CBSE Class 12 Biology Sample Paper 2026 27 with Solutions PDF Download

Section - A

Q. Nos. 1 to 12 are multiple choice questions. Only one of the choices is correct. Select and write the correct choice as well as the answer to these questions.

1. A researcher observes that in a particular flowering plant, the pistil matures several days before the stamens in all flowers of the plant. What is the most significant consequence of this timing difference? [1 Mark]
A. It guarantees self-pollination
B. It promotes autogamy
C. It ensures cross-pollination
D. It leads to geitonogamy

Answer: C. It ensures cross-pollination

Teacher's Note:
a) When the pistil matures before the stamens, the stigma is receptive before pollen of the same flower is released.
b) This is an outbreeding device, so it prevents self-pollination (autogamy) and favours cross-pollination.

2. A male patient undergoes a medical procedure where a small section of the vas deferens is cut and tied at both ends. Which of the following effects is the intended and direct result of this procedure? [1 Mark]
A. Blockage of sperm transport into the ejaculatory duct and urethra.
B. Inhibition of testosterone production by the testes.
C. Suppression of the formation of seminal plasma by accessory glands.
D. Prevention of spermatogenesis in the seminiferous tubules.

Answer: A. Blockage of sperm transport into the ejaculatory duct and urethra.

Teacher's Note:
a) This procedure is vasectomy, a surgical method of contraception in males.
b) The testes still make sperms and testosterone; only the path of the sperms is blocked.

3. During the early stage of human embryonic development, the blastomeres are arranged into an outer layer and an inner group of cells known as the Inner Cell Mass (ICM). What is the fundamental significance of the outer layer, the trophoblast? [1 Mark]
A. It is responsible for forming the primary germ layers (ectoderm, mesoderm, endoderm).
B. It provides essential nutrients to developing embryo and helps in its implantation.
C. It secretes colostrum which contains antibodies.
D. It gives rise to all the organs of the future embryo (organogenesis).

Answer: B. It provides essential nutrients to developing embryo and helps in its implantation.

Teacher's Note:
a) The trophoblast layer attaches to the endometrium, which helps in implantation.
b) The germ layers and all organs of the embryo come from the inner cell mass, not the trophoblast.

4. Select the option that gives the correct description of the process of natural selection with respect to the length of the neck of the giraffe. [1 Mark]
A. Stabilising selection as giraffes with longer neck lengths are selected further.
B. Disruptive selection as giraffes with smaller and longer neck lengths are selected.
C. Directional selection as giraffes with longer neck lengths are selected.
D. Stabilising selection as giraffes with medium neck lengths are selected.

[Figure: A graph with Number of giraffes on the y-axis and Neck length in giraffe on the x-axis, showing a single bell-shaped curve.]

Answer: C. Directional selection as giraffes with longer neck lengths are selected.

Teacher's Note:
a) In directional selection, individuals at one end of the range (here, longer necks) are favoured, so the population shifts in that direction.
b) Remember the three types: stabilising favours the middle value, directional favours one end, disruptive favours both ends.

For Visually Impaired Candidates (in lieu of Q. 4)

Evolution of antibiotic-resistant bacterial population signifies the information that [1 Mark]
A. Acquired traits are inherited.
B. Nature selects for fitness.
C. Genetic variations are not a prerequisite factor for natural selection.
D. The theory of spontaneous generation of life holds true.

Answer: B. Nature selects for fitness.

Teacher's Note:
a) Bacteria that already carry resistance survive the antibiotic and multiply, so the resistant type becomes common.
b) This is an example of natural selection in action, not inheritance of acquired traits.

5. The initiation step during the process of transcription in bacteria is shown below. Identify A, B, C and D by selecting the option: [1 Mark]
A. A-RNA polymerase, B-Promoter, C-Sigma factor, D-DNA helix
B. A-Promoter, B-Sigma factor, C-RNA polymerase, D-DNA helix
C. A-Promoter, B-RNA polymerase, C-DNA helix, D-Sigma factor
D. A-Promoter, B-RNA polymerase, C-Sigma factor, D-DNA helix

[Figure: A DNA double helix with ends marked 3' and 5' on the upper strand and 5' and 3' on the lower strand. A points to a short region at the left end of the DNA; B points to an oval enzyme lying below the DNA near the left end; C points to a small circle marked σ on that enzyme; D points to the DNA helix at the right side.]

Answer: D. A-Promoter, B-RNA polymerase, C-Sigma factor, D-DNA helix

Teacher's Note:
a) In bacteria, RNA polymerase binds to the promoter with the help of the sigma (σ) factor to start transcription.
b) The σ symbol drawn on the enzyme is the clue for identifying the sigma factor.

For Visually Impaired Candidates (in lieu of Q. 5)

The promoter site and the terminator site for transcription are located at: [1 Mark]
A. 3' (downstream) end and 5' (upstream) end, respectively of the transcription unit
B. 5' (upstream) end and 3' (downstream) end, respectively of the transcription unit
C. the 5' (upstream) end
D. the 3' (downstream) end

Answer: B. 5' (upstream) end and 3' (downstream) end, respectively of the transcription unit

Teacher's Note:
a) The promoter lies towards the 5' end (upstream) of the structural gene, with reference to the coding strand.
b) The terminator lies towards the 3' end (downstream) and marks the end of transcription.

