed by the toxin, however the toxin although produced by the bacteria does not affect it. Explain giving reason.

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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

In bacteria toxin is produced in inactive form/protoxin, insect ingests it, protoxin/inactive protein becomes active due to alkaline pH of gut (solubilize), toxin binds to epithelial cells (of midgut) and creates pores, causes swelling, lysis and death. [½×6]

[Total: 3 marks]

Q56.2022 Main 57/2/1 · Q123 marks

An application of biotechnology in agriculture involves production of pest resistant plants using 'cry' genes from a bacterium. Name the bacterium. Explain why the 'cry' gene kills the pest but not the bacterium itself.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

Bacillus thuringiensis [1]

Bacillus thuringiensis forms protein crystals which exist as inactive toxin / protoxin, therefore do not kill it. [1]

The ingested inactive toxin in insect gets solubilised / becomes active, due to alkaline pH of insect's gut thereby killing the insect. [½ + ½]

[Total: 3 marks]

Q57.2022 Main 57/3/1 · Q93 marks

What are cry-proteins? With the help of a suitable example, explain how it acts as a biological pesticide.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

Toxic insecticidal proteins / pro-toxin / crystal proteins produced by cry genes of Bacillus thuringiensis. [1]

e.g., in Bt cotton / Bt corn / rice / tomato / soyabean (any other correct example of any crop or any insect) [½]

Bt toxin protein exists as inactive protoxins, which get activated inside the alkaline pH of the gut of insect, causes perforations / swelling / lysis of midgut ultimately killing the insect. [½ × 3]

[Total: 3 marks]

Q58.2022 Main 57/4/1 · Q103 marks

With the help of a storyboard or a flowchart only, explain all the stages of how human insulin is produced through r-DNA technology.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

Preparation of two DNA sequences corresponding to A, and B chains of human insulin (using rDNA technology) [½+½]

↓

Introduction in plasmids of E. coli [½]

↓

Chain A and B were separately produced [½]

↓

Chain A and B were extracted [½]

↓

Both chains were combined by creating di-sulphide bonds to form human insulin. [½]

//

Same marking to be followed when explained with the help of Storyboard

[Total: 3 marks]

Q59.2022 Main 57/5/1 · Q113 marks

(a) How does a gene therapy involving direct modification of the cells, in order to achieve a therapeutic goal is used in the treatment of ADA deficiency? Explain.

(b) A host cell must be made competent, before it is able to receive an rDNA. Justify.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

(a) Isolation of lymphocytes from the patients, culturing of lymphocytes, introduction of functional ADA cDNA into lymphocytes, modified lymphocytes are injected back to the patient. [½×4]

(b) DNA molecule being hydrophilic cannot pass through cell membrane [1]

[Total: 3 marks]

Q60.2022 Supp. 57/6/1 · Q113 marks

(a) One of the potential uses of genetic engineering is in correction of a gene defect that has been diagnosed in a child/embryo. Explain how gene therapy is of help in ADA deficiency.

OR

(b) A schematic diagram of matured human insulin is given below:

Figure

How is the process of its formation naturally in the human body different from that of its formation by rDNA technology? Explain.

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

(a) Normal ADA gene is inserted into patient’s cell / tissue / embryo to treat a disease. [½]

It is done by isolation of lymphocytes from the blood of the patient and culturing of lymphocytes outside the body, introduction of functional ADA cDNA into lymphocyte using retroviral vector, modified lymphocytes are injected back to the patient, if gene isolated from marrow cells producing ADA is introduced into the cells at early embryonic stage, it is a permanent cure. [½×5]

OR

(b) ● Insulin production in human body:

– Synthesised naturally in the form of proinsulin consisting of polypeptide chain A and polypeptide chain B, linked together by disulphide bonds and an extra stretch called C-peptide

– The C-peptide is removed during processing and proinsulin matures into functional insulin. [½+½]

● Insulin production by rDNA technology

– Two DNA sequences corresponding to chain A and chain B of human insulin are synthesised

– They are introduced into two different plasmids of E. coli

– Chain A and chain B are produced separately,

– extracted and combined by disulphide bond to form human insulin. [½×4]

[Total: 3 marks]

Q61.2025 Main 57/4/1 · Q325 marks
  1. Name and explain the biotechnological strategy wherein the infection by the nematode Meloidegyne incognitia can be prevented using Agrobacterium vectors in the roots of tobacco plant by RNA interference.

OR

  1. Explain the amplification of gene of interest using the technique of Polymerase chain reaction (PCR).
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Answer not available: no official marking scheme published.

Q62.2025 Main 57/5/1 · Q315 marks
  1. Define transgenic animals. Explain in detail any four areas where they can be used for human benefit.

OR

  1. Describe the structure and working of a sparged stirred-tank bioreactor.
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Answer not available: no official marking scheme published.

Q63.2025 Main 57/6/1 · Q315 marks
    1. Explain how some strains of Bacillus thuringiensis produce proteins that kill certain insects such as lepidopterans but do not kill the Bacillus.
    2. How is the above mechanism exploited for the production of Bt cotton plant by biotechnologists ?

OR

    1. Explain how the amplification of gene of interest is done using PCR.
    2. State two applications of the desired amplified fragment of DNA.
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Answer not available: no official marking scheme published.

Q64.2022 Main 57/1/1 · Q135 marks

To save the crop plant from the attack of various insect pests the biotechnologists have developed many pest resistant plants. One such example is Bt corn plant. In this plant 'cry' genes were introduced which produces cry-proteins in the plant that has toxic effect on the pest (corn borer). Thus saves the corn plant from the attack of the corn borer. An experimental field study was conducted by the scientists to see the efficacy of the Bt corn plant against the attack of corn borers. Three different species of corn borers namely 'A', 'B', 'C' were collected and were independently fed on non Bt corn plants and Bt corn plants separately for the same period. The extent of the damage caused to the leaf area of the plant was observed and noted down. With the help of the observations and data collected the following bar graph was plotted. Study the graph and answer the questions that follow.

Figure

  1. Identify the species of the corn borer that was most successfully controlled by Bt corn plant. Give appropriate reason for your inference.
  2. Identify the species of the corn borers which shows least impact of toxin produced by Bt genes.
  3. What would be your advise as a Scientist, to the farmers for growing this particular Bt corn variety in the area which is infested by species-'B' of corn borers?
  4. Name one Bt gene that encodes protein in corn plants to control corn borers.

OR

A gene was identified in a fungus by a research worker in a lab which was considered to be of a great importance in the field of agriculture. As a student of biotechnology, write the steps you would suggest to (i) Isolate this gene of interest from the fungus and (ii) amplify this gene for further experimentation and research.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1
  1. Species A [1]; The leaf area damaged by species A in Bt-corn is the least [1]
  2. Species B [1]
  3. Not to grow Bt variety as seeds are expensive and of not much benefit (productivity wise) / advise to grow Bt corn with its proper justification [1]
  4. Cry IAb. (No marks if only cry gene is written) [1]

OR

(i) (Isolation of genetic material) Fungal cell treated with chitinase, RNA to be removed by treating with RNAase, protein removed by treating with protease, and then addition of chilled ethanol [½ × 4]; Cutting of DNA at specific location by restriction enzymes [½]; Fragments are separated by gel electrophoresis [½]

(ii) Multiple copies of separated genes of interest is synthesized by following steps of the method given below: PCR (polymerase chain reaction) – Denaturation, Annealing, Extension (followed by amplification) [½ × 4]

// (alternative: PCR shown as a diagram, ½ mark each for the four labelled parts)

Figure

//

(ii) Multiple copies of separated genes of interest are synthesized by following the given below method: rDNA technology, same restriction enzyme cutting both foreign DNA and vector DNA at specific point, ligases join foreign DNA to Plasmid, transformation (cell divides and helps in multiplication of genes) [½ × 4]

// (alternative: rDNA technology shown as a diagram, ½ mark each)

Figure

[Total: 5 marks]

Q65.2022 Main 57/4/1 · Q135 marks

(a) There are two different farm lands, one where Bt-cotton crop was cultivated and the other where non Bt-cotton crop (indigenous) was cultivated. Farmers responsible for this experimental cultivation were free to use the farming practices of their choice. During the cultivation period, the data was collected with respect to the amount of pesticide used, water required for irrigation and at harvesting time, the crop productivity. Based on the data collected, a bar graph was plotted which is shown below.

Figure

Answer the following questions:

  1. Write your interpretation, with reason, on the basis of the three parameters plotted in the graph.
  2. Which one of the crops would you like to cultivate in your farm and why?
  3. Which one out of these two crops would a farmer from Rajasthan like to cultivate and why?

OR

(b) Gene of interest/alien gene is introduced by a cloning vector into a host cell to bring about a desired phenotypic expression in a host cell. The cloning vectors used are plasmid and bacteriophages. Biotechnologists in their labs, for desired results engineered specialised cloning vectors. One such vector is pBR322. Study the diagram carefully and answer the questions that follow.

Figure

  1. What do ‘EcoR I’, ‘BamH I’ and ‘Hind III’ represent? State their functions.
  2. Identify the gene you would select for the role of a selectable marker in pBR322. Explain why.
  3. Write the property/characteristic of plasmid and bacteriophage that makes them efficient cloning vectors.
  4. Biotechnologists always insert ‘ori’ gene in their engineered cloning vector. Justify the statement.
  5. Will the experiment be successful if the alien DNA is ligated at Hind III restriction site? Give reason in support of your answer.
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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

(a) (i) ● Crop productivity is quite high in Bt crop as compared to non-Bt crop. ● Pesticide use has reduced in Bt crop. It saves resources and is environment friendly. ● Water is required more for Bt crop as compared to non-Bt crop. [1+1+1]

(ii) Weighing all the parameters it is concluded that Bt crop will be beneficial to farmers. (Less pesticide, more output) [1]

(iii) Usually in Rajasthan there is shortage of water hence farmers would like to cultivate non Bt-cotton crop. [1]

OR

(b) (i) ● All are restriction endonucleases [½] ● act as a molecular scissors / cut at a specific site within DNA [½]

(ii) ampR / tetR, help in identification and selection of transformants from non-transformants / identification of recombinant from non-recombinant [½+½]

(iii) Have the ability to replicate within bacterial cells independent of the control of chromosomal DNA / Autonomously replicating / bacteriophages and some bacteria have high copy number per cell, can replicate the desirable gene into large number of copies, presence of selectable marker, presence of cloning sites, presence of ori. [1] (Any one)

(iv) It is the sequence where replication begins / It controls copy number of vector [1]

(v) No, identification and selection of transformants from non-transformants / identification of recombinant from non-recombinant would not be possible [½+½]

[Total: 5 marks]

Unit X · Ecology and Environment

11. Organisms and Populations

Q1.2026 Main 57/2/1 · Q41 mark

What could be the reason of extinction of Abingdon Tortoise from Galapagos Islands ?

  1. Intraspecific Competition
  2. Predation
  3. Parasitism
  4. Interspecific Competition
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(D) Interspecific Competition [1 mark]

Q2.2026 Main 57/4/1 · Q111 mark

Connell’s elegant field experiments on the rocky seacoast of Scotland, found that larger barnacles Balanus dominated the intertidal area and removed the smaller barnacles Chathamalus. This happened due to :

  1. Parasitism
  2. Predation
  3. Mutualism
  4. Competition
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(D) Competition [1 mark]

Q3.2026 Main 57/5/1 · Q91 mark

Choose a correct option that shows an interaction in which one species benefits while the other is neither harmed nor benefitted.

  1. Mycorrhizal associations between fungi and roots of higher plants
  2. Sea anemone and clownfish
  3. Abingdon tortoises and goats in the Galapagos Island
  4. Cuscuta on hedge plants
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(B) Sea anemone and clownfish. [1 mark]

Q4.2026 Supp. 57/7/1 · Q101 mark

If a new habitat is being colonized, out of the four processes affecting population growth, which are the two factors that contribute to an increase in population density ?

  1. Immigration
  2. Natality
  3. Emigration
  4. Mortality

Options :

  1. (i) and (ii)
  2. (ii) and (iii)
  3. (iii) and (iv)
  4. (i) and (iv)
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(A) (i) and (ii) [1 mark]

Q5.2025 Main 57/7/1 · Q101 mark

A biologist studied the population of rats in a granary. He found the average natality was 280, average mortality was 200, immigration was 40 and emigration was 50. The net increase in population is :

  1. 80
  2. 70
  3. 10
  4. 90
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Answer not available: no official marking scheme published.

Q6.2025 Supp. 57/S/1 · Q161 mark

Assertion (A): Competition is a type of interspecific interaction between two or more species when the resources necessary for them are limited and adversely affect them.

Reason (R): Special adaptation may allow the competing species to coexist in the same geographical area.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
Reveal answerHide answer

Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.

Q7.2024 Main 57/3/1 · Q121 mark

The population growth curve applicable for a population growing in a geometric fashion, when the resources are not limiting in the habitat will be :

Options (A)–(D) are shown in the figure.

Figure

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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(C) [1 mark]

Figure

Q8.2024 Main 57/4/1 · Q131 mark

Assertion (A): A given fig species can be pollinated only by its partner wasp.

