Identify the plant in which emasculation is not required for artificial hybridization.
- Rice
- Wheat
- Pea
- Papaya
Reveal answerHide answer
(D) Papaya [1 mark]
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.
Identify the plant in which emasculation is not required for artificial hybridization.
(D) Papaya [1 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.
(B) a – Meiosis I, b – Meiosis II, c – No division, d – Mitosis [1 mark]
Assertion (A): Apomixis leads to production of genetically identical offsprings.
Reason (R): Apomixis involves fertilization between two genetically similar plants.
(C) Assertion (A) is true, Reason (R) is false. [1 mark]
Select the correct option for the pair of plants pollinated by water.
(A) Vallisneria and Hydrilla [1 mark]
Choose the option that gives the correct number of pollen grains that will be formed after 325 microspore mother cells undergo microsporogenesis.
(C) 1300 [1 mark]
Assertion (A): Cleistogamous flowers have assured seed set.
Reason (R): Cleistogamous flowers are invariably autogamous and do not depend on any pollinating agents.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A) [1 mark]
Identify the parts P, Q, R, S in the figure given below and select the option in the correct order.
| P | Q | R | S | |
|---|---|---|---|---|
| (A) | Seed | Thalamus | Mesocarp | Endocarp |
| (B) | Thalamus | Seed | Endocarp | Mesocarp |
| (C) | Seed | Thalamus | Endocarp | Mesocarp |
| (D) | Thalamus | Seed | Mesocarp | Endocarp |
(B) P – Thalamus, Q – Seed, R – Endocarp, S – Mesocarp [1 mark]
The undifferentiated sheath that encloses the radicle and root cap in a monocot embryo, is known as :
(B) Coleorrhiza [1 mark]
Assertion (A): Apomictic embryos are genetically identical to the parent plant.
Reason (R): Apomixis is the production of seeds without fertilisation.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]
The diagram given below shows labelling of four parts of a dicot embryo during its development as P, Q, R and S.
Choose the option that indicates correct labelling of ‘P’, ‘Q’, ‘R’ and ‘S’ of embryo in different stages of its development :
Answer not available: no official marking scheme published.
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 ?
Answer not available: no official marking scheme published.
Given below is a diagram of T.S. of a monocot seed with parts I, II & III labelled :
Choose the option where parts I, II and III are identified correctly.
Answer not available: no official marking scheme published.
Select the statements that are true for a typical dicotyledonous embryo from the given options.
Choose the correct answer :
Answer not available: no official marking scheme published.
Select the statements that are true for pollination mechanism in flowering plants from the given options.
Choose the correct answer :
Answer not available: no official marking scheme published.
Select the statements that are true for the seed of angiosperm from the given options :
Choose the correct option :
Answer not available: no official marking scheme published.
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.
Options :
Answer not available: no official marking scheme published.
Choose the option that correctly describes the gynoecium of Michelia :
Answer not available: no official marking scheme published.
Identify the various parts, ‘a’, ‘b’ and ‘c’, in the given diagram of ovule :
(C) a – Hilum, b – Funicle, c – Embryo sac [1 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.
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
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 ?
(C) Maize [1 mark]
An angiosperm embryo sac is located within the :
(C) Nucellus [1 mark]
The part of the ovule that develops into protective coats of a seed after fertilization in a typical flowering plant is :
(C) integuments [1 mark]
In which one of the following floral plants are many embryos formed in the seeds without fertilisation of the egg cell ?
(D) Citrus [1 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.
(D) Assertion (A) is false, but Reason (R) is true. [1 mark]
In a fertilized ovule of an angiosperm, the cells in which n, 2n and 3n conditions respectively occur are :
(A) antipodal, zygote and endosperm [1 mark]
Study the following diagram of Transverse Section of a young anther of an angiosperm :
Select the option where parts ‘A’, ‘B’ and ‘C’ are correctly identified.
(A) A – Connective, B – Endothecium, C – Pollen grain [1 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 :
(A) Pseudocopulation [1 mark]
Identify the correct labellings in the figure of a fertilised embryo sac of an angiosperm given below :
(B) A – degenerating synergids, B – zygote, C – PEN, D – degenerating antipodals [1 mark]
Long ribbon-like pollen grains are seen in some :
(A) Aquatic plants [1 mark]
Choose the option that gives the correct number of pollen grains that will be formed after 325 microspore mother cells undergo microsporogenesis.
(C) 1300 [1 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.
(C) A is true, but R is false. [1 mark]
Which of the following seeds have remained alive for the longest period ?
(A) Phoenix dactylifera [1 mark]
Select the option that shows the correctly identified ‘U’, ‘X’, ‘Y’ and ‘Z’ in a developing dicot embryo.
(C) X – Suspensor (2n), Y – Cotyledon (2n), Z – Radicle (2n), U – Plumule (2n). [1 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.
(A) P – Thalamus [1 mark]
Assertion (A): Banana is a parthenocarpic fruit.
Reason (R): It develops only from the ovary.
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). [1 mark]
The male gametes in plants are formed by:
(C) Mitotic division of nucleus of generative cell [1 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?
(C) It ensures cross-pollination [1 mark]
The primary endosperm nucleus is formed by fusion of which of the following?
(B) A male gamete and two polar nuclei [1 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.
(C) A is true but R is false [1 mark]
(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.
(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 seed | Castor seed |
|---|---|
| Non endospermic seed | Endospermic seed |
| Non-albuminous seed | Albuminous 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]
(a) Write one way by which apomictic seeds can be developed.
(b) Why do farmers prefer apomictic seeds over hybrid seeds ?
(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]
(a) Mention the functions of each of the following :
OR
(b) Farmers prefer apomictic seeds over hybrid seeds. Give any two reasons.
(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]
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.
(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]
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’.
Answer not available: no official marking scheme published.
(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 ?
(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]
Given below is a transverse sectional view of an apple. Label the following parts (i) and (ii) and write down their functions.
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
Identify A, B, C and D in the table given below :
| Terms | Part of the plant it represents |
|---|---|
| Pericarp | ‘A’ |
| ‘B’ | Cotyledon in seed of grass family |
| Embryonal axis | ‘C’ |
| ‘D’ | Remains of nucellus in a seed |
[½ × 4 = 2 marks]
OR
(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]
(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]
(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]
The male and female flowers of corn plant (maize) are well adapted for pollination by wind. Justify.
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]
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?
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)
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.
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]
(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 ?
(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]
Draw a neat diagram of megasporangium of an angiosperm and label any six parts.
(Label any six parts) [½ × 6]
[Total: 3 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.
(Award ½ mark for each correct labelling) [½ × 4]
[Total: 3 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]
(a)
(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]
Draw a neat diagram of a maize grain showing the internal structure and label any five parts.
(Diagram = ½, Any correct labelling = ½ × 5)
[Total: 3 marks]
Differentiate between wind-pollinated and insect-pollinated flowers. Give an example of each type.
(Any four differences) [½ × 4]
| Wind pollinated flowers | Insect 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 pollens | Stigma is often non feathery. |
(Any other correct difference)
Example: (Any one) [½+½]
| Wind pollinated flowers | Insect Pollinated flowers |
|---|---|
| Maize, Cannabis, Grasses | Yucca, Sunflower |
(Any other correct example)
[Total: 3 marks]
List any three outbreeding devices that flowering plants have developed for cross-pollination and explain how they help to encourage cross-pollination.
(Any three) [1×3 = 3 marks]
(a) Draw a diagram of L.S. of an anatropous ovule of an angiosperm. Label the following parts :
(b) How does endosperm in angiosperms become triploid ?
(a)
(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]
(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.
Answer not available: no official marking scheme published.
Given below is a flower with its characteristic features specialised for the most common type of abiotic pollination.
Answer the following questions based on the above diagram :
Answer not available: no official marking scheme published.
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 :
Answer not available: no official marking scheme published.
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 grains | Pollen grains of Plant A | Pollen grains of Plant B | Pollen 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.
Answer not available: no official marking scheme published.
Draw a longitudinal section of pistil of a flower showing growth of the pollen tube. Label the part :
(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]
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).
[Vegetative cell, generative cell, exine, intine, germ pore] (Any four labelling to be considered) [½ × 4]
[Marks to be given to two correct labelling with their respective correct function only]
[Total: 3 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.
[Total: 3 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.
[1×3] (Half mark each if only name of process Emasculation and Bagging is mentioned)
[Total: 3 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.
(a)
(½ mark each for any 4 correct labellings) [½ × 4]
(b) Guide pollen tube into synergids. [1]
[Total: 3 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.
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]
The following question is based on pollination. Study the figures carefully and answer the questions that follow.
OR
Answer not available: no official marking scheme published.
Given below is a set of information about some fruits and seeds.
| Fruit | Fruit and seed formation |
|---|---|
| P | Nucellar cells surrounding the embryo sac develop into embryos. |
| Q | Ovary develops into the fruit by the application of growth hormones. |
| R | Thalamus contributes to fruit formation. |
| S | Ovary 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.
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)
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]
(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]
(a) (i) Stigma has the ability to recognise the right or compatible and wrong or incompatible type of pollen [1]
(ii) [1 × 4]
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)
[½+½]
[Total: 5 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.
OR
(B) Given below is the diagram of human ovum surrounded by a few sperms. Observe the diagram and answer the questions that follows :
Answer not available: no official marking scheme published.
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 :
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
(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 :
(ii) Write the function of Tapetum.
(iii) Explain the following giving reasons :
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.
(a) (i)
(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]
(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 :
(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.
(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)
[½ × 5]
OR
(b)
(i)
[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]
(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.
(a)
(i)
(ii) To provide assured nutrition to the developing embryo. [1]
OR
(b)
(i)
(ii)
[Total: 5 marks]
OR
(a) (i)
(ii) To prevent inbreeding depression / to have more variations for better adaptation. [1]
OR
(b) Menstrual cycle in a normal human female
| Phase | Uterine event | Ovarian event | Marks | |
|---|---|---|---|---|
| (i) | Proliferative phase / Follicular phase | Endometrium 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 phase | Maintenance 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 phase | Breakdown 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]
(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.
(a) (i)
(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)
[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]
OR
Numbers indicate the days of the menstrual cycle
(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)
| Days | Ovarian hormones | Pituitary hormones | |
|---|---|---|---|
| 1 | 8–12 | Follicular growth/proliferation of endometrial cells. | Stimulates follicular development/secretion of estrogen by growing follicles |
| 2 | 13–15 | Maturation of ovarian follicles/formation of graafian follicles/thickening of endometrium. | Rupture of graafian follicle to release ovum. |
| 3 | 16–28 | Maintenance of endometrium | Secretion of progesterone from corpus luteum. |
[½×6]
[Total: 5 marks]
OR
(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]
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]
[Total: 5 marks]
Select the option that shows correct statements :
Options :
(B) (i) and (iii) are correct [1 mark]
Which connective tissue connects ovary to pelvic wall and uterus ?
(B) Ligaments [1 mark]
Foetal ejection reflex in human female triggers the release of which hormone ?
(D) Oxytocin from maternal pituitary gland [1 mark]
Select the phase of menstrual cycle in which the remaining parts of Graafian follicle transform into the corpus luteum.
(A) Secretory phase [1 mark]
The foetal ejection reflex in human triggers the release of _________ hormone from __________.
Fill in the blanks by choosing right option.
(B) oxytocin, maternal pituitary [1 mark]
In a human female, menstrual cycle of 28 days is represented by the diagram given below.
Select the correct statement related with this diagram.
(D) Both (A) and (B) [1 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.
(C) Assertion (A) is true but Reason (R) is false. [1 mark]
The path of the sperm transport is shown below. Identify P and Q and select the correct option.
(C) P – Vasa efferentia, Q – Vas deferens [1 mark]
Match the following items of Column-I with that of Column-II :
| Column-I | Column-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.
Answer not available: no official marking scheme published.
The number of autosomes present in a human secondary spermatocyte
Answer not available: no official marking scheme published.
Assertion (A): Corpus luteum secretes the hormone, progesterone.
Reason (R): Hormone Progesterone is essential for maintenance of the endometrium.
Answer not available: no official marking scheme published.
