Chapter 6 — Molecular Basis of Inheritance — is the molecular genetics chapter covering DNA structure, replication, transcription, translation, gene regulation, and the Human Genome Project. This carries 10-12 marks — one of the highest-weightage chapters in Biology. Master DNA replication, central dogma, and the lac operon.
Key Concepts
DNA Structure (Watson-Crick Model)
Double helix: Two antiparallel polynucleotide strands (5’→3′ and 3’→5′)
Base pairing: A=T (2 H-bonds), G≡C (3 H-bonds) → Chargaff’s rule: A=T, G=C
Sugar-phosphate backbone: Deoxyribose sugar + phosphodiester bonds
Pitch: 3.4 nm per turn; 10 bp per turn; each bp = 0.34 nm apart
Diameter: 2 nm; Right-handed helix
Base pairing: A=T (2 H-bonds), G≡C (3 H-bonds) → Chargaff’s rule: A=T, G=C
Sugar-phosphate backbone: Deoxyribose sugar + phosphodiester bonds
Pitch: 3.4 nm per turn; 10 bp per turn; each bp = 0.34 nm apart
Diameter: 2 nm; Right-handed helix
DNA Replication
Semiconservative (proved by Meselson and Stahl, 1958 using E. coli + ¹⁵N/¹⁴N)
Key enzymes:
Helicase → unwinds double helix
SSB proteins → stabilise single strands
Primase → synthesises RNA primer
DNA Polymerase III → adds nucleotides (5’→3′ only)
DNA Polymerase I → removes RNA primer, fills gaps
Ligase → joins Okazaki fragments
Leading strand: Continuous synthesis (5’→3′ towards replication fork)
Lagging strand: Discontinuous (Okazaki fragments, 5’→3′ away from fork)
Key enzymes:
Helicase → unwinds double helix
SSB proteins → stabilise single strands
Primase → synthesises RNA primer
DNA Polymerase III → adds nucleotides (5’→3′ only)
DNA Polymerase I → removes RNA primer, fills gaps
Ligase → joins Okazaki fragments
Leading strand: Continuous synthesis (5’→3′ towards replication fork)
Lagging strand: Discontinuous (Okazaki fragments, 5’→3′ away from fork)
Central Dogma
DNA →(Transcription) mRNA →(Translation) Protein
Transcription (DNA → mRNA)
Template strand: 3’→5′ (read by RNA polymerase)
Coding/sense strand: 5’→3′ (same sequence as mRNA, except T→U)
RNA polymerase synthesises mRNA in 5’→3′ direction
In prokaryotes: single RNA polymerase
In eukaryotes: RNA Pol I (rRNA), RNA Pol II (mRNA), RNA Pol III (tRNA)
Post-transcriptional processing (eukaryotes):
Capping (5′ cap) + Polyadenylation (3′ poly-A tail) + Splicing (remove introns, join exons)
Coding/sense strand: 5’→3′ (same sequence as mRNA, except T→U)
RNA polymerase synthesises mRNA in 5’→3′ direction
In prokaryotes: single RNA polymerase
In eukaryotes: RNA Pol I (rRNA), RNA Pol II (mRNA), RNA Pol III (tRNA)
Post-transcriptional processing (eukaryotes):
Capping (5′ cap) + Polyadenylation (3′ poly-A tail) + Splicing (remove introns, join exons)
Genetic Code
| Property | Details |
|---|---|
| Triplet | 3 bases = 1 codon = 1 amino acid |
| Degenerate | More than one codon for most amino acids (61 sense codons for 20 amino acids) |
| Universal | Same code in almost all organisms |
| Non-overlapping | Codons read sequentially without overlap |
| Non-ambiguous | One codon codes for only one amino acid |
| Start codon | AUG (methionine) — initiates translation |
| Stop codons | UAA, UAG, UGA — terminate translation |
Translation (mRNA → Protein)
Occurs on ribosomes. tRNA brings amino acids.
