BiologyCBSE Class 12 · Sample paper, session 2025-26

CBSE Class 12 Biology: sample paper 2025-26

The sample question paper CBSE published for the 2025-26 session, as printed: 33 questions for 70 marks, in sections, with each OR choice shown as an alternative. Try each question first, then open our solution.

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

Source: official sample question paper from cbseacademic.nic.in: https://cbseacademic.nic.in/web_material/SQP/ClassXII_2025_26/Biology-SQP.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
C	22-28
D	29-30	case-based
E	31-33

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

Section A

MCQ 1-12, assertion-reason 13-16

Q1 · 1 mark · MCQSexual Reproduction in Flowering Plants
The male gametes in plants are formed by:
1. (A)Mitotic division of nucleus of vegetative cell
2. (B)Meiotic division of nucleus of vegetative cell
3. (C)Mitotic division of nucleus of generative cell
4. (D)Meiotic division of nucleus of generative cell
Try it first, then show our solutionSolution
1. In the 2-celled pollen grain, the generative cell divides by mitosis (it is already haploid) to give two male gametes; the vegetative cell forms the pollen tube. Hence option (C).
  1 mark
Answer: (C) Mitotic division of nucleus of generative cell
Where the marks are
- Selects option C (1 mark)
Q2 · 1 mark · MCQSexual Reproduction in Flowering Plants
The primary endosperm nucleus is formed by fusion of which of the following?
1. (A)A male gamete and a female gamete
2. (B)A male gamete and two polar nuclei
3. (C)A female gamete and two synergids
4. (D)Two male gametes and an egg cell
Try it first, then show our solutionSolution
1. In triple fusion, one male gamete fuses with the two polar nuclei of the central cell, giving the triploid (3n) primary endosperm nucleus. Hence (B).
  1 mark
Answer: (B) A male gamete and two polar nuclei
Where the marks are
- Selects option B (1 mark)
Q3 · 1 mark · MCQHuman Reproduction
During the menstrual cycle of a human female, formation of graafian follicle is stimulated by secretion of which of the following gonadotropin hormones?
1. (A)Estrogen and progesterone
2. (B)FSH and Estrogen
3. (C)FSH and LH
4. (D)Progesterone and LH
Try it first, then show our solutionSolution
1. Gonadotropins are pituitary hormones FSH and LH; estrogen and progesterone are ovarian steroids, so only (C) fits.
2. FSH and LH stimulate follicular development into the Graafian follicle in the follicular phase. Answer (C).
  1 mark
Answer: (C) FSH and LH
Where the marks are
- Selects option C (1 mark)
Open Q3 on its own page
Q4 · 1 mark · MCQMolecular Basis of Inheritance
The experimental proof on the thermal stability of genetic material was first provided by experiments of
1. (A)Hershey and Chase
2. (B)Meselson and Stahl
3. (C)Frederick Griffith
4. (D)Jacob an Monod
Try it first, then show our solutionSolution
1. Griffith showed that heat-killed S strain could still transform R strain, implying the genetic material survived heating — first evidence of its thermal stability. Answer (C).
  1 mark
Answer: (C) Frederick Griffith
Where the marks are
- Selects option C (1 mark)
Q5 · 1 mark · MCQBiotechnology: Principles and Processes
Short stretches of DNA used to identify complementary sequences in a sample are called
1. (A)Probes
2. (B)Markers
3. (C)Primers
4. (D)Minisatellites
Try it first, then show our solutionSolution
1. Probes are short, labelled single-stranded DNA/RNA that hybridise with complementary sequences to detect them (e.g. in DNA fingerprinting). Answer (A).
  1 mark
Answer: (A) Probes
Where the marks are
- Selects option A (1 mark)
Q6 · 1 mark · MCQEvolution
Select the incorrect statement among the following.
1. (A) $p^{2} + 2 pq + q^{2} = 1$ . This is binomial expansion of $(p + q)^{2}$ .
2. (B)When frequency measured differs from expected values, the difference (direction) indicates the extent of evolutionary change.
3. (C)Hardy-Weinberg principle says that phenotype frequencies in a population are stable and are constant from generation to generation.
4. (D)The gene pool (total genes and their alleles in a population) remains constant. This is called genetic equilibrium. Sum total of all the allelic frequencies is 1.
Try it first, then show our solutionSolution
1. Hardy-Weinberg principle states that allele (gene) frequencies remain stable across generations, not phenotype frequencies. So statement (C) is incorrect; A, B, D are correct.
  1 mark
Answer: (C)
Where the marks are
- Identifies C as incorrect (1 mark)
Q7 · 1 mark · MCQPrinciples of Inheritance and Variation
Albinism is known to be due to an autosomal recessive mutation. The first child of a couple with normal skin pigmentation was an albino. What is the probability that their second child will also be an albino?
1. (A)100%
2. (B)25%
3. (C)50%
4. (D)75%
Try it first, then show our solutionSolution
1. Normal parents having an albino (aa) child must both be carriers, Aa.
2. $A a \times A a$ gives AA : Aa : aa = 1 : 2 : 1, so probability of aa for each child = 1/4 = 25% (independent of the first child). Answer (B).
  1 mark
Answer: (B) 25%
Where the marks are
- Deduces both parents Aa and gets 25% (1 mark)
Open Q7 on its own page
Q8 · 1 mark · MCQCompetency-basedEvolution
"In Cricket species, the sound produced by rubbing the wings or legs together play a crucial role in attracting mates, any change in the morphology of Cricket legs could potentially affect their ability to produce sound".
A mutant Cricket had thicker hind legs. What would you expect for this cricket species?
1. (A)The leg mutation will not lead to speciation if they diversify into new habitats.
2. (B)The leg mutation will have little effect on other external features, and therefore have little effect on speciation.
3. (C)The leg mutation will have no effect on behavior, and thus have little effect on speciation.
4. (D)The leg mutation might lead to reproductive isolation and speciation due to an effect on the mating call.
Try it first, then show our solutionSolution
1. Leg morphology affects sound production, which is the mating call; an altered call may not attract normal females, causing behavioural (reproductive) isolation that can lead to speciation. Answer (D).
  1 mark
Answer: (D)
Where the marks are
- Selects option D (1 mark)

Q9 · 1 mark · MCQCompetency-basedHuman Health and Disease

Plasmodium is a pathogen that causes malaria. Identify the correct sequence of transmission of the pathogen.

