CBSE Class 12 Biology Sample Paper 2026 (with Solutions) — Free PDF
This is a full-length CBSE Class 12 Biology paper built to the current board blueprint: 33 questions across five sections for 70 marks in 3 hours, the rema
TL;DR: This is a full-length CBSE Class 12 Biology paper built to the current board blueprint: 33 questions across five sections for 70 marks in 3 hours, the…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
This is a full-length CBSE Class 12 Biology paper built to the current board blueprint: 33 questions across five sections for 70 marks in 3 hours, the remaining 30 marks being the practical component assessed in school. It covers reproduction, genetics and evolution, biology in human welfare, biotechnology and ecology in the proportions the board uses. Every question is followed by a complete worked answer written the way a full-credit board answer is expected to read.
Board: CBSE · Class: Class 12 · Subject: Biology · Total: 70 marks · Time: 3 hours
Section A — Objective (1 mark each) — 16 marks
Questions 1–14 are multiple choice. Questions 15 and 16 are Assertion–Reason: mark (a) if both A and R are true and R explains A, (b) if both are true but R does not explain A, (c) if A is true and R is false, (d) if A is false and R is true.
Q1. The part of the flower that develops into the fruit is the (1 mark)
Answer: The ovary. After fertilisation the ovary wall becomes the pericarp of the fruit, while the ovules become the seeds.
Q2. Double fertilisation in angiosperms produces (1 mark)
Answer: A diploid zygote and a triploid primary endosperm nucleus. One male gamete fuses with the egg, the other with the two polar nuclei.
Q3. In humans, the site of fertilisation is the (1 mark)
Answer: The ampullary–isthmic junction of the fallopian tube, not the uterus.
Q4. The hormone responsible for the milk ejection reflex is (1 mark)
Answer: Oxytocin, released from the posterior pituitary; prolactin governs milk production rather than its ejection.
Q5. A cross between a tall heterozygous pea plant (Tt) and a short one (tt) gives a phenotypic ratio of (1 mark)
Answer: 1 tall : 1 short. This is a test cross, and the offspring are half Tt and half tt.
Q6. Haemophilia is an example of (1 mark)
Answer: An X-linked recessive disorder. It appears far more often in males, who carry only one X chromosome and so cannot mask the allele.
Q7. The enzyme that joins DNA fragments during replication is (1 mark)
Answer: DNA ligase. It seals the nicks between Okazaki fragments on the lagging strand by forming phosphodiester bonds.
Q8. In the lac operon, the inducer is (1 mark)
Answer: Lactose (specifically allolactose). It binds the repressor, which then cannot bind the operator, so transcription proceeds.
Q9. The genetic material of the HIV virus is (1 mark)
Answer: Single-stranded RNA. HIV is a retrovirus and uses reverse transcriptase to make DNA from its RNA.
Q10. Which of these is a primary lymphoid organ: spleen, thymus, or tonsils? (1 mark)
Answer: The thymus. Primary lymphoid organs are where lymphocytes mature; the spleen and tonsils are secondary organs where they encounter antigens.
Q11. The bacterium used to produce the Bt toxin is (1 mark)
Answer: Bacillus thuringiensis. Its cry genes are transferred into crops such as Bt cotton to give insect resistance.
Q12. Restriction endonucleases cut DNA at (1 mark)
Answer: Specific palindromic sequences, usually producing staggered cuts that leave sticky ends able to pair with complementary fragments.
Q13. An organism's ecological niche is best described as (1 mark)
Answer: Its functional role in the ecosystem — what it eats, what eats it, and the range of conditions it tolerates — as opposed to its habitat, which is merely where it lives.
Q14. In an ecological pyramid, the pyramid of energy is always (1 mark)
Answer: Upright. Energy is lost as heat at every transfer, so each level necessarily holds less energy than the one below it.
Q15. Assertion (A): Pollen grains of many species cause severe allergies. Reason (R): Pollen grains have a highly resistant outer wall made of sporopollenin. (1 mark)
Answer: (b) Both are true but R does not explain A. The allergy is caused by proteins carried on the pollen, not by the durability of sporopollenin.
Q16. Assertion (A): DNA is a better hereditary material than RNA. Reason (R): DNA is chemically less reactive and structurally more stable. (1 mark)
Answer: (a) Both true and R explains A. The 2'-OH group absent in DNA makes RNA more reactive, and the thymine in place of uracil aids repair — together these make DNA the more stable store of information.
Section B — Very Short Answer (2 marks each) — 10 marks
Answer all 5 questions. Each carries 2 marks.
Q1. Define parthenocarpy and give one example. (2 marks)
Answer: Parthenocarpy is the development of a fruit without fertilisation, so the fruit is seedless. Bananas are the standard example, and the effect can be induced artificially by applying growth hormones such as auxins to unpollinated flowers.
Q2. What is the difference between spermatogenesis and oogenesis in terms of the number of gametes produced? (2 marks)
Answer: Spermatogenesis produces four functional sperm from each primary spermatocyte. Oogenesis produces only one functional ovum from each primary oocyte, the other three nuclei being discarded as polar bodies, so that nearly all the cytoplasm is retained by the single egg.
