BiologyCBSE Class 12 · Sample paper, session 2024-25

CBSE Class 12 Biology: sample paper 2024-25

The sample question paper CBSE published for the 2024-25 session, as printed: 33 questions for 70 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

Every question here has our step-by-step solution, split into the steps that earn marks. Each one was checked by an independent re-solve from the question paper.

Source: official sample question paper from cbseacademic.nic.in: https://cbseacademic.nic.in/web_material/SQP/ClassXII_2024_25/Biology-SQP.pdf. Questions are transcribed from that PDF; figures are described in words, so check the official paper for them. Questions printed only for visually impaired candidates are left out (7).

Practise a paper like this, timedMy board paper sets a full paper in this exact format from real board questions, with a timer.Get your written answers markedThe Answer Coach marks a photo of your handwritten answer step by step, like the examiner.

How the paper is set

Section	Questions	What it has
A	1-16	MCQ 1-12, assertion-reason 13-16
B	17-21
C	22-28
D	29-30	case-based
E	31-33

Instructions in brief: All questions compulsory; no overall choice, internal choices in some questions; neat labelled diagrams wherever necessary.

Section A

MCQ 1-12, assertion-reason 13-16

Q1 · 1 mark · MCQHuman Reproduction
Signals for parturition in human female originate from
1. (A)Fully developed foetus only
2. (B)Both placenta as well as fully developed foetus
3. (C)Placenta only
4. (D)Oxytocin released from maternal pituitary
Try it first, then show our solutionSolution
1. At the end of gestation, the signals for parturition come from the fully developed foetus and the placenta; together they induce mild uterine contractions (the foetal ejection reflex), which then trigger oxytocin release from the maternal pituitary. Oxytocin is therefore a response to the signal, not its source. Correct option: (B).
  1 mark
Answer: (B) Both placenta as well as fully developed foetus
Where the marks are
- Identifies foetus and placenta as the source of the parturition signal (1 mark)
Q2 · 1 mark · MCQSexual Reproduction in Flowering Plants
To produce 1600 seeds, the number of meiotic divisions required will be
1. (A)2400
2. (B)2000
3. (C)1600
4. (D)1800
Try it first, then show our solutionSolution
1. Each seed needs 1 pollen grain and 1 egg. One meiosis of a microspore mother cell gives 4 pollen grains, so 1600 pollen grains need $1600/4 = 400$ meioses.
  ½ mark
2. One meiosis of a megaspore mother cell gives only 1 functional megaspore, hence 1 egg. So 1600 eggs need 1600 meioses. Total $= 400 + 1600 = 2000$ . Correct option: (B).
  ½ mark
Answer: 2000, option (B)
Where the marks are
- Calculates 400 meioses for the pollen grains and 1600 for the eggs, total 2000 (1 mark)
Open Q2 on its own page
Q3 · 1 mark · MCQMolecular Basis of Inheritance
A sample of normal double-stranded DNA was found to have thymine content of 27%. What will be the expected proportion of guanine in this strand?
1. (A)23%
2. (B)32%
3. (C)36%
4. (D)73%
Try it first, then show our solutionSolution
1. By Chargaff's rule, A = T = 27%, so A + T = 54%. This leaves G + C = 100 − 54 = 46%. Since G = C, G = 23%. Correct option: (A).
  1 mark
Answer: 23%, option (A)
Where the marks are
- Applies Chargaff's rule to obtain G = 23% (1 mark)
Q4 · 1 mark · MCQMolecular Basis of Inheritance
Observe the schematic diagram that depicts a small section of nucleic acid. The bases in two strands are paired through hydrogen bonds that are shown by the dark lines. Identify the correct sequence of nucleotide in the 5’-3’ direction.
1. (A)GCAT
2. (B)CGTA
3. (C)TAGC
4. (D)ATCG
Figure in the paper: Schematic of a short double-stranded nucleic acid section with four base pairs; sugar-phosphate backbones on both sides, bases joined by dark lines representing hydrogen bonds (triple lines for some pairs, double for others); OH marked at the free end of each strand (top right and bottom left); phosphate groups shown as green circles. See it in the official paper.
Try it first, then show our solutionSolution
1. The 3' end of each strand carries a free OH. Reading the strand from its phosphate (5') end towards its OH (3') end, and using the pairing rules (triple bonds for G≡C, double bonds for A=T), the sequence is 5'-CGTA-3'. Correct option: (B).
  1 mark
Answer: CGTA, option (B)
Where the marks are
- Reads the strand from the 5' end to the 3'-OH end to get CGTA (1 mark)
Q5 · 1 mark · MCQHuman Health and Disease
Suresh and Rajesh have defective haemoglobin due to genetic disorders. In Suresh, the problem is qualitative as he is having incorrectly functioning globin molecules while in Rajesh the problem is quantitative as he is having very few globin molecules. Identify the disorder they are suffering from.
1. (A)Suresh: Thalassemia - Autosomal Dominant blood disorder; Rajesh: Sickle Cell Anaemia - Autosomal linked Recessive trait
2. (B)Suresh: Sickle Cell Anaemia - Autosomal linked Dominant trait; Rajesh: Thalassemia - Autosomal Recessive blood disorder
3. (C)Suresh: Sickle Cell Anaemia – Autosomal linked Recessive trait; Rajesh: Thalassemia – Autosomal Recessive blood disorder
4. (D)Suresh: Thalassemia - Autosomal Dominant blood disorder; Rajesh: Sickle Cell Anaemia - Autosomal linked Dominant trait
Try it first, then show our solutionSolution
1. Sickle cell anaemia is a qualitative defect: a substitution of Glu by Val gives an abnormal globin. It is autosomal recessive, which fits Suresh. Thalassemia is a quantitative defect: too few globin chains are made. It is also autosomal recessive, which fits Rajesh. Correct option: (C).
  1 mark
Answer: Option (C)
Where the marks are
- Matches sickle cell anaemia (qualitative) to Suresh and thalassemia (quantitative) to Rajesh, both autosomal recessive (1 mark)
Q6 · 1 mark · MCQPrinciples of Inheritance and Variation
In E.coli, the lac operon gets switched on when lactose is
1. (A)present in the medium and it binds to the repressor.
2. (B)not present in the medium and the repressor binds to the operator.
3. (C)not present in the medium and RNA polymerase binds to the operator.
4. (D)Active lactose present in the medium binds to RNA polymerase.
Try it first, then show our solutionSolution
1. Lactose acts as the inducer. When present in the medium, it binds to the repressor and inactivates it. The repressor can then no longer bind the operator, so RNA polymerase transcribes the structural genes. Correct option: (A).
  1 mark
Answer: Option (A)
Where the marks are
- States that lactose binds and inactivates the repressor (1 mark)
Q7 · 1 mark · MCQPrinciples of Inheritance and Variation
Which of the following features shows the mechanism of sex determination in honey-bee?
(i) An offspring formed from the union of a sperm and egg develops as a female.
