Chapter 4 — Molecular Genetics
Key Concepts Quick Recap
| Term | Definition |
|---|
| DNA | Deoxyribonucleic acid — genetic material; double helix (Watson & Crick, 1953) |
| Nucleotide | Monomer of DNA: phosphate + deoxyribose sugar + nitrogenous base |
| Chargaff's rule | A = T and G = C in double-stranded DNA |
| Replication | DNA duplication; semi-conservative (Meselson-Stahl, 1958) |
| Transcription | DNA → mRNA (in nucleus) |
| Translation | mRNA → Protein (at ribosome) |
| Codon | 3-base sequence on mRNA coding for one amino acid |
| Anticodon | 3-base sequence on tRNA complementary to codon |
| Start codon | AUG (methionine) — initiates translation |
| Stop codons | UAA, UAG, UGA — terminate translation |
| Operon | Functional unit of gene regulation in prokaryotes |
| Lac operon | E. coli model of gene regulation — induced by lactose |
I. Choose the Best Answer (MCQs)
1. DNA double helix model was proposed by: (a) Chargaff (b) Griffith (c) Watson and Crick (d) Meselson and Stahl
Answer(c) Watson and Crick James Watson and Francis Crick proposed the double helix model of DNA in 1953, using X-ray crystallography data from Rosalind Franklin and Maurice Wilkins. They were awarded the Nobel Prize in 1962.
2. Semi-conservative replication of DNA was experimentally proved by: (a) Watson and Crick (b) Meselson and Stahl (c) Kornberg (d) Chargaff
Answer(b) Meselson and Stahl Matthew Meselson and Franklin Stahl (1958) used ¹⁵N/¹⁴N labelled DNA in E. coli and density-gradient centrifugation (CsCl) to prove DNA replication is semi-conservative — each daughter DNA molecule has one old and one new strand.
3. The enzyme that synthesises RNA from a DNA template is: (a) DNA polymerase (b) RNA primase (c) RNA polymerase (d) Reverse transcriptase
Answer(c) RNA polymerase RNA polymerase transcribes the DNA template strand (3'→5') to produce mRNA (5'→3'). In prokaryotes, a single RNA polymerase does all transcription. In eukaryotes: RNA Pol I (rRNA), RNA Pol II (mRNA), RNA Pol III (tRNA).
4. The start codon in protein synthesis is: (a) UAA (b) AUG (c) UGA (d) UAG
Answer(b) AUG AUG codes for methionine (Met) and serves as the start codon — all proteins begin with methionine (may be cleaved later). UAA, UAG, UGA are stop codons — they do not code for any amino acid and terminate translation.
5. The genetic code is described as degenerate because: (a) One codon codes for more than one amino acid (b) Multiple codons can code for the same amino acid (c) The code is not universal (d) Codons overlap
Answer(b) Multiple codons can code for the same amino acid Degeneracy: 64 codons but only 20 amino acids → multiple codons per amino acid. Example: Leucine has 6 codons (UUA, UUG, CUU, CUC, CUA, CUG). Third base of codon often varies without changing amino acid (wobble position).
6. In the lac operon, the inducer is: (a) Glucose (b) RNA polymerase (c) Allolactose (d) Repressor protein
Answer(c) Allolactose Allolactose (a derivative of lactose) is the inducer — it binds to the repressor protein, inactivating it. With the repressor removed from the operator, RNA polymerase transcribes the structural genes (lacZ, lacY, lacA) for lactose metabolism.
7. Which of the following is NOT a property of the genetic code? (a) It is triplet (b) It is overlapping (c) It is degenerate (d) It is universal
Answer(b) It is overlapping The genetic code is NON-overlapping — each nucleotide is part of only one codon. The other three (triplet, degenerate, universal) are genuine properties of the genetic code.
II. Short Answer Questions
Q1. What are Chargaff's rules? How do they support the double helix model?
Answer
Erwin Chargaff (1950) observed in double-stranded DNA:
- A = T (adenine = thymine, linked by 2 hydrogen bonds)
- G = C (guanine = cytosine, linked by 3 hydrogen bonds)
- Therefore, A + G = T + C (purines = pyrimidines)
Support for double helix: Watson and Crick used Chargaff's rules to establish base pairing in their double helix model — A always pairs with T and G always pairs with C on opposite strands. This anti-parallel complementary base pairing (A-T, G-C) is the structural basis of DNA.