Given below are a few statements with reference to the accessory ducts of the human male reproductive system :
Choose the option with all true statements from the given options :
Answer not available: no official marking scheme published.
Given below are few statements with reference to the uterus in the female reproductive system :
Choose the option with all true statements from the given options :
Answer not available: no official marking scheme published.
Given below are a few statements with respect to spermatogenesis in a human male.
Choose the option with all true statements from the given options :
Answer not available: no official marking scheme published.
Which of the following statements are true in respect of chorionic villi in humans ?
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
Which one of the following hormones is secreted by the human placenta that helps in the maintenance of pregnancy ?
(B) Human Chorionic Gonadotropin [1 mark]
In humans, the secondary oocyte completes meiotic division when :
(C) it is penetrated by the sperm cell. [1 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 :
(C) Ovum – A, Morula – E, Blastocyst – G [1 mark]
Estrogen is secreted by :
(B) Membrane granulosa of Graafian follicle [1 mark]
In which one of the following options does the endocrine gland correctly match with its hormonal secretion and its function ?
(C) Leydig cells – Androgen – Initiate the production of sperms. [1 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 Hormone | Site of Production | ||
|---|---|---|---|
| P. | Oxytocin | i. | Placenta |
| Q. | Relaxin | ii. | Corpus Luteum |
| R. | hCG | iii. | Pituitary Gland |
| S. | Progesterone | iv. | Ovaries |
(C) P-iii, Q-iv, R-i, S-ii [1 mark]
During the menstrual cycle of a human female, formation of graafian follicle is stimulated by secretion of which of the following gonadotropin hormones?
(C) FSH and LH [1 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?
(B) It provides essential nutrients to developing embryo and helps in its implantation. [1 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.
(A) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). [1 mark]
(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 :
(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 Spermatogenesis | Meiosis I in Oogenesis |
|---|---|
| Starts at puberty | Starts during embryonic stage |
| Results in production of two equal haploid secondary spermatocyte | Results in production of unequal haploid secondary oocyte and first polar body |
[Total: 2 marks]
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
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 :
What does the graph with respect to woman ‘X’ and woman ‘Y’ indicate ? Give suitable reason.
[Total: 2 marks]
State the cellular nature and functions of myometrium and endometrium.
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]
The diagram given below shows a developmental stage of human embryo. Answer the following questions with reference to it :
(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]
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.
(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]
Differentiate between Follicular phase and Luteal phase of the Menstrual cycle in human females on the basis of the following criteria :
(½ mark to be awarded for every correct difference)
| Follicular phase | Luteal phase | |
|---|---|---|
| (a) Days of their occurrence in the cycle | 6th–13th / 6th–14th day | 15th–28th / 15th–29th day |
| (b) Stage of the follicle | Development of primary follicle into Graafian follicle | Transformation of Graafian follicle into corpus luteum |
| (c) Hormones influencing the phases | LH / FSH / Estrogen | Progesterone |
| (d) State of endometrium | Regeneration of endometrium through proliferation. | Endometrium further proliferates and thickens. |
[½ × 4 = 2 marks]
Explain the process of formation of placenta in a human female after the implantation of the blastocyst in the endometrium of the uterus.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Answer not available: no official marking scheme published.
Given below is the diagram showing a part of the reproductive system of a human female.
(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.
(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]
(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]
Given below is a diagrammatic sectional view of a seminiferous tubule. State the developmental process of :
(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]
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:
(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)
Study the graphs given below for Case 1 and Case 2 showing different levels of certain hormones and answer the question that follows :
(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 ?
Answer not available: no official marking scheme published.
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 :
during the process of spermatogenesis.
(c) Name the accessory ducts that the sperms travel through from seminiferous tubules to reach the epididymis.
(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]
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 :
OR
(a)
| Hormone | Source Organ |
|---|---|
| FSH | Pituitary gland |
| LH | Pituitary gland |
| Estrogen | ovary / graafian follicle |
| Progesterone | ovary / 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]
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)
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]
(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.
(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]
(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]
(a) (i)
(Diagram given in the scheme:)
(ii) Luteinizing hormone (LH) and Follicle stimulating hormone (FSH) [½+½]
OR
(b) (i)
[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]
(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.
(a)
OR
(b) (i)
(½ mark for any 6 correct labelling) [½ × 6]
(ii) (any one difference) [1]
| Perisperm | Endosperm |
|---|---|
| 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]
| Perisperm | Endosperm |
|---|---|
| Eg. Black pepper / beet / any other example | eg. Castor / Wheat / Maize / Barley / Sunflower / Coconut / any other example |
[Total: 5 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]
(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:
OR
(b) (i) Hormones in sequence of their secretion in a pregnant woman: FSH; LH; hCG; Relaxin. [1]
(ii) [½+½ each]
[Total: 5 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.
(a)
[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]
Answer the following questions :
OR
Answer not available: no official marking scheme published.
OR
Observe the picture of Commelina plant bearing two types of flowers given above.
(a) (i) Spermatogenesis [½]
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 / blossom | closed flower / do not blossom |
| (2) | Autogamy as well as geitonogamy and xenogamy / cross pollination | only autogamy / self-pollination |
[1 + 1]
(iii) Any two:
To prevent inbreeding depression / to discourage self-pollination and to encourage cross pollination / to develop genetic variation. [1]
[Total: 5 marks]
OR
(a) (i)
(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)
(Eight labellings) [½ × 8 = 4]
(ii) Providing nutrition to the developing embryo [1]
[Total: 5 marks]
OR
(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]
(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.
(a)
(i)
(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)
(Any four points) [½ × 4]
(ii)
(1) ‘A’ – X, ‘B’ – P [½ + ½]
(2)
[Total: 5 marks]
How many of the following features are related to Nirodh – a popular brand of condoms ?
(B) (i) and (iii) [1 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 :
(C) IUT [1 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.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). [1 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.
Answer not available: no official marking scheme published.
Assertion (A): ‘Saheli’, an oral contraceptive inhibits ovulation and increases phagocytosis of sperms.
Reason (R): It is a non-steroidal preparation.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
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. | 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 female |
| c. | ZIFT | iii. | Injecting sperms directly into the ovum |
| d. | IUI | iv. | Transfer of early embryos into the fallopian tube |
Select the correct match :
(D) a– i, b – iii, c – iv, d – ii [1 mark]
The periodic abstinence by a couple for family planning should be from :
(C) Day 10 to 17 of menstrual cycle [1 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.
(C) Assertion (A) is true, but Reason (R) is false. [1 mark]
Study the table given below :
| Contraceptive / Contraceptive Method | Mode of Action |
|---|---|
| A. The pill | I. Prevent sperm reaching cervix |
| B. Condom | II. Prevent implantation |
| C. Vasectomy | III. Inhibits ovulation |
| D. Copper-T | IV. Semen contains no sperm |
Select the option where contraceptive/contraceptive method are correctly matched with their mode of action.
(C) A – III, B – I, C – IV, D – II [1 mark]
Observe the schematic representation of assisted reproductive technology given below :
Identify the most appropriate technique depicted in the above diagram.
(C) ICSI [1 mark]
Amniocentesis is a technique that is used to :
(B) Determine any genetic disorder of the foetus [1 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.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). [1 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 loop | i. | Inhibition of ovulation |
| Q. | Multiload 375 | ii. | Phagocytosis of sperms in uterus |
| R. | Subcutaneous Norplant | iii. | Causes thickening of cervical mucous |
| S. | Saheli | iv. | Makes cervix hostile to sperms |
(A) P– (ii), Q – (iv), R – (iii), S – (i) [1 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.
(C) Point Q [1 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 ART | S. 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.
(D) (A) – (i), (B) – (iii), (C) – (iv), (D) – (ii). [1 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.
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). [1 mark]
Which one of the following techniques is employed in test tube baby programme ?
(A) Intra Cytoplasmic Sperm Injection
OR
(D) Zygote Intra Fallopian Transfer [1 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.
(C) (A) is true, but (R) is false. [1 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?
(A) Blockage of sperm transport into the ejaculatory duct and urethra [1 mark]
How do ‘implants’ act as an effective method of contraception in human females ? Mention one advantage of implants over contraceptive pills.
[Total: 2 marks]
Define the term ‘amniocentesis’. Why has the government imposed a statutory ban in spite of its importance in the medical field ?
[Total: 2 marks]
How can childless couples get help by the following assisted reproductive technologies ?
(a) GIFT
(b) ICSI
(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]
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
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 ?
[Total: 2 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.
(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)
(Any two devices with explanation) [½ × 4 = 2 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 ?
[Total: 2 marks]
(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]
How do ‘implants’ act as an effective method of contraception in human females ? Mention their one advantage over contraceptive pills.
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]
Study the diagram given below :
(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]
(a)
(b) Saheli is different from other pills in term of:
[Total: 3 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]
(a) Steps in IVF
(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]
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 ?
Answer not available: no official marking scheme published.
(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.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
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 ?
(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]
Explain the mode of action of contraceptive pills taken by human females. Mention the schedule to be followed for effective outcome.
[Total: 3 marks]
(a) Why is “in vitro fertilization (IVF)” so named ? State its importance.
(b) Distinguish between GIFT and ZIFT.
(a)
(b)
| GIFT | ZIFT |
|---|---|
| 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]
Explain the IUI and IUT methods of assisted reproductive technologies.
[Total: 3 marks]
OR
(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)
| Perisperm | Pericarp |
|---|---|
| Persistent nucellus in some seeds | The wall of ovary develops into wall of fruit. |
[1]
(ii)
| Syncarpous | Apocarpous |
|---|---|
| fused pistils. | free pistils. |
[1]
(iii)
| Plumule | Radicle |
|---|---|
| Future stem/terminal part of epicotyl/shoot tip of embryonal axis | Future root/terminal part of hypocotyl/root tip of embryonal axis |
[1]
[Total: 3 marks]
Expand and explain the following techniques used in the ‘Test Tube Baby’ programme :
(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]
Name and explain a surgical contraceptive method that can be adopted by the male partner of a couple.
• Vasectomy [1]
• A small part of the vas deferens is cut/removed, tied-up to block sperm transport [1×2]
[Total: 3 marks]
Suggest a suitable contraceptive device for the following cases with justification.
(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)
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.
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]
In the following figure, two ways of pairing of two homologous pairs of chromosomes are shown. Which of the following phenomena is expressed ?
(B) Independent assortment of genes [1 mark]
In a dihybrid cross, 2400 individuals are produced in F2 generation. Approximately how many will be phenotypically similar to parents ?
(B) 1500 [1 mark]
The smallest part of DNA molecule that can be changed by point mutation is :
(D) Nucleotide [1 mark]
Select the correct statement. All genes located on the same chromosome :
(B) form one linkage group [1 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.
(B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of the Assertion(A) [1 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 ?
(D) 100% [1 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 :
(A) Did not segregate independent of each other [1 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 ?
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
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 :
Answer not available: no official marking scheme published.
In a pedigree chart the symbol
represents :
Answer not available: no official marking scheme published.
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.
Choose the option that indicates the correct inheritance of trait in the above pedigree chart :
Answer not available: no official marking scheme published.
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 :
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Which of the following is not an example of aneuploidy ?
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Study the pedigree chart of a family sharing the inheritance of sickle cell anemia.
The trait traced in the above pedigree chart is :
Answer not available: no official marking scheme published.
Which of the following do not follow the law of independent assortment ?
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Which of the following conditions correctly describes the manner of determining the sex in the given options ?
Answer not available: no official marking scheme published.
Assertion (A): A male individual always inherits haemophilia from his mother.
Reason (R): The gene for haemophilia lies on X-chromosome.
Answer not available: no official marking scheme published.
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 ?
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
What would be the genotype of parents if the F1 offspring have the phenotypes in 1 : 1 proportion ?
(D) Aa × aa [1 mark]
Which of the following animals exhibit male heterogamety ?
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
Select the incorrect match from the following :
| Human Karyotype | Characters | |
|---|---|---|
| (A) | 45 + XX | Broad palm with characteristic palm crease |
| (B) | 44 + XXY | Overall feminine development |
| (C) | 44 + XO | Sterile females as ovaries are rudimentary |
| (D) | 44 + XY | Normal male |
(B) 44 + XXY – Overall feminine development [1 mark]
Assertion (A): Linked genes do not show dihybrid F2 ratio 9 : 3 : 3 : 1.
