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

SECTION A

1. 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 man gets his Y chromosome from his father, so he is normal (X⁺Y); the woman is a carrier (X⁺X^c) as she got X^c from her mother.
b) Sons get their X only from the mother, so half the sons get X^c and are colour-blind.

2. 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 Government body that decides on the validity of GM research and the safety of introducing GM organisms for public use.
b) Learn the full form exactly, as options differ by only one or two words.

3. 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 : [1 Mark]
(A) (i)-(b), (ii)-(d), (iii)-(c), (iv)-(a)
(B) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a)
(C) (i)-(d), (ii)-(a), (iii)-(b), (iv)-(c)
(D) (i)-(c), (ii)-(b), (iii)-(a), (iv)-(d)

Answer: (B) (i)-(c), (ii)-(d), (iii)-(b), (iv)-(a)

Teacher's Note:
a) Ladybird beetles feed on aphids; Glomus forms mycorrhiza; Trichoderma is a fungal biocontrol agent.
b) Methanobacterium is the methanogen that produces methane in biogas plants.

4. The process of mineralization by microorganisms help in the release of : [1 Mark]
(A) inorganic nutrients from humus.
(B) both organic and inorganic nutrients from detritus.
(C) organic nutrients from humus.
(D) inorganic nutrients from detritus and formation of humus.

Answer: (A) inorganic nutrients from humus.

Teacher's Note:
a) Humus is degraded slowly by some microbes, and this releases inorganic nutrients; this step is called mineralisation.
b) Do not confuse it with humification, which is the formation of humus from detritus.

5. 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 caused by T-lymphocytes, which is cell mediated immunity (CMI).
b) Humoral immunity works through antibodies made by B-lymphocytes, not through T-cells.

6. 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. [1 Mark]
(A) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
(B) (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
(C) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
(D) (a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)

Answer: (A) (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

Teacher's Note:
a) The trophoblast is the outer layer of the blastocyst, and the inner cell mass forms the embryo.
b) A morula is the 8-16 celled stage, formed before the blastocyst.

7. 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 joined by 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 curved lobes and S pointing to the notch between them.]

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

Teacher's Note:
a) The embryo develops from the zygote, and the suspensor is the row of cells that attaches it.
b) In the mature dicot embryo the two large lobes are cotyledons and the plumule lies between them.

8. 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 of denaturation, annealing and extension can copy the DNA about a billion times.
b) A thermostable DNA polymerase (Taq polymerase) from Thermus aquaticus is used.

9. 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) The mRNA has the same sequence and polarity as the coding strand, with U in place of T.
b) A common mistake is to write the complement of the coding strand; that is the template strand.

10. How many pollen grains and ovules are likely to be formed in the anther and the ovary of an angiosperm bearing 50 microspore mother cells and 50 megaspore mother cells respectively ? [1 Mark]
(A) 100, 25
(B) 200, 50
(C) 50, 50
(D) 200, 100

Answer: (B) 200, 50

Teacher's Note:
a) Each microspore mother cell divides by meiosis to give a tetrad of 4 pollen grains, so \( 50 \times 4 = 200 \).
b) Each megaspore mother cell gives only one functional megaspore, so 50 ovules are formed.

11. 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 are older ape-like ancestors and come before Australopithecines.
b) Neanderthal man came after Homo erectus and before modern Homo sapiens.

12. RNA interference (RNAi) helps in making tobacco-plant resistant to a nematode (Meloidegyne incognitia)
Choose the correct option that shows how RNAi is achieved : [1 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) The dsRNA formed in the host binds to the specific mRNA of the parasite and silences it, so the mRNA is not translated.
b) RNAi acts at the mRNA level; it does not affect DNA replication or transcription.

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) : Saheli is the World's first non-steroidal oral contraceptive pill.
Reason (R) : It has been developed by National Institute of Immunology, New Delhi. [1 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) Saheli is a once-a-week non-steroidal pill with a high contraceptive value.
b) It was developed by the Central Drug Research Institute (CDRI), Lucknow, not by NII, New Delhi.

