BiologyCBSE Class 12 · Board exam 2023 (session 2022-23)

CBSE Class 12 Biology: 2023 board paper, set 57/5/1

The question paper set in the CBSE Class 12 board exam of 2023, 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/2023/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 1 question in Section B, 1 in C, 2 in D and 1 in E.

Section A

MCQ 1-12, assertion-reason 13-16

Q1 · 1 mark · MCQEvolution
At which stage during evolution did human use hides to protect their bodies and buried their dead ?
1. (A)Homo habilis
2. (B)Neanderthal man
3. (C)Java man
4. (D)Homo erectus
Try it first, then show our solutionSolution
1. Neanderthal man (1400 cc brain, lived about 1,00,000–40,000 years ago in the Near East and Central Asia) used hides to protect the body and buried the dead. Hence option (B).
  1 mark
Answer: (B) Neanderthal man
Where the marks are
- Identifies Neanderthal man (1 mark)

Q2 · 1 mark · MCQReproductive Health

Given below are Column A with a list of certain Assisted Reproductive Technologies (ART) and in Column B the procedures followed during ART :

S. No.	Names of ART	S. No.	Procedures
(A)	GIFT	(i)	Transfer of ovum from a donor into the fallopian tube of another female.
(B)	ICSI	(ii)	Transfer of semen from the donor into the vagina of the female.
(C)	ZIFT	(iii)	Injecting sperms directly into the ovum.
(D)	IUI	(iv)	Transfer of early embryos into the fallopian tube.

Choose the option where ART correctly matches with the procedure.

(A)(A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
(B)(A)-(iv), (B)-(i), (C)-(ii), (D)-(iii)
(C)(A)-(iv), (B)-(iii), (C)-(i), (D)-(ii)
(D)(A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)

Try it first, then show our solutionSolution

GIFT transfers an ovum from a donor into the fallopian tube, so (A)–(i). ICSI injects a sperm directly into the ovum, so (B)–(iii).
½ mark
ZIFT transfers early embryos (up to 8 blastomeres) into the fallopian tube, so (C)–(iv). IUI places donor semen in the female tract, so (D)–(ii). Hence option (D).
½ mark

Answer: (D) (A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)

Where the marks are

Correct matching, option (D) (1 mark)

