Question Bank — Molecular Basis of Inheritance

NCERT Class 12 Biology · Chapter 5 · CBSE Board Pattern

Class 12 Biology Chapter 5 Question Bank CBSE Board Pattern

Section A — MCQs (10 questions, 1 mark each)

  1. The two strands of DNA are held together by hydrogen bonds between base pairs. How many hydrogen bonds are formed between adenine and thymine?
    (a) One
    (b) Two
    (c) Three
    (d) Four

  2. In the Hershey-Chase experiment, bacteriophages were grown in media containing radioactive phosphorus and radioactive sulfur. Which of the following was observed?
    (a) Bacteria infected with radioactive phosphorus phages became radioactive
    (b) Bacteria infected with radioactive sulfur phages became radioactive
    (c) Both became radioactive
    (d) Neither became radioactive

  3. Which of the following is the correct sequence of the central dogma of molecular biology?
    (a) DNA → Protein → RNA
    (b) RNA → DNA → Protein
    (c) DNA → RNA → Protein
    (d) Protein → RNA → DNA

  4. In a double-stranded DNA, if the percentage of cytosine is 20%, the percentage of adenine is:
    (a) 20%
    (b) 30%
    (c) 40%
    (d) 60%

  5. The discontinuous fragments synthesised on the lagging strand during DNA replication are joined by:
    (a) DNA polymerase
    (b) DNA ligase
    (c) Helicase
    (d) RNA polymerase

Assertion-Reason Questions

  1. Assertion (A): DNA is chemically and structurally more stable than RNA and is preferred as genetic material.
    Reason (R): DNA lacks the 2'-OH group present in RNA and contains thymine instead of uracil.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

  2. Assertion (A): In transcription, only one strand of DNA is copied into RNA.
    Reason (R): If both strands are transcribed, they would code for two different RNA molecules, complicating genetic information transfer.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

  3. The enzyme that catalyses the formation of peptide bonds during translation is:
    (a) DNA polymerase
    (b) RNA polymerase
    (c) Ribozyme (23S rRNA)
    (d) DNA ligase

  4. Which of the following codons functions both as an initiator codon and codes for methionine?
    (a) UAA
    (b) AUG
    (c) UGA
    (d) UAG

  5. In the lac operon, the structural gene z codes for:
    (a) Permease
    (b) Transacetylase
    (c) β-galactosidase
    (d) Repressor

Section B — Very Short Answer (6 questions, 2 marks each)

  1. Differentiate between euchromatin and heterochromatin.
  2. What is the role of the origin of replication in DNA replication?
  3. Name the three types of RNA and state one function of each.
  4. Define the term “genetic code”. List any two salient features of the genetic code.
  5. Why is RNA considered the first genetic material?
  6. What are VNTRs? How do they form the basis of DNA fingerprinting?

Section C — Short Answer (5 questions, 3 marks each)

  1. Describe the salient features of the double-helix structure of DNA.
  2. Explain the transformation experiment of Griffith. How did Avery, MacLeod and McCarty identify the biochemical nature of the transforming principle?
  3. What is a transcription unit? Describe the three regions of a transcription unit.
  4. Explain the process of charging of tRNA (aminoacylation) and its significance in translation.
  5. Calculate the length of DNA in a typical mammalian cell if the haploid content is 3.3 × 10⁹ bp and the distance between two consecutive base pairs is 0.34 nm. Show the steps.

Section D — Long Answer (3 questions, 5 marks each)

  1. Describe the semiconservative replication of DNA with the help of a labelled diagram of the replicating fork. Explain the roles of DNA polymerase, DNA ligase and replication fork. (Diagram required)
  2. Explain the process of transcription in eukaryotes. How does it differ from transcription in prokaryotes? Describe the post-transcriptional modifications of hnRNA.
  3. (i) Calculate the number of base pairs in E. coli DNA if its length is given as 1.36 mm. (Distance between two consecutive base pairs = 0.34 nm). Show full working.
    (ii) How is such a long DNA packaged in a prokaryotic cell? Describe the role of histone proteins in eukaryotic DNA packaging.

Section E — Case/Source-Based (2 questions, 4 marks each)

Case 1:

In 1952, Alfred Hershey and Martha Chase conducted experiments with bacteriophages to determine whether protein or DNA is the genetic material. They labelled phages with radioactive ³²P (which labels DNA) and ³⁵S (which labels protein). After infection, they used a blender to separate the phage coats from the bacteria and centrifuged the mixture.

Sub-questions:
(a) Why were radioactive phosphorus and sulfur chosen for labelling?
(b) What observation was made regarding radioactivity in the bacterial pellet?
(c) What conclusion was drawn from the experiment?
(d) How does this experiment support that DNA, and not protein, is the genetic material?