6. Given below is the illustration of the different steps of experiments conducted by MacLeod, McCarty and Avery to find the chemical nature of the 'transforming principle' as DNA. Select the option that incorrectly depicts the step of the experiment. [1 Mark]

[Figure: A flask labelled Heat killed S bacteria gives Bacterial Filtrate, which is divided into four tubes. A: RNase, then R bacteria (live), then No transformation. B: Protease, then R bacteria (live), then Transformation. C: DNase, then R bacteria (live), then No transformation. D: Lipase, then R bacteria (live), then Transformation.]

Answer: A. No transformation

Teacher's Note:
a) RNase digests only RNA, so the DNA stays intact and transformation should occur; showing no transformation is wrong.
b) Only DNase stopped transformation, which proved that DNA is the transforming principle.

For Visually Impaired Candidates (in lieu of Q. 6)

In the Griffith experiment, which of the following, when injected into mice, did not cause pneumonia? [1 Mark]
A. Heat-killed S strain only
B. Live R strain only
C. Both (A) and (B)
D. Live S strain only

Answer: C. Both (A) and (B)

Teacher's Note:
a) The R strain has no capsule and is non-virulent, and heat-killed S strain cannot multiply, so the mice lived in both cases.
b) Only live S strain, or heat-killed S mixed with live R, killed the mice.

7. A scientist was culturing E.coli in a ¹⁵NH₄Cl medium initially, then he shifted E.coli to a ¹⁴NH₄Cl medium. The DNA was extracted from E.coli after 60 minutes of transfer to ¹⁴NH₄Cl Medium and performed density gradient centrifugation to check the isotope distribution in newly replicated DNA. What will be the densities of DNA molecules formed? [1 Mark]
A. 25% hybrid and 75% light
B. 50% hybrid and 50% light
C. 75% hybrid and 25% light
D. 85% hybrid and 15% light

Answer: A. 25% hybrid and 75% light

Teacher's Note:
a) E. coli divides every 20 minutes, so 60 minutes means 3 generations and 8 DNA molecules.
b) Due to semiconservative replication, only 2 of the 8 molecules are hybrid (2/8 = 25%) and 6 are light (75%).

8. Study the given pedigree and identify the disease represented [1 Mark]
A. Myotonic Dystrophy
B. Sickle cell anaemia
C. Haemophilia
D. Colour blindness

[Figure: Pedigree chart. Generation I: an unaffected male (square) and an unaffected female (circle). Generation II: an unaffected male, an affected male (shaded square), an unaffected male joined to an unaffected female, an affected female (shaded circle) and an unaffected male. Generation III (from the joined couple): an affected male and two unaffected females.]

Answer: B. Sickle cell anaemia

Teacher's Note:
a) Unaffected parents have affected children, so the trait is recessive.
b) An affected daughter is born to an unaffected father, so it cannot be X-linked recessive; it is autosomal recessive, like sickle cell anaemia.

For Visually Impaired Candidates (in lieu of Q. 8)

If a genetic disease is transferred from a phenotypically normal but carrier female to only some of the male progeny the disease is: [1 Mark]
A. Autosomal dominant
B. Autosomal recessive
C. Sex-linked dominant
D. Sex-linked recessive

Answer: D. Sex-linked recessive

Teacher's Note:
a) A carrier mother passes her affected X chromosome to about half of her sons, who show the disease.
b) Haemophilia and colour blindness follow this pattern.

9. A patient with fever, chest pain, cough and breathing difficulty undergoes a chest X-ray. The film shows patchy opacities and reduced clarity of lung fields. Which observation best supports pneumonia? [1 Mark]
A. Enlarged heart size
B. Inflammation or fluid buildup in the lungs
C. Fractured ribs
D. Air pockets in the lungs

Answer: B. Inflammation or fluid buildup in the lungs

Teacher's Note:
a) In pneumonia, the alveoli get filled with fluid, which shows as patchy opacities on the X-ray.
b) Link the symptoms (fever, cough, breathing difficulty) with infection of the lungs.

10. A stretch of an euchromatin has 200 nucleosomes, how many base pairs (bp) will be there in the stretch and what would be the length of the typical euchromatin? [1 Mark]
A. 20,000 bp and 13,000x10^-9 m
B. 10,000 bp and 10,000x10^-9 m
C. 40,000 bp and 13,600x10^-9 m
D. 40,000 bp and 13,900x10^-9 m

Answer: C. 40,000 bp and 13,600x10^-9 m

Teacher's Note:
a) A typical nucleosome has 200 bp of DNA, so \( 200 \times 200 = 40,000 \) bp.
b) Distance between two base pairs is 0.34 nm, so length \( = 40,000 \times 0.34 = 13,600 \) nm \( = 13,600 \times 10^{-9} \) m.

11. A biotechnology company grew Streptomyces in a fermenter to produce an antibiotic. It was observed that Day 7 onwards, the actual yield drops sharply, even though nutrient level, temperature, and aeration remain unchanged. Also, Microscopy revealed new colonies of bacteria which were not present in the original culture. So the fermenter was stopped. What measures should be taken to prevent the fall in yield in future fermentations? [1 Mark]
A. Increasing temperature only at the end of fermentation to kill contaminants.
B. Adding antibiotics to the culture so unwanted microbes cannot grow.
C. Ensuring sterile handling, sterilised equipment, and aseptic transfer before and during fermentation.
D. Using a genetically modified strain designed to grow faster than contaminants.

Answer: C. Ensuring sterile handling, sterilised equipment, and aseptic transfer before and during fermentation.

