CBSE Class 12 Biology Question Paper 2025 Solved Code 57-1-3

SECTION A

1. What type of ecological pyramid would be obtained with the following data?
Primary producer = 10 g
Secondary consumer = 120 g
Primary consumer = 60 g [1 Mark]
(A) Upright pyramid of biomass
(B) Upright pyramid of number
(C) Inverted pyramid of biomass
(D) Upright pyramid of energy

Answer: (C) Inverted pyramid of biomass

Teacher's Note:
a) Biomass increases from producer to secondary consumer here, so the pyramid is inverted.
b) This type of pyramid is commonly seen in pond ecosystems.

2. Flowers which have single ovule in the ovary and are packed into inflorescence are usually pollinated by [1 Mark]
(A) Water
(B) Bat
(C) Bee
(D) Wind

Answer: (D) Wind

Teacher's Note:
a) Wind-pollinated flowers usually have single ovules and are clustered in inflorescences to increase chance of pollination.
b) Example - grasses.

3. 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: Diagram showing stages of dicot embryo development - single celled stage P, developing into globular embryo, heart-shaped embryo with Q labelled at top, and mature embryo with R labelled and S labelled below it]

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

Teacher's Note:
a) The zygote divides to form proembryo which develops through globular and heart shaped stages into a mature embryo.
b) The suspensor helps in nutrient transfer to the developing embryo.

4. 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) Remember the evolutionary sequence: Dryopithecus/Ramapithecus, Australopithecines, Homo erectus, Neanderthal man, Homo sapiens.
b) Homo erectus had a larger brain capacity than Australopithecines.

5. 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 is \(X^{C}Y\) (normal, since X comes from his mother, not his colour-blind father).
b) The woman is a carrier \(X^{C}X^{c}\) (X^c from her colour-blind mother, X^C from her normal father).
c) Crossing \(X^{C}Y \times X^{C}X^{c}\) gives half the sons as \(X^{c}Y\) (colour-blind).

6. Endosperm is completely consumed by the developing embryo in which of the following? [1 Mark]
(A) Maize and Castor
(B) Castor and Groundnut
(C) Maize and Pea
(D) Pea and Groundnut

Answer: (D) Pea and Groundnut

Teacher's Note:
a) Seeds like pea and groundnut are non-endospermic as food is stored in cotyledons.
b) Maize and castor are endospermic seeds where endosperm persists till maturity.

7. 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 decides the validity of GM research and its safety for the environment.
b) This is now called Genetic Engineering Appraisal Committee in updated syllabus.

8. 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) T-lymphocytes recognise foreign tissue antigens and cause graft rejection.
b) This is why immunosuppressants are given to transplant patients.

9. 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) This is achieved through PCR technique using Taq polymerase.
b) Repeated denaturation, annealing and extension cycles cause exponential amplification.

10. 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) RNAi involves complementary dsRNA binding to specific mRNA and silencing it.
b) This is used to make tobacco resistant to Meloidogyne incognita.

11. 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 except T is replaced by U.
b) mRNA is synthesised complementary to the template strand only.

12. Which of the following is correct for the condition when plant YyRr is back crossed with the double recessive parent? [1 Mark]
(A) 9 : 3 : 3 : 1 ratio of phenotypes only
(B) 9 : 3 : 3 : 1 ratio of genotypes only
(C) 1 : 1 : 1 : 1 ratio of phenotypes only
(D) 1 : 1 : 1 : 1 ratio of phenotypes and genotypes

Answer: (D) 1 : 1 : 1 : 1 ratio of phenotypes and genotypes

Teacher's Note:
a) A test cross with double recessive parent always gives 1:1:1:1 ratio of both phenotype and genotype.
b) This is used to determine the genotype of an unknown dominant individual.

13. Assertion (A): The mammary glands secrete milk for the nourishment of the young ones.
Reason (R): These are modified sweat glands. [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) Both statements are factually correct but R does not explain why milk is secreted.
b) Mammary glands are indeed modified sweat glands.

14. Assertion (A): Scaffold proteins are non-histone chromosomal proteins.
Reason (R): They are rich in lysine and arginine. [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) Being rich in lysine and arginine is a property of histone proteins, not scaffold (non-histone) proteins.
b) Scaffold proteins help maintain the higher order packaging of chromatin.

15. 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) Higher parental type is due to linkage, not crossing over.
b) Crossing over produces recombinant (non-parental) gene combinations in lower frequency.

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) Stirred tank bioreactors are indeed the most widely used type.
b) But this fact does not explain why bioreactors provide optimal growth conditions.

