CBSE Class 12 Biology Question Paper 2025 Solved Code 57-1-2

SECTION A

1. Mohit performed an analysis of two different soil samples from two areas A and B. He recorded these results :
Soil Sample A : Lignin 40%, Sugar 10%, Chitin 45%, Nitrogen 5%
Soil Sample B : Lignin 5%, Sugar 35%, Chitin 15%, Nitrogen 45%
Which of these is true about their rate of decomposition in both the soil ? [1 Mark]
(A) Soil A has a slower rate of decomposition than Soil B.
(B) Soil A has a faster rate of decomposition than Soil B.
(C) Both have the same rate of decomposition.
(D) No decomposition occurs in A and B.

Answer: (A) Soil A has a slower rate of decomposition than Soil B.

Teacher's Note:
a) Detritus rich in lignin and chitin decomposes slowly.
b) Detritus rich in nitrogen and water-soluble substances like sugars decomposes quickly.

2. Diameter of the pollen grain generally is [1 Mark]
(A) 5 to 10 micrometer
(B) 10 to 15 micrometer
(C) 25 to 50 micrometer
(D) 50 to 100 micrometer

Answer: (C) 25 to 50 micrometer

Teacher's Note:
a) Pollen grains are generally spherical and about 25 to 50 micrometers in diameter.
b) This is a direct NCERT fact, so learn such numerical values carefully.

3. Menstrual cycle in human females consists of various events.
Select the option that indicates the correct sequence of these events of menstrual cycle. [1 Mark]
(A) Menstrual phase, Follicular phase, Luteal phase, Ovulatory phase
(B) Luteal phase, Follicular phase, Ovulatory phase, Menstrual phase
(C) Menstrual phase, Follicular phase, Ovulatory phase, Luteal phase
(D) Follicular phase, Luteal phase, Menstrual phase, Ovulatory phase

Answer: (C) Menstrual phase, Follicular phase, Ovulatory phase, Luteal phase

Teacher's Note:
a) The cycle starts with menstruation (days 1 to 5), followed by the follicular phase.
b) Ovulation occurs around day 14, and the luteal phase comes after ovulation.

4. A Dihybrid cross is done between two parent pea plants (pure line) who differ in two pairs of contrasting traits : Seed colour and seed shape.
In the F₂ generation the number of phenotypes and genotypes will be : [1 Mark]
(A) phenotypes = 4; genotypes = 16
(B) phenotypes = 9; genotypes = 14
(C) phenotypes = 4; genotypes = 8
(D) phenotypes = 4; genotypes = 9

Answer: (D) phenotypes = 4; genotypes = 9

Teacher's Note:
a) A dihybrid F₂ gives 4 phenotypes in the ratio 9:3:3:1.
b) Do not confuse 16 combinations of gametes with 9 different genotypes.

5. RNA interference (RNAi) helps in making tobacco-plant resistant to a nematode (Meloidegyne incognita)
Choose the correct option that shows how RNAi is achieved : [1 Mark]
(A) Preventing the process of translation of mRNA of the nematode.
(B) Preventing the process of replication of DNA of the nematode.
(C) Preventing the process of transcription of DNA of the plant.
(D) Preventing the process of replication of DNA of the plant.

Answer: (A) Preventing the process of translation of mRNA of the nematode.

Teacher's Note:
a) In RNAi, dsRNA binds to the specific mRNA of the parasite and silences it.
b) As a result, the nematode cannot survive in the transgenic host.

6. The sequence of nitrogenous bases in a segment of a coding strand of DNA is 5' - AATGCTAGGCAC - 3'. Choose the option that shows the correct sequence of nitrogenous bases in the mRNA transcribed by the DNA. [1 Mark]
(A) 5' - UUACGAACCGAG - 3'
(B) 5' - AAUGCUAGGCAC - 3'
(C) 5' - UUACGUACCGUG - 3'
(D) 5' - AACGUAGGCAGC - 3'

Answer: (B) 5' - AAUGCUAGGCAC - 3'

Teacher's Note:
a) mRNA has the same sequence as the coding strand, with U in place of T.
b) Keep the same polarity (5' to 3') as the coding strand.

