Section 15 — Board Exam-Pattern Questions (CBSE Class 12 Biology · Chapter 5)
Chapter 5 (Molecular Basis of Inheritance) is one of the highest-yield chapters in the CBSE Class 12 Biology Board exam. Its parent unit (Genetics and Evolution) carries about 20 marks, and a substantial share typically comes from this chapter — spread across MCQs, 2/3-mark short answers, and frequently a 5-mark question on DNA structure, replication, transcription–translation or the lac operon.
This section curates 24 high-yield Board exam-pattern questions (modelled on the CBSE style — not year-tagged official questions) in CBSE answer-writing format and mark-weighting:
| Marks | Number of Qs | Indicative frequency in papers |
|---|---|---|
| 1 mark | 7 questions (Q1–Q7) | 2–3 / paper |
| 2 marks | 6 questions (Q8–Q13) | 2 / paper |
| 3 marks | 6 questions (Q14–Q19) | 1–2 / paper |
| 5 marks | 5 questions (Q20–Q24) | 1 / paper (very high probability) |
| Total | 24 questions | 62 marks of practice |
How to use: Cover the answer, attempt it yourself scaled to the marks, then check against the model answer.
No quiz at the end: Pair with Sections 14 (Solved Examples) and 16 (NEET-Pattern Practice Questions).
What CBSE Keeps Asking
Most CBSE questions on Chapter 5 fall into 6 recurring themes:
1. DNA structure & Chargaff (1- or 2-mark) — base pairing, hydrogen bonds, antiparallel strands, A=T/G=C sums.
2. The classic experiments (2- or 3-mark) — Griffith's transforming principle, Avery–MacLeod–McCarty, Hershey–Chase, and especially Meselson–Stahl band interpretation.
3. Replication & enzymes (3- or 5-mark) — semiconservative proof, leading vs lagging strand, Okazaki fragments, DNA polymerase and ligase.
4. Transcription, code & translation (3- or 5-mark) — transcription unit, template vs coding strand, features of the genetic code, charging of tRNA, ribosome as ribozyme.
5. The lac operon (3- or 5-mark) — structure (i, p, o, z, y, a), the repressor, lactose as inducer, ON/OFF logic, negative regulation.
6. HGP & DNA fingerprinting (2-, 3- or 5-mark) — salient features of the human genome, VNTR, steps and applications.
1-Mark Questions (Q1–Q7)
Q1. [1 mark] How many hydrogen bonds are present between adenine and thymine in a DNA double helix?
Answer: Two hydrogen bonds (A=T); guanine and cytosine are held by three (G≡C).
Q2. [1 mark] Name the three stop (termination) codons of the genetic code.
Answer: UAA, UAG and UGA.
Q3. [1 mark] Which radioactive isotope was used to label the DNA of bacteriophages in the Hershey–Chase experiment?
Answer: Radioactive phosphorus, (protein was labelled with ).
Q4. [1 mark] What is the role of DNA ligase in replication?
Answer: It joins (seals) the Okazaki fragments of the lagging strand into a continuous strand.
Q5. [1 mark · Assertion–Reason]
Assertion (A): AUG is called the initiator codon. Reason (R): AUG also codes for the amino acid methionine.
Options: (a) Both A and R true and R explains A. (b) Both true but R doesn't explain A. (c) A true, R false. (d) A false, R true.
Answer: (b) — both statements are true, but coding for methionine is a separate fact and does not by itself explain why AUG is the start signal.
Q6. [1 mark] Which enzyme produced by the z gene of the lac operon breaks down lactose?
Answer: Beta-galactosidase, which splits lactose into galactose and glucose.
Q7. [1 mark] Expand VNTR as used in DNA fingerprinting.
Answer: Variable Number of Tandem Repeats.
2-Mark Questions (Q8–Q13)
Q8. [2 marks] State Chargaff's rule. If a DNA sample has 18% cytosine, what is the percentage of adenine?
Answer: Chargaff's rule: in double-stranded DNA, A = T and G = C. Here C = G = 18%, so G + C = 36%, leaving 64% for A + T. Since A = T, adenine = 32%.
Q9. [2 marks] Distinguish between the template strand and the coding strand of a transcription unit.
Answer: The template strand has polarity 3'→5' and is the one actually read by RNA polymerase to make RNA. The coding strand runs 5'→3', is not transcribed, and has the same sequence as the mRNA except that T replaces U.
Q10. [2 marks] Why is DNA replication called semiconservative? Name the scientists who proved it in E. coli.
Answer: It is semiconservative because each daughter DNA keeps one parental (old) strand and one newly synthesised strand. It was proved in E. coli by Meselson and Stahl (1958) using density-gradient centrifugation.