Reason (R): The wasp pollinates the fig inflorescence while searching for suitable egg laying sites.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]

Q9.2024 Supp. 57/S/1 · Q121 mark

The population interaction where one species is harmed and the other is unaffected is :

  1. Amensalism
  2. Commensalism
  3. Parasitism
  4. Predation
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(A) Amensalism [1 mark]

Q10.2023 Main 57/1/1 · Q31 mark

The status of the human population reflected in the human age pyramid given below is :

Figure

  1. Declining population
  2. Stable population
  3. Expanding population
  4. Extinct population
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(A) Declining population. [1 mark]

Q11.2023 Main 57/1/1 · Q141 mark

Assertion (A): A cattle egret and grazing cattle in close association is a classic example of commensalism.

Reason (R): As grazing cattle move through the field, they stir up and flush out insects from the vegetation that otherwise might be difficult for egrets to find and catch.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(A) A and R are true and R is the correct explanation of A. [1 mark]

Q12.2023 Main 57/3/1 · Q91 mark

Identify the option that does not exhibit a parasitic relationship.

  1. Head lice in humans
  2. Cuscuta on a mango tree
  3. Female Anopheles
  4. Ticks on dogs
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(C) Female Anopheles [1 mark]

Q13.2023 Main 57/3/1 · Q161 mark

Assertion (A): The Mediterranean orchid Ophrys uses sexual deceit to get pollinated by a species of bee.

Reason (R): The female bee changes its colour depending on the temperature of the area.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(C) A is true but R is false. [1 mark]

Q14.2023 Main 57/5/1 · Q51 mark

A Tight one-to-one relationship between many species of fig tree and certain wasps is an example of –

  1. Commensalism
  2. Parasitism
  3. Amensalism
  4. Mutualism
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(D) Mutualism. [1 mark]

Q15.2023 Main 57/5/1 · Q111 mark

Important attributes belonging to a population but not to an individual are :

  1. Birth rate and death rate
  2. Male and female
  3. Birth and death
  4. Sex-ratio

Select the correct option from the given options :

  1. (i) only
  2. (ii) only
  3. (ii) and (iii)
  4. (i) and (iv)
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(D) (i) and (iv). [1 mark]

Q16.2023 Supp. 57/C/1 · Q101 mark

In asymptote state, population is :

  1. Increasing
  2. Decreasing
  3. Stabilized
  4. Changing
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(C) Stabilized [1 mark]

Q17.SQP 2025-26 · Q141 mark

Assertion (A): Some organisms are better adapted to survive in an otherwise hostile environment.

Reason (R): Adaptive ability is inherited and has a genetic basis.

  1. Both A and R are true and R is the correct explanation of A.
  2. Both A and R are true and R is not the correct explanation of A.
  3. A is true but R is false.
  4. A is false but R is true.
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Official CBSE Sample Paper Marking Scheme · 2025-26

(A) Both A and R are true and R is the correct explanation of A [1 mark]

Q18.2026 Main 57/2/1 · Q202 marks

(a) In some cases, measuring population density by counting individuals may not be feasible. Mention two other parameters that can be used and justify their use with suitable examples.

OR

(b) What factors contribute to the Amazonian rain forests in South America being the most biodiverse regions on Earth ?

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) (Any two)

  • Percent cover / biomass: In an area if there are 200 carrot grass or Parthenium plants but only single huge banyan tree with a large canopy, stating that population density of banyan is low amounts to underestimating the enormous role of Banyan.
  • Relative density: laboratory culture in a petri dish / fish caught per trap
  • Indirect estimate: tiger census in national parks and tiger reserves is often based on pug marks and fecal pellets.

[½ × 4 = 2 marks]

OR

(b) (Any two)

  • Amazonian Rain Forests in South America has remained relatively undisturbed for millions of years.
  • It is less seasonal, more constant and predictable.
  • More solar energy available in tropical areas that contributes to higher productivity and more diversity

[1+1 = 2 marks]

Q19.2025 Main 57/2/1 · Q212 marks

Student to attempt either option (A) or (B).

(A) Identify the type of pyramid given below and write two identifying features of such a pyramid :

Figure

OR

(B) (i) Construct an ideal pyramid of energy when 10,00,000 Joules of sunlight is available.

(ii) Mention the energy obtained by the fourth level of this pyramid.

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Answer not available: no official marking scheme published.

Q20.2025 Main 57/4/1 · Q212 marks
  1. Explain how the interaction between sea anemone and clownfish is one of the best examples of commensalism in nature.

OR

  1. Correctly depict (also indicate the trophic level) and describe the ecological pyramid of biomass in sea with 40 standing crop of phytoplankton supporting 90 standing crop of zooplankton which further supports 120 small fishes.
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Answer not available: no official marking scheme published.

Q21.2025 Main 57/5/1 · Q212 marks
  1. Explain how the interaction between a fig tree and its tight one-to-one relationship with the pollinator species of wasp is one of the best examples of mutualism.

OR

  1. Correctly depict (also indicate the trophic level) and describe the ecological pyramid of number with 32 birds dependent on 20 insects feeding on one banyan tree.
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Answer not available: no official marking scheme published.

Q22.2025 Main 57/6/1 · Q172 marks
  1. How is the interaction between Ophrys and its specific bee pollinator one of the best examples of co-evolution ? Explain.

OR

  1. Arrange the given important steps of decomposition in their correct order of occurrence in the breakdown of complex organic matter and explain the fourth step in the process.

    Catabolism, Leaching, Fragmentation, Mineralisation, Humification

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Answer not available: no official marking scheme published.

Q23.2024 Main 57/5/1 · Q192 marks

Observe the population growth curve and answer the questions given below :

Figure

  1. State the conditions under which growth curve ‘A’ and growth curve ‘B’ plotted in the graph are possible.
  2. Mention what does ‘K’ in the graph represent.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a) Growth curve ‘A’ – unlimited resources (food and space) or limited competition or in absence of checks or in absence of environmental resistance [½]

Growth curve ‘B’ – limited resources or more competition or in presence of checks or in presence of environmental resistance [½]

(b) ‘K’ is the carrying capacity or maximum number of individuals of a population a given habitat can accommodate (beyond which no further growth is possible). [1]

[Total: 2 marks]

Q24.2024 Supp. 57/S/1 · Q212 marks

(a) Comment on the interaction between a fig tree and wasp. Mention the phenomenon that operates in their relationship.

OR

(b) Explain “brood parasitism” with the help of an example.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a)

They show mutualism, where the female wasp uses the fruit not only as an oviposition (egg laying) site but uses the developing seeds within the fruit for nourishing its larvae; in return the wasp pollinates the fig inflorescence. [½ × 3]

Co-evolution is the phenomenon that operates in their relationship. [½]

OR

(b)

Brood parasitism – the parasitic bird lays its egg in the nest of its host and lets the host incubate them. [1]

Example: cuckoo (koel) lays its eggs in the nest of crow, which resemble the host’s egg in size and colour / any other correct example. [1]

[Total: 2 marks]

Q25.2023 Main 57/5/1 · Q182 marks

“Cattle and goats do not browse the Calotropis plant.” Justify the statement giving reasons.

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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

Calotropis plant produces highly poisonous cardiac glycosides. [1×2]

[Total: 2 marks]

Q26.2022 Main 57/1/1 · Q52 marks

Study the graph given below, showing the population growth curves 'A' and 'B' respectively. Answer the following questions:

Figure

(a) What is 'Carrying Capacity' in respect of Curve 'B' indicative of?

(b) Mention the action of possible natural forces that could have lead to curve 'B'.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

(a) Carrying capacity is the maximum possible number of organisms (beyond which no further growth is possible) in a habitat. [1]

(b) Limited resources lead to competition, fittest survives and reproduces [½ + ½]

[Total: 2 marks]

Q27.2022 Main 57/2/1 · Q52 marks

Explain the two physiological attributes which enable a kangaroo rat to survive in desert conditions.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

Internal fat oxidation (water is by product) [1]

Concentrated urine / minimal volume of water is used to remove excretory products [1]

[Total: 2 marks]

Q28.2022 Main 57/3/1 · Q52 marks

Mammals are capable of maintaining homeostasis. But, very few smaller animals can survive in polar regions. Give reasons.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

Heat loss or gain is a function of surface area, since small animals have large surface area compared to their volume, they tend to lose body heat very fast during severe winters, have to expend much energy to generate body heat through metabolism so they undergo hibernation or migration. [½ × 4]

[Total: 2 marks]

Q29.2022 Main 57/3/1 · Q62 marks

(a)

Figure

Observe the schematic representation given above and answer the following questions:

  1. Identify A and B.
  2. Calculate the growth rate of bacteria in a curd sample, where 1 million bacteria increased to two million, within a period of one hour.

OR

(b)

Figure

Identify the type of pyramid given above. Write the identifying feature on the basis of which you identified it.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

(a) (i) A – Mortality / No. of deaths in population; B – Natality / No. of births in population [½ + ½]

(ii) 1 bacteria per bacteria per hour [1] [Growth rate = (2 million – 1 million) / 1 million]

OR

(b) Expanding age pyramid [½ + ½]; the number of individuals in the 'pre-reproductive' age group is more than the 'reproductive' age group. [1]

[Total: 2 marks]

Q30.2022 Main 57/4/1 · Q52 marks

The figures given below show the results of a lab experiment in which two microbial species A and B belonging to same genus were grown in three petri dishes having same culture medium. In Petri dish-I, Species-A was grown alone for 8 weeks. In Petri dish-II, Species-B was grown alone for 8 weeks. In Petri dish-III, both the species were grown together with the same number as Petri dish-I and Petri dish-II for 8 weeks.

Figure

What conclusion will you draw from the graphs in terms of impact of interaction on growth? Explain, giving reason.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

● Both the species (Species A and B) show better growth when grown together. [1]

● Both are benefitted from each other / They show mutualism. [1]

[Total: 2 marks]

Q31.2022 Main 57/4/1 · Q62 marks

(a) The graphs given below, A and B, represent population of elephants in two different National Parks (a hypothetical situation) at different times.

Figure

Study the graphs and comment upon the pattern of growth observed. Mention the possible reason for such patterns seen in nature.

OR

(b) As per a recent newspaper report, a particular country showed declining population growth rate. According to you, what could be the two most possible reasons for this decline and why?

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

(a) ● Graph A depicts exponential growth / geometric growth, due to unlimited resources and absence of checks / environmental resistances [½+½]

● Graph B shows extinction, due to lack of resources and environmental resistances. [½+½]

OR

(b) Possible reasons for decline in human population: Increased emigration, increased mortality, decreased birth rate / natality, use of contraceptives, spread of epidemic or any other relevant reason [1+1] (any two reasons)

//

Possible reasons for decline in plants, animals / other organisms’ population: Deforestation, decreased food availability, predation pressure, adverse weather, competition or any other relevant reason [1+1] (any two reasons)

[Total: 2 marks]

Q32.2022 Main 57/5/1 · Q52 marks

Write the best method to measure the population density of a single Banyan tree in comparison to 20 Parthenium weeds in a forest by an ecologist. Explain and justify your answer.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

● Biomass / Percent Cover in an area [1]

● Canopy or Biomass or Percent Cover of single Banyan tree is more than 20 Parthenium weed plants in an area. [1]

[Total: 2 marks]

Q33.2022 Main 57/5/1 · Q62 marks

(a) (i) Write the observations made at the end of Connell’s field experiment on barnacles on the rocky sea coasts of Scotland.

(ii) Name any two categories of organisms that in general are adversely affected by competition.

OR

(b) The graphs (X) and (Y) given below depict the diurnal variations in the solar radiations in the month of June (Summer) and in December (Winters):

Figure

(i) Which of the two graphs depicts tropical region and temperate regions respectively?

(ii) Which of the two regions (X) or (Y) will show high biological diversity and why?

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

(a) (i) The larger and competitively superior barnacle (Balanus) dominates the intertidal area, and excludes the smaller barnacle (Chathamalus) from the zone [½+½]

(ii) Herbivores, plants [½+½]

OR

(b) (i) Note: Since graphs shown are not conclusive enough for drawing any interpretation, so one mark to be given to students, if attempted. [1]

(ii) Tropical region has high biological diversity due to less seasonal fluctuations / predictable climate / relatively more constant environments / constant environment promotes niche specialisation / more solar energy available / long evolutionary time for species diversification [1] (any one point)

[Total: 2 marks]

Q34.2022 Supp. 57/6/1 · Q52 marks

Explain the pollination mechanism involved in co-evolution of the two species, namely Ophrys (orchid) and its insect pollinator bees (& bumble bees).

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

– Ophrys employs sexual deceit (to get pollination done) [½]

– one petal of its flower (has uncanny) resemblance to the female bee in size / colour / markings [½]

– The male bee is attracted and ‘pseudocopulates’ with the female flower achieving pollination [½]

– When this same bee ‘pseudocopulates’ with another flower, it transfers pollen to it and pollinates the flower. [½]

[Total: 2 marks]

Q35.2022 Supp. 57/6/1 · Q62 marks

(a) The given graph shows diagrammatic representation of organismic response to abiotic factors:

Figure

  1. Which of the two lines, (A) or (B), depicts the relationship of organismic response to temperature in cat and lizard respectively?
  2. Explain how animal (A) is different from animal (B).