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Q3 · 1 mark · MCQHuman Health and Disease
The decrease in the T-lymphocytes count in human blood will result in :
1. (A)Decrease in antigens
2. (B)Decrease in antibodies
3. (C)Increase in antibodies
4. (D)Increase in antigens
Try it first, then show our solutionSolution
1. T-lymphocytes help B-cells to produce antibodies and also kill pathogens directly. If T-cells decrease, pathogens are not removed, so antigens accumulate in the body. Hence option (D).
  1 mark
Answer: (D) Increase in antigens
Where the marks are
- Identifies increase in antigens (1 mark)
Q4 · 1 mark · MCQMolecular Basis of Inheritance
Given below is a sequence of bases in mRNA of a bacterial cell. Identify the amino acid that would be incorporated at codon position 3 and codon position 5 during the process of its translation.
3' AUCAGGUUUGUGAUGGUACGA 5'
1. (A)Phenylalanine, Methionine
2. (B)Cysteine, Glycine
3. (C)Alanine, Proline
4. (D)Serine, Valine
Try it first, then show our solutionSolution
1. Read the mRNA 5′→3′: AGC AUG GUA GUG UUU GGA CUA. Codon 3 is GUA (Valine) and codon 5 is UUU (Phenylalanine).
  ½ mark
2. Read as written instead: AUC AGG UUU GUG AUG … gives codon 3 = UUU (Phe) and codon 5 = AUG (Met), which is option (A). Translation is always 5′→3′, so no option is strictly correct. The board awards the mark for any attempt.
  ½ mark
Answer: No correct option; mark awarded for attempting
Where the marks are
- Award 1 mark for attempting the question (1 mark)
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Q5 · 1 mark · MCQOrganisms and Populations
A Tight one-to-one relationship between many species of fig tree and certain wasps is an example of -
1. (A)Commensalism
2. (B)Parasitism
3. (C)Amensalism
4. (D)Mutualism
Try it first, then show our solutionSolution
1. The fig is pollinated only by its partner wasp. The wasp in turn uses the fig fruit for egg-laying and its larvae feed on the seeds. Both partners benefit, so this is mutualism. Hence option (D).
  1 mark
Answer: (D) Mutualism
Where the marks are
- Identifies mutualism (1 mark)
Q6 · 1 mark · MCQHuman Health and Disease
Select the pathogen mismatched with the symptoms of disease caused by it from the list given below :
1. (A)Entamoeba histolytica : Constipation, abdominal pain.
2. (B)Epidermophyton : Dry scaly lesions on nail.
3. (C)Wuchereria bancrofti : Chronic inflammation of lymphatic vessels of lower limb.
4. (D)Haemophilus influenzae : Blockage of the intestinal passage.
Try it first, then show our solutionSolution
1. Haemophilus influenzae causes pneumonia, which affects the alveoli of the lungs (fever, chills, cough). It does not cause intestinal blockage, which is a symptom of Ascaris. So option (D) is the mismatch.
  1 mark
Answer: (D)
Where the marks are
- Identifies H. influenzae pair as mismatched (1 mark)
Q7 · 1 mark · MCQEcosystem
The primary productivity in an ecosystem is expressed as :
1. (A) $g m^{- 2} y r^{- 1}$
2. (B) $g m^{- 2} yr$
3. (C) $K cal m^{- 2} y r^{- 1}$
4. (D) $K cal m^{- 2}$
Try it first, then show our solutionSolution
1. Primary productivity is the biomass produced per unit area over a time period. It is expressed in $g m^{- 2} y r^{- 1}$ (weight) or $kcal m^{- 2} y r^{- 1}$ (energy).
  ½ mark
2. The board keys option (A), $g m^{- 2} y r^{- 1}$ , the weight unit given in NCERT.
  ½ mark
Answer: (A) $g m^{- 2} y r^{- 1}$
Where the marks are
- Chooses $g m^{- 2} y r^{- 1}$ (1 mark)
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Q8 · 1 mark · MCQBiotechnology: Principles and Processes
Given below is the restriction site of a restriction endonuclease Pst-I and the cleavage sites on a DNA molecule.
5' C-T-G-C-A ↓ G 3'
3' G ↑ A-C-G-T-C 5'
(The arrows mark the cleavage sites: on the upper strand between A and G, on the lower strand between G and A.)
Choose the option that gives the correct resultant fragments by the action of the enzyme Pst-I.
1. (A)5' C-T-G C-A-G 3'
  3' G-A-C-G-T C 5'
2. (B)5' C-T G-C-A-G 3'
  3' G-A-G-C T-C 5'
3. (C)5' C-T-G-C A-G 3'
  3' G-A-C-G T-C 5'
4. (D)5' C-T-G-C-A G 3'
  3' G A-C-G-T- C 5'
Figure in the paper: Restriction site of Pst-I on double-stranded DNA with two arrows showing cleavage positions on the two strands. Options show pairs of resulting fragments. See it in the official paper.
Try it first, then show our solutionSolution
1. Pst-I cuts the upper strand between A and G, giving 5′ CTGCA and G 3′.
  ½ mark
2. It cuts the lower strand between G and A, giving 3′ G and ACGTC 5′. The resulting fragments match option (D).
  ½ mark
Answer: (D)
Where the marks are
- Correct fragments, option (D) (1 mark)
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Q9 · 1 mark · MCQBiodiversity and Conservation
The IUCN Red Data List (2004) in the last 500 years documents the extinction of nearly 784 species including :
1. (A)330 invertebrates
2. (B)338 invertebrates
3. (C)359 invertebrates
4. (D)362 invertebrates
Try it first, then show our solutionSolution
1. The 784 extinctions in the IUCN (2004) list comprise 338 vertebrates, 359 invertebrates and 87 plants. Hence option (C).
  1 mark
Answer: (C) 359 invertebrates
Where the marks are
- 359 invertebrates (1 mark)
Q10 · 1 mark · MCQMicrobes in Human Welfare
Given below are the list of the commercially important products and their source organisms. Select the option that gives the correct matches.
S. No. Bioactive Products S. No. Microbes (Source Organism)
(A) Cyclosporin A (i) Streptococcus
(B) Statins (ii) Tricoderma polysporum
(C) Streptokinase (iii) Penicillium notatum
(D) Penicillin (iv) Monascus purpureus
