CBSE Class 12 Biology: sample paper 2026-27

Section A

MCQ 1-12, assertion-reason 13-16

1. Q1 · 1 mark · MCQCompetency-basedSexual Reproduction in Flowering Plants

   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. (A)It guarantees self-pollination
   2. (B)It promotes autogamy
   3. (C)It ensures cross-pollination
   4. (D)It leads to geitonogamy

   Try it first, then show our solutionSolution

   1. When the pistil matures before the stamens (protogyny, a form of dichogamy), the stigma is receptive while the anthers of the same plant have not yet released pollen.

   2. The stigma can therefore only receive pollen from another plant, so self-pollination is prevented and cross-pollination is ensured. Option (C).

      1 mark

   Answer: (C) It ensures cross-pollination

   Where the marks are

   • Identifies that the time gap between pistil and stamen maturity prevents selfing and ensures cross-pollination (1 mark)

   Open Q1 on its own page

2. Q2 · 1 mark · MCQHuman Reproduction

   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. (A)Blockage of sperm transport into the ejaculatory duct and urethra.
   2. (B)Inhibition of testosterone production by the testes.
   3. (C)Suppression of the formation of seminal plasma by accessory glands.
   4. (D)Prevention of spermatogenesis in the seminiferous tubules.

   Try it first, then show our solutionSolution

   1. The procedure described is vasectomy, a surgical method of male contraception.

   2. Cutting and tying the vas deferens blocks the passage of sperms from the epididymis to the ejaculatory duct and urethra. Testosterone production, spermatogenesis and seminal plasma secretion continue normally. Option (A).

      1 mark

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

   Where the marks are

   • Recognises vasectomy blocks sperm transport only (1 mark)

   Open Q2 on its own page

3. Q3 · 1 mark · MCQHuman Reproduction

   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. (A)It is responsible for forming the primary germ layers (ectoderm, mesoderm, endoderm).
   2. (B)It provides essential nutrients to developing embryo and helps in its implantation.
   3. (C)It secretes colostrum which contains antibodies.
   4. (D)It gives rise to all the organs of the future embryo (organogenesis).

   Try it first, then show our solutionSolution

   1. The trophoblast layer of the blastocyst attaches to the endometrium, which brings about implantation.

   2. The trophoblast also forms chorionic villi that help nourish the embryo. The germ layers and organs arise from the inner cell mass, not from the trophoblast. Option (B).

      1 mark

   Answer: (B) It provides nutrients to the developing embryo and helps in implantation

   Where the marks are

   • Identifies the role of the trophoblast in implantation and nourishment (1 mark)

   Open Q3 on its own page

4. Q4 · 1 mark · MCQCompetency-basedEvolution

   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. (A)Stabilising selection as giraffes with longer neck lengths are selected further.
   2. (B)Disruptive selection as giraffes with smaller and longer neck lengths are selected.
   3. (C)Directional selection as giraffes with longer neck lengths are selected.
   4. (D)Stabilising selection as giraffes with medium neck lengths are selected.

   Figure in the paper: Bell-shaped curve graph: x-axis 'Neck length in giraffe', y-axis 'Number of giraffes'; the curve peaks at medium neck length and falls at both extremes. See it in the official paper.

   Try it first, then show our solutionSolution

   1. In giraffes, individuals with longer necks could reach higher foliage, so they survived and reproduced more.

   2. Over generations the mean neck length shifted towards one extreme, the longer side. This is directional selection. Option (C).

      1 mark

   Answer: (C) Directional selection

   Where the marks are

   • Identifies directional selection favouring longer necks (1 mark)

   Open Q4 on its own page

5. Q5 · 1 mark · MCQMolecular Basis of Inheritance

   The initiation step during the process of transcription in bacteria is shown below. Identify A, B, C and D by selecting the option:

   1. (A)A-RNA polymerase, B-Promoter, C-Sigma factor, D-DNA helix
   2. (B)A-Promoter, B-Sigma factor, C-RNA polymerase, D-DNA helix
   3. (C)A-Promoter, B-RNA polymerase, C-DNA helix, D-Sigma factor
   4. (D)A-Promoter, B-RNA polymerase, C-Sigma factor, D-DNA helix

   Figure in the paper: DNA double helix with 3' and 5' ends labelled on both strands; an oval structure (B) bound near the left end with a smaller circle marked sigma (C) on it; A points to the left DNA region where the strands are separated (yellow); D points to the DNA helix at the right. See it in the official paper.

   Try it first, then show our solutionSolution

   1. Transcription initiation in bacteria works as follows: RNA polymerase (B), with its sigma factor (C) attached, binds at the promoter (A), where the strands open. The rest of the molecule is the DNA helix (D).

   2. This gives A – Promoter, B – RNA polymerase, C – Sigma factor, D – DNA helix. Option (D).

      1 mark

   Answer: (D)

   Where the marks are

   • Correctly labels the promoter, RNA polymerase, sigma factor and DNA helix (1 mark)

   Open Q5 on its own page

6. Q6 · 1 mark · MCQMolecular Basis of Inheritance

   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. (A)A
   2. (B)B
   3. (C)C
   4. (D)D

   Figure in the paper: Flowchart: heat-killed S bacteria, bacterial filtrate treated with four enzymes, then mixed with live R bacteria: A RNase → No transformation; B Protease → Transformation; C DNase → No transformation; D Lipase → Transformation. See it in the official paper.

   Try it first, then show our solutionSolution

   1. Only DNase destroyed the transforming principle. RNase, protease and lipase did not stop transformation.

   2. So the RNase step should show 'Transformation'. Step A, which shows 'No transformation' after RNase, is therefore incorrect. Option (A).

