BiologyCBSE Class 12 · Board exam 2024 (session 2023-24)

CBSE Class 12 Biology: 2024 board paper, set 57/5/1

The question paper set in the CBSE Class 12 board exam of 2024, 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 question paper from cbse.gov.in: https://www.cbse.gov.in/cbsenew/question-paper/2024/XII/BIOLOGY.zip (a zip; this paper is the file 57-5-1_Biology.pdf). Questions are transcribed from that PDF; figures are described in words, so check the official paper for them.

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	very short answer
C	22-28	short answer
D	29-30	case-based
E	31-33	long answer

Instructions in brief: All questions compulsory; no overall choice; internal choice in sections B, C and D (and E questions 31-33 as printed).

Section A

MCQ 1-12, assertion-reason 13-16

Q1 · 1 mark · MCQPrinciples of Inheritance and Variation
A single gene that controls the expression of more than one trait is said to show
1. (A)Multiple allelism
2. (B)Polygenic inheritance
3. (C)Incomplete dominance
4. (D)Pleiotropism
Try it first, then show our solutionSolution
1. When one gene influences several phenotypic traits (e.g. the phenylketonuria gene affecting mental ability and skin/hair pigmentation), the gene is called pleiotropic and the phenomenon is pleiotropism. So the answer is option (D).
  1 mark
Answer: (D) Pleiotropism
Where the marks are
- Identifies pleiotropism (1 mark)
Q2 · 1 mark · MCQPrinciples of Inheritance and Variation
A person with trisomy of 21st chromosome shows
(i) Furrowed tongue (ii) Characteristic palm crease (iii) Rudimentary ovaries (iv) Gynaecomastia
Select the correct option, from the choices given below :
1. (A)(ii) and (iv)
2. (B)(i), (ii) and (iv)
3. (C)(ii) and (iii)
4. (D)(i) and (ii)
Try it first, then show our solutionSolution
1. Down's syndrome (trisomy 21) shows a furrowed tongue, partly open mouth and a characteristic palm crease. Rudimentary ovaries belong to Turner's syndrome and gynaecomastia to Klinefelter's syndrome. So (i) and (ii) are correct: option (D).
  1 mark
Answer: (D) (i) and (ii)
Where the marks are
- Selects furrowed tongue and palm crease (1 mark)
Q3 · 1 mark · MCQReproductive Health
Observe the schematic representation of assisted reproductive technology given below. Identify the most appropriate technique depicted in the above diagram.
1. (A)IUT
2. (B)IUI
3. (C)ICSI
4. (D)ZIFT
Figure in the paper: An egg held by a holding tool on the left; an injection needle injects a sperm into the cytoplasm of the egg using a fine needle. Labels: Egg, Injection Needle, Holding tool, 'Sperm being injected into the cytoplasm of the egg using a fine needle'. See it in the official paper.
Try it first, then show our solutionSolution
1. The figure shows a single sperm being injected directly into the cytoplasm of the egg with a fine needle. This is intracytoplasmic sperm injection (ICSI): option (C).
  1 mark
Answer: (C) ICSI
Where the marks are
- Identifies ICSI (1 mark)
Q4 · 1 mark · MCQHuman Health and Disease
Interferons are proteins secreted by
1. (A)RBC
2. (B)WBC
3. (C)Bacteria infected cell
4. (D)Virus infected cell
Try it first, then show our solutionSolution
1. Interferons are proteins released by virus-infected cells. They protect nearby non-infected cells from further viral infection, which is a cytokine barrier of innate immunity. So the answer is option (D).
  1 mark
Answer: (D) Virus infected cell
Where the marks are
- Virus-infected cells secrete interferons (1 mark)
Q5 · 1 mark · MCQMicrobes in Human Welfare
During biological treatment of sewage, the masses of bacteria held together by fungal filaments to form mesh like structures are called
1. (A)primary sludge
2. (B)flocs
3. (C)activated sludge
4. (D)anaerobic sludge
Try it first, then show our solutionSolution
1. In the aeration tanks of secondary treatment, aerobic bacteria and fungal filaments form mesh-like masses called flocs. These flocs consume the organic matter and lower the BOD of the effluent. So the answer is option (B).
  1 mark
Answer: (B) flocs
Where the marks are
- Identifies flocs (1 mark)
Q6 · 1 mark · MCQBiotechnology: Principles and Processes
Which one of the following statements is correct in the context of observing DNA separation by agarose gel electrophoresis ?
1. (A)DNA can be seen in visible light.
2. (B)DNA can be seen without staining in visible light.
3. (C)Ethidium bromide stained DNA can be seen in visible light
4. (D)Ethidium bromide stained DNA can be seen under UV light.
Try it first, then show our solutionSolution
1. DNA bands cannot be seen in visible light. After staining with ethidium bromide and exposure to UV light, the DNA bands glow bright orange. So the answer is option (D).
  1 mark
Answer: (D)
Where the marks are
- EtBr-stained DNA is visible under UV (1 mark)
Q7 · 1 mark · MCQSexual Reproduction in Flowering Plants
A phenomenon where a male insect mistakenly identified the patterns of a orchid flower as the female insect partner, and tries to copulate and thereby pollinates the flower is said to be :
1. (A)Pseudocopulation
2. (B)Pseudopollination
3. (C)Pseudoparthenocarpy
4. (D)Pseudofertilisation
Try it first, then show our solutionSolution
1. In Ophrys, a petal resembles the female bee. The male bee tries to copulate with it ('pseudocopulation') and in doing so carries pollen to the flower. So the answer is option (A).
  1 mark
Answer: (A) Pseudocopulation
Where the marks are
- Identifies pseudocopulation (1 mark)
Q8 · 1 mark · MCQSexual Reproduction in Flowering Plants
Identify the correct labellings in the figure of a fertilised embryo sac of an angiosperm given below :
1. (A)A – zygote, B – degenerating synergids, C – degenerating antipodals, D – PEN
2. (B)A – degenerating synergids, B – zygote, C – PEN, D – degenerating antipodals
3. (C)A – degenerating antipodals, B – PEN, C – degenerating synergids, D – zygote
4. (D)A – degenerating synergids, B – zygote, C – degenerating antipodals, D – PEN
Figure in the paper: Fertilised embryo sac: A marks cells at the top (micropylar end, two cells), B marks a cell just below at the top, C marks the large nucleus in the centre labelled Primary Endosperm Cell (PEC), D marks cells at the bottom (chalazal end). See it in the official paper.