1. (A)(A)-(i), (B)-(ii), (C)-(iii), (D)-(iv)
2. (B)(A)-(iii), (B)-(iv), (C)-(ii), (D)-(i)
3. (C)(A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)
4. (D)(A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)
Try it first, then show our solutionSolution
1. Cyclosporin A comes from Trichoderma polysporum, so (A)–(ii). Statins come from the yeast Monascus purpureus, so (B)–(iv).
  ½ mark
2. Streptokinase comes from Streptococcus, so (C)–(i). Penicillin comes from Penicillium notatum, so (D)–(iii). Hence option (D).
  ½ mark
Answer: (D)
Where the marks are
- Correct matching, option (D) (1 mark)
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Q11 · 1 mark · MCQOrganisms and Populations
Important attributes belonging to a population but not to an individual are :
(i) Birth rate and death rate
(ii) Male and female
(iii) Birth and death
(iv) Sex-ratio
Select the correct option from the given options :
1. (A)(i) only
2. (B)(ii) only
3. (C)(ii) and (iii)
4. (D)(i) and (iv)
Try it first, then show our solutionSolution
1. An individual is born and dies and is male or female, but rates and ratios describe only a group. So birth rate, death rate and sex ratio are population attributes, i.e. (i) and (iv). Hence option (D).
  1 mark
Answer: (D) (i) and (iv)
Where the marks are
- Chooses (i) and (iv) (1 mark)
Q12 · 1 mark · MCQSexual Reproduction in Flowering Plants
Select the option that shows the correctly identified 'U', 'X', 'Y' and 'Z' in a developing dicot embryo.
1. (A)X – Plumule (2n), Y – Suspensor (n), Z – Cotyledon (2n), U – Radicle (2n).
2. (B)X – Plumule (2n), Y – Suspensor (2n), Z – Radicle (2n), U – Cotyledon (2n).
3. (C)X – Suspensor (2n), Y – Cotyledon (2n), Z – Radicle (2n), U – Plumule (2n).
4. (D)X – Cotyledon (2n), Y – Radicle (n), Z – Plumule (n), U – Suspensor (n).
Figure in the paper: Photograph of a developing dicot embryo (heart/curved stage): a small rounded structure at the top labelled X connected by a thin stalk to the long main body labelled Z; a curved tip on the right labelled Y; a point at the bottom left labelled U. See it in the official paper.
Try it first, then show our solutionSolution
1. In a dicot embryo, the stalk-like structure attaching the embryo is the suspensor (X). The curved lobes are the cotyledons (Y), the radicle lies at the micropylar end (Z) and the plumule is the shoot apex (U).
  ½ mark
2. All embryo parts develop from the zygote, so all are diploid (2n). This rules out the options with n. Hence option (C).
  ½ mark
Answer: (C)
Where the marks are
- Correct identification, option (C) (1 mark)
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Q13 · 1 mark · Assertion-reasonCompetency-basedReproductive Health
Assertion (A) : Determining the sex of an unborn child followed by MTP is an illegal practice.
Reason (R) : Amniocentesis is a practice to test the presence of genetic disorders also.
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. Sex determination of the foetus followed by MTP is banned by law to prevent female foeticide, so (A) is true.
  ½ mark
2. Amniocentesis does detect genetic disorders, so (R) is true. However, it does not explain why the practice is illegal. Hence option (B).
  ½ mark
Answer: (B)
Where the marks are
- Both true, R not the explanation (1 mark)
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Q14 · 1 mark · Assertion-reasonCompetency-basedBiotechnology: Principles and Processes
Assertion (A) : Synthetic oligonucleotide polymers are used during Annealing in a PCR.
Reason (R) : The primers bind to the double stranded DNA at their complementary regions.
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 primers used in annealing are synthetic oligonucleotides, so (A) is true.
  ½ mark
2. Primers bind to single-stranded, denatured DNA, not to double-stranded DNA, so (R) is false. Hence option (C).
  ½ mark
Answer: (C)
Where the marks are
- A true, R false (1 mark)
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Q15 · 1 mark · Assertion-reasonCompetency-basedEcosystem
Assertion (A) : Decomposition process is slower if detritus is rich in lignin and cutin.
Reason (R) : Decomposition is largely an oxygen requiring process.
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. Lignin and chitin-rich detritus decomposes slowly, so (A) is true. Decomposition is also largely an oxygen-requiring process, so (R) is true.
  ½ mark
2. The slow rate in (A) is due to the chemical nature of the detritus, not the need for oxygen. So (R) does not explain (A). Hence option (B).
  ½ mark
Answer: (B)
Where the marks are
- Both true, R not explanation (1 mark)
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Q16 · 1 mark · Assertion-reasonCompetency-basedPrinciples of Inheritance and Variation
Assertion (A) : In Thalassemia an abnormal myoglobin chain is synthesized due to a gene defect.
Reason (R) : α-Thalassemia is controlled by genes HBA1 and HBA2 on chromosome 16.
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. Thalassemia is a quantitative defect, i.e. too few globin chains of haemoglobin are made. It does not involve an abnormal myoglobin chain, so (A) is false.
  ½ mark
2. α-thalassemia is controlled by the genes HBA1 and HBA2 on chromosome 16, so (R) is true. Hence option (D).
  ½ mark
Answer: (D)
Where the marks are
- A false, R true (1 mark)
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S. No.	Bioactive Products	S. No.	Microbes (Source Organism)
(A)	Cyclosporin A	(i)	Streptococcus
(B)	Statins	(ii)	Tricoderma polysporum
(C)	Streptokinase	(iii)	Penicillium notatum
(D)	Penicillin	(iv)	Monascus purpureus