(ii) Males have half the number of chromosomes than that of female.
(iii) The males are haploid having 32 chromosomes.
(iv) All workers and males are diploid having 16 chromosomes
1. (A)(i) and (ii)
2. (B)(ii) and (iii)
3. (C)(i) and (iv)
4. (D)(ii) and (iv)
Try it first, then show our solutionSolution
1. Honey-bees show a haplodiploid system. A fertilised egg develops into a female (diploid, 32 chromosomes), while an unfertilised egg develops into a male (haploid, 16 chromosomes). So statements (i) and (ii) are true, while (iii) and (iv) give wrong chromosome numbers. Correct option: (A).
  1 mark
Answer: (i) and (ii), option (A)
Where the marks are
- Identifies statements (i) and (ii) as correct (1 mark)
Q8 · 1 mark · MCQMolecular Basis of Inheritance
The following diagram shows a fragment of DNA which is going to be transcribed, the upper strand with polarity 3’ to 5’ is the template strand:
3´ ATTGCC 5´
5´ TAACGG 3´
After transcription the mRNA can be represented by:
1. (A)5´ AUUGCC 3´
2. (B)5´ AUUGCC 3’
3. (C)5´ UAACGG 3´
4. (D)5´ GGCAAU 3´
Try it first, then show our solutionSolution
1. The mRNA is complementary to the template strand 3'-ATTGCC-5'. It therefore matches the coding strand 5'-TAACGG-3', with T replaced by U: 5'-UAACGG-3'. Correct option: (C).
  1 mark
Answer: 5' UAACGG 3', option (C)
Where the marks are
- Writes the mRNA as the coding strand with U in place of T (1 mark)
Q9 · 1 mark · MCQMicrobes in Human Welfare
Idli – dosa dough rises due to production of which of the following gas?
1. (A)CO
2. (B) $C O_{2}$
3. (C)NO
4. (D) $N O_{2}$
Try it first, then show our solutionSolution
1. Microbes in the dough, mainly bacteria, ferment it and release CO₂. The trapped gas makes the dough puff up. Correct option: (B).
  1 mark
Answer: CO₂, option (B)
Where the marks are
- Identifies CO₂ from fermentation (1 mark)
Q10 · 1 mark · MCQEvolution
Adaptive radiation leads to which of the following?
1. (A)Increased competition among species
2. (B)Decreased speciation rates
3. (C)Limited morphological diversity among species
4. (D)Rapid divergence of traits among populations inhabiting a given geographical area.
Try it first, then show our solutionSolution
1. Adaptive radiation is the evolution of different forms from a common ancestor, radiating into various habitats within one geographical area, for example Darwin's finches. This means rapid divergence of traits. Correct option: (D).
  1 mark
Answer: Option (D)
Where the marks are
- Recognises adaptive radiation as rapid divergence from a common ancestor in one area (1 mark)
Q11 · 1 mark · MCQBiotechnology: Principles and Processes
Eco R1 cuts the DNA between bases G and A only when the sequence of GAATTC is present. The number of nucleotides present in the resultant sticky ends that will be formed in each of the two strands of DNA after this enzyme cuts the DNA will be:
Vector DNA Foreign DNA
A. 1 & 5 5 & 1
B. 2 & 4 4 & 2
C. 2 & 5 5 & 2
D. 3 & 4 4 & 3
1. (A)Vector DNA 1 & 5; Foreign DNA 5 & 1
2. (B)Vector DNA 2 & 4; Foreign DNA 4 & 2
3. (C)Vector DNA 2 & 5; Foreign DNA 5 & 2
4. (D)Vector DNA 3 & 4; Foreign DNA 4 & 3
Try it first, then show our solutionSolution
1. EcoRI cuts the sequence G^AATTC on both strands. Each cut end therefore has fragments of 1 nucleotide (G) and 5 nucleotides (AATTC) in the two strands. The vector shows 1 & 5 and the complementary foreign DNA shows 5 & 1. Correct option: (A).
  1 mark
Answer: Option (A)
Where the marks are
- Counts 1 and 5 nucleotides on either side of the cut (1 mark)
Q12 · 1 mark · MCQMicrobes in Human Welfare
During the secondary treatment of sewage, which of the following change in the effluent occur due to flocs?
1. (A)Reduction in BOD
2. (B)Increase in BOD
3. (C)Decrease in DO
4. (D)No change in DO or BOD
Try it first, then show our solutionSolution
1. Flocs are masses of aerobic bacteria and fungal filaments. They consume most of the organic matter in the effluent, which significantly lowers its BOD. Correct option: (A).
  1 mark
Answer: Reduction in BOD, option (A)
Where the marks are
- States that flocs reduce BOD (1 mark)
Q13 · 1 mark · Assertion-reasonCompetency-basedSexual Reproduction in Flowering Plants
Assertion (A): Cells of tapetum have more than one nucleus.
Reason (R): They undergo meiosis without cytokinesis.
1. (A)Both A and R are true and R is the correct explanation of A.
2. (B)Both A and R are true and 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. Tapetal cells are often multinucleate, so the Assertion is true. However, this arises from nuclear division without cytokinesis in a mitotic process, not meiosis; tapetum does not undergo meiosis. The Reason is false. Correct option: (C).
  1 mark
Answer: Option (C)
Where the marks are
- States that A is true and R is false (1 mark)
Q14 · 1 mark · Assertion-reasonCompetency-basedMolecular Basis of Inheritance
Assertion (A): Deoxyribonucleoside triphosphates serve dual purposes.
Reason (R): They act as proof readers and provide energy.
1. (A)Both A and R are true and R is the correct explanation of A.
2. (B)Both A and R are true and 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. dNTPs serve two roles: they act as substrates for DNA synthesis and they provide energy for polymerisation. So the Assertion is true. They do not proofread; proofreading is done by DNA polymerase. The Reason is false. Correct option: (C).
  1 mark
Answer: Option (C)
Where the marks are
- States that A is true and R is false (1 mark)
Q15 · 1 mark · Assertion-reasonCompetency-basedMicrobes in Human Welfare
Assertion (A): A floating cover placed over the slurry in a biogas plant keeps on rising.
Reason (R): This cover keeps on rising due to the gas produced in the tank by the microbial activity.
1. (A)Both A and R are true and R is the correct explanation of A.
2. (B)Both A and R are true and 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. Methanogens in the slurry produce biogas. As the gas accumulates in the tank, it pushes the floating cover upward. Both statements are true, and R explains A. Correct option: (A).
  1 mark
Answer: Option (A)
Where the marks are
- States that both are true and R explains A (1 mark)
Q16 · 1 mark · Assertion-reasonCompetency-basedBiotechnology: Principles and Processes
Assertion (A): DNA fragments can be isolated by Gel electrophoresis on the basis of their size.
Reason (R): The larger the fragment size, the faster it moves.
1. (A)Both A and R are true and R is the correct explanation of A.
2. (B)Both A and R are true and 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. Gel electrophoresis separates DNA fragments by size, so the Assertion is true. The agarose matrix sieves the fragments, so smaller fragments move farther and larger fragments move more slowly. The Reason is false. Correct option: (C).
  1 mark
Answer: Option (C)
Where the marks are
- States that A is true and R is false (1 mark)