Try it first, then show our solutionSolution
1. At the micropylar end lie the degenerating synergids (A), and just below them is the zygote (B). The central large nucleus is the PEN (C), and the chalazal cells are the degenerating antipodals (D). So the answer is option (B).
  1 mark
Answer: (B)
Where the marks are
- Correct labelling sequence (1 mark)
Q9 · 1 mark · MCQPrinciples of Inheritance and Variation
Study the pedigree chart of a family showing the inheritance pattern of a certain disorder. Select the option that correctly identifies the nature of the trait depicted in the pedigree chart.
1. (A)Dominant X – linked
2. (B)Recessive X – linked
3. (C)Autosomal dominant
4. (D)Autosomal recessive
Figure in the paper: Pedigree: unaffected parents (square male, circle female) have five children: unaffected son, affected son (filled square), unaffected daughter, affected daughter (filled circle), unaffected son. The unaffected daughter marries an unaffected male and has three children: an affected son (filled square) and two unaffected daughters. See it in the official paper.
Try it first, then show our solutionSolution
1. Unaffected parents have affected children, so the trait is recessive.
2. An affected daughter is born to an unaffected father, which rules out X-linked recessive inheritance. The trait is therefore autosomal recessive: option (D).
  1 mark
Answer: (D) Autosomal recessive
Where the marks are
- Concludes autosomal recessive with reasoning (1 mark)
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Q10 · 1 mark · MCQMolecular Basis of Inheritance
Match the following genes of the lac operon listed in column 'A' with their respective products listed in column 'B' :
A (Gene) B (Products)
a. 'i' gene (i) β-galactosidase
b. 'z' gene (ii) lac permease
c. 'a' gene (iii) repressor
d. 'y' gene (iv) transacetylase
Select the correct option :
1. (A)a-(i), b-(iii), c-(ii), d-(iv)
2. (B)a-(iii), b-(i), c-(ii), d-(iv)
3. (C)a-(iii), b-(i), c-(iv), d-(ii)
4. (D)a-(iii), b-(iv), c-(i), d-(ii)
Try it first, then show our solutionSolution
1. The genes code for these products: i → repressor, z → β-galactosidase, a → transacetylase, y → permease. This matches option (C).
  1 mark
Answer: (C) a-(iii), b-(i), c-(iv), d-(ii)
Where the marks are
- Correct matching (1 mark)
Q11 · 1 mark · MCQPrinciples of Inheritance and Variation
If both the parents are carriers for thalassaemia, the chances of an afflicted child to be born to them is :
1. (A)25%
2. (B)50%
3. (C)75%
4. (D)100%
Try it first, then show our solutionSolution
1. Thalassaemia is autosomal recessive. The cross Tt × Tt gives ¼ TT : ½ Tt : ¼ tt, so ¼ of the children (tt) are affected, i.e. 25%: option (A).
  1 mark
Answer: (A) 25%
Where the marks are
- Monohybrid cross gives ¼ affected (1 mark)
Q12 · 1 mark · MCQMolecular Basis of Inheritance
If the sequence of nitrogen bases of the coding strand in a transcription unit is 5' – ATGAATG – 3', the sequence of bases in its RNA transcript would be
1. (A)5' – AUGAAUG – 3'
2. (B)5' – UACUUAC – 3'
3. (C)5' – CAUUCAU – 3'
4. (D)5' – GUAAGUA – 3'
Try it first, then show our solutionSolution
1. The RNA transcript has the same sequence as the coding strand, with U in place of T. It is 5'-AUGAAUG-3': option (A).
  1 mark
Answer: (A) 5'–AUGAAUG–3'
Where the marks are
- Replaces T by U in the coding-strand sequence (1 mark)
Q13 · 1 mark · Assertion-reasonCompetency-basedHuman Health and Disease
Assertion (A) : AIDS is a syndrome caused by HIV.
Reason (R) : HIV is a virus that damages the immune system with DNA as its genetic material.
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. The assertion is true: AIDS is caused by HIV.
2. The reason is false, because HIV is a retrovirus with RNA, not DNA, as its genetic material. So the answer is option (C).
  1 mark
Answer: (C)
Where the marks are
- (A) true, (R) false because HIV has RNA as its genome (1 mark)
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Q14 · 1 mark · Assertion-reasonCompetency-basedBiotechnology: Principles and Processes
Assertion (A) : In molecular diagnosis, single stranded DNA or RNA tagged with radioactive molecule is called a probe.
Reason (R) : A probe always searches and hybridises with its complementary DNA in a clone of cells.
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. A probe is a radioactively labelled single-stranded DNA or RNA. It hybridises with its complementary DNA in a clone of cells, and this hybridisation is what makes it useful for diagnosis. Both statements are true and (R) explains (A), so the answer is option (A).
  1 mark
Answer: (A)
Where the marks are
- Both true and R explains A (1 mark)
Q15 · 1 mark · Assertion-reasonCompetency-basedPrinciples of Inheritance and Variation
Assertion (A) : In birds the sex of the offspring is determined by males.
Reason (R) : Males are homogametic while females are heterogametic.
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. In birds, females are ZW (heterogametic) and males are ZZ (homogametic). The reason is therefore true.
2. Because the female produces two kinds of eggs, the female decides the sex of the offspring. The assertion is false, so the answer is option (D).
  1 mark
Answer: (D)
Where the marks are
- (A) false, (R) true due to the ZW system (1 mark)
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Q16 · 1 mark · Assertion-reasonCompetency-basedBiodiversity and Conservation
Assertion (A) : Communities that comprise of more species tend to be more stable.
Reason (R) : A higher number of species results in less year to year variation in total biomass.
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. Tilman's experiments showed that plots with more species had less year-to-year variation in total biomass. That steadiness is what makes species-rich communities more stable. Both statements are true and (R) explains (A), so the answer is option (A).
  1 mark
Answer: (A)
Where the marks are
- Both true and R explains A (1 mark)

A (Gene)	B (Products)
a. 'i' gene	(i) β-galactosidase
b. 'z' gene	(ii) lac permease
c. 'a' gene	(iii) repressor
d. 'y' gene	(iv) transacetylase