OR

(b) Given below an equation describing the growth pattern of a population:

dN/dt = rN

  1. Mention the type of growth model or growth pattern of population described by the given equation.
  2. What does ‘r’ in the equation signify?
  3. Mention the type of growth curve that will be obtained if the population density (N) is plotted against time (t).
  4. According to you, will the resource availability be limited or unlimited for this type of growth in a given population?
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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

(a) (i) A – cat [½]   B – lizard [½]

(ii) ‘A’ (Regulator) can maintain homeostasis or constancy in body temperature, but only over a limited range of environmental conditions [½]

‘B’ (Conformer) changes its body temperature in accordance with the external temperature (as shown in the graph, range 35° – 45° C, beyond which they simply conform) [½]

OR

(b) (i) Exponential growth model / Geometric growth pattern [½]

(ii) ‘r’ – intrinsic rate of natural increase [½]

(iii) ‘J’ shaped curve [½]

(iv) Unlimited resources [½]

[Total: 2 marks]

Q36.2026 Main 57/3/1 · Q243 marks

(a) What is the carrying capacity of a species in a habitat ? [1]

(b) Explain the growth curve that takes this capacity into account. [2]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) Carrying capacity: It denotes the limit of the maximum number of individuals of a species (population) that a habitat can support with the available resources. [1]

(b) Growth Curve is called Verhulst Pearl Logistic Growth. A population growing in a habitat with limited resources shows a lag phase initially, it is followed by acceleration and deceleration and finally an asymptote. We get a Sigmoid Curve [½+½]

dN/dt = rN ((K – N)/K) [1]

N = Population at time t; r = Intrinsic rate of natural increase; K = Carrying capacity

//

Figure

Verhulst Pearl Logistic Growth

Lag phase = ½, Asymptote = ½, K = ½, Equation = ½

[Total: 3 marks]

Q37.2026 Main 57/4/1 · Q273 marks

(a) Study the graph given below and answer the questions that follow :

Figure

  1. Write the status of food and space in the curves a and b.
  2. In the absence of predators, which one of the two curves would appropriately depict the prey population ?
  3. ‘Time’ has been shown on x-axis and there is a parallel dotted line above it. Give the significance of this dotted line.

OR

(b) Answer the following questions :

  1. Define ‘predation’.
  2. Why is predation required in a community of different organisms ? Give any two points.
  3. Why are predators ‘prudent’ in nature ?
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) (i) Curve (a) – The food and space are unlimited. [½]

Curve (b) – The food and space are limited. [½]

(ii) Curve ‘a’ [1]

(iii) It represents the carrying capacity which means the maximum number of individuals that a given environment can support and beyond which further growth is not possible. [1]

OR

(b) (i) Predation is the interspecific interaction in which a large animal called predator kills and consumes a smaller animals called prey. [1]

(ii) (any two) [½ × 2]

  • It keeps the prey population under control.
  • It helps in maintaining species diversity in a community by reducing the intensity of competition among competing prey species.
  • It acts as conduit for energy transfer across trophic level.
  • Biological control methods adopted in agricultural pest control are based on the ability of the predator to regulate prey population.

(iii) If a predator is too efficient and over exploits its prey then the prey might become extinct and following it the predator will also become extinct for lack of food. [1]

[Total: 3 marks]

Q38.2026 Main 57/5/1 · Q273 marks

Explain Gause’s ‘Competitive Exclusion Principle’ with the help of a suitable example.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1
  • Gause’s competitive exclusion principle states that two closely related species competing for the same limited resources cannot co-exist indefinitely, and competitively inferior one will be eliminated eventually / If resources are limited the stronger competitor will dominate, driving the weaker one to extinction. [1+1]
  • Example: In some shallow American lakes visiting flamingoes and resident fishes compete for their common food i.e. zooplankton in the lake / The Abingdon tortoise in Galapagos islands became extinct within a decade after the goats were introduced into the island probably due to the greater browsing efficiency of the goat or the feeding efficiency of one species (tortoise) is reduced by the interfering presence of the other species (goat) / any other correct example. [1]

[Total: 3 marks]

Q39.2026 Supp. 57/7/1 · Q273 marks

(a) Name the type of interaction seen in each of the following examples.

  1. An orchid growing on branch of mango tree
  2. Wasp pollinating fig inflorescence

(b) Also explain the above interactions shown by (i) and (ii).

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) (i) Commensalism [1]

(ii) Mutualism [1]

(b) (i) Commensalism – It is an interspecific interaction in which one of the species is benefitted and the other is neither benefitted nor harmed / Orchid grows as an epiphyte while mango is neither benefitted nor harmed [½]

(ii) Mutualism – It is an interspecific interaction in which both the partner species are benefitted / Wasp pollinates fig inflorescence while fig offers egg laying site and some of its developing seeds as food for developing wasp larvae. [½]

[Total: 3 marks]

Q40.2025 Main 57/2/1 · Q283 marks

(a) Draw a graph for a population whose population density has reached the carrying capacity.

(b) Out of the two population growth curves, which one is considered a more realistic for most populations ? Why ?

(c) Draw a growth curve where resources are not limiting for the growth of a population and give its equation.

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Answer not available: no official marking scheme published.

Q41.2025 Supp. 57/S/1 · Q273 marks

(a) How does the Mediterranean orchid ‘Ophrys’ ensure its pollination by bees without offering any floral rewards ?

(b) Name the type of interaction seen in each of following examples :

  1. Ascaris worms living in the intestine of humans.
  2. Clown fish living amongst the tentacles of sea anemone.
Reveal answerHide answer
Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(a) Mediterranean orchid Ophrys uses ‘sexual deceit’ to get the pollination done by bees; one of the petals of the orchid resembles the female bee in size or colour or markings; the male bee visits the petal as if it were a female bee (pseudocopulates) and gets dusted with pollen from the flower; when the same bee pseudocopulates with another flower it transfers the pollen to that flower. [½ × 4]

(b) (i) Parasitism [½]

(ii) Commensalism [½]

[Total: 3 marks]

Q42.2024 Main 57/1/1 · Q283 marks

(a) Explain how it is ensured that the orchid Ophrys is pollinated by a specific species of bee.

(b) Describe co-evolution with the help of this example.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a)

  • Orchid Ophrys employs ‘sexual deceit’ to get pollinated by a species of bee
  • One petal of flower resembles female of bee in size, colour and markings
  • Male bee attracted and pseudocopulates the flower and gets dusted with pollens
  • When same bee ‘pseudocopulates’ with other flower, it transfers the pollens to it

[½ × 4]

(b) If female bee pattern changes during evolution the flower needs to co-evolve to resemble the female bee to get pollinated. [1]

[Total: 3 marks]

Q43.2024 Main 57/2/1 · Q283 marks

Predation is referred to as a detrimental interaction. Explain any three positive roles, supported by an example each, that a predator plays in an ecosystem.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1
  • Predators act as conduits for energy transfer across trophic levels. For example: Grass → Goat → Lion / Lion (predator) transfers the energy fixed by plants to the ecosystem. (or any other correct example) [½ + ½]
  • Predators keep prey populations under control. e.g. Cactus feeding predator (moth) control the spreading of the prickly pear cactus. (or any other correct example) [½ + ½]
  • Predators help in maintaining species diversity by reducing the intensity of competition among competing prey species. e.g. extinction of more than 10 species of invertebrates due to removal of starfish Pisaster (predator). (or any other correct example) [½ + ½]

[Total: 3 marks]

Q44.2024 Main 57/3/1 · Q283 marks

Describe how fig tree and wasp relationship is a spectacular example of mutualism.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1
  • The female wasp uses the fruit not only as an oviposition (egg laying) site,
  • But uses the developing seeds within the fruit for nourishing the larvae,
  • The wasp pollinates the fig inflorescence or flower, while reaching for the suitable egg laying sites. [1 × 3]

[Total: 3 marks]

Q45.2023 Main 57/3/1 · Q283 marks

Explain the following population interactions with the help of one example each :

  1. Brood Parasitism
  2. Co-evolution of mutualists
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) Parasitic bird lays resembling eggs in the nest of host bird [½]

Cuckoo (Koel) lay eggs in the nest of crow [1]

//

Eggs of cuckoo (Koel) have evolved in time to resemble the eggs of the crow, koel lays eggs in the nest of the crow and lets them be hatched there, cuckoo is the parasitic bird here exhibiting brood parasitism. [½×3]

(Any other correct example)

(b) When evolution of one species is tightly linked with the evolution of other species [½]

Plant pollinator interaction/fig species and wasp/any other relevant example [1]

[Total: 3 marks]

Q46.SQP 2025-26 · Q283 marks

The population pyramids of Japan for the years 1950, 2007 and on the basis of the estimated value for 2050 have been shown below to answer the questions that follow:

Figure

  1. Comment upon the growth status of the population on the basis of shape of the age pyramids.
  2. What insights can you gain about their population dynamics?
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Official CBSE Sample Paper Marking Scheme · 2025-26

1950 – Expanding - The population structure in 1950 exhibits a broad-based pyramid with a wider base, indicating a higher percentage of young individuals. This suggests a population with a higher birth rate. [1]

2007 – Stable - The narrowing of the pyramid towards the top signifies a lower proportion of elderly individuals. By 2007, the population was more stable with the number of pre reproductive and reproductive age nearly being the same. Bell shape indicates static population. [1]

2050 – Declining - By 2050 urn shaped pyramid shows a declining population where birth rates are very less and the populations of elderly people will increase. [1]

Q47.SQP 2026-27 · Q283 marks

The nematode Myrmeconema neotropicum lives inside the abdomen of certain tropical ants (Cephalotes atratus). Inside the ant's body, the nematode develops slowly without killing it. Once mature, it changes the color of the ant's abdomen to bright red, making it resemble a berry. Birds, mistaking it for a fruit, eat the ant. The nematode's eggs survive in the bird's digestive system and pass out in droppings, which other ants contact while foraging, starting the cycle again.

  1. Why would harming the ant too early be disadvantageous for the nematode? (1)
  2. How is natural selection playing a vital role in maintaining this interaction? (1)
  3. What kind of ecological interaction is seen between nematode and ant? Define it in one sentence. (1)
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Official CBSE Sample Paper Marking Scheme · 2026-27

A. If the ant dies too soon, the nematode won't complete its development and won't be transmitted to new hosts. [1]

B. Natural selection favors nematodes that keep the ant alive until it can be eaten by a bird, ensuring better dispersal and survival of the nematode. The birds eat only the red-bellied ant. This ensures complete maturation of nematode and its dispersal and protects the other ants from predation. [1]

C. Type: Parasitism, Definition: An interaction where one organism benefits and the other is harmed. [1]

Q48.2026 Main 57/1/1 · Q294 marks

The diagram below shows the distribution of two barnacle species, Chthamalus and Balanus on a rocky sea shore.

When Balanus is experimentally removed, Chthamalus expands its range in lower intertidal zone.

Figure

(a) Identify and define the ecological phenomenon demonstrated by this observation.

OR

(a) State the principle that explains the elimination of one species.

(b) Two different species can compete for the same resource. Give another example of it.

(c) How do species avoid competition in nature ? Explain with an example.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) • Competitive release. [½]

• A species whose distribution is restricted to small geographical area due to the presence of competitively superior species expands its distributional range when competing superior species is removed. [½]

OR

(a) Gause’s competitive exclusion principle states that two closely related species cannot co-exist indefinitely and the competitively inferior one will be eliminated eventually. [½ + ½]

(b) In shallow South American lakes visiting flamingos and resident fishes compete for common food zooplankton / any other correct example [1]

(c) • By “resource partitioning” [1]

• They choose different times for feeding or different foraging patterns e.g. five closely related species of warblers avoid competition and coexist due to behavioural differences in their foraging activities / any other correct example. [1]

[Total: 4 marks]

Q49.2025 Main 57/1/1 · Q304 marks

Read the following passage and answer the questions that follow :

In nature, we rarely find isolated, single individuals of any species; majority of them live in groups in a well-defined geographical area, share or compete for similar resources, potentially interbreed and thus constitute a population. The population has certain attributes whereas, an individual organism does not. A population at a given time is composed of individuals of different ages. The size of the population tells us a lot about its status in the habitat. Whatever ecological processes we wish to investigate in a population, be it the outcome of competition with another species, the impact of the predator or the effect of pesticide application, we always evaluate in terms of any change in the population size. The size, in nature, could be low or go into millions. Population size, technically called population density (N) need not necessarily be measured in numbers only. The size of a population for any species is not a static parameter. It keeps on changing with time depending on various factors including food availability, predation pressure and adverse weather.

(a) The Monarch butterfly is highly distasteful to its predator because of a special chemical present in its body. How does the butterfly acquire this chemical ?

(b) If population density at a time t + 1 is 800, Emigration = 100, Immigration = 200, Natality = 200 and Mortality = 150, calculate the population density at time t and comment upon the type of age pyramid that will be formed in this case.

Student to attempt either sub-part (c) or (d) :

(c) What is the difference in a method of measuring population density in an area if there are 200 carrot grass plants to only single huge banyan tree ?