Teacher's Note:
a) The new colonies show contamination, and contaminants compete with Streptomyces and lower the yield.
b) Sterile (aseptic) conditions must be kept throughout, which is a key feature of a bioreactor.

12. A rice farmer's field, which typically thrives in waterlogged conditions and requires soil rich in nutrients like phosphorus, is experiencing a severe drought this year. The crop is also susceptible to root infestation by pathogens. Which of the following techniques is best to address all these issues? [1 Mark]
A. Adding nitrogen-fixing bacteria to the soil
B. Applying chemical fertilizers
C. Adding spores of Glomus to the soil
D. Flooding the field artificially to mimic normal conditions

Answer: C. Adding spores of Glomus to the soil

Teacher's Note:
a) Glomus forms mycorrhiza, which absorbs phosphorus from soil and passes it to the plant.
b) Mycorrhizal plants also show tolerance to drought and resistance to root-borne pathogens, so all three problems are solved.

Question No. 13 to 16 consist of two statements - Assertion (A) and Reason (R). Answer these questions 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): Spermatogenesis is an ongoing process throughout the reproductive life of a human male, while oogenesis is a discontinuous process.
Reason (R): Oogenesis begins during the embryonic stage, gets arrested at prophase-I, resumes after puberty, and is arrested again at metaphase-II. [1 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) Oogenesis starts before birth and stops at two points, which makes it discontinuous.
b) Spermatogenesis starts at puberty and continues without such breaks.

14. Assertion (A): There is no variation in the percentage recombination due to loosely or tightly linked genes.
Reason (R): Linked genes result in occurrence of higher proportion of parental gene combinations. [1 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: D. A is False but R is true.

Teacher's Note:
a) Tightly linked genes show very low recombination and loosely linked genes show higher recombination, so A is false.
b) Linked genes tend to stay together, so parental combinations are more frequent; R is true.

15. Assertion (A): The primary goal of sewage treatment is to eliminate all organic matter from the effluent.
Reason (R): High BOD indicates a low level of organic matter in the water, which can lead to increased oxygen consumption by microorganisms, potentially harming aquatic life. [1 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) High BOD means a high amount of organic matter in water, not a low amount, so R is false.
b) Sewage treatment aims to lower the organic matter (BOD) of the effluent before it is released into water bodies.

16. Assertion (A): Tissue culture is used when traditional breeding technique is insufficient to keep pace with the demand of crop improvement.
Reason (R): Totipotency is the ability of a single plant cell or explant to regenerate into a whole plant under sterile conditions in nutrient media. [1 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 correct facts about tissue culture.
b) R only defines totipotency; it does not explain why tissue culture is needed when traditional breeding falls short.

Section - B

17. Attempt either option A or B.

A. Geitonogamy can be compared with cross-pollination as well as self-pollination. Comment. [2 Marks]

Answer:
1. Geitonogamy is the transfer of pollen grains from the anther of one flower to the stigma of another flower of the same plant.
2. Functionally it is like cross-pollination, because it needs a pollinating agent. Genetically it is like self-pollination (autogamy), because the pollen grains come from the same plant.

Teacher's Note:
a) Write both sides clearly: "functionally cross-pollination" and "genetically autogamy".
b) Start with the definition of geitonogamy, as it carries part of the marks.

OR

B. Mohit wants to produce hybrid seeds in a plant in his field in which male and female flowers mature at the same time and are close together. Suggest one strategy to promote only desired cross pollination and justify your answer. [2 Marks]

Answer:
1. Strategy: Mohit can remove the anthers (emasculation) from the selected flowers before the pollen is released. He should then pollinate them by hand with pollen from the chosen male parent and cover them with bags (bagging).
2. Justification: This prevents self-pollination and entry of unwanted pollen, so only the desired cross produces seeds.

Teacher's Note:
a) Use the keywords emasculation, bagging and artificial (hand) pollination.
b) Emasculation must be done before the anthers dehisce, otherwise self-pollination may already have happened.

18. The movement of landmasses over geological time changed the distribution and evolution of animals. Explain this idea using two examples from different regions of the world. [2 Marks]

Answer:
1. South America: It had mammals resembling horse, hippopotamus, bear and rabbit. Due to continental drift, when South America joined North America, these animals were overridden by North American fauna.
2. Australia: Due to continental drift, many different Australian marsupials evolved from one ancestral stock within the island continent. These pouched mammals survived because there was no competition from other mammals.

Teacher's Note:
a) Mention "continental drift" in both examples, as it is the key idea.
b) The Australian marsupials are also an example of adaptive radiation.

19. The following is a realistic example dataset and not a real world patient dataset:
Sample ID | Tumor size (mm³) | Glucose level mM/litre | Oxygen level (mmHg) | Neighboring cell's viability/ vitality (%)
1 | 150 | 2.0 | 15 | 60
2 | 50 | 4.5 | 30 | 85
3 | 200 | 1.3 | 10 | 50
4 | 100 | 3.0 | 20 | 70
5 | 250 | 0.9 | 7 | 40
A. Interpret the relationship between tumor size and glucose level.
B. Evaluate the data to determine how tumor progression may affect the viability/vitality of neighboring cells. Provide a brief explanation. [2 Marks]

Answer:
A. As tumor size increases, the glucose level decreases. Larger tumors use more glucose because of their higher metabolic demand, which may show a more aggressive growth.
B. As tumor size increases, the viability of neighbouring cells decreases. The growing tumor competes for nutrients and less oxygen is available, so it creates a hostile environment that stresses or kills the surrounding normal cells.

