CBSE Class 12 Biology Board Paper 1 — MCQs with Answers
Original paper · CBSE patternWritten by our subject team, not a reprint of an official sample paper.
- Questions
- 20
- Time
- 30 min
- Marking
- +1, no negative marking
- Mix
- 4 Easy · 8 Medium · 8 Hard
Board Paper 1 — CBSE Class 12 Biology is a full 20-question objective paper for Class 12 Biology, written in the CBSE board pattern and marked the way the board marks: +1 for a correct answer and 0 for a wrong one — no negative marking, exactly like the real board paper. It follows Section A of a real CBSE paper: Q1–Q16 are single-correct MCQs, Q17–Q18 are case-based questions built on a short source, and Q19–Q20 are assertion–reason items using the board's own four option strings.
The difficulty sits deliberately a notch above the real board — 4 Easy · 8 Medium · 8 Hard — because a paper you can clear comfortably tells you nothing on exam day. You get 30 minutes — about 90 seconds a question, the pace the real paper demands. 8 of the questions are solved in full below, with the correct option and a worked explanation for each; the remaining 12 are timed and scored when you sign in free.
Sit it under exam conditions rather than open-book. The score you get on a timed full-length paper is the only honest signal of whether your Biology revision is holding together across chapters — which is precisely what the board tests and what chapter-wise practice cannot tell you.
8 solved questions from this paper
Answer and worked explanation shown for each. The remaining 12 are timed and scored when you sign in free.
- Q1Easy
In a typical angiosperm ovule the pollen tube most often enters the embryo sac through the narrow pore left by the integuments at the tip of the ovule. This pore is the:
A.hilumB.micropyle✓ CorrectC.chalazaD.funicleAnswer: B. micropyle
Explanation: The micropyle is the opening left where the integuments fail to meet, and pollen tube entry through it, called porogamy, is the commonest route in angiosperms. The hilum is merely the scar marking where the funicle is attached to the body of the ovule, so it plays no part in fertilisation.
- Q2Easy
Ringworm, one of the commonest infectious diseases in humans, is caused by members of the genera Microsporum, Trichophyton and Epidermophyton. These organisms are:
A.bacteriaB.protozoansC.fungi✓ CorrectD.helminthsAnswer: C. fungi
Explanation: All three genera are dermatophytic fungi that live on the keratin of skin, nails and scalp and produce dry, scaly lesions with intense itching. Heat and moisture favour their growth, which is why the infection travels on shared towels and clothing, and antibacterial drugs have no effect on them.
- Q3Easy
The amount of organic matter produced by autotrophs per unit area per unit time, counted before any of it is used up in respiration, is called:
A.gross primary productivity✓ CorrectB.net primary productivityC.secondary productivityD.standing cropAnswer: A. gross primary productivity
Explanation: Gross primary productivity is the total rate at which producers fix energy as organic matter; subtracting the respiratory loss R gives net primary productivity, and it is the net figure that becomes available to herbivores. Standing crop is the biomass present at one moment and is not a rate at all.
- Q4Medium
In the human testis, spermiation is the process in which:
A.mature spermatozoa are detached from the Sertoli cells and shed into the lumen of the seminiferous tubules✓ CorrectB.spermatids are transformed into spermatozoaC.primary spermatocytes divide to form secondary spermatocytesD.semen is expelled through the urethraAnswer: A. mature spermatozoa are detached from the Sertoli cells and shed into the lumen of the seminiferous tubules
Explanation: Spermiogenesis is the reshaping of spermatids into spermatozoa; spermiation is the later step in which those sperm heads, until then embedded in Sertoli cells, are released into the tubular lumen. Expulsion of semen through the urethra is ejaculation, an event far downstream of both.
- Q5Medium
E. coli grown for many generations on ¹⁵NH₄Cl was transferred to a medium containing only ¹⁴NH₄Cl. After exactly two rounds of replication in the light medium, the DNA obtained on CsCl density-gradient centrifugation would be:
A.entirely of intermediate (hybrid) densityB.25% hybrid and 75% lightC.50% hybrid and 50% light✓ CorrectD.entirely of light densityAnswer: C. 50% hybrid and 50% light
Explanation: Meselson and Stahl showed replication to be semiconservative, so each parental strand is conserved: after one round the single molecule gives two molecules, both hybrid. In the second round those two hybrids give four molecules, two hybrid and two entirely light, which is one half hybrid and one half light; an all-hybrid result belongs to the first generation only.
- Q6Medium
The immunosuppressant cyclosporin A, which made organ transplantation practicable, is obtained from:
A.Monascus purpureusB.Aspergillus nigerC.Streptococcus thermophilusD.Trichoderma polysporum✓ CorrectAnswer: D. Trichoderma polysporum
Explanation: Cyclosporin A is a cyclic peptide obtained from the fungus Trichoderma polysporum and it suppresses T-cell activation, so the graft is not rejected. Monascus purpureus yields statins, which lower blood cholesterol by competitively inhibiting the enzyme that makes it, a completely different clinical use.
- Q7Hard
A man of blood group AB marries a woman of blood group O whose parents were both of blood group O. The probability that their first child will be of blood group B is:
A.1/2✓ CorrectB.1/4C.0D.3/4Answer: A. 1/2
Explanation: The father is Iᴬ Iᴮ and the mother is ii, so his gametes carry Iᴬ or Iᴮ with equal chance while hers all carry i. Every child is therefore Iᴬi, which is group A, or Iᴮi, which is group B, in equal proportion, giving a one-half chance of group B; groups AB and O cannot appear at all in this cross.
- Q8Hard
A dihybrid female whose two genes lie 20 map units apart on the same chromosome is test-crossed with a fully recessive male. Out of 1,000 offspring, the number expected to be of recombinant types is:
A.100B.200✓ CorrectC.400D.500Answer: B. 200
Explanation: One map unit is defined as 1% recombination, so 20 map units means 20% of the gametes are recombinant: 0.20 × 1,000 = 200 offspring, split roughly equally between the two recombinant classes, leaving 800 parental types. A figure of 500 would arise only if the two genes were unlinked and assorted independently.
Sit the full paper — free
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Attempt this paper free →More Class 12 Biology board papers
- Board Paper 1 — you are here
- Board Paper 2 — 20 questions
- Board Paper 3 — 20 questions
- Board Paper 4 — 20 questions
- Board Paper 5 — 20 questions
Prefer chapter-by-chapter revision first? Class 12 Biology notes & chapter MCQs →
Questions students ask about this paper
Is this an official CBSE sample paper?
No — and that is deliberate. This is an original paper written in the CBSE board pattern by our subject team, not a copy of CBSE's official Sample Question Paper. It means you get fresh questions you have not already seen on a dozen other sites, with instant scoring and worked solutions rather than a PDF.
How is this Class 12 Biology paper marked?
+1 for a correct answer and 0 for a wrong one — no negative marking, exactly like the real board paper. So attempt every question — there is no penalty for a wrong answer on a board paper, and leaving a question blank can only cost you.
How long should this paper take?
30 minutes for 20 questions, roughly 90 seconds each. The timer runs whether or not you are watching it, which is the point: board marks are lost to pacing at least as often as to gaps in knowledge.
Do I need to pay to attempt it?
No. Sign in free with Google and the full paper opens with a live timer, automatic scoring and a worked solution for every question — including the ones not shown on this page.