	I Stage of pathogen as it is transferred by vector bite	II First site in the host body where the pathogens infect and proliferates	III Second site in the host body where the pathogen infects and manifests clinical symptoms	IV Stage of pathogen as it is transferred to a new vector
A	Sporozoites	Erythrocyte infection	Liver infection	Gametocytes
B	Gametocytes	Erythrocyte infection	Liver infection	Sporozoites
C	Gametocytes	Liver infection	Erythrocyte infection	Sporozoites
D	Sporozoites	Liver infection	Erythrocyte infection	Gametocytes

(A)Row A
(B)Row B
(C)Row C
(D)Row D

Try it first, then show our solutionSolution

Female Anopheles injects sporozoites; they first multiply in liver cells, then attack RBCs (causing chills and fever via haemozoin); gametocytes formed in blood are taken up by a new mosquito. This matches Row D.
1 mark

Answer: (D) Sporozoites, Liver, Erythrocytes, Gametocytes

Where the marks are

Selects row D (1 mark)

Q10 · 1 mark · MCQMolecular Basis of Inheritance
Which mRNA will be translated to a polypeptide chain containing 8 amino acids?
1. (A)AUGUUAAUAGACGAGUAGCGACGAUGU
2. (B)AUGAGACGGACUGCAUUCCCAACCUGA
3. (C)AUGCCCAACCGUUAUUCAUGCUAG
4. (D)AUGUCGACAGUCUAAAACAGCGGG
Try it first, then show our solutionSolution
1. Read in triplets from AUG. (B): AUG AGA CGG ACU GCA UUC CCA ACC UGA — 8 codons before stop UGA, so 8 amino acids.
  1 mark
2. Others: (A) stops after 5 (UAG), (C) after 7 (UAG), (D) after 4 (UAA).
Answer: (B)
Where the marks are
- Splits into codons and counts 8 before UGA (1 mark)
Open Q10 on its own page
Q11 · 1 mark · MCQBiotechnology: Principles and Processes
In order to isolate genetic material of a bacterium, the cell must be treated with
1. (A)Lysozyme, ribonuclease, protease, chilled ethanol
2. (B)Cellulase, ribonuclease, protease, chilled ethanol
3. (C)Chitinase, ribonuclease, chilled ethanol, water
4. (D)Ribonuclease, protease, chilled ethanol, water
Try it first, then show our solutionSolution
1. Bacterial cell wall is broken by lysozyme (cellulase is for plants, chitinase for fungi); RNA removed by ribonuclease, proteins by protease, and DNA precipitated by chilled ethanol. Answer (A).
  1 mark
Answer: (A)
Where the marks are
- Selects option A (1 mark)
Q12 · 1 mark · MCQMicrobes in Human Welfare
Integrated Pest Management involves
I. Using pesticides/insecticides judiciously
II. Using biocontrol agents
III. Engaging in organic farming
1. (A)Only I
2. (B)Only II
3. (C)Both I and II
4. (D)Only III
Try it first, then show our solutionSolution
1. IPM combines judicious use of chemical pesticides with biocontrol agents to keep pests below damaging levels; organic farming avoids chemicals altogether and is a separate system. Answer (C).
  1 mark
Answer: (C) Both I and II
Where the marks are
- Selects option C (1 mark)
Q13 · 1 mark · Assertion-reasonCompetency-basedSexual Reproduction in Flowering Plants
Assertion (A): The ability of the pistil to recognise the pollen is the result of a continuous dialogue between pollen and pistil.
Reason (R): This electrical dialogue allows only compatible pollen to germinate.
1. (A)Both A and R are true and R is the correct explanation of A.
2. (B)Both A and R are true and R is not the correct explanation of A.
3. (C)A is true but R is false.
4. (D)A is False but R is true.
Try it first, then show our solutionSolution
1. A is true: pollen–pistil recognition involves continuous dialogue.
2. R is false: the dialogue is mediated by chemical components of pollen and pistil, not electrical signals. Answer (C).
  1 mark
Answer: (C) A true, R false
Where the marks are
- Identifies dialogue as chemical, so R false (1 mark)
Open Q13 on its own page
Q14 · 1 mark · Assertion-reasonCompetency-basedEvolution
Assertion (A): Some organisms are better adapted to survive in an otherwise hostile environment.
Reason (R): Adaptive ability is inherited and has a genetic basis.
1. (A)Both A and R are true and R is the correct explanation of A.
2. (B)Both A and R are true and R is not the correct explanation of A.
3. (C)A is true but R is false.
4. (D)A is False but R is true.
Try it first, then show our solutionSolution
1. Variation means some individuals are better adapted to hostile conditions; this adaptive ability is genetically based and inherited, which explains why it persists and is selected. Both true, R explains A. Answer (A).
  1 mark
Answer: (A)
Where the marks are
- Recognises both true with R explaining A (1 mark)
Q15 · 1 mark · Assertion-reasonCompetency-basedHuman Health and Disease
Assertion (A): Excess dose of coke or crack produces a sense of euphoria, increased energy and causes hallucinations.
Reason (R): It interferes with the transport of dopamine
1. (A)Both A and R are true and R is the correct explanation of A.
2. (B)Both A and R are true and R is not the correct explanation of A.
3. (C)A is true but R is false.
4. (D)A is False but R is true.
Try it first, then show our solutionSolution
1. Cocaine (coke/crack) interferes with dopamine transport, stimulating the CNS and producing euphoria and increased energy; excess dose causes hallucinations. Both true, R explains A. Answer (A).
  1 mark
Answer: (A)
Where the marks are
- Recognises dopamine interference explains effects (1 mark)
Q16 · 1 mark · Assertion-reasonCompetency-basedBiotechnology and its Applications
Assertion (A): Rosie was the first transgenic cow to make more nutritionally balanced milk for consumption by human babies.
Reason (R): The milk of Rosie cow contained human beta- lactalbumin which made the milk rich in protein.
1. (A)Both A and R are true and R is the correct explanation of A.
2. (B)Both A and R are true and R is not the correct explanation of A.
3. (C)A is true but R is false.
4. (D)A is False but R is true.
Try it first, then show our solutionSolution
1. A is true: Rosie (1997) produced more nutritionally balanced milk for human babies.
2. R is false: her milk contained human alpha-lactalbumin (2.4 g/L), not beta-lactalbumin. Answer (C).
  1 mark
Answer: (C) A true, R false
Where the marks are
- Identifies alpha-lactalbumin, so R false (1 mark)
Open Q16 on its own page