	Vector DNA	Foreign DNA
A.	1 & 5	5 & 1
B.	2 & 4	4 & 2
C.	2 & 5	5 & 2
D.	3 & 4	4 & 3

Section B

Answer one of these 2 alternatives.
Q17(A) · 2 marks · Very short answerCompetency-basedHuman Reproduction
(i) A blood test reported negative for hCG. What does negative hCG imply? Name the tissue which produces hCG?
(ii) If a blood test reported positive for hCG in a person, then which other hormones would also be secreted by the tissue secreting hCG?
Try it first, then show our solutionSolution
1. (i) hCG is secreted only once an embryo has implanted, so a negative hCG test means the person is not pregnant.
  ½ mark
2. (i) hCG (human chorionic gonadotropin) is produced by the placenta.
  ½ mark
3. (ii) A positive hCG test means a placenta is present. Besides hCG, the placenta also secretes human placental lactogen (hPL), estrogens, progestogens and relaxin. Any two of these are enough.
  1 mark
Answer: (i) Negative hCG means no pregnancy; hCG is produced by the placenta. (ii) hPL, estrogens, progestogens, relaxin (any two).
Where the marks are
- States that negative hCG means no pregnancy (0.5 mark)
- Names placenta as the tissue producing hCG (0.5 mark)
- Names any two other placental hormones: hPL, estrogens, progestogens, relaxin (0.5 × 2 = 1 mark)
Open Q17(A) on its own page
OR
Q17(B) · 2 marks · Very short answerHuman Reproduction
(i) The human male ejaculates about 200 to 300 million sperm during a coitus, however the ovum is fertilized by only one sperm. How does the ovum block the entry of additional sperms?
(ii) All copulations will not lead to fertilization. Why?
Try it first, then show our solutionSolution
1. (i) When the first sperm contacts the zona pellucida, it induces changes in that layer. The cortical reaction hardens the zona pellucida, and this blocks the entry of any additional sperms, so only one sperm fertilises the ovum.
  1 mark
2. (ii) Fertilisation happens only in the ampullary region of the fallopian tube. It succeeds only if the ovum and the sperms reach the ampulla at the same time, which happens when coitus occurs close to ovulation. Copulation at other times will not lead to fertilisation.
  1 mark
Answer: (i) Sperm contact changes the zona pellucida, which then blocks further sperm entry. (ii) Fertilisation needs the ovum and sperms to reach the ampullary region simultaneously.
Where the marks are
- Explains that sperm contact changes the zona pellucida and blocks entry of other sperms (1 mark)
- States that ovum and sperms must be transported simultaneously to the ampullary region (1 mark)
Open Q17(B) on its own page
Answer one of these 2 alternatives.
Q18(A) · 2 marks · Very short answerMolecular Basis of Inheritance
The schematic representation given below shows a DNA strand and two types of mutations in the DNA strand.
(i) Identify the type of mutation exhibited in I and II.
(ii) Which of the above mutation is more harmful? Give reason.
Figure in the paper: Three sequences. Original template: A U G C A G A C A U C U U A G with codons Met, Gln, Thr, Ser, Stop. Mutation I: A U G A A G A C A U C U U A G with Met, Lys, Thr, Ser, Stop. Mutation II: A U G A G A C A U C U U A G (one base deleted) with Met, Arg, His, Leu, and no stop shown. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) In I, a single base is substituted (C → A in the second codon, so CAG/Gln becomes AAG/Lys) and the rest is unchanged. This is a point mutation (base substitution).
  ½ mark
2. (i) In II, one base is deleted, so the reading frame shifts and every codon after it changes (Arg, His, Leu…). This is a frameshift mutation (deletion).
  ½ mark
3. (ii) Mutation II (frameshift) is more harmful, because it changes all the codons downstream of the deletion, not just one.
  ½ mark
4. (ii) Reason: many incorrect amino acids are inserted and the stop codon is lost or misread, so the polypeptide length and composition change. This gives a non-functional protein with severe phenotypic effects.
  ½ mark
Answer: (i) I – point (substitution) mutation; II – frameshift (deletion) mutation. (ii) II is more harmful, as it alters all downstream codons and gives a non-functional protein.
Where the marks are
- Identifies I as point mutation and II as frameshift mutation (0.5 + 0.5 = 1 mark)
- States II is more harmful (0.5 mark)
- Gives a valid reason, e.g. many codons altered or premature/absent stop giving a non-functional protein (0.5 mark)
Open Q18(A) on its own page
OR
Q18(B) · 2 marks · Very short answerCompetency-basedMolecular Basis of Inheritance
Given below is a schematic representation of a mRNA strand
5' AGGAGGUAUGAUCUCGUAAAAUAAA 3'
(i) In the above sequence identify the translational unit in mRNA.
(ii) Where are UTRs found and what is their significance?
Figure in the paper: A line labelled mRNA from 5' to 3' with the base sequence AGGAGGUAUGAUCUCGUAAAAUAAA written below. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The translational unit is the stretch from the start codon AUG to the stop codon UAA, which codes for a polypeptide:
  5′ AGGAGGU AUG AUC UCG UAA AAUAAA 3′
  Translational unit = AUG AUC UCG UAA.
  1 mark
2. (ii) UTRs (untranslated regions) are found at the 5′ end, before the start codon (AGGAGGU here), and at the 3′ end, after the stop codon (AAUAAA here).
  ½ mark
3. (ii) Significance: UTRs are not translated, but they are required for efficient translation of the mRNA.
  ½ mark
Answer: (i) AUG AUC UCG UAA; (ii) UTRs lie at the 5′ end before AUG and at the 3′ end after the stop codon; they are needed for efficient translation.
Where the marks are
- Identifies the translational unit AUG AUC UCG UAA, flanked by start and stop codons (1 mark)
- States the location of UTRs at the 5′ end (before start codon) and 3′ end (after stop codon) (0.5 mark)
- States that UTRs are required for efficient translation (0.5 mark)
Open Q18(B) on its own page
Q19 · 2 marks · Very short answerCompetency-basedHuman Health and Disease
Given below is the relationship between the HIV levels in the blood and helper T cell count in a person detected with AIDS. Study the relationship and answer the questions that follow.