Reason (R): Linked genes do not undergo independent assortment.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). [1 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 :
(C) 1-(iii), 2-(iv), 3-(i), 4-(ii) [1 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 :
(A) 25% [1 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 :
(C) 50% pink : 50% white [1 mark]
In which of the given chromosomal disorders does the individual have tall stature with feminised character ?
(A) Klinefelter’s syndrome [1 mark]
Identify the category of genetic disorder depicted in the pedigree chart given below :
(C) Autosomal recessive [1 mark]
Turner’s syndrome in humans occurs due to
(A) Aneuploidy [1 mark]
A single gene that controls the expression of more than one trait is said to show
(D) Pleiotropism [1 mark]
A person with trisomy of 21st chromosome shows
Select the correct option, from the choices given below :
(D) (i) and (ii) [1 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.
(D) Autosomal recessive [1 mark]
If both the parents are carriers for thalassaemia, the chances of an afflicted child to be born to them is :
(A) 25 % [1 mark]
Assertion (A): In birds the sex of the offspring is determined by males.
Reason (R): Males are homogametic while females are heterogametic.
(D) (A) is false, but (R) is true. [1 mark]
In humans, non-disjunction of the 21st pair of chromosomes leads to :
(D) Down’s Syndrome [1 mark]
Which one of the following processes results in the production of recombinants in future generations ?
(B) (ii) and (iv) [1 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.
(A) A and R are true and R is correct explanation of A. [1 mark]
Choose the set of sex determining chromosomes that indicates the correct sex of the respective organism.
(B) XO type of sex chromosome determines male sex in grasshopper. [1 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.
(A) Both A and R are true and R is correct explanation of A. [1 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.
(D) (A) is false, but (R) is true. [1 mark]
When round yellow seed bearing heterozygous pea plant is selfed, the frequency of occurrence of RrYY genotype among the F2 offsprings is :
(D) 2/16 [1 mark]
In a pedigree analysis, the symbol shown below represents :
(C) Consanguineous mating [1 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?
(B) 25% [1 mark]
Study the given pedigree and identify the disease represented
(B) Sickle cell anaemia [1 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.
(D) A is false but R is true. [1 mark]
(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.
(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 nucleotide | RNA nucleotide |
|---|---|
| Has deoxyribose sugar | Has ribose sugar |
| Types of nitrogen bases are A, G, C, T | Types 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]
(a) The diagram below shows the sequence of amino acids in part of haemoglobin molecule.
(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.
(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)
[1+1 = 2 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 ?
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]
Study the cross given below :
Identify the abnormalities ‘1’ and ‘2’ in the offsprings of a cross done between a couple and distinguish between them.
Answer not available: no official marking scheme published.
Study the pedigree chart given below, showing the inheritance pattern of blood group in a family :
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’ ?
Answer not available: no official marking scheme published.
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.
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.
Answer not available: no official marking scheme published.
A karyotype of a human suffering from a certain disorder is given below :
Answer not available: no official marking scheme published.
(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’.
OR
(b) Observe the diagram given below and answer the following questions.
(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
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]
Although Haemophilia and sickle cell anemia are two blood related Mendelian disorders, yet, they differ in their pattern of inheritance. State any two differences.
| Haemophilia | Sickle cell anaemia |
|---|---|
| Sex linked / X-linked recessive disorder | Autosomal 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]
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.
Case I – Homozygous dominant
GG × gg (Green Pods × Yellow Pods)
| G | G | |
|---|---|---|
| g | Gg | Gg |
| g | Gg | Gg |
Phenotype: All green coloured pod [½+½]
Case II – Heterozygous dominant
Gg × gg (Green Pods × Yellow Pods)
| G | g | |
|---|---|---|
| g | Gg | gg |
| g | Gg | gg |
Phenotype: 50% green & 50% yellow colour pod [½+½]
[Total: 2 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.
(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]
(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]
(a) Drosophila melanogaster [1]
(b) (Any four)
[½ × 4 = 2 marks; Total: 3 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]
(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]
Define test cross. How does test cross help to determine the genotype of an individual ? Explain with the help of an example.
• It is a cross to know the genotype of an unknown dominant trait by crossing it with homozygous recessive trait. [1]
• e.g. :
(Award 1 mark to each cross) [1 + 1]
[Total: 3 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.
Answer not available: no official marking scheme published.
Gregor Mendel conducted hybridisation experiments in garden pea for seven years and proposed the law of inheritance.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Study the diagram above and answer the following questions :
Answer not available: no official marking scheme published.
Answer the following questions with respect to the sex determining mechanism observed in honey bee.
| Parent | Male |
|---|---|
| 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) ____________ |
Answer not available: no official marking scheme published.
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.
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]
[1]
[Total: 3 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.
(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]
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, axial | 6/16 white, axial 2/16 white, terminal 6/16 violet, axial 2/16 violet, axial |
| (b) Violet, axial × white, terminal | 1/4 violet, axial 1/4 violet, terminal 1/4 white, axial 1/4 white, terminal |
| (c) Violet, axial × violet, axial | 3/4 violet, axial 1/4 white, axial |
Find the genotypes of each of the parental pairs of crosses ‘a’, ‘b’ and ‘c’.
(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]
(a)
(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]
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.
(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.
(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]
Study the given pedigree chart and answer the questions that follow.
(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]
(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.
(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]
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.
BbGg male is crossed with BbGg female which are both heterozygous of both the characters for eye and coat colour. [1]
Punnett square [1]
1/16 or 6.25 % - blue eyes and white coat [1]
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.
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
| RB | Rb | rB | rb | |
|---|---|---|---|---|
| Rb | RRBb | RRbb | RrBb | Rrbb |
| Rb | RRBb | RRbb | RrBb | Rrbb |
| rb | RrBb | Rrbb | rrBb | rrbb |
| rb | RrBb | Rrbb | rrBb | rrbb |
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]
How can a single gene defect caused by the mutation result in Phenylketonuria (PKU)? Explain the molecular mechanism underlying pleiotropy.
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]
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 :
(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.
(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]
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 :
(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 ?
(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]
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.
(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]
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.
OR
(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]
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.
OR
(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]
(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.
(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]
| Thalassemia | Sickle 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 affected | Only β chain is affected |
OR
(b) (i) Major goals of HGP: (any four) [½ × 4]
(ii) Methodologies:
(iii) Yeast artificial chromosomes; It is used as vector. [½+½]
[Total: 5 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.
(a) In Pea plants: [2½ for the cross (½ each: gametes, F1, F2 ...)]
Phenotypic ratio: violet : white = 3 : 1 [½]
Conclusion: This is the case of dominance [½]
In Antirrhinum plant [2½]
Phenotypic ratio: Red : Pink : white = 1 : 2 : 1 [½]
Conclusion: This is a case of incomplete dominance [½]
OR
(b) (i) [1]
| Mendelian disorder | Chromosomal disorder |
|---|---|
| Occurs due to mutation / alteration in single gene / follow mendelian inheritance pattern | Occurs 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]
Karyotype – 44+XO (45) [½]
[Total: 5 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 :
Answer not available: no official marking scheme published.
(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 :
(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 ?
(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:
[½ × 2]
OR
(b) (i)
[½ × 8]
(ii) Repressor binds to operator to inhibit the gene expression [1]
[Total: 5 marks]
OR
(a)
(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
OR (diagram, each labelled point [½ × 8]):
(ii) Repressor binds to the operator to inhibit gene expression therefore it is referred to be negatively regulated. [1]
[Total: 5 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.
(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)
| mRNA | tRNA |
|---|---|
| Provides template for protein synthesis | Acts as adaptor molecule / brings amino acids and reads genetic code |
[1]
[Total: 5 marks]
To separate repetitive DNA from bulk genomic DNA, which of the following method is used ?
(D) Density Gradient Centrifugation [1 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 ?
(B) 50 [1 mark]
Four different transcription units are shown below. Choose the option with the correct image.
Options (A)–(D) are shown in the figure.
(C) [1 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.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A) [1 mark]
Choose the correct statements :
Options :
(C) (ii) and (iv) [1 mark]
A DNA molecule is 160 base pairs long. It has 30% Guanine. How many Adenine bases are present in this DNA molecule ?
(B) 64 [1 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.
(B) Both Assertion (A) & Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A) [1 mark]
Select the option that has only termination codons.
(A) UAA, UGA, UAG [1 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.
Answer not available: no official marking scheme published.
Assertion (A): One of the property of genetic code is degeneracy.
Reason (R): Some amino acids can be coded by more than one codon.
Answer not available: no official marking scheme published.
The phosphoester linkage in the nucleotides is between
Answer not available: no official marking scheme published.
Given below is a heterogeneous RNA formed during Eukaryotic transcription :
How many introns and exons respectively are present in the hnRNA ?
Answer not available: no official marking scheme published.
Study the items of Column-I and those of Column-II :
| Column-I | Column-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 :
Answer not available: no official marking scheme published.
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 ?
Answer not available: no official marking scheme published.
In its extended ‘beads-on-string’ form of chromatin, the ‘beads’ in the string represent :
Answer not available: no official marking scheme published.
The substrate used during DNA replication by the enzyme DNA-dependent DNA polymerase is :
Answer not available: no official marking scheme published.
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 : | |
|---|---|
| AUC | Isoleucine |
| AUG | Methionine |
| UAC | Tyrosine |
| GUA | Valine |
Options :
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
The histone core in a nucleosome of chromatin thread is a/an :
Answer not available: no official marking scheme published.
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 :
Answer not available: no official marking scheme published.
What would happen if a gene encoding a polypeptide of 50 amino acids, 25th Codon (UAU) is mutated to “UAA” ?
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
The process of splicing in eukaryotes represents the dominance of the :
Answer not available: no official marking scheme published.
Regulation of lac operon by repressor is referred to as :
Answer not available: no official marking scheme published.
SNPs in Human Genome Project refers to :
Answer not available: no official marking scheme published.
What are minisatellites ?
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
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.
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
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.
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
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 :
(C) 2 [1 mark]
The type of bond represented by the dotted line ‘– – – – –’ in a schematic polynucleotide chain is :
(C) N-glycosidic linkage [1 mark]
In the double helical structure of DNA molecule, the strands are :
(C) anti-parallel and complementary [1 mark]
In a ‘transcription unit’, the ‘terminator’ is located towards the :
(D) 3′ end of the coding strand. [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 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 ?
(C) 1720 [1 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.
(C) Hybrid : Heavy, 1 : 31 [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]
Which was the last of the 24 human chromosome to be completely sequenced ?
(A) Chromosome – 1 [1 mark]
Which of the options has correct identification of ‘P’, ‘Q’ and ‘R’ in the illustration of ‘Central Dogma’ given below ?
(D) P – Transcription, Q – mRNA, R – Translation [1 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 :
(C) (iii), (i), (iv), (ii) [1 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
(A) 5′ – AUGAAUG – 3′ [1 mark]
Which one of the following codons has dual function ?
(A) AUG [1 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.
(C) X = Promoter, Y = Sigma factor, Z = RNA polymerase. [1 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’ ?
(C) 1 and 3 [1 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.
(C) (iii) → (iv) → (i) → (ii) → (v) [1 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.
(A) Individual 1 and 3 [1 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′
Answer not available: the official marking scheme gives no answer key for this question and awards the mark to everyone who attempts it.
Automated DNA sequences work on the principle of a method developed by :
(D) Frederick Sanger [1 mark]
The experimental proof on the thermal stability of genetic material was first provided by experiments of
(C) Frederick Griffith [1 mark]
Short stretches of DNA used to identify complementary sequences in a sample are called
(A) Probes [1 mark]
The initiation step during the process of transcription in bacteria is shown below. Identify A, B, C and D by selecting the option:
(D) A - Promoter, B - RNA polymerase, C - Sigma factor, D - DNA helix [1 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.
(A) No transformation [1 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?
(A) 25% hybrid and 75% light [1 mark]
Which mRNA will be translated to a polypeptide chain containing 8 amino acids?
(B) AUGAGACGGACUGCAUUCCCAACCUGA [1 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?
(C) 40,000 bp and 13,600×10-9 m [1 mark]
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.