7. A man whose father was colour-blind marries a woman who had a colour-blind mother and normal father. What percentage of male children of this couple will be colour-blind ? [1 Mark]
(A) 25%
(B) 0%
(C) 50%
(D) 75%

Answer: (C) 50%

Teacher's Note:
a) The woman is a carrier because she got one defective X from her colour-blind mother.
b) Sons get their X only from the mother, so half the sons will be colour-blind.

8. Evolution of modern man involves the following man-like primates. Choose the correct series of human evolution. [1 Mark]
(A) Dryopithecus \( \rightarrow \) Homo erectus \( \rightarrow \) Australopithecines \( \rightarrow \) Homo sapiens
(B) Australopithecines \( \rightarrow \) Homo erectus \( \rightarrow \) Neanderthal \( \rightarrow \) Homo sapiens
(C) Australopithecines \( \rightarrow \) Ramapithecus \( \rightarrow \) Dryopithecus \( \rightarrow \) Homo sapiens
(D) Homo erectus \( \rightarrow \) Australopithecines \( \rightarrow \) Homo sapiens \( \rightarrow \) Neanderthal

Answer: (B) Australopithecines \( \rightarrow \) Homo erectus \( \rightarrow \) Neanderthal \( \rightarrow \) Homo sapiens

Teacher's Note:
a) Dryopithecus and Ramapithecus came much earlier than Australopithecines.
b) Neanderthal man lived just before modern Homo sapiens appeared.

9. GEAC stands for [1 Mark]
(A) Genome Engineering Action Committee
(B) Ground Environment Action Committee
(C) Genetic and Environment Approval Committee
(D) Genetic Engineering Approval Committee

Answer: (D) Genetic Engineering Approval Committee

Teacher's Note:
a) GEAC is the Indian body that decides on the validity of GM research and the safety of GM organisms for public use.
b) Remember the exact full form, as options differ only slightly.

10. Transplantation of tissues/organs to some patients often fails due to rejection of such tissues/organs by the body of the patient. Which type of immune response is responsible for such rejections ? [1 Mark]
(A) Autoimmune response
(B) Humoral immune response
(C) Physiological immune response
(D) Cell mediated immune response

Answer: (D) Cell mediated immune response

Teacher's Note:
a) Graft rejection is mediated by T-lymphocytes, so it is a cell mediated immune response.
b) This is why patients take immunosuppressants after organ transplantation.

11. Amplification of gene of interest by using DNA polymerase may go upto [1 Mark]
(A) 0.1 million times
(B) 1 million times
(C) 1 billion times
(D) 1 trillion times

Answer: (C) 1 billion times

Teacher's Note:
a) In PCR, repeated cycles with thermostable DNA polymerase can amplify a DNA segment about 1 billion times.
b) Taq polymerase from Thermus aquaticus is used because it stays active at high temperatures.

12. 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 : [1 Mark]
(A) P - Egg, Q - Suspensor, R - Radicle, S - Cotyledon
(B) P - Zygote, Q - Suspensor, R - Cotyledon, S - Plumule
(C) P - Egg, Q - Radicle, R - Suspensor, S - Cotyledon
(D) P - Zygote, Q - Suspensor, R - Cotyledon, S - Radicle

[Figure: Stages of development of a dicot embryo shown left to right with arrows: a single cell labelled P, a two-celled stage, a globular embryo on a row of cells, a heart-shaped embryo with Q pointing to the row of cells above it, and a mature embryo with R pointing to the two folded lobes of the mature embryo and S (written below the heart-shaped embryo) with a line pointing into the mature embryo. Labels: Globular embryo, Heart-shaped embryo, Mature embryo.]

Answer: (B) P - Zygote, Q - Suspensor, R - Cotyledon, S - Plumule

Teacher's Note:
a) The embryo develops from the zygote, not directly from the egg.
b) The suspensor is the row of cells that attaches the young embryo to the embryo sac wall.

For Question numbers 13 to 16, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Answer these questions by selecting the appropriate option given below :
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true, but (R) is false.
(D) (A) is false, but (R) is true.

13. Assertion (A) : Perisperm is a diploid tissue.
Reason (R) : Perisperm is the remains of nucellus which surrounds the embryo in certain seeds. [1 Mark]
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true, but (R) is false.
(D) (A) is false, but (R) is true.

Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).

Teacher's Note:
a) Nucellus is a diploid maternal tissue, so its persistent remains (perisperm) are also diploid.
b) Examples of seeds with perisperm are black pepper and beet.

14. Assertion (A) : While working on Staphylococci, Alexander Fleming observed that Penicillium notatum inhibits the growth of bacteria.
Reason (R) : The inhibiting chemical was commercially extracted and its full potential was established by Alexander Fleming. [1 Mark]
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true, but (R) is false.
(D) (A) is false, but (R) is true.

Answer: (C) (A) is true, but (R) is false.

Teacher's Note:
a) The full potential of penicillin as an antibiotic was established later by Ernest Chain and Howard Florey.
b) Fleming only made the original observation of bacterial growth being inhibited by the mould.

15. Assertion (A) : One of the property of genetic code is degeneracy.
Reason (R) : Some amino acids can be coded by more than one codon. [1 Mark]
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true, but (R) is false.
(D) (A) is false, but (R) is true.

Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).

Teacher's Note:
a) Degeneracy means one amino acid can be coded by more than one codon.
b) So the Reason is itself the definition of degeneracy, which explains the Assertion.

16. Assertion (A) : A bioreactor provides the optimal conditions for achieving the desired product by providing optimum growth conditions.
Reason (R) : The most commonly used bioreactors are of stirring type. [1 Mark]
(A) Both (A) and (R) are true and (R) is the correct explanation of (A).
(B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(C) (A) is true, but (R) is false.
(D) (A) is false, but (R) is true.

Answer: (B) Both (A) and (R) are true, but (R) is not the correct explanation of (A).

Teacher's Note:
a) Bioreactors provide optimum temperature, pH, substrate, salts, vitamins and oxygen.
b) The stirring type is only a common design; it does not explain why optimal conditions are provided.

SECTION B

17. Study the two different crosses given below :
Identify the abnormalities 1 and 2 in the offsprings of given crosses and distinguish between them. [2 Marks]

[Figure: Cross I - Parents: 44 + XX crossed with 44 + XY. Gametes of the female: 23 + X and 21 + X; gametes of the male: 22 + X and 22 + Y. The 23 + X gamete is joined by lines to both 22 + X and 22 + Y, giving Offspring 1. Cross II - Parents: 44 + XX crossed with 44 + XY. Gametes of the female: 22 + XX and 22 + O; gametes of the male: 22 + X and 22 + Y. The 22 + XX gamete is joined to 22 + Y, giving Offspring 2.]

Answer:
1. Abnormality 1 is Down's syndrome (45 + XX or 45 + XY), a trisomy of autosome. Abnormality 2 is Klinefelter's syndrome (44 + XXY).
2. Difference: Down's syndrome is caused by an extra copy of autosome 21 (trisomy 21), and the affected person is short statured with a small round head, furrowed tongue and retarded mental development; Klinefelter's syndrome is caused by an extra sex chromosome (X), and the affected person is a sterile male with overall masculine development but some feminine features such as gynaecomastia.

Teacher's Note:
a) Count the autosomes in the zygote: 23 + 22 = 45 autosomes means trisomy of an autosome.
b) Half a mark is for each correct name, and one mark is for one clear difference.

18. Study the pedigree analysis of Myotonic dystrophy given below :
Answer the following questions :
(a) Write genotype of A, B, C and D
(b) Identify whether the trait is
(i) Sex linked or autosomal
(ii) Dominant or recessive [2 Marks]

[Figure: Pedigree chart. Parents: an affected (shaded) female A married to an unaffected male B. Generation I: an unaffected female, an affected (shaded) male C, an unaffected male married to an unaffected female, and an affected (shaded) female married to an unaffected male. Generation II: children of the unaffected couple - an unaffected female, an unaffected female D and an unaffected male; children of the affected female - an unaffected male, an affected (shaded) female and an unaffected male.]

Answer:
1. Genotypes: A = Aa, B = aa, C = Aa, D = aa.
2. (b) (i) The trait is autosomal. (ii) The trait is dominant.