SECTION B

17. Study the pedigree analysis given below:
Parents: a (unfilled square) - b (unfilled circle)
Generation I: unfilled square, c (filled/shaded square), unfilled square - unfilled circle, filled/shaded circle, unfilled square
Generation II: filled/shaded square, d (unfilled circle), unfilled circle
Answer the following questions: [2 Marks]
(A) Write genotypes of c and d.
(B) Identify whether the trait is
(i) Sex-linked or autosomal
(ii) Dominant or recessive

[Figure: Pedigree chart with unaffected parents a and b, five children in Generation I (one affected male c, one affected female shown shaded, three unaffected), a mating between an unaffected member of Generation I and an affected member producing Generation II with one affected male, one unaffected female d and one unaffected female]

Answer:
1. Genotype of c = aa (affected, homozygous recessive) and genotype of d = AA or Aa (unaffected).
2. The trait is autosomal and recessive, since unaffected parents produce affected offspring and both sexes are equally affected.

Teacher's Note:
a) Unaffected parents having an affected child confirms a recessive trait.
b) Since both sons and daughters are affected equally, the trait is autosomal, not sex-linked.

18. Study the two different crosses given below:
Cross I
Parents: 44 + XX | 44 + XY
Gametes: 22 + O, 22 + XX | 22 + X, 22 + Y
Offspring: 1
Cross II
Parents: 44 + XY | 44 + XX
Gametes: 22 + Y, 22 + X | 23 + X, 21 + X
Offspring: 2
Identify the abnormalities '1' and '2' in the offsprings of given crosses and distinguish between them. [2 Marks]

[Figure: Two pedigree crosses showing non-disjunction of sex chromosomes producing offspring 1 with genotype 44+XO from combination of 22+O gamete with 22+X gamete, and offspring 2 with genotype 45+XY from combination of 22+Y gamete with 23+X gamete]

Answer:
1. Offspring '1' has Turner's Syndrome (44 + XO), caused by non-disjunction of the X chromosome resulting in absence of one X chromosome; such individuals are sterile females with rudimentary ovaries and underdeveloped secondary sexual characters.
2. Offspring '2' has Down's Syndrome (45 + XY / 45 + XX), caused by trisomy of chromosome 21 due to non-disjunction; affected individuals have short stature, a small round head, furrowed tongue and partially open mouth with retarded mental and physical development.

Teacher's Note:
a) Turner's syndrome is due to sex chromosome non-disjunction (monosomy X), while Down's syndrome is autosomal trisomy of chromosome 21.
b) Both arise from errors during meiosis (non-disjunction).

19. 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 derived from the poppy plant.
2. Morphine is a very effective sedative and painkiller, while heroin is a depressant that slows down body functions.

Teacher's Note:
a) Remember heroin = diacetylmorphine, chemically modified morphine.
b) Both are addictive and abused as narcotics.

OR

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

Answer:
1. With distillation: Whisky / Brandy / Rum.
2. Without distillation: Wine / Beer.
3. Cyanobacteria fix atmospheric nitrogen, add organic matter to the soil, and increase soil fertility, thereby acting as natural bio-fertilizers.

Teacher's Note:
a) Distilled drinks have higher alcohol content than fermented ones like wine and beer.
b) Cyanobacteria such as Nostoc and Anabaena are common examples of nitrogen-fixing bio-fertilizers.

20. 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 more productive as it contains more biomass and higher biodiversity, which traps and stores more solar radiation.
2. The young forest is still developing with fewer trees, so it captures less solar radiation and has lower productivity, while the shallow polluted lake has fewer producers and more dead organic matter, making it the least productive of the three.

Teacher's Note:
a) Productivity depends on the amount of standing biomass and photosynthetic surface area.
b) Pollution reduces the number of primary producers, lowering productivity.

OR

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

Answer:
1. Gross Primary Productivity (GPP) is the total rate of production of organic matter during photosynthesis by an ecosystem.
2. Net Primary Productivity (NPP) is GPP minus the respiratory losses (R), and represents the biomass available for consumption by heterotrophs.

Teacher's Note:
a) NPP = GPP - R is a key formula to remember.
b) NPP is what is actually available to herbivores and decomposers.

21. 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 cells, so they need to be periodically replaced in the patient's body.
2. A permanent cure could be achieved if the ADA gene isolated from bone marrow cells is introduced into cells (lymphocytes) at an early embryonic stage.

Teacher's Note:
a) Gene therapy attempts a permanent correction by targeting stem cells rather than mature lymphocytes.
b) ADA deficiency causes severe combined immunodeficiency (SCID).