A. What kind of relationship is observed in the virus levels and the immune response after some days of the initial infection?
B. Does it completely clear the virus from the body permanently? Give reason for your answer.
Figure in the paper: Graph of concentration in blood (y-axis) vs time (x-axis, months then years). Virus cultured from blood (blue) rises sharply in first weeks, falls sharply, stays low for a long period, then rises steadily over years. T cell count (red) falls initially, recovers to a plateau, stays steady, then declines steadily over the years, crossing the virus curve. See it in the official paper.
Try it first, then show our solutionSolution
1. A. A reciprocal (inverse) relationship is seen. As the adaptive immune response builds up, the helper T-cell count recovers and the virus level in blood falls sharply. As helper T levels rise, virus levels decline.
  1 mark
2. B. No, the virus is not cleared permanently. HIV survives inside macrophages, which act as an 'HIV factory', and inside helper T-lymphocytes. Over the years it keeps replicating and destroys helper T cells, so their number falls progressively while virus levels rise again. If untreated, the person loses the adaptive immune response, including the ability to make antibodies, and develops AIDS.
  1 mark
Answer: A. Reciprocal relationship: as helper T cells rise, virus level falls. B. No; HIV persists in macrophages and T cells and progressively destroys helper T cells, so virus levels rise again.
Where the marks are
- Describes the reciprocal relationship: rise in helper T cells with fall in virus level (1 mark)
- States No, with reason: HIV persists and progressively destroys helper T cells, causing loss of immunity (1 mark)
Open Q19 on its own page
Q20 · 2 marks · Very short answerBiotechnology: Principles and Processes
A culture plate of Lactobacillus shows blue-coloured colonies and colourless colonies. Explain the principle involved in the formation of such variance in the colour of colonies.
Try it first, then show our solutionSolution
1. The difference in colony colour is due to the principle of insertional inactivation (blue–white selection).
  ½ mark
2. The foreign DNA is inserted within the coding sequence of the enzyme β-galactosidase in the vector, and this inactivates the enzyme.
  ½ mark
3. The medium contains a chromogenic substrate. Bacteria whose plasmid has no insert have an active β-galactosidase, which acts on the substrate and produces blue colonies (non-recombinants).
  ½ mark
4. Bacteria with an insert have an inactivated β-galactosidase, cannot act on the substrate and form colourless (white) colonies. These are the recombinant colonies.
  ½ mark
Answer: Insertional inactivation of β-galactosidase: blue colonies are non-recombinants; colourless colonies are recombinants.
Where the marks are
- Names insertional inactivation (0.5 mark)
- Explains that the insert in the β-galactosidase coding sequence inactivates the enzyme (0.5 mark)
- Blue colonies with chromogenic substrate when there is no insert (0.5 mark)
- Colourless colonies are recombinants (0.5 mark)
Open Q20 on its own page
Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerCompetency-basedEcosystem
(i) It was estimated that if an evergreen forest has a GPP of 400 J/m²/day and 150 J/m²/day worth of carbon dioxide flows out of that forest, what is the NPP in that forest?
(ii) Explain why pyramids of energy must always be upright.
Try it first, then show our solutionSolution
1. (i) The CO₂ flowing out represents respiration losses (R).
  Formula: NPP = GPP − R
  Substitution: NPP = 400 J/m²/day − 150 J/m²/day
  NPP = 250 J/m²/day
  1 mark
2. (ii) Energy flows one way, from each trophic level to the next. At every level a large part of the energy is lost as heat in respiration, and only about 10% is passed on. Each higher level therefore always has less energy than the one below it, so the pyramid of energy is always upright and can never be inverted.
  1 mark
Answer: (i) NPP = 250 J/m²/day; (ii) Energy is lost as heat at each trophic level, so higher levels always have less energy.
Where the marks are
- Uses NPP = GPP − R and gets 250 J/m²/day (1 mark)
- Explains that energy is lost as heat at each trophic level, so the pyramid is always upright (1 mark)
Open Q21(A) on its own page
OR
Q21(B) · 2 marks · Very short answerCompetency-basedEcosystem
(i) Assume that, GPP Forest A = GPP Forest B = GPP Forest C, If Forest A has NPP = 1254 J/m²/day; Forest B, NPP =2157 J/m²/day; and Forest C, NPP = 779 J/m²/day, which one of these forests has maximum energy loss by respiration? Give reason.
(ii) Draw an ecological pyramid of number of the following food chains
a. Grass — Animal —Fleas on the host animal
b. Tree — Insects — Woodpecker
Try it first, then show our solutionSolution
1. (i) NPP = GPP − R, so R = GPP − NPP. With GPP equal in all three forests, the forest with the smallest NPP has the largest respiratory loss. Forest C has the lowest NPP (779 J/m²/day), so Forest C loses the maximum energy by respiration.
  1 mark
2. (ii)(a) Grass → Animal → Fleas (parasitic chain): draw three stacked horizontal bars. The bottom bar (grass) is broad because the producers are numerous. The middle bar (one host animal) is very narrow. The top bar (fleas) is broad because many fleas live on one host. The pyramid is therefore partly inverted at the top.
  ½ mark
3. (ii)(b) Tree → Insects → Woodpecker: the bottom bar (one tree) is narrow, the middle bar (many insects) is broad, and the top bar (a few woodpeckers) is narrow. This gives a spindle-shaped pyramid of numbers.
  ½ mark
Answer: (i) Forest C (lowest NPP, so highest respiration since R = GPP − NPP). (ii) (a) broad–narrow–broad, inverted at top; (b) narrow–broad–narrow, spindle-shaped.
Where the marks are
- Uses R = GPP − NPP to identify Forest C with reason (1 mark)
- Correct pyramid of numbers for grass–animal–fleas (0.5 mark)
- Correct spindle-shaped pyramid for tree–insects–woodpecker (0.5 mark)
Open Q21(B) on its own page