      1 mark

   Answer: (A)

   Where the marks are

   • Identifies that RNase treatment still allows transformation (1 mark)

   Open Q6 on its own page

7. Q7 · 1 mark · MCQCompetency-basedMolecular Basis of Inheritance

   A scientist was culturing $E.coli$ in a ${}^{15}\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$ medium initially, then he shifted $E.coli$ to a ${}^{14}\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$ medium. The DNA was extracted from $E.coli$ after 60 minutes of transfer to ${}^{14}\mathrm{N}{\mathrm{H}}_4\mathrm{Cl}$ 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. (A)25% hybrid and 75% light
   2. (B)50% hybrid and 50% light
   3. (C)75% hybrid and 25% light
   4. (D)85% hybrid and 15% light

   Try it first, then show our solutionSolution

   1. E. coli divides every 20 min, so 60 min means 3 generations, giving $2^3=8$ DNA molecules.

   2. Replication is semiconservative, so only 2 of these 8 molecules still carry an old ${}^{15}N$ strand (hybrid). Hybrid DNA is $2/8=25\%$ and light DNA is $6/8=75\%$. Option (A).

      1 mark

   Answer: (A) 25% hybrid, 75% light

   Where the marks are

   • Works out 3 generations and applies semiconservative replication to get 25% hybrid and 75% light (1 mark)

   Open Q7 on its own page

8. Q8 · 1 mark · MCQCompetency-basedPrinciples of Inheritance and Variation

   Study the given pedigree and identify the disease represented

   1. (A)Myotonic Dystrophy
   2. (B)Sickle cell anaemia
   3. (C)Haemophilia
   4. (D)Colour blindness

   Figure in the paper: Pedigree chart: normal parents (male square, female circle) have five children: normal male, affected male, normal male, normal female (carrier-like, unshaded), affected female (shaded circle), normal male. The normal male and normal female in the second generation... the unshaded female (marries a normal male) has an affected son (shaded square) and two normal daughters. See it in the official paper.

   Try it first, then show our solutionSolution

   1. Unaffected parents have both an affected son and an affected daughter, so the trait is recessive and not sex-linked.

   2. An autosomal recessive disease is sickle cell anaemia. Myotonic dystrophy is dominant, while haemophilia and colour blindness are X-linked. Option (B).

      1 mark

   Answer: (B) Sickle cell anaemia

   Where the marks are

   • Deduces autosomal recessive inheritance and identifies sickle cell anaemia (1 mark)

   Open Q8 on its own page

9. Q9 · 1 mark · MCQCompetency-basedHuman Health and Disease

   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. (A)Enlarged heart size
   2. (B)Inflammation or fluid buildup in the lungs
   3. (C)Fractured ribs
   4. (D)Air pockets in the lungs

   Try it first, then show our solutionSolution

   1. In pneumonia the alveoli become filled with fluid and inflammatory cells.

   2. This fluid shows up as patchy opacities on the X-ray. Option (B).

      1 mark

   Answer: (B) Inflammation or fluid buildup in the lungs

   Where the marks are

   • Links the X-ray opacities to fluid-filled alveoli in pneumonia (1 mark)

10. Q10 · 1 mark · MCQMolecular Basis of Inheritance

    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. (A)20,000 bp and 13,000x10${}^{-9}$ m
    2. (B)10,000 bp and 10,000x10${}^{-9}$ m
    3. (C)40,000 bp and 13,600x10${}^{-9}$ m
    4. (D)40,000 bp and 13,900x10${}^{-9}$ m

    Try it first, then show our solutionSolution

    1. A typical nucleosome has about 200 bp, so 200 nucleosomes contain $200\times 200=40,000$ bp.

    2. Length $=40,000\times 0.34\times 10^{-9}$ m $=13,600\times 10^{-9}$ m. Option (C).

       1 mark

    Answer: (C) 40,000 bp and 13,600×10⁻⁹ m

    Where the marks are

    • Calculates the base pairs and the length correctly (1 mark)

    Open Q10 on its own page

11. Q11 · 1 mark · MCQCompetency-basedBiotechnology: Principles and Processes

    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. (A)Increasing temperature only at the end of fermentation to kill contaminants.
    2. (B)Adding antibiotics to the culture so unwanted microbes cannot grow.
    3. (C)Ensuring sterile handling, sterilised equipment, and aseptic transfer before and during fermentation.
    4. (D)Using a genetically modified strain designed to grow faster than contaminants.

    Try it first, then show our solutionSolution

    1. The fall in yield was caused by contaminating bacteria, not by any change in growth conditions.

    2. Contamination is best prevented by aseptic technique: sterilised equipment and media, and sterile transfer before and during fermentation. Option (C).

       1 mark

    Answer: (C)

    Where the marks are

    • Identifies aseptic technique as the way to prevent contamination (1 mark)

    Open Q11 on its own page

12. Q12 · 1 mark · MCQCompetency-basedMicrobes in Human Welfare

    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. (A)Adding nitrogen-fixing bacteria to the soil
    2. (B)Applying chemical fertilizers
    3. (C)Adding spores of Glomus to the soil
    4. (D)Flooding the field artificially to mimic normal conditions

    Try it first, then show our solutionSolution

    1. Glomus is a mycorrhizal fungus.

    2. It absorbs phosphorus for the plant and also gives tolerance to drought and resistance to root pathogens, so it addresses all three problems. Option (C).

       1 mark

    Answer: (C) Adding spores of Glomus

    Where the marks are

    • Recognises that mycorrhiza provides phosphorus uptake, drought tolerance and pathogen resistance (1 mark)

    Open Q12 on its own page

13. Q13 · 1 mark · Assertion-reasonCompetency-basedHuman Reproduction

    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. (A)Both A and R are true and R is the correct explanation of A.
    2. (B)Both A and R are true but R is not the correct explanation of A.
    3. (C)A is true but R is false.
    4. (D)A is False but R is true.

    Try it first, then show our solutionSolution

    1. Spermatogenesis continues from puberty throughout life, so A is true.

    2. Oogenesis starts in the foetus, is arrested at prophase I, resumes at puberty, and is arrested again at metaphase II until fertilisation. R is true and explains why oogenesis is discontinuous. Option (A).