14. 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 that one amino acid can be coded by more than one codon, which is exactly what R states.
b) Do not confuse degeneracy with ambiguity; the code is unambiguous, as one codon codes for only one amino acid.

15. 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 for the desired product.
b) The fact that stirred-tank bioreactors are most common is true, but it does not explain why bioreactors give optimal conditions.

16. Assertion (A) : When the two genes in a dihybrid cross are situated on the same chromosome, the proportion of parental gene combinations is much higher than non-parental type.
Reason (R) : Higher parental gene combinations can be attributed to crossing over between two genes. [1 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) Genes on the same chromosome are linked, so parental combinations are much more frequent.
b) Crossing over produces non-parental (recombinant) combinations, so R is false.

SECTION B

17. Student to attempt either option-(A) or (B) : [2 Marks]

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

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

Teacher's Note:
a) The keyword "acetylation of morphine" carries the full mark for the relation.
b) Write one separate effect for each drug to get both half marks.

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/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; wine and beer are made without distillation.
b) Examples like Anabaena, Nostoc and Oscillatoria are common cyanobacterial biofertilisers in paddy fields.

18. Student to attempt either option-(A) or (B) : [2 Marks]

(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 to trap and store solar energy 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.

Teacher's Note:
a) Name the most productive ecosystem clearly first, then compare it with 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; a considerable part of it is used by plants in respiration.
2. Net primary productivity (NPP) is GPP minus respiration losses (R), that is \( NPP = GPP - R \); it is the biomass available for consumption by heterotrophs.

Teacher's Note:
a) Writing the relation \( NPP = GPP - R \) makes the difference clear at once.
b) Remember that only NPP is available to herbivores and decomposers.

19. 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. [2 Marks]

[Figure: Parents 44 + XY and 44 + XX. Gametes of the male parent: 22 + XY and 22 + O; gametes of the female parent: 22 + X and 22 + X. Lines join the gametes to give offsprings labelled 1 (from 22 + XY with 22 + X) and 2 (from 22 + O with 22 + X).]

Answer:
1. Offspring 1 is Klinefelter's syndrome (44 + XXY) and offspring 2 is Turner's syndrome (44 + XO).
2. Klinefelter's syndrome has an extra X chromosome (47 chromosomes, trisomy of sex chromosomes) and gives a sterile male, while Turner's syndrome lacks one X chromosome (45 chromosomes, monosomy of sex chromosomes) and gives a sterile female.

Teacher's Note:
a) Write the chromosome formula (XXY and XO) with each name to get the half marks.
b) Other valid differences: Klinefelter's shows masculine development with gynaecomastia; Turner's shows rudimentary ovaries and short stature.

20. 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' ? [2 Marks]

[Figure: Pedigree chart. Parents: male 1 (blood group B, square) and female 2 (blood group A, circle). Generation I: female 3 (B) married to male 4 (A); their siblings are male 5 (AB), female 6 (B) and male 7 (A). Generation II (children of 3 and 4): female 8 (O) and male 9 (B).]

Answer:
1. (a) Individual 1 = I^Bi and individual 2 = I^Ai.
2. (b) Individual 5 (AB) has both antigen A and antigen B on the RBCs; individual 8 (O) has neither antigen A nor antigen B.

Teacher's Note:
a) Parents 1 and 2 must each carry the recessive allele i, because their grandchild 8 is group O and got one i from 4.
b) Blood group O means no A or B antigen on the RBC surface.

21. Student to attempt either option-(A) or (B) : [2 Marks]

(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 they die after some time and the patient needs periodic infusion.
2. A permanent cure is possible if the gene producing ADA, isolated from bone marrow cells, is introduced into cells at early embryonic stages.

Teacher's Note:
a) The key phrase for the first mark is "lymphocytes are not immortal".
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 (gene gun): plant cells are bombarded with high velocity micro-particles of gold or tungsten coated with DNA.

Teacher's Note:
a) Name the technique and describe it briefly; each carries half a mark.
b) Heat shock (ice, then \( 42^{\circ} \)C, then ice again) is also accepted as a vectorless method.