(a) • Francis Crick [½]
(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]
[Total: 2 marks]
Assume that the given mRNA (start site is not depicted) is theoretically translated in two reading frames.
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.
Answer not available: no official marking scheme published.
Given below are the diagrammatic representations of the replicating fork of DNA in E. coli. Study the diagrams and answer the questions that follow.
Answer not available: no official marking scheme published.
Study the given molecular structure of double-stranded polynucleotide chain of DNA and answer the questions that follow.
Answer not available: no official marking scheme published.
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′
(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]
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.
[Total: 2 marks]
(a) Given below is a representation of the “Central dogma” :
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.
(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]
List the different criteria (4 points) that a molecule must fulfil to act as a genetic material.
A molecule that can act as a genetic material must fulfil the following criteria:
[½ × 4 = 2 marks]
How is the rate of initiation of RNA polymerase at a given promoter in a transcriptional unit of prokaryotes regulated?
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]
(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]
(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]
(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]
(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]
(a) Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical coding strand given below :
(b) Write the RNA strand transcribed from the above transcription unit along with its polarity.
(a) [½ × 4]
(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]
(b) The RNA transcribed as 5′ G A U A C G U A C G U A 3′ [1]
[Total: 3 marks]
Draw a labelled diagram of a ‘replicating fork’ showing the polarity. Why does DNA replication occur within such ‘fork’ ?
[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]
Study the diagram given below and answer the questions that follows :
(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.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Explain how the addition of lactose in the medium regulates the switching on of the lac operon in bacteria.
Answer not available: no official marking scheme published.
A segment of DNA, ‘TTG AGG GGG ATG’ was translated into an oligopeptide with the amino acids, Lysine – Serine – Proline – Tyrosine.
Answer not available: no official marking scheme published.
Draw a labelled schematic sketch of replication fork of DNA. Explain the role of any two enzymes involved in DNA replication.
(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]
(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.
(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]
(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 ?
(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]
(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]
(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]
Explain the processing of heterogeneous nuclear RNA (hnRNA) into a fully functional mRNA in eukaryotes. Where does this processing occur in the cell ?
[Total: 3 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.
(a) RNA polymerase III / RNA polymerase [1]
(b) [½ + ½]
(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]
OR
(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]
Human Genome Project (HGP) was a mega project launched in the year 1990 with some important goals.
(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]
(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.
(a)
| ‘R’ Strain | ‘S’ Strain |
|---|---|
| Non-virulent | Virulent |
| No polysaccharide coat | Have 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]
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]
(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]
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]
(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]
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 :
(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.
Answer not available: no official marking scheme published.
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.
OR
(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]
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.
OR
(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)
[½×4]
OR
(d) (In presence of inducer)
[½×4]
[Total: 4 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 ?
(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]
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]
(a) Observe the segment of mRNA given below and answers the questions given below :
(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
(a) (i)
(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]
| R | r | |
|---|---|---|
| R | RR Red coat | Rr Red coat |
| r | Rr 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]
| R | r | |
|---|---|---|
| R | RR Red coat | Rr Pink coat |
| r | Rr Pink coat | rr White coat |
F2 phenotypic ratio: 1 Red : 2 Pink : 1 White [½]
[Total: 5 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 :
(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]
(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]
(II) Carrier female × Normal male [½+½]
(III) Carrier female × Haemophilic male [½+½]
(ii) Conclusions
[Total: 5 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.
(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)
[½ × 6]
(iii) Lactose is the inducer / it inactivates repressor. [1]
[Total: 5 marks]
OR
Answer not available: no official marking scheme published.
(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.
(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
| Gametes | TV | tV |
|---|---|---|
| TV | TTVV Tall Violet | TtVV Tall Violet |
| tV | TtVV Tall Violet | ttVV 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
| Gametes | TV | Tv | tV | tv |
|---|---|---|---|---|
| TV | TTVV Tall violet | TTVv Tall Violet | TtVV Tall Violet | TtVv Tall Violet |
| Tv | TTVv Tall Violet | TTvv Tall white | TtVv Tall Violet | Ttvv Tall white |
| tV | TtVV Tall violet | TtVv Tall Violet | ttVV dwarf violet | ttVv dwarf Violet |
| tv | TtVv 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
| Gametes | TV | Tv |
|---|---|---|
| TV | TTVV Tall Violet | TTVv Tall Violet |
| Tv | TTVv 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]
OR
(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)
[½ × 4]
[Total: 5 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.
(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]
(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]
OR
(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]
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]
OR
(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]
OR
(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]
| R | r | |
|---|---|---|
| R | RR (Red) | Rr (Pink) |
| r | Rr (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]
A. Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical template strand given below.
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.
A.
[2]
B. [3]
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]
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.
OR
B.
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.
[1]
4th triplet codon codes for phenylalanine. [0.5]
The idea of use and disuse of organs for evolution of organism was proposed by
(D) Lamarck [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A). [1 mark]
What conclusion can be drawn if the forelimbs of an aquatic animal and a terrestrial animal have similar anatomical structure ?
(B) Share common ancestors [1 mark]
Appearance of antibiotic-resistant bacteria is an example of evolution due to :
(D) Anthropogenic action [1 mark]
Identify the statements that correctly describe Darwin’s Theory of Evolution.
Choose the correct option.
(A) (i), (ii), (iv) and (v) are correct [1 mark]
Name the extinct ancestor of humans who ate only fruits and hunted with stone weapons.
(B) Australopithecus [1 mark]
The given diagram shows the skulls of two different mammals.
Which of the following options accurately describes the differences between these skulls ?
(D) Skull A is of an adult chimpanzee and skull B is of human [1 mark]
Arrange the following in the correct sequence of their evolution and select the correct option :
Options :
(C) (ii), (iii), (i) [1 mark]
Select the odd option in the context of convergent evolution.
(D) Thorns of Bougainvillea and Tendrils of Cucurbita. [1 mark]
Choose the correct option that refers to adaptive radiation.
(A) Evolution of different species from a common ancestor to other geographical areas [1 mark]
Evolution of modern man involves the following man-like primates. Choose the correct series of human evolution.
Answer not available: no official marking scheme published.
Which one of the following options shows the correct evolutionary order of the plants mentioned below ?
Choose the correct option.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
About 15 mya during human evolution, the primates which used to walk like gorillas and chimpanzees were :
Answer not available: no official marking scheme published.
Which of the following combinations is a correct example of convergent evolution in Australian marsupials and Placental mammals ?
| Australian Marsupials | Placental Mammals | |
|---|---|---|
| (A) | Lemur | Spotted cuscus |
| (B) | Tasmanian tiger cat | Anteater |
| (C) | Bobcat | Lemur |
| (D) | Numbat | Anteater |
Answer not available: no official marking scheme published.
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 :
Answer not available: no official marking scheme published.
In his observations of small black birds in the Galapagos Islands, Darwin found that all the finches arose from the original ancestor :
Answer not available: no official marking scheme published.
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 ?
Choose the correct option :
Options (A)–(D) are shown in the figure.
Answer not available: no official marking scheme published.
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 :
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
Which one of the following statements is not true ?
(D) Homologous structures are a result of convergent evolution. [1 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 :
(C) 0·48 [1 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 :
(B) less in number [1 mark]
S.L. Miller in 1953, to support the theory of chemical evolution, created conditions in the closed flask that included :
(B) CH4, H2, NH3, H2O vapour at 800°C [1 mark]
Who proposed the mutation theory in favour of organic evolution ?
(D) Hugo de Vries [1 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 ?
(D) CH4, H2, NH3 and water vapour at 800 °C [1 mark]
The most primitive ancestor of humans is :
(C) Ramapithecus [1 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.
(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).)
Identify the option that gives the correct type of evolution exhibited by the two animals shown, living in the same habitat in Australia.
(A) convergent evolution [1 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.
(C) A is true but R is false. [1 mark]
At which stage during evolution did human use hides to protect their bodies and buried their dead ?
(B) Neanderthal man. [1 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.
(C) (A) is true, but (R) is false. [1 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.
(C) Directional selection as giraffes with longer neck lengths are selected. [1 mark]
Select the incorrect statement among the following.
(C) Hardy-Weinberg principle says that phenotype frequencies in a population are stable and are constant from generation to generation. [1 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?
(D) The leg mutation might lead to reproductive isolation and speciation due to an effect on the mating call. [1 mark]
(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 + 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 :
(Working to be shown) [½ + ½ + ½ + ½ = 2]
(a) (i) (Any two) [½+½]
(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]
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 :
Answer not available: no official marking scheme published.
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.
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]
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.
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]
(a) How is Hardy-Weinberg expression (p2 + 2pq + q2) = 1 derived ?
(b) List any two factors that disturb the genetic equilibrium.
(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]
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]
(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]
Explain the three different ways in which natural selection operates. Explain with the help of graphs.
[½]
[½]
[½]
// (alternative: combined diagram)
[½ × 6]
[Total: 3 marks]
Study the given chart showing evolution of plants. Answer the following questions :
(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]
Explain the Hardy-Weinberg principle with the help of an algebric equation.
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]
Compare and contrast convergent and divergent evolution.
Answer not available: no official marking scheme published.
How does the process of Natural Selection affect Hardy-Weinberg equilibrium ? Explain with the help of graphs.
Answer not available: no official marking scheme published.
Answer not available: no official marking scheme published.
(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 ?
(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]
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).
(a) Hardy–Weinberg Equilibrium: p2 + 2pq + q2 = 1; AA + 2Aa + aa = 1
(b) Natural selection [1]
[Total: 3 marks]
(a) A – Modern human being; B – Baby chimpanzee; C – Adult chimpanzee [½ × 3]
(b) (i) Dryopithecus (ii) Ramapithecus [½ + ½]
[Total: 3 marks]
Differentiate between the explanations given by Darwin and de Vries respectively on the mechanism of evolution. Write any three differences.
| Darwin’s Evolution | de 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 speciation | Single step mutation causes speciation |
| Variations are small and directional | Mutations are random and directionless |
| Theory is based on natural selection | Theory is based on mutation |
(Any three points) [1 × 3]
[Total: 3 marks]
(a) Dryopithecus, Ramapithecus [½+½]
(b) Time period: 2 million years ago [½]
Place: East African grasslands [½]
(c)
| Homo habilis | Homo 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]
(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]
Industrial melanism in England after 1850 is an excellent example of Natural selection. Explain how ?
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]
Explain with the help of a diagrammatic representation, how natural selection operates on different traits in a population.
(½ mark for each figure with correct labelling) [½ × 6 = 3 marks]
Shown below are certain aquatic vertebrates, where natural selection has favoured them to develop certain characteristics which enable them to live in water.
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]
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. | Animal | No. of amino acid differences in haemoglobin with that of humans |
|---|---|---|
| 1. | Macaque | 08 |
| 2. | Dog | 32 |
| 3. | Bird | 45 |
| 4. | Frog | 67 |
| 5. | Lamprey | 125 |
OR
Answer not available: no official marking scheme published.
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 :
| Groups | Nucleotide substitution in the gene of Cytochrome-C | Millions of years ago |
|---|---|---|
| Human/Kangaroo | 100 | 125 mya |
| Human/Cow | 75 | 120 mya |
| Human/Rodent | 60 | 75 mya |
| Options | Time of separation of two groups during evolution | Number of nucleotide substitutions |
|---|---|---|
| (i) | Lesser | Greater |
| (ii) | Greater | Lesser |
| (iii) | Greater | Greater |
OR
Answer not available: no official marking scheme published.
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. | Vertebrates | Percentage similarities |
|---|---|---|
| 1. | Chimpanzee | 100 |
| 2. | Gibbon | 94 |
| 3. | Rhesus Monkey | 88 |
| 4. | Lemur | 47 |
| 5. | Treeshrew | 28 |
| 6. | Mouse | 21 |
| 7. | Hedgehog | 19 |
| 8. | Chicken | 10 |
OR
Answer not available: no official marking scheme published.
(a) (i)
Natural selection operates in different ways in nature.
(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 :
(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 Vries | Darwin |
|---|---|
| 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 sudden | 3. 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)
(I) Tyr at the amino acid site [1]; (II) anticodon at its correct site [1]
[Total: 5 marks]
OR
(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]
Which of the following is not a component of immune system ?