       1 mark

    Answer: (A)

    Where the marks are

    • Recognises both statements are true and R explains A (1 mark)

    Open Q13 on its own page

14. Q14 · 1 mark · Assertion-reasonCompetency-basedPrinciples of Inheritance and Variation

    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. (A)Both A and R are true and R is the correct explanation of A.
    2. (B)Both A and R are true but R is not the correct explanation of A.
    3. (C)A is true but R is false.
    4. (D)A is False but R is true.

    Try it first, then show our solutionSolution

    1. Tightly linked genes show low recombination and loosely linked genes show higher recombination, so A is false.

    2. Linked genes do produce mostly parental combinations, so R is true. Option (D).

       1 mark

    Answer: (D)

    Where the marks are

    • Judges A false and R true (1 mark)

    Open Q14 on its own page

15. Q15 · 1 mark · Assertion-reasonCompetency-basedMicrobes in Human Welfare

    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. (A)Both A and R are true and R is the correct explanation of A.
    2. (B)Both A and R are true but R is not the correct explanation of A.
    3. (C)A is true but R is false.
    4. (D)A is False but R is true.

    Try it first, then show our solutionSolution

    1. Sewage treatment aims to remove organic matter and lower the BOD of the effluent.

    2. R is false, because a high BOD indicates a high level of organic matter, not a low one. Option (C).

       1 mark

    Answer: (C)

    Where the marks are

    • Identifies that R wrongly links high BOD with low organic matter (1 mark)

    Open Q15 on its own page

16. Q16 · 1 mark · Assertion-reasonCompetency-basedBiotechnology: Principles and Processes

    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. (A)Both A and R are true and R is the correct explanation of A.
    2. (B)Both A and R are true but R is not the correct explanation of A.
    3. (C)A is true but R is false.
    4. (D)A is False but R is true.

    Try it first, then show our solutionSolution

    1. Tissue culture (micropropagation) is used because conventional breeding cannot keep pace with food demand, so A is true.

    2. R correctly defines totipotency, but it does not explain why tissue culture is needed. Option (B).

       1 mark

    Answer: (B)

    Where the marks are

    • Recognises both statements are true but R does not explain A (1 mark)

    Open Q16 on its own page

Section B

1. Answer one of these 2 alternatives.

   Q17(A) · 2 marks · Very short answerSexual Reproduction in Flowering Plants

   Geitonogamy can be compared with cross-pollination as well as self-pollination. Comment.

   Try it first, then show our solutionSolution

   1. Geitonogamy is the transfer of pollen grains from the anther of one flower to the stigma of another flower on the same plant.

      1 mark

   2. Like cross-pollination (functionally): the pollen moves between two different flowers, so it needs a pollinating agent (wind, insects and so on). Like self-pollination/autogamy (genetically): the pollen comes from the same plant, so the gametes are genetically identical and no new genetic variation results.

      1 mark

   Answer: Geitonogamy is functionally cross-pollination (it needs a pollinating agent between two flowers) but genetically similar to autogamy (the pollen comes from the same plant).

   Where the marks are

   • Defines geitonogamy as pollen transfer between different flowers of the same plant (1 mark)
   • States it is functionally like cross-pollination (needs an agent) but genetically like autogamy (same plant) (1 mark)

   Open Q17(A) on its own page

   OR

   Q17(B) · 2 marks · Very short answerCompetency-basedSexual Reproduction in Flowering Plants

   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.

   Try it first, then show our solutionSolution

   1. Strategy (emasculation and bagging): Mohit should remove the anthers from the flowers chosen as the female parent before they shed pollen. He should then cover these flowers with butter-paper bags. When the stigma becomes receptive, he should dust it with pollen from the chosen male parent and re-bag the flower.

      1 mark

   2. Justification: Emasculation prevents self-pollination, and bagging keeps out unwanted foreign pollen. As a result, only the desired cross fertilises the ovules, giving true hybrid seeds.

      1 mark

   Answer: Emasculate the selected flowers, bag them, and pollinate them by hand with pollen from the chosen male parent. This prevents self-pollination and contamination, so only the desired cross forms seeds.

   Where the marks are

   • Suggests emasculation followed by bagging and artificial pollination with the desired pollen (1 mark)
   • Justifies: prevents self/unwanted pollination, ensuring only the desired hybrid seeds form (1 mark)

   Open Q17(B) on its own page

2. Q18 · 2 marks · Very short answerEvolution

   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.

   Try it first, then show our solutionSolution

   1. South America: South America once had native mammals resembling the horse, hippopotamus, bear and rabbit. When continental drift joined South America to North America, these animals were overridden and replaced by the North American fauna.

      1 mark

   2. Australia: Australia stayed isolated after continental drift. Many different marsupials evolved from a common ancestral stock within this island continent (adaptive radiation). They survived because there was no competition from other (placental) mammals.

      1 mark

   Answer: (1) Joining of South and North America let North American fauna override South American mammals. (2) Isolation of Australia allowed marsupials to radiate from one ancestor without competition from other mammals.

   Where the marks are

   • South American mammals overridden by North American fauna after the continents joined (1 mark)
   • Australian marsupials radiated from a common ancestor in isolation, with no competition (1 mark)

   Open Q18 on its own page

3. Q19 · 2 marks · Very short answerCompetency-basedHuman Health and Disease

   The following is a realistic example dataset and not a real world patient dataset:

   Sample ID	Tumor size (mm${}^3$)	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.

   Try it first, then show our solutionSolution

   1. A. Tumour size and glucose level are inversely related: as tumour size rises from 50 to 250 mm³, glucose falls from 4.5 to 0.9 mM/L. Larger tumours consume more glucose because their rapidly dividing cells have a high metabolic demand, which suggests more aggressive growth.

      1 mark

   2. B. As the tumour grows, the viability of neighbouring cells falls from 85% to 40%, and oxygen falls from 30 to 7 mmHg. The growing tumour competes for nutrients and oxygen. This creates a hostile microenvironment that starves and stresses the surrounding normal cells, leading to their death.

      1 mark

   Answer: A. Inverse relation: bigger tumours use more glucose. B. Tumour growth lowers neighbouring cell viability (85%→40%) because it competes for nutrients and oxygen.