Teacher's Note:
a) Quote the data to support your point, for example tumor size 250 mm³ has glucose 0.9 mM/litre and viability 40%.
b) Say the direction of the relationship clearly: one value rises while the other falls.

20. Attempt either option A or B.

A. The table below shows hypothetical data comparing α-lactalbumin concentration in three types of milk, including that of Rosie, a transgenic cow designed to express human milk proteins.
Source of Milk | α-lactalbumin Content approximately (g/L)
Normal cow's milk | 0.1
Rosie cow's milk | 2.4
Human milk | 3.5
I. Assuming all other nutritional components are identical, which milk would provide the lowest nutritional contribution from α-lactalbumin?.
II. If further genetic modification increases α-lactalbumin levels in Rosie to 3.5 g/L, predict one ethical concern that might arise from such a development. [2 Marks]

Answer:
I. Normal cow's milk (about 0.1 g/L) gives the lowest contribution, because its α-lactalbumin content is very low compared to human milk.
II. Ethical concern: Going beyond morality, genetic modification of organisms can have long-term, unpredictable results.

Teacher's Note:
a) Other accepted concerns are patenting of the resource, loss of genetic diversity of natural cows and economic exploitation; write any one.
b) In part I, give the reason with the value from the table.

OR

B. A farmer replaces the GM crop with a conventional variety. Predict any two challenges she/he may face. Support your answer with two reasons. [2 Marks]

Answer:
1. More pesticide will be needed, because the conventional variety lacks pest resistance.
2. Yield will be lower under abiotic stress conditions such as drought or cold, because the conventional crop is not made tolerant to such stress.

Teacher's Note:
a) Another accepted challenge is faster depletion of soil nutrients due to less efficient mineral usage.
b) Each challenge must have a reason, since the question asks you to support your answer.

21. Attempt either option A or B.

A. The table below is a hypothetical data on Seasonal Temperatures seen over two different zones
Region | Temp. in Spring (°C) | Temp. in Summer (°C) | Temp. in Autumn (°C) | Temp. in Winter (°C) | Primary Productivity (GPP) (g/m²/year) | Respiration (R) (g/m²/year)
Region X | 27 | 28 | 27 | 26 | 3500 | 1300
Region Y | 12 | 20 | 14 | 4 | 2000 | 800
I. Justify which region will have greater species diversity based on the given data?
II. Calculate the net primary productivity (NPP) of the region mentioned in (I) and give its significance? [2 Marks]

Answer:
I. Region X will have greater species diversity (like a tropical region). Its temperature is nearly constant across seasons (27, 28, 27, 26 °C), which gives stable conditions, and its productivity is higher (GPP 3500) than that of Region Y.
II. NPP of Region X = GPP - R = 3500 - 1300 = 2200 g/m²/year.
Significance: NPP is the biomass available for consumption by heterotrophs (herbivores and decomposers).

Teacher's Note:
a) Always write the formula NPP = GPP - R before substituting values.
b) Do not forget the unit g/m²/year in the final answer.

OR

B. Following is a hypothetical data on Plant Pollinator interaction
Plant | Main Pollinator | Seed set (%) with pollinator | Seed set (%) without pollinator
P1 | H1 (Bee) | 80 | 40
P2 | H2 (Butterfly) | 75 | 45
P3 | H3 (Moth) | 85 | 35
I. From the data above, justify which plant shows the highest dependence on its main pollinator?
II. What kind of ecological interaction is shown in the table above? Define it in one sentence. [2 Marks]

Answer:
I. P3 shows the highest dependence. Its seed set drops by 50 percentage points (85% to 35%) without the pollinator, while the other plants show a smaller drop (40 for P1 and 30 for P2).
II. Mutualism: an interaction in which both species benefit; here the plant gets pollination and the pollinator gets nectar/food.

Teacher's Note:
a) Compare the drop in seed set for all three plants to justify your choice.
b) Do not confuse mutualism (both gain) with commensalism (one gains, the other is unaffected).

Section - C

22. A couple, A and B, visit a fertility clinic. Examination reveals that female A has a blockage in her fallopian tubes, but her ovaries can produce viable ova. Male B produces sperm with a very low sperm count, making natural fertilization unlikely.
Which two different Assisted Reproductive Technologies (ARTs) would a doctor most likely recommend for this couple to increase their chances of conceiving? Justify your selection by briefly stating the principle behind each chosen technique concerning the couple's specific issues. [3 Marks]

Answer:
1. ART 1 - Intracytoplasmic Sperm Injection (ICSI): It addresses Male B's low sperm count. In this procedure a sperm is directly injected into the ovum to form the embryo in vitro.
2. ART 2 - In vitro fertilisation followed by Intrauterine Transfer (IUT): It addresses Female A's blocked fallopian tubes.
3. Since the natural path of the egg is blocked, the embryo formed in the laboratory (by IVF/ICSI) is transferred directly into the uterus.

Teacher's Note:
a) Link each ART to one problem: ICSI for low sperm count, IVF with IUT for blocked tubes.
b) Write the full forms of ICSI and IUT at least once.