Section C

Q22 · 3 marks · Short answerCompetency-basedSexual Reproduction in Flowering Plants
The image below shows two germinated seeds X and Y which belong to the same species. Seed X is produced by apomixis whereas seed Y is a product of sexual reproduction.
A. Write the number of embryo(s), embryo sac(s) and ovules in the ovary of seed X.
B. How multiples embryos are formed in citrus fruits?
C. What advantage will plants developed from seed Y have over seed X?
Figure in the paper: Two germinated seeds. Seed X (left) has three seedlings labelled PLANT 1, PLANT 2, PLANT 3 growing from the one seed. Seed Y (right) has a single seedling labelled PLANT 1. See it in the official paper.
Try it first, then show our solutionSolution
1. A. Seed X has 3 embryos (three seedlings), 1 embryo sac and 1 ovule, because one seed develops from one ovule containing one embryo sac.
  1½ marks
2. B. In citrus, some nucellar cells surrounding the embryo sac divide mitotically, protrude into the embryo sac and develop into embryos. These are formed by asexual means without fertilisation, so one seed contains several embryos (polyembryony).
  ½ mark
3. C. The plant from seed Y comes from fertilisation, so it shows genetic variation and greater adaptability to changing environmental conditions. Plants from seed X are clones, lack variation and may succumb to stress. Also, the single plant in seed Y uses the entire endosperm, whereas the three embryos of seed X must share it and may be undernourished.
  1 mark
Answer: A. 3 embryos, 1 embryo sac, 1 ovule; B. Nucellar cells divide mitotically to form extra embryos; C. Seed Y plant has genetic variation (adaptability) and uses the whole endosperm.
Where the marks are
- States 3 embryos, 1 embryo sac and 1 ovule (0.5 × 3 = 1.5 marks)
- Explains that nucellar cells divide mitotically to form embryos asexually (0.5 mark)
- Gives an advantage of seed Y: genetic variation/adaptability or no sharing of endosperm (1 mark)
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Q23 · 3 marks · Short answerHuman Reproduction
Name the place in human ovary where the first meiotic division is completed during oogenesis. What are the products of this division? Give the chromosome number of each type of cells involved in the process.
Try it first, then show our solutionSolution
1. The first meiotic division is completed by the primary oocyte within the tertiary (Graafian) follicle of the ovary, just before ovulation.
  1 mark
2. Products: the primary oocyte divides unequally to form a large haploid secondary oocyte and a tiny first polar body.
  1 mark
3. Chromosome numbers: primary oocyte = 46 (diploid); secondary oocyte = 23 (haploid); first polar body = 23 (haploid).
  1 mark
Answer: In the primary oocyte inside the tertiary/Graafian follicle; products: secondary oocyte and first polar body; primary oocyte 46, secondary oocyte 23, polar body 23.
Where the marks are
- Names the primary oocyte in the tertiary/Graafian follicle as the site (1 mark)
- Names the products: secondary oocyte and first polar body (1 mark)
- Gives chromosome numbers 46, 23 and 23 (1 mark)
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Q24 · 3 marks · Short answerMolecular Basis of Inheritance
The schematic representation given below shows the concept of Central Dogma.
A. During the process of replication and transcription the pairing of nitrogenous bases is not similar. Explain.
B. How is the above process modified in a retrovirus? Name the process.
C. Justify why during the process of transcription only a segment of DNA is copied into RNA.
Figure in the paper: Central dogma scheme: DNA with a circular arrow labelled replication; DNA → mRNA labelled transcription; mRNA → protein labelled translation. See it in the official paper.
Try it first, then show our solutionSolution
1. A. In replication (DNA → DNA), adenine pairs with thymine (A=T), and G pairs with C.
  ½ mark
2. A. In transcription (DNA → RNA), adenine of the DNA template pairs with uracil (A=U), because RNA has uracil in place of thymine. G still pairs with C.
  ½ mark
3. B. In retroviruses the genetic material is RNA, which is used as a template to synthesise DNA (RNA → DNA) by the enzyme reverse transcriptase.
  ½ mark
4. B. The process is called reverse transcription.
  ½ mark
5. C. Copying the whole DNA would waste a great deal of energy, since transcription is an energy-expensive process. Only the gene (transcription unit) coding for the protein needed at that time is transcribed. Also, if both strands were copied, two complementary RNAs would form a double strand and could not be translated.
  1 mark
Answer: A. A pairs with T in replication but with U in transcription; B. RNA → DNA by reverse transcriptase – reverse transcription; C. Transcribing all DNA wastes energy; only the required gene is transcribed.
Where the marks are
- A pairs with T during replication (0.5 mark)
- A pairs with U during transcription (0.5 mark)
- In retroviruses RNA is used to make DNA (0.5 mark)
- Names reverse transcription (0.5 mark)
- Justifies on energy cost or need-based transcription of only specific genes (1 mark)
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Q25 · 3 marks · Short answerBiotechnology: Principles and Processes
Describe the steps involved in Southern blot hybridization using radiolabeled VNTR as a probe.
Try it first, then show our solutionSolution
1. 1. Isolation of DNA from the sample cells (e.g. blood or hair root).
  ½ mark
2. 2. Digestion of the DNA with restriction endonucleases into fragments.
  ½ mark
3. 3. Separation of the DNA fragments by size using gel electrophoresis.
  ½ mark
4. 4. Blotting: transfer of the separated fragments onto a synthetic membrane such as nitrocellulose or nylon.
  ½ mark
5. 5. Hybridisation of the membrane with the radiolabelled VNTR probe, which binds to complementary sequences.
  ½ mark
6. 6. Detection of the hybridised fragments by autoradiography, which shows a pattern of dark bands (the DNA fingerprint).
  ½ mark
Answer: Isolation of DNA → restriction digestion → gel electrophoresis → blotting on nitrocellulose/nylon → hybridisation with labelled VNTR probe → autoradiography.
Where the marks are
- Isolation of DNA (0.5 mark)
- Digestion by restriction endonucleases (0.5 mark)
- Separation by electrophoresis (0.5 mark)
- Blotting onto nitrocellulose/nylon membrane (0.5 mark)
- Hybridisation with labelled VNTR probe (0.5 mark)
- Detection by autoradiography (0.5 mark)
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Q26 · 3 marks · Short answerBiotechnology and its Applications
Bio-fertilisers are organisms that enrich the nutrient quality in the soil. Explain the role of three main sources of bio-fertilisers.
Try it first, then show our solutionSolution
1. Bacteria: Rhizobium lives symbiotically in the root nodules of leguminous plants and fixes atmospheric nitrogen into organic forms that the plant uses. Free-living bacteria such as Azospirillum and Azotobacter fix nitrogen in the soil and so enrich its nitrogen content.
  1 mark
2. Fungi: Fungi form symbiotic mycorrhizal associations with plant roots, e.g. members of the genus Glomus. The fungus absorbs phosphorus from the soil and passes it to the plant. It also gives resistance to root-borne pathogens, tolerance to salinity and drought, and better overall growth.
  1 mark
3. Cyanobacteria: These are autotrophic microbes, many of which fix atmospheric nitrogen, e.g. Anabaena, Nostoc and Oscillatoria. They are important biofertilisers in paddy fields and also add organic matter to the soil, increasing its fertility.
  1 mark
Answer: Bacteria (Rhizobium, Azotobacter, Azospirillum – nitrogen fixation), fungi (mycorrhiza/Glomus – phosphorus absorption), cyanobacteria (Anabaena, Nostoc – nitrogen fixation and organic matter in paddy fields).
Where the marks are
- Role of bacteria in nitrogen fixation with examples (1 mark)
- Role of mycorrhizal fungi in phosphorus uptake and other benefits (1 mark)
- Role of cyanobacteria in nitrogen fixation and adding organic matter (1 mark)
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Q27 · 3 marks · Short answerBiotechnology: Principles and Processes
Explain how PCR technique can be used for amplification of a small amount of DNA template.
Try it first, then show our solutionSolution
1. PCR (Polymerase Chain Reaction) makes many copies of a gene or DNA of interest in vitro. It uses two sets of primers (small, chemically synthesised oligonucleotides complementary to the ends of the target region), DNA polymerase, nucleotides and the genomic DNA as template.
  1 mark
2. Each cycle has three steps:
  Denaturation: heating separates the double-stranded DNA into single strands.
  Annealing: the primers bind to their complementary regions on each strand.
  Extension: DNA polymerase extends the primers using the nucleotides, copying the template.
  1 mark
3. The cycle is repeated many times, and the DNA doubles each time, so the segment is amplified about a billion times. This is possible because a thermostable DNA polymerase (Taq polymerase, from Thermus aquaticus) stays active through the high temperature of denaturation. The amplified DNA can then be ligated into a vector for cloning.
  1 mark
Answer: PCR amplifies DNA in vitro using two primers and thermostable Taq polymerase through repeated cycles of denaturation, annealing and extension, giving about a billion copies.
Where the marks are
- Explains PCR as in vitro synthesis of copies using primers, DNA polymerase and template (1 mark)
- Names the three steps: denaturation, annealing, extension (1 mark)
- Repeated cycles give about a billion copies; thermostable Taq polymerase used (1 mark)
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Q28 · 3 marks · Short answerCompetency-basedOrganisms and Populations
A. Diagram given below depicts different species of Warbler birds feeding on different regions on a Spruce tree. Explain the mechanism which helps them to co-exist.
B. What does Gause‘s exclusion principle state? Does it apply in the case shown above? Explain.
Figure in the paper: Spruce tree with feeding height scale (0 to 18 m). Cape May Warbler feeds at the tips of branches near the top of the tree; Bay-Breasted Warbler feeds in the middle part of the tree; Yellow-Rumped Warbler feeds in the lower part of the tree and at the bases of the middle branches. See it in the official paper.
Try it first, then show our solutionSolution
1. A. The warblers co-exist by resource partitioning.
  ½ mark
2. A. Species competing for the same resource avoid competition by using different foraging patterns: they feed at different times or in different zones of the same tree.
  ½ mark
3. A. MacArthur showed that five closely related warbler species on the same spruce tree avoided competition through behavioural differences in foraging. For example, the Cape May warbler feeds at the top branch tips, the Bay-breasted in the middle and the Yellow-rumped in the lower part.
  ½ mark
4. B. Gause's Competitive Exclusion Principle states that two closely related species competing for the same resources cannot co-exist indefinitely,
  ½ mark
5. B. ...and the competitively inferior species will eventually be eliminated.
  ½ mark
6. B. No, it does not apply here. The warblers partition the resource by feeding in different regions of the tree, so they are not competing for exactly the same resource and can co-exist.
  ½ mark
Answer: A. Resource partitioning: different foraging zones on the tree (MacArthur's warblers). B. Two closely related species competing for the same resource cannot co-exist indefinitely and the inferior one is eliminated; No, it does not apply because of resource partitioning.
Where the marks are
- Names resource partitioning (0.5 mark)
- Explains different foraging patterns to avoid competition (0.5 mark)
- Cites MacArthur's warbler study (0.5 mark)
- States that closely related species competing for the same resource cannot co-exist indefinitely (0.5 mark)
- States that the inferior competitor is eliminated (0.5 mark)
- Says No, with reason (0.5 mark)
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Section D