Section B

very short answer

Answer one of these 2 alternatives.
Q17(a) · 2 marks · Very short answerSexual Reproduction in Flowering Plants
"Farmers prefer apomictic seeds to hybrid seeds." Justify giving two reasons.
Try it first, then show our solutionSolution
1. Reason 1: Hybrid seeds lose their hybrid character because the progeny segregate in the next generation. Farmers must therefore buy fresh hybrid seed every year. Apomictic seeds form without meiosis and fertilisation, so the plants raised from them stay genetically identical to the parent and keep the desirable hybrid traits generation after generation.
  1 mark
2. Reason 2: Producing hybrid seed is expensive, which makes it costly for farmers. If the hybrids are made apomictic, the farmer can save and re-sow the seed, which is much cheaper.
  1 mark
Answer: Apomictic seeds do not segregate, so hybrid traits are retained and seed need not be bought every year; they are also cheaper than costly hybrid seed.
Where the marks are
- States that hybrid progeny segregate, so hybrid seed must be bought every year, whereas apomictic seeds retain the parental characters (1 mark)
- States that hybrid seed production is costly while apomictic seed is cheap and can be re-used (1 mark)
Open Q17(a) on its own page
OR
Q17(b) · 2 marks · Very short answerReproductive Health
Mention one advantage and one disadvantage of amniocentesis.
Try it first, then show our solutionSolution
1. Advantage: Amniocentesis tests the foetal cells present in the amniotic fluid. It can detect chromosomal abnormalities or genetic disorders in the foetus, such as Down's syndrome, haemophilia or sickle-cell anaemia, and helps assess the survivability of the foetus.
  1 mark
2. Disadvantage: The same test reveals the sex of the foetus. It is misused for sex determination, which can lead to female foeticide. For this reason it is legally banned for that purpose.
  1 mark
Answer: Advantage: detects genetic or chromosomal disorders (e.g. Down's syndrome) in the foetus. Disadvantage: misused for foetal sex determination, leading to female foeticide.
Where the marks are
- One correct advantage, e.g. detecting genetic disorders such as Down's syndrome (1 mark)
- Disadvantage: misuse for sex determination leading to female foeticide (1 mark)
Open Q17(b) on its own page
Q18 · 2 marks · Very short answerBiotechnology: Principles and Processes
5' – G↓AATTC – 3'
3' – CTTAA↑G – 5'
(the arrows mark a cut on each strand: after G on the top strand and before G on the bottom strand)
1. (a)
  Name the restriction enzyme that recognises the given specific sequence of bases. What are such sequence of bases referred to as ?
  1 mark
2. (b)
  What are the arrows in the given figure indicating ? Write the result obtained thereafter.
  1 mark
Figure in the paper: Double-stranded sequence 5'-GAATTC-3' / 3'-CTTAAG-5' with a downward arrow after G on the top strand and an upward arrow between A and G on the bottom strand. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) The enzyme that recognises 5'-GAATTC-3' is the restriction endonuclease EcoRI.
  ½ mark
2. (a) Such sequences read the same on both strands in the 5'→3' direction. They are called palindromic nucleotide sequences (palindromes).
  ½ mark
3. (b) The arrows mark the restriction (recognition) sites, i.e. the exact points where EcoRI cuts the sugar-phosphate backbone of each strand, between G and A.
  ½ mark
4. (b) The cuts are slightly away from the centre of the palindrome. This leaves single-stranded overhanging stretches called sticky ends, which can join complementary ends with the help of DNA ligase.
  ½ mark
Answer: (a) EcoRI; palindromic nucleotide sequence. (b) The cutting (restriction) sites of EcoRI on the two strands; cutting produces sticky ends.
Where the marks are
- Names EcoRI (½ mark)
- Calls the sequence a palindrome or palindromic sequence (½ mark)
- Arrows show the cutting or restriction sites of the enzyme (½ mark)
- Result: sticky ends are formed (½ mark)
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Q19 · 2 marks · Very short answerCompetency-basedOrganisms and Populations
Observe the population growth curve and answer the questions given below.
1. (a)
  State the conditions under which growth curve 'A' and growth curve 'B' plotted in the graph are possible.
  1 mark
2. (b)
  Mention what does 'K' in the graph represent.
  1 mark
Figure in the paper: Graph of population density (N) vs time (t). Curve A rises steeply exponentially, labelled dN/dt = rN. Curve B is sigmoidal and levels off at a dotted horizontal line marked K, labelled dN/dt = rN((K-N)/K). See it in the official paper.
Try it first, then show our solutionSolution
1. (a) Curve A (exponential, $\frac{d N}{d t} = r N$ ) is possible when resources such as food and space are unlimited and there is no environmental resistance.
  ½ mark
2. (a) Curve B (logistic/sigmoid, $\frac{d N}{d t} = r N (\frac{K - N}{K})$ ) is possible when resources are limited, so competition and environmental resistance check further growth.
  ½ mark
3. (b) K is the carrying capacity: the maximum number of individuals of a population that a given habitat can support. Beyond this limit no further growth is possible.
  1 mark
Answer: (a) A: unlimited resources, no environmental resistance; B: limited resources, with environmental resistance. (b) K = carrying capacity.
Where the marks are
- Curve A: unlimited resources or no environmental resistance (½ mark)
- Curve B: limited resources or environmental resistance present (½ mark)
- K is the carrying capacity, the maximum population a habitat can support (1 mark)
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Q20 · 2 marks · Very short answerMicrobes in Human Welfare
Explain how are plants benefitted by their association with "Glomus species".
Try it first, then show our solutionSolution
1. Glomus is a fungus that forms mycorrhiza with plant roots. Its hyphae absorb phosphorus from the soil and pass it to the plant.
  1 mark
2. Plants with this association also gain resistance to root-borne pathogens, tolerance to salinity and drought, and an overall increase in growth and development.
  1 mark
Answer: Glomus (mycorrhiza) absorbs phosphorus for the plant and gives resistance to root pathogens, tolerance to salinity and drought, and better overall growth.
Where the marks are
- Absorbs phosphorus from the soil and passes it to the plant (1 mark)
- Any other benefit: resistance to root-borne diseases, tolerance to salinity or drought, or increased growth (1 mark)
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Q21 · 2 marks · Very short answerMolecular Basis of Inheritance
If the base adenine constitutes 31% of an isolated DNA fragment, then write what will be the expected percentage of the base cytosine in it. Explain how did you arrive at the answer given.
Try it first, then show our solutionSolution
1. By Chargaff's rule for double-stranded DNA, A = T and G = C, i.e. A/T = G/C = 1 and A + G = C + T.
  ½ mark
2. Since A = 31%, T = 31%. So A + T = 31% + 31% = 62%.
  ½ mark
3. G + C = 100% − 62% = 38%.
  ½ mark
4. Since G = C, cytosine = $\frac{38}{2}$ = 19%.
  ½ mark
Answer: Cytosine = 19%
Where the marks are
- States Chargaff's rule A = T and G = C (½ mark)
- T = 31%, so A + T = 62% (½ mark)
- G + C = 38% (½ mark)
- C = 19% (½ mark)
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Section C