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) Heat shock method is a third alternative technique, using calcium treatment and incubation on ice followed by 42°C.
b) These are all vector-less DNA transfer methods used in biotechnology.

SECTION C

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

Answer:
1. Divergent evolution occurs when the same structure develops along different directions due to adaptation to different needs, producing homologous structures.
2. Convergent evolution occurs when different structures evolve for the same function and become similar in appearance, resulting in analogous structures.
3. Both types of evolution are influenced by the environment and both contribute to the evolution of species.

Teacher's Note:
a) Divergent evolution indicates common ancestry (homology).
b) Convergent evolution indicates similar habitat leading to similar adaptive features (analogy), not common ancestry.

23. 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. Amniocentesis can be suggested, in which the amniotic fluid surrounding the developing embryo is analysed for foetal cells and dissolved substances to detect the presence of genetic disorders.
2. Medical Termination of Pregnancy (MTP) is a possible option; yes, it is safe for Radha at the current gestational age since MTP is comparatively safe up to 12 weeks (the first trimester) of pregnancy.
3. This process becomes illegal when it is performed by unqualified persons (quacks), or when it is used for female foeticide after determining the sex of a normal female foetus.

Teacher's Note:
a) Amniocentesis is legally banned for sex determination in India under the PNDT Act.
b) MTP within the first trimester carries lower health risk to the mother.
c) Misuse of prenatal diagnostic techniques for sex selection is a punishable offence.

24. (a) Write the palindromic nucleotide sequence for the 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. The palindromic complementary sequence is 3' - CTTAAG - 5'.
2. The restriction endonuclease that recognizes this sequence is EcoRI.
3. The restriction enzyme cuts the DNA strand a little away from the centre of the palindrome site but between the same two bases on the opposite strands, leaving single stranded overhangs called sticky ends; these sticky ends form hydrogen bonds with their complementary cut ends and help in joining vector DNA with foreign DNA, facilitating the action of DNA ligase.

Teacher's Note:
a) EcoRI always cuts between G and A of the GAATTC sequence.
b) Sticky ends make ligation of vector and foreign DNA more efficient than blunt ends.

25. 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: Life cycle diagram of Plasmodium showing an oocyst labelled P bursting to release sporozoites, an arrow indicating "Fertilization and development" leading down to a mosquito biting human skin, and Q labelled pointing to gametocytes released into the mosquito's gut]

Answer:
1. The infective stage stored in the female Anopheles mosquito is the sporozoite.
2. Fertilization and development of the parasite take place in the gut of the female Anopheles mosquito.
3. P is the salivary gland (containing sporozoites) and Q is the gametocytes.
4. The asexual phase occurs in humans and the sexual phase occurs in the female Anopheles mosquito.

Teacher's Note:
a) Sporozoites are injected into humans through the mosquito's saliva during a bite.
b) Human is the intermediate host and mosquito is the definitive host for Plasmodium.

26. 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: Diagram showing a spherical bead-like structure with DNA (labelled P) wound around it and a flat structure (labelled Q) representing the histone octamer core]

Answer:
1. The structure shown is a nucleosome.
2. P is DNA and Q is the histone octamer.
3. Histone proteins are basic in nature (positively charged) due to their richness in lysine and arginine residues.
4. Euchromatin is loosely packed, lightly stained and transcriptionally active chromatin; Heterochromatin is densely packed, darkly stained and transcriptionally inactive chromatin.

Teacher's Note:
a) Positively charged histones bind tightly to negatively charged DNA.
b) Darker staining under microscope always indicates heterochromatin.

27. (A) Who is having more species diversity - Columbia or Greenland and why?
(B) Explain the concept proposed by Paul Ehrlich. [3 Marks]

Answer:
1. Columbia has more species diversity because it is located near the equator, and as we move from the equator towards the poles, biodiversity decreases.
2. Paul Ehrlich proposed the rivet popper hypothesis, comparing an ecosystem to an airplane held together by thousands of rivets (species); if every passenger starts popping a rivet to take home, the loss of a few species may not immediately affect flight safety (ecosystem functioning).
3. However, as more and more rivets (species) are removed, the plane becomes dangerously weak over time, and removal of rivets on the wings (key species) poses a more serious threat to flight safety, showing that loss of key species is far more damaging to an ecosystem.

Teacher's Note:
a) Species richness generally follows a latitudinal gradient, highest near the equator.
b) The rivet popper hypothesis highlights the importance of keystone species in maintaining ecosystem stability.