case-based

Q29 · 4 marks · Case-basedCompetency-basedEvolution
Assuming that within a population of beetles where Hardy Weinberg conditions are met, the colour black (B) is dominant over the colour red (b). 40% of all beetles are red (bb).
Given this information, answer the questions below:
1. A.
  What is the frequency of red beetles?
  1 mark
2. B.
  Calculate is the percentage of beetles in the population that are heterozygous.
  2 marks
3. C.
  What is the frequency of homozygous dominant individuals?
  1 mark
4. OR
  D.
  Assuming that Hardy Wienberg conditions are met in the beetle population consisting of 1500 beetles. How many beetles would you expect to be black and red in colour respectively?
  1 mark
Try it first, then show our solutionSolution
1. A. Red beetles are bb, so the frequency of red beetles is q² = 0.40. The frequency of the recessive allele b is therefore q = √0.40 ≈ 0.63.
  1 mark
2. B. Since p + q = 1, the frequency of allele B is p = 1 − 0.63 = 0.37.
  1 mark
3. B. Frequency of heterozygotes = 2pq = 2 × 0.37 × 0.63 ≈ 0.47. So about 47% of the beetles are heterozygous (Bb).
  1 mark
4. C. Frequency of homozygous dominant (BB) = p² = (0.37)² ≈ 0.14.
  1 mark
5. D. Red beetles = 40% of 1500 = 0.4 × 1500 = 600.
  ½ mark
6. D. Black beetles (BB + Bb) = 1500 − 600 = 900.
  ½ mark
Answer: A. q² = 0.4, so q ≈ 0.63; B. p = 0.37, 2pq ≈ 0.47, i.e. 47% heterozygous; C. p² ≈ 0.14; D. 900 black and 600 red.
Where the marks are
- Finds q = √0.4 ≈ 0.63 from q² = 0.4 (1 mark)
- Finds p = 1 − q = 0.37 (1 mark)
- Calculates 2pq ≈ 0.47 (47%) (1 mark)
- C: p² = 0.14 (1 mark) OR D: 600 red (0.5 mark) and 900 black (0.5 mark)
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Q30 · 4 marks · Case-basedCompetency-basedHuman Health and Disease
Given below is the pattern of temperature in a person suffering from a non-viral disease transmitted by mosquitoes. Study the graph and answer the questions that follow:
1. A.
  Explain the factor(s) responsible for this pattern of temperature.
  1 mark
2. B.
  How does this pathogen multiply in the human body?
  2 marks
3. C.
  How is this infection transmitted to humans?
  1 mark
4. OR
  D.
  Which stages of the life cycle of this pathogen are completed in the mosquito’s gut?
  1 mark
Figure in the paper: Line graph of temperature (°C, 35 to 41) vs time (days, 0 to 13). Temperature fluctuates repeatedly in peaks and troughs, with peaks rising over time: about 38.7 at day 3, 38.8 at day 5, 40 at day 6.7, 40.3 at day 8.2, 40.8 at day 10.2, falling to about 36-37 between peaks, ending near 36.8 at day 12. See it in the official paper.
Try it first, then show our solutionSolution
1. The disease is malaria, caused by the protozoan Plasmodium and spread by mosquitoes.
  A. The repeated fever spikes come from the rupture of RBCs, which releases a toxic substance called haemozoin. This causes chills and high fever that recur every 3–4 days, matching each cycle of RBC bursting.
  1 mark
2. B. Sporozoites injected by the mosquito reach the liver. There the parasite multiplies asexually, the liver cells burst, and the parasites are released into the blood.
  1 mark
3. B. The parasites then attack red blood cells and multiply asexually again. The RBCs rupture and the released parasites infect new RBCs. Some parasites in the RBCs develop into the sexual stages (gametocytes).
  1 mark
4. C. Infection is transmitted by the bite of an infected female Anopheles mosquito, which injects sporozoites (the infectious stage) into the human body.
  1 mark
5. D. In the mosquito's gut, fertilisation of the gametes and further development take place. The sporozoites formed then move to the salivary glands.
Answer: A. RBC rupture releasing haemozoin causes chills and fever every 3–4 days; B. Asexual multiplication in liver cells, then in RBCs, with gametocytes forming in RBCs; C. Bite of female Anopheles injecting sporozoites; D. Fertilisation and development.
Where the marks are
- Links fever pattern to RBC rupture and haemozoin release (1 mark)
- Asexual multiplication in liver cells with release into blood (1 mark)
- Asexual multiplication in RBCs and formation of gametocytes (1 mark)
- C: bite of female Anopheles introducing sporozoites (1 mark) OR D: fertilisation and development in the mosquito gut (1 mark)
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Section E