Teacher's Note:
a) Myotonic dystrophy is an autosomal dominant trait, so affected persons with unaffected children are heterozygous (Aa).
b) The trait appears in every generation and in both sexes, which points to autosomal dominant inheritance.
c) One mark is awarded if any two genotypes are correct.

19. 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. [2 Marks]

Answer:
1. The genetically engineered lymphocytes are not immortal, so the patient needs periodic infusion of such cells.
2. A permanent cure is possible if the gene isolated from bone marrow cells producing ADA is introduced into cells at early embryonic stages.

Teacher's Note:
a) The keyword "not immortal" is the value point for the first part.
b) For the permanent cure, mention both "bone marrow cells" and "early embryonic stages".

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. [2 Marks]

Answer:
1. Micro-injection: recombinant DNA is directly injected into the nucleus of an animal cell.
2. Biolistics or gene gun: plant cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA.

Teacher's Note:
a) Give the name of the technique and one line on how it works for full marks.
b) Heat shock treatment (ice, then 42 degrees C, then ice) is also accepted as a technique.

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

(A) How are morphine and heroin related ? Mention their effect on the human body. [2 Marks]

Answer:
1. Heroin is obtained by acetylation of morphine; both are opioids.
2. Effects: morphine is a very effective sedative and painkiller, while heroin is a depressant that slows down body functions.

Teacher's Note:
a) Heroin is chemically diacetylmorphine, a white, odourless, bitter crystalline compound.
b) Both drugs bind to specific opioid receptors in the central nervous system and gastrointestinal tract.

OR

(B) (i) Name an alcoholic drink which is produced by the help of microbes :
(1) With distillation
(2) Without distillation
(ii) Explain how cyanobacteria can be used as bio-fertilizer. [2 Marks]

Answer:
1. (i) (1) With distillation: Whisky (or brandy or rum). (2) Without distillation: Wine (or beer).
2. (ii) Cyanobacteria fix atmospheric nitrogen and add organic matter to the soil, which increases soil fertility.

Teacher's Note:
a) Distilled drinks have a higher alcohol content than drinks made without distillation.
b) Anabaena, Nostoc and Oscillatoria are common cyanobacteria used as biofertilisers in paddy fields.

21. 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. [2 Marks]

Answer:
1. The natural old forest will be the most productive, as it has more biomass and high biodiversity, which traps and stores more solar radiation as biomass.
2. A young forest is still developing and has fewer trees, so it captures less solar radiation; a shallow polluted lake has fewer producers and a large amount of dead organic matter, so its productivity is lower than that of the natural old forest.

Teacher's Note:
a) First name the most productive ecosystem clearly, then give a reason for each of the other two.
b) Link productivity to the amount of producers (biomass) that can trap sunlight.

OR

(B) Differentiate between Net primary productivity and Gross primary productivity in an ecosystem. [2 Marks]

Answer:
1. Gross primary productivity (GPP) is the rate of production of organic matter during photosynthesis, while net primary productivity (NPP) is GPP minus respiration losses (R), that is, NPP = GPP - R.
2. A considerable amount of GPP is used by plants in respiration, while NPP is the biomass available for consumption by heterotrophs.

Teacher's Note:
a) Remember the relation NPP = GPP - R; it carries the key value point.
b) Write the difference in pairs, one point for GPP against the matching point for NPP.

SECTION C

22. (a) Explain the importance of 'ori' and 'rop' in the E. coli cloning vector shown in the above given diagram.
(b) Differentiate between exonucleases and endonucleases. [3 Marks]

[Figure: Circular map of the E. coli cloning vector pBR322 showing the genes amp^R, tet^R, ori and rop, and the restriction sites EcoR I, Cla I, Hind III, BamH I, Sal I, Pvu II, Pst I and Pvu I.]

Answer:
1. 'ori' is the sequence in the cloning vector from where replication starts; it also controls the copy number of the linked DNA.
2. 'rop' codes for the proteins involved in the replication of the plasmid.
3. Exonucleases remove nucleotides from the ends (5' or 3' terminal region) of DNA, while endonucleases make cuts at specific positions within the DNA strand.

Teacher's Note:
a) Mention "copy number" when writing the role of ori; it is a key value point.
b) Remember: "exo" means outside (ends) and "endo" means inside the strand.