23. In the post-fertilization embryo sac shown in figure, structure 'P' results from the fusion of one male gamete with the egg cell, and structure 'Q' results from the fusion of the other male gamete with the polar nuclei.
A. Identify structure 'P' and structure 'Q' and state their ploidy.
B. Evaluate the following statement: "The division of 'P' always precedes the division of 'Q'." Justify your evaluation based on the nutritional requirements for seed development.
C. Which characteristic feature of endosperm helps in its nutritive role? [3 Marks]

[Figure: Fertilised embryo sac in a flowering plant. Labels: Degenerating synergids and P near the upper (narrow) end, Q as a large nucleus in the centre, and Degenerating antipodals at the lower end.]

Answer:
A. P is the zygote, which is diploid (2n). Q is the Primary Endosperm Nucleus (PEN), which is triploid (3n).
B. The statement is false. The PEN ('Q') divides much earlier than the zygote ('P') to form the endosperm tissue, which provides nutrition to the embryo.
C. The cells of the endosperm are filled with reserve food materials, which act as the main source of nutrition for the developing embryo.

Teacher's Note:
a) Syngamy forms the zygote and triple fusion forms the PEN; both together are called double fertilisation.
b) Remember the order: endosperm develops before the embryo, so the embryo gets food when it starts growing.

For Visually Impaired Candidates (in lieu of Q. 23)

A. Which two products are formed after double fertilization in the embryo sac of a flowering plant?
B. State the ploidy of these two structures.
C. Which of these two structures develops first during embryonic development? State its significance. [3 Marks]

Answer:
A. Zygote (which forms the embryo) and Primary Endosperm Nucleus (which forms the endosperm).
B. The zygote/embryo is diploid (2n) and the Primary Endosperm Nucleus/endosperm is triploid (3n).
C. The Primary Endosperm Nucleus develops first; it divides much earlier than the zygote to form the endosperm. The endosperm cells are filled with reserve food, which is the main source of nutrition for the developing embryo.

Teacher's Note:
a) One male gamete fuses with the egg (syngamy) and the other fuses with the two polar nuclei (triple fusion).
b) In part C, state both the structure and its significance to get the full mark.

24. In mice, the traits for running ability and hair colour assort independently. Running (R) is dominant over walking in circles (r).
Black hair (B) is dominant over brown hair (b). A heterozygous running, brown-haired female mouse was artificially inseminated using sperm from a heterozygous running, heterozygous black male.

A. Write the genotypes of both parents and draw a Punnett square showing all possible offspring combinations. [1 Mark]

Answer:
Female: heterozygous running (Rr) and brown hair (bb), so genotype = Rrbb. Gametes: Rb, Rb, rb, rb.
Male: heterozygous running (Rr) and heterozygous black (Bb), so genotype = RrBb. Gametes: RB, Rb, rB, rb.
Punnett square (female gametes in rows, male gametes in columns):
Female / Male | RB | Rb | rB | rb
Rb | RRBb | RRbb | RrBb | Rrbb
Rb | RRBb | RRbb | RrBb | Rrbb
rb | RrBb | Rrbb | rrBb | rrbb
rb | RrBb | Rrbb | rrBb | rrbb

Teacher's Note:
a) Write the gametes of both parents before drawing the square.
b) The female can form only two kinds of gametes (Rb and rb) because she is homozygous bb.

B. Using the Punnett square, state the phenotypic ratio for: Running-Black, Running-Brown, Walking-Black, Walking-Brown. [1 Mark]

Answer:
Running-Black (RRBb x 2, RrBb x 4) = 6/16
Running-Brown (RRbb x 2, Rrbb x 4) = 6/16
Walking-Black (rrBb x 2) = 2/16
Walking-Brown (rrbb x 2) = 2/16
Ratio = 6 : 6 : 2 : 2 or 3 : 3 : 1 : 1

Teacher's Note:
a) Check that the four fractions add up to 16/16.
b) Quick check: Rr x Rr gives 3 running : 1 walking and bb x Bb gives 1 black : 1 brown, so the ratio is 3 : 3 : 1 : 1.

C. If 64 offspring were produced, calculate how many would be expected to show the Running-Black phenotype. [1 Mark]

Answer: Running-Black = \( \frac{6}{16} \times 64 = 24 \). So 24 offspring will show the Running-Black phenotype.

Teacher's Note:
a) Multiply the fraction of the phenotype by the total number of offspring.
b) Show the working line; the final number alone may not get full credit.

25. How can a single gene defect caused by the mutation result in Phenylketonuria (PKU)? Explain the molecular mechanism underlying pleiotropy. [3 Marks]

Answer:
1. PKU is an inborn error of metabolism caused by mutation in a single gene. The affected person lacks the enzyme that converts the amino acid phenylalanine into tyrosine.
2. So phenylalanine accumulates and is converted into phenylpyruvic acid and other derivatives. Their accumulation in the brain results in mental retardation.
3. Pleiotropy: a single gene shows multiple phenotypic expressions because its product affects metabolic pathways in different ways.

Teacher's Note:
a) Name the enzyme's job clearly: phenylalanine to tyrosine.
b) PKU itself is the textbook example of pleiotropy, so link the two parts of your answer.

26. Drug-use disorders often begin with occasional consumption but may progress into a compulsive state.
Analyse how addiction differs from dependence in both origin and consequence, and justify why long-term drug intake forces the user to increase dose over time. [3 Marks]

Answer:
1. Addiction: It is a psychological attachment to the effects of drugs/alcohol. It is driven by the desire for euphoria and a temporary feeling of well-being, even when the drug is not needed.
2. Dependence: It is the physical tendency of the body to show withdrawal symptoms (anxiety, shakiness, nausea, sweating) when the regular dose of the drug is stopped.
3. Increase in dose: Repeated use of the drug raises the tolerance level of the body's receptors, so the receptors respond only to higher doses to give the same effect.