28. (A) Differentiate between Recombinant DNA and cDNA.
(B) Explain the method to increase the competency of the bacterial cell membrane to take up recombinant DNA. [3 Marks]

Answer:
1. Recombinant DNA is formed by joining together two DNA fragments from two different sources.
2. cDNA (complementary DNA) is formed by reverse transcription of mRNA.
3. Competency is increased by treating bacterial cells with a divalent cation such as calcium, which increases the efficiency with which DNA enters the bacterium through pores in its cell wall; the cells are then incubated on ice, followed by briefly placing them at 42°C (heat shock), and then putting them back on ice.

Teacher's Note:
a) cDNA lacks introns since it is made from mature mRNA.
b) Heat shock treatment is essential for the bacterial cell to take up the recombinant DNA.

SECTION D

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?
(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. [4 Marks]

Answer:
1. The Monarch butterfly acquires the distasteful chemical during its caterpillar stage by feeding on a poisonous weed.
2. Using the formula \(N_{t+1} = N_t + [(B+I) - (D+E)]\):
\(800 = N_t + [(200+200) - (150+100)]\)
\(800 = N_t + (400 - 250)\)
\(800 = N_t + 150\)
\(N_t = 800 - 150 = 650\)
Comment: Since the population density is increasing with time, the age pyramid formed would be that of an expanding population.
3. For sub-part (c): A single huge banyan tree is measured in terms of biomass or percent cover, while 200 carrot grass plants are measured in terms of percent cover only (not by counting individual plants directly, due to differences in size).

Teacher's Note:
a) The formula \(N_{t+1} = N_t + (B+I) - (D+E)\) must be rearranged correctly to find \(N_t\).
b) An expanding population shows a broad base age pyramid indicating high proportion of young individuals.
c) For part (d): the two methods to measure tiger population density are pug marks and faecal pellet counts.

30. Study the graphs given below for Case 1 and Case 2 showing different levels of certain hormones and answer the questions 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? [4 Marks]

[Figure: Two graphs showing hormone levels across follicular and luteal phases. Both graphs show hormone P peaking sharply just before ovulation and hormone Q rising afterward. Case 2 additionally shows an HCG curve rising in the luteal phase, indicating pregnancy]

Answer:
1. Luteinizing Hormone (LH) is responsible for the peak observed in both Case 1 and Case 2; it induces rupturing of the graafian follicle, helping in ovulation.
2. During the follicular phase, in the ovary the follicles mature (maturation of follicles), while in the uterus there is proliferation of the endometrium lining.
3. For sub-part (c): Hormone Q in Case 2 is progesterone; it maintains pregnancy by supporting the endometrium.

Teacher's Note:
a) LH surge is the trigger for ovulation in the menstrual cycle.
b) Presence of HCG in Case 2 indicates pregnancy, so the corpus luteum is maintained by HCG.
c) For part (d): the corpus luteum remains functional in Case 2, formed from the graafian follicle after ovulation due to stimulation by HCG.

SECTION E

31. Student to attempt either option-(A) or (B).
(A) (i) You are given axial pea flowers with violet colour whose genotypes are unknown. How would you find the genotype of these plants? Explain with the help of cross.
(ii) Explain Haplodiploidy in honey-bees. [5 Marks]

Answer:
1. To determine the unknown genotype, a test cross is performed by crossing the plant with axial violet flowers (dominant traits) with a double recessive plant (terminal white flowers, aavv).
2. If the plant is homozygous (AAVV), the cross AAVV \(\times\) aavv produces 100% axial violet offspring (AaVv), confirming a homozygous dominant genotype.
3. If the plant is heterozygous for one gene, for example AaVV \(\times\) aavv, the offspring show 50% axial violet and 50% terminal violet, confirming genotype AaVV.
4. Similarly, if heterozygous for both genes (AaVv \(\times\) aavv), offspring show 25% each of four phenotype classes (axial violet, axial white, terminal violet, terminal white), confirming genotype AaVv.
5. In honey-bees, males (drones) are haploid, developing from unfertilised eggs (parthenogenesis), while females (queens and workers) are diploid, developing from fertilised eggs; this system is called haplodiploidy and determines the sex of the offspring.

Teacher's Note:
a) A test cross always uses a double recessive individual to reveal hidden alleles.
b) In haplodiploidy, males have no father and cannot have sons, but do have grandsons.