23. The great German naturalist and geographer Alexander Von Humboldt observed that within a region species richness increased with increasing explored area, but only upto a limit.
(a) For the above situation, construct a graph and write an equation.
(b) Also write various values of Z (regression coefficient). [3 Marks]

Answer:
1. Graph: species richness is plotted on the Y-axis and area on the X-axis. The curve \( S = CA^{Z} \) rises steeply at first and then levels off (rectangular hyperbola); on a log-log scale the relationship becomes a straight line.
2. Equation: \( S = CA^{Z} \) or \( \log S = \log C + Z \log A \), where S = species richness, A = area, C = Y-intercept and Z = slope of the line (regression coefficient).
3. Values of Z: 0.1 to 0.2 for small areas, regardless of the taxonomic group or the region; 0.6 to 1.2 for very large areas such as entire continents; 1.15 for frugivorous birds and mammals in the tropical forests.

Teacher's Note:
a) Label both axes of the graph; one correctly drawn line (curve or log-log line) earns the mark.
b) Write either form of the equation and explain what each symbol stands for.
c) Remember that Z is much steeper for whole continents than for small regions.

24. 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) is illegal ? [3 Marks]

Answer:
1. (a) Amniocentesis: some of the amniotic fluid of the developing foetus is taken out, and the foetal cells and dissolved substances in it are analysed to test for the presence of genetic disorders.
2. (b) They can consider Medical Termination of Pregnancy (MTP). Yes, it is safe for Radha, because MTP is comparatively safe up to 12 weeks (the first trimester) of pregnancy, and she is in her second month.
3. (c) MTP is illegal when it is performed by unqualified quacks, or when it is done for female foeticide after finding that the foetus is a normal female.

Teacher's Note:
a) Name the test and also explain what is analysed to get full marks for part (a).
b) Link safety to the first trimester (up to 12 weeks); second month falls within it.
c) Sex determination using amniocentesis is legally banned in India.

25. (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. [3 Marks]

Answer:
1. (a) 3' - CTTAAG - 5'
2. (b) EcoRI
3. (c) The restriction enzyme cuts each DNA strand a little away from the centre of the palindrome, between the same two bases (G and A for EcoRI) on the opposite strands. This leaves single stranded overhanging stretches at the ends, called sticky ends. Role: sticky ends form hydrogen bonds with their complementary cut counterparts, which helps in joining vector DNA and foreign DNA; this stickiness helps the action of the enzyme DNA ligase.

Teacher's Note:
a) The palindrome reads the same on both strands when read in the same direction (5' to 3').
b) Always write the polarity (5' and 3') with the complementary strand.
c) Mention DNA ligase when writing the role of sticky ends.

26. 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. [3 Marks]

[Figure: Part of the life cycle of Plasmodium. A mosquito is shown biting a human hand; an arrow leads from the hand down to a group of small cells labelled Q. An arrow leads from Q up to a single cell with a tail-like projection, and then an arrow marked "Fertilization and development" leads up to a rounded structure labelled P, which contains many thread-like bodies and a few round bodies.]

Answer:
1. (a) Sporozoites.
2. (b) Fertilisation and development take place in the gut (stomach) of the female Anopheles mosquito. (c) P is the salivary glands; Q is the gametocytes.
3. (d) The asexual phase takes place in human, and the sexual phase takes place in mosquito.

Teacher's Note:
a) Sporozoites are stored in the salivary glands and enter the human body when the mosquito bites.
b) Gametocytes are taken up by the mosquito from human blood; they fuse in the mosquito gut.
c) The mosquito is the definitive host because sexual reproduction occurs in it.

27. Compare and contrast convergent and divergent evolution. [3 Marks]

Answer:
1. Divergent evolution occurs when the same structure develops along different directions due to adaptations to different needs; convergent evolution occurs when different structures evolve for the same function and hence show similarities.
2. Divergent evolution produces homologous structures and indicates common ancestry; convergent evolution results in analogous structures and indicates that a similar habitat has led to selection of similar adaptive features.
3. Similarities: both are influenced by the environment, and both contribute to the evolution of species.

Teacher's Note:
a) "Compare and contrast" needs both differences and at least one similarity.
b) Give examples if time allows: forelimbs of whale, bat and man (divergent); wings of butterfly and bird (convergent).