OR

(d) Name two methods to measure the population density of tigers.

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Answer not available: no official marking scheme published.

Q50.2024 Main 57/3/1 · Q304 marks

Study the figures given below that depict the comparative age distribution of human populations in Sweden and Rwanda. (International Data Base 2003) and answer the questions that follow :

Figure

  1. What can be inferred from the very broad base of Rwanda’s age pyramid ? Support your answer with the data provided in the figure.
  2. Sweden has an age distribution that is approximately of the same width near its base as at the apex. What does this indicate ?
  3. Name the type of age pyramid shown above for Sweden.

OR

  1. Name the type of age pyramid shown above for Rwanda.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) Rwanda – very broad base of Rwanda’s age distribution indicates a rapidly growing population / population explosion / increasing population / expanding population. Because large number of individuals are in pre-reproduction age group. [1 + 1]

(b) It indicates that number of individuals in pre-reproductive and post-reproductive age groups are same. [1]

(c) Declining age pyramid. [1]

OR

(c) Expanding age pyramid [1]

[Total: 4 marks]

Q51.2024 Main 57/4/1 · Q294 marks

Populations evolve to maximise their reproductive fitness in the habitat in which they live. Ecologists suggest, the life history of organisms have evolved in relations to the constraints imposed by the biotic and abiotic components of the habitat in which they live. This gets reflected in the population growth pattern of all organisms including humans.

Study the population growth curves shown in the given graph and answer the questions that follow :

Figure

  1. Identify the growth curves ‘A’ and ‘B’.
  2. Mention what does the dotted line in the graph indicate and state its importance also.

OR

  1. Growth curve ‘B’ shows a different pattern from that of growth curve ‘A’. Justify giving one reason.
    1. Which one of the two curves is more “realistic” and why ?
    2. Which one of the two curves is relevant in present days with respect to human population in our country and why ?
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) A – exponential growth curve / J-shaped curve [½]; B – Logistic growth curve / S-shaped curve / sigmoid curve [½]

(b) It represents carrying capacity (K) [½]; maximum possible number of individuals beyond which no growth of population is observed. [½]

OR

(b) Growth curve B is formed when resources (food and space) in nature are limited, have environmental checks, while growth curve A is formed when resources are unlimited with no environmental checks. [1]

(c) (i) ‘B’ / Logistic growth curve [½]; as resources are never infinite in nature. [½]

(ii) J-shaped curve / exponential growth [½]; it is a continuously growing population. [½]

[Total: 4 marks]

Q52.2025 Main 57/4/1 · Q335 marks
    1. Describe the population growth curve applicable in a population of any species in nature that has limited resources at its disposal.
    2. Give the equation of this growth curve.
    3. Name the growth curve and depict a graphical plot for this type of population growth.

OR

    1. Explain the Species-Area relationship within a natural forest and also predict the nature of graph when species richness is plotted against the area for a wide variety of taxa.
    2. Depict the graphical relationship between species richness and area.
    3. Give the equation of the Species-Area relationship for a wide variety of taxa on a logarithmic scale.
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Answer not available: no official marking scheme published.

Q53.2025 Main 57/5/1 · Q325 marks
    1. Describe the population growth curve applicable in a population of any species in nature that has unlimited resources at its disposal.
    2. Explain the equation of this growth curve.
    3. Name the growth curve and depict a graphical plot for this type of population growth.

OR

    1. Explain the conclusion drawn by Alexander von Humboldt during his extensive explorations in the wilderness of South American jungles.
    2. Give the equation of the Species-Area relationship.
    3. Draw a graphical representation of the relation between species richness and area for a wide variety of taxa such as birds, bats, etc.
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Answer not available: no official marking scheme published.

Q54.SQP 2025-26 · Q335 marks

Justify the following statements with suitable proof/examples:

  1. ‘competition is not limited to closely related species’
  2. ‘competition is not always dependent on resources being limiting’
  3. ‘competitive exclusion occurs in nature’
  4. ‘competing species may evolve mechanisms for co-existence’
  5. ‘competition in nature comes from what is called ‘competitive release’’

OR

A. How does a simple food chain exemplify the First Law of Thermodynamics?

B. The table below shows the number of species in different parts of the world.

Name of PlaceNumber of Bird species
Columbia1400
India1200
Northern South America1300
New York105
Denmark504

Identify the common factor in regions with a higher number of bird species and suggest at least two reasons for this greater diversity.

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Official CBSE Sample Paper Marking Scheme · 2025-26

A. Totally unrelated species could also compete for the same resource. For instance, in some shallow South American lakes, visiting flamingoes and resident fishes compete for their common food, the zooplankton in the lake. [1]

B. Resources need not be limiting for competition to occur; in interference competition, the feeding efficiency of one species might be reduced due to the interfering and inhibitory presence of the other species, even if resources (food and space) are abundant. [1]

C. Gause and other experimental ecologists believed when resources are limited the competitively superior species will eventually eliminate the other species. The Abingdon tortoise in Galapagos Islands became extinct within a decade after goats were introduced on the island, apparently due to the greater browsing efficiency of the goats. [1]

D. One such mechanism is ‘resource partitioning’. If two species compete for the same resource, they could avoid competition by choosing, for instance, different times for feeding or different foraging patterns. MacArthur showed that five closely related species of warblers living on the same tree were able to avoid competition and co-exist due to behavioural differences in their foraging activities. [1]

E. A species whose distribution is limited to a smaller area because of the presence of a competitively superior organism, is found to increase its range when the competing species is experimentally removed. For example, in the coasts of Scotland the superior barnacle (Balanus) dominates the intertidal region and excludes the smaller barnacle from that zone. [1]

OR

A. Food chain and the first law of thermodynamics. The solar energy trapped by the plants is transformed to chemical energy through the process of photosynthesis. When the plants are consumed by the animals, the organic matter (chemical energy) is transferred to the animal, some amount of this organic matter is lost as heat through the process of respiration. Death of the producers or consumers will result in the organic matter getting transferred to detritivores. As demonstrated, energy is not created nor destroyed but transformed from one form to another. [2]

B. Greater diversity is seen in regions closer to equator/ tropics (latitudinal range of 23.5° N to 23.5° S) harbour more species. [1]

Reasons for this greater diversity are: (Any two reasons)

  • Speciation is generally a function of time, unlike temperate regions subjected to frequent glaciations in the past, tropical latitudes have remained relatively undisturbed for millions of years and thus, had a long evolutionary time for species diversification
  • Tropical environments, unlike temperate ones, are less seasonal, relatively more constant and predictable. Such constant environments promote niche specialisation and lead to a greater species diversity
  • There is more solar energy available in the tropics, which contributes to higher productivity; this in turn might contribute indirectly to greater diversity.

[2, any two]

Unit X · Ecology and Environment

12. Ecosystem

Q1.2026 Main 57/1/1 · Q61 mark

In an ecosystem, different species occupy different levels and vertical distribution of species is found. This is called _________.

  1. Stratification
  2. Layering
  3. Fragmentation
  4. Population
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(A) Stratification [1 mark]

Q2.2026 Main 57/1/1 · Q161 mark

Assertion (A): In terrestrial ecosystem much larger fraction of energy flows through grazing food chain.

Reason (R): Grazing food chain may be connected to Detritus food chain at some levels.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(D) Assertion (A) is false, but Reason (R) is true. [1 mark]

Q3.2026 Main 57/2/1 · Q121 mark

Observe the contents 1, 2, 3 and 4 of soil samples I, II and III shown in the graph. If the temperature and moisture of all soil samples are identical, which soil sample(s) will show faster decomposition ?

Figure

  1. Soil Sample I
  2. Soil Sample II
  3. Soil Sample III
  4. Both Soil Samples II and III
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(C) Soil Sample III [1 mark]

Q4.2026 Main 57/3/1 · Q21 mark

Which of the following is not a functional unit of an ecosystem ?

  1. Energy flow
  2. Decomposition
  3. Stratification
  4. Productivity
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(C) Stratification [1 mark]

Q5.2026 Main 57/3/1 · Q91 mark

Every trophic level has a certain mass of living material at a particular time. What is it called ?

  1. Standing crop
  2. Primary productivity
  3. Standing state
  4. Ecological efficiency
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(A) Standing crop [1 mark]

Q6.2026 Main 57/3/1 · Q141 mark

Assertion (A): Ecosystems require a constant supply of energy to synthesize the molecules they require.

Reason (R): This is to counteract the universal tendency towards increasing disorderliness as per Second Law of Thermodynamics.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of the Assertion (A) [1 mark]

Q7.2026 Main 57/4/1 · Q161 mark

Assertion (A): The pyramid of biomass in a pond aquatic ecosystem is always inverted.

Reason (R): The biomass of consumers like fishes far exceeds that of the producers like phytoplanktons.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A) [1 mark]

Q8.2026 Supp. 57/7/1 · Q111 mark

During the decomposition process, mineralisation by micro-organisms leads to the release of which nutrients into the soil ?

  1. Inorganic nutrients from humus
  2. Both organic and inorganic nutrients from detritus
  3. Organic nutrients from humus
  4. Inorganic nutrients from detritus and formation of humus
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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(A) Inorganic nutrients from humus [1 mark]

Q9.2025 Main 57/1/1 · Q41 mark

The process of mineralization by microorganisms help in the release of :

  1. inorganic nutrients from humus.
  2. both organic and inorganic nutrients from detritus.
  3. organic nutrients from humus.
  4. inorganic nutrients from detritus and formation of humus.
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Answer not available: no official marking scheme published.

Q10.2024 Main 57/1/1 · Q121 mark

The rate of formation of new organic matter by consumers, and the biomass available for consumption of herbivores as well as decomposers are referred to as :

  1. Gross primary productivity and Net primary productivity
  2. Net primary productivity and Secondary productivity
  3. Secondary productivity and Net primary productivity
  4. Secondary productivity and Gross primary productivity
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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(C) Secondary productivity and Net primary productivity [1 mark]

Q11.2024 Main 57/2/1 · Q121 mark

The pyramid of biomass in sea is generally inverted because in sea :

  1. Biomass of fishes exceeds that of phytoplankton.
  2. Number of phytoplanktons is more.
  3. Number of phytoplanktons is less.
  4. Large fishes feed on small fishes.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(A) Biomass of fish exceeds that of phytoplankton [1 mark]

Q12.2024 Main 57/4/1 · Q161 mark

Assertion (A): Some aquatic ecosystems have inverted biomass pyramids.

Reason (R): More energy is required by the organisms occupying higher trophic levels.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). [1 mark]

Q13.2024 Supp. 57/S/1 · Q161 mark

Assertion (A): Gross primary productivity is always less than net primary productivity.

Reason (R): Rate of synthesis of organic matter by consumers is known as secondary productivity.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(D) Assertion (A) is false, but Reason (R) is true. [1 mark]

Q14.2023 Main 57/5/1 · Q71 mark

The primary productivity in an ecosystem is expressed as :

  1. gm−2 yr−1
  2. gm−2 yr
  3. K cal m−2 yr−1
  4. K cal m−2
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(A) gm−2 yr−1. [1 mark]

Q15.2023 Main 57/5/1 · Q151 mark

Assertion (A): Decomposition process is slower if detritus is rich in lignin and cutin.

Reason (R): Decomposition is largely an oxygen requiring process.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). [1 mark]

Q16.2026 Main 57/1/1 · Q212 marks

(a) Given below is a pyramid found in an ecosystem, where each bar represents the standing crop available in the trophic level.

Figure

(i) Identify the kind of pyramid and with the help of an example explain the conditions where this kind of pyramid is possible in nature.

(ii) Write any two limitations of ecological pyramids.

OR

(b) In an ecosystem there was loss of biodiversity due to some project in that area.

(i) How will biodiversity be affected ? (2 points)

(ii) List two major causes of loss of biodiversity.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(a) (i) • Inverted pyramid of biomass [½]

• Such pyramids are seen in aquatic conditions where a small standing crop of phytoplanktons supports a large standing crop of zooplankton or fish / in terrestrial ecosystem – when a large number of insects are feeding on a single tree / or any other correct example [½]

(ii) • Ecological pyramids do not take into account the same species belonging to two or more trophic levels.

• It does not accommodate a food web.

• Saprophytes are not given any place in ecological pyramids even though they play a vital role in the ecosystem.

(Any two) [½ + ½]

OR

(b) (i) • Decline in plant production.

• Lowered resistance to environmental perturbations as drought.

• Increased variability in certain ecosystem processes such as water use and pest and disease cycles

• Any other correct point (any two points) [½ + ½]

(ii) • Habitat loss and fragmentation. • Alien species invasion. • Over exploitation • Co-extinction (any two) [½ + ½]

[Total: 2 marks]

Q17.2026 Main 57/5/1 · Q212 marks

(a) Differentiate between grazing food chain and detritus food chain.

OR

(b) Ecological pyramids are widely accepted but they still have some limitations. Write any two limitations.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(a) (any other relevant difference, any two) [1×2]

Grazing food chainDetritus food chain
Starts from producers.Starts from Detritus / decomposers / dead organic matter.
It is the major conduit for energy flow in aquatic ecosystemIt is the major conduit for energy flow in terrestrial ecosystem
Less fraction of energy flow (because food chain is large)Larger fraction of energy flow (because food chain is smaller)

OR

(b) (Any two) [1×2]

  • It does not take into account the same species belonging to two or more trophic levels.
  • It assumes a simple food chain something that almost never exists in nature
  • It does not accommodate a food web.
  • Saprophytes are not given any place in ecological pyramids even though they play a vital role in the ecosystem.