Section B

Answer one of these 2 alternatives.
Q17(A) · 2 marks · Very short answerSexual Reproduction in Flowering Plants
During artificial hybridization, it is important to ensure that only desired pollen grains are used for pollination. How is it ensured?
Try it first, then show our solutionSolution
1. Emasculation: If the female parent has bisexual flowers, the anthers are removed from the flower bud with forceps before they dehisce. This stops self-pollination by the flower's own pollen.
  1 mark
2. Bagging: The emasculated flower is covered with a bag of suitable size, usually made of butter paper. This keeps unwanted pollen off the stigma. When the stigma becomes receptive, the desired pollen is dusted on it and the flower is re-bagged.
  1 mark
Answer: By emasculation (removing anthers before dehiscence) followed by bagging (covering with butter-paper bag) to prevent contamination by unwanted pollen.
Where the marks are
- Explains emasculation – removal of anthers before dehiscence (1 mark)
- Explains bagging – covering the flower with butter-paper bag to avoid unwanted pollen (1 mark)
Open Q17(A) on its own page
OR
Q17(B) · 2 marks · Very short answerSexual Reproduction in Flowering Plants
Why is manual transfer of desired pollen grains relatively easy in pea plants for hybridization, but in wheat, pollen grains are often utilized from pollen banks for hybridization purposes?
Try it first, then show our solutionSolution
1. Pea: Pea flowers are large and bisexual, so manual emasculation and transfer of pollen are easy. Pea belongs to Leguminosae, whose pollen stays viable for months. Fresh or stored pollen can therefore be used directly.
  1 mark
2. Wheat: Wheat pollen loses its viability within about 30 minutes of release. Viable pollen is therefore stored in pollen banks by cryopreservation in liquid nitrogen at $- 19 6^{\circ}$ C and used when needed for hybridisation.
  1 mark
Answer: Pea pollen remains viable for months and its flowers are easy to emasculate; wheat pollen loses viability within ~30 min, so it is stored in pollen banks (cryopreserved).
Where the marks are
- Pea: easy emasculation and pollen viable for months (Leguminosae) (1 mark)
- Wheat: pollen viable only about 30 minutes, hence stored in pollen banks (1 mark)
Open Q17(B) on its own page
Q18 · 2 marks · Very short answerMolecular Basis of Inheritance
How is the rate of initiation of RNA polymerase at a given promoter in a transcriptional unit of prokaryotes regulated?
Try it first, then show our solutionSolution
1. The ability of RNA polymerase to start transcription at a promoter depends on accessory (regulatory) proteins. These proteins help the polymerase recognise the start site.
  1 mark
2. The regulatory proteins can act positively (activators) or negatively (repressors). In an operon, they work by interacting with the operator, which lies next to the promoter. A repressor bound to the operator blocks RNA polymerase, as in the lac operon.
  1 mark
Answer: By accessory regulatory proteins (activators/repressors) that act on the operator adjacent to the promoter and control recognition of start sites by RNA polymerase.
Where the marks are
- Accessory proteins regulate the recognition of start sites by RNA polymerase (1 mark)
- Positive (activator) and negative (repressor) regulation through the operator next to the promoter (1 mark)
Open Q18 on its own page
Q19 · 2 marks · Very short answerCompetency-basedHuman Health and Disease
The table below shows a hypothetical blood report of a patient.
Test description Observed value Unit Reference range
Leucocytes
Total leucocyte count 1100 Per Microliter 4400-11000
Neutrophils 31 % 55-70
Lymphocytes 25 % 20-40
Basophils 0.5 % 0.5 - 1
Eosinophils 02 % 1-4
Monocytes 0 % 1-8
A. Looking at the values suggest which defense mechanism/ immunity is affected and state how this defense mechanism provides immunity.
B. Name the barrier with least count and enumerate its role in providing immunity.
Try it first, then show our solutionSolution
1. A. Total leucocyte count (1100), neutrophils (31%) and monocytes (0%) are all below normal. These are phagocytic cells of the cellular barrier, so innate (non-specific) immunity, present from birth, is affected. Innate immunity gives physical, physiological, cellular and cytokine barriers that stop foreign agents entering the body and destroy microbes.
  1 mark
2. B. The cellular barrier cell with the least count is the monocyte (0%). Monocytes, and the macrophages formed from them, phagocytose and destroy microbes, preventing microbial growth.
  1 mark
Answer: A. Innate immunity – barriers prevent entry of and destroy foreign agents. B. Cellular barrier – monocytes (0%), which phagocytose and destroy microbes.
Where the marks are
- Identifies innate (non-specific) immunity and how it works through barriers (1 mark)
- Names monocytes of the cellular barrier and their phagocytic role (1 mark)
Open Q19 on its own page
Answer one of these 2 alternatives.
Q20(A) · 2 marks · Very short answerCompetency-basedMicrobes in Human Welfare
A cheese maker claims to be a biotechnologist. How will you support the same?
Try it first, then show our solutionSolution
1. Biotechnology means using living organisms, or their products or enzymes, to make products and processes useful to humans. Anyone who does this with technology can be called a biotechnologist.
  1 mark
2. A cheese maker uses microbes such as lactic acid bacteria and fungi (e.g., Propionibacterium shermanii for Swiss cheese, Penicillium for Roquefort) to ferment milk into cheese commercially. So his claim is justified.
  1 mark
Answer: Yes – he uses living microbes (bacteria, fungi) to produce cheese commercially, which fits the definition of biotechnology.
Where the marks are
- Defines a biotechnologist as one using organisms or their products with technology (1 mark)
- Links cheese making to the use of microbes for commercial production (1 mark)
Open Q20(A) on its own page
OR
Q20(B) · 2 marks · Very short answerBiotechnology: Principles and Processes
Why does the 'insertional inactivation' method to detect recombinant DNA is preferred to 'antibiotic resistance' procedure?
Try it first, then show our solutionSolution
1. Selecting recombinants by inactivating an antibiotic resistance gene is cumbersome. It needs simultaneous replica plating on two plates with different antibiotics (e.g., ampicillin and tetracycline).
  1 mark
2. In insertional inactivation, the foreign DNA is inserted within the coding sequence of β-galactosidase, which inactivates the enzyme. On a medium with a chromogenic substrate, non-recombinants give blue colonies, while recombinants give colourless/white colonies. Recombinants are thus identified on a single plate.
  1 mark
Answer: Antibiotic-based selection needs two plates (cumbersome); insertional inactivation of β-galactosidase gives blue (non-recombinant) vs colourless (recombinant) colonies on one plate.
Where the marks are
- States antibiotic resistance method is cumbersome due to plating on two antibiotic plates (1 mark)
- Explains β-galactosidase inactivation and blue/colourless colony selection (1 mark)
Open Q20(B) on its own page
Answer one of these 2 alternatives.
Q21(A) · 2 marks · Very short answerCompetency-basedEcosystem
(i) Compare the two ecological pyramids of biomass I and II given below and explain the situations in which this is possible.
(ii) Construct an ideal pyramid of energy if 200,000 joules of sunlight are available.
(For Visually impaired students: (i) Compare the upright and inverted ecological pyramids of biomass and explain the situations in which this is possible. (ii) Construct an ideal pyramid of energy if 200,000 joules of sunlight is available.)
Figure in the paper: Two biomass pyramids with trophic level on left and dry weight (kg m^-2) on right. Fig. I (upright): PP 809, PC 37, SC 11, TC 1.5 (widest at base, narrowing upward). Fig. II (inverted): PC 21 (wide, top) above PP 4 (narrow, bottom). See it in the official paper.
Try it first, then show our solutionSolution
1. (i) Pyramid I (upright) belongs to a terrestrial ecosystem such as a forest or grassland. Biomass is maximum at the producer level (809 kg m $^{- 2}$ ) and falls through PC (37) and SC (11) to TC (1.5).
  Pyramid II (inverted) belongs to an aquatic (sea) ecosystem. A small standing crop of phytoplankton (4 kg m $^{- 2}$ ) supports a larger standing crop of zooplankton (21 kg m $^{- 2}$ ). This is possible because phytoplankton multiply rapidly and have a high turnover.
  1 mark
2. (ii) Using about 1% capture of sunlight by producers and the 10% law at each transfer:
  Trophic level Energy
  Tertiary consumer (TC) 2 J
  Secondary consumer (SC) 20 J
  Primary consumer (PC) 200 J
  Producer (PP) 2,000 J
  Sunlight 200,000 J
  The pyramid is always upright: a broad base of producers narrowing to the top carnivores.
  1 mark
Answer: (i) I – upright, terrestrial; II – inverted, aquatic (small phytoplankton crop supports larger zooplankton). (ii) Upright energy pyramid: PP 2000 J → PC 200 J → SC 20 J → TC 2 J.
Where the marks are
- Compares upright (terrestrial) and inverted (aquatic) biomass pyramids with reason (1 mark)
- Draws an upright energy pyramid with values falling by 10% at each level (1 mark)
Open Q21(A) on its own page
OR
Q21(B) · 2 marks · Very short answerCompetency-basedBiodiversity and Conservation
A tropical rainforest in South America is home to more than 40,000 species of plants, 3,000 of fishes, 1,300 of birds, 427 of mammals, 427 of amphibians, 378 of reptiles and 1,25,000 insects, snails and worms.
From the given data, calculate the total number of known vertebrate species in the rainforest.
(i) Give a reason to justify the huge difference in the number of plant and animal species.
Try it first, then show our solutionSolution
1. Vertebrates are fishes, birds, mammals, amphibians and reptiles:
  $3000 + 1300 + 427 + 427 + 378 = 5532$ vertebrate species.
  1 mark
2. (i) Reason: Animals are mobile and have complex nervous systems and receptors. They can migrate away from harsh conditions and respond adaptively, so fewer separate species are needed. Plants are fixed and must adapt locally to water, minerals and sunlight in every microhabitat. Also, energy available falls at higher trophic levels (10% law), so producers can support only a limited number of vertebrate species.
  1 mark
Answer: (i) 5532 vertebrate species; difference due to mobility/adaptive responses of animals vs fixed plants (and reduced energy at higher trophic levels).
Where the marks are
- Adds fishes, birds, mammals, amphibians, reptiles = 5532 (1 mark)
- Any one valid reason, e.g. mobility and adaptive responses of animals vs sessile plants (1 mark)
Open Q21(B) on its own page