short answer

Q22 · 3 marks · Short answerBiotechnology and its Applications
Identify a, b, c, d, e and f in the table given below :
Sl. No. Organism Bioactive Molecule Use
1. Monascus purpureus a b
2. c d Antibiotic
3. e Cyclosporin A f
Try it first, then show our solutionSolution
1. a – Statins (produced by the yeast Monascus purpureus).
  ½ mark
2. b – Blood-cholesterol lowering agent. Statins competitively inhibit the enzyme of cholesterol synthesis.
  ½ mark
3. c – Penicillium notatum (any other correct antibiotic-producing organism is acceptable).
  ½ mark
4. d – Penicillin (the antibiotic matching the organism named in c).
  ½ mark
5. e – Trichoderma polysporum (a fungus).
  ½ mark
6. f – Immunosuppressive agent, used in organ-transplant patients.
  ½ mark
Answer: a – Statin; b – blood-cholesterol lowering agent; c – Penicillium notatum; d – Penicillin; e – Trichoderma polysporum; f – immunosuppressant (organ transplants).
Where the marks are
- Each of a, b, c, d, e and f correctly identified (½ mark each, ½ × 6)
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Answer one of these 2 alternatives.
Q23(a) · 3 marks · Short answerBiodiversity and Conservation
Tropical regions harbour more species than the temperate regions. How have biologists tried to explain this in their own ways ? Explain.
Try it first, then show our solutionSolution
1. Hypothesis 1 – Long evolutionary time: Temperate regions went through frequent glaciations in the past. Tropical regions have stayed relatively undisturbed for millions of years, so they have had a long evolutionary time for species diversification.
  1 mark
2. Hypothesis 2 – Stable environment: Tropical environments are less seasonal, relatively constant and predictable. Such constant conditions promote niche specialisation, which leads to greater species diversity.
  1 mark
3. Hypothesis 3 – More solar energy: Tropical regions receive more solar energy. This contributes to higher productivity, which can support more species.
  1 mark
Answer: Three explanations: (1) no past glaciations, giving a long evolutionary time; (2) a less seasonal, stable environment promoting niche specialisation; (3) more solar energy, giving higher productivity.
Where the marks are
- Absence of frequent glaciation, giving long evolutionary time for diversification (1 mark)
- Less seasonal, constant and predictable environment promoting niche specialisation (1 mark)
- More solar energy leading to higher productivity (1 mark)
Open Q23(a) on its own page
OR
Q23(b) · 3 marks · Short answerEcosystem
(i) What does an ecological pyramid represent ?
(ii) The Ecological pyramids may have an 'upright' or an 'inverted' shape. Justify with the help of suitable examples.
Try it first, then show our solutionSolution
1. (i) An ecological pyramid is a diagrammatic representation of the relationship between organisms at successive trophic levels, expressed in terms of number, biomass or energy, with producers at the base.
  1 mark
2. (ii) Upright pyramid: this shape results when the number, biomass or energy is greatest at the producer level and decreases at each higher trophic level.
  ½ mark
3. Example of an upright pyramid: the pyramid of number in a grassland, where grasses outnumber herbivores, which outnumber carnivores. (A pyramid of energy is always upright.)
  ½ mark
4. Inverted pyramid: this shape results when the number or biomass at the producer level is less than that of the consumers.
  ½ mark
5. Examples of an inverted pyramid: the pyramid of biomass in the sea, where a small standing crop of phytoplankton supports a larger standing crop of zooplankton; or the pyramid of number of a single big tree supporting many insects.
  ½ mark
Answer: (i) The relationship between trophic levels in terms of number, biomass or energy. (ii) Upright when producers exceed consumers (e.g. grassland pyramid of number); inverted when producers are fewer or have less biomass (e.g. sea pyramid of biomass, tree–insect pyramid of number).
Where the marks are
- Definition of an ecological pyramid in terms of number, biomass or energy at trophic levels (1 mark)
- Reason for the upright shape (½ mark) with a correct example (½ mark)
- Reason for the inverted shape (½ mark) with a correct example (½ mark)
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Q24 · 3 marks · Short answerBiotechnology and its Applications
(a) What are transgenic animals ?
(b) Name the transgenic animal having the largest number amongst all the existing transgenic animals.
(c) State any 3 reasons for which these types of animals are being produced.
Try it first, then show our solutionSolution
1. (a) Transgenic animals are animals whose DNA has been manipulated to possess and express an extra (foreign) gene.
  1 mark
2. (b) Mice – over 95% of all existing transgenic animals are mice.
  ½ mark
3. (c) Reason 1: to study how genes are regulated and how they affect normal physiology and development.
  ½ mark
4. (c) Reason 2: to study how genes contribute to the development of diseases, so that transgenic models of human diseases (e.g. cancer, cystic fibrosis) can be used to find new treatments.
  ½ mark
5. (c) Reason 3: to obtain useful biological products, such as α-1-antitrypsin, or human-protein-enriched milk from the transgenic cow Rosie. Other valid reasons are vaccine safety testing and chemical safety (toxicity) testing.
  ½ mark
Answer: (a) Animals with manipulated DNA that possess and express a foreign gene. (b) Mice. (c) Any three of: study of normal physiology and development, study of disease, biological products, vaccine safety testing, chemical or toxicity testing.
Where the marks are
- Correct definition of transgenic animals (1 mark)
- Mice (½ mark)
- Any three correct reasons (½ mark each)
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Q25 · 3 marks · Short answerSexual Reproduction in Flowering Plants
If the cells in the leaves of a maize plant contain 10 chromosomes each, write the number of chromosomes in its endosperm and zygote. Name and explain the process by which an endosperm and a zygote are formed in maize.
Try it first, then show our solutionSolution
1. Leaf cells are diploid (2n = 10), so n = 5. The endosperm is triploid (3n), so it has 15 chromosomes.
  ½ mark
2. The zygote is diploid (2n), so it has 10 chromosomes.
  ½ mark
3. The endosperm is formed by triple fusion.