   Where the marks are

   • Inverse relation between tumour size and glucose, explained by higher metabolic demand (1 mark)
   • Viability of neighbouring cells decreases due to nutrient/oxygen competition (1 mark)

   Open Q19 on its own page

4. Answer one of these 2 alternatives.

   Q20(A) · 2 marks · Very short answerCompetency-basedBiotechnology and its Applications

   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.

   Try it first, then show our solutionSolution

   1. I. Normal cow's milk (about 0.1 g/L) gives the lowest nutritional contribution from α-lactalbumin. Its content is far lower than that of Rosie's milk (2.4 g/L) and human milk (3.5 g/L).

      1 mark

   2. II. One ethical concern: further genetic modification of animals may have long-term unpredictable effects on the animal's health and welfare and on consumers. Other acceptable concerns are loss of genetic diversity in native cows, patenting of the modified organism, and economic exploitation.

      1 mark

   Answer: I. Normal cow's milk. II. Unpredictable long-term effects of genetic modification on animal welfare and health (or loss of genetic diversity, patenting or exploitation).

   Where the marks are

   • Identifies normal cow's milk, giving the reason of lowest α-lactalbumin (1 mark)
   • Any one valid ethical concern (1 mark)

   Open Q20(A) on its own page

   OR

   Q20(B) · 2 marks · Very short answerCompetency-basedBiotechnology and its Applications

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

   Try it first, then show our solutionSolution

   1. Challenge 1: more pesticide use. The conventional variety lacks built-in pest resistance (for example, the Bt toxin). The farmer will therefore need more chemical pesticides, which raises costs and pollution.

      1 mark

   2. Challenge 2: lower yield or faster soil depletion. The conventional crop is less tolerant of abiotic stresses such as drought, cold, salt and heat, so it gives lower yields under stress. It also uses minerals less efficiently, so it may deplete soil fertility faster.

      1 mark

   Answer: (1) More pesticide use, because there is no pest resistance. (2) Lower yield under abiotic stress (or faster soil nutrient depletion), because there is no stress tolerance or efficient mineral use.

   Where the marks are

   • First challenge with reason, e.g. more pesticide due to lack of pest resistance (1 mark)
   • Second challenge with reason, e.g. lower yield under stress or soil nutrient depletion (1 mark)

   Open Q20(B) on its own page

5. Answer one of these 2 alternatives.

   Q21(A) · 2 marks · Very short answerCompetency-basedEcosystem

   The table below is a hypothetical data on Seasonal Temperatures seen over two different zones

   Region	Temp. in Spring (${}^{\circ }$C)	Temp. in Summer (${}^{\circ }$C)	Temp. in Autumn (${}^{\circ }$C)	Temp. in Winter (${}^{\circ }$C)	Primary Productivity (GPP) (g/m${}^2$/year)	Respiration (R) (g/m${}^2$/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?

   Try it first, then show our solutionSolution

   1. I. Region X (a tropical region) will have greater species diversity. Its temperature is nearly constant through the year (26–28 °C), giving stable, predictable conditions that favour niche specialisation. Its productivity is also higher (GPP 3500 vs 2000 g/m²/year).

      1 mark

   2. II. NPP = GPP − R = 3500 − 1300 = 2200 g/m²/year.
      Significance: NPP is the biomass available for consumption by heterotrophs (herbivores and decomposers).

      1 mark

   Answer: I. Region X. II. NPP = 2200 g/m²/year; it is the biomass available to heterotrophs.

   Where the marks are

   • Region X, justified by near-constant temperature or higher productivity (1 mark)
   • NPP = 3500 − 1300 = 2200 g/m²/year, with its significance (1 mark)

   Open Q21(A) on its own page

   OR

   Q21(B) · 2 marks · Very short answerCompetency-basedOrganisms and Populations

   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.

   Try it first, then show our solutionSolution

   1. I. P3 shows the highest dependence on its pollinator. Its seed set falls from 85% to 35%, a drop of 50 percentage points. The drops for P1 (40 points) and P2 (30 points) are smaller.

      1 mark

   2. II. The interaction is mutualism: an interaction in which both species benefit. Here the plant gets pollinated, and the pollinator obtains nectar or food.

      1 mark

   Answer: I. P3 (seed set drops by 50 percentage points). II. Mutualism: both species benefit.

   Where the marks are

   • P3, justified by the largest drop in seed set (1 mark)
   • Mutualism, with a correct one-line definition (1 mark)

   Open Q21(B) on its own page

Section C

1. Q22 · 3 marks · Short answerCompetency-basedReproductive Health

   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.

   Try it first, then show our solutionSolution

   1. ART 1: Intracytoplasmic Sperm Injection (ICSI). This addresses male B's very low sperm count. In ICSI, a single sperm is injected directly into the cytoplasm of the ovum in the laboratory to form the embryo in vitro. A large number of sperm is therefore not needed.

      1½ marks

   2. ART 2: IVF followed by embryo transfer into the uterus (IUT). This addresses female A's blocked fallopian tubes. Her ova are collected and fertilised outside the body (by IVF/ICSI). Because the tubes are blocked, the embryo (with more than 8 blastomeres) is transferred directly into the uterus, bypassing the tubes so that it can implant.

      1½ marks

   Answer: ICSI (for the low sperm count) and IVF with intrauterine transfer, IUT (for the blocked fallopian tubes).

   Where the marks are

   • Names ICSI, justified by direct injection of sperm into the ovum for the low sperm count (1.5 marks)
   • Names IVF–IUT, justified by transferring the embryo directly to the uterus, bypassing the blocked tubes (1.5 marks)

   Open Q22 on its own page

2. Q23 · 3 marks · Short answerCompetency-basedSexual Reproduction in Flowering Plants

   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?

   Figure in the paper: Figure: Fertilised embryo sac in a flowering plant. Elongated sac with degenerating synergids at the top, structure P (small nucleus) near the top, structure Q (large nucleus) at the centre, and degenerating antipodals at the bottom. See it in the official paper.