Test description	Observed value	Unit	Reference range
Leucocytes
Total leucocyte count	1100	Per Microliter	4400-11000
Neutrophils	31	%	55-70
Lymphocytes	25	%	20-40
Basophils	0.5	%	0.5 - 1
Eosinophils	02	%	1-4
Monocytes	0	%	1-8

Trophic level	Energy
Tertiary consumer (TC)	2 J
Secondary consumer (SC)	20 J
Primary consumer (PC)	200 J
Producer (PP)	2,000 J
Sunlight	200,000 J

Section C

Q22 · 3 marks · Short answerCompetency-basedReproductive Health
Suggest a suitable contraceptive device for the following cases with justification.
(i) Mohini does not want to take the risk of conception and sexually transmitted infections (STIs).
(ii) Lalita has two children and does not want any more children.
(iii) Geeta wants a contraceptive that she can take till she wants to avoid conception and can resume back to her fertile life without the intervention of the doctor. Also, it should have a lower failure rate.
Try it first, then show our solutionSolution
1. (i) Mohini – Condoms. A condom is a barrier method that stops sperm entering the vagina or cervix. It also blocks the transfer of microbes, so it protects against both pregnancy and STIs.
  1 mark
2. (ii) Lalita – Surgical sterilisation: tubectomy for her or vasectomy for her husband. It is a terminal, highly effective method, though poorly reversible, which suits a couple wanting no more children.
  1 mark
3. (iii) Geeta – Oral contraceptive pills (progestogen alone or progestogen–estrogen combination, e.g., Saheli). They inhibit ovulation and implantation and have a very low failure rate. They can be taken as long as conception is to be avoided, and stopping them restores fertility without a doctor's intervention.
  1 mark
Answer: (i) Condoms; (ii) Tubectomy/vasectomy (sterilisation); (iii) Oral contraceptive pills.
Where the marks are
- Condom with reason – blocks sperm and STI microbes (1 mark)
- Tubectomy/vasectomy – highly effective terminal method (1 mark)
- Oral pills – effective, reversible, user-controlled (1 mark)
Open Q22 on its own page
Q23 · 3 marks · Short answerCompetency-basedHuman Reproduction
Given below is a figure showing transport of ovum, fertilisation and passage of growing embryo through fallopian tube in a human female. Answer the questions that follow:
(i) What will be the ploidy of cells shown in (a) and (c) stage in the figure given below?
(ii) What will happen if component L as shown in the figure (g) given below does not attach properly to the endometrium?
(iii) In a pregnant mother (case X), during early pregnancy, the fertilised egg splits into two embryos at stage C shown in the figure given below, resulting in the formation of twins. Will the genome of cells of these two embryos exhibit variation? Justify your answer.
(For Visually impaired students: Explain the events and journey of an ovum in the fallopian tube of a mother from fertilisation stage to implantation.)
Figure in the paper: Figure of human female reproductive tract showing ovum (a), fertilisation by sperms (b), cleavage stages (c), (d), morula (e), movement through fallopian tube with points A-H, blastocyst implantation in uterus, and blastocyst (g) with labelled parts L (outer layer), M and N (inner cell mass region); (f) is an early stage. See it in the official paper.
Try it first, then show our solutionSolution
1. (i) (a) The ovum is haploid (n). (c) The blastomeres formed by cleavage of the zygote are diploid (2n).
  1 mark
2. (ii) L is the trophoblast. If it does not attach properly to the endometrium, implantation fails. The embryo cannot get nourishment and the placenta does not form, leading to pregnancy loss or other complications.
  1 mark
3. (iii) No variation. Both embryos come from the same zygote by mitotic division, so they have identical genomes. They become identical (monozygotic) twins.
  1 mark
4. (For visually impaired students)
  Fertilisation occurs in the ampullary region. The zygote then undergoes mitotic cleavage while moving through the isthmus towards the uterus, forming 2, 4, 8 and 16 blastomeres. The 8–16 cell embryo is the morula.
  The morula becomes the blastocyst, with an outer trophoblast and an inner cell mass.
  The trophoblast attaches to the endometrium and the uterine cells cover the blastocyst, so it becomes embedded (implantation). The inner cell mass then differentiates into the embryo.
Answer: (i) (a) n, (c) 2n; (ii) Implantation fails – pregnancy loss; (iii) No, identical genome as both arise from the same zygote by mitosis.
Where the marks are
- Ploidy: ovum n, blastomeres 2n (1 mark)
- Failure of trophoblast attachment leads to failed implantation or pregnancy loss (1 mark)
- No variation – both embryos from one zygote (identical twins) (1 mark)
Open Q23 on its own page
Q24 · 3 marks · Short answerPrinciples of Inheritance and Variation
In guinea pigs, black coat colour (G) dominates over white (g) and brown eyes (B) dominate over blue (b). The alleles for coat colour and eye colour are not linked. What will be the probability of the offspring having blue eyes and a white coat if both parents are heterozygous for eye and coat colour? Find the probability using a Punnett square
Try it first, then show our solutionSolution
1. Both parents are heterozygous for both traits: BbGg × BbGg. Since the genes are not linked, they assort independently and each parent forms four gametes: BG, Bg, bG, bg.
  1 mark
2. Punnett square:
  ♂ \ ♀ BG Bg bG bg
  BG BBGG BBGg BbGG BbGg
  Bg BBGg BBgg BbGg Bbgg
  bG BbGG BbGg bbGG bbGg
  bg BbGg Bbgg bbGg bbgg
  Phenotypic ratio = 9 brown eyes, black coat : 3 brown eyes, white coat : 3 blue eyes, black coat : 1 blue eyes, white coat.
  1 mark
3. Blue eyes and white coat needs genotype bbgg, which appears in only 1 of 16 boxes.
  Probability $= \frac{1}{4} \times \frac{1}{4} = \frac{1}{16} = 6.25%$ .
  1 mark
Answer: Probability of blue-eyed, white-coated offspring (bbgg) = 1/16 = 6.25%.
Where the marks are
- Writes the cross BbGg × BbGg with the four gametes (1 mark)
- Correct 4×4 Punnett square (1 mark)
- Probability of bbgg = 1/16 (6.25%) (1 mark)
Open Q24 on its own page
Q25 · 3 marks · Short answerCompetency-basedEvolution
Shown below are certain aquatic vertebrates, where natural selection has favoured them to develop certain characteristics which enable them to live in water.
A. Name and explain the phenomenon exhibited by the above animals.
B. Which one of the above is the most primitive one? What is its significance?
(For Visually impaired students: A. What are the basic conceptual similarities and differences between Lamarckian and Darwinian theory of evolution? B. Give an example that supports any one of the above two theories.)
Figure in the paper: Line drawings of six aquatic vertebrates, labelled salmon, shark, dolphin, ichthyosaur, blue whale and seal, all with streamlined bodies. See it in the official paper.
Try it first, then show our solutionSolution