(C) Red Blood Cells [1 mark]
What is the role of histamine released by mast cells ?
(B) Causes allergic symptoms [1 mark]
Which of the following is not a symptom of Ascariasis ?
(D) Skin ulcers [1 mark]
Some statements regarding drugs and alcohol are given below :
Choose the option with correct statement(s).
(B) Statements (i), (ii) and (iii) are correct [1 mark]
Nicotine affects blood pressure and heart rate due to release of certain adrenal hormones. Following are some of the adrenal hormones :
Select the correct option :
(D) (i) and (iii) [1 mark]
Tetanus antitoxin, when injected into the human body, immediately provides which kind of immunity ? Choose the correct option.
(B) Passive immunity [1 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.
(D) Assertion (A) is false, but Reason (R) is true. [1 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 ?
Answer not available: no official marking scheme published.
The decrease in the T-Lymphocytes count in human blood will finally result in
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
In which of the following human diseases does the body’s self-defence mechanism attack self-cells ?
Answer not available: no official marking scheme published.
The technique for the early detection of a disease based on the principle of antigen-antibody interaction is :
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Colostrum secreted by the mother’s mammary glands in a human female during the initial days of lactation is rich in antibody :
Answer not available: no official marking scheme published.
A characteristic property that distinguishes a malignant tumor from a benign tumor is :
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
The following information is about drugs and tobacco. Select the correct statement from the options given below.
Answer not available: no official marking scheme published.
Which one of the following immune system components does not correctly match with its respective role ?
Answer not available: no official marking scheme published.
Diseases like pneumonia and common cold are caused by pathogens. Choose the option that correctly indicates the difference between Pneumonia and Common cold :
(C) Pneumonia pathogen infects alveoli of the lungs, whereas common cold affects nose and respiratory passage but not lungs [1 mark]
In which of the following conditions/diseases is there a substantial increase in the activity of mast cells observed in the human body ?
(B) Allergy [1 mark]
The vector for dengue fever is :
(A) Female Aedes mosquito [1 mark]
Assertion (A): Virus-infected cells produce interferons.
Reason (R): Interferons can cause inflammation of virus-infected cells.
(C) Assertion (A) is true, but Reason (R) is false [1 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 ?
(C) Typhoid [1 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.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). [1 mark]
ELISA technique is based on the principle of
(B) Antigen-antibody interaction [1 mark]
Interferons are proteins secreted by
(D) Virus infected cell [1 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.
(C) (A) is true, but (R) is false. [1 mark]
The mosquito-borne disease in humans causing chronic inflammation of the lymphatic vessels is :
(A) Elephantiasis [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). [1 mark]
Select the options which is/are incorrect statement(s) with respect to T-lymphocytes in the human body.
(B) (iii) only [1 mark]
//
(D) (iii) and (iv) only [1 mark]
(The scheme accepts either (B) or (D).)
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.
(A) Both A and R are true and R is correct explanation of A. [1 mark]
The decrease in the T-lymphocytes count in human blood will result in :
(D) Increase in antigens. [1 mark]
Select the pathogen mismatched with the symptoms of disease caused by it from the list given below :
(D) Haemophilus influenzae: Blockage of the intestinal passage. [1 mark]
Choose the correct option in which the diagram of the flowering plant shown below is correctly matched with the drug obtained from it :
(A) Dhatura – Hallucinogen [1 mark]
Assertion (A): Cocaine produces a sense of euphoria and energy.
Reason (R): It interferes with the transport of neurotransmitter dopamine.
(A) Both (A) and (R) are true, and (R) is the correct explanation of (A). [1 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 | |
|---|---|---|---|---|
| A | Sporozoites | Erythrocyte infection | Liver infection | Gametocytes |
| B | Gametocytes | Erythrocyte infection | Liver infection | Sporozoites |
| C | Gametocytes | Liver infection | Erythrocyte infection | Sporozoites |
| D | Sporozoites | Liver infection | Erythrocyte infection | Gametocytes |
(D) Sporozoites, Liver infection, Erythrocyte infection, Gametocytes [1 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?
(B) Inflammation or fluid buildup in the lungs [1 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.
(A) Both A and R are true and R is the correct explanation of A [1 mark]
(a) Define Humoral immune response.
(b) Name any two types of antibodies found to give humoral immune response in humans.
(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]
(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.
[½ + 1 + ½ = 2]
OR
(b) A girl develops sneezing, watery eyes and difficulty in breathing every time she visits a flower garden.
[½ + ½ + ½ + ½ = 2]
(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]
Name the lymphoid organ that acts as a large reservoir of erythrocytes. Explain its role as a lymphoid organ.
[Total: 2 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.
(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]
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 :
(ii) Explain how cyanobacteria can be used as bio-fertilizer.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
(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.
OR
(b) Why do organic farmers not recommend complete eradication of insect pests ? Explain giving reason.
(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]
Identify A, B, C and D in the following table :
| Scientific name of the plant | Drug | Effect on the human body/human system | |
|---|---|---|---|
| (a) | Papaver somniferum | A | Depressant/slows down body function |
| (b) | Cannabis sativa | Cannabinoids | B |
| (c) | Erythroxylum coca | C | D |
[½ × 4 = 2 marks]
What is a vaccine ? Write the basis on which it acts when administered in the body.
[Total: 2 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.
(a) Bone marrow [1]
(b) T-Lymphocytes; Maturation / Proliferation / become antigen sensitive / become effector cells. [½ + ½]
[Total: 2 marks]
Mention the parts of human body that get affected by Pneumonia and common cold infections. Write the causative agents of the two diseases.
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]
What is the importance of female Anopheles mosquitoes in the life of a malarial Parasite Plasmodium?
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)
[½ × 4]
[Total: 2 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.
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]
(a) A person suffering from AIDS dies of opportunistic infections (ARC) i.e., infections that could have been otherwise overcome.
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.
(a)
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]
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?
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]
(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.
(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]
Explain, giving two reasons, how immune response by “vaccine” is different from that by “antitoxin” in humans.
| Immune response by vaccine | Immune response by antitoxin |
|---|---|
| Immune response is generated by antibodies produced in response to weakened antigen introduced in the body | Immune response is generated by pre-formed antibodies or antitoxin is injected |
| Response takes sometime | Quick immune response |
| Memory cells are generated / long lasting | One time response |
| Vaccine provides active immunity | Antitoxin provides passive immunity |
[1+1] (Any two differences)
[Total: 2 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.
● 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]
(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.
(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]
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.
● 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]
(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?
(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]
The table below shows a hypothetical blood report of a patient.
| Test description | Observed value | Unit | Reference range |
|---|---|---|---|
| Leucocytes | |||
| Total leucocyte count | 1100 | Per Microliter | 4400-11000 |
| Neutrophils | 31 | % | 55-70 |
| Lymphocytes | 25 | % | 20-40 |
| Basophils | 0.5 | % | 0.5 - 1 |
| Eosinophils | 02 | % | 1-4 |
| Monocytes | 0 | % | 1-8 |
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]
The following is a realistic example dataset and not a real world patient dataset:
| Sample ID | Tumor size (mm3) | Glucose level mM/litre | Oxygen level (mmHg) | Neighboring cell's viability/ vitality (%) |
|---|---|---|---|---|
| 1 | 150 | 2.0 | 15 | 60 |
| 2 | 50 | 4.5 | 30 | 85 |
| 3 | 200 | 1.3 | 10 | 50 |
| 4 | 100 | 3.0 | 20 | 70 |
| 5 | 250 | 0.9 | 7 | 40 |
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]
(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.
(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]
(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]
(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]
[Total: 3 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 ?
(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]
(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.
(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]
Study a part of life cycle of Plasmodium given below :
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.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
A person is suffering from high grade fever. Which symptoms will help to identify if he/she is suffering from Typhoid, Pneumonia or Malaria ?
Answer not available: no official marking scheme published.
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.
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]
(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.
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
Identify A, B, C, D, E and F in the table given below :
| Name of Human Disease | Causative Organism | Symptoms |
|---|---|---|
| Pneumonia | Streptococcus | ‘A’ |
| Typhoid | ‘B’ | High fever, weakness, headache, stomach pain |
| Common Cold | Rhino virus | ‘C’ |
| Ringworm | ‘D’ | Dry scaly lesions on body parts, redness, itching |
| Ascariasis | Ascaris | ‘E’ |
| ‘F’ | Entamoeba histolytica | Constipation, cramps, stools with mucous and blood clots |
[½ × 6 = 3 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.
(a)
| Humoral immune response | Cell-mediated immune response | |
|---|---|---|
| (i) | Mediated by B-lymphocytes | Mediated 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)
(Any correct four labels) [½ × 4]
[Total: 3 marks]
OR
(a) (i) Cannabinoids [1]
(ii) Inhalation, oral ingestion [½ + ½]
(iii) Brain / Heart (cardiovascular system) [1]
OR
(b)
[Four labelling – antigen binding site, light chain, heavy chain, C-terminal, N-terminal, disulphide bridge – ½ mark each] [½ × 4]
[Total: 3 marks]
(a)
| S. No. | Malignant tumor | Benign tumor |
|---|---|---|
| 1 | Cells 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 |
| 2 | Show metastasis | Do 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]
(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]
OR
(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]
//
[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]
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.
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)
[Total: 3 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.
(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]
(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.
(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]
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.
(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]
(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?
(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]
‘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.
● 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]
(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.
(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]
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.
● 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]
Given below is a diagrammatic representation of immune system of the human body:
(a) Identify ‘X’ and ‘Y’ in the given diagram.
(b) Explain two major functions of the organs that you have identified.
(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]
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.
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]
Expand ELISA. On what principle is the ELISA test based? List two ways by which an infection can be detected by this test.
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]
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.
(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]
(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]
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]
(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 Immunity | Passive 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]
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.
OR
(d) How is active immunity different from passive immunity ?
Answer not available: no official marking scheme published.
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.
OR
Answer not available: no official marking scheme published.
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.
OR
Answer not available: no official marking scheme published.
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.
OR
Answer not available: no official marking scheme published.
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.
(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]
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.”
OR
(a) (i) Sporozoites, (ii) gametocytes [½ + ½]
(b)
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]
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.
(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]
The diagram shows the life cycle of a pathogenic protozoan.
OR
(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]
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 :
OR
(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)
[½×4]
(c) • Period of illness [½]
• B lymphocytes/B cells [½]
[Total: 4 marks]
The graph below shows the Antibody concentration in young calves. Study the graph and answer the questions that follow:
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)
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]
(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.
(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]
(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 :
(a) (i) Disease: Acquired Immuno Deficiency Syndrome / AIDS [1]
Causative Agent: Human Immuno deficiency Virus / HIV [1]
(ii) [½ × 4]
// (alternatively, the following diagram)
(iii) (Any Two) [½+½]
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]
(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]
(a) (i) (Any four labellings) [½ × 4]
(ii) (any three correct difference) [1×3]
| Active Immunity | Passive Immunity |
|---|---|
| When antibodies are produced in our body in response to an antigen then it is called active immunity | When 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 response | It gives quick response |
| It lasts for longer period | It lasts for shorter period |
| Memory cells are formed | Memory 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]
Student to attempt either option-(A) or (B) :
(A) Answer the following questions :
OR
(B) (i) Study the following table & fill ‘H’, ‘I’, ‘J’, ‘K’, ‘L’ and ‘M’ in following table with suitable words :
| Chemical / Bioactive Molecule | Micro-organism | Category | Use | |
|---|---|---|---|---|
| (a) | Butyric acid | H | I | Important applications in food, chemical & pharma industry |
| (b) | J | Monoscus purpureus | K | Inhibit cholesterol biosynthesis pathway |
| (c) | Cyclosporin A | L | Fungus | M |
(ii) Why are baculoviruses used as biological control agents ?
Answer not available: no official marking scheme published.
(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 :
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
(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 :
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)
(ii) Name any two organisms commonly used as biofertilisers, belonging to different kingdoms. Write how each one acts as a biofertiliser.
(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]
OR
(a) (i) Life cycle of the Plasmodium from the period it enters the human body till a female Anopheles mosquito bites an infected person.