Teacher's Note:
a) Addiction is psychological and dependence is physical; this contrast is the key point.
b) Use the keyword "tolerance" to explain the increasing dose.

27. Pathogens are often considered harmful, yet they have been adapted as powerful tools in genetic engineering. Explain how pathogens have been modified into useful vectors. Describe the role of a similarly modified plant vector and animal vector. [3 Marks]

Answer:
1. Some pathogens naturally transfer their DNA into host cells. Scientists removed their disease-causing genes but kept their ability to deliver DNA, which made them useful cloning vectors.
2. Animal vector: Retroviruses have been disarmed by removing their cancer-causing genes, while keeping their ability to integrate DNA into the host genome. So they are used to deliver desired genes into animal cells.
3. Plant vector: Using the Ti plasmid of Agrobacterium tumefaciens, nematode-specific genes were introduced into plants. These produced dsRNA, which triggered RNA interference (RNAi) and silenced the nematode mRNA, protecting the plant from the parasite.

Teacher's Note:
a) The word "disarmed" (pathogenic genes removed) is the keyword for all such vectors.
b) Name the exact organism: Agrobacterium tumefaciens (Ti plasmid) for plants and retroviruses for animals.

28. The nematode Myrmeconema neotropicum lives inside the abdomen of certain tropical ants (Cephalotes atratus). Inside the ant's body, the nematode develops slowly without killing it. Once mature, it changes the color of the ant's abdomen to bright red, making it resemble a berry. Birds, mistaking it for a fruit, eat the ant. The nematode's eggs survive in the bird's digestive system and pass out in droppings, which other ants contact while foraging, starting the cycle again.

A. Why would harming the ant too early be disadvantageous for the nematode? [1 Mark]

Answer: If the ant dies too soon, the nematode will not complete its development and will not be transmitted to new hosts.

Teacher's Note:
a) The nematode depends on the living ant until it is mature.
b) Mention both points: incomplete development and no transmission.

B. How is natural selection playing a vital role in maintaining this interaction? [1 Mark]

Answer: Natural selection favours nematodes that keep the ant alive until it is eaten by a bird, which ensures better dispersal and survival of the nematode. Birds eat only the red-bellied ants, so the nematode matures fully and disperses, and the other ants are protected from predation.

Teacher's Note:
a) Link natural selection to survival and dispersal of the nematode.
b) Keep the answer to the case given; do not add new organisms.

C. What kind of ecological interaction is seen between nematode and ant? Define it in one sentence. [1 Mark]

Answer: Parasitism - an interaction in which one organism benefits and the other is harmed.

Teacher's Note:
a) Here the nematode benefits and the ant is harmed.
b) Write both the name of the interaction and its one-sentence definition.

Section - D

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

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

Answer: Anaemia, due to excessive blood loss.

Teacher's Note:
a) Fatigue and weakness in the case are clues that point to anaemia.
b) Write the cause (excessive blood loss) along with the name of the condition.

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

Answer:
1. Without a distinct LH surge, ovulation may not occur, so no corpus luteum forms to secrete progesterone. This explains the irregular cycles.
2. The endometrial lining then grows excessively under unopposed estrogen and is shed unpredictably and heavily, causing heavy periods.

Teacher's Note:
a) Follow the chain: no LH surge - no ovulation - no corpus luteum - low progesterone.
b) Explain both parts of the question: irregularity and heaviness.

Attempt either subpart C or D.

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

Answer: It suggests a lack of ovulation due to a hormonal imbalance affecting ovarian function.

Teacher's Note:
a) A normal cycle is about 28 days; wide variation points to irregular or absent ovulation.
b) Use the keywords "lack of ovulation" and "hormonal imbalance".

OR

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

Answer: GnRH secretion from the hypothalamus and the resulting LH/FSH secretion from the pituitary would be disrupted.

Teacher's Note:
a) GnRH from the hypothalamus controls the release of LH and FSH from the anterior pituitary.
b) Name both levels: hypothalamus (GnRH) and pituitary (LH and FSH).

30. A tribal farmer is cultivating Brassica oleracea (cabbage) in a forest region rich in pollinators and rare butterflies. Recently, his crops have been heavily damaged by cabbage looper caterpillars. To manage the pest outbreak, he applies Nucleopolyhedrovirus (NPV). After a few weeks, the pest population decreases significantly without affecting other insects or animals in the area.

A. Justify with reasons why the use of Nucleopolyhedrovirus (NPV) was a suitable choice for controlling cabbage looper caterpillars. [1 Mark]

Answer: NPV is species-specific and a narrow spectrum insecticide. It affects only the target insect pests and causes no harm to plants, pollinators, mammals, birds, fish or non-target insects.

Teacher's Note:
a) The region is rich in pollinators and rare butterflies, so a species-specific agent is needed.
b) Use the keywords "species-specific" and "narrow spectrum".

B. Suggest another IPM strategy suitable for controlling caterpillar pests, and briefly describe its mode of action. [2 Marks]

Answer:
1. Strategy: Spraying spores of the bacterium Bacillus thuringiensis (Bt) on the crop.
2. Mode of action: When the caterpillars eat the spores, the bacterial toxin is released in the gut of the larvae and the larvae get killed.