1. A. The animals show convergent evolution. Salmon and shark (fishes), ichthyosaur (reptile), and dolphin, whale and seal (mammals) are not closely related. Living in the same aquatic habitat, they evolved similar adaptive features: streamlined bodies, fins and flippers.
  1 mark
2. These structures do the same job (swimming) but are not anatomically or ancestrally similar, so they are analogous structures. Natural selection in a similar habitat favoured similar adaptations in different groups, and analogy is the result of convergent evolution.
  1 mark
3. B. The ichthyosaur is the most primitive (oldest extinct form) among these. Its significance: some land reptiles went back into water and evolved into fish-like reptiles (ichthyosaurs) about 200 mya, showing that reptiles re-adapted to aquatic life.
  1 mark
4. (For visually impaired students) Similarity: Both Lamarck and Darwin held that organisms have heritable traits and that useful traits become more common over generations. Difference: Lamarck proposed inheritance of acquired characters, e.g. a giraffe stretching its neck passes a longer neck to its offspring. Darwin proposed natural selection acting on random, heritable variations, e.g. giraffes that were naturally long-necked survived and reproduced more.
Answer: A. Convergent evolution – analogous structures (streamlined body, fins/flippers) in unrelated animals sharing an aquatic habitat. B. Ichthyosaur – land reptiles returned to water ~200 mya as fish-like reptiles.
Where the marks are
- Names convergent evolution (1 mark)
- Explains analogous structures arising from similar habitat and function (1 mark)
- Ichthyosaur with significance – reptiles returning to water about 200 mya (1 mark)
Open Q25 on its own page
Q26 · 3 marks · Short answerMicrobes in Human Welfare
The Biological Oxygen Demand (BOD) of a primary effluent during sewage treatment is reduced. Enlist the process of how this is achieved.
Try it first, then show our solutionSolution
1. The primary effluent is passed into large aeration tanks, where it is constantly agitated mechanically and air is pumped in. This lets useful aerobic microbes grow into flocs: masses of bacteria held together by slime and fungal filaments in mesh-like structures.
  1 mark
2. The microbes consume most of the organic matter, turning it into microbial biomass and releasing minerals. This sharply reduces the BOD. When the BOD has fallen to a small fraction of that of raw sewage, the effluent goes to a settling tank, where the flocs sediment. The sediment is called activated sludge.
  1 mark
3. Part of the activated sludge is pumped back into the aeration tank as inoculum. The rest goes to anaerobic sludge digesters, where anaerobic bacteria digest the bacteria and fungi in the sludge, producing biogas (methane, H $_{2}$ S, CO $_{2}$ ). The treated effluent is released into rivers or streams, and the leftover sludge can be used as manure.
  1 mark
Answer: Secondary (biological) treatment: aeration tanks with flocs consume organic matter (BOD falls) → settling tank (activated sludge) → part recycled as inoculum, rest digested anaerobically in sludge digesters producing biogas.
Where the marks are
- Aeration tank and formation of flocs (1 mark)
- Microbial consumption of organic matter lowers BOD; settling gives activated sludge (1 mark)
- Activated sludge used as inoculum and in anaerobic sludge digesters (1 mark)
Open Q26 on its own page
Q27 · 3 marks · Short answerHuman Health and Disease
Expand ELISA. On what principle is the ELISA test based? List two ways by which an infection can be detected by this test.
Try it first, then show our solutionSolution
1. ELISA = Enzyme Linked Immunosorbent Assay.
  1 mark
2. Principle: ELISA is based on the specific antigen–antibody interaction.
  1 mark
3. An infection can be detected in two ways:
  (a) by detecting antigens of the pathogen (proteins, glycoproteins, etc.) in the sample;
  (b) by detecting antibodies the body has made against the pathogen.
  1 mark
Answer: Enzyme Linked Immunosorbent Assay; antigen–antibody interaction; detects pathogen antigens or antibodies produced against the pathogen.
Where the marks are
- Correct expansion of ELISA (1 mark)
- Principle – antigen–antibody interaction (1 mark)
- Two ways – detection of antigens and of antibodies (1 mark)
Open Q27 on its own page
Q28 · 3 marks · Short answerCompetency-basedOrganisms and Populations
The population pyramids of Japan for the years 1950, 2007 and on the basis of the estimated value for 2050 have been shown below to answer the questions that follow:
A. Comment upon the growth status of the population on the basis of shape of the age pyramids.
B. What insights can you gain about their population dynamics?
(For visually impaired students: The population pyramids of Japan for the years 1950, 2007 and on the basis of the estimated value for 2050 have broad-based, inverted bell and urn shape, respectively. Answer the following questions: A. Comment upon the growth status of the population on the basis of shape of the age pyramids. B. What insights can you gain about their population dynamics?)
Figure in the paper: Three age-sex pyramids titled 'Changes in the Population Pyramid' for 1950, 2007, 2050 (males left, females right, age 0-90, millions). 1950: broad base; 65 and over 4.9%, 15-64 59.6%, 0-14 35.4%. 2007: inverted bell; 65 and over 21.5%, 15-64 65.0%, 0-14 13.5%. 2050: urn/inverted shape; 65 and over 39.6%, 15-64 51.8%, 0-14 8.6%. See it in the official paper.
Try it first, then show our solutionSolution
1. 1950 – Expanding (growing) population. The broad base (0–14 years = 35.4%) shows a large share of young people, so the birth rate is high and the population is growing.
  1 mark
2. 2007 – Stable population. The base has narrowed (0–14 years = 13.5%) and pre-reproductive and reproductive age groups are nearly balanced (15–64 years = 65%). The bell-like shape shows the population was nearly static, with falling birth rate and rising elderly share (21.5%).
  1 mark
3. 2050 – Declining population. The urn-shaped pyramid (0–14 years = 8.6%, 65+ = 39.6%) shows very low birth rates and a growing elderly population, so the population will shrink. Insight: Japan is moving from high to low birth rate with rising life expectancy. This means an ageing population, a shrinking workforce and more dependence on fewer working-age people.
  1 mark
Answer: 1950 – expanding (broad base, high birth rate); 2007 – stable (bell shape); 2050 – declining (urn shape, low birth rate, ageing population).
Where the marks are
- 1950 expanding – broad base, high birth rate (1 mark)
- 2007 stable – nearly equal pre-reproductive and reproductive groups (1 mark)
- 2050 declining – urn shape, low birth rate, more elderly (1 mark)
Open Q28 on its own page

♂ \ ♀	BG	Bg	bG	bg
BG	BBGG	BBGg	BbGG	BbGg
Bg	BBGg	BBgg	BbGg	Bbgg
bG	BbGG	BbGg	bbGG	bbGg
bg	BbGg	Bbgg	bbGg	bbgg

Section D

case-based

Q29 · 4 marks · Case-basedCompetency-basedSexual Reproduction in Flowering Plants

Given below is a set of information about some fruits and seeds.