[Total: 2 marks]

Q18.2026 Supp. 57/7/1 · Q202 marks

(a) Define stratification in an ecosystem.

(b) How is stratification represented in a forest ecosystem ?

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

(a) Stratification in an ecosystem refers to vertical distribution of different species occupying different levels. [½]

(b) In a forest the trees occupy the upper / top most vertical strata, shrubs the second, the herbs and the grasses the bottom layers. [½ × 3]

[Total: 2 marks]

Q19.2025 Main 57/1/1 · Q182 marks

Student to attempt either option-(A) or (B) :

(A) Analyse the following ecosystems and discuss, which will be more productive in terms of primary productivity :

A young forest, a natural old forest, a shallow polluted lake.

OR

(B) Differentiate between Net primary productivity and Gross primary productivity in an ecosystem.

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Answer not available: no official marking scheme published.

Q20.2025 Supp. 57/S/1 · Q212 marks

(a) (i) Draw a pyramid of numbers where a large number of insects are feeding on the leaves of a tree. What is the shape of this pyramid ?

(ii) Will the pyramid of energy also be of the same shape in this situation ? Give reason for your response.

OR

(b) (i) What could be the reason for the fast rate of decomposition in the tropics ?

(ii) Identify ‘x’ and ‘y’ in the figure given below representing proportionate number of major vertebrate taxa.

Figure

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(a) (i) Pyramid of numbers:

Figure

[½]

Shape is inverted [½]

(ii) No, pyramid of energy is always upright as some amount of energy is always lost (as heat) or decreases on moving from one trophic level to the next trophic level. [½ × 2]

OR

(b) (i) Warm and moist environment favours the rate of decomposition [1]

(ii) x – Mammals, y – Amphibians. [½ × 2]

[Total: 2 marks]

Q21.2024 Main 57/1/1 · Q212 marks

Study the diagram of a pyramid of biomass given below. Name the two standing crops that could be occupying level ‘A’ and level ‘B’ in it. Name this type of pyramid and the ecosystem in which it is found.

Figure

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1
  • A – Zooplankton
  • B – Phytoplankton
  • Inverted pyramid of biomass
  • Sea ecosystem / Aquatic ecosystem

[½ × 4 = 2 marks]

Q22.2024 Main 57/2/1 · Q212 marks

How is the rate of decomposition affected by the nature of detritus and temperature ?

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1
  • Detritus rich in lignin and chitin – slow decomposition; rich in nitrogen and water soluble substances like sugar – decomposition rate is faster. [½ + ½]
  • Warm environment – favour decomposition; low temperature – inhibit decomposition. [½ + ½]

[Total: 2 marks]

Q23.2024 Main 57/3/1 · Q212 marks
  1. Write important features of ‘humus’ formed during the decomposition cycle in a terrestrial ecosystem.

OR

    1. Graphically represent the relationship between species richness and area on a log-log scale for bats and fishes.
    2. Write the equation for the relationship as on a logarithmic scale.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/3/1

(a) Humus – Dark coloured, amorphous, highly resistant to microbial activities, undergoes decomposition at slow rate, colloidal, reservoir of nutrients. (Any four features) [½ × 4]

OR

(b) (i)

Figure

[1]

(ii) Log S = Log C + Z Log A [1]

[Total: 2 marks]

Q24.2024 Main 57/4/1 · Q202 marks
  1. Differentiate between grazing food chain and detritus food chain.

OR

  1. Explain Brood parasitism with the help of a suitable example.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a)

Grazing Food ChainDetritus Food Chain
1.It starts with producers.Starts from dead organic matter / detritus
2.2nd trophic level is occupied by herbivores.2nd trophic level is occupied by detritivores.
3.In aquatic ecosystem a major fraction of energy flows through this type of food chain.In terrestrial ecosystem much larger fraction of energy flows through such type of food chain.

(Any two corresponding points) [1 + 1]

OR

(b)

  • Brood parasitism is a method of parasitism in which the parasitic bird lays its eggs in the nest of the host bird and lets the host bird incubate them. [1]
  • Example – The cuckoo bird lays its eggs in the nest of crow; egg of cuckoo bird resembles crow’s egg in size and colour to reduce the chance of host bird detecting and ejecting them from nest. [½ + ½]

[Total: 2 marks]

Q25.2023 Main 57/1/1 · Q202 marks

Ecological pyramids give important information about the ecological system, but do have some limitations. List any two limitations of ecological pyramids.

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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1
  • Doesn’t take into account same species belonging to two or more trophic levels.
  • Assumes a simple food chain which never exists in nature/does not accommodate a food web.
  • Saprophytes are not given any place though they play an important ecological role.

(Any two points) [1×2]

[Total: 2 marks]

Q26.2023 Main 57/3/1 · Q192 marks

Given below is a food web that involves nine organisms.

Figure

  1. Identify two producers and two carnivores shown in the food web.
  2. Is it possible to make an ecological pyramid depicting this food web ? Give reason in support of your answer.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(a) • Producers: 1/2/3 [½]

• Carnivores: 6/7/8/9 [½]

(Half mark for any one correct producer and half mark for any one correct carnivore)

(b) • No [½]

• Pyramid does not accommodate food web [½]

//

• Yes [½]

• In the given food web no organism occupy more than one trophic level [½]

[Total: 2 marks]

Q27.SQP 2025-26 · Q212 marks

Attempt either option A or B.

A.

  1. Compare the two ecological pyramids of biomass I and II given below and explain the situations in which this is possible.

Figure

  1. Construct an ideal pyramid of energy if 200,000 joules of sunlight are available.

OR

B. A tropical rainforest in South America is home to more than 40,000 species of plants, 3,000 of fishes, 1,300 of birds, 427 of mammals, 427 of amphibians, 378 of reptiles and 1,25,000 insects, snails and worms.

From the given data, calculate the total number of known vertebrate species in the rainforest.

  1. Give a reason to justify the huge difference in the number of plant and animal species.
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Official CBSE Sample Paper Marking Scheme · 2025-26

A.

(i) The first (upright) pyramid of biomass corresponds to a terrestrial ecosystem. Producers have maximum biomass, decreasing with herbivores (primary consumer), secondary consumer and tertiary consumer. Second (inverted) pyramid refers to a small standing crop of phytoplankton supporting a large standing crop of zooplankton/aquatic ecosystem. [1]

(ii)

Figure

[1]

OR

B.

(i) 3000+1300+427+427+378 = 5532 [1]

(ii)

  • (a) Animals are mobile and can migrate to escape harsh conditions or explore new areas.
  • (b) Animals have adapted to changing environments, developing complex nervous systems and receptors. Their responses are adaptive and ensure survival. Plants, being fixed, have fewer evolutionary adaptations for water, minerals, and sunlight.

[Anyone, 1]

Q28.2025 Main 57/7/1 · Q283 marks

Name the type of food chains responsible for the flow of larger fraction of energy in an aquatic ecosystem and a terrestrial ecosystem, respectively. Mention any two differences between the two food chains.

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Answer not available: no official marking scheme published.

Q29.2024 Supp. 57/S/1 · Q273 marks

Differentiate between :

(a) Grazing food chain and Detritus food chain (Any three differences)

(b) Upright pyramid and Inverted pyramid (Any three differences)

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a)

Grazing food chainDetritus food chain
Starts with green plants called producers as first trophic levelStarts with dead organic matter and decomposers called saprotrophs
A large fraction of energy flows through aquatic ecosystemA much larger fraction of energy flows through terrestrial ecosystem
Energy for food chain comes from sunEnergy for food chain comes from organic remains or detritus

[½ × 3]

(b)

Upright pyramidInverted pyramid
It has wide base and narrow apexIt has narrow base and wide apex
Producers are more in number and biomass than herbivoresProducers are less in number and biomass than herbivores
Pyramid of energy is always uprightPyramid of biomass in sea is generally inverted

(Any other correct difference) [½ × 3]

[Total: 3 marks]

Q30.2023 Supp. 57/C/1 · Q263 marks

(a) Name the different components of a pond ecosystem and explain how they function as a unit.

OR

(b) (i) How does the Mediterranean orchid ‘Ophyrs’ ensure its pollination by bees without offering any floral reward ?

(ii) Explain the kind of population interaction observed in the following diagram.

Figure

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(a)

Abiotic components (water, soil, sunlight) [½]

Biotic components {Phytoplanktons (producers), Zooplanktons, free swimming forms, bottom dwellers and decomposers (Fungi, Bacteria, Flagellates)} [½]

Conversion of inorganic into organic material with the help of the radiant energy of the sun by autotroph like phytoplanktons (Productivity), Consumptions of autotrophs by heterotrophs like zooplanktons / free swimming forms / bottom dwellers (flow of energy), decomposition and mineralization of the dead organic matter, to release them back for reuse by the autotrophs (Cycling of nutrients). [½ × 4]

OR

(b)

(i) The Mediterranean orchid Ophrys employs ‘sexual deceit’ to get pollination done by a species of bee. One petal of its flower bears an uncanny resemblance to the female of the bee in size, colour and markings. The male bee is attracted to what it perceives as a female, ‘pseudocopulates’ with the flower, and during this process is dusted with pollen from the flower and transfers them to another flower. [½ × 4]

(ii) Brood parasitism; the parasitic bird lays its eggs in the nest of its host and lets the host incubate them. [½ + ½]

[Total: 3 marks]

Q31.2026 Main 57/2/1 · Q304 marks

Below is shown a pond that represents a self-sustaining natural ecosystem with both biotic and abiotic components interacting in a balanced food web.

Figure

Based on the diagram, answer the questions :

(a) Mention two abiotic components of a pond ecosystem. [1]

(b) Differentiate between the roles of producers and decomposers in such an ecosystem. [2]

(c) Draw a simple food chain operating in such an ecosystem and label each trophic level. [1]

OR

(c) Why is transfer of energy not 100% efficient between trophic levels ? [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(a) (Any two) Water, Soil, sunlight, temperature, organic substances, inorganic substances [½+½]

(b)

Role of producersRole of decomposers
Conversion of inorganic molecules into organic material with the help of sunlight through the process of photosynthesisBreak down of dead and decayed organic material into simpler inorganic substances through the process of decomposition

[2]

(c) Food Chain: Phytoplanktons → Zooplanktons → Fish

Trophic Levels: Producers / 1st trophic level → Primary consumer / 2nd trophic level → Secondary consumer / 3rd trophic level [1]

OR

(c) Some energy is dissipated and lost as heat to the environment / only 10% of energy being transferred to the next trophic level. [1]

[Total: 4 marks]

Unit X · Ecology and Environment

13. Biodiversity and Conservation

Q1.2026 Main 57/1/1 · Q41 mark

Select the mismatched pair :

  1. Mega diversity countries in the world – 12
  2. Genetically different strains of rice in India – Less than 1000
  3. Robert May’s estimation on global species diversity – 7 million species worldwide
  4. Zoological parks – Ex-situ conservation
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(B) Genetically different strains of rice in India – less than 1000 [1 mark]

Q2.2026 Main 57/1/1 · Q71 mark

Which of the following sacred groves is found in Meghalaya ?

  1. Jaintia hills
  2. Bastar
  3. Chanda
  4. Sarguja
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Official CBSE Marking Scheme · 2026 (Main), Set 57/1/1

(A) Jaintia hills [1 mark]

Q3.2026 Main 57/2/1 · Q21 mark

Select the correct option :

  1. Tropics between 23.5° N – 23.5° S – Higher biodiversity than temperate areas
  2. Columbia – 105 bird species
  3. New York – 1400 bird species
  4. India – 2000 bird species
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(A) Tropics between 23.5°N – 23.5°S – Higher biodiversity than temperate areas [1 mark]

Q4.2026 Main 57/2/1 · Q151 mark

Assertion (A): IUCN Red List helps in identifying and protecting endangered species worldwide.

Reason (R): The list includes the species facing risk of extinction.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation for Assertion (A).
  3. Assertion (A) is true, Reason (R) is false.
  4. Assertion (A) is false, Reason (R) is true.
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Official CBSE Marking Scheme · 2026 (Main), Set 57/2/1

(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A). [1 mark]

Q5.2026 Main 57/5/1 · Q121 mark

Which of the following fish led to the extinction of an ecologically unique assemblage of more than 200 species of cichlid fish in Lake Victoria of East Africa ?

  1. Catla Catla
  2. Dogfish
  3. Nile Perch
  4. African Catfish
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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1

(C) Nile perch [1 mark]

Q6.2025 Main 57/7/1 · Q111 mark

India has only 2·4% of the world’s land area but its share of the global species diversity is :

  1. 8·1%
  2. 12·9%
  3. 7·3%
  4. 5·1%
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Answer not available: no official marking scheme published.

Q7.2024 Main 57/2/1 · Q161 mark

Assertion (A): Loss of biodiversity can occur due to overexploitation of resources.