Section B

Very short answer

Q17 · 2 marks · Very short answerCompetency-basedHuman Reproduction
The graph given below shows the number of primordial follicles per ovary in women at different ages. Study the graph and answer the questions that follow.
1. (a)
  What is the average age of the women at the onset of menopause ?
  1 mark
2. (b)
  At what age are maximum primordial follicles present in the ovary, according to the given graph ?
  1 mark
Figure in the paper: Scatter graph: x-axis Age (y) from 0 to 60 (marks at 10,20,30,40,50,60); y-axis Primordial follicles/ovary on a log scale (0,1,10,100,1000,10,000,100,000). Filled squares (premenopausal women, regular menses) cluster near 100,000 at young ages and decline with age; open squares (perimenopausal women, irregular menses for at least one year) lie around ages 40-55 at lower values; triangles (post menopausal women, no menses for at least one year) lie near 0-1 at ages about 45-55. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) Read the age where the open squares (perimenopausal women) give way to the triangles (postmenopausal women, follicle count near 0–1). This happens at about 47 years, so the average age at the onset of menopause is about 47 years (roughly 45–49 years).
  1 mark
2. (b) The filled squares reach their highest point, about 100,000 primordial follicles per ovary, in early childhood at about 6–8 years of age. After that the count falls steadily with age.
  1 mark
Answer: (a) About 47 years (45–49 years); (b) About 6–8 years of age
Where the marks are
- States the average age at onset of menopause as about 47 years; any value from 45 to 49 is accepted (1 mark)
- States the age of maximum primordial follicles as about 6–8 years (1 mark)
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Q18 · 2 marks · Very short answerOrganisms and Populations
“Cattle and goats do not browse the Calotropis plant.” Justify the statement giving reasons.
Try it first, then show our solutionSolution
1. Calotropis produces highly poisonous cardiac glycosides, which are chemicals that affect the heart.
  1 mark
2. This is a chemical defence against herbivores. Cattle and goats that ate the plant could be poisoned or killed, so they have learnt to avoid browsing it.
  1 mark
Answer: Calotropis produces highly poisonous cardiac glycosides (a chemical defence against herbivores), so cattle and goats avoid it.
Where the marks are
- Names the chemical produced as highly poisonous cardiac glycosides (1 mark)
- Explains this as a defence against herbivores: grazing animals avoid the plant because it is toxic (1 mark)
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Q19 · 2 marks · Very short answerPrinciples of Inheritance and Variation
By using Punnett square depict the genotypes and phenotypes of test crosses (where green pod colour (G) is dominant over yellow pod colour (g)) in Garden pea with unknown genotype.
Try it first, then show our solutionSolution
1. In a test cross, the green-podded plant of unknown genotype is crossed with the homozygous recessive yellow-podded plant (gg).
  Case I – unknown plant is homozygous dominant: GG (green) × gg (yellow).
2. Punnett square for Case I:
  G G
  g Gg Gg
  g Gg Gg
  Genotype: all Gg.
  ½ mark
3. Phenotype for Case I: all (100%) green pods.
  ½ mark
4. Case II – unknown plant is heterozygous: Gg (green) × gg (yellow).
  G g
  g Gg gg
  g Gg gg
  Genotype: Gg : gg = 1 : 1.
  ½ mark
5. Phenotype for Case II: 50% green : 50% yellow pods (1 : 1).
  ½ mark
Answer: GG × gg → all Gg, all green pods; Gg × gg → 1 Gg : 1 gg, 50% green : 50% yellow pods
Where the marks are
- Punnett square for GG × gg giving all Gg (½ mark)
- States the phenotype as all green pods (½ mark)
- Punnett square for Gg × gg giving Gg : gg in a 1 : 1 ratio (½ mark)
- States the phenotype as 50% green : 50% yellow pods (½ mark)
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Answer one of these 2 alternatives.
Q20(a) · 2 marks · Very short answerMicrobes in Human Welfare
Answer the following.
1. (i)
  Give an example of a genus of virus used as narrow spectrum insecticidal biocontrol agent.
  0 marks
2. (ii)
  How does its use serve as an aid in overall integrated pest management programme ?
  0 marks
Try it first, then show our solutionSolution
1. (i) Nucleopolyhedrovirus (a baculovirus) is used as a narrow-spectrum insecticidal biocontrol agent.
  1 mark
2. (ii) It is species-specific, with narrow-spectrum action. It attacks only the target insect pest and has no harmful effect on plants, mammals, birds, fish or non-target insects.
  ½ mark
3. Because beneficial insects are spared, it is ideal for integrated pest management, especially in ecologically sensitive areas.
  ½ mark
Answer: (i) Nucleopolyhedrovirus (Baculovirus); (ii) It is species-specific, harms no non-target organisms and conserves beneficial insects, so it suits IPM in ecologically sensitive areas.
Where the marks are
- Names Nucleopolyhedrovirus as the genus (1 mark)
- States it is species-specific / narrow-spectrum in action (½ mark)
- States it does not harm non-target organisms, conserves beneficial insects, or suits ecologically sensitive areas (½ mark)
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OR
Q20(b) · 2 marks · Very short answerHuman Health and Disease
Why a malignant tumour considered to be more damaging than a benign tumour ? Explain.
Try it first, then show our solutionSolution
1. A benign tumour stays confined to its original site and causes little damage. Malignant tumour cells, in contrast, grow very rapidly, compete with normal cells for vital nutrients and starve them. They also invade and damage the surrounding normal tissues.
  1 mark
2. Malignant cells break away, travel through the blood and set up new tumours in distant parts of the body. This is called metastasis, and it is why malignant tumours are far more dangerous.
  1 mark
Answer: A malignant tumour grows rapidly, starves and invades normal tissues, and spreads to other organs by metastasis, whereas a benign tumour stays localised.
Where the marks are
- Rapid growth that starves normal cells by competing for nutrients, or invasion and damage of surrounding tissue (1 mark)
- Metastasis: spread to distant body parts through blood (1 mark)
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Q21 · 2 marks · Very short answerBiotechnology: Principles and Processes
Answer the following.
1. (a)
  Write the scientific name of the source organism of the themostable DNA polymerase used in PCR.
  0 marks
2. (b)
  State the advantage of using Thermostable DNA polymerase.
  0 marks
Try it first, then show our solutionSolution
1. (a) The source organism is the bacterium Thermus aquaticus, and the enzyme is called Taq polymerase.
  1 mark
2. (b) Thermostable DNA polymerase remains active at the high temperature (about 94 °C) used to denature double-stranded DNA in each PCR cycle. So it need not be added again after every denaturation step, and the cycles can run repeatedly.
  1 mark
Answer: (a) Thermus aquaticus; (b) It stays active at high denaturation temperatures, so it need not be added afresh in every cycle.
Where the marks are
- Writes the scientific name Thermus aquaticus (1 mark)
- States it remains active during high-temperature denaturation, so no re-addition is needed each cycle (1 mark)
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	G	G
g	Gg	Gg
g	Gg	Gg