Fruit	Fruit and seed formation
P	Nucellar cells surrounding the embryo sac develop into embryos.
Q	Ovary develops into the fruit by the application of growth hormones.
R	Thalamus contributes to fruit formation.
S	Ovary matures into a fruit after fertilisation.

On the basis of the information provided above, answer the following questions with justification for each answer.

A
How many embryo sacs will be present in each ovule of S before maturation and how many egg(s) will be present in each embryo sac when the embryo sac is developed from a single megaspore?
1 mark
B
(i) Which of these fruits exhibits polyembryony? Will their embryos exhibit genetic variation? Justify.
(ii) What will be ploidy of the embryonic cells in the above case?
2 marks
C
Which of these fruits can be considered as parthenocarpic? Give a reason.
1 mark
OR
D
Which of the fruits P, Q, R or S is a true fruit with seeds? Give reason.
1 mark

Try it first, then show our solutionSolution

A. Each ovule of S has one embryo sac. When the embryo sac develops from a single functional megaspore (monosporic development), it is 7-celled and 8-nucleate and has one egg cell in the egg apparatus.
1 mark
B(i) P shows polyembryony, because nucellar cells develop into several embryos in one seed. These embryos form by apomixis, an asexual process involving only mitosis, so they show no genetic variation: they are clones of the mother plant.
1 mark
B(ii) Nucellar cells are diploid somatic cells, so the embryonic cells are diploid (2n).
1 mark
C. Q is parthenocarpic. Its ovary develops into a fruit under growth hormones without fertilisation, so the fruit is seedless.
1 mark
D (alternative to C). S is a true fruit with seeds. It develops from the ovary after fertilisation, and the fertilised ovules mature into seeds. (R is a false fruit, since the thalamus contributes to it.)

Answer: A. One embryo sac, one egg; B(i) P – no genetic variation (apomictic); (ii) 2n; C. Q (formed without fertilisation, seedless); D. S (from ovary after fertilisation).

Where the marks are

One embryo sac per ovule and one egg per embryo sac (1 mark)
P shows polyembryony; no variation as formed by apomixis (1 mark)
Embryonic cells diploid (2n) (1 mark)
C: Q parthenocarpic as formed without fertilisation / D: S true fruit from fertilised ovary (1 mark)

Open Q29 on its own page

Q30 · 4 marks · Case-basedCompetency-basedHuman Health and Disease
The graph below shows the Antibody concentration in young calves. Study the graph and answer the questions that follow:
(For visually impaired students: The antibody concentration in a young calf was studied. It was found that the antibodies derived from colostrum (passive immunity) decreased from day 1 to 14, while the antibodies derived from immune cells (active immunity) increased between day 1 to day 14 and remained steady thereafter.)
1. A
  What do you think is the difference between passive and active immunity in this case?
  1 mark
2. B
  What kind of immunity will be observed when a vaccine is administered to the calf and why?
  2 marks
3. C
  What kind of trend does passive immunity show and why?
  1 mark
4. OR
  D
  Why does active immunity start after 14 days as per the above graph?
  1 mark
Figure in the paper: Graph of antibody concentration (y-axis) vs time (x-axis, marked 1 d and 14 d). Passive immunity (colostrum) curve rises sharply to a peak at about day 1 then falls steeply to a low level. Active immunity (immune system) curve stays low until ~14 d then rises steadily to a high plateau. The period between 1 d and 14 d where both are low is marked 'High risk period'. See it in the official paper.
Try it first, then show our solutionSolution
1. A. Passive immunity comes from ready-made antibodies in colostrum. It is immediate but transient and creates no memory cells. Active immunity comes from antibodies made by the calf's own immune system. It starts after a lag but is long-lasting and memory-based.
  1 mark
2. B. A vaccine gives active immunity. It contains inactivated or weakened pathogens (or antigenic proteins) that cannot cause disease but are still recognised as foreign.
  1 mark
3. The calf's B- and T-lymphocytes are activated, and antibodies are produced against the pathogen. Memory B- and T-cells are also formed, so on later exposure the calf mounts a quick, strong response, giving long-term protection.
  1 mark
4. C. Passive immunity peaks at about day 1 and then falls steeply. The colostrum antibodies are used up or broken down, and no memory cells are made to replace them.
  1 mark
5. D (alternative to C). Active immunity is slow to develop. The calf's immune system needs time to meet antigens, activate lymphocytes and build up enough antibodies, so the response becomes significant only after about 14 days.
Answer: A. Passive – ready-made colostrum antibodies, quick, short-lived, no memory; active – own antibodies, slow, long-lasting, memory. B. Active immunity – vaccine antigens induce antibodies and memory cells. C. Peaks at day 1 then declines (antibodies used up, no memory). D. Active immunity is slow; it takes time to build up.
Where the marks are
- Distinguishes passive and active immunity (1 mark)
- Vaccine gives active immunity (1 mark)
- Reason – weakened antigen activates lymphocytes, antibodies and memory cells form (1 mark)
- C: declining trend as antibodies are used up with no memory / D: active immunity is slow to develop (1 mark)
Open Q30 on its own page