Reason (R): Introduction of Clarias gariepinus in Indian rivers has led to a decline in native Indian fishes.

  1. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  2. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  3. Assertion (A) is true, but Reason (R) is false.
  4. Assertion (A) is false, but Reason (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). [1 mark]

Q8.2024 Main 57/5/1 · Q161 mark

Assertion (A): Communities that comprise of more species tend to be more stable.

Reason (R): A higher number of species results in less year to year variation in total biomass.

  1. Both (A) and (R) are true and (R) is the correct explanation of (A).
  2. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
  3. (A) is true, but (R) is false.
  4. (A) is false, but (R) is true.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(A) Both (A) and (R) are true and (R) is the correct explanation of (A). [1 mark]

Q9.2023 Main 57/3/1 · Q81 mark

Human settlement often leads to habitat loss which leads to fragmentation, forming smaller patches of habitats. Select the statements that describe how a small patch differs from a large patch of the same habitat.

  1. Invasive species will never be seen here.
  2. Population of large animals decreases.
  3. Biodiversity decreases.
  4. Competition from surrounding habitats increases.
  1. (ii), (iii) and (iv) only
  2. (ii) and (iv) only
  3. (i) and (iii) only
  4. (i), (ii) and (iii) only
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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

(A) (ii), (iii) and (iv) only [1 mark]

Q10.2023 Main 57/5/1 · Q91 mark

The IUCN Red Data List (2004) in the last 500 years documents the extinction of nearly 784 species including :

  1. 330 invertebrates
  2. 338 invertebrates
  3. 359 invertebrates
  4. 362 invertebrates
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(C) 359 invertebrates. [1 mark]

Q11.2023 Supp. 57/C/1 · Q111 mark

Choose the pair that is incorrectly matched.

  1. Khasi and Jaintia Hills – Meghalaya
  2. Aravalli Hills – Karnataka
  3. Western Ghats – Maharashtra
  4. Chanda and Bastar areas – Madhya Pradesh
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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(B) Aravalli Hills – Karnataka [1 mark]

Q12.2026 Main 57/4/1 · Q212 marks

(a) Biodiversity must be conserved as it plays an important role in many ecosystem services that nature provides. Explain any two ecosystem services.

OR

(b) (i) State, what does ‘standing crop’ of a trophic level represent.

(ii) List any two ways of measuring the standing crop of a trophic level.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) (Any two) [1+1]

  • 20% of the total oxygen in the earth’s atmosphere produced by Amazon forest.
  • Pollination without which plants cannot give us fruits and seeds is another service which ecosystem provides through honeybees or bumblebees or birds or and bats.
  • Provide aesthetic cultural and spiritual values.
  • Provide wildlife habitat and help in their sustenance.
  • Maintain balance in various components of ecosystem through photosynthesis, cycling of nutrients or generating fertile soil or mitigating droughts or controlling flood.
  • Obtaining products of economic importance.

OR

(b) (i) ‘Standing crop’ of a trophic level represents the mass of living matter or biomass at a given time. [1]

(ii) Measured as

  • biomass i.e. mass of living organisms (dry weights) in an unit area. [½]
  • number per unit area. [½]

[Total: 2 marks]

Q13.2024 Main 57/4/1 · Q212 marks
  1. Biodiversity hotspots cover less than 2% of Earth’s land area. Strict protection of these areas can reduce the rate of ongoing extinctions. Explain.
  2. Name any two hotspots in India.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/4/1

(a) Biodiversity hotspots are the regions with very high level of species richness, and high degree of endemism. [½ + ½]

(b) Western Ghats, The Himalayas, Indo-Burma region (Any two) [½ + ½]

[Total: 2 marks]

Q14.2023 Main 57/1/1 · Q212 marks

India is the seventh largest country in the world in terms of total land area including land and water. Write the value of the land area of our country (in terms of percentage) of the world. Mention then, what makes India one of the 12 mega diversity countries of the world.

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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

2.4 percent [1]

8.1 percent share of the global species diversity [1]

[Total: 2 marks]

Q15.2023 Main 57/3/1 · Q202 marks

Illustrate with the help of an example how introduction of an alien species turns invasive and causes decline of an indigenous species.

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Official CBSE Marking Scheme · 2023 (Main), Set 57/3/1

Introduction of Nile perch introduced into Lake Victoria, led eventually to the extinction of more than 200 species of Cichlid fish [1+1]

//

African catfish/Clarias gariepinus, causes threat to indigenous catfish [1+1]

//

Parthenium/carrot grass/Lantana/water hyacinth/Eichhornia sp., causes threat to native species [1+1]

[Total: 2 marks]

Q16.2022 Main 57/1/1 · Q62 marks

The histogram given below representing the data for annual shark harvest in the great barrier reef / coral reef located on the east coast of Queensland, Australia. Study the histogram and answer the questions that follow.

Figure

(a) Write your interpretation of the data given.

(b) Write the impact on the biodiversity of the area that you can interpret on the basis of given data.

OR

"Stability of community depends upon its species richness." How did David Tilman show this experimentally?

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

(a) Shark harvest increases over the years from 1992 to 2003.

(b) Biodiversity decreases due to over exploitation / excessive harvesting [1 + 1]

OR

Tilman conducted experiments in out door plots, plots with more species shows less year to year variation in biomass (more stability). [1 + 1]

[Total: 2 marks]

Q17.2022 Main 57/2/1 · Q62 marks

(a) Study the given graphs (X) and (Y) depicting the annual variation in solar radiation on Earth from January to December and answer the undermentioned questions.

Figure

  1. Identify the graph that depicts (I) tropical region, and (II) temperate region, respectively.
  2. Which of the two regions, (X) or (Y), will show high biological diversity and why? Give two reasons.

OR

(b) (i) When prickly pear cactus was introduced into Australia in early 1920s, it caused havoc and ecosystem instability by achieving very high population densities. Explain the reason for its rapid spread into millions of hectares of rangeland.

(ii) State the importance of 'Prey-predator' relationship in a habitat.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

(a)

  1. (I) Y, (II) X [½]
    Note: ½ mark to be awarded if student writes either one or both parts correctly in 6(a)(i).
  2. Y will show high biological diversity [½]; Reason: less seasonal / constant and predictable environment, more solar energy available for species diversification [½ + ½]

OR

(b) (i) Does not have its natural predators. [1]

(ii) Predators act as conduits for energy transfer across trophic levels, they keep prey population under control, predators help in maintaining species diversity in a community by reducing competition among prey species, if a predator is too efficient and over exploits its prey then the prey become extinct followed by predator / predators are prudent in nature (any two) [½ + ½]

[Total: 2 marks]

Q18.SQP 2026-27 · Q212 marks

Attempt either option A or B.

A. The table below is a hypothetical data on Seasonal Temperatures seen over two different zones

RegionTemp. in Spring (°C)Temp. in Summer (°C)Temp. in Autumn (°C)Temp. in Winter (°C)Primary Productivity (GPP) (g/m2/year)Respiration (R) (g/m2/year)
Region X2728272635001300
Region Y12201442000800
  1. Justify which region will have greater species diversity based on the given data?
  2. Calculate the net primary productivity (NPP) of the region mentioned in (I) and give its significance?

OR

B. Following is a hypothetical data on Plant Pollinator interaction

PlantMain PollinatorSeed set (%) with pollinatorSeed set (%) without pollinator
P1H1 (Bee)8040
P2H2 (Butterfly)7545
P3H3 (Moth)8535
  1. From the data above, justify which plant shows the highest dependence on its main pollinator?
  2. What kind of ecological interaction is shown in the table above? Define it in one sentence.
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Official CBSE Sample Paper Marking Scheme · 2026-27

A.

I. Region X will have greater species diversity/Tropical region because Temperatures are near-constant across seasons in X (27–28–27–26 °C), which supports stable conditions / higher diversity/ Higher productivity/GPP 3500/ higher NPP in X vs lower productivity in Y. [1]

II. NPP of Region X = GPP − R = 3500 − 1300 = 2200 g/m2/year. Net primary productivity is the available biomass for the consumption to heterotrophs (herbivores and decomposers) [1]

OR

B.

I. P3 → drop of over 50 percentage points without main pollinator/ other plants show comparatively less drop in seed set without pollinator. [1]

II. The relationship is Mutualism as both species benefit; here, plants gain pollination and pollinators obtain nectar/food. [1]

Q19.2026 Main 57/4/1 · Q283 marks

There are many animals, which have become extinct in the wild, but continue to be maintained in zoological parks. Answer the questions that follow :

(a) What type of bio-diversity conservation is observed in this case ? [½]

(b) Explain any other two ways which help in this type of conservation. [2]

(c) Besides zoological parks, mention one place that serves this purpose. [½]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/4/1

(a) Ex-situ conservation. [½]

(b) (Any two) [1+1]

  • Cryopreservation – (–196°C) sperms or eggs or animal tissues or gametes and embryos of threatened species can be stored.
  • In vitro using tissue culture methods plants can be propagated
  • Seed banks in which seeds of different genetic strains of commercially important plants can be kept.
  • Botanical Gardens in which threatened plants can be conserved in special settings.
  • Any other correct option with explanation.

(c) Botanical Gardens, Wildlife safari parks (Any one) [½]

[Total: 3 marks]

Q20.2026 Main 57/5/1 · Q283 marks

Explain the level of biodiversity at genetic, species and ecological levels with the help of one example each.

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Official CBSE Marking Scheme · 2026 (Main), Set 57/5/1
  • Genetic diversity – It refers to the high diversity of a species at the genetic level over its distributional range, e.g. India has more than 50,000 genetically different strains of rice / The genetic variation shown by the medicinal plant Rauwolfia vomitoria growing in different Himalayan ranges might be in terms of the potency and concentration of the active chemical (reserpine) that the plant produces / India has about 1000 varieties of mango / any other correct example. [½+½]
  • Species diversity – Diversity at species level / the number of different species in a given region, e.g. Western Ghats have a greater amphibian species diversity than Eastern Ghats / any other correct example. [½+½]
  • Ecological diversity – Diversity at the ecosystem level / presence of different types of ecosystems in a particular area. e.g. India has a number of ecosystems like rain forests, coral reefs, deserts / any other correct example. [½+½]

[Total: 3 marks]

Q21.2026 Supp. 57/7/1 · Q283 marks

‘Indo-Burma’ is considered as a hotspot. What is meant by the term ‘hotspots’ in biodiversity ? List two criteria used for determining a ‘hotspot’. Name any two hotspots of India.

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Official CBSE Marking Scheme · 2026 (Supplementary/Compartment), Set 57/7/1

A hotspot is a region with very high levels of species richness and high degree of endemism. [1]

Criteria used for determining hotspot – high levels of species richness [½]; high degree of endemism [½]

Western Ghats, Eastern Himalayas, Indo-Burma (any two) [½ × 2]

[Total: 3 marks]

Q22.2025 Main 57/1/1 · Q243 marks

(a) Indiscriminate human activities such as alien species invasion, fragmentation and habitat loss have accelerated the loss of biodiversity. Justify by taking one example for each.

(b) State the importance of the following :

  1. IUCN Red data list
  2. Hot spots in conservation of biodiversity
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Answer not available: no official marking scheme published.

Q23.2025 Main 57/4/1 · Q273 marks

According to a recent wildlife report, the biggest threat to the tiger’s survival in Mudumalai Tiger Reserve (MTR) was found to be a small, beautiful flower, Lantana camara, a tropical American shrub, that invaded 40% of India’s tiger range. Tamil Nadu department’s Lantana weed eradication drive helped to restore the dying MTR thereby also reducing human-wildlife conflicts. MTR is home to 25 species of grasses and legumes.

Answer the given questions based on the information given above.

  1. Explain how did the removal of Lantana help in restoring the dying Mudumalai Tiger Reserve.
  2. Why is the invasion of Lantana camara a cause of concern in MTR.
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Answer not available: no official marking scheme published.

Q24.2025 Main 57/5/1 · Q273 marks

Explain how the loss of habitat and fragmentation drives plants and animals to extinction with the help of an example of habitat loss in the Tropical Rain Forest. Also write the effect of fragmentation of a habitat on the population decline.

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Answer not available: no official marking scheme published.

Q25.2025 Main 57/6/1 · Q233 marks
  1. Alien species are highly invasive and are a threat to indigenous species. Substantiate this statement with the help of any two examples.
  2. State any two criteria for determining biodiversity hotspots.
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Answer not available: no official marking scheme published.

Q26.2024 Main 57/2/1 · Q273 marks

Given below is a pie chart representing global diversity of vertebrates.

Figure

(a) Redraw the pie chart identifying the groups ‘A’, ‘B’, ‘C’ and ‘D’ in their respective positions.

(b) Mention two examples of recently extinct animals.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/2/1

(a)

Figure

(A – Mammals, B – Birds, C – Reptiles, D – Amphibians) [½ × 4]

(b) Dodo, Quagga, Thylacine, Steller’s Sea Cow, 3 subspecies – Bali / Javan / Caspian of tiger (or any other correct example – any two) [½ + ½]

[Total: 3 marks]

Q27.2024 Main 57/5/1 · Q233 marks
  1. Tropical regions harbour more species than the temperate regions. How have biologists tried to explain this in their own ways ? Explain.