SECTION C

22. (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) The complementary palindromic sequence is 3' - CTTAAG - 5'.
2. (b) The restriction endonuclease is 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) on the opposite strands. This leaves single stranded overhanging stretches at the ends, called sticky ends.
4. Role: sticky ends form hydrogen bonds with their complementary cut counterparts, so they help to join vector DNA and foreign DNA with the help of DNA ligase in rDNA technology.

Teacher's Note:
a) A palindrome reads the same on both strands when read in the same direction (5' to 3').
b) A labelled diagram of EcoRI cutting between G and A is also accepted for the part on sticky ends.
c) Always mention DNA ligase when explaining the role of sticky ends.

23. 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 female mosquito biting a human hand, with small bodies labelled Q taken in from the blood; an arrow leads to a structure marked "Fertilization and development", and an upward arrow leads to an oval structure containing many thread-like bodies, labelled P.]

Answer:
1. (a) Sporozoites.
2. (b) Fertilisation and development take place in the gut (stomach) of the female Anopheles mosquito.
3. (c) P - Salivary gland of female Anopheles mosquito; Q - Gametocytes.
4. (d) The asexual phase takes place in the human and the sexual phase takes place in the mosquito.

Teacher's Note:
a) Sporozoites are stored in the salivary glands and enter the human body through the mosquito bite.
b) Gametocytes are formed in the human RBCs and are taken up by the mosquito when it sucks blood.
c) Do not mix up the hosts: the mosquito is the host for the sexual phase.

24. (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 :
(i) IUCN Red data list
(ii) Hot spots in conservation of biodiversity [3 Marks]

Answer:
1. (a) Alien species invasion: the Nile perch introduced into Lake Victoria in East Africa led to the extinction of more than 200 species of cichlid fish.
2. Habitat loss and fragmentation: the Amazon rain forest, which harbours probably millions of species, is being cut and cleared for soya bean cultivation and for grasslands to raise cattle. When large habitats are broken into small fragments, mammals and birds needing large territories are badly affected and their populations decline.
3. (b) (i) The IUCN Red data list gives information about species under threat of extinction, which helps to identify and protect species at high risk and to plan conservation.
4. (ii) Hot spots have very high levels of species richness and a high degree of endemism, so protecting them conserves a large number of species.

Teacher's Note:
a) Give one clear, named example for each cause; Parthenium, Lantana and water hyacinth are also accepted for alien species.
b) For hot spots, the two keywords are "species richness" and "endemism".

25. 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 spherical core with a thick coiled thread (labelled P) wrapped around it; Q points to the spherical core. A thin vertical oval disc stands across the thread on one side.]

Answer:
1. (a) The structure is a nucleosome.
2. (b) P - DNA; Q - Histone octamer.
3. (c) Histones are basic proteins; they are positively charged (rich in lysine and arginine).
4. (d) Euchromatin is loosely packed, lightly stained and transcriptionally active, whereas heterochromatin is densely packed, darkly stained and transcriptionally inactive.

Teacher's Note:
a) The positive charge of histones lets them bind the negatively charged DNA.
b) One correct difference between euchromatin and heterochromatin is enough for the 1 mark.

26. 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 ? [3 Marks]

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

Teacher's Note:
a) In part (b), give the option, the answer "Yes" and the reason; each carries half a mark.
b) Remember that amniocentesis is misused for sex determination, which is banned in India.

27. Explain the basis on which gel electrophoresis technique works. Write any two ways the products obtained through this technique can be utilised. [3 Marks]

Answer:
1. DNA fragments are negatively charged, so under an electric field they move towards the anode through a medium or matrix (agarose gel).
2. The fragments separate according to their size due to the sieving effect of the gel; the smaller the fragment, the farther it moves.
3. Uses of the separated DNA fragments: (i) in recombinant DNA technology (joining with cloning vectors), and (ii) in DNA fingerprinting.

Teacher's Note:
a) The keywords are "negatively charged", "towards the anode" and "sieving effect".
b) Use in PCR is also an accepted answer for the utilisation part.