Section E

Answer one of these 2 alternatives.
Q31(A) · 5 marks · Long answerMolecular Basis of Inheritance
A. Construct a complete transcription unit with promoter and terminator on the basis of the hypothetical template strand given below.
B. How is transcription a more complex process in eukaryotic cells? Explain the additional processes that a precursor mRNA has to undergo in these organisms.
(For visually impaired students: A. Why are both the strands not copied during transcription? B. How is transcription a more complex process in eukaryotic cells? Explain the additional processes that a precursor mRNA has to undergo in these organisms.)
Figure in the paper: Template strand with bases A T G C A T G C A T A C written in a row, with arrow at left end of the strand. See it in the official paper.
Try it first, then show our solutionSolution
1. A. The template strand reads 3′→5′. The coding strand is complementary to it and runs 5′→3′:
  Template: 3′ – A T G C A T G C A T A C – 5′
  Coding: 5′ – T A C G T A C G T A T G – 3′
  The mRNA formed has the coding-strand sequence with U for T: 5′-UACGUACGUAUG-3′.
  1 mark
2. Transcription unit:
  5′ —[Promoter]— TACGTACGTATG —[Terminator]— 3′ (coding strand)
  3′ —[Promoter]— ATGCATGCATAC —[Terminator]— 5′ (template strand)
  The promoter lies upstream, at the 5′ end of the coding strand, and is where RNA polymerase binds. The terminator lies downstream, at the 3′ end of the coding strand, and ends transcription.
  1 mark
3. B. Transcription in eukaryotes uses three RNA polymerases in the nucleus:
  RNA polymerase I transcribes rRNAs (28S, 18S, 5.8S).
  RNA polymerase II transcribes hnRNA, the precursor of mRNA.
  RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs.
  1 mark
4. Splicing: The primary transcript (hnRNA) contains both exons (coding) and introns (non-coding). The introns are removed and the exons are joined in a defined order.
  1 mark
5. Capping and tailing: In capping, an unusual nucleotide, methyl guanosine triphosphate, is added to the 5′ end of hnRNA. In tailing, about 200–300 adenylate residues are added at the 3′ end without a template. The fully processed hnRNA, now mRNA, is transported out of the nucleus for translation.
  1 mark
Answer: A. Coding strand 5′-TACGTACGTATG-3′ with promoter at its 5′ (upstream) end and terminator at its 3′ (downstream) end. B. Three RNA polymerases; hnRNA undergoes splicing, 5′ capping and 3′ poly-A tailing.
Where the marks are
- Correct complementary coding strand with polarity (1 mark)
- Promoter upstream (5′ of coding strand) and terminator downstream (1 mark)
- Three RNA polymerases and their products (1 mark)
- Splicing – removal of introns, joining of exons (1 mark)
- Capping (methyl guanosine triphosphate at 5′) and tailing (poly-A at 3′) (1 mark)
Open Q31(A) on its own page
OR
Q31(B) · 5 marks · Long answerMolecular Basis of Inheritance
A. Explain the process of aminoacylation of tRNA. Mention its role in translation.
B. How do ribosomes in the cells act as factories for protein synthesis?
C. Given below is a strand of mRNA undergoing the process of translation, what will be the sequence of Amino acids that will be translated? Name the triplet codons that should be added to bring to the end of translation at X.
(For visually impaired students: C. Explain the phenomenon that forms the genetic basis to prove that codon is a triplet and it is read in a contiguous manner.)
Figure in the paper: mRNA strand with codons CGA, ACA (in ribosome, with tRNA UGU carrying Threonine and Arginine), CAC, CUU, CCG, GGA, CCG, CCG, CCG, CUU, AGU; region marked X at the end. Free tRNAs shown with anticodons and amino acids: Serine (UCA), Leucine (GAA), Glycine (CCU), Proline (GGC), Histidine (GUG), Leucine (GAG), Arginine (GCU). See it in the official paper.
Try it first, then show our solutionSolution
1. A. Aminoacylation (charging) of tRNA: Each amino acid is activated by ATP and attached to its specific tRNA. The enzyme aminoacyl-tRNA synthetase catalyses this, forming aminoacyl-tRNA.
  Role: The charged tRNA carries the correct amino acid to the matching codon. When two charged tRNAs come close on the ribosome, peptide bond formation between them becomes energetically favourable.
  1 mark
2. B. The ribosome is the cell's protein-synthesising factory. When inactive, it exists as separate large and small subunits. Translation begins when the small subunit meets the mRNA and the large subunit then joins.
  1 mark
3. The large subunit has two sites where successive aminoacyl-tRNAs bind, holding the amino acids close enough for a peptide bond. The ribosome moves codon by codon along the mRNA. Its rRNA (23S rRNA in bacteria) acts as a ribozyme that catalyses peptide bond formation.
  1 mark
4. C. Reading the codons CGA–ACA–CAC–CUU–CCG–GGA–CCG–CCG–CCG–CUU–AGU gives the amino acid sequence:
  Arginine – Threonine – Histidine – Leucine – Proline – Glycine – Proline – Proline – Proline – Leucine – Serine.
  1 mark
5. To end translation at X, a stop (termination) codon must be added: UAA, UAG or UGA. A release factor binds the stop codon and releases the polypeptide.
  1 mark
6. (For visually impaired students, C) Frameshift mutations: Inserting or deleting one or two bases changes the reading frame from that point onward. Inserting or deleting three bases (or multiples of three) inserts or removes a whole codon and leaves the rest of the frame intact. This proves the codon is a triplet read contiguously, without punctuation.
Answer: A. Amino acids activated by ATP are linked to tRNA by aminoacyl-tRNA synthetase; makes peptide bond formation favourable. B. Ribosome subunits assemble on mRNA; large subunit holds two aminoacyl-tRNAs; 23S rRNA ribozyme forms peptide bonds. C. Arg-Thr-His-Leu-Pro-Gly-Pro-Pro-Pro-Leu-Ser; stop codon UAA/UAG/UGA at X.
Where the marks are
- Aminoacylation by aminoacyl-tRNA synthetase with ATP and its role in peptide bond formation (1 mark)
- Ribosome subunits assemble on mRNA; two binding sites in the large subunit (1 mark)
- rRNA (23S) acts as ribozyme catalysing peptide bonds (1 mark)
- Correct amino acid sequence (1 mark)
- Stop codon UAA/UAG/UGA at X (1 mark)
Open Q31(B) on its own page
Answer one of these 2 alternatives.
Q32(A) · 5 marks · Long answerBiotechnology: Principles and Processes
Some plant and animal pathogens serve as one of the tools of recombinant DNA (rDNA) technology.'
A. Name one animal and one plant pathogen and discuss the pathogenic nature of both. State how they serve as a tool in rDNA technology.
B. What are the enzymes needed for rDNA technology?
C. A farmer owns a cotton farm land which is getting infested with coleopteran pests. He is not willing to use the microbes to protect his farm.
i. Name an alternate method to introduce the gene of interest the pathogen would have otherwise delivered and discuss how the alternate method would deliver the gene.
ii. State how this gene would control the pest.
Try it first, then show our solutionSolution
1. A. Animal pathogen – retroviruses: They can transform normal animal cells into cancerous cells. Plant pathogen – Agrobacterium tumefaciens: It causes crown gall disease by transferring a piece of DNA called T-DNA (from its Ti plasmid) into plant cells, turning normal cells into tumour cells.
  1 mark
2. Use in rDNA technology: The pathogenic genes are removed, so 'disarmed' retroviruses and the disarmed Ti plasmid can no longer cause disease. They are used as cloning vectors to deliver desirable genes into animal and plant cells.
  1 mark
3. B. The key enzymes are restriction endonucleases (molecular scissors that cut DNA at specific sites) and DNA ligase (joins the cut DNA fragments). DNA polymerase, used in PCR, is also needed.
  1 mark
4. C(i) Biolistics (gene gun): Plant cells are bombarded with high-velocity micro-particles of gold or tungsten coated with DNA carrying the cry gene. The DNA enters the cells and integrates into the plant genome, so no microbial vector is needed.
  1 mark
5. C(ii) The cry gene codes for the Bt toxin, made as an inactive protoxin in the plant. When the pest eats the plant, the alkaline pH of its gut converts the protoxin into the active toxin. The active toxin binds the midgut epithelial cells and creates pores, so the cells swell and lyse and the pest dies.
  1 mark
Answer: A. Retrovirus (cancer-causing) and Agrobacterium tumefaciens (crown gall via T-DNA); disarmed forms used as vectors. B. Restriction endonuclease and DNA ligase. C(i) Biolistics/gene gun – DNA-coated gold/tungsten particles; (ii) cry gene → Bt protoxin activated in alkaline gut, pores in gut epithelium, pest dies.
Where the marks are
- Names retrovirus and Agrobacterium with their pathogenic nature (1 mark)
- States their use as disarmed cloning vectors (1 mark)
- Restriction enzyme and ligase (1 mark)