Q3. State the law of independent assortment. (2 marks)
Answer: When two pairs of contrasting characters are considered together, the alleles of one pair segregate independently of the alleles of the other pair during gamete formation. Mendel's dihybrid cross gave the 9 : 3 : 3 : 1 ratio that follows from this. The law holds for genes on different chromosomes or far apart on the same one.
Q4. Write the full form of PCR and name the enzyme that makes it possible. (2 marks)
Answer: PCR is the polymerase chain reaction, used to amplify a specific DNA segment into millions of copies. It depends on Taq polymerase, a heat-stable DNA polymerase from the thermophilic bacterium Thermus aquaticus, which survives the repeated denaturation step at about 94 °C.
Q5. Distinguish between innate and acquired immunity. (2 marks)
Answer: Innate immunity is present from birth, is non-specific, and acts the same way against any pathogen — skin, mucus, phagocytes and inflammation. Acquired immunity develops after exposure to a specific pathogen, is specific to it, and carries memory, so the second response to the same antigen is faster and stronger.
Section C — Short Answer (3 marks each) — 21 marks
Answer all 7 questions. Each carries 3 marks.
Q1. Describe the structure of a mature embryo sac in a flowering plant, naming its cells. (3 marks)
Answer: A typical mature embryo sac is 7-celled and 8-nucleate. At the micropylar end sits the egg apparatus: one egg cell flanked by two synergids bearing the filiform apparatus that guides the pollen tube. At the opposite chalazal end lie three antipodal cells. In the centre is one large central cell containing two polar nuclei, which after fertilisation form the triploid endosperm.
Q2. Explain the roles of FSH, LH and progesterone in the human menstrual cycle. (3 marks)
Answer: FSH from the anterior pituitary stimulates the growth of the ovarian follicle and its secretion of oestrogen during the follicular phase. A sharp surge of LH at about day 14 triggers ovulation, the release of the secondary oocyte, and then converts the ruptured follicle into the corpus luteum. Progesterone secreted by the corpus luteum thickens and maintains the endometrium for implantation; if fertilisation does not occur, the corpus luteum degenerates, progesterone falls and menstruation follows.
Q3. A colour-blind man marries a woman with no history of colour blindness in her family. Work out the probability of colour blindness in their children. (3 marks)
Answer: Colour blindness is X-linked recessive. The father is X^c Y and the mother is X^X (homozygous normal). All daughters receive X^c from the father and X from the mother, so every daughter is X^c X — a carrier with normal vision. All sons receive the father's Y and the mother's normal X, so every son is X Y and is normal. Therefore none of the children is colour blind, but every daughter is a carrier.
Q4. Describe the process of transcription in prokaryotes, naming the three stages. (3 marks)
Answer: In initiation, RNA polymerase with its sigma factor binds the promoter and unwinds the DNA. In elongation, the polymerase moves along the template strand in the 3'→5' direction, adding ribonucleotides to the growing RNA at its 3' end following complementary base pairing, with uracil in place of thymine. In termination, the polymerase reaches the terminator sequence, the rho factor or a hairpin structure causes release of the transcript, and the enzyme detaches. In prokaryotes the transcript needs no processing and can be translated even before transcription is finished.
Q5. What is biofortification? Give two examples and state why it matters. (3 marks)
Answer: Biofortification is the breeding of crops for higher levels of vitamins, minerals or proteins, so that nutrition improves through the ordinary diet rather than through supplements. Examples include the wheat variety Atlas 66 rich in protein, maize hybrids with twice the lysine and tryptophan, and iron-fortified rice. It matters because it reaches poor populations who cannot afford supplements, using seed they already plant.
Q6. Explain the process of gel electrophoresis and state why DNA moves towards the anode. (3 marks)
Answer: DNA fragments obtained by restriction digestion are loaded into wells in an agarose gel and an electric field is applied. The fragments move through the pores of the gel, smaller fragments travelling faster and therefore further, so the mixture separates by size. The separated bands are stained with ethidium bromide and viewed under UV light as bright orange bands. DNA migrates towards the anode because the phosphate groups of its backbone carry a net negative charge at the pH used.
Q7. Define population density and describe any two ways it can be measured. (3 marks)
Answer: Population density is the number of individuals of a species per unit area or volume at a given time. Direct counting works for large, visible or sessile organisms, such as counting trees in a marked quadrat. For mobile animals the mark–recapture method is used: a sample is captured, marked and released, and the proportion of marked individuals in a later sample gives an estimate of the total. Where counting individuals is impractical, percentage cover or biomass serves as a relative measure.
Section D — Case-based questions (4 marks each) — 8 marks
Answer both case-based questions. Each has sub-parts totalling 4 marks.