  ½ mark
4. In triple fusion, the second male gamete fuses with the two polar nuclei (or the secondary nucleus) of the central cell to form the triploid primary endosperm nucleus.
  ½ mark
5. The zygote is formed by syngamy.
  ½ mark
6. In syngamy, one male gamete fuses with the egg cell to form the diploid zygote. Syngamy together with triple fusion makes up double fertilisation.
  ½ mark
Answer: Endosperm: 15 chromosomes (triple fusion of a male gamete with two polar nuclei). Zygote: 10 chromosomes (syngamy of a male gamete with the egg).
Where the marks are
- Endosperm has 15 chromosomes (½ mark)
- Zygote has 10 chromosomes (½ mark)
- Names triple fusion (½ mark) and explains it as the fusion of a male gamete with two polar nuclei (½ mark)
- Names syngamy (½ mark) and explains it as the fusion of a male gamete with the egg (½ mark)
Open Q25 on its own page
Q26 · 3 marks · Short answerMolecular Basis of Inheritance
(a) Why does DNA replication occur within a replication fork and not in its entire length simultaneously ?
(b) "DNA replication is continuous and discontinuous on the two strands within the replication fork." Explain with the help of a schematic representation.
Try it first, then show our solutionSolution
1. (a) Replication is an energetically very expensive process. The two strands of the long DNA cannot be separated along their entire length at once, because of the very high energy needed. So replication proceeds in a small opening of the helix called the replication fork.
  1 mark
2. (b) DNA-dependent DNA polymerase can polymerise nucleotides only in the 5'→3' direction.
  ½ mark
3. On the template strand with 3'→5' polarity, the new strand (the leading strand) is synthesised continuously, towards the fork.
  ½ mark
4. On the template strand with 5'→3' polarity, the new strand (the lagging strand) is synthesised discontinuously as short Okazaki fragments, away from the fork. These fragments are later joined by DNA ligase.
  ½ mark
5. Schematic (described): Draw a Y-shaped replication fork with the parental strands labelled 3'→5' and 5'→3'. Along the 3'→5' template, draw one continuous arrow (5'→3') moving into the fork, labelled 'leading strand – continuous'. Along the 5'→3' template, draw several short arrows, each 5'→3', pointing away from the fork, labelled 'lagging strand – Okazaki fragments (discontinuous)'. The polarity must be shown correctly.
  ½ mark
Answer: (a) Unwinding the whole DNA at once would need very high energy, so replication occurs in a small fork. (b) Polymerase works only 5'→3', so the strand on the 3'→5' template is made continuously and the strand on the 5'→3' template is made discontinuously (Okazaki fragments, joined by ligase).
Where the marks are
- States that replication is energetically very expensive, hence the fork (1 mark)
- Polymerase acts only 5'→3' (½ mark)
- Continuous synthesis on the 3'→5' template (½ mark)
- Discontinuous synthesis on the 5'→3' template (½ mark)
- Correctly labelled schematic with polarity shown (½ mark; no marks if polarity is wrong)
Open Q26 on its own page
Q27 · 3 marks · Short answerMolecular Basis of Inheritance
Explain the processing of heterogeneous nuclear RNA (hnRNA) into a fully functional mRNA in eukaryotes. Where does this processing occur in the cell ?
Try it first, then show our solutionSolution
1. The primary transcript (hnRNA) made by RNA polymerase II contains both coding exons and non-coding introns. It is non-functional until processed.
2. Capping: an unusual nucleotide, methyl guanosine triphosphate (m⁷Gppp), is added to the 5'-end of hnRNA.
  ½ mark
3. Tailing: adenylate residues (200–300) are added to the 3'-end in a template-independent manner to form the poly-A tail.
  ½ mark
4. Splicing: the introns are removed.
  ½ mark
5. The exons are then joined together in a defined order.
  ½ mark
6. The fully processed hnRNA is the mature, functional mRNA, which is transported to the cytoplasm for translation.
  ½ mark
7. Site: this processing occurs in the nucleus.
  ½ mark
Answer: hnRNA undergoes 5' capping (m⁷Gppp), 3' tailing (poly-A) and splicing (introns removed, exons joined) to form mRNA; this takes place in the nucleus.
Where the marks are
- Capping at the 5' end with methyl guanosine triphosphate (½ mark)
- Tailing at the 3' end with poly-A (adenylate residues) (½ mark)
- Splicing: removal of introns (½ mark)
- Joining of exons in a defined order (½ mark)
- Formation of functional mRNA (½ mark)
- Site of processing: nucleus (½ mark)
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Q28 · 3 marks · Short answerBiodiversity and Conservation
The world is facing accelerated rates of species extinction largely due to human activities. Explain any three human activities responsible for accelerated rates of species extinction.
Try it first, then show our solutionSolution
1. 1. Habitat loss and fragmentation – this is the most important cause of extinction.
  ½ mark
2. For example, tropical rain forests such as the Amazon are cleared for soybean cultivation and cattle pastures. Large habitats get broken into fragments, so animals needing large territories and migratory animals decline.
  ½ mark
3. 2. Over-exploitation – humans take natural resources at a rate faster than they can be replenished.
  ½ mark
4. For example, Steller's sea cow and the passenger pigeon became extinct due to over-hunting, and many marine fish populations are over-harvested.
  ½ mark
5. 3. Alien species invasion – introduced species can drive indigenous species towards extinction.
  ½ mark
6. For example, the Nile perch introduced into Lake Victoria caused the extinction of more than 200 species of cichlid fish. Similarly, the illegally introduced African catfish Clarias gariepinus threatens native catfishes. (Co-extinction or pollution, with an example, are also acceptable.)
  ½ mark
Answer: Any three: habitat loss and fragmentation (Amazon forest clearing), over-exploitation (Steller's sea cow, passenger pigeon), alien species invasion (Nile perch in Lake Victoria); also co-extinction or pollution.
Where the marks are
- Each of three human activities named (½ mark each)
- A correct explanation or example for each activity (½ mark each)
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Sl. No.	Organism	Bioactive Molecule	Use
1.	Monascus purpureus	a	b
2.	c	d	Antibiotic
3.	e	Cyclosporin A	f