Steps:
1. Initiation: Small ribosomal subunit + mRNA + initiator tRNA (with Met) → start at AUG
2. Elongation: Aminoacyl-tRNA enters A-site → peptide bond formed → ribosome translocates
3. Termination: Stop codon (UAA/UAG/UGA) → release factor → polypeptide released
Steps:
1. Initiation: Small ribosomal subunit + mRNA + initiator tRNA (with Met) → start at AUG
2. Elongation: Aminoacyl-tRNA enters A-site → peptide bond formed → ribosome translocates
3. Termination: Stop codon (UAA/UAG/UGA) → release factor → polypeptide released
Regulation of Gene Expression — Lac Operon
Jacob and Monod model (E. coli):
Structural genes: lacZ, lacY, lacA (encode β-galactosidase, permease, transacetylase)
Promoter: RNA polymerase binds here
Operator: Repressor protein binds here
Regulator gene (i): produces repressor
Without lactose: Repressor binds to operator → blocks transcription → genes OFF
With lactose: Lactose (inducer) binds to repressor → repressor falls off → RNA polymerase transcribes → genes ON
This is an inducible operon (switched on by inducer).
Structural genes: lacZ, lacY, lacA (encode β-galactosidase, permease, transacetylase)
Promoter: RNA polymerase binds here
Operator: Repressor protein binds here
Regulator gene (i): produces repressor
Without lactose: Repressor binds to operator → blocks transcription → genes OFF
With lactose: Lactose (inducer) binds to repressor → repressor falls off → RNA polymerase transcribes → genes ON
This is an inducible operon (switched on by inducer).
Human Genome Project (HGP)
- 1990-2003, coordinated by NIH (USA) and Wellcome Trust (UK)
- Total human genome: ~3.2 billion base pairs
- ~20,000-25,000 genes (only ~2% of genome codes for proteins)
- Chromosome 1 has most genes; Y chromosome has fewest
- Used BAC and YAC vectors; Sanger’s method and shotgun sequencing
DNA Fingerprinting
Based on VNTRs (Variable Number Tandem Repeats) — unique to each individual
Steps: Isolation → Restriction digestion → Gel electrophoresis → Southern blotting → Hybridisation with probe → Autoradiography
Applications: paternity testing, forensics, immigration disputes
Steps: Isolation → Restriction digestion → Gel electrophoresis → Southern blotting → Hybridisation with probe → Autoradiography
Applications: paternity testing, forensics, immigration disputes
Solved Examples
Example 1
Q: The sequence of one strand of DNA is 3′-ATGCATGCATGC-5′. Write the mRNA sequence.
Solution: mRNA is synthesised using the template (3’→5′) strand. mRNA = 5′-UACGUACGUACG-3′ (complementary, with U replacing T).
Example 2
Q: Explain why the lac operon is called an inducible operon.
Solution: The lac operon is normally switched OFF (repressor blocks operator). It is switched ON only when lactose (inducer) is present — lactose binds to repressor, changes its shape, repressor detaches from operator, allowing transcription. Since it needs an inducer to turn on, it’s called an inducible operon.
Important Questions
3 Mark
- Describe the semiconservative nature of DNA replication.
- What is the central dogma? Draw a flow chart.
- Explain the regulation of lac operon in E. coli.
5 Mark
- Describe the process of translation with a diagram.
- Explain the Watson-Crick model of DNA. How did Meselson and Stahl prove semiconservative replication?
Quick Revision Points
- DNA: double helix, antiparallel, A=T (2 H-bonds), G≡C (3 H-bonds), 3.4 nm pitch
- Replication: semiconservative (Meselson-Stahl); leading (continuous) + lagging (Okazaki fragments)
- Transcription: template 3’→5′; RNA Pol II makes mRNA in eukaryotes
- Genetic code: 64 codons (61 sense + 3 stop); AUG = start; degenerate, universal
- Translation: initiation (AUG) → elongation → termination (UAA/UAG/UGA)
- Lac operon: inducible; lactose = inducer; repressor blocks operator when no lactose
- HGP: 3.2 billion bp, ~20,000 genes, 2% coding
- DNA fingerprinting: VNTRs, Southern blotting, used in forensics
Chapter Navigation
Previous: Principles of Inheritance and Variation Class 12 Notes
Next: Evolution Class 12 Notes
Related Chapters in Class 12 Biology
- Principles of Inheritance and Variation Class 12 Notes
- Evolution Class 12 Notes
- Biotechnology Principles and Processes Class 12 Notes
Practice What You Learned
Test yourself with our NEET Mock Test Set 1 to see how well you’ve mastered the concepts.