28. 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 look similar.
2. Divergent evolution produces homologous structures and shows common ancestry; convergent evolution produces analogous structures and shows that a similar habitat has selected 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 similarities and differences; missing the similarity loses 1 mark.
b) Forelimbs of whale, bat and man are homologous; wings of butterfly and bird are analogous.

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

[Figure: Two graphs of hormone levels against time, each divided into Follicular Phase and Luteal Phase. Case 1: curve P rises to a sharp Peak at ovulation (shown by a dashed line and a symbol of an ovum being released) and then falls; curve Q rises after ovulation in the luteal phase and then falls back. Case 2: curve P shows the same sharp Peak at ovulation; curve Q rises after ovulation and keeps rising to the end of the graph, and another curve labelled HCG starts 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) causes the peak; it induces rupture of the Graafian follicle and so causes ovulation.

Teacher's Note:
a) The LH surge in the middle of the cycle (about day 14) is what triggers ovulation.
b) Name the hormone and give its function; each carries half a mark.

(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 and proliferates (thickens).

Teacher's Note:
a) The follicular phase is also called the proliferative phase because of the growth of the endometrium.
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 the endometrium and hence maintains pregnancy.

Teacher's Note:
a) In Case 2 progesterone keeps rising along with HCG, which shows that pregnancy has occurred.
b) A fall in progesterone (as in Case 1) leads to menstruation.

OR

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

Answer: The corpus luteum remains functional; it is formed from the ruptured Graafian follicle after ovulation.

Teacher's Note:
a) The corpus luteum secretes large amounts of progesterone during pregnancy.
b) In the absence of fertilisation (Case 1) the corpus luteum degenerates.

30. 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 butterfly acquires this chemical during its caterpillar stage by feeding on a poisonous weed.

Teacher's Note:
a) This is an example of a chemical defence used by prey against predators (birds).
b) Mention the "caterpillar stage" clearly, as the adult does not feed on the weed.

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

Teacher's Note:
a) Births and immigration add to the population; deaths and emigration reduce it.
b) Since \( N_{t+1} \) is greater than \( N_t \), the population is growing, which gives a triangular (expanding) 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, and the carrot grass is measured in terms of percent cover, since counting numbers alone would underestimate the huge role of the banyan tree.

Teacher's Note:
a) Numbers are not always the best measure; biomass or percent cover is more meaningful when individuals differ greatly in size.
b) This example is taken directly from the NCERT chapter on Organisms and Populations.

OR

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

Answer: The population density of tigers can be measured by counting their pug marks and faecal pellets.

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

SECTION E

31. Student to attempt either option-(A) or (B) : [5 Marks]

(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 Immuno-Sorbent 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; the thymus and bone marrow provide the micro-environment for the development and maturation of lymphocytes (T-lymphocytes mature in the thymus), so immature lymphocytes become antigen-sensitive lymphocytes.

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

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 ? [5 Marks]

Answer:
1. (i) H = Clostridium butylicum; I = Bacteria.
2. J = Statin; K = Fungi (Yeast).
3. L = Trichoderma polysporum; M = Immunosuppressant, used to suppress the immune system in patients with newly transplanted organs.
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, so they are ideal for IPM and ecologically sensitive areas.

Teacher's Note:
a) Statins act by competitively inhibiting the enzyme of cholesterol synthesis.
b) Baculoviruses of the genus Nucleopolyhedrovirus are the common example to quote.
c) Write scientific names in full with correct spelling.

32. Student to attempt either option-(A) or (B) : [5 Marks]

(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 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 in method '3'. [5 Marks]

[Figure: Two plants. On the right plant, arrow 1 shows pollen moving from the anther to the stigma of the same flower. The left plant has two flowers on branches of the same stem; arrow 2 shows pollen moving from one flower to the other flower of the same plant. Arrow 3 shows pollen moving from a flower of the left plant to a flower of the right 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 an irregular shaped nucleus, and helps to form the pollen tube; the generative cell is small, floats in the cytoplasm of the vegetative cell and forms the two male gametes.
2. (b) (i) 1 = Autogamy; 2 = Geitonogamy; 3 = Xenogamy.
3. (ii) (a) Water lily is pollinated by insects or wind (its flowers emerge above the water surface).
4. (b) Vallisneria is pollinated by water.
5. (iii) Advantages of xenogamy: it brings genetic variation and gives healthier offspring (no inbreeding depression).