Q1. CASE STUDY 1 — A couple has been unable to conceive for four years. Investigation shows the husband produces very few sperm, while the wife's reproductive tract is normal. (i) Name the condition in the husband. (ii) Suggest a suitable assisted reproductive technique and explain it. (iii) Name one technique used when the sperm count is extremely low. (iv) State one ethical concern with these technologies. (4 marks)
Answer: (i) Oligospermia — an abnormally low sperm count. (ii) Intra-uterine insemination, in which semen is collected, concentrated and introduced directly into the uterus, bypassing the need for a high count in the cervix; alternatively IVF, where egg and sperm are fused outside the body and the embryo transferred. (iii) ICSI, intracytoplasmic sperm injection, in which a single sperm is injected directly into the ovum. (iv) These techniques raise concerns about the fate of surplus embryos, the high cost placing them out of reach for most couples, and their potential misuse for sex selection, which is illegal in India.
Q2. CASE STUDY 2 — A lake near a town receives untreated sewage. Over several months the water turns green, fish die in large numbers and the water begins to smell. (i) Name the phenomenon. (ii) Explain why the fish die, referring to BOD. (iii) What does a high BOD indicate? (iv) Suggest one remedy. (4 marks)
Answer: (i) Eutrophication — nutrient enrichment of the water body, here caused by the nitrates and phosphates in sewage. (ii) The nutrients cause an algal bloom; when the algae die, decomposer bacteria multiply and consume dissolved oxygen to break them down. The oxygen level falls so far that fish suffocate. (iii) A high biochemical oxygen demand means a large amount of biodegradable organic matter, and therefore heavy pollution — clean water has a low BOD. (iv) Treat the sewage before discharge in a sewage treatment plant, so that the organic load is oxidised before the effluent reaches the lake.
Section E — Long Answer (5 marks each) — 15 marks
Answer all 3 questions. Each carries 5 marks.
Q1. (a) Describe the structure of DNA as proposed by Watson and Crick. (b) State Chargaff's rule. (c) Explain what is meant by semi-conservative replication and name the experiment that proved it. (5 marks)
Answer: (a) DNA is a double helix of two antiparallel polynucleotide chains, one running 5'→3' and the other 3'→5', wound round a common axis in a right-handed sense. The sugar–phosphate backbones lie outside and the nitrogenous bases point inwards, paired by hydrogen bonds — adenine with thymine by two bonds and guanine with cytosine by three. The helix has a pitch of 3.4 nm with about ten base pairs per turn, so adjacent pairs are 0.34 nm apart. (b) Chargaff's rule states that in any double-stranded DNA the amount of adenine equals that of thymine and the amount of guanine equals that of cytosine, so A + G = T + C. (c) In semi-conservative replication each daughter molecule keeps one parental strand and gains one newly synthesised strand. Meselson and Stahl proved it by growing E. coli in heavy nitrogen ¹⁵N and transferring it to ¹⁴N: after one generation all the DNA was of intermediate density, and after two generations half was intermediate and half light — exactly what the semi-conservative model predicts.
Q2. (a) What is meant by an ecosystem's productivity? Distinguish gross and net primary productivity. (b) Explain the 10 per cent law with an example. (c) Why are pyramids of biomass in an aquatic ecosystem often inverted? (5 marks)
Answer: (a) Productivity is the rate at which biomass is produced per unit area in unit time. Gross primary productivity is the total rate at which producers capture and fix chemical energy through photosynthesis. A part of it is spent in the producers' own respiration, and what remains as new biomass is the net primary productivity: NPP = GPP − R. Only NPP is available to herbivores. (b) The 10 per cent law states that only about a tenth of the energy at one trophic level is transferred to the next, the rest being lost as heat and in metabolism. So if producers hold 10,000 J, herbivores obtain about 1,000 J and primary carnivores about 100 J. (c) In a pond the producers are tiny, short-lived phytoplankton with a very high turnover rate. At any instant their standing biomass is small, yet over time they support a much larger biomass of longer-lived zooplankton and fish. The pyramid measures biomass at a single moment, so it appears inverted even though the energy pyramid remains upright.
Q3. (a) What is recombinant DNA technology? List the three basic steps. (b) Name the tools required and state the function of each. (c) Give one application in medicine. (5 marks)
Answer: (a) Recombinant DNA technology is the set of methods by which a gene from one organism is isolated and introduced into another, so that the host expresses the desired product. The three basic steps are the isolation of the gene of interest, its insertion into a suitable vector to form recombinant DNA, and the transfer of that construct into a host cell where it multiplies and is expressed. (b) Restriction endonucleases cut DNA at specific palindromic sites, producing sticky ends. DNA ligase joins the fragment to the vector by sealing the phosphodiester backbone. A vector, usually a plasmid such as pBR322 or a bacteriophage, carries the gene into the host and provides an origin of replication and selectable markers. A host cell, commonly E. coli, replicates and expresses the construct. Polymerases amplify the gene, and gel electrophoresis separates and checks the fragments. (c) Human insulin, marketed as Humulin, is produced by inserting the genes for the A and B chains into E. coli, growing the bacteria, then extracting and joining the chains — giving a supply free of the immune reactions that animal insulin sometimes caused.
FAQs
What is the total marks for Class 12 Biology board exam?
70 marks for theory and 30 marks for practicals; total 100 marks.
Which topics have highest weightage?
Genetics, molecular biology, evolution, and ecology carry significant weightage in the exam.
Are diagrams essential for biology answers?
Yes, labeled diagrams are required for processes like photosynthesis, respiration, and cell division.
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