Q11. [2 marks] Why can DNA polymerase synthesise the lagging strand only in pieces? What are these pieces called?
Answer: DNA polymerase works only in the 5'→3' direction. On the strand whose template runs the opposite way, synthesis must proceed away from the fork in short stretches, called Okazaki fragments, which are later joined by DNA ligase.
Q12. [2 marks] Name the inducer of the lac operon and state what it does to the repressor.
Answer: Lactose is the inducer. It binds and inactivates the repressor, so the repressor can no longer hold the operator; RNA polymerase then transcribes the z, y and a genes.
Q13. [2 marks] Mention any two salient features of the human genome revealed by the Human Genome Project.
Answer: Any two, e.g.: the genome has about 3164.7 million base pairs; it contains roughly 30,000 genes; 99.9% of bases are identical between any two people; less than 2% codes for protein; chromosome 1 has the most genes (2968) and the Y the fewest (231).
3-Mark Questions (Q14–Q19)
Q14. [3 marks] Describe the Hershey–Chase experiment and state its conclusion.
Answer: Hershey and Chase (1952) grew bacteriophages on media so that some had DNA labelled with and others had protein labelled with . The phages were allowed to infect E. coli, then the cultures were agitated in a blender (to detach viral coats) and centrifuged. The (DNA) was found inside the bacteria, while the (protein coat) remained outside.
Conclusion: Only the DNA entered the bacteria to direct the production of new phages, proving that DNA is the genetic material.
Q15. [3 marks] List any six salient features of the Watson–Crick double-helix model of DNA.
Answer:
- Two antiparallel polynucleotide chains, one 5'→3', the other 3'→5'.
- Sugar–phosphate backbone outside; nitrogenous bases project inward.
- Base pairing by hydrogen bonds: A=T (2 bonds), G≡C (3 bonds).
- A purine always pairs with a pyrimidine, giving a uniform helix width.
- The helix is right-handed, pitch 3.4 nm, about 10 bp per turn (0.34 nm between base pairs).
- Base stacking, along with hydrogen bonds, stabilises the helix.
Q16. [3 marks] Explain the Meselson–Stahl experiment and how its results support semiconservative replication.
Answer: E. coli was grown for many generations in heavy medium so that all DNA was heavy, then shifted to light medium. DNA extracted at each generation was separated by CsCl density-gradient centrifugation.
- After one generation: a single hybrid (intermediate) band — ruling out the conservative model.
- After two generations: equal amounts of hybrid and light DNA — ruling out the dispersive model.
Only semiconservative replication (each daughter = one old + one new strand) explains both results.
Q17. [3 marks] List any six features of the genetic code.
Answer: The genetic code is: (1) triplet — three bases code one amino acid; (2) degenerate — most amino acids have more than one codon; (3) unambiguous and specific — one codon codes for only one amino acid; (4) nearly universal — the same codon means the same amino acid in almost all organisms; (5) comma-less (non-overlapping) — read continuously, three bases at a time; (6) it has a start codon (AUG) and three stop codons (UAA, UAG, UGA).
Q18. [3 marks] Describe the structure of the lac operon, naming each component and its function.
Answer: The lac operon has:
- A regulatory gene (i) — codes for the repressor protein.
- A promoter (p) — the binding site for RNA polymerase.
- An operator (o) — the site where the repressor binds.
- Three structural genes: z → beta-galactosidase (splits lactose into galactose + glucose), y → permease (lactose entry into the cell), a → transacetylase.
Together these make and regulate the enzymes needed for lactose metabolism.
Q19. [3 marks] Write the steps of DNA fingerprinting in their correct sequence.
Answer: (1) Isolation of DNA from the sample; (2) digestion of DNA with restriction endonucleases; (3) separation of fragments by gel electrophoresis according to size; (4) blotting — transferring fragments onto a nylon/nitrocellulose membrane (Southern blotting); (5) hybridisation with a labelled VNTR probe; (6) detection of the band pattern by autoradiography.
5-Mark Questions (Q20–Q24)
Q20. [5 marks] Describe Griffith's transformation experiment with Streptococcus pneumoniae, and explain how Avery, MacLeod and McCarty identified the transforming principle.
Answer:
Griffith (1928) worked with two strains: the S strain (smooth, capsuled, virulent — kills mice) and the R strain (rough, no capsule, non-virulent). His results were:
- Live S → mouse dies.
- Live R → mouse lives.
- Heat-killed S → mouse lives.
- Heat-killed S + live R → mouse dies, and live S bacteria are recovered from it.
He concluded that a 'transforming principle' had passed from the dead S cells, converting harmless R cells into virulent S — but its chemical nature was unknown.