28. 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. [3 Marks]

[Figure: A thread-like strand, labelled P, coiled around a large spherical protein core, labelled Q, with the strand continuing out on both sides.]

Answer:
1. (a) Nucleosome. (b) P = DNA; Q = Histone octamer.
2. (c) Histones are basic proteins; they are positively charged.
3. (d) Euchromatin is loosely packed, lightly stained and transcriptionally active chromatin, whereas heterochromatin is densely packed, darkly stained and transcriptionally inactive chromatin.

Teacher's Note:
a) Histones are positively charged because they are rich in basic amino acids lysine and arginine.
b) One correct difference between euchromatin and heterochromatin earns one mark.

SECTION D

Question Nos. 29 and 30 are case based questions. Each question has 3 sub-questions with internal choice in one sub-question.

29. 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 ? [1 Mark]

Answer: The Monarch butterfly acquires this chemical during its caterpillar stage by feeding on a poisonous weed.

Teacher's Note:
a) Mention both "caterpillar stage" and "poisonous weed" for the full mark.
b) This is an example of a chemical defence of prey against predators.

(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. [2 Marks]

Answer:
1. Using \( N_{t+1} = N_t + [(B + I) - (D + E)] \):
\( 800 = N_t + [(200 + 200) - (150 + 100)] \)
\( 800 = N_t + (400 - 250) = N_t + 150 \)
\( N_t = 800 - 150 = 650 \)
2. Comment: the population density is increasing with time, so the age pyramid would be of an expanding (growing) population.

Teacher's Note:
a) Natality and immigration add to the population, while mortality and emigration reduce it.
b) An expanding population has a broad-based, triangular age pyramid.

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 ? [1 Mark]

Answer: The single huge banyan tree is measured in terms of biomass or percent cover, whereas the carrot grass plants are measured in terms of percent cover.

Teacher's Note:
a) Counting numbers is misleading here, because one banyan tree is far larger than 200 carrot grass plants.
b) Percent cover or biomass is a more meaningful measure of population size in such cases.

OR

(d) Name two methods to measure the population density of tigers. [1 Mark]

Answer: Pug marks and faecal pellets.

Teacher's Note:
a) These are indirect methods, because tigers cannot easily be counted directly.
b) Tiger census in national parks is based on pug marks and faecal pellets.

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

[Figure: Two graphs of hormone levels. Case 1: curve P rises sharply during the follicular phase to a peak, with ovulation marked by a dotted line just after the peak, and then falls; curve Q rises during the luteal phase to a hump and then falls. Case 2: curve P shows the same sharp peak followed by ovulation; curve Q rises in the luteal phase and keeps on increasing; a third curve labelled HCG begins rising late in the luteal phase.]

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

Answer: Luteinising hormone (LH) is responsible for the peak. It helps in ovulation by inducing rupture of the Graafian follicle.

Teacher's Note:
a) The mid-cycle surge of LH is called the LH surge.
b) Ovulation takes place just after the LH peak, around day 14.

(b) Write changes that take place in the ovary and uterus during follicular phase. [2 Marks]

Answer:
1. Ovary: the primary follicles grow and mature into a fully mature Graafian follicle.
2. Uterus: the endometrium lining regenerates through proliferation.

Teacher's Note:
a) Write the change in the ovary and the uterus separately; each carries one mark.
b) These changes are brought about by rising levels of FSH, LH and oestrogen.

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

(c) Name the hormone Q of Case 2. Write one function of hormone Q. [1 Mark]

Answer: Q is Progesterone. It maintains pregnancy by maintaining the endometrium.

Teacher's Note:
a) In Case 2 progesterone keeps rising because pregnancy has occurred.
b) The endometrium is needed for implantation and for other events of pregnancy.

OR

(d) Which structure in the ovary will remain functional in Case 2 ? How is it formed ? [1 Mark]

Answer: The corpus luteum will remain functional. It is formed when the Graafian follicle transforms into the corpus luteum after ovulation.

Teacher's Note:
a) The rising HCG in Case 2 shows pregnancy, which keeps the corpus luteum active.
b) If fertilisation does not occur, the corpus luteum degenerates and menstruation follows.