Answer one of these 2 alternatives.
Q31(A) · 5 marks · Long answerCompetency-basedHuman Reproduction
Cryptorchidism is a condition in which the testes fail to descend into the scrotum. It can also lead to compromised Sertoli cell function and has an impact on Leydig cell function.
(i) Identify at least 3 parameters of male fertility which get affected due to cryptorchidism.
(ii) Which process will be affected if mature spermatids are not released from Sertoli cells?
(iii) Name and explain one assisted reproductive technology (ART process) in which the sperm/semen is used to assist fertilization.
(iv) Name and explain the assisted reproductive technology that should be used to complete the development of embryos I and II shown in the figure given below.
Figure in the paper: Two embryos: Embryo I with 8 blastomeres and Embryo II with 16 blastomeres, each a circle containing cells with nuclei; blastomeres labelled. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Testes kept at body temperature (higher than scrotal temperature) cannot function normally, so:
  Sperm count falls, because spermatogenesis is impaired.
  Spermiogenesis is affected, because faulty Sertoli cells cannot nourish spermatids to become spermatozoa.
  Androgen (testosterone) secretion falls, because the Leydig cells are affected.
  1½ marks
2. (ii) If mature spermatids (sperms) are not released from the Sertoli cells, spermiation is affected.
  ½ mark
3. (iii) Artificial insemination (AI): semen from the husband or a healthy donor is artificially introduced into the vagina or into the uterus (IUI – intra-uterine insemination) of the female. It is used when the male cannot inseminate or has a low sperm count.
  (Alternatively, ICSI – intra-cytoplasmic sperm injection: a sperm is injected directly into the ovum to form an embryo in the laboratory.)
  1 mark
4. (iv) Embryo I (8 blastomeres): use ZIFT (zygote intra-fallopian transfer). The zygote or early embryo with up to 8 blastomeres is transferred into the fallopian tube to complete its development.
  1 mark
5. (iv) Embryo II (16 blastomeres): use IUT (intra-uterine transfer). An embryo with more than 8 blastomeres is transferred into the uterus to complete its development.
  1 mark
Answer: (i) Reduced sperm count/spermatogenesis, impaired spermiogenesis, reduced androgen secretion; (ii) Spermiation; (iii) AI/IUI (or ICSI); (iv) Embryo I – ZIFT; Embryo II – IUT.
Where the marks are
- Any three fertility parameters affected: sperm count/spermatogenesis, spermiogenesis, androgen secretion (0.5 × 3 = 1.5 marks)
- Names spermiation (0.5 mark)
- Names and explains AI/IUI or ICSI (1 mark)
- ZIFT for the 8-blastomere embryo, with explanation (1 mark)
- IUT for the 16-blastomere embryo, with explanation (1 mark)
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OR
Q31(B) · 5 marks · Long answerCompetency-basedSexual Reproduction in Flowering Plants
(i) Explain the significance of each of the following features present in plants given below:
a) In rose-bay plant the stamens ripen before the stigma.
b) In certain species of primrose, the flowers have short stamen and long style.
c) The bisexual flower of mustard exhibits rejection of self-pollen grain.
(ii) Explain how autogamy is prevented in castor and papaya plant respectively.
Try it first, then show our solutionSolution
1. (i)(a) In rose-bay the stamens and stigma mature at different times (dichogamy – protandry). When the pollen is released, the stigma is not yet receptive, so the pollen cannot germinate on the same flower's stigma. This prevents autogamy.
  1 mark
2. (i)(b) In primrose the anthers and stigma are placed at different positions (short stamens, long style – heterostyly). Pollen cannot easily reach the stigma of the same flower, so self-pollination (autogamy) is prevented.
  1 mark
3. (i)(c) Mustard shows self-incompatibility, a genetic mechanism of pollen–pistil interaction. The pistil recognises self-pollen and prevents it from germinating or forming a pollen tube, so self-fertilisation is prevented.
  1 mark
4. (ii) Castor is monoecious: male and female flowers are on the same plant but are separate (unisexual), so autogamy is prevented, though geitonogamy can still occur.
  1 mark
5. (ii) Papaya is dioecious: male and female flowers are on different plants, which prevents both autogamy and geitonogamy.
  1 mark
Answer: (i)(a) Dichogamy prevents autogamy; (b) different positions of anther and stigma (heterostyly) prevent self-pollination; (c) self-incompatibility rejects self-pollen. (ii) Castor is monoecious (unisexual flowers); papaya is dioecious (male and female flowers on different plants).
Where the marks are
- Explains that different maturation times prevent autogamy in rose-bay (1 mark)
- Explains that different positions of anther and stigma prevent autogamy in primrose (1 mark)
- Explains self-incompatibility as a genetic mechanism preventing pollen tube growth (1 mark)
- Castor: monoecious with unisexual flowers (1 mark)
- Papaya: dioecious (1 mark)
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Answer one of these 2 alternatives.
Q32(A) · 5 marks · Long answerCompetency-basedBiotechnology and its Applications
Explain how advent of biotechnology has helped in preventing infestation by nematodes and thereby increasing crop yield.
Try it first, then show our solutionSolution
1. Many nematodes parasitise plants and animals. The nematode Meloidogyne incognita infects the roots of tobacco plants and greatly reduces yield.
  ½ mark
2. A biotechnological strategy based on RNA interference (RNAi) was used to prevent this infestation.
  ½ mark
3. RNAi occurs in all eukaryotic organisms as a method of cellular defence.
  ½ mark
4. In RNAi, a specific mRNA is silenced when a complementary dsRNA molecule binds to it and prevents its translation.
  ½ mark
5. In nature, this complementary RNA may come from viruses with RNA genomes or from transposons that replicate via an RNA intermediate.
  ½ mark
6. Using Agrobacterium vectors, nematode-specific genes were introduced into the host (tobacco) plant.
  ½ mark
7. The DNA was introduced so that it produced both sense and anti-sense RNA in the host cells.
  ½ mark
8. These two complementary RNAs joined to form double-stranded RNA (dsRNA), which started RNAi.
  ½ mark
9. The dsRNA silenced the specific mRNA of the nematode after it fed on the plant.
  ½ mark
10. As a result, the parasite could not survive in the transgenic host expressing the interfering RNA. The plant was protected from infestation and crop yield increased.
  ½ mark
Answer: RNAi: Agrobacterium-mediated transfer of nematode-specific genes into tobacco produces sense and antisense RNA, which form dsRNA. This silences the mRNA of Meloidogyne incognita, so the nematode cannot survive and yield is protected.
Where the marks are
- Names Meloidogyne incognita infecting tobacco roots and reducing yield (0.5 mark)
- Names RNA interference (RNAi) as the strategy (0.5 mark)
- RNAi as a cellular defence in eukaryotes (0.5 mark)
- Silencing of a specific mRNA by complementary dsRNA (0.5 mark)
- Source of complementary RNA: RNA viruses or transposons (0.5 mark)
- Agrobacterium vectors used to introduce nematode-specific genes (0.5 mark)
- Production of sense and antisense RNA (0.5 mark)
- Formation of dsRNA that initiates RNAi (0.5 mark)
- Silencing of nematode mRNA (0.5 mark)
- Parasite cannot survive, so the plant is protected (0.5 mark)
Open Q32(A) on its own page
OR
Q32(B) · 5 marks · Long answerCompetency-basedBiotechnology and its Applications
In the future, genetic therapies may be used to prevent, treat, or cure certain inherited disorders in humans. Justify the statement with a suitable example.