	G	g
g	Gg	gg
g	Gg	gg

Section C

Short answer

Q22 · 3 marks · Short answerReproductive Health
Name and explain a surgical contraceptive method that can be adopted by the male partner of a couple.
Try it first, then show our solutionSolution
1. The surgical contraceptive method (sterilisation) for males is vasectomy.
  1 mark
2. In vasectomy, a small part of the vas deferens is cut or removed through a small incision on the scrotum.
  1 mark
3. The cut ends are tied up, which blocks the transport of sperms. The semen therefore contains no sperms, and conception is prevented.
  1 mark
Answer: Vasectomy: a small part of the vas deferens is removed or cut and tied, blocking sperm transport.
Where the marks are
- Names vasectomy (1 mark)
- States that a small part of the vas deferens is cut or removed (1 mark)
- States that the ends are tied, blocking sperm transport (1 mark)
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Q23 · 3 marks · Short answerBiotechnology and its Applications
Human Genome Project (HGP) was a mega project launched in the year 1990 with some important goals.
1. (a)
  Enlist any four prime goals of HGP.
  0 marks
2. (b)
  Name any one common non-human animal model organism which has also been sequenced thereafter.
  0 marks
Try it first, then show our solutionSolution
1. (a) Goal 1: Identify all the approximately 20,000–25,000 genes in human DNA.
  ½ mark
2. Goal 2: Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
  ½ mark
3. Goal 3: Store this information in databases.
  ½ mark
4. Goal 4: Improve tools for data analysis and transfer the related technologies to other sectors, such as industries. Addressing the ethical, legal and social issues (ELSI) arising from the project is also a goal.
  ½ mark
5. (b) The nematode Caenorhabditis elegans has also been sequenced; Drosophila is another example.
  1 mark
Answer: (a) Identify all 20,000–25,000 genes; sequence 3 billion base pairs; store data in databases; improve data-analysis tools, transfer technology, address ELSI. (b) Caenorhabditis elegans (or Drosophila)
Where the marks are
- Any four correct HGP goals (½ × 4 = 2 marks)
- Names a non-human model organism such as C. elegans or Drosophila (1 mark)
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Q24 · 3 marks · Short answerEvolution
Industrial melanism in England after 1850 is an excellent example of Natural selection. Explain how ?
Try it first, then show our solutionSolution
1. Before industrialisation (pre-1850): tree trunks and walls were covered with a thick growth of almost white lichens.
  ½ mark
2. White-winged moths were camouflaged against the pale lichens, so they escaped predators and increased in number.
  ½ mark
3. Dark-winged moths were easily seen and picked out by predatory birds, so their numbers stayed low.
  ½ mark
4. After industrialisation: smoke and soot darkened the tree trunks, and the lichens could not grow because of pollution.
  ½ mark
5. Dark-winged moths were now camouflaged, so they survived and increased in number.
  ½ mark
6. White-winged moths became conspicuous and were eaten by predators, so they declined. Thus the environment 'selected' the better-adapted form, which shows natural selection.
  ½ mark
Answer: Before industrialisation, white moths were camouflaged on lichen-covered trunks and dark moths were eaten. After soot darkened the trunks, dark moths were camouflaged and increased while white moths were eaten. This change in frequency is natural selection.
Where the marks are
- White lichen cover on tree trunks before industrialisation (½ mark)
- White moths camouflaged and increased (½ mark)
- Dark moths picked out by predators and decreased (½ mark)
- Trunks darkened by soot and lichens absent after industrialisation (½ mark)
- Dark moths camouflaged and increased (½ mark)
- White moths eaten and decreased; nature selects the fitter form (½ mark)
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Q25 · 3 marks · Short answerSexual Reproduction in Flowering Plants
One of the major approaches of crop improvement programme is Artificial Hybridisation. Explain the steps involved in making sure that only the desired pollen grain pollinate the stigma of a bisexual flower by a plant breeder.
Try it first, then show our solutionSolution
1. Emasculation: the anthers are removed from the flower bud with forceps before they dehisce. This prevents self-pollination of the bisexual flower.
  1 mark
2. Bagging: the emasculated flower is covered with a bag of suitable size, usually butter paper. This protects its stigma from contamination by unwanted pollen.
  1 mark
3. Dusting and rebagging: when the stigma becomes receptive, mature pollen collected from the desired male parent is dusted on it. The flower is then rebagged and left to develop fruit.
  1 mark
Answer: Emasculation (remove anthers before dehiscence), bagging (cover the flower to prevent foreign pollen), then dusting desired pollen on the receptive stigma and rebagging.
Where the marks are
- Emasculation explained: removing anthers before dehiscence (1 mark)
- Bagging explained: covering the flower to prevent unwanted pollen (1 mark)
- Dusting desired pollen on the stigma and rebagging (1 mark)
- Only ½ mark each if emasculation and bagging are merely named
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Answer one of these 2 alternatives.
Q26(a) · 3 marks · Short answerHuman Health and Disease
“Plasmodium protozoan needs both a mosquito and a human host for its continuity.” Explain.
Try it first, then show our solutionSolution
1. When an infected female Anopheles mosquito bites a person, the infectious stage, sporozoites, enters the human blood.
  ½ mark
2. The sporozoites reach the liver cells and multiply there asexually.
  ½ mark
3. The parasites then attack RBCs, multiply asexually and rupture them, releasing haemozoin, which causes chills and high fever. This is the asexual phase in humans.
  ½ mark
4. Some parasites in the RBCs develop into gametocytes, which are taken up by a female Anopheles when it bites an infected person.
  ½ mark
5. Fertilisation of gametes and further development (the sexual phase) take place in the mosquito's gut and produce sporozoites.
  ½ mark
6. The sporozoites are stored in the mosquito's salivary glands and passed to a new human at the next bite. Since the asexual phase occurs in humans and the sexual phase in the mosquito, Plasmodium needs both hosts to complete its life cycle.
  ½ mark
Answer: Asexual multiplication (liver, RBCs) and gametocyte formation occur in humans; fertilisation and sporozoite formation occur in the mosquito, which also transmits the sporozoites. So both hosts are needed.
Where the marks are
- Sporozoites enter humans through the bite of a female Anopheles (½ mark)
- Asexual multiplication in liver cells (½ mark)
- Asexual multiplication in and rupture of RBCs (½ mark)
- Gametocytes formed in RBCs and taken up by the mosquito (½ mark)
- Fertilisation and development into sporozoites in the mosquito gut (sexual phase) (½ mark)
- Sporozoites stored in salivary glands and transmitted, so both hosts are needed (½ mark)
Open Q26(a) on its own page
OR
Q26(b) · 3 marks · Short answerHuman Health and Disease
We all must work towards maintaining good health because ‘health is wealth’. Enlist any six ways of achieving good health.
Try it first, then show our solutionSolution
1. Eat a balanced diet that supplies all nutrients in proper amounts.
  ½ mark
2. Maintain personal hygiene, such as bathing and washing hands.
  ½ mark
3. Do regular exercise or yoga.
  ½ mark
4. Get timely vaccination/immunisation against infectious diseases.
  ½ mark
5. Ensure proper disposal of waste and excreta, and control vectors such as mosquitoes by draining stagnant water.
  ½ mark
6. Consume hygienic food and clean drinking water, and avoid drugs, alcohol and smoking.
  ½ mark
Answer: Balanced diet, personal hygiene, regular exercise/yoga, vaccination, proper waste disposal and vector control, hygienic food and water (and avoiding drugs/alcohol).
Where the marks are
- Any six valid ways of maintaining good health (½ × 6 = 3 marks)
Open Q26(b) on its own page
Q27 · 3 marks · Short answerBiotechnology and its Applications
On spraying Bacillus thuringiensis on an infected cotton crop field the pests are killed by the toxin, however the toxin although produced by the bacteria does not affect it. Explain giving reason.
Try it first, then show our solutionSolution
1. Bacillus thuringiensis produces the Bt toxin as an inactive protoxin, in crystal form. So the toxin does not harm the bacterium itself.
  ½ mark
2. The insect pest ingests the inactive protoxin while feeding on the sprayed crop.
  ½ mark
3. The alkaline pH of the insect's gut solubilises the crystals and converts the protoxin into the active toxin.
  ½ mark
4. The active toxin binds to the surface of the midgut epithelial cells.
  ½ mark
5. It creates pores in these cells.
  ½ mark
6. The cells swell and lyse, and the insect dies.
  ½ mark
Answer: The toxin is made as an inactive protoxin, so the bacterium is safe. The alkaline insect gut activates it, and the active toxin binds midgut epithelial cells, forms pores, and causes swelling, lysis and death of the pest.
Where the marks are
- Toxin produced as an inactive protoxin, so the bacterium is unaffected (½ mark)
- Protoxin ingested by the insect (½ mark)
- Activated by the alkaline pH of the gut (½ mark)
- Binds to midgut epithelial cells (½ mark)
- Creates pores (½ mark)
- Swelling, lysis and death of the insect (½ mark)
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Q28 · 3 marks · Short answerBiodiversity and Conservation
“Biodiversity plays a major role in many ecosystem services that nature provides.”
1. (a)
  Describe any two broadly utilatarian arguments to justify the given statement.
  0 marks
2. (b)
  State one ethical reason of conserving biodiversity.
  0 marks
Try it first, then show our solutionSolution
1. (a) Argument 1: Forests and other biodiversity-rich ecosystems carry out photosynthesis. For example, the Amazon forest is estimated to produce about 20% of the oxygen in Earth's atmosphere, which is a service we cannot pay for.
  1 mark
2. Argument 2: Pollination of crops and other plants by bees, birds and bats depends on biodiversity; we could not get fruits and seeds without it. Biodiversity also mitigates droughts and floods, cycles nutrients, forms fertile soil, stores carbon and gives aesthetic pleasure (woods, bird watching).
  1 mark
3. (b) Ethical argument: every species has an intrinsic value, even if it has no economic use to us. We therefore have a moral duty to care for its well-being and pass on our biological legacy in good order to future generations.
  1 mark
Answer: (a) Oxygen production through photosynthesis; pollination (also flood/drought control, nutrient cycling, fertile soil, aesthetic pleasure). (b) Every species has intrinsic value, so we have a moral duty to conserve it for future generations.
Where the marks are
- First utilitarian ecosystem service described, e.g. oxygen production (1 mark)
- Second utilitarian ecosystem service described, e.g. pollination (1 mark)
- Ethical reason: intrinsic value of every species and our moral duty towards future generations (1 mark)
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Section D