SECTION E

31. 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.
(i) The given diagram shows three methods of pollen transfer in plants. Examine them carefully and write the technical terms used for pollen transfer in methods '1', '2' and '3'.
(ii) How do the following plants achieve pollination successfully ?
(a) Water lily
(b) Vallisneria
(iii) Write advantages of pollen transfer method '3'. [5 Marks]

[Figure: Two flowering plants. The plant on the left bears two flowers; the plant on the right bears one flower. Curved arrow 1 goes from the anther to the stigma within the single flower on the right. Curved arrow 2 goes from one flower to the other flower on the same left plant. Curved arrow 3 goes from the flower of the right plant to a flower of the left plant.]

Answer:
1. (a) The two cells are the vegetative cell and the generative cell. The vegetative cell is bigger, has abundant food reserve and a large irregularly shaped nucleus, and helps in forming the pollen tube. The generative cell is small, floats in the cytoplasm of the vegetative cell, and divides to form two male gametes.
2. (b) (i) Method 1 = Autogamy; Method 2 = Geitonogamy; Method 3 = Xenogamy.
3. (ii) (a) Water lily is pollinated by insects or wind. (b) Vallisneria is pollinated by water.
4. (iii) Advantages of xenogamy: it brings genetic variation, and it produces healthier offspring (it also avoids inbreeding depression).

Teacher's Note:
a) Half a mark is for naming both cells and one mark is for a correct difference.
b) Geitonogamy is functionally cross-pollination but genetically similar to autogamy, as both flowers are on the same plant.
c) Water lily is an aquatic plant, but its flowers emerge above water and are pollinated by insects or wind.

OR

(B) Given below is the diagram of human ovum surrounded by a few sperms. Observe the diagram and answer the questions that follows :
(i) Compare the fate of sperms 'P', 'Q' and 'R' shown in the diagram.
(ii) Write the role of Zona pellucida in this process.
(iii) Analyse the changes occurring in the ovum after the entry of sperm.
(iv) How acrosome and middle piece are able to play an important role in human fertilization ? [5 Marks]

[Figure: A human ovum surrounded by the zona pellucida and an outer layer of cells of the corona radiata, with the perivitelline space labelled. Three sperms are shown: sperm P has passed through the corona radiata and reached the zona pellucida, sperm Q is at the outer surface of the corona radiata, and sperm R is outside the corona radiata.]

Answer:
1. (i) Sperm P is able to penetrate and fertilise the ovum, whereas sperms Q and R are unable to penetrate or fertilise it.
2. (ii) When a sperm comes in contact with the zona pellucida layer of the ovum, it induces changes in the membrane that block the entry of additional sperms.
3. (iii) Entry of the sperm induces completion of the meiotic division of the secondary oocyte, forming the second polar body and a haploid ovum (ootid).
4. (iv) Acrosome: it is filled with enzymes that help the sperm to enter into the cytoplasm of the ovum.
5. Middle piece: it has numerous mitochondria, which produce energy for the movement of the tail; this gives the sperm the motility needed for fertilisation.

Teacher's Note:
a) Only one sperm fertilises the ovum, because the zona pellucida blocks polyspermy.
b) Mention "second polar body" and "ootid" as key words in part (iii).
c) Link acrosome with enzymes and middle piece with mitochondria (energy).

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

(A) Answer the following questions :
(i) State what do you understand by "MALT" ? Where it is located inside our body ?
(ii) Explain cytokine barriers.
(iii) Name the diagnostic test for AIDS. On what principle does it work ?
(iv) Bone marrow and thymus play an important role in human immune system. Explain how are they able to achieve this. [5 Marks]

Answer:
1. (i) MALT is Mucosa Associated Lymphoid Tissue. It is located within the lining of the major tracts such as the respiratory, digestive and urogenital tracts.
2. (ii) Cytokine barriers: virus-infected cells secrete proteins called interferons, which protect non-infected cells from further viral infection.
3. (iii) The diagnostic test for AIDS is ELISA (Enzyme Linked Immunosorbent Assay).
4. ELISA works on the principle of antigen-antibody interaction.
5. (iv) Bone marrow is the main lymphoid organ where all blood cells, including lymphocytes, are produced; some lymphocytes migrate to the thymus. Both organs provide the micro-environment for the development and maturation of lymphocytes, where immature lymphocytes differentiate into antigen-sensitive lymphocytes.