Avery, MacLeod and McCarty purified the biochemicals from heat-killed S and treated them with specific enzymes. Transformation was destroyed by DNase but not by protease or RNase, proving the transforming principle is DNA.
Q21. [5 marks] Explain the process of transcription in prokaryotes, mentioning the transcription unit, the enzyme involved and the roles of the initiation and termination factors.
Answer:
Transcription is the synthesis of RNA from a DNA template. A transcription unit has three parts along the DNA: a promoter, the structural gene, and a terminator. Of the two DNA strands, only the template strand (3'→5') is copied; the other is the coding strand.
The enzyme is DNA-dependent RNA polymerase, which synthesises RNA in the 5'→3' direction, with A pairing to U.
- Initiation: RNA polymerase binds the promoter; the sigma (σ) factor helps it start.
- Elongation: the polymerase moves along the template, adding ribonucleotides.
- Termination: the rho (ρ) factor helps the polymerase and the completed RNA fall off at the terminator.
In bacteria a single RNA polymerase makes all RNA types, and because there is no nucleus, transcription and translation can occur together.
Q22. [5 marks] Explain the regulation of the lac operon in E. coli in the presence and absence of lactose. Why is it called negative regulation?
Answer:
The i gene continuously produces a repressor protein. Whether the operon runs depends on lactose, the inducer.
In the absence of lactose (operon OFF): the repressor binds the operator and physically blocks RNA polymerase, so the structural genes z, y and a are not transcribed and no lactose-metabolising enzymes are made.
In the presence of lactose (operon ON): lactose binds the repressor and inactivates it; the inactivated repressor leaves the operator. RNA polymerase can now move from the promoter and transcribe z, y and a, producing beta-galactosidase, permease and transacetylase.
It is called negative regulation because the controlling protein — the repressor — works by blocking (switching off) transcription; removing its action turns the operon on. The model was worked out by Jacob and Monod.
Q23. [5 marks] Describe the process of translation, explaining charging of tRNA, the role of the ribosome, and the three stages of polypeptide synthesis.
Answer:
Translation is the synthesis of a polypeptide whose amino-acid sequence is dictated by the mRNA codons.
Charging (aminoacylation): each amino acid is first activated using ATP and then attached to its specific tRNA, forming aminoacyl-tRNA. This is the charging of tRNA.
Ribosome: the site of protein synthesis, made of two subunits. Its rRNA acts as a ribozyme (a catalytic RNA) that forms the peptide bonds.
Three stages:
- Initiation: the small subunit binds the mRNA and the initiator tRNA pairs at the AUG start codon; the large subunit then joins.
- Elongation: tRNAs bring amino acids codon by codon; peptide bonds link them as the ribosome moves along the mRNA.
- Termination: at a stop codon (UAA/UAG/UGA) a release factor binds (no tRNA reads a stop codon), and the completed polypeptide is released.
The untranslated regions (UTRs) at the 5' and 3' ends are needed for efficient translation.
Q24. [5 marks] What were the goals of the Human Genome Project? List its salient findings, and outline how DNA fingerprinting uses genome variation.
Answer:
Goals of the HGP (1990–2003): identify all the genes (about 20,000–25,000); determine the sequence of the ~3 billion base pairs; store the information in databases; develop tools for data analysis; transfer related technologies; and address the ethical, legal and social issues (ELSI).
Salient findings:
- The genome has about 3164.7 million base pairs; an average gene is ~3000 bp.
- There are roughly 30,000 genes, fewer than expected.
- 99.9% of the bases are identical in all people.
- Less than 2% of the genome codes for proteins; much is repetitive.
- Chromosome 1 has the most genes (2968) and the Y the fewest (231); about 1.4 million SNPs were identified.
DNA fingerprinting exploits the tiny 0.1% variation, which lies mostly in non-coding VNTR (Variable Number of Tandem Repeats) regions. Because the copy number of these repeats differs between people, cutting and probing the DNA gives a person-specific band pattern, used in forensics, paternity disputes and population studies. The technique was developed by Alec Jeffreys.
End of Section 15
You have now worked through 24 high-yield CBSE Board questions spanning DNA structure, the classic experiments, replication, transcription, the genetic code, translation, the lac operon, the Human Genome Project and DNA fingerprinting. If you can answer the five 5-markers (Q20–Q24) from memory with their key points, you have effectively secured the chapter's contribution to your Board paper.
Final tip: On Board day, draw and label diagrams (the double helix, the replication fork, the lac operon, the steps of fingerprinting) wherever the question allows — examiners give marks for clear, labelled structure. Tabulate comparison-type answers (template vs coding strand, leading vs lagging strand).