Case-based

Q29 · 4 marks · Case-basedCompetency-basedHuman Health and Disease
When a microorganism invades a host, a definite sequence of events usually occur leading to infection and disease, causing suffering to the host. This process is called pathogenesis. Once a microorganism overcomes the defense system of the host, development of the disease follows a certain sequence of events as shown in the graph. Study the graph given below for the sequence of events leading to appearance of a disease and answer the questions that follow :
1. (a)
  In which period, according to the graph there are maximum chances of a person transmitting a disease / infection and why ?
  1 mark
2. (b)
  Study the graph and write what is an incubation period. Name a sexually transmitted disease that can be easily transmitted during this period. Name the specific type of lymphocytes that are attacked by the pathogen of this disease.
  2 marks
3. OR
  (b)
  Draw a schematic labelled diagram of an antibody.
  2 marks
4. (c)
  In which period, the number of immune cells forming antibodies will be the highest in a person suffering from pneumonia ?
  Name the immune cells that produce antibodies.
  1 mark
Figure in the paper: Graph: x-axis Time, y-axis Number of microorganisms. A bell-shaped curve rises and falls. Labelled regions: Incubation period (initial left part), Period of illness (rising part, dark shaded), Period of decline (falling part, light shaded), 'Death if immune responses and/or medical treatments fail' pointing to the peak. Two horizontal bands at the bottom: 'Mild signs or symptoms' and 'No signs or symptoms'. See it in the official paper.
Try it first, then show our solutionSolution
1. (a) The period of illness gives the maximum chance of transmission.
  ½ mark
2. In this period the number of microorganisms is at its highest, and they are released in large numbers through coughing and sneezing (droplet infection). (The incubation period, when there are no symptoms so the person takes no precautions, is also accepted.)
  ½ mark
3. (b) The incubation period is the time between the entry of the pathogen (infection) and the appearance of the first signs or symptoms of the disease.
  1 mark
4. AIDS can be easily transmitted during this long symptomless period, which may last 5–10 years.
  ½ mark
5. HIV attacks helper T-lymphocytes (T cells).
  ½ mark
6. (b) OR – Antibody diagram, described: a Y-shaped molecule (H₂L₂) with two heavy chains forming the stem and arms, and two light chains on the outer side of the arms. Disulphide (S–S) bonds join the chains. The tips of the arms are the antigen-binding sites (variable regions), and the rest is the constant region.
7. (c) Antibody-forming immune cells are highest during the period of illness.
  ½ mark
8. Antibodies are produced by B-lymphocytes (B cells, which become plasma cells).
  ½ mark
Answer: (a) Period of illness (highest microbe load, droplet spread); (b) Incubation period = time from infection to appearance of symptoms; AIDS; helper T-lymphocytes [OR antibody diagram: 2 heavy + 2 light chains, S–S bonds, antigen-binding sites]; (c) Period of illness; B-lymphocytes
Where the marks are
- Names the period of illness (½ mark)
- Reason: maximum microorganisms, spread by droplets (½ mark); incubation period with a valid reason is also accepted
- Defines incubation period as the time from infection to appearance of symptoms (1 mark)
- Names AIDS (½ mark)
- Names T-lymphocytes (helper T cells) (½ mark)
- OR (b): antibody diagram with any four correct labels, e.g. heavy chain, light chain, antigen-binding site, S–S bond (½ × 4 = 2 marks)
- (c) Names the period of illness (½ mark) and B-lymphocytes (½ mark)
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Q30 · 4 marks · Case-basedCompetency-basedPrinciples of Inheritance and Variation
The chromosome number is fixed for all normal organisms leading to species specification whereas any abnormality in the chromosome number of an organism results into abnormal individuals. For example, in humans 46 is the fixed number of chromosomes both in male and female. In male it is ‘44 + XY’ and in female it is ‘44 + XX’. Thus the human male is heterogametic, in other words produces two different types of gametes one with ‘22 + X’ chromosomes and the other with ‘22 + Y’ chromosomes respectively. Human female, on the other hand is homogametic i.e. produces only one type of gamete with ‘22 + X’ chromosomes only.
Sometimes an error may occur during meiosis of cell cycle, where the sister chromatids fail to segregate called nondisjunction, leading to the production of abnormal gametes with altered chromosome number. On fertilisation such gametes develop into abnormal individuals.
1. (a)
  State what is aneuploidy.
  1 mark
2. (b)
  If during spermatogenesis, the chromatids of sex chromosomes fail to segregate during meiosis, write only the different types of gametes with altered chromosome number that could possibly be produced.
  1 mark
3. (c)
  A normal human sperm (22 + Y) fertilises an ovum with karyotype ‘22 + XX’. Name the disorder the offspring thus produced would suffer from and write any two symptoms of the disorder.
  2 marks
4. OR
  (c)
  Name a best known and most common autosomal aneuploid abnormality in human and write any two symptoms.
  2 marks
Try it first, then show our solutionSolution
1. (a) Aneuploidy is the gain or loss of one or more chromosomes. It results from failure of chromatids (or chromosomes) to segregate during cell division (nondisjunction).
  1 mark
2. (b) If the sex chromosomes fail to separate during spermatogenesis, the abnormal gametes formed are 22 + XY and 22 + 0 (no sex chromosome).
  1 mark
3. (c) 22 + Y (sperm) + 22 + XX (ovum) gives 44 + XXY. The disorder is Klinefelter's syndrome.
  1 mark
4. Symptoms: the individual has overall masculine development but shows gynaecomastia (development of breasts), is tall, and is sterile.
  1 mark
5. (c) OR: The commonest autosomal aneuploidy is Down's syndrome (trisomy of chromosome 21). Symptoms: short stature with a small round head, furrowed tongue and partially open mouth, broad palm with a characteristic palm crease, and physical, psychomotor and mental retardation.
Answer: (a) Gain or loss of chromosome(s) due to nondisjunction; (b) 22 + XY and 22 + 0; (c) Klinefelter's syndrome (44 + XXY): gynaecomastia, sterility [OR Down's syndrome: short stature with small round head, mental retardation]
Where the marks are
- Correct definition of aneuploidy (1 mark)
- Gametes 22 + XY and 22 + 0 (½ × 2 = 1 mark)
- Names Klinefelter's syndrome (1 mark) with any two symptoms (½ × 2 = 1 mark)
- OR: names Down's syndrome (1 mark) with any two symptoms (½ × 2 = 1 mark)
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Section E