Section D

case-based

Q29 · 4 marks · Case-basedCompetency-basedHuman Reproduction
In a human female, the reproductive phase starts on the onset of puberty and ceases around middle age of the female. Study the graph given below regarding menstrual cycle and answer the questions that follow :
1. (a)
  Name the hormones and their source organ, which are responsible for menstrual cycle at puberty.
  1 mark
2. (b)
  For successful pregnancy, at what phase of the menstrual cycle an early embryo (upto 3 blastomeres) should be Implanted in the Uterus (IUT) of a human female who has opted for Assisted Reproductive Technology (ART) ? Support your answer with a reason.
  1 mark
3. (c)
  Name the hormone and its source organ responsible for the events occurring during proliferative phase of menstrual cycle. Explain the event.
  2 marks
4. OR
  (c)
  In a normal human female, why does menstruation only occurs if the released ovum is not fertilised ? Explain.
  2 marks
Figure in the paper: Graph of the 28-day menstrual cycle with four panels: pituitary hormone levels (FSH, LH with LH surge around day 14), ovarian events (developing follicle, mature follicle, ovulation at day ~14, developing corpus luteum, regressing corpus luteum), ovarian hormone levels (estrogen peaks before ovulation, progesterone peaks in luteal phase), and uterine events (menses, endometrium thickening). Days 1-29; phases: Menstruation, Follicular phase (Proliferative phase), Luteal phase (Secretory phase), Next cycle begins. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) FSH and LH – source: anterior pituitary gland.
  ½ mark
2. (a) Estrogen – source: ovary (Graafian follicle); progesterone – source: ovary (corpus luteum). Any two hormones with their correct sources earn the full mark.
  ½ mark
3. (b) The embryo should be transferred during the luteal (secretory) phase.
  ½ mark
4. (b) Reason: under progesterone, the endometrium is thick, well developed and vascular, so it is ready for implantation of the early embryo.
  ½ mark
5. (c) [First alternative] Hormone: estrogen; source: ovary (growing or Graafian follicles).
  1 mark
6. (c) [First alternative] Event: rising estrogen causes the endometrium of the uterus to regenerate through proliferation after menstruation.
  1 mark
7. (c) [OR alternative] If the ovum is not fertilised, the corpus luteum degenerates.
  ½ mark
8. (c) [OR alternative] As a result, the progesterone level falls.
  ½ mark
9. (c) [OR alternative] Without progesterone, the endometrium and its blood vessels disintegrate.
  ½ mark
10. (c) [OR alternative] The disintegrated endometrium is shed as the menstrual flow. If fertilisation occurs, the corpus luteum is maintained, so menstruation does not occur.
  ½ mark
Answer: (a) FSH and LH (pituitary); estrogen and progesterone (ovary). (b) Luteal/secretory phase, because the endometrium is thick and ready for implantation. (c) Estrogen from ovarian follicles, which regenerates the endometrium by proliferation; OR without fertilisation the corpus luteum degenerates, progesterone falls and the endometrium breaks down as menstrual flow.
Where the marks are
- (a) Any two hormones with correct source organs (½ + ½)
- (b) Luteal/secretory phase (½) with the reason that the endometrium is ready for implantation (½)
- (c) Estrogen and ovary (½ + ½), and explains regeneration or proliferation of the endometrium (1)
- (c) OR: corpus luteum degenerates, progesterone falls, endometrium disintegrates, menstrual flow (½ × 4)
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Q30 · 4 marks · Case-basedCompetency-basedHuman Health and Disease
Read the following passage and answer the questions that follow :
"Mosquitoes are drastically affecting the human health in almost all the developing tropical countries. Different species of mosquitoes cause very fatal diseases so much so that many humans loose their life and if they survive, are unable to put in productive hours to sustain their life. With the result the health index of the country goes down."
1. (a)
  Name the form in which Plasmodium gains entry into (i) human body (ii) the female Anopheles body.
  1 mark
2. (b)
  Why do the symptoms of malaria not appear in a person immediately after being bitten by an infected female Anopheles ? Give one reason. Explain when and how do the symptoms of the disease would appear.
  2 marks
3. OR
  (b)
  Explain the events which occur within a female Anopheles mosquito after it has sucked blood from a malaria patient.
  2 marks
4. (c)
  Name a species of mosquito other than female Anopheles and the disease, for which it carries the pathogen.
  1 mark
Try it first, then show our solutionSolution
1. (a) (i) Plasmodium enters the human body as sporozoites, injected with the saliva of an infected female Anopheles.
  ½ mark
2. (a) (ii) It enters the female Anopheles as gametocytes, taken in with the blood meal.
  ½ mark
3. (b) [First alternative] Symptoms do not appear immediately because the sporozoites first multiply asexually in the liver cells and then in the RBCs. This takes time (the incubation period).
  1 mark
4. (b) [First alternative] When the RBCs rupture, they release a toxic substance, haemozoin.
  ½ mark
5. (b) [First alternative] Haemozoin is responsible for the chill and high fever recurring every 3–4 days.
  ½ mark
6. (b) [OR alternative] The male and female gametocytes in the blood meal enter the mosquito.
  ½ mark
7. (b) [OR alternative] Fertilisation takes place in the gut (stomach) of the mosquito.
  ½ mark
8. (b) [OR alternative] The parasites develop further, and sporozoites are formed that escape from the gut.
  ½ mark
9. (b) [OR alternative] The sporozoites migrate to the salivary glands, ready to infect the next human bitten.
  ½ mark
10. (c) Mosquito species: Aedes (or Culex).
  ½ mark
11. (c) Disease: dengue or chikungunya (or filariasis/elephantiasis for Culex).
  ½ mark
Answer: (a) (i) Sporozoites; (ii) gametocytes. (b) Asexual multiplication in the liver and RBCs takes time; RBC rupture releases haemozoin, causing chill and high fever every 3–4 days. OR: gametocytes fertilise in the mosquito gut, sporozoites form and migrate to the salivary glands. (c) Aedes – dengue/chikungunya (or Culex – filariasis).
Where the marks are
- (a) Sporozoites (½) and gametocytes (½)
- (b) Reason: asexual multiplication in liver and RBCs first (1); RBC rupture releases haemozoin (½); this causes chill and high fever every 3–4 days (½)
- (b) OR: gametocytes taken in, fertilisation in gut, sporozoites formed and escape, migrate to salivary glands (½ × 4)
- (c) Correct mosquito (½) with its disease (½)
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Section E