   Try it first, then show our solutionSolution

   1. A. P = zygote, which is diploid (2n). Q = primary endosperm nucleus (PEN), which is triploid (3n).

      1 mark

   2. B. The statement is false. The PEN (Q) divides before the zygote (P) to form the endosperm. The endosperm must be ready first because it provides nourishment to the developing embryo.

      1 mark

   3. C. The cells of the endosperm are filled with reserve food materials. These act as the main source of nutrition for the developing embryo.

      1 mark

   Answer: A. P = zygote (2n), Q = PEN (3n). B. False: the PEN divides before the zygote, to provide nutrition. C. Endosperm cells are filled with reserve food.

   Where the marks are

   • Identifies the zygote (2n) and PEN (3n) (1 mark)
   • Statement false: the PEN divides earlier to nourish the embryo (1 mark)
   • Endosperm cells are rich in reserve food materials (1 mark)

   Open Q23 on its own page

3. Q24 · 3 marks · Short answerCompetency-basedPrinciples of Inheritance and Variation

   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)
   B. Using the Punnett square, state the phenotypic ratio for: Running–Black, Running–Brown, Walking–Black, Walking–Brown. (1)
   C. If 64 offspring were produced, calculate how many would be expected to show the Running–Black phenotype. (1)

   Try it first, then show our solutionSolution

   1. A. Female (heterozygous running, brown) = Rrbb, with gametes Rb and rb. Male (heterozygous running, heterozygous black) = RrBb, with gametes RB, Rb, rB and rb.

      ♀ \ ♂	RB	Rb	rB	rb
      Rb	RRBb	RRbb	RrBb	Rrbb
      rb	RrBb	Rrbb	rrBb	rrbb

      1 mark

   2. B. Out of 8 boxes:

      • Running–Black (RRBb, RrBb ×2) = 3/8
      • Running–Brown (RRbb, Rrbb ×2) = 3/8
      • Walking–Black (rrBb) = 1/8
      • Walking–Brown (rrbb) = 1/8

      Phenotypic ratio = 3 : 3 : 1 : 1 (written as 6 : 6 : 2 : 2 on a 4×4 square).

      1 mark

   3. C. Running–Black offspring = $\frac{3}{8}\times 64=24$.

      1 mark

   Answer: Parents Rrbb × RrBb; ratio 3:3:1:1 (Running-Black : Running-Brown : Walking-Black : Walking-Brown); 24 Running–Black offspring.

   Where the marks are

   • Correct parental genotypes, gametes and Punnett square (1 mark)
   • Phenotypic ratio 3:3:1:1 (1 mark)
   • 64 × 3/8 = 24 (1 mark)

   Open Q24 on its own page

4. Q25 · 3 marks · Short answerPrinciples of Inheritance and Variation

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

   Try it first, then show our solutionSolution

   1. PKU is an autosomal recessive inborn error of metabolism caused by a mutation in a single gene. The affected person lacks the enzyme (phenylalanine hydroxylase) that converts the amino acid phenylalanine into tyrosine.

      1 mark

   2. As a result, phenylalanine accumulates and is converted into phenylpyruvic acid and other derivatives. These build up in the brain, causing mental retardation, and are also excreted in the urine because of poor absorption by the kidneys.

      1 mark

   3. Molecular basis of pleiotropy: a single gene product (an enzyme) acts in a metabolic pathway. When it is defective, many dependent reactions are affected, so one gene shows multiple phenotypic effects. PKU itself is an example, causing mental retardation and reduced hair and skin pigmentation.

      1 mark

   Answer: A mutant gene fails to make the enzyme that converts phenylalanine to tyrosine. Phenylalanine and phenylpyruvic acid accumulate in the brain and cause mental retardation. Pleiotropy arises because one gene product affects metabolic pathways in different ways, giving multiple phenotypic effects.

   Where the marks are

   • Mutation causes lack of the enzyme converting phenylalanine to tyrosine (1 mark)
   • Accumulation of phenylalanine/phenylpyruvic acid in the brain causes mental retardation (1 mark)
   • Pleiotropy: a single gene affects metabolic pathways, giving multiple phenotypic expressions (1 mark)

   Open Q25 on its own page

5. Q26 · 3 marks · Short answerCompetency-basedHuman Health and Disease

   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.

   Try it first, then show our solutionSolution

   1. Addiction: its origin is psychological. It is an attachment to certain effects of the drug, such as euphoria and a temporary feeling of well-being. Its consequence is that the person keeps taking the drug even when it is not needed, which leads to compulsive use.

      1 mark

   2. Dependence: its origin is physical. The body comes to rely on the drug. Its consequence is that withdrawal symptoms (anxiety, shakiness, nausea, sweating) appear when regular intake is abruptly stopped.

      1 mark

   3. Why the dose increases: repeated use raises the tolerance level of the body's receptors. The receptors then respond only to higher doses to give the same effect, so the user is forced to increase the dose over time.

      1 mark

   Answer: Addiction is a psychological attachment to euphoria; dependence is physical, shown by withdrawal symptoms. Increased tolerance of the receptors makes higher doses necessary.

   Where the marks are

   • Addiction as a psychological attachment to euphoric effects (1 mark)
   • Dependence as physical, shown by withdrawal symptoms on stopping (1 mark)
   • Increased receptor tolerance means higher doses are needed for the same effect (1 mark)

   Open Q26 on its own page

6. Q27 · 3 marks · Short answerBiotechnology: Principles and Processes

   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.

   Try it first, then show our solutionSolution

   1. Modification of pathogens: some pathogens naturally transfer their DNA into host cells. Scientists remove their disease-causing genes but keep this ability to deliver DNA, so the disarmed pathogen becomes a useful vector.

      1 mark

   2. Plant vector: the Ti plasmid of Agrobacterium tumefaciens normally causes tumours in dicots. It has been disarmed into a cloning vector that is no longer pathogenic but still delivers desired genes into plants. For example, nematode-specific genes introduced through it produce dsRNA. This triggers RNA interference, which silences the mRNA of Meloidogyne incognita and protects tobacco plants.