(or as in the diagram, each labelled stage [½ × 6])
(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 tumor | Benign 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)
(Any three techniques with explanation) [½ + ½] × 3
[Total: 5 marks]
OR
(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)
OR
(b) (i)
(ii) [½ × 5]
[Total: 5 marks]
| Type of immunity | Production of antibodies | Presence of memory cells | Mode attained | |
|---|---|---|---|---|
| (1) | Natural, active | Yes | ‘P’ | ‘Q’ |
| (2) | Natural, passive | No | ‘R’ | Across the placenta during pregnancy/breast feeding |
| (3) | Acquired, active | ‘S’ | Yes | Getting a vaccine during breast feeding |
| (4) | Acquired, passive | ‘T’ | No | Getting an injection of antibodies |
OR
(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]
(a) Explain the following with reference to ‘Cancer’ :
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.
(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)
(iv)
| Benign tumours | Malignant tumours |
|---|---|
| Remain confined to their original location | Not remain confined to their original location / Show property of metastasis |
| Cause little damage | Damage surrounding tissue and starve normal cells by competing for vital nutrients |
(1 mark for each correct difference) [1 + 1]
OR
(b)
(i)
(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]
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.
(C) Assertion (A) is true, but Reason (R) is false. [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A). [1 mark]
Assertion (A): Activated sludge contains large population of bacteria.
Reason (R): These bacteria help in anaerobic digestion of human waste in biogas plant.
(C) Assertion (A) is true, but Reason (R) is false [1 mark]
Microbes used for biocontrol of pest butterfly caterpillars is :
(B) Bacillus thuringiensis [1 mark]
Assertion (A): Nucleopolyhedrovirus are useful as they are good biocontrol agents.
Reason (R): They are species-specific, narrow spectrum bioinsecticides.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A) [1 mark]
Monascus purpureus is a yeast used commercially in the production of which one of the following ?
(D) Statins [1 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.
(B) Both Assertion (A) & Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A) [1 mark]
Identify the microbe that is responsible for producing large-holes in Swiss cheese.
(A) Propionibacterium sharmanii [1 mark]
Match the items in Column-A with that of Column-B :
| Column-A | Column-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 :
Answer not available: no official marking scheme published.
For commercial and industrial production of citric acid, which one of the following microbes is used ?
Answer not available: no official marking scheme published.
Bioactive molecule Cyclosporin A used for human welfare is derived from :
Answer not available: no official marking scheme published.
Large scale industrial production of Butyric acid for human welfare is done using the microbe :
Answer not available: no official marking scheme published.
Assertion (A): Biogas plants are more often built in rural areas.
Reason (R): The excreta or ‘gobar’ of cattle is rich in Methanobacterium.
Answer not available: no official marking scheme published.
Bottled fruit juices are clearer as compared to those made at home, as they are clarified by the use of :
Answer not available: no official marking scheme published.
Assertion (A): Swiss cheese is characterized by large holes due to CO2 production.
Reason (R): It is ripened by growing a specific fungi.
Answer not available: no official marking scheme published.
Lactobacillus that sets milk into curd is categorised as :
(D) Heterotrophic bacteria [1 mark]
Which one of the following pairs is not correctly matched ?
(C) Monascus purpureus – Citric Acid [1 mark]
Study the following list of bioactive substances and their action :
| Bioactive Substance | Role |
|---|---|
| A. Statin | I. Removal of oil stains |
| B. Cyclosporin A | II. Removal of clots from blood vessels |
| C. Streptokinase | III. Lowering of blood cholesterol |
| D. Lipase | IV. Immuno-suppressive agent |
Select the option in which the bioactive substances are correctly matched with their action.
(B) A – III, B – IV, C – II, D – I [1 mark]
During biological treatment of sewage, the masses of bacteria held together by fungal filaments to form mesh like structures are called
(B) Flocs [1 mark]
In plants, which one of the following helps in the absorption of phosphorus from soil ?
(A) Glomus [1 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 ?
(A) Point (i) [1 mark]
Identify the fungus that ripens the famous ‘Roquefort’ cheese :
Answer not available: the official marking scheme gives no answer key for this question and awards the mark to everyone who attempts it.
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 Products | S. 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 |
(D) A – (ii), B – (iv), (C) – (i), (D) – (iii). [1 mark]
Integrated Pest Management involves
(C) Both I and II [1 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?
(C) Adding spores of Glomus to the soil [1 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.
(C) A is true but R is false. [1 mark]
After significant reduction of BOD during secondary treatment of sewage, what further processes are carried out ? Mention any one important outcome of these processes.
[Total: 2 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 ?
(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]
Explain how are plants benefitted by their association with “Glomus species”.
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]
(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]
OR
(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]
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.
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]
(a) Give an example of viral biocontrol agent.
(b) Why are they considered to be desirable when an ecologically sensitive area is being treated?
(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]
"Curd is easier to digest by humans than milk." Justify, giving two suitable reasons.
LAB produce acids that coagulate milk [1]
Partially digests milk proteins [1]
[Total: 2 marks]
Study the given diagrams of Biogas plants – 'A' and 'B' and answer the questions that follow:
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.
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]
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.
(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]
Organic farmer use Trichoderma and Baculovirus as biological control agents. Explain.
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]
(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.
(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]
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.
● 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]
Different species belonging to genus Trichoderma are useful to humans as well as to plants. Justify their roles by giving one instance of each.
Trichoderma act as biocontrol agent for several plant pathogens, Trichoderma produces immunosuppressive agent / cyclosporin A / immunosuppressant useful for humans [1+1]
[Total: 2 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.
● 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]
Study the given diagram of Sewage Treatment Plant (S.T.P.) and answer the questions that follow:
(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”.
(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]
A particular cyanobacteria is spread by the farmers in their fields while growing paddy. Name the cyanobacteria used and give two advantages of it.
● 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]
Study the given diagrams of Sewage Treatment Plants (STP) and answer the questions that follow:
(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?
(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]
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]
(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]
Mention a product of human welfare obtained with the help of each one of the following :
(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]
Explain the biological treatment of primary effluent when passed into the large aeration tanks in a sewage treatment plant (STP).
Answer not available: no official marking scheme published.
Answer not available: no official marking scheme published.
Answer not available: no official marking scheme published.
OR
(a) (i)
(ii) Used as manure / fertilizer (organic) [1]
OR
(b)
| Bioactive molecule | Microbial source | ||
|---|---|---|---|
| (i) | Myocardial infarction | Streptokinase | Streptococcus |
| (ii) | High blood cholesterol level | Statins | Monascus purpureus |
| (iii) | Organ transplantation | Cyclosporin A | Trichoderma polysporum |
[½ × 2] for each of (i), (ii), (iii)
[Total: 3 marks]
Explain the role of the following during the sewage treatment :
(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]
Identify a, b, c, d, e and f in the table given below :
| Sl. No. | Organism | Bioactive Molecule | Use |
|---|---|---|---|
| 1. | Monascus purpureus | a | b |
| 2. | c | d | Antibiotic |
| 3. | e | Cyclosporin A | f |
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]
Treatment of wastewater is done in a sewage treatment plant to make it less polluting. Explain the following with reference to this treatment process :
(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]
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.
[Total: 3 marks]
Enumerate the main sources of bio-fertilisers giving one example of each.
Bacteria – Rhizobium / Azospirillum / Azotobacter [½ + ½]
Fungi – Mycorrhiza / Glomus [½ + ½]
Cyanobacteria – Anabaena / Nostoc / Oscillatoria [½ + ½]
[Total: 3 marks]
Explain giving reason the action plan followed by organic farmers that support their key belief "biodiversity furthers health of crop lands".
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]
Bacillus thuringiensis plays an important role in Integrated Pest Management strategy. Explain how. Name any two crops that are protected efficiently from pests.
● 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]
The Biological Oxygen Demand (BOD) of a primary effluent during sewage treatment is reduced. Enlist the process of how this is achieved.
‘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.
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 :
(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 ?
(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]
Study the figure given below and answer the questions that follow.
OR
Answer not available: no official marking scheme published.
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)
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]
(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.
(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]
Match Column-I with Column-II and choose the correct option :
| Column-I | Column-II | ||
|---|---|---|---|
| a. | Biolistic gun | i. | Bacterial cell |
| b. | Chitinase | ii. | Tumour inducing |
| c. | Ti | iii. | Animal cell |
| d. | Ca++ | iv. | Fungal cell |
| v. | Plant cell |
Options :
(C) a – v, b – iv, c – ii, d – i [1 mark]
Which of the following statements about plasmids is incorrect ?
(B) Their replication is controlled by chromosomal DNA [1 mark]
Which of the following is used to visualize DNA bands under UV light after gel electrophoresis ?
(C) Ethidium Bromide [1 mark]
Assertion (A): DNA Ligase is used to join DNA fragments.
Reason (R): It catalyses the formation of glycosidic bonds between nucleotides.
(C) Assertion (A) is true, Reason (R) is false. [1 mark]
Match Column I with Column II and select the correct option :
| Column I | Column II |
|---|---|
| a. Primers | i. β-galactosidase |
| b. Insertional inactivation | ii. To separate and purify products |
| c. Bioreactor | iii. Chemically synthesized oligonucleotides |
| d. Downstream processing | iv. Large scale production of specific product of activity of microbes |
Options :
(A) a-iii, b-i, c-iv, d-ii [1 mark]
Assertion (A): DNA cannot pass through cell membrane of a bacterial cell.
Reason (R): DNA is a hydrophilic molecule.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of the Assertion (A) [1 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.
(D) 5′ C-T-G-C-A G 3′ / 3′ G A-C-G-T-C 5′ [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A) [1 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.
(C) P – Deoxynucleotide(s), Q – Thermostable DNA polymerase [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]
Amplification of gene of interest by using DNA polymerase may go upto
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Select the correct statement from the following biotechnological procedures :
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Isolation of DNA from a fungal cell can be achieved by using :
Answer not available: no official marking scheme published.
The cloning site present in the tetracycline resistance gene of E. coli cloning vector pBR322 is :
Answer not available: no official marking scheme published.
Identify the incorrect statement regarding PCR.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Identify the material that helps in the precipitation of purified DNA during the process of isolation of DNA as shown in the given diagram.
(D) Chilled Ethanol [1 mark]
What is the process of binding a primer to the denatured strand called ?
(B) Annealing [1 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.
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
Restriction Endonuclease Hind II always cuts DNA molecules at a particular point by recognising a specific sequence of :
(A) Six base pairs [1 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.
(D) Assertion (A) is false, but Reason (R) is true [1 mark]
Which one of the following is not a feature of plasmids ?
(C) Single stranded [1 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.
(C) Assertion (A) is true, but Reason (R) is false [1 mark]
Which native plasmid did Stanley Cohen and Herbert Boyer use for the construction of the first recombinant DNA ?
(A) Salmonella typhimurium [1 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.
(B) [1 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.
(C) Assertion (A) is true, but Reason (R) is false. [1 mark]
The ‘molecular scissors’ fall in the category of :
(D) Restriction enzymes [1 mark]
Assertion (A): Plasmids are autonomously replicating circular extra-chromosomal DNA.
Reason (R): Plasmids are usually present in Eukaryotic cells.
(C) Assertion (A) is true, but Reason (R) is false. [1 mark]
Which one of the following statements is correct in the context of observing DNA separation by agarose gel electrophoresis ?
(D) Ethidium bromide stained DNA can be seen under UV light [1 mark]
The sequence that controls the copy number of linked DNA in the vector is termed :
(B) Ori site [1 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 Ligase | i. | Removes nucleotides from ends of DNA |
| Q. | Restriction exonuclease | ii. | Extends primer on a DNA template |
| R. | Taq polymerase | iii. | Joins the DNA fragments |
| S. | Restriction endonuclease | iv. | Cuts DNA at a specific position |
(D) P - (iii), Q - (i), R - (ii), S - (iv) [1 mark]
The organism used in construction of the first artificial recombinant DNA by Cohen and Boyer in 1972 was :
(B) Salmonella typhimurium. [1 mark]
The given schematic illustration shows three steps ‘P’, ‘Q’ and ‘R’ of the polymerase chain reaction.
Which of the following statements are correct with reference to the illustration given above ?