Q2. Explain semi-conservative replication of DNA.
Answer
Semi-conservative replication means each daughter DNA molecule retains one parental (template) strand and has one newly synthesised strand.
Meselson and Stahl experiment (1958): 1. Grew E. coli in heavy nitrogen (¹⁵N) medium — DNA became heavy (¹⁵N-¹⁵N). 2. Shifted to light ¹⁴N medium — cells replicated for one generation. 3. After 1st generation: all DNA was intermediate density (¹⁵N-¹⁴N hybrid) → one old strand + one new strand. 4. After 2nd generation: half hybrid + half light (¹⁴N-¹⁴N).
This pattern matches semi-conservative replication prediction — conclusively proving the model.
Q3. Describe the steps of translation (protein synthesis).
Answer
Translation is the synthesis of protein from mRNA at ribosomes.
Steps:
Initiation:
- Small ribosomal subunit binds to mRNA near 5' cap.
- Initiator tRNA (met-tRNA) binds to start codon (AUG) at the P site.
- Large subunit joins → complete 80S ribosome (eukaryote).
Elongation:
- A site: next aminoacyl-tRNA binds (anticodon recognises mRNA codon).
- Peptide bond forms between amino acids (peptidyl transferase).
- Ribosome translocates 3 nucleotides in 5'→3' direction.
- Used tRNA exits via E site.
- Process repeats for each codon.
Termination:
- Ribosome reaches a stop codon (UAA, UAG, UGA).
- No aminoacyl-tRNA binds to stop codon — release factor binds instead.
- Polypeptide chain is released; ribosome dissociates.
Q4. Explain the lac operon model of gene regulation.
Answer
The lac operon (Jacob and Monod, 1961) is a cluster of genes in E. coli that controls lactose metabolism.
Components:
- Structural genes: lacZ (β-galactosidase), lacY (permease), lacA (transacetylase)
- Operator (O): DNA sequence between promoter and structural genes
- Promoter (P): RNA polymerase binding site
- Repressor gene (lacI): Produces repressor protein
When lactose is ABSENT:
- Repressor protein binds to operator → blocks RNA polymerase → structural genes are NOT transcribed (OFF).
When lactose is PRESENT:
- Lactose → allolactose (inducer).
- Allolactose binds repressor → changes shape → repressor leaves operator.
- RNA polymerase transcribes lacZ, lacY, lacA → enzymes for lactose metabolism produced (ON).
Significance: Operon model showed that gene expression is regulated — genes are switched ON/OFF based on need. Foundation of molecular biology and genetic engineering.
III. Worked Examples
Example 1 — Transcription and Translation
Template DNA strand (3'→5'): 3'-TAC-ACC-GTT-ATT-5'
mRNA (5'→3'): 5'-AUG-UGG-CAA-UAA-3'
Translation:
- AUG = Met (start)
- UGG = Trp
- CAA = Gln
- UAA = Stop
Polypeptide: Met-Trp-Gln (3 amino acids, methionine may be removed)
Example 2 — Chargaff's calculation
A DNA sample has 30% adenine. Find % of each base.
A = 30% → T = 30% (A=T)
A + T + G + C = 100%
30 + 30 + G + C = 100%
G + C = 40% → G = C = 20%
Answer: A=30%, T=30%, G=20%, C=20%
Common Mistakes
- Forgetting RNA polymerase reads template strand 3'→5' and synthesises mRNA 5'→3'.
- Confusing codons (on mRNA) with anticodons (on tRNA).
- Saying replication is conservative — it is SEMI-conservative.
- Forgetting AUG is both start codon AND codes for methionine.
- Mixing up lac operon: lactose INDUCES (turns ON); repressor turns OFF.
Exam Checklist
- Chargaff's rules: A=T, G=C; calculate % from given data.
- Semi-conservative replication: Meselson-Stahl experiment.
- Central dogma: DNA → RNA → Protein.
- Start codon AUG; stop codons UAA, UAG, UGA.
- Genetic code properties: triplet, degenerate, non-overlapping, universal, unambiguous.
- Lac operon: inducer = allolactose; repressor gene lacI; structural genes lacZ, lacY, lacA.