Teacher's Note:
a) MALT makes up about 50 percent of the lymphoid tissue in the human body.
b) PCR (based on amplification of nucleic acid) is also accepted as the diagnostic test in part (iii).
c) Bone marrow and thymus are primary lymphoid organs.

OR

(B) (i) 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 ? [5 Marks]

Answer:
1. (i) (a) H = Clostridium butylicum; I = Bacteria.
2. (b) J = Statin; K = Fungi (yeast).
3. (c) L = Trichoderma polysporum; M = Immunosuppressant agent used in patients who have had organ transplants.
4. (ii) Baculoviruses are species-specific and have narrow spectrum insecticidal properties.
5. They have no negative impact on non-target organisms such as plants, mammals, birds or fish.

Teacher's Note:
a) Write scientific names in italics or underline them when handwritten.
b) Statins lower blood cholesterol by competitively inhibiting the enzyme needed for cholesterol synthesis.
c) Baculoviruses of the genus Nucleopolyhedrovirus are ideal for IPM programmes.

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

(A) (i) Haemophilia and red green colourblindness is usually observed in men. Why ?
(ii) Perform a cross(es) where haemophilic daughter(s) and haemophilic son(s) are produced in same ratio. [5 Marks]

Answer:
1. (i) Both are sex-linked (X-linked) recessive disorders. Males have only a single X chromosome, so a single recessive allele on the X chromosome is easily expressed.
2. (ii) Cross 1: Carrier woman (X^hX) \( \times \) Haemophilic man (X^hY). Gametes of woman: X^h and X; gametes of man: X^h and Y.
3. Offspring: X^hX^h (haemophilic daughter), X^hX (carrier daughter), X^hY (haemophilic son), XY (normal son). Ratio of haemophilic daughter : haemophilic son = 1 : 1.
4. Cross 2: Haemophilic father (X^hY) \( \times \) Haemophilic mother (X^hX^h). Gametes of father: X^h and Y; gametes of mother: X^h and X^h.
5. Offspring: X^hX^h, X^hX^h (haemophilic daughters) and X^hY, X^hY (haemophilic sons). Ratio of haemophilic daughters : haemophilic sons = 2 : 2 or 1 : 1.

Teacher's Note:
a) One mark is for the correct genotypes of the parents and one mark is for the correct cross.
b) A daughter can be haemophilic only if her father is haemophilic and her mother is at least a carrier.
c) Always show the gametes clearly before writing the offspring.

OR

(B) (i) Where do transcription and translation occur in bacteria and eukaryotes respectively ?
(ii) Draw a labelled schematic sketch of replication fork of DNA.
(iii) A DNA segment has a total of 1000 nucleotides, out of which 240 of them are Adenine containing nucleotides. How many pyrimidine bases this segment possesses ? [5 Marks]

Answer:
1. (i) In bacteria, both transcription and translation occur in the cytoplasm. In eukaryotes, transcription occurs in the nucleus and translation occurs in the cytoplasm.
2. (ii) The sketch shows the two parental template DNA strands (5' to 3' and 3' to 5') separating at a Y-shaped fork. On one template, the new strand is formed by continuous synthesis (leading strand, in the 5' to 3' direction towards the fork). On the other template, the new strand is formed by discontinuous synthesis in short fragments (lagging strand, also 5' to 3' but away from the fork). The newly synthesised strands are labelled with their polarity.
3. (iii) Total nucleotides = 1000; Adenine (A) = 240. Since A = T, Thymine (T) = 240, so A + T = 480.
4. C + G = 1000 - 480 = 520. Since C = G, Cytosine (C) = \( 520 \div 2 = 260 \).
5. Total pyrimidines = T + C = 240 + 260 = 500.

Teacher's Note:
a) In the sketch, half a mark each goes to template DNA, continuous synthesis, discontinuous synthesis and newly synthesised strands with correct polarity.
b) Pyrimidines are thymine and cytosine; purines are adenine and guanine.
c) Quick check: in double stranded DNA, pyrimidines are always half of the total bases.

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