      1 mark

   3. Animal vector: retroviruses normally transform normal cells into cancerous cells. They have been engineered by removing the cancer-causing genes while keeping their ability to integrate DNA into the host genome. This makes them effective vectors for delivering genes into animal cells.

      1 mark

   Answer: Pathogens are disarmed by removing their disease genes but keeping DNA transfer. Plant vector: the Ti plasmid of Agrobacterium (e.g. for RNAi-based nematode resistance). Animal vector: disarmed retroviruses.

   Where the marks are

   • Pathogens disarmed: disease genes removed, DNA-transfer ability retained (1 mark)
   • Ti plasmid of Agrobacterium as a plant vector, with its role (1 mark)
   • Disarmed retroviruses as animal vectors (1 mark)

   Open Q27 on its own page

7. Q28 · 3 marks · Short answerCompetency-basedOrganisms and Populations

   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)
   B. How is natural selection playing a vital role in maintaining this interaction? (1)
   C. What kind of ecological interaction is seen between nematode and ant? Define it in one sentence. (1)

   Try it first, then show our solutionSolution

   1. A. If the ant dies too early, the nematode cannot complete its development or reach maturity. The ant's abdomen will also not be eaten by a bird, so the nematode's eggs will not be dispersed to new hosts.

      1 mark

   2. B. Natural selection favours nematodes that keep the ant alive until it is eaten by a bird. This ensures complete maturation and wide dispersal through bird droppings, so such nematodes leave more offspring. Only the red-bellied infected ants are eaten, which also spares the uninfected ants.

      1 mark

   3. C. The interaction is parasitism: an interaction in which one organism (the parasite) benefits while the other (the host) is harmed.

      1 mark

   Answer: A. The nematode would not mature or be transmitted. B. Selection favours nematodes that keep the host alive until a bird eats it. C. Parasitism: one benefits, the other is harmed.

   Where the marks are

   • Early death prevents development and transmission (1 mark)
   • Selection favours nematodes that keep the ant alive until bird predation (1 mark)
   • Parasitism, correctly defined (1 mark)

   Open Q28 on its own page

Section D

case-based

1. Q29 · 4 marks · Case-basedCompetency-basedReproductive Health

   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.

   1. A.

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

      1 mark
   2. 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
   3. C.

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

      1 mark

   4. OR

      D.

      If a doctor suspects hypothalamic-pituitary dysfunction in Madhuri, which hormone secretion pattern would you expect to be disrupted?

      1 mark

   Try it first, then show our solutionSolution

   1. A. The likely complication is anaemia due to excessive blood loss, which causes fatigue and weakness.

      1 mark

   2. B. The LH surge triggers ovulation. Without it, ovulation fails, so no corpus luteum forms and progesterone stays low. This explains the irregular cycles.

      1 mark

   3. The endometrium keeps growing under unopposed estrogen without progesterone to stabilise it. It is then shed unpredictably and heavily, giving prolonged, heavy bleeding.

      1 mark

   4. C. The irregularity indicates anovulation (lack of regular ovulation) caused by hormonal imbalance affecting ovarian function.

      1 mark

   5. D. (Alternative) The pulsatile secretion of GnRH from the hypothalamus would be disrupted, and with it the secretion of LH and FSH from the anterior pituitary.

   Answer: A. Anaemia. B. No LH surge means no ovulation, no corpus luteum and low progesterone; unopposed estrogen thickens the endometrium, which is shed irregularly and heavily. C. Lack of ovulation (anovulation). D. GnRH, and in turn LH/FSH, secretion disrupted.

   Where the marks are

   • Anaemia from blood loss (1 mark)
   • No LH surge leads to no ovulation, no corpus luteum and low progesterone (1 mark)
   • Unopposed estrogen causes excess endometrial growth and irregular, heavy shedding (1 mark)
   • C: anovulation/ovarian dysfunction OR D: GnRH and LH/FSH disrupted (1 mark)

   Open Q29 on its own page

2. Q30 · 4 marks · Case-basedCompetency-basedBiotechnology and its Applications

   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.

   1. A.

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

      1 mark
   2. B.

      Suggest another IPM strategy suitable for controlling caterpillar pests, and briefly describe its mode of action.

      2 marks
   3. C.

      Name the group of pathogens to which Nucleopolyhedrovirus belong.

      1 mark

   4. OR

      D.

      Justify why the use of NPV is considered a biocontrol method.

      1 mark

   Try it first, then show our solutionSolution

   1. A. NPV is species-specific and narrow-spectrum. It kills only the target insect pest and does not harm plants, pollinators, rare butterflies, birds, mammals or other non-target insects. This makes it ideal for an ecologically sensitive forest area.

      1 mark

   2. B. Another IPM strategy is to spray spores of the bacterium Bacillus thuringiensis (Bt) on the crop. The caterpillars eat them along with the leaves.

      1 mark

   3. In the alkaline gut of the larvae, the inactive protoxin is converted to the active toxin. This damages the gut lining and kills the larvae.

      1 mark

   4. C. NPV belongs to the Baculoviruses (genus Nucleopolyhedrovirus).

      1 mark

   5. D. (Alternative) NPV is considered biocontrol because it uses a living natural enemy (a pathogen) of the pest, instead of chemicals, to control the pest population.

   Answer: A. Species-specific, narrow-spectrum, harmless to non-targets. B. Bt spores: the toxin is activated in the larval gut and kills the larvae. C. Baculoviruses. D. It uses a living organism, not chemicals, to control pests.

   Where the marks are

   • NPV is species-specific and narrow-spectrum, with no harm to non-target organisms (1 mark)
   • Names Bacillus thuringiensis spores as an IPM strategy (1 mark)
   • Toxin is released in the larval gut and kills the larvae (1 mark)
   • C: Baculovirus OR D: a biological agent is used instead of chemicals (1 mark)

   Open Q30 on its own page

Section E

1. Answer one of these 2 alternatives.

   Q31(A) · 5 marks · Long answerCompetency-basedMolecular Basis of Inheritance

   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?