Case 2:

In the absence of glucose, when lactose is added to the growth medium of E. coli, the lac operon is induced. The operon consists of a regulatory gene (i) and three structural genes (z, y, a) controlled by a common promoter and operator. The repressor protein binds to the operator and prevents transcription in the absence of lactose.

Sub-questions:
(a) What is the product of the i gene and its function?
(b) Name the enzymes coded by z, y and a genes.
(c) Explain how lactose acts as an inducer.
(d) Why is regulation of the lac operon considered both negative and inducible regulation?

▶ Answer Key Attempt all questions first,
then tap to reveal

Section A

  1. (b) Two — 1 mark
  2. (a) Bacteria with ³²P became radioactive — 1 mark
  3. (c) DNA → RNA → Protein — 1 mark
  4. (b) 30% (A = T, G = C; A = 30%) — 1 mark
  5. (b) DNA ligase — 1 mark
  6. (a) Both true and R explains A — 1 mark
  7. (a) Both true and R explains A — 1 mark
  8. (c) Ribozyme (23S rRNA) — 1 mark
  9. (b) AUG — 1 mark
  10. (c) β-galactosidase — 1 mark

Section B

  1. Euchromatin: loosely packed, transcriptionally active, light staining. Heterochromatin: densely packed, inactive, dark staining. (1+1)
  2. Site where replication begins; provides origin for vectors in rDNA technology. (1+1)
  3. mRNA (template for protein), tRNA (adapter, brings amino acids), rRNA (structural + catalytic in ribosome). (Any two correct with function)
  4. Genetic code: triplet sequence of bases in mRNA that codes for amino acids. Features: degenerate, universal, unambiguous, read continuously, AUG initiator, UAA/UGA/UAG stop. (Definition + any two)
  5. RNA acted as both genetic material and catalyst (ribozyme); DNA evolved later for stability.
  6. VNTRs = variable number tandem repeats (mini-satellites); highly polymorphic, produce unique banding pattern after hybridisation.

Section C

  1. Two antiparallel strands, right-handed helix, pitch 3.4 nm, 10 bp/turn, A-T (2 H-bonds), G-C (3 H-bonds), uniform distance due to purine-pyrimidine pairing, stacking stability. (Any 3 points)
  2. Griffith: live R + heat-killed S → live S recovered (transformation). Avery et al.: only DNA from S strain transformed R to S; proteases/RNases ineffective, DNase inhibited.
  3. Promoter (upstream, RNA pol binding), structural gene (coding sequence), terminator (downstream, transcription stop).
  4. Amino acid + ATP + tRNA → aminoacyl-tRNA (charged tRNA) catalysed by aminoacyl-tRNA synthetase. Provides activated amino acid for peptide bond formation.
  5. Length = 3.3 × 10⁹ × 2 × 0.34 × 10⁻⁹ m = 2.244 m (≈ 2.2 m). (Calculation shown)

Section D

  1. Semiconservative: each daughter duplex has one parental + one new strand. Diagram of replication fork showing leading (continuous 5'→3') and lagging (Okazaki fragments) strands, helicase, primase, DNA pol III, ligase. Roles as per text.
  2. Eukaryotic transcription: three RNA pols (I = rRNA, II = hnRNA, III = tRNA/5S rRNA). Post-transcriptional: splicing (remove introns), capping (5' methyl-G), tailing (3' poly-A). Prokaryotes: single RNA pol, no processing, coupled transcription-translation.
  3. (i) 1.36 mm = 1.36 × 10⁻³ m; bp = (1.36 × 10⁻³) / (0.34 × 10⁻⁹) = 4 × 10⁶ bp. (Working shown)
    (ii) Prokaryotes: nucleoid with looped DNA + proteins. Eukaryotes: nucleosome (DNA + histone octamer, 200 bp), beads-on-string → chromatin fibres → chromosomes; NHC proteins; euchromatin/heterochromatin.

Section E

Case 1

(a) ³²P labels DNA (phosphorus present), ³⁵S labels protein (sulfur present).
(b) ³²P radioactivity entered bacteria; ³⁵S remained outside.
(c) DNA is the genetic material.
(d) Only DNA entered bacterial cell and directed viral reproduction.

Case 2

(a) i gene codes repressor protein that binds operator and blocks transcription.
(b) z = β-galactosidase, y = permease, a = transacetylase.
(c) Lactose/allolactose binds repressor, inactivates it, allows RNA polymerase to transcribe.
(d) Negative (repressor control) + inducible (substrate lactose induces).

← Revision Summary All Chapters

All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.