Long answer

Answer one of these 2 alternatives.
Q31(a) · 5 marks · Long answerSexual Reproduction in Flowering Plants
Answer the following.
1. (i)
  Explain the monosporic development of embryo sac in the ovule of an angiosperm.
  3 marks
2. (ii)
  Draw a diagram of the mature embryo sac of an angiospermic ovule and label any four parts in it.
  2 marks
Try it first, then show our solutionSolution
1. (i) The megaspore mother cell (2n) in the micropylar region of the nucellus divides by meiosis to form a linear tetrad of four haploid (n) megaspores.
  ½ mark
2. Three megaspores degenerate, and only one functional megaspore remains. The embryo sac develops from this single megaspore, hence 'monosporic' development.
  ½ mark
3. The nucleus of the functional megaspore divides by mitosis to form two nuclei, which move to opposite poles; this is the 2-nucleate embryo sac.
  ½ mark
4. Two more successive mitotic divisions give the 4-nucleate and then the 8-nucleate stage. These are free nuclear divisions, meaning nuclear division is not followed by cell wall formation.
  ½ mark
5. Cell walls then form around six of the nuclei: three cells group at the micropylar end (egg apparatus) and three at the chalazal end (antipodals).
  ½ mark
6. The remaining two polar nuclei lie in the large central cell. The mature embryo sac is therefore 7-celled and 8-nucleate.
  ½ mark
7. (ii) Diagram (described): an oval embryo sac. At the micropylar end is the egg apparatus, consisting of one egg cell and two synergids with filiform apparatus. At the chalazal end are three antipodal cells. In the middle is the large central cell with two polar nuclei.
  2 marks
Answer: MMC (2n) undergoes meiosis to form 4 megaspores; 3 degenerate and 1 is functional; three free-nuclear mitotic divisions give 8 nuclei; walls form, giving a 7-celled, 8-nucleate embryo sac (egg, 2 synergids, 3 antipodals, central cell with 2 polar nuclei).
Where the marks are
- MMC divides by meiosis to form a tetrad of megaspores (½ mark)
- Three megaspores degenerate, one functional megaspore (½ mark)
- Mitosis of the functional megaspore nucleus (½ mark)
- Free nuclear divisions up to the 8-nucleate stage (½ mark)
- Cell wall formation organising egg apparatus and antipodals (½ mark)
- 7-celled, 8-nucleate embryo sac with polar nuclei in the central cell (½ mark)
- Diagram with any four correct labels (½ × 4 = 2 marks)
Open Q31(a) on its own page
OR
Q31(b) · 5 marks · Long answerHuman Reproduction
Answer the following.
1. (i)
  Explain the formation of placenta after the implantation in a human female.
  3 marks
2. (ii)
  Draw a diagram showing human foetus within the uterus and label any four parts in it.
  2 marks
Try it first, then show our solutionSolution
1. (i) After implantation, finger-like projections called chorionic villi appear on the trophoblast.
  1 mark
2. These villi are surrounded by uterine tissue and maternal blood. The chorionic villi and the uterine tissue become interdigitated with each other.
  1 mark
3. Together they form the placenta, a structural and functional unit between the developing embryo (foetus) and the maternal body. It is connected to the embryo by the umbilical cord and supplies oxygen and nutrients, removes CO₂ and wastes, and secretes hormones such as hCG, hPL and progesterone.
  1 mark
4. (ii) Diagram (described): a section of the uterus with the foetus lying in the amniotic cavity, which is filled with amniotic fluid. The umbilical cord joins the foetus to the placenta on the uterine wall. A yolk sac and the plug of mucus in the cervix may also be shown, with the endometrium and myometrium of the uterine wall.
  2 marks
Answer: Chorionic villi form on the trophoblast and interdigitate with the uterine tissue and maternal blood, forming the placenta, the structural and functional link between embryo and mother. Diagram labels: placenta, umbilical cord, amniotic cavity, uterine wall, plug of mucus in cervix, yolk sac.
Where the marks are
- Chorionic villi appear on the trophoblast (1 mark)
- Villi surrounded by uterine tissue and maternal blood and interdigitated with it (1 mark)
- Together form the placenta, the structural and functional unit between embryo and mother (1 mark)
- Diagram with any four correct labels (½ × 4 = 2 marks)
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Answer one of these 2 alternatives.
Q32(a) · 5 marks · Long answerBiotechnology and its Applications
Name and describe the steps involved in the technique widely used in forensics that serves as the basis of paternity testing in case of disputes.
Try it first, then show our solutionSolution
1. The technique is DNA fingerprinting. It compares the repetitive DNA, VNTRs (variable number of tandem repeats), of different individuals.
  1 mark
2. Step 1 – Isolation of DNA from the sample (blood, hair root, semen, etc.).
  ½ mark
3. Step 2 – Digestion of DNA into fragments by restriction endonucleases.
  ½ mark
4. Step 3 – Separation of the DNA fragments by size using gel electrophoresis.
  1 mark
5. Step 4 – Transfer (blotting) of the separated fragments to a synthetic membrane such as nitrocellulose or nylon (Southern blotting).
  ½ mark
6. Step 5 – Hybridisation of the fragments with a labelled (radioactive) VNTR probe.
  1 mark
7. Step 6 – Detection of the hybridised fragments by autoradiography. This gives a band pattern unique to each person. A child's bands come half from the mother and half from the father, so comparing the patterns settles paternity.
  ½ mark
Answer: DNA fingerprinting: isolate DNA, digest it with restriction endonucleases, separate fragments by electrophoresis, blot them onto a nitrocellulose/nylon membrane, hybridise with a labelled VNTR probe, detect by autoradiography.
Where the marks are
- Names DNA fingerprinting (1 mark)
- Isolation of DNA (½ mark)
- Digestion by restriction endonuclease (½ mark)
- Separation of fragments by electrophoresis (½ × 2 = 1 mark)
- Transfer to a nitrocellulose or nylon membrane (½ mark)
- Hybridisation using a labelled VNTR probe (½ × 2 = 1 mark)
- Detection by autoradiography (½ mark)
Open Q32(a) on its own page
OR
Q32(b) · 5 marks · Long answerPrinciples of Inheritance and Variation
It is sometimes observed that the $F_{1}$ progeny has a phenotype that does not resemble either of the two parents and has intermediate phenotype. Explain by taking a suitable example and working out the cross upto $F_{2}$ progeny.