long answer

Answer one of these 2 alternatives.
Q31(a) · 5 marks · Long answerBiotechnology: Principles and Processes
(i) Draw a schematic diagram of the cloning vector pBR 322 and label (1) Bam HI site (2) gene for ampicillin resistance (3) 'ori' (4) 'rop' gene.
(ii) State the role of 'rop' gene.
(iii) A cloning vector does not have a selectable marker. How will it affect the process of cloning ?
(iv) Why is insertional inactivation preferred over the use of selectable markers in cloning vectors ?
Try it first, then show our solutionSolution
1. (i) Diagram (described): Draw a circular double-stranded plasmid labelled pBR322 (4361 bp). Label (1) the BamHI site within the tet^R gene; (2) the amp^R gene (ampicillin resistance) as a separate segment; (3) ori, the origin of replication; and (4) the rop gene placed next to ori. Other sites such as EcoRI, HindIII, PstI, PvuI, PvuII and ClaI may also be shown.
  2 marks
2. (ii) The rop gene codes for the proteins involved in the replication of the plasmid, which control its copy number.
  1 mark
3. (iii) Without a selectable marker it would not be possible to differentiate transformants from non-transformants (or recombinants from non-recombinants). Identifying cells carrying the vector would become very difficult.
  1 mark
4. (iv) Selection by antibiotic markers requires simultaneous plating on two plates with different antibiotics, which is cumbersome. Insertional inactivation, e.g. inserting the DNA into the lacZ gene coding for β-galactosidase, lets recombinants be identified in one step by colour (blue–white screening): recombinant colonies are white and non-recombinant colonies are blue in the presence of a chromogenic substrate.
  1 mark
Answer: (i) pBR322 with BamHI (in tet^R), amp^R, ori and rop labelled. (ii) rop codes for proteins for plasmid replication. (iii) Transformants cannot be distinguished from non-transformants. (iv) Insertional inactivation gives easy one-step colour (blue–white) selection instead of cumbersome plating on two antibiotic plates.
Where the marks are
- Four correct labels on the pBR322 diagram (½ mark each)
- rop codes for proteins involved in plasmid replication (1 mark)
- Transformants or recombinants cannot be identified without a marker (1 mark)
- Insertional inactivation allows simple one-step blue–white selection (1 mark; the scheme awards it if attempted)
Open Q31(a) on its own page
OR
Q31(b) · 5 marks · Long answerBiotechnology and its Applications
(i) Name the nematode (scientific name) that infects the roots of tobacco plant and reduces its yield.
(ii) Name the vector that is used to introduce nematode-specific genes into the host plant (tobacco).
(iii) How do sense and anti-sense RNAs function ?
(iv) Why could parasite not survive in a transgenic tobacco plant ?
Try it first, then show our solutionSolution
1. (i) The nematode is Meloidogyne incognita.
  1 mark
2. (ii) The vector is Agrobacterium tumefaciens (its Ti plasmid).
  1 mark
3. (iii) Nematode-specific genes are introduced so that the plant produces both sense and anti-sense RNA. The two are complementary to each other.
  1 mark
4. (iii) Being complementary, they pair to form double-stranded RNA (dsRNA), which initiates RNA interference (RNAi).
  1 mark
5. (iv) The dsRNA binds to the specific mRNA of the nematode and silences it.
  ½ mark
6. (iv) This prevents translation of that mRNA, so the parasite cannot survive in the transgenic host.
  ½ mark
Answer: (i) Meloidogyne incognita. (ii) Agrobacterium tumefaciens (Ti plasmid). (iii) Sense and anti-sense RNA are complementary and form dsRNA, which starts RNAi. (iv) The dsRNA silences the nematode's specific mRNA, preventing its translation, so the parasite dies.
Where the marks are
- Meloidogyne incognita (1 mark)
- Agrobacterium tumefaciens (1 mark)
- Sense and anti-sense RNA are complementary (1 mark) and form dsRNA (1 mark)
- dsRNA binds the specific mRNA (½ mark) and prevents its translation, silencing it (½ mark)
Open Q31(b) on its own page
Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerHuman Reproduction
(i) Draw a diagram of a human sperm. Label any four parts and write their functions. (4)
(ii) In a human female, probability of an ovum to get fertilized by more than one sperm is impossible. Give reason. (1)
Try it first, then show our solutionSolution
1. (i) Diagram (described): Draw an elongated sperm with, in order, a head (an oval nucleus capped anteriorly by an acrosome), a neck, a middle piece packed with spiral mitochondria, and a long tail, all enclosed by the plasma membrane.
2. Acrosome – a cap-like structure filled with enzymes that help the sperm penetrate the ovum during fertilisation.
  1 mark
3. Head / nucleus – contains the haploid chromosomal (genetic) material.
  1 mark
4. Middle piece – contains numerous mitochondria, which produce energy for the movement of the tail.
  1 mark
5. Tail – provides motility, letting the sperm swim towards the ovum. (The plasma membrane, which envelops and protects the sperm, is also acceptable.)
  1 mark
6. (ii) When a sperm comes in contact with the zona pellucida of the ovum, it induces changes in the membrane.
  ½ mark
7. These changes block the entry of any additional sperm, so fertilisation by more than one sperm (polyspermy) is prevented.
  ½ mark
Answer: (i) Sperm diagram labelled with acrosome (enzymes for penetration), head/nucleus (genetic material), middle piece (mitochondria for energy) and tail (motility). (ii) Contact of a sperm with the zona pellucida induces membrane changes that block entry of other sperms.
Where the marks are
- Four correct labels on the diagram (½ mark each)
- A correct function for each labelled part (½ mark each)
- Sperm contact with the zona pellucida induces membrane changes (½ mark)
- These changes block entry of additional sperms (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerSexual Reproduction in Flowering Plants
(i) With the help of labelled diagram only, show the different stages of embryo development in a dicot plant. (4)
(ii) Endosperm development precedes embryo development. Justify. (1)
Try it first, then show our solutionSolution
1. (i) Diagram: stages of embryo development in a dicot (e.g. Capsella). Draw five figures in sequence, left to right, joined by arrows, with a heading for each stage (½ mark per correct label, 8 labels):
  1. Zygote: a single cell at the micropylar end of the embryo sac, surrounded by endosperm nuclei.