Teacher's Note:
a) Bt is the standard textbook example of a microbial biocontrol agent for caterpillars.
b) One mark is for the strategy and one for explaining how it works.

Attempt either subpart C or D.

C. Name the group of pathogens to which Nucleopolyhedrovirus belong. [1 Mark]

Answer: Baculovirus (baculoviruses).

Teacher's Note:
a) Nucleopolyhedroviruses belong to the genus Nucleopolyhedrovirus of the baculoviruses.
b) Spell the name correctly, as it is a one-word answer.

OR

D. Justify why the use of NPV is considered a biocontrol method. [1 Mark]

Answer: Because it uses a biological agent (a virus) to control plant diseases and pests, instead of chemicals.

Teacher's Note:
a) Biocontrol means using living organisms or their products to control pests.
b) Contrast it with chemical pesticides to make the justification clear.

Section - E

31. Attempt either option A or B.

A. A short stretch of DNA strand that codes for a polypeptide is shown below:
3'-TACAAAAAGAAGAAAAAGAAGATT--5'
Due to exposure to harmful UV radiation two different types of errors occurred during DNA replication affecting their sequences of bases in the DNA in two different cells.
I. What is the term given to a segment of DNA that codes for a polypeptide?
II. Which type of mutation could have occurred in each of the following due to the error during replication?
a) 3'-- TACAAAAAGAAGAAAAAGAAAATT--5'
b) 3'-- TACAAAAAGAAGAAAAAGAGATT--5'
III. If DNA strand (a) codes for Methionine and sequences of Phenylalanine, mention the sequence of amino acids that will constitute the polypeptide chain after translation.
How many amino acids will be translated from this strand?
IV. If all the codons of the stretch code for Phenylalanine excluding the initiator codon which feature does it depict? [5 Marks]

Answer:
I. Cistron.
II. (a) 3'-- TACAAAAAGAAGAAAAAGAAAATT--5' - Point mutation (one base, G, is replaced by A).
(b) 3'-- TACAAAAAGAAGAAAAAGAGATT--5' - Deletion / Frameshift mutation (one base is lost, so the reading frame shifts).
III. mRNA of strand (a): AUG UUU UUC UUC UUU UUC UUU UAA.
Polypeptide: Methionine - Phenylalanine - Phenylalanine - Phenylalanine - Phenylalanine - Phenylalanine - Phenylalanine.
Number of amino acids translated = 7 (UAA is a stop codon).
IV. Degenerate (degeneracy of the genetic code), since more than one codon (UUU and UUC) codes for the same amino acid.

Teacher's Note:
a) The template strand is read 3' to 5', so write the mRNA 5' to 3' with complementary bases (A pairs with U in RNA).
b) Substitution of one base is a point mutation, while loss or gain of a base causes a frameshift.
c) Do not count the stop codon as an amino acid.

OR

B. I. If GUA is an intron how many amino acids will be translated from the mature or processed RNA. Pick out the untranslated region from the following messenger RNA and mention their location and sequence. What is its significance?
5'ACGUCGAUGACCGUAGCGUUUGUAUCUUUAGUAGUGGUAUUAGUAGGCUAAAAA3'
II. If AAA is the anticodon of phenylalanine, draw the tRNA adaptor molecule, Mention the position of amino acid Phenylalanine in the sequence of polypeptide counting the triplet codons. [5 Marks]

Answer:
I. Coding region from AUG, with the GUA introns: AUGACCGUAGCGUUUGUAUCUUUAGUAGUGGUAUUAGUAGGC
After removing all GUA introns, the processed RNA reads AUG ACC GCG UUU UCU UUA GUG UUA GGC, followed by the stop codon UAA.
So nine amino acids will be translated from the mature or processed RNA.
Untranslated regions (UTRs): one is at the 5' end with the sequence 5'ACGUCG, and the other is at the 3' end with the sequence UAAAAA3'.
Significance: UTRs are required for an efficient translation process.
II. tRNA adaptor molecule (described in words): the tRNA is drawn as a folded (clover-leaf / inverted L) molecule with a 5' end and a 3' end. Phenylalanine (Phe) is attached at the 3' end (amino acid acceptor end). The anticodon loop carries the anticodon AAA, which pairs with the codon UUU on the mRNA.
The 4th triplet codon (UUU) of the processed RNA codes for phenylalanine, so Phe is the 4th amino acid of the polypeptide.

Teacher's Note:
a) Start reading codons only from the start codon AUG; the bases before it form the 5' UTR.
b) The 3' UTR strictly lies after the stop codon UAA; the scheme writes the 3' untranslated sequence as UAAAAA.
c) In the tRNA drawing, label the anticodon (AAA), the codon (UUU), the amino acid (Phe) and the 3' end.

32. Attempt either option A or B.

A. A scientist is screening bacterial transformants engineered with DNA fragments of the β-globin gene. A radioactive DNA probe complementary to the normal β-globin sequence is used. After autoradiography:
Plates A and C show strong and week radioactive signals respectively.
Plate B shows no radioactivity.
Answer the following:
I. Why does plate B fail to show radioactivity, while A and C do? [2 Marks]
II. How can this experiment help in distinguishing a normal individual, a carrier, and a patient of sickle cell anaemia? [2 Marks]
III. Name one other commonly used technique (apart from autoradiography) to differentiate recombinants from non-recombinants. [1 Mark]

Answer:
I. The probe could not hybridise with the DNA on plate B, which shows that plate B has the mutated β-globin gene (sickle mutation). Plates A and C show signals because their DNA has the normal sequence complementary to the probe.
II. Normal individual: the probe binds fully, so strong radioactivity is seen.
Carrier (heterozygote): both normal and mutated sequences are present, so there is partial hybridisation (mixed pattern, weaker signal).
Patient (homozygous mutant): no hybridisation with the probe, so no radioactivity is seen.
III. Insertional inactivation.