(B) (ii) and (iii) only [1 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 by the action of the enzyme Pst-I.
Options (A)–(D) are shown in the figure.
(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]
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.
(C) (A) is true, but (R) is false. [1 mark]
The vector for ‘T-DNA’ is :
(B) Agrobacterium tumefaciens [1 mark]
Plasmids are suitable vectors for gene cloning because :
(B) They are small circular DNA molecules with their own origin of replication site. [1 mark]
In order to isolate genetic material of a bacterium, the cell must be treated with
(A) Lysozyme, ribonuclease, protease, chilled ethanol [1 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?
(C) Ensuring sterile handling, sterilised equipment, and aseptic transfer before and during fermentation. [1 mark]
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 ?
(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]
How do scientists use microinjection and gene gun in biotechnology ?
[Total: 2 marks]
(a) (i) Given below is the schematic representation of the process of electrophoresis. Identify the alphabets representing the :
(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+1 = 2]
(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]
Write the role of ‘ori’ and ‘restriction’ site in a cloning vector pBR322.
[Total: 2 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.
(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]
[Total: 2 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’ ?
Answer not available: no official marking scheme published.
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-1 | Vector DNA (cut using Restriction Enzyme EcoR I) + Alien DNA (cut using Restriction Enzyme EcoR I) |
| ↓ Enzyme A | |
| Step-2 | Recombinant DNA molecule |
| ↓ | |
| Step-3 | Transfer of recombinant DNA molecule in E. coli (Host) |
| ↓ | |
| Step-4 | Replication of recombinant DNA molecule in E. coli |
Answer not available: no official marking scheme published.
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-1 | Plasmid DNA (cut using Restriction Enzyme EcoR I) + Foreign DNA (cut using Restriction Enzyme EcoR I) |
| ↓ DNA ligase | |
| Step-2 | Recombinant DNA molecule |
| ↓ | |
| Step-3 | Transfer of recombinant DNA molecule to the host cell |
| ↓ | |
| Step-4 | Replication of recombinant DNA molecule in the host cell |
Answer not available: no official marking scheme published.
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 1 | Vector DNA (Plasmid) (cut using Restriction Enzyme) + Alien DNA (cut using Restriction Enzyme) |
| ↓ | |
| Step 2 | Recombinant DNA molecule |
| ↓ | |
| Step 3 | Transfer of recombinant DNA molecule in E. coli (Host) |
| ↓ | |
| Step 4 | Replication of the recombinant DNA molecule in E. coli to form multiple copies of the alien gene |
Answer not available: no official marking scheme published.
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 :
(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.
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
Write the role of ‘ori’ and restriction site in the cloning vector pBR322.
[Total: 2 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 :
(a) Set-up A [½]
(b) I [½]
[Total: 2 marks]
(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]
[Total: 2 marks]
Why is making host cells ‘competent’ essential for rDNA technology ? Mention any two ways by which this can be achieved.
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]
OR
(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]
OR
(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]
(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]
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?
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]
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 :
(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.
What changes were made to the previous design to get a band at ‘X’. Why did the band appear at ‘X’ ?
(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]
(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.
Answer not available: no official marking scheme published.
Explain the basis on which gel electrophoresis technique works. Write any two ways the products obtained through this technique can be utilised.
Answer not available: no official marking scheme published.
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.
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]
Study the steps shown below, that are carried during a specific technique :
(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]
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.
[½×6]
[Total: 3 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.
(a) Name the technique and its purpose.
(b) Explain the procedure carried out under the following steps :
(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]
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.
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]
OR
What are plasmids? How are they different from cloning vectors? Give one example each for a viral and a bacterial cloning vector.
OR
[Total: 3 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.
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]
(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
(b) What is elution? State the importance of elution in this process.
(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]
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.
(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)
[Total: 3 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.
● 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]
(½ mark each for denaturation, annealing, extension and repetition as shown in the figure)
[Total: 3 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.
(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]
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.
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]
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]
(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]
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 :
(a) Why are plasmids and bacteriophages used as cloning vectors ? [1]
(b) (i) Identify : [½ + ½ = 1]
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]
(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]
Study the diagram given below that shows the steps involved in the procedure of selecting transformed bacteria and answer the questions that follow :
(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.
(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]
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.
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′
OR
(a)
(b) Alu I; Alu I site is present in the given sequence and BamH I site is not given. [1]
(c)
[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]
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.
(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]
An illustration given below was on the display board in the class. Study the illustration and answer the questions that follow.
(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.
(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]
(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]
(a) (i)
(ii)
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:
[Total: 5 marks]
(a) (i) EcoR I has played a very important role in rDNA technology.
(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]
(a) (i) (I) Naming of restriction enzymes (EcoRI):
(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)
(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]
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.
(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]
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.
Answer not available: no official marking scheme published.
(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.
(a) (i)
(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]
OR
(a) (i)
[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]
OR
(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]
Answer the following questions w.r.t. the given paragraph :
OR
(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]
(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:
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:
(a)
I. BamHI or SalI; II. PvuI or PstI; III. Ori (any two correct labellings)
OR
(b)
[Total: 5 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.
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.
(a)
OR
(b)
[Total: 5 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:
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:
(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]
(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.
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:
(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]
‘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.
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.
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]
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]
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:
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.
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.
[2]
III. Insertional inactivation [1]
OR
B.
I.
[2]
II. The further steps are
[2]
III. Spooled DNA can be used in genetic engineering, PCR amplification, DNA fingerprinting, or sequencing. [1]
Pomato was produced by fusing protoplasts of
(A) Tomato and Potato [1 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.
(C) Assertion (A) is true, but Reason (R) is false. [1 mark]
Genetically modified tobacco plants were produced using Agrobacterium vectors. Which property of Agrobacterium helps in carrying nematode specific genes into the host plant ?
(B) Infectious for dicot plant; has Ti plasmid. [1 mark]
Golden rice is a promising transgenic crop. When released for cultivation, it will help in :
(B) Reduction of Vitamin A deficiency in humans [1 mark]
In RNAi, genes are silenced using which of the following biomolecules ?
(C) dsRNA [1 mark]
Protein encoded by gene Cry-I Ab controls the infestation of which of the following ?
(C) Corn borer [1 mark]
Transgenic animals have their genetic material manipulated. Select the correct option in reference to transgenic animals.
(C) Foreign DNA is present in all of its cells. [1 mark]
Cry proteins coded by the gene Cry IAb controls which one of the following pests ?
(B) Corn borer [1 mark]
Genetically modified tobacco plants were produced using Agrobacterium vectors. Which properties of Agrobacterium help in carrying nematode-specific genes into the host plant ?
(B) Infectious for dicot plant; has Ti plasmid [1 mark]
GEAC stands for
Answer not available: no official marking scheme published.
RNA interference (RNAi) helps in making tobacco-plant resistant to a nematode (Meloidegyne incognitia).
Choose the correct option that shows how RNAi is achieved :
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Assertion (A): Isolated single cells can be fused to produce somatic hybrids.
Reason (R): Cells selected for somatic hybridisation have desirable characters.
Answer not available: no official marking scheme published.
Which one of the following is not the product of transgenic experiments ?
Answer not available: no official marking scheme published.
Out of the following, select the correct match :
Answer not available: no official marking scheme published.
Golden rice is a promising transgenic crop. When released for cultivation, it will help in :
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
Which one of the following transgenic animals is being used to test the safety of the polio vaccine ?
(D) Mice [1 mark]
The improved trait that is found in the genetically modified transgenic crop Golden rice is :
(D) High vitamin – A content [1 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.
(C) Assertion (A) is true, but reason (R) is false. [1 mark]
Assertion (A): Patents are granted by government to an inventor.
Reason (R): Patents prevents others from commercial use of an invention.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 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.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A). [1 mark]
Crystals of Bt toxin produced by some bacteria, do not kill the bacteria producing them because :
(C) toxin is inactive [1 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.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). [1 mark]
Who among the following challenged the patent right granted to the University of Mississippi Medical Centre for ‘use of turmeric in wound healing’ ?
(D) Dr. R.A. Mashelkar. [1 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.
(C) A is true, but R is false. [1 mark]
Nematode resistant tobacco plants have been developed by introduction of the DNA that produces :
(C) Sense and Antisense RNA [1 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.
(C) A is true but R is false [1 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.
(B) Both A and R are true but R is not the correct explanation of A. [1 mark]
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.
(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]
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.
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
(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]
(a) How have transgenic animals proved to be beneficial in the study of :
OR
(b) Mention four applications of plants whose genes have been altered by manipulation.
(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)
(Any four applications) [½ × 4 = 2 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 Milk | α-lactalbumin Content approximately (g/L) |
|---|---|
| Normal cow's milk | 0.1 |
| Rosie cow's milk | 2.4 |
| Human milk | 3.5 |
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.
A.
I. Normal cow's milk (~0.1 g/L) is least suitable. Reason: Very low α-lactalbumin compared to human milk. [1]
II.
[Any one, 1]
OR
B.
[Any two, 1+1]
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.
(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]
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]
(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]
(a) How is insulin different from proinsulin ?
(b) Explain all the stages of how human insulin is produced through rDNA technology.
[1+2 = 3]
(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]
[Total: 3 marks]
Transgenic animals are produced by man. Explain any three ways in which such animals can be beneficial to humans.
(Any three with correct explanation) [1×3 = 3 marks]
Explain any three different approaches used in the treatment of a person suffering from Adenosine Deaminase deficiency.
The three different approaches include –
[Total: 3 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 ?
Answer not available: no official marking scheme published.
Enlist one advantage and two disadvantages of green revolution.
Answer not available: no official marking scheme published.
Enlist three advantages of genetically modified plants.
Answer not available: no official marking scheme published.
Explain the beneficial role of the following, produced as a result of the processes of biotechnology, to mankind :
Answer not available: no official marking scheme published.
Answer not available: no official marking scheme published.
Answer not available: no official marking scheme published.
Biotechnology has helped farmers to get pest-resistant cotton crops. Explain the technique adopted along with its mode of action.
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]
(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.
(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):
[½ × 4]
(ii)
[½ + ½]
[Total: 3 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.
Explain how this transformation was achieved in the tobacco plant.
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]
(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]
Explain the role of transgenic animals in :
(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]
(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]
[Total: 3 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.
(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]
(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]
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.
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]
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.
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]
What are cry-proteins? With the help of a suitable example, explain how it acts as a biological pesticide.
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]
With the help of a storyboard or a flowchart only, explain all the stages of how human insulin is produced through r-DNA technology.
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]
(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.
(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]
(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:
How is the process of its formation naturally in the human body different from that of its formation by rDNA technology? Explain.
(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]
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
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.
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.
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)
//
(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)
[Total: 5 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.
Answer the following questions:
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.
(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]
What could be the reason of extinction of Abingdon Tortoise from Galapagos Islands ?
(D) Interspecific Competition [1 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 :
(D) Competition [1 mark]
Choose a correct option that shows an interaction in which one species benefits while the other is neither harmed nor benefitted.
(B) Sea anemone and clownfish. [1 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 ?
Options :
(A) (i) and (ii) [1 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 :
Answer not available: no official marking scheme published.
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.
Answer not available: official marking scheme supplied is for Set 57/S/3; this Set 1 question could not be matched to it.
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.
(C) [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). [1 mark]
The population interaction where one species is harmed and the other is unaffected is :
(A) Amensalism [1 mark]
The status of the human population reflected in the human age pyramid given below is :
(A) Declining population. [1 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.
(A) A and R are true and R is the correct explanation of A. [1 mark]
Identify the option that does not exhibit a parasitic relationship.
(C) Female Anopheles [1 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.
(C) A is true but R is false. [1 mark]
A Tight one-to-one relationship between many species of fig tree and certain wasps is an example of –
(D) Mutualism. [1 mark]
Important attributes belonging to a population but not to an individual are :
Select the correct option from the given options :
(D) (i) and (iv). [1 mark]
In asymptote state, population is :
(C) Stabilized [1 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.
(A) Both A and R are true and R is the correct explanation of A [1 mark]
(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 ?
(a) (Any two)
[½ × 4 = 2 marks]
OR
(b) (Any two)
[1+1 = 2 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 :
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.
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
OR
Catabolism, Leaching, Fragmentation, Mineralisation, Humification
Answer not available: no official marking scheme published.