Teacher's Note:
a) Geitonogamy is functionally cross-pollination but genetically similar to autogamy, since pollen comes from the same plant.
b) Other accepted advantages of xenogamy: disease resistance, elimination of recessive traits, heterosis and evolution.
c) Name both cells first; half a mark is for the names and one mark for the difference.

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 of a human sperm are able to play an important role in human fertilization ? [5 Marks]

[Figure: A human ovum surrounded by the zona pellucida, the perivitelline space and the cells of the corona radiata. Three sperms are shown: P on the left with its head at the zona pellucida, Q at the top among the corona cells, and R at the upper right 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 the cytoplasm of the ovum.
5. Middle piece: it has numerous mitochondria, which produce energy for the movement of the tail, giving sperm motility needed for fertilisation.

Teacher's Note:
a) The zona pellucida ensures that only one sperm fertilises an ovum (prevents polyspermy).
b) The secondary oocyte completes meiosis II only after sperm entry; remember the term "ootid".
c) Link each sperm part to its function: acrosome - enzymes, middle piece - mitochondria and energy.

33. Student to attempt either option-(A) or (B) : [5 Marks]

(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. [5 Marks]

Answer:
1. (i) Cross: Parents Hb^AHb^S \( \times \) Hb^AHb^S; gametes Hb^A, Hb^S from each parent. F₁ progeny: Hb^AHb^A (unaffected), Hb^AHb^S (carrier), Hb^AHb^S (carrier), Hb^SHb^S (affected).
2. Ratio of carrier : disease free : diseased = 2 : 1 : 1. Adenine is substituted by thymine (GAG changes to GTG), so glutamic acid is replaced by valine at the sixth position of the beta globin chain.
3. (ii) In garden pea, flower colour follows the Law of Dominance, because violet colour is completely dominant over white.
4. Cross: WW (violet) \( \times \) ww (white); gametes W and w; F₁ all Ww, all violet.
5. In snapdragon, flower colour shows incomplete dominance, because red is not completely dominant over white. Cross: RR (red) \( \times \) rr (white); gametes R and r; F₁ all Rr, all pink coloured flowers.

Teacher's Note:
a) Show parents, gametes and progeny (a Punnett square is best) in every cross; the cross itself carries marks.
b) In incomplete dominance the F₁ phenotype is intermediate (pink), unlike the dominant parent's phenotype in pea.
c) The sickle cell mutation is a single base substitution (point mutation).

OR

(B) Explain with the help of well-labelled diagrams how lac operon operates in E. coli :
(i) In presence of an inducer.
(ii) In absence of an inducer. [5 Marks]

Answer:
1. (i) In presence of inducer: the lac operon has the genes p (promoter of i), i (regulator), p (promoter), o (operator) and the structural genes z, y and a. The i gene is transcribed into repressor mRNA, which is translated into the repressor protein.
2. The inducer (lactose) binds to the repressor and makes it inactive, so the inactive repressor cannot bind to the operator.
3. RNA polymerase now transcribes the structural genes into lac mRNA, which is translated into beta-galactosidase (z), permease (y) and transacetylase (a).
4. (ii) In absence of inducer: the i gene still produces the repressor protein through repressor mRNA.
5. The active repressor binds to the operator region (o) and prevents RNA polymerase from transcribing the operon, so the structural genes are switched off.

Teacher's Note:
a) Diagrams must show the genes p, i, p, o, z, y, a in order; each correct label in the first diagram earns half a mark.
b) Lactose acts as the inducer; the lac operon is an example of negative regulation.
c) Beta-galactosidase breaks lactose into glucose and galactose, permease increases permeability of the cell to lactose.

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