OR

    1. What does an ecological pyramid represent ?
    2. The Ecological pyramids may have an ‘upright’ or an ‘inverted’ shape. Justify with the help of suitable examples.
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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

(a)

  • The tropical region has less seasonal and relatively more constant and predictable environment that promotes niche specialisation leads to greater species diversity. [1]
  • In the tropical region more availability of solar energy which contributes to higher productivity. [1]
  • Tropical regions are not subjected to frequent glaciation in the past and remain undisturbed for millions of years hence had a long evolutionary time for species diversification. [1]

OR

(b) (i) Ecological pyramid is the diagrammatic representation of relationship between organisms at different trophic levels (in terms of energy / biomass / number of organisms in an ecosystem). [1]

(ii)

  • Pyramid is upright because number or biomass or energy is more in producers than in the consumers [½]
  • Example: pyramid in grassland ecosystem / any other relevant example to support the shape of pyramid [½]
  • Pyramid is inverted because number or biomass is less in producers than in the consumers [½]
  • Example: in terms of biomass, standing crop of phytoplankton supports large standing crop of zooplanktons, or large number of insects feeding on a big tree [½]

(Diagrammatic representation of any correct example in relevance to the shape of pyramids should be considered.)

[Total: 3 marks]

Q28.2024 Main 57/5/1 · Q283 marks

The world is facing accelerated rates of species extinction largely due to human activities. Explain any three human activities responsible for accelerated rates of species extinction.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/5/1

Any three human activities with relevant explanation: [½ × 6]

  • Habitat loss and fragmentation – degradation of habitat such as cutting and clearing of (Amazon rain) forest for developing cultivable land leads to the fragmentation of habitat will affect the population of organisms living in that area / any other correct explanation
  • Overexploitation of natural resources – Steller’s Sea cow or passenger pigeon or marine fish / any other correct explanation
  • Alien species invasion – Nile perch introduced in lake Victoria in east Africa led eventually to the extinction of 200 species of Cichlid fish, or introduction of Clarias gariepinus (African catfish) for aquaculture purposes is posing a threat to indigenous catfishes, or Parthenium (carrot grass), Lantana or Eichhornia (water hyacinth) introduced unintentionally causing decline of indigenous species / (any other correct explanation with example can be considered)
  • Co-extinctions of species – extinction of any organism or plant or animal or plant-pollinator mutualism or fish-parasite relationship will lead to extinction of other (obligatory) species
  • Degradation of habitat by pollution – threatens the survival of many species / mammals / (migratory) birds.
  • The colonization of (tropical Pacific) islands by humans led to extinction of (more than 2000 species) native birds.

[Total: 3 marks]

Q29.2024 Supp. 57/S/1 · Q283 marks

(a) (i) What are “biodiversity hotspot” regions ? Mention any two criteria used for determining any region as a “hotspot”.

(ii) Name any two hotspots of India.

OR

(b) Explain giving three reasons, why tropics show greatest level of species diversity.

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Official CBSE Marking Scheme · 2024 (Supplementary/Compartment), Set 57/S/1

(a) (i) Biodiversity hotspots are the regions with very high level of species richness and high degree of endemism. [1]

Criteria used for determining any region as a hotspot – species richness, endemism. [½ + ½]

(ii) Western Ghats and Sri Lanka, Indo-Burma, Himalayas. (Any two) [½ + ½]

OR

(b) Tropical latitudes have remained undisturbed and have had a long evolutionary time for species diversification; tropical environments have less seasonal variations with more constant and predictable environmental conditions to promote niche specialisation for greater diversity; there is more availability of solar energy which contributes to higher productivity. [1 × 3]

[Total: 3 marks]

Q30.2023 Main 57/5/1 · Q283 marks

“Biodiversity plays a major role in many ecosystem services that nature provides.”

  1. Describe any two broadly utilatarian arguments to justify the given statement.
  2. State one ethical reason of conserving biodiversity.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/5/1

(a) Oxygen production, pollination, aesthetic pleasure of thick woods and flowers, bird watching, mitigate droughts and floods, cycle nutrients, generate fertile soils, provide wildlife habitat, maintain biodiversity, provide storage site for carbon, provide aesthetic/cultural/spiritual values, any other relevant point (Any two) [1×2]

(b) Philosophical or spiritual realisation that every species has an intrinsic value/Moral duty to care for their well-being/pass on our biological legacy in good order to future generations. [1]

[Total: 3 marks]

Q31.2022 Main 57/1/1 · Q73 marks

The data collected based on the survey conducted for species richness of group of mammals, in three different climatic regions of the world is shown in the bar graph given below.

Panama has nearly 560 species of mammals, Canada has nearly 301 species of mammals and Denmark has 67 species of mammals.

Figure

  1. Based on the species, richness, identify the location of these countries in the respective climatic regions given.
  2. Plants and animals do not have a uniform diversity in the world. Write the term given to this pattern of diversity and why?
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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1
  1. Panama – Tropic; Canada – Temperate; Denmark – Polar [½ × 3]
  2. Latitudinal gradient, species diversity decreases as we move away from equator toward poles / species diversity vary due to different climatic conditions present at different places of world. [½ + 1]

[Total: 3 marks]

Q32.2022 Main 57/1/1 · Q83 marks

Bio-diversification of life started to occur almost 3 billion years ago. Since then new species have been evolving and then disappearing en masse from earth.

(a) How many episodes of mass extinctions of species have already taken place and which one is in progress in the current era?

(b) How is current episode in progress different from the previous episodes and why? Explain.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/1/1

(a) Five extinctions have already occurred [½]; sixth is in progress [½]

(b) Sixth extinction is much faster / sixth extinction is 100 to 1000 times faster [1]; Human activities like industrialisation, loss of habitat, over exploitation, land reforms (any two) [½ + ½]

[Total: 3 marks]

Q33.2022 Main 57/2/1 · Q103 marks

(a) "Some species in an ecosystem exert greater influence in driving major ecosystem services than others." How did Paul Ehrlich explain this concept in his 'rivet popper hypothesis'?

OR

(b) (i) How has the introduction of Nile Perch into Lake Victoria in East Africa had a devastating effect on the indigenous species in the lake?

(ii) Name two invasive weed species which have posed a threat to the existence of our native species.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

(a) Paul Ehrlich compared airplane with ecosystem, in an airplane (ecosystem) all parts are joined together using thousands of rivets (species), if every passenger travelling in it starts popping a rivet to take home (causing a species to become extinct), it may not affect flight safety (proper functioning of ecosystem) initially, if more and more rivets are removed then the plane becomes dangerously weak over a period of time, loss of rivets on the wings (key species that drives major ecosystem functions) is a more serious threat to flight [½ × 6]

OR

(b) (i) Extinction of an ecologically unique assemblage of more than 200 species of cichlid fish in the lake [1]

(ii) Carrot grass / (Parthenium), Lantana, water hyacinth / (Eichhornia) (or any correct example) [1 + 1]

[Total: 3 marks]

Q34.2022 Main 57/2/1 · Q113 marks

Given below is a picture of a natural large habitat that is being continuously broken into smaller patches. This process is continuing, increasing and expanding over time due to human impact, thereby posing a great risk to the existence of species in the habitat.

Figure

(a) Do you agree with the above statement? Give two suitable reasons in support of your answer.

(b) Write the ecological term used for such patches formed in an original habitat.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/2/1

(a) Yes [1]; Degradation of habitats by fragmentation threatens the survival of many species / mammals / birds / migratory birds which require large territories (are badly affected), leading to decline of population (or any other relevant reason) [½ + ½]

(b) Fragments / Fragmentation [1]

[Total: 3 marks]

Q35.2022 Main 57/3/1 · Q103 marks

(a) Write the inference drawn by Alexander von Humboldt after his extensive explorations of South Americal jungle.

(b) Study the graph given below:

Figure

As per Alexander von Humboldt, what do the symbols S, A, Z and C in the graph stand for, in respect of a species and area relationship?

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

(a) Species richness increased with increasing explored area up to a limit [1]

(b) S = Species richness; A = Area; Z = Slope of the line / regression coefficient; C = Y-intercept [½ × 4]

[Total: 3 marks]

Q36.2022 Main 57/3/1 · Q113 marks

(a) Explain the concept of "co-extinction" by taking two examples.

OR

(b) "Forests provide intangible benefits to us." Explain by taking three different areas, how.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/3/1

(a) When a species becomes extinct, the plant and animal species associated with it in an obligatory way also becomes extinct. [1]

Examples (any two, or any other correct example) [1 + 1]:

  • When a host (fish) species becomes extinct, the plant and animal species associated with it in an obligatory way also become extinct.
  • Coevolved plant-pollinator mutualism where extinction of one leads to the extinction of other (any other example) / The coevolved orchid Ophrys and bee pollinator.

OR

(b) Any three of the following [1 + 1 + 1]:

  • Source of oxygen in the earth's atmosphere;
  • Pollination (without which plants cannot produce fruits or seeds) is another service provided by forest through different pollinators;
  • Source of aesthetic pleasures (walking through thick woods / watching spring flowers in full bloom / waking up of bulbul's song in the morning);
  • Human derive countless economic benefits from nature – food (cereals, pulses, fruits);
  • Source of firewood / fibre / construction material / industrial products (tannins / lubricants / dyes / resins / perfumes);
  • Contributes to the traditional medicines used by native people around the world;
  • Philosophically or spiritually – we need to realize that every species has an intrinsic value – even if it may not be of current or any economic value to us;
  • This is the ethical argument for conserving forests which relates to what we owe to the millions of plants / animals / microbe species with whom we share this planet.

Note: According to Hindi version, all benefits of forest are included.

[Total: 3 marks]

Q37.2022 Main 57/4/1 · Q73 marks

(a) The graph given below shows species-area relationship of a certain region.

Figure

  1. Study the graph and explain what it represents.
  2. After a while, a small area was taken for constructing a road which divided the region into two. Write the impact this construction would have on species richness of the region.

OR

(b) How did Dr. David Tilman relate experimentally, the stability of a community and its species richness? Explain.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

(a) (i) “Within a region species richness increased with increasing explored area, but only up to a limit.” [1+1]

(ii) Very gradually impact on species diversity will be visible / species diversity will be gradually reduced / negative impact on species diversity. [1]

OR

(b) Dr. Tilman carried out long-term experiments in outdoor plots, he observed plots with more species showed less year-to-year variation in biomass, he showed increased diversity (species richness) contributed to higher productivity. [1+1+1]

[Total: 3 marks]

Q38.2022 Main 57/4/1 · Q83 marks

What are the consequences of loss of biodiversity in a region? Explain.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/4/1

Decline in flora/fauna, lower resistance to environmental perturbations such as drought, increased variability in certain ecosystem process such as plant productivity / water use / pest / disease cycle [1×3] (Any three)

[Total: 3 marks]

Q39.2022 Main 57/5/1 · Q103 marks

(a) Given below is an equation describing the Species-Area relationship between species richness and area for a wide variety of taxa as angiosperm plants, birds, bats etc.

S = CAZ

  1. Give a graphical representation of the given equation showing Species-Area relationship.
  2. What does ‘S’ represent in the given equation?
  3. What is the value of ‘Z’ (regression coefficient) for frugivorous birds and mammals in the tropical forests of different continents?

OR

(b) Given below is a ‘pie chart’ representing the global biodiversity: proportionate number of species of major taxa.

Figure

  1. Identify (X) and (Y) in the given ‘pie chart’.
  2. “Extinction of species across taxa are not random.” Which group amongst the vertebrates is more vulnerable to extinction.
  3. Give one example each of recent extinctions of species in Russia, Mauritius and Australia.
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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

(a) (i) (Graph: rectangular hyperbola of species richness against area, curve rising and levelling off) [1]

Figure

(ii) S = Species richness [1]

(iii) 1·15 [1]

OR

(b) (i) X → Fishes [½]   Y → Amphibians [½]

(ii) Y → Amphibians [½]

(iii) Russia — Steller’s sea cow [½]; Mauritius — Dodo [½]; Australia — Thylacine [½] (any other correct example)

[Total: 3 marks]

Q40.2022 Main 57/5/1 · Q123 marks

(a) Enlist two criteria that are used to identify a region for maximum protection as ‘Biodiversity hotspots’.

(b) Name any two “hotspot” regions in our country.

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Official CBSE Marking Scheme · 2022 (Main), Set 57/5/1

(a) – High degree of endemism [1] – high levels of species richness [1]

(b) Western Ghats, Himalaya, Indo-Burma (Any two) [½+½]

[Total: 3 marks]

Q41.2022 Supp. 57/6/1 · Q103 marks

Study the pie chart given below, representing the global diversity: proportionate number of species of major taxa and answer the following questions based on it.

Figure

(a) Identify ‘X’ and ‘Y’ in the given pie chart.

(b) Which one of the two, ‘X’ or ‘Y’, is the most species-rich taxonomic group and by what percentage?

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

(a) X – Insects [1]   Y – Molluscs [1]

(b) ● X – makes most species rich taxonomic group [½]

● more than 70% of the total [½]

[Total: 3 marks]

Q42.2022 Supp. 57/6/1 · Q123 marks

How have the following factors affected the biodiversity of an area? Explain.