OR

(B) Explain the steps of DNA fingerprinting that will help in processing of the two blood samples R and S picked up from the crime scene. [5 Marks]

Answer:
1. Isolation of DNA from both the blood samples R and S.
2. Digestion of the DNA of both samples using the same restriction endonucleases.
3. Separation of DNA fragments by gel electrophoresis, placing them in different wells of the agarose gel.
4. Transferring (blotting) of the separated DNA fragments onto synthetic membranes such as nitrocellulose or nylon.
5. Hybridisation using a labelled VNTR probe, followed by detection of the hybridised DNA fragments by autoradiography, which allows comparison of the DNA fingerprint patterns of samples R and S.

Teacher's Note:
a) VNTR (Variable Number of Tandem Repeats) probes are the key to DNA fingerprinting technology.
b) Southern blotting is the technique used for transferring DNA fragments onto the membrane.

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. MALT stands for Mucosa Associated Lymphoid Tissue; it is located within the lining of major tracts such as the respiratory, digestive or urogenital tract.
2. Cytokine barriers: virus-infected cells secrete proteins called interferons, which protect non-infected cells from further viral infection.
3. The diagnostic test for AIDS is ELISA (Enzyme Linked Immuno-Sorbent Assay); it works on the principle of antigen-antibody interaction.
4. Bone marrow is the main lymphoid organ where all blood cells including lymphocytes are produced, and some of these immature lymphocytes migrate to the thymus, where they differentiate into antigen-sensitive mature T-lymphocytes.

Teacher's Note:
a) MALT constitutes a significant part of the body's innate immune system.
b) PCR can also be accepted as a diagnostic test based on nucleic acid amplification.

OR

(B) (i) Fill 'H', 'I', 'J', 'K', 'L' and 'M' in following table with suitable words:
Chemical/Bioactive Molecule: Butyric acid | Micro-organism: H | Category: I | Use: Important applications in food, chemical & pharma industry
Chemical/Bioactive Molecule: J | Micro-organism: Monoscus purpureus | Category: K | Use: Inhibit cholesterol biosynthesis pathway
Chemical/Bioactive Molecule: Cyclosporin A | Micro-organism: L | Category: Fungus | Use: M
(ii) Why are baculoviruses used as biological control agents? [5 Marks]

Answer:
1. 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. Baculoviruses are used as biological control agents because they are species-specific and have narrow spectrum insecticidal properties, and have no negative impact on non-target species like plants, mammals, birds or fishes.

Teacher's Note:
a) Statins are used to lower blood cholesterol by inhibiting HMG-CoA reductase.
b) Cyclosporin A obtained from Trichoderma polysporum is used as an immunosuppressive agent.

33. 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 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: Diagram showing three flowers with arrows depicting different modes of pollen transfer - method 1 shows pollen moving within the same flower, method 2 shows pollen moving between different flowers on the same plant, and method 3 shows pollen moving between flowers on two different plants]

Answer:
1. The vegetative cell is big with abundant food reserve and an irregular shaped nucleus, and it helps in the formation of the pollen tube; the generative cell is small, floats in the cytoplasm of the vegetative cell, and forms two male gametes.
2. Method 1 = Autogamy, Method 2 = Geitonogamy, Method 3 = Xenogamy.
3. Water lily achieves pollination with the help of insects or wind, while Vallisneria achieves pollination through water.
4. The advantages of xenogamy (method 3) include genetic variation, healthier offspring, elimination of recessive traits, disease resistance and no inbreeding depression.

Teacher's Note:
a) Xenogamy is the only true form of cross pollination between genetically different plants.
b) Geitonogamy is genetically similar to autogamy since both flowers belong to the same plant.

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: Diagram of a human ovum surrounded by cells of the corona radiata, with a zona pellucida layer and perivitelline space labelled, and three sperms labelled P, Q and R approaching the ovum, with P shown penetrating the zona pellucida]

Answer:
1. Sperm P is able to penetrate and fertilise the ovum, whereas sperms Q and R are unable to penetrate or fertilize it.
2. When a sperm comes in contact with the zona pellucida layer of the ovum, it induces changes in the membrane that blocks the entry of additional sperms.
3. Entry of the sperm induces completion of the meiotic division of the secondary oocyte, resulting in the formation of the second polar body and a haploid ovum (ootid).
4. The acrosome is filled with enzymes that help the sperm to penetrate into the cytoplasm of the ovum, and the middle piece contains numerous mitochondria that produce energy for the movement of the tail, facilitating sperm motility needed for fertilisation.

Teacher's Note:
a) Only one sperm is normally allowed to fertilise the ovum due to the zona reaction.
b) The acrosomal enzymes help dissolve the zona pellucida for sperm entry, while mitochondria in the middle piece power sperm motility.

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