Try it first, then show our solutionSolution
1. Gene therapy is a set of methods that corrects a gene defect diagnosed in a child or embryo. Genes are inserted into a person's cells and tissues to treat a disease.
  ½ mark
2. Correction means delivering a normal (functional) gene into the individual or embryo, so that it takes over the function of the defective, non-functional gene.
  ½ mark
3. Example: the first clinical gene therapy was given in 1990 to a 4-year-old girl with adenosine deaminase (ADA) deficiency. The ADA enzyme is essential for the immune system to work.
  ½ mark
4. This disorder is caused by deletion of the gene that codes for adenosine deaminase.
  ½ mark
5. Conventional treatments for ADA deficiency are bone marrow transplantation and enzyme replacement therapy, in which functional ADA is injected into the patient.
  ½ mark
6. Neither of these approaches is completely curative.
  ½ mark
7. Gene therapy procedure, step 1: lymphocytes are taken from the patient's blood and grown in a culture outside the body.
  ½ mark
8. Step 2: a functional ADA cDNA is introduced into these lymphocytes using a retroviral vector. The engineered lymphocytes are then returned to the patient.
  ½ mark
9. These lymphocytes are not immortal, so the patient needs periodic infusions of genetically engineered lymphocytes.
  ½ mark
10. If the ADA gene isolated from marrow cells is introduced into cells at early embryonic stages, it could give a permanent cure. This justifies the statement that gene therapy may prevent, treat or cure inherited disorders in future.
  ½ mark
Answer: Gene therapy inserts a normal gene to replace or compensate for a defective one. Example: ADA deficiency (1990). The patient's lymphocytes are cultured, given functional ADA cDNA through a retroviral vector and returned to the body, with periodic infusions needed. Introducing the gene at early embryonic stages could give a permanent cure.
Where the marks are
- Defines gene therapy as correction of a gene defect by inserting genes into cells (0.5 mark)
- States that a normal gene is delivered to compensate for the non-functional gene (0.5 mark)
- Names ADA deficiency, the first clinical gene therapy in 1990 on a 4-year-old girl, and the role of ADA in immunity (0.5 mark)
- States the cause: deletion of the ADA gene (0.5 mark)
- Mentions bone marrow transplant or enzyme replacement therapy (0.5 mark)
- Notes that these conventional methods are not completely curative (0.5 mark)
- Patient's lymphocytes are cultured outside the body (0.5 mark)
- Functional ADA cDNA is introduced using a retroviral vector and the cells are returned to the patient (0.5 mark)
- Cells are not immortal, so periodic infusion is needed (0.5 mark)
- Gene introduced at early embryonic stage could give a permanent cure (0.5 mark)
Open Q32(B) on its own page
Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerBiodiversity and Conservation
(i) Why is there a need to conserve biodiversity? (Any two reasons)
(ii) Name and explain any two causes that are responsible for the loss of biodiversity.
Try it first, then show our solutionSolution
1. (i) Reason 1: Narrowly utilitarian. Humans get direct economic benefits from biodiversity. These include food, firewood, fibre, construction material, industrial products (resins, gums, dyes, tannins) and medicinally important products.
  1 mark
2. (i) Reason 2: Broadly utilitarian. Biodiversity sustains many ecosystem services. For example, the Amazon forest is estimated to produce about 20% of the earth's atmospheric oxygen, and pollinators such as bees, birds and bats help plants form fruits and seeds. Biodiversity also gives aesthetic pleasures such as bird watching.
  (Alternatively, ethical reason: every species has intrinsic value even without economic value, so it is our moral duty to protect all living creatures for future generations.)
  1 mark
3. (ii) Cause 1: Habitat loss and fragmentation. This is the primary cause of extinction. Examples include clearing of tropical rainforests. When large habitats are broken up by human settlements, roads and canals, populations of animals needing large territories, and of migratory animals, decline.
  1½ marks
4. (ii) Cause 2: Over-exploitation. When humans over-harvest natural resources, species are degraded and driven to extinction. Examples: Steller's sea cow, the passenger pigeon and many marine fishes.
  (Other accepted causes: alien species invasion, e.g. Nile perch introduced into Lake Victoria wiped out the native cichlid fishes; co-extinction, e.g. parasites that depend only on a host fish die out when the host becomes extinct.)
  1½ marks
Answer: (i) Narrowly utilitarian (direct products such as food, fibre and medicines) and broadly utilitarian (ecosystem services such as oxygen and pollination); the ethical reason is also accepted. (ii) Habitat loss and fragmentation, and over-exploitation (alien species invasion and co-extinction are also accepted).
Where the marks are
- First reason for conservation explained with examples (1 mark)
- Second reason for conservation explained with examples (1 mark)
- First cause of biodiversity loss named and explained with an example (1.5 marks)
- Second cause of biodiversity loss named and explained with an example (1.5 marks)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerBiodiversity and Conservation
(i) Name the two types of desirable approaches to conserve biodiversity? Explain with examples bringing out the difference between the two types.
(ii) State the features of a stable biological community?
Try it first, then show our solutionSolution
1. (i) In situ conservation (on site): the species is protected in its natural habitat, so the whole ecosystem is conserved. Examples include biodiversity hotspots, national parks, wildlife sanctuaries, biosphere reserves and sacred groves (e.g. Khasi and Jaintia hills in Meghalaya).
  1 mark
2. (i) Ex situ conservation (off site): threatened species are taken out of their natural habitat and protected in special settings. Examples include zoological parks, botanical gardens, wildlife safari parks, cryopreservation of gametes, seed banks, and in vitro fertilisation and tissue culture.
  Difference: in situ protects species together with their habitat and natural interactions. Ex situ protects individual species in artificial surroundings, mainly for those on the verge of extinction.
  1 mark
3. (ii) Feature 1: a stable community does not show too much variation in productivity from year to year.
  1 mark
4. (ii) Feature 2: it is resistant or resilient to occasional natural or man-made disturbances.
  1 mark
5. (ii) Feature 3: it is resistant to invasion by alien species. (Tilman's experiments showed that plots with more species had less year-to-year variation in total biomass.)
  1 mark
Answer: (i) In situ conservation protects species in their natural habitat (national parks, sanctuaries, biosphere reserves, sacred groves). Ex situ conservation protects them outside their natural habitat (zoos, botanical gardens, seed banks, cryopreservation). (ii) A stable community has little year-to-year variation in productivity, is resistant or resilient to disturbances, and resists alien species invasion.
Where the marks are
- In situ conservation named and explained with examples (1 mark)
- Ex situ conservation named and explained with examples, with the difference brought out (1 mark)
- Little variation in productivity from year to year (1 mark)
- Resistant or resilient to occasional disturbances (1 mark)
- Resistant to invasion by alien species (1 mark)
Open Q33(B) on its own page

Source: official sample paper, cbseacademic.nic.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.