   Try it first, then show our solutionSolution

   1. I. A segment of DNA that codes for a polypeptide is called a cistron.

      ½ mark

   2. II(a). Only one base has changed: the 7th codon AAG has become AAA (G replaced by A), and the length is unchanged. This is a point mutation (base substitution).

      1 mark

   3. II(b). The strand has one base fewer than the original (23 instead of 24), because a single base has been deleted. This is a deletion, which causes a frameshift mutation in the reading frame.

      1 mark

   4. III. Template (a): 3'-TAC AAA AAG AAG AAA AAG AAA ATT-5'.
      mRNA: 5'-AUG UUU UUC UUC UUU UUC UUU UAA-3'.
      Polypeptide: Met–Phe–Phe–Phe–Phe–Phe–Phe (UAA is a stop codon). 7 amino acids are translated.

      2 marks

   5. IV. Different codons (UUU and UUC) code for the same amino acid, phenylalanine. This depicts the degeneracy of the genetic code.

      ½ mark

   Answer: I. Cistron. II. (a) Point mutation (substitution); (b) deletion/frameshift mutation. III. Met–Phe–Phe–Phe–Phe–Phe–Phe; 7 amino acids. IV. Degeneracy of the code.

   Where the marks are

   • Cistron (0.5 mark)
   • (a) Point mutation (1 mark)
   • (b) Deletion/frameshift mutation (1 mark)
   • Met followed by 6 Phe, giving 7 amino acids, with UAA as stop (2 marks)
   • Degenerate code (0.5 mark)

   Open Q31(A) on its own page

   OR

   Q31(B) · 5 marks · Long answerCompetency-basedMolecular Basis of Inheritance

   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.

   Try it first, then show our solutionSolution

   1. I. The coding region runs from AUG to just before the stop codon: AUG ACC GUA GCG UUU GUA UCU UUA GUA GUG GUA UUA GUA GGC (14 codons). Removing the five GUA introns leaves the mature RNA AUG ACC GCG UUU UCU UUA GUG UUA GGC, so 9 amino acids will be translated.

      1 mark

   2. Untranslated regions (UTRs):

      • At the 5' end, before the start codon AUG: 5'-ACGUCG.
      • At the 3' end, from the stop codon onwards: UAAAAA-3'.
      2 marks

   3. Significance: the UTRs are not translated but are required for an efficient translation process.

      ½ mark

   4. II. The tRNA adaptor is drawn as a clover leaf. Its anticodon loop carries the anticodon 3'-AAA-5', which pairs with the mRNA codon 5'-UUU-3'. The amino acid acceptor end (3' CCA end) carries phenylalanine. The other loops/arms are also shown.

      1 mark

   5. In the mature mRNA (AUG, ACC, GCG, UUU, …), phenylalanine is coded by the 4th triplet codon.

      ½ mark

   Answer: I. 9 amino acids; UTRs: 5'-ACGUCG at the 5' end and UAAAAA-3' at the 3' end; needed for efficient translation. II. Clover-leaf tRNA with anticodon AAA carrying Phe at the 3' CCA end; Phe is at the 4th codon.

   Where the marks are

   • Correct count of 9 amino acids after removing the GUA introns (1 mark)
   • Locates and writes the 5' UTR (ACGUCG) and 3' UTR (UAAAAA) (2 marks)
   • UTRs are required for efficient translation (0.5 mark)
   • Labelled clover-leaf tRNA with anticodon AAA and Phe at the amino acid acceptor end (1 mark)
   • Phe at the 4th codon position (0.5 mark)

   Open Q31(B) on its own page

2. Answer one of these 2 alternatives.

   Q32(A) · 5 marks · Long answerCompetency-basedBiotechnology: Principles and Processes

   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:

   1. I.

      Why does plate B fail to show radioactivity, while A and C do?

      2 marks
   2. II.

      How can this experiment help in distinguishing a normal individual, a carrier, and a patient of sickle cell anaemia?

      2 marks
   3. III.

      Name one other commonly used technique (apart from autoradiography) to differentiate recombinants from non-recombinants.

      1 mark

   Try it first, then show our solutionSolution

   1. I. The radioactive probe is complementary to the normal β-globin sequence. In plate B, the DNA carries the mutated (sickle-cell) β-globin sequence, so the probe cannot hybridise and no radioactivity shows on autoradiography.

      1 mark

   2. Plates A and C contain the normal complementary sequence, so the probe hybridises and gives a signal. In A the signal is strong, with more normal sequence; in C it is weak, with less normal sequence.

      1 mark

   3. II. The strength of probe hybridisation tells the three genotypes apart:

      • Normal individual (HbAHbA): the probe binds fully, giving a strong radioactive signal.
      • Carrier (HbAHbS): both normal and mutant alleles are present, so hybridisation is partial and the signal is weaker or mixed.
      1 mark
   4. • Patient (HbSHbS): only the mutant sequence is present, so there is no hybridisation and no radioactivity.
      1 mark

   5. III. Insertional inactivation, for example blue–white screening using the lacZ (β-galactosidase) gene: recombinants give white colonies and non-recombinants give blue colonies.

      1 mark

   Answer: I. Plate B has the mutant (sickle) sequence, so the probe does not hybridise; A and C have the normal sequence. II. Strong signal = normal; weak/partial = carrier; none = patient. III. Insertional inactivation (blue–white screening).

   Where the marks are

   • Plate B carries the mutated sequence, so there is no probe hybridisation (1 mark)
   • A and C have the normal complementary sequence, so the probe binds and gives a signal (1 mark)
   • Normal = strong signal; carrier = partial/weak signal (1 mark)
   • Patient = no hybridisation and no signal (1 mark)
   • Insertional inactivation/blue–white selection (1 mark)

   Open Q32(A) on its own page

   OR

   Q32(B) · 5 marks · Long answerCompetency-basedBiotechnology: Principles and Processes

   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.

   1. I.

      Predict and explain in which trial(s) DNA isolation would fail, and why? Also suggest how it could be corrected.

      2 marks
   2. II.

      After rectifying the mistake made in 'a', outline the steps to obtain visible spooled DNA from the sample.

      2 marks
   3. III.