  2. Proembryo (2-celled and 4-celled): the zygote divides into a terminal (apical) cell and a basal cell. The basal cell forms a row of cells, the suspensor.
  3. Globular embryo: a spherical mass of cells borne at the tip of the suspensor.
  4. Heart-shaped embryo: two lobes, the developing cotyledons, appear on the suspensor.
  5. Mature embryo: shows two cotyledons, the plumule (stem tip) between them, the embryonal axis with hypocotyl below the cotyledons, the radicle (root tip) at the lower end, and the remains of the suspensor.
  4 marks
2. (ii) The primary endosperm nucleus divides before the zygote does, so endosperm forms first. Endosperm is a nutritive tissue that stores food. The developing embryo absorbs this food as it grows, so the embryo can develop only after the endosperm is ready.
  1 mark
Answer: (i) Labelled diagrams: zygote → proembryo (suspensor) → globular → heart-shaped → mature embryo (plumule, cotyledons, hypocotyl, radicle). (ii) Endosperm forms first because it provides nutrition to the developing embryo.
Where the marks are
- Any eight correct labels on the diagrams of embryo stages, such as zygote, proembryo, suspensor, globular embryo, heart-shaped embryo, cotyledons, plumule, hypocotyl, radicle, mature embryo (½ × 8 = 4 marks)
- States that endosperm develops first to supply nutrition to the developing embryo (1 mark)
Open Q32(b) on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerCompetency-basedEvolution
Natural selection operates in different ways in nature.
(i) Identify the type of natural selection depicted in the graph above. (1)
(ii) In England after industrialisation, the population of dark winged moths were more favoured than white winged moth. Explain. (2)
(iii) Anthropogenic action can enhance the rate of evolution. Explain with the help of an example. (2)
Figure in the paper: Graph of number of individuals vs phenotype/trait showing two overlapping bell curves: left curve labelled 'White Winged moth' and right curve labelled 'Dark Winged moth', the two peaks distinct (disruptive/directional shift of distribution). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) The graph shows directional selection. Individuals at one extreme of the trait (dark-winged moths) are favoured, so the population peak shifts towards that extreme.
  1 mark
2. (ii) Before industrialisation, tree trunks were covered with pale lichens. White-winged moths were camouflaged against them, while dark moths were easily seen and eaten by predators (birds).
  ½ mark
3. After industrialisation, smoke and soot darkened the tree trunks. Lichens, being sensitive to pollution, could not grow on the trunks.
  ½ mark
4. White moths now stood out against the dark bark, so predators picked them off and their numbers fell.
  ½ mark
5. Dark (melanised) moths were camouflaged and survived to reproduce more. Natural selection therefore increased their proportion in the population.
  ½ mark
6. (iii) Human activities create strong new selection pressures, so favoured variants rise in number within a few generations instead of over geological time.
  1 mark
7. Example: heavy use of herbicides, pesticides, antibiotics or drugs kills the susceptible individuals. The few resistant weeds, insects, microbes or cancer cells survive and multiply, so resistant varieties become common within months or years. Industrial melanism in moths is a similar example.
  1 mark
Answer: (i) Directional selection; (ii) soot-darkened, lichen-free trunks camouflaged dark moths, while white moths were eaten by predators; (iii) e.g. excessive use of pesticides or antibiotics selects resistant organisms in a short time.
Where the marks are
- Identifies directional selection (1 mark)
- Smoke and soot darkened the tree trunks and lichens could not grow (½ mark)
- White moths were no longer camouflaged and were eaten by predators (½ mark)
- Their population decreased (½ mark)
- Dark moths were camouflaged and selected, so their population increased (½ mark)
- Explains that anthropogenic action speeds up selection of favoured or resistant forms (1 mark)
- Gives a correct example, such as pesticide, herbicide or antibiotic resistance, or industrial melanism (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerMolecular Basis of Inheritance
(i) Why did Hershey and Chase use ' $^{35}$ S' and ' $^{32}$ P' in their experiment ? Explain. (1)
(ii) State the importance of (1) blending and (2) centrifugation in their experiment. (2)
(iii) Write the conclusion they arrived at the end of their experiment. (2)
Try it first, then show our solutionSolution
1. (i) Sulphur is present in protein but not in DNA, so $^{35}$ S labelled the viral protein coat. Phosphorus is present in DNA but not in protein, so $^{32}$ P labelled the viral DNA.
  ½ mark
2. Labelling each molecule separately let them trace which one, protein or DNA, entered the bacteria (E. coli) on infection.
  ½ mark
3. (ii)(1) Blending: shaking (agitating) the culture in a blender detached the empty viral coats from the surface of the bacteria after infection.
  1 mark
4. (ii)(2) Centrifugation: spinning separated the heavier bacteria, which formed a pellet, from the lighter free viral coats, which stayed in the supernatant. The radioactivity of each fraction could then be checked.
  1 mark
5. (iii) Observation: bacteria infected with $^{32}$ P-labelled phages were radioactive, while the $^{35}$ S label stayed in the supernatant and not in the bacteria. So only the viral DNA entered the bacteria.
  1 mark
6. Conclusion: DNA, not protein, is the genetic material that passes from the virus to the bacterium.
  1 mark
Answer: (i) $^{35}$ S labels protein and $^{32}$ P labels DNA, to trace which enters the bacteria; (ii) blending removes viral coats from bacteria, and centrifugation separates bacteria from the viral coats; (iii) DNA is the genetic material.
Where the marks are
- States that $^{35}$ S labels protein (it is absent from DNA) and $^{32}$ P labels DNA (it is absent from protein) (½ mark)
- Explains that the labels trace which molecule passes into the bacteria (½ mark)
- Blending detaches viral coats from the bacteria (1 mark)
- Centrifugation separates bacteria (pellet) from viral particles or coats (supernatant) (1 mark)
- Concludes that DNA is the genetic material, supported by the radioactivity observation (2 marks)
Open Q33(b) on its own page

Source: official question paper, cbse.gov.in. Chapter tags and solutions are ours. Spotted a transcription error? Write tosupport@questionbuddy.in.