(a) Habitat loss and fragmentation

(b) Over-exploitation

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Official CBSE Marking Scheme · 2022 (Supplementary/Compartment), Set 57/6/1

(a) ● loss of habitat leads to loss of biodiversity and threatens the survival of plants and animals to extinction. [1]

● Mammals and birds requiring large territories and certain animals with migratory habits are badly affected due to fragmentation, leading to population decline. [1]

(b) Many commercially important species are overharvested, endangering their existence which may lead to their extinction. [1]

[Total: 3 marks]

Q43.2026 Main 57/3/1 · Q304 marks

Read the following passage and answer the questions that follow :

India is one of the megadiverse countries housing around 8·1 per cent of global species diversity, although its land area is only 2·4 per cent of the world’s land area. Many of the species are highly threatened due to human activities like deforestation, mining and habitat fragmentation. Laws like Wildlife (Protection) Act, 1972 were enacted by the Government of India to preserve our biological wealth. Various conservation measures are being implemented to save the threatened species.

The following bar graph shows the number of species conserved under different biodiversity conservation methods.

Figure

Study the graph and answer the questions.

(a) Which method conserves the highest number of species ? Is it ex situ or in situ conservation ? [1]

(b) Which other methods shown in the diagram are opposite to the one identified by you in question (a) ? How are these two conservation approaches different ? [2]

(c) (i) Write two features of Biodiversity hotspots. [1]

OR

(c) (ii) To which category do sacred groves belong and how do they help in bio-conservation ? [1]

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Official CBSE Marking Scheme · 2026 (Main), Set 57/3/1

(a) Wild life sanctuaries conserve highest number of species, In-situ conservation. [½+½]

(b) Zoological Parks and Botanical Gardens are different, They are ex-situ. [½+½]

(Any one difference) [1]

In situ ConservationEx situ Conservation
1) Conserves the whole ecosystem and biodiversity at all levels is protected.Conserves animals or plants that are endangered/threatened and given urgent measures to save them from extinction.
2) Organisms are protected in their natural habitat.Organisms are taken out from their natural habitat and placed in special settings where they can be protected and given special care.

(Any other relevant difference)

(c) (i) High level of species richness, high degree of endemism [½+½]

OR

(ii) In-situ Conservation. [½]

The tracts of forests were set aside and all trees and wild life within were venerated and given total protection. [½]

[Total: 4 marks]

Q44.2025 Supp. 57/S/1 · Q304 marks

Biodiversity shows latitudinal and altitudinal gradients. There is little biodiversity at poles. It increases in temperate areas but reaches the maximum in tropics (latitudinal range of 23·5° N to 23·5° S). It is because the tropical rainforests have favourable environmental conditions not only for speciation, but also for supporting both variety and number of organisms. Harsh conditions exist in temperate areas during the cold season only. Number of vascular species is 118 – 236/0·1 ha in tropical forests and 21 – 48 species/0·1 ha in temperate forests. Their number would be 0·0 – 10·0 species/0·1 ha in arctic regions. Such a correlation occurs also in case of other taxonomic groups like ants, butterflies, birds, etc.

Figure

Based on the above information and graph, answer the following questions :

(a) What does the graph depict ?

(b) With the help of graph, show the species-area relationship for a wide variety of taxa in a forest. Write an equation that shows this relationship on a logarithm scale.

(c) (i) Explain the correlation between latitudinal gradient and number of plant species.

OR

(c) (ii) Tropics are considered to be more stable where species continue to flourish and evolve undisturbed for millions of years. Give any two reasons.

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Official CBSE Marking Scheme · 2025 (Supplementary/Compartment), Set 57/S/1

(a) Graph represents that as the latitude increases from 0° N to 80° N the number of ant species decreases from 200 to 25 species / latitudinal gradient of biodiversity / tropical regions have higher biodiversity than temperate regions [1]

(b)

Figure

[1]

Equation: log S = log C + Z log A [1]

S = Species richness; Z = Slope of the line, C = Y-intercept

(c) (i) • Plant species number is highest at the low latitudes and decreases towards the poles

• There is little biodiversity at poles though it increases in temperate areas but reaches the maximum in tropics

• Number of plant species are 118–236 / 0.1 ha in tropical forest and decreases to 21–48 species / 0.1 ha in temperate forest and further very low 0.0–10 species / 0.1 ha in arctic regions

(Any one correlation) [1]

OR

(c) (ii) Tropical latitudes remained undisturbed for many years; tropics are less seasonal; constant environment; more solar energy; higher productivity; greater biodiversity (any two) [½ + ½]

[Total: 4 marks]

Q45.2024 Main 57/1/1 · Q294 marks

Read the following passage and answer the questions that follow.

Isn’t it incredible that India’s land area is only 2.4 per cent of the world’s total land area whereas its share of the global species diversity is an impressive 8.1 per cent ! However, in these estimates of species, prokaryotes do not figure anywhere.

Biologists are always keen on collecting data with respect to species diversity observed in different regions of the world. The data collected based on the survey conducted for species richness of groups of mammals in three different regions of the world is shown in the bar graph given below :

Figure

(a) Why is the species richness maximum in Region III in the bar graph ?

OR

(a) Why is the species richness minimum in Region I in the bar graph ?

(b) Plants and animals do not have uniform diversity in the world but show rather uneven distribution. Mention what this kind of diversity is referred to as.

(c) Why is it that prokaryotes do not have an estimated number of their species diversity as seen in plants and animals ? Explain.

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Official CBSE Marking Scheme · 2024 (Main), Set 57/1/1

(a) This region is less seasonal with constant and more predictable environment / More solar energy so higher productivity and higher diversity / it represents tropical latitudes which remain relatively undisturbed for millions of years and had a long evolutionary time for species diversification. [1]

OR

(a) Region I represents temperate region subjected to frequent glaciation and gets lesser evolutionary time for species diversification / has more seasonal with less constant and less predictable environment which lead to lower species diversification / has lower solar energy available which reduces productivity and in turn contributes to lesser diversity. [1]

(b) Latitudinal gradient in diversity [1]

(c) Conventional taxonomic methods are not suitable for identifying microbial species, and many species are not culturable under laboratory conditions. [1 + 1]

[Total: 4 marks]

Q46.2023 Main 57/1/1 · Q304 marks

Study the diagrammatic representation given below of the Earth with regions marked ‘A’ and ‘B’ respectively. Answer the questions that follow.

Figure

  1. Write the observations made regarding the species diversity when moving from region ‘A’ to region ‘B’. Give two reasons also.
  2. Stating the reason, mention the approximate number of bird species recorded in India.

OR

  1. Name the region in the world that records the greatest biodiversity and mention why.
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Official CBSE Marking Scheme · 2023 (Main), Set 57/1/1

(a) Species diversity decreases as we move from region A to region B. [1]

Reasons: less constant mean annual temperature, lesser habitable land area, availability of lesser solar energy, lesser productivity, any other correct reason in ‘B’ region. (Any two) [1+1]

(b) More than 1200 species of birds, Indian land mass being largely in the tropical latitudes. [½+½]

OR

(b) Amazonian rainforest (in South America), mainly being in tropical region. [½+½]

[Total: 4 marks]

Q47.2023 Supp. 57/C/1 · Q304 marks

The diversity of plants and animals is not uniform throughout the world but shows a rather uneven distribution. Study carefully the illustrations ‘A’ and ‘B’ given below. Answer the questions based on them.

Figure

(i) (a) Mention the pattern of biodiversity that you observe in Figure A.

OR

(b) Figure B shows number of bird species. Identify the countries marked ‘X’ and ‘Y’.

(ii) Explain the three reasons for the pattern of biodiversity that is observed in Figure A.

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Official CBSE Marking Scheme · 2023 (Supplementary/Compartment), Set 57/C/1

(i)

(a) Biodiversity increases as we move from arctic towards tropics / biodiversity decreases as we move from tropics towards arctic [1]

OR

(b) X – Columbia; Y – India / South America (or any other correct name) [½ × 2]

(ii) Tropics harbor more biological diversity:

  • Speciation is generally a function of time; unlike temperate regions subjected to frequent glaciations in the past, tropical latitudes have remained relatively undisturbed for millions of years and thus had a long evolutionary time for species diversification.
  • Tropical environments are less seasonal, relatively more constant and predictable, which promote niche specialization and lead to a greater species diversity.
  • There is more solar energy available in the tropics, which contributes to higher productivity that contributes indirectly to greater diversity.

[1 × 3]

[Total: 4 marks]

Q48.2025 Main 57/2/1 · Q335 marks

Student to attempt either option (A) or (B).

(A) (i) Explain giving three reasons why tropics show greatest levels of species diversity.

(ii) Draw a graph showing species-area relationship. Name the naturalist who studied such relationship. Write the observation made by him.

OR

(B) (i) The world is facing the accelerated rate of species extinctions due to human activities. Explain any three major causes of biodiversity losses.

(ii) Describe ‘Ex situ’ approach for conserving biodiversity. Give any two examples.

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Answer not available: no official marking scheme published.

Q49.2025 Main 57/6/1 · Q335 marks
    1. Describe the Species-Area relationship as observed by Alexander von Humboldt, for a wide variety of taxa in nature.
    2. Draw the graph showing Species-Area relationship for S = CAZ. What is the significance of ‘Z’ in Species-Area relationship ?

OR

    1. Describe the logistic population growth curve with the help of a suitable graphical representation.
    2. Write the equation of Verhulst-Pearl logistic growth curve and explain what ‘K’ and ‘r’ suggest in the given equation.
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Answer not available: no official marking scheme published.

Q50.2025 Main 57/7/1 · Q335 marks
    1. How does alien species invasion cause a decline in biodiversity ? Explain.

      How have the following contributed to biodiversity loss ?

      1. Nile Perch
      2. Lantana and Eichhornia
      3. Clarias gariepinus
    2. Why have certain regions been declared as biodiversity hotspots by environmentalists of the world ? Name any two such regions in India.

OR

    1. Write an equation for Verhulst-Pearl Logistic Growth Curve where :

      N = Population density at time ‘t’

      r = Intrinsic rate of natural increase

    2. Draw a graph for a population whose population density has reached carrying capacity.
    3. Draw a growth curve where resources are non-limiting to growth of population.
    4. Which growth curve is considered more realistic and why ? Explain.
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Answer not available: no official marking scheme published.

Q51.SQP 2026-27 · Q335 marks

Attempt either option A or B.

A. A Green Valley X, home to rare medicinal plants and several migratory bird species, is facing multiple environmental challenges.

  1. A luxury brand of cosmetics with little regulation has been sourcing wild medicinal plants from the valley.
  2. The company has introduced an ornamental grass species from another country, which has taken over large areas of native grassland.
  3. There is also a reported drastic decline in the endemic trees.
  4. Large scale forest fire in a part of the valley.

a) From the above hypothetical case, identify and explain the four major biodiversity threats. (4)

b) Suggest one in-situ conservation practice that could help protect the biodiversity of Green Valley X, and justify your choice. (1)

OR

B. At a composting ground, after a community clean-up drive, large amounts of vegetable peels, garden leaves, and other organic waste are piled up. Over the next few weeks, the volume of the waste decreases noticeably.

  1. Explain the sequence of processes that could have led to this transformation, and name the major organisms involved at each step. (4)
  2. If a prolonged dry spell occurs during this composting period, explain how it could affect the rate of the above processes. (1)
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Official CBSE Sample Paper Marking Scheme · 2026-27

A.

(a)

Biodiversity ThreatExplanation of the effect it will have
OverexploitationUnsustainable sourcing of wild medicinal plants by the cosmetic brand threatens plant species/may lead to depletion and extinction of valuable species. [1]
Invasion by alien speciesOrnamental grass from another country spreading aggressively into native grassland/outcompetes native grasses, reduces local biodiversity. [1]
Co-extinctionDisappearance of nesting tree species causing sharp decline in migratory bird population/ disrupts ecological balance, loss of niches. [1]
Habitat loss and fragmentationForest fires lead to habitat loss and fragmentation by destroying forest vegetation and breaking continuous habitats into smaller, isolated patches. This forces species to migrate, reduces population sizes, and alters species distribution, often resulting in local extinction and decline in biodiversity. [1]

(b) Example: Declaration of Green Valley X as a protected area (wildlife sanctuary/national park/biosphere reserve/sacred grove/hotspot).
Justification: Maintains species in their natural habitat, preserves ecological interactions, and allows long-term protection of genetic, species, and ecosystem diversity. [1]

OR

B.

I.

Step / ProcessDescription
FragmentationBreaking down of waste into smaller particles by detritivores (e.g., earthworms, termites). [1]
LeachingSoluble inorganic nutrients dissolve in water and percolate into the composting soil. [1]
CatabolismEnzymatic breakdown of complex organic matter into simpler inorganic substances by bacteria/fungi. [1]
Humification & MineralisationFormation of humus and release of inorganic nutrients into the compost by microbes. [1]

II. A dry spell would lower moisture content, slowing microbial activity and enzymatic breakdown, thereby reducing the rate of decomposition and delaying the reduction in waste volume. [1]