- Biolistics/gene gun with gold/tungsten micro-particles (1 mark)
- Bt toxin activated in alkaline gut, damages gut epithelium (1 mark)
Open Q32(A) on its own page
OR
Q32(B) · 5 marks · Long answerBiotechnology: Principles and Processes
BamH1 is a restriction enzyme which recognizes the sequence- 5' - GGATCC - 3'. The restriction activity of this enzyme is between G and G.
A. Construct the palindrome for the above sequence.
B. Draw a labeled diagram to show the formation of recombinant DNA (rDNA) using BamH1.
C. PBR322 is a plasmid that has a restriction site for this enzyme at the tetracycline resistant gene. If BamH1 were to be used, how will it impact the response of the transformant with rDNA to antibiotics- ampicillin and tetracycline. Justify.
Try it first, then show our solutionSolution
1. A. Palindrome: A palindromic sequence reads the same on both strands when each is read in the same direction (5'→3'). The complementary strand of GGATCC, read 5'→3', is also GGATCC.
  5' – G G A T C C – 3'
  3' – C C T A G G – 5'
  1 mark
2. B. Diagram (described in words): formation of rDNA using BamH1.
  1. Draw the vector DNA (plasmid) and the foreign DNA, each carrying the site 5'-GGATCC-3' / 3'-CCTAGG-5'. Label: vector DNA, foreign DNA, BamH1 recognition site.
  2. Show BamH1 cutting each strand between G and G, giving fragments with single-stranded overhangs 5'-GATC-3'. Label: BamH1 (restriction endonuclease) and sticky ends:
  5'–G GATCC–3'
  3'–CCTAG G–5'
  3. Show the sticky ends of the foreign DNA fragment pairing by complementary bases with the sticky ends of the cut vector. Label: complementary base pairing.
  4. Show DNA ligase sealing the sugar-phosphate backbone. Label: DNA ligase, recombinant DNA (rDNA).
  2 marks
3. C. Ampicillin: BamH1 cuts pBR322 inside the tetracycline resistance gene ( $t e t^{R}$ ), so the ampicillin resistance gene ( $am p^{R}$ ) stays intact. The transformant carrying rDNA therefore grows on a medium containing ampicillin.
  1 mark
4. C. Tetracycline: The foreign DNA inserted into $t e t^{R}$ breaks up its coding sequence (insertional inactivation), so tetracycline resistance is lost. The recombinant transformant cannot grow (dies) on a medium containing tetracycline. Non-recombinants grow on both media, which lets recombinants be selected.
  1 mark
Answer: A. 5'-GGATCC-3' / 3'-CCTAGG-5'. B. Labelled diagram: BamH1 cuts vector and foreign DNA leaving GATC sticky ends, which pair and are sealed by DNA ligase to form rDNA. C. The transformant survives on ampicillin ( $am p^{R}$ intact) but dies on tetracycline ( $t e t^{R}$ inactivated by the insert).
Where the marks are
- Writes the palindrome with the complementary strand 3'-CCTAGG-5' and correct polarity (1 mark)
- Diagram shows BamH1 cutting vector and foreign DNA, sticky ends and their joining (1 mark)
- Labels: BamH1, sticky ends, DNA ligase, recombinant DNA (1 mark)
- Transformant resistant to ampicillin because $am p^{R}$ is intact (1 mark)
- Transformant sensitive to tetracycline because of insertional inactivation of $t e t^{R}$ (1 mark)
Open Q32(B) on its own page
Answer one of these 2 alternatives.
Q33(A) · 5 marks · Long answerOrganisms and Populations
Justify the following statements with suitable proof/examples: -
A. 'competition is not limited to closely related species'
B. 'competition is not always dependent on resources being limiting'
C. 'competitive exclusion occurs in nature'
D. 'competing species may evolve mechanisms for co-existence'
E. 'competition in nature comes from what is called 'competitive release''
Try it first, then show our solutionSolution
1. A. Even totally unrelated species can compete for a common resource. In some shallow South American lakes, visiting flamingoes and resident fishes compete for the same food, the zooplankton.
  1 mark
2. B. Competition can occur even when resources are plentiful. This is called interference competition: the feeding efficiency of one species is reduced by the interfering, inhibitory presence of another species, even though food and space are abundant.
  1 mark
3. C. Gause's competitive exclusion principle states that two closely related species competing for the same limited resource cannot co-exist indefinitely; the competitively inferior one is eliminated. Example: the Abingdon tortoise of the Galapagos Islands became extinct within a decade of goats being introduced, apparently because the goats browsed more efficiently.
  1 mark
4. D. Competing species can co-exist through resource partitioning, for example by feeding at different times or using different foraging patterns. MacArthur showed that five closely related warbler species living on the same tree avoided competition because they foraged differently.
  1 mark
5. E. A species whose range is restricted by a competitively superior species expands its range when the competitor is experimentally removed; this is competitive release. On the rocky coasts of Scotland, the larger barnacle Balanus dominates the intertidal zone and excludes the smaller barnacle Chthamalus. When Balanus is removed, Chthamalus spreads into that zone.
  1 mark
Answer: A. Flamingoes and fishes compete for zooplankton. B. Interference competition occurs even when resources are abundant. C. Abingdon tortoise was eliminated by goats (Gause's principle). D. Resource partitioning among MacArthur's warblers. E. Balanus and Chthamalus on the Scottish coast.
Where the marks are
- Unrelated species compete: flamingoes and fishes for zooplankton (1 mark)
- Interference competition even when resources are abundant (1 mark)
- Competitive exclusion: Abingdon tortoise and goats (1 mark)
- Resource partitioning: MacArthur's five warbler species (1 mark)
- Competitive release: Balanus and Chthamalus on the Scottish coast (1 mark)
Open Q33(A) on its own page
OR
Q33(B) · 5 marks · Long answerCompetency-basedEcosystem
A. How does a simple food chain exemplify the First Law of Thermodynamics?
B. The table below shows the number of species in different parts of the world.
Name of Place Number of Bird species
Columbia 1400
India 1200
Northern South America 1300
New York 105
Denmark 504
Identify the common factor in regions with a higher number of bird species and suggest at least two reasons for this greater diversity.
Try it first, then show our solutionSolution
1. A. The First Law of Thermodynamics states that energy can be neither created nor destroyed, only changed from one form to another. In a food chain, producers convert solar energy into chemical energy stored in organic matter through photosynthesis.
  1 mark
2. When herbivores and then carnivores eat organisms of the previous level, this chemical energy passes along the chain. Part of it is released as heat during respiration, and on death the remaining organic matter passes to decomposers (detritivores). At every stage energy is only transformed or transferred, never created or destroyed, which illustrates the First Law.
  1 mark
3. B. Common factor: Colombia, India and Northern South America, which have the most bird species, all lie close to the equator in the tropics (about 23.5° N to 23.5° S). New York and Denmark are temperate regions at higher latitudes and have far fewer species. Diversity falls as we move from the equator towards the poles (latitudinal gradient).
  1 mark
4. Reason 1: Tropical regions have stayed relatively undisturbed for millions of years, while temperate regions suffered frequent glaciations. Since speciation is a function of time, the tropics have had a longer evolutionary time for species to diversify.
  1 mark
5. Reason 2: Tropical environments are less seasonal, more constant and predictable, which promotes niche specialisation and greater species diversity. Also acceptable: the tropics receive more solar energy, which raises productivity and may indirectly support greater diversity.
  1 mark
Answer: A. Energy is transformed from solar to chemical energy and to heat along the food chain, never created or destroyed. B. High-diversity regions lie in the tropics, near the equator. Reasons: long undisturbed evolutionary time (no glaciation), less seasonal and more constant environment, and more solar energy leading to higher productivity.
Where the marks are
- States the First Law and the conversion of solar to chemical energy by producers (1 mark)
- Explains transfer through consumers, loss as heat and passage to detritivores, with energy only transformed (1 mark)
- Identifies tropical or near-equatorial location as the common factor (1 mark)
- Any two correct reasons for greater tropical diversity (2 marks)
Open Q33(B) on its own page

Name of Place	Number of Bird species
Columbia	1400
India	1200
Northern South America	1300
New York	105
Denmark	504

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