      Suggest one application of the spooled DNA in biotechnology.

      1 mark

   Try it first, then show our solutionSolution

   1. I. Prediction: Isolation will fail in test tube A (bacteria) and test tube C (fungus), and will succeed in test tube B (plant). Cellulase can only digest cellulose, which is found in plant cell walls. The bacterial wall is made of peptidoglycan and the fungal wall is made of chitin, so cellulase cannot break these cells open and their DNA is never released.

      1 mark

   2. Correction: Treat each sample with the enzyme that suits its wall: lysozyme for bacteria (tube A), cellulase for plant tissue (tube B) and chitinase for fungi (tube C).

      1 mark

   3. II. Steps to obtain spooled DNA:
      1. Once the wall is digested, the cells break open. DNA is released along with RNA, proteins, polysaccharides and lipids.
      2. Add ribonuclease to digest the RNA and protease to digest the proteins. Remove polysaccharides and lipids by suitable treatments.

      1 mark

   4. 3. Add chilled ethanol to the purified solution. The DNA precipitates as fine white threads, which can be wound (spooled) around a glass rod.

      1 mark

   5. III. Application: The spooled DNA can be cut with restriction enzymes and joined to a vector to make recombinant DNA, which is genetic engineering. It can also be used for PCR amplification, DNA fingerprinting or sequencing. (Any one.)

      1 mark

   Answer: I. Fails in A (bacteria, peptidoglycan wall) and C (fungus, chitin wall); succeeds in B (plant). Correct by using lysozyme for A and chitinase for C (cellulase for B). II. Lyse cells → RNase + protease treatment → remove polysaccharides/lipids → add chilled ethanol → spool white DNA threads. III. Recombinant DNA/genetic engineering (or PCR, DNA fingerprinting, sequencing).

   Where the marks are

   • Identifies failure in A and C and gives the reason based on wall composition (1 mark)
   • Suggests the correct enzymes: lysozyme, cellulase and chitinase (1 mark)
   • Cell lysis followed by RNase and protease treatment and removal of other contaminants (1 mark)
   • Precipitation with chilled ethanol and spooling of DNA threads (1 mark)
   • Any one valid biotechnological application (1 mark)

   Open Q32(B) on its own page

3. Answer one of these 2 alternatives.

   Q33(A) · 5 marks · Long answerCompetency-basedBiodiversity and Conservation

   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)
   b) Suggest one in-situ conservation practice that could help protect the biodiversity of Green Valley $X$, and justify your choice. (1)

   Try it first, then show our solutionSolution

   1. (a) I – Over-exploitation: The cosmetic brand is collecting wild medicinal plants unsustainably. Their populations fall faster than they can regenerate, which can push these rare species towards extinction.

      1 mark

   2. II – Invasion by alien species: The exotic ornamental grass spreads aggressively. It outcompetes native grasses and the species that depend on them, reducing local biodiversity. This is similar to Parthenium and Lantana in India.

      1 mark

   3. III – Co-extinction: When the endemic trees decline, the migratory birds that depend on them for nesting also decline sharply. Loss of one species therefore drags down an associated species and upsets the ecological balance.

      1 mark

   4. IV – Habitat loss and fragmentation: The forest fire destroys vegetation and breaks the continuous habitat into small, isolated patches. Species are forced to migrate, populations shrink, and local extinctions can follow.

      1 mark

   5. (b) In-situ practice: Declare Green Valley X a protected area, such as a national park, wildlife sanctuary or biosphere reserve. Justification: This protects the species in their natural habitat and keeps their ecological interactions intact, for example between trees and nesting birds. It also gives long-term protection to genetic, species and ecosystem diversity, and legally stops plant collection and habitat destruction.

      1 mark

   Answer: (a) I – Over-exploitation; II – Alien species invasion; III – Co-extinction; IV – Habitat loss and fragmentation. (b) Declare the valley a protected area (national park/wildlife sanctuary/biosphere reserve), which conserves species in their natural habitat.

   Where the marks are

   • Over-exploitation named and explained (1 mark)
   • Alien species invasion named and explained (1 mark)
   • Co-extinction named and explained (1 mark)
   • Habitat loss and fragmentation named and explained (1 mark)
   • Suggests an in-situ protected area with a justification (1 mark)

   Open Q33(A) on its own page

   OR

   Q33(B) · 5 marks · Long answerCompetency-basedEcosystem

   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)
   II. If a prolonged dry spell occurs during this composting period, explain how it could affect the rate of the above processes. (1)

   Try it first, then show our solutionSolution

   1. I. Fragmentation: Detritivores such as earthworms and termites break the vegetable peels and leaves into smaller particles. This increases the surface area available for microbes to act on.

      1 mark

   2. Leaching: Water-soluble inorganic nutrients dissolve and percolate down into the soil layers, where some are precipitated as unavailable salts.

      1 mark

   3. Catabolism: Bacteria and fungi release enzymes that break the complex organic matter down into simpler inorganic substances.

      1 mark

   4. Humification and mineralisation: Microbes convert the remains into dark, amorphous humus, which resists microbial action and decomposes very slowly. Further microbial degradation of humus releases inorganic nutrients (mineralisation). Together these processes reduce the waste volume.

      1 mark

   5. II. Effect of a dry spell: Low moisture slows the activity of microbes and detritivores and reduces their enzymatic breakdown. Decomposition therefore slows down, and the waste volume falls more slowly.

      1 mark

   Answer: I. Fragmentation (detritivores such as earthworms and termites) → Leaching (soluble nutrients percolate down) → Catabolism (enzymes of bacteria and fungi) → Humification and mineralisation (microbes form humus and release inorganic nutrients). II. A dry spell lowers moisture, reduces microbial activity and so slows decomposition.

   Where the marks are

   • Fragmentation by detritivores, with examples (1 mark)
   • Leaching of soluble nutrients (1 mark)
   • Catabolism by bacterial and fungal enzymes (1 mark)
   • Humification and mineralisation by microbes (1 mark)
   • Dry spell lowers moisture and so slows microbial decomposition (1 mark)

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