Observe the population growth curve and answer the questions given below :
(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]
(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.
(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]
“Cattle and goats do not browse the Calotropis plant.” Justify the statement giving reasons.
Calotropis plant produces highly poisonous cardiac glycosides. [1×2]
[Total: 2 marks]
Study the graph given below, showing the population growth curves 'A' and 'B' respectively. Answer the following questions:
(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'.
(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]
Explain the two physiological attributes which enable a kangaroo rat to survive in desert conditions.
Internal fat oxidation (water is by product) [1]
Concentrated urine / minimal volume of water is used to remove excretory products [1]
[Total: 2 marks]
Mammals are capable of maintaining homeostasis. But, very few smaller animals can survive in polar regions. Give reasons.
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]
(a)
Observe the schematic representation given above and answer the following questions:
OR
(b)
Identify the type of pyramid given above. Write the identifying feature on the basis of which you identified it.
(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]
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.
What conclusion will you draw from the graphs in terms of impact of interaction on growth? Explain, giving reason.
● 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]
(a) The graphs given below, A and B, represent population of elephants in two different National Parks (a hypothetical situation) at different times.
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?
(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]
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.
● 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]
(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):
(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?
(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]
Explain the pollination mechanism involved in co-evolution of the two species, namely Ophrys (orchid) and its insect pollinator bees (& bumble bees).
– 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]
(a) The given graph shows diagrammatic representation of organismic response to abiotic factors:
OR
(b) Given below an equation describing the growth pattern of a population:
dN/dt = rN
(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]
(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]
(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
//
Verhulst Pearl Logistic Growth
Lag phase = ½, Asymptote = ½, K = ½, Equation = ½
[Total: 3 marks]
(a) Study the graph given below and answer the questions that follow :
OR
(b) Answer the following questions :
(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]
(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]
Explain Gause’s ‘Competitive Exclusion Principle’ with the help of a suitable example.
[Total: 3 marks]
(a) Name the type of interaction seen in each of the following examples.
(b) Also explain the above interactions shown by (i) and (ii).
(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]
(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.
Answer not available: no official marking scheme published.
(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 :
(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]
(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.
(a)
[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]
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.
[Total: 3 marks]
Describe how fig tree and wasp relationship is a spectacular example of mutualism.
[Total: 3 marks]
Explain the following population interactions with the help of one example each :
(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]
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:
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]
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.
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]
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.
(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.
(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]
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.
Answer not available: no official marking scheme published.
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 :
OR
(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]
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 :
OR
(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]
OR
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
Justify the following statements with suitable proof/examples:
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 Place | Number of Bird species |
|---|---|
| Columbia | 1400 |
| India | 1200 |
| Northern South America | 1300 |
| New York | 105 |
| Denmark | 504 |
Identify the common factor in regions with a higher number of bird species and suggest at least two reasons for this greater diversity.
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)
[2, any two]
In an ecosystem, different species occupy different levels and vertical distribution of species is found. This is called _________.
(A) Stratification [1 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.
(D) Assertion (A) is false, but Reason (R) is true. [1 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 ?
(C) Soil Sample III [1 mark]
Which of the following is not a functional unit of an ecosystem ?
(C) Stratification [1 mark]
Every trophic level has a certain mass of living material at a particular time. What is it called ?
(A) Standing crop [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of the Assertion (A) [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A) [1 mark]
During the decomposition process, mineralisation by micro-organisms leads to the release of which nutrients into the soil ?
(A) Inorganic nutrients from humus [1 mark]
The process of mineralization by microorganisms help in the release of :
Answer not available: no official marking scheme published.
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 :
(C) Secondary productivity and Net primary productivity [1 mark]
The pyramid of biomass in sea is generally inverted because in sea :
(A) Biomass of fish exceeds that of phytoplankton [1 mark]
Assertion (A): Some aquatic ecosystems have inverted biomass pyramids.
Reason (R): More energy is required by the organisms occupying higher trophic levels.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). [1 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.
(D) Assertion (A) is false, but Reason (R) is true. [1 mark]
The primary productivity in an ecosystem is expressed as :
(A) gm−2 yr−1. [1 mark]
Assertion (A): Decomposition process is slower if detritus is rich in lignin and cutin.
Reason (R): Decomposition is largely an oxygen requiring process.
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A). [1 mark]
(a) Given below is a pyramid found in an ecosystem, where each bar represents the standing crop available in the trophic level.
(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.
(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]
(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.
(a) (any other relevant difference, any two) [1×2]
| Grazing food chain | Detritus food chain |
|---|---|
| Starts from producers. | Starts from Detritus / decomposers / dead organic matter. |
| It is the major conduit for energy flow in aquatic ecosystem | It 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]
[Total: 2 marks]
(a) Define stratification in an ecosystem.
(b) How is stratification represented in a forest ecosystem ?
(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]
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.
Answer not available: no official marking scheme published.
(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.
(a) (i) Pyramid of numbers:
[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]
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.
[½ × 4 = 2 marks]
How is the rate of decomposition affected by the nature of detritus and temperature ?
[Total: 2 marks]
OR
(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)
[1]
(ii) Log S = Log C + Z Log A [1]
[Total: 2 marks]
OR
(a)
| Grazing Food Chain | Detritus 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)
[Total: 2 marks]
Ecological pyramids give important information about the ecological system, but do have some limitations. List any two limitations of ecological pyramids.
(Any two points) [1×2]
[Total: 2 marks]
Given below is a food web that involves nine organisms.
(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]
Attempt either option A or B.
A.
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.
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)
[1]
OR
B.
(i) 3000+1300+427+427+378 = 5532 [1]
(ii)
[Anyone, 1]
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.
Answer not available: no official marking scheme published.
Differentiate between :
(a) Grazing food chain and Detritus food chain (Any three differences)
(b) Upright pyramid and Inverted pyramid (Any three differences)
(a)
| Grazing food chain | Detritus food chain |
|---|---|
| Starts with green plants called producers as first trophic level | Starts with dead organic matter and decomposers called saprotrophs |
| A large fraction of energy flows through aquatic ecosystem | A much larger fraction of energy flows through terrestrial ecosystem |
| Energy for food chain comes from sun | Energy for food chain comes from organic remains or detritus |
[½ × 3]
(b)
| Upright pyramid | Inverted pyramid |
|---|---|
| It has wide base and narrow apex | It has narrow base and wide apex |
| Producers are more in number and biomass than herbivores | Producers are less in number and biomass than herbivores |
| Pyramid of energy is always upright | Pyramid of biomass in sea is generally inverted |
(Any other correct difference) [½ × 3]
[Total: 3 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.
(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]
Below is shown a pond that represents a self-sustaining natural ecosystem with both biotic and abiotic components interacting in a balanced food web.
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]
(a) (Any two) Water, Soil, sunlight, temperature, organic substances, inorganic substances [½+½]
(b)
| Role of producers | Role of decomposers |
|---|---|
| Conversion of inorganic molecules into organic material with the help of sunlight through the process of photosynthesis | Break 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]
Select the mismatched pair :
(B) Genetically different strains of rice in India – less than 1000 [1 mark]
Which of the following sacred groves is found in Meghalaya ?
(A) Jaintia hills [1 mark]
Select the correct option :
(A) Tropics between 23.5°N – 23.5°S – Higher biodiversity than temperate areas [1 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.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation for Assertion (A). [1 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 ?
(C) Nile perch [1 mark]
India has only 2·4% of the world’s land area but its share of the global species diversity is :
Answer not available: no official marking scheme published.
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.
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). [1 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.
(A) Both (A) and (R) are true and (R) is the correct explanation of (A). [1 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.
(A) (ii), (iii) and (iv) only [1 mark]
The IUCN Red Data List (2004) in the last 500 years documents the extinction of nearly 784 species including :
(C) 359 invertebrates. [1 mark]
Choose the pair that is incorrectly matched.
(B) Aravalli Hills – Karnataka [1 mark]
(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.
(a) (Any two) [1+1]
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
[Total: 2 marks]
(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]
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.
2.4 percent [1]
8.1 percent share of the global species diversity [1]
[Total: 2 marks]
Illustrate with the help of an example how introduction of an alien species turns invasive and causes decline of an indigenous species.
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]
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.
(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?
(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]
(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.
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.
(a)
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]
Attempt either option A or B.
A. The table below is a hypothetical data on Seasonal Temperatures seen over two different zones
| Region | Temp. 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 X | 27 | 28 | 27 | 26 | 3500 | 1300 |
| Region Y | 12 | 20 | 14 | 4 | 2000 | 800 |
OR
B. Following is a hypothetical data on Plant Pollinator interaction
| Plant | Main Pollinator | Seed set (%) with pollinator | Seed set (%) without pollinator |
|---|---|---|---|
| P1 | H1 (Bee) | 80 | 40 |
| P2 | H2 (Butterfly) | 75 | 45 |
| P3 | H3 (Moth) | 85 | 35 |
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]
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]
(a) Ex-situ conservation. [½]
(b) (Any two) [1+1]
(c) Botanical Gardens, Wildlife safari parks (Any one) [½]
[Total: 3 marks]
Explain the level of biodiversity at genetic, species and ecological levels with the help of one example each.
[Total: 3 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.
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]
(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 :
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
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.
Answer not available: no official marking scheme published.
Answer not available: no official marking scheme published.
Given below is a pie chart representing global diversity of vertebrates.
(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.
(a)
(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]
OR
(a)
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)
(Diagrammatic representation of any correct example in relevance to the shape of pyramids should be considered.)
[Total: 3 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.
Any three human activities with relevant explanation: [½ × 6]
[Total: 3 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.
(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]
“Biodiversity plays a major role in many ecosystem services that nature provides.”
(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]
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.
[Total: 3 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.
(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]
(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.
(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]
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.
(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.
(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]
(a) Write the inference drawn by Alexander von Humboldt after his extensive explorations of South Americal jungle.
(b) Study the graph given below:
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?
(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]
(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.
(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]:
OR
(b) Any three of the following [1 + 1 + 1]:
Note: According to Hindi version, all benefits of forest are included.
[Total: 3 marks]
(a) The graph given below shows species-area relationship of a certain region.
OR
(b) How did Dr. David Tilman relate experimentally, the stability of a community and its species richness? Explain.
(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]
What are the consequences of loss of biodiversity in a region? Explain.
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]
(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
OR
(b) Given below is a ‘pie chart’ representing the global biodiversity: proportionate number of species of major taxa.
(a) (i) (Graph: rectangular hyperbola of species richness against area, curve rising and levelling off) [1]
(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]
(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.
(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]
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.
(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?
(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]
How have the following factors affected the biodiversity of an area? Explain.
(a) Habitat loss and fragmentation
(b) Over-exploitation
(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]
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.
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]
(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 Conservation | Ex 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]
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.
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.
(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)
[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]
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 :
(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.
(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]
Study the diagrammatic representation given below of the Earth with regions marked ‘A’ and ‘B’ respectively. Answer the questions that follow.
OR
(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]
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.
(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.
(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:
[1 × 3]
[Total: 4 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.
Answer not available: no official marking scheme published.
OR
Answer not available: no official marking scheme published.
How have the following contributed to biodiversity loss ?
OR
N = Population density at time ‘t’
r = Intrinsic rate of natural increase
Answer not available: no official marking scheme published.
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.
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.
A.
(a)
| Biodiversity Threat | Explanation of the effect it will have |
|---|---|
| Overexploitation | Unsustainable 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 species | Ornamental grass from another country spreading aggressively into native grassland/outcompetes native grasses, reduces local biodiversity. [1] |
| Co-extinction | Disappearance of nesting tree species causing sharp decline in migratory bird population/ disrupts ecological balance, loss of niches. [1] |
| Habitat loss and fragmentation | Forest 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 / Process | Description |
|---|---|
| Fragmentation | Breaking down of waste into smaller particles by detritivores (e.g., earthworms, termites). [1] |
| Leaching | Soluble inorganic nutrients dissolve in water and percolate into the composting soil. [1] |
| Catabolism | Enzymatic breakdown of complex organic matter into simpler inorganic substances by bacteria/fungi. [1] |
| Humification & Mineralisation | Formation 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]