Teacher's Note:
a) A probe gives a signal only when it finds a complementary sequence to hybridise with.
b) Match the three plates to the three conditions: strong (normal), weak (carrier), none (patient).
c) In insertional inactivation, the insert inactivates a marker gene, for example the β-galactosidase gene giving blue and white colonies.

OR

B. A student placed DNA from three different sources: bacteria, plant tissue, and fungal cells in a test tube A, B and C respectively. During the process of isolating and purifying DNA, the student added cellulase to all test tubes.
I. Predict and explain in which trial(s) DNA isolation would fail, and why? Also suggest how it could be corrected. [2 Marks]
II. After rectifying the mistake made in 'a', outline the steps to obtain visible spooled DNA from the sample. [2 Marks]
III. Suggest one application of the spooled DNA in biotechnology. [1 Mark]

Answer:
I. Test tube A (bacteria): isolation will fail, because the peptidoglycan cell wall of bacteria needs lysozyme, not cellulase.
Test tube B (plant): isolation will succeed, because cellulase digests the cellulose wall of plant cells.
Test tube C (fungus): isolation will fail, because the chitin wall of fungi needs chitinase.
Correction: use lysozyme for bacteria, cellulase for plants and chitinase for fungi.
II. 1. Break open the cells; DNA is released along with RNA, proteins, polysaccharides and lipids.
2. Remove RNA by treating with ribonuclease and remove proteins by treating with protease.
3. Remove polysaccharides and lipids by specific treatments.
4. Add chilled ethanol; the purified DNA precipitates as fine white threads, which are spooled out on a glass rod.
III. Spooled DNA can be used in genetic engineering (for example PCR amplification, DNA fingerprinting or sequencing).

Teacher's Note:
a) Match each enzyme to its cell wall: lysozyme - bacteria, cellulase - plant, chitinase - fungus.
b) Chilled ethanol is the step that makes DNA visible; do not miss it.
c) For part III, one clear application is enough.

33. Attempt either option A or B.

A. A Green Valley X, home to rare medicinal plants and several migratory bird species, is facing multiple environmental challenges.
I. A luxury brand of cosmetics with little regulation has been sourcing wild medicinal plants from the valley.
II. The company has introduced an ornamental grass species from another country, which has taken over large areas of native grassland.
III. There is also a reported drastic decline in the endemic trees.
IV. Large scale forest fire in a part of the valley.
a) From the above hypothetical case, identify and explain the four major biodiversity threats. [4 Marks]
b) Suggest one in-situ conservation practice that could help protect the biodiversity of Green Valley X, and justify your choice. [1 Mark]

Answer:
a) 1. Over-exploitation: unsustainable sourcing of wild medicinal plants by the cosmetic brand may lead to depletion and extinction of valuable plant species.
2. Invasion by alien species: the ornamental grass from another country spreads aggressively, outcompetes native grasses and reduces local biodiversity.
3. Co-extinction: disappearance of the (nesting) endemic trees causes a sharp decline in the migratory bird population that depends on them, and disturbs the ecological balance through loss of niches.
4. Habitat loss and fragmentation: forest fire destroys vegetation and breaks continuous habitats into small, isolated patches. Species are forced to migrate, population sizes fall, and local extinction may occur.
b) Declare Green Valley X a protected area, such as a wildlife sanctuary, national park or biosphere reserve. Justification: it keeps species in their natural habitat, preserves ecological interactions and gives long-term protection to genetic, species and ecosystem diversity.

Teacher's Note:
a) These four threats match the "Evil Quartet": habitat loss and fragmentation, over-exploitation, alien species invasions and co-extinctions.
b) Link each threat to the correct line of the case (I to IV) for full marks.
c) In-situ means conservation in the natural habitat; zoos and seed banks are ex-situ.

OR

B. At a composting ground, after a community clean-up drive, large amounts of vegetable peels, garden leaves, and other organic waste are piled up. Over the next few weeks, the volume of the waste decreases noticeably.
I. Explain the sequence of processes that could have led to this transformation, and name the major organisms involved at each step. [4 Marks]
II. If a prolonged dry spell occurs during this composting period, explain how it could affect the rate of the above processes. [1 Mark]

Answer:
I. 1. Fragmentation: detritivores such as earthworms and termites break down the waste into smaller particles.
2. Leaching: soluble inorganic nutrients dissolve in water and percolate down into the composting soil.
3. Catabolism: enzymes of bacteria and fungi break down the complex organic matter into simpler inorganic substances.
4. Humification and mineralisation: microbes form humus and release inorganic nutrients into the compost.
II. A dry spell lowers the moisture content, which slows down microbial activity and enzymatic breakdown. So the rate of decomposition falls and the reduction in waste volume is delayed.

Teacher's Note:
a) Write the steps in the correct order: fragmentation, leaching, catabolism, humification and mineralisation.
b) Decomposition is faster when the conditions are warm and moist, and slower when they are dry.