Try it first, then show our solutionSolution
1. The phenomenon is incomplete dominance. Neither allele is completely dominant over the other, so the heterozygous $F_{1}$ shows a phenotype intermediate between the two parents.
  ½ mark
2. Example: flower colour in snapdragon (dog flower, Antirrhinum sp.).
  ½ mark
3. Parental cross: pure red $R R$ × pure white $r r$ .
  ½ mark
4. Gametes: the red parent forms only $R$ gametes and the white parent forms only $r$ gametes.
  ½ mark
5. ** $F_{1}$ :** all plants are $R r$ with pink flowers. This intermediate colour resembles neither parent, because one $R$ allele cannot make enough red pigment.
  ½ mark
6. ** $F_{1}$ selfed:** $R r$ (pink) × $R r$ (pink). Each parent forms $R$ and $r$ gametes in equal proportion.
  ½ mark
7. ** $F_{2}$ Punnett square:**
  ♀ \ ♂ $R$ $r$
  $R$ $R R$ (Red) $R r$ (Pink)
  $r$ $R r$ (Pink) $r r$ (White)
  ½ mark
8. ** $F_{2}$ progeny:** 1 $R R$ red, 2 $R r$ pink and 1 $r r$ white.
  ½ mark
9. Phenotypic ratio: Red : Pink : White = $1 : 2 : 1$ .
  ½ mark
10. Genotypic ratio: $R R : R r : r r = 1 : 2 : 1$ . The phenotypic ratio equals the genotypic ratio, unlike Mendel's $3 : 1$ monohybrid ratio. The $R$ and $r$ alleles still segregate unchanged, which shows that the factors do not blend.
  ½ mark
Answer: Incomplete dominance, e.g. snapdragon flower colour. $R R$ (red) × $r r$ (white) gives $F_{1}$ $R r$ (pink). $F_{2}$ phenotypic and genotypic ratios are both $1 : 2 : 1$ (Red : Pink : White; $R R : R r : r r$ ).
Where the marks are
- Names incomplete dominance (½ mark)
- Gives snapdragon / Antirrhinum as the example (½ mark)
- Parental genotypes RR × rr with phenotypes (½ mark)
- Gametes R and r (½ mark)
- F1 Rr pink (½ mark)
- Selfing of F1 Rr × Rr (½ mark)
- F2 gametes / Punnett square (½ mark)
- F2 progeny RR red, Rr pink, rr white (½ mark)
- Phenotypic ratio 1:2:1 (½ mark)
- Genotypic ratio 1:2:1 (½ mark)
Open Q32(b) on its own page
Answer one of these 2 alternatives.
Q33(a) · 5 marks · Long answerCompetency-basedBiotechnology: Principles and Processes
Bioreactors are the containment vehicles of any biotechnology-based production process. For large scale production and for economic reasons the final success of biotechnological process depends on the efficiency of the bioreactor.
Answer the following questions w.r.t. the given paragraph :
1. (i)
  List the operational guidelines that must be adhered to so as to achieve optimisation of the bioreactor system. Enlist any four.
  0 marks
2. (ii)
  Mention the phase of the growth we refer to in the statement “Optimisation of growth and metabolic activity of the cells”.
  0 marks
3. (iii)
  Is the biological product formed in the bioreactor suitable for the intended use immediate ? Give reason in support of your answer.
  0 marks
Try it first, then show our solutionSolution
1. (i) A bioreactor is optimised by keeping the cultures at their optimum growth conditions. Four such parameters are:
  optimum temperature
  optimum pH
  adequate substrate (nutrient supply)
  required salts and vitamins, plus adequate oxygen (aeration)
  2 marks
2. (ii) The statement refers to the log (exponential) phase. In this phase the cells divide most rapidly and are most active metabolically.
  1 mark
3. (iii) No, the product is not immediately suitable for its intended use.
  1 mark
4. Reason: the product must first undergo downstream processing. This means separation and purification, then formulation with suitable preservatives, and strict quality-control testing. For drugs, clinical trials are also required before the product can be marketed.
  1 mark
Answer: (i) Optimum temperature, pH, substrate, salts, vitamins and oxygen (any four). (ii) Log/exponential phase. (iii) No; it needs downstream processing (separation, purification, formulation with preservatives, quality control/clinical trials).
Where the marks are
- Any four of: temperature, pH, substrate, salts, vitamins, oxygen (½ × 4 = 2 marks)
- Log/exponential phase (1 mark)
- States 'No' (1 mark)
- Reason: downstream processing, i.e. separation, purification, formulation, quality control or clinical trials (1 mark)
Open Q33(a) on its own page
OR
Q33(b) · 5 marks · Long answerBiotechnology: Principles and Processes
Answer the following.
1. (i)
  ‘EcoRI’ has played very significant role in r-DNA technology.
  (I) Explain the convention for naming EcoRI.
  (II) Write the recognition site and the cleavage sites of this restriction endonuclease.
  3 marks
2. (ii)
  What are the protruding and hanging stretches of DNA produced by these restriction enzymes called ? Describe their role in formation of r-DNA.
  2 marks
Try it first, then show our solutionSolution
1. (i)(I) Naming convention of EcoRI:
  E = genus Escherichia
  co = species coli
  R = strain RY 13
  I (Roman numeral) = the order in which the enzyme was isolated from that strain (the first one)
  2 marks
2. (i)(II) Recognition site: the palindromic sequence
  5′–G A A T T C–3′
  3′–C T T A A G–5′
  ½ mark
3. Cleavage site: EcoRI cuts between G and A on both strands (5′–G↓AATTC–3′ and 3′–CTTAA↑G–5′). This leaves 5′ AATT overhangs.
  ½ mark
4. (ii) The protruding, single-stranded hanging stretches are called sticky ends.
  1 mark
5. Role in r-DNA formation:
  The vector and the foreign DNA are cut with the same enzyme, so their sticky ends are complementary. They pair with each other through hydrogen bonds.
  This stickiness holds the fragments together and makes it easier for the enzyme DNA ligase to join them, which forms the recombinant DNA.
  1 mark
Answer: (i)(I) E = Escherichia (genus), co = coli (species), R = strain RY13, I = first enzyme isolated from that strain. (II) Recognition site 5′GAATTC3′ / 3′CTTAAG5′; cut between G and A on both strands. (ii) Sticky ends; they hydrogen-bond with complementary ends and help DNA ligase join the DNA to form r-DNA.
Where the marks are
- E = genus, co = species, R = strain, I = order of isolation (½ × 4 = 2 marks)
- Correct palindromic recognition site GAATTC/CTTAAG (½ mark)
- Cleavage between G and A on both strands (½ mark)
- Names sticky ends (1 mark)
- Sticky ends hydrogen-bond with complementary cut ends (½ mark)
- This facilitates the action of DNA ligase to form r-DNA (½ mark)
Open Q33(b) on its own page

♀ \ ♂	$R$	$r$
$R$	$R R$ (Red)	$R r$ (Pink)
$r$	$R r$ (Pink)	$r r$ (White)

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