In Class 12 Biology, "Molecular Basis of Inheritance" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
If you unraveled and tied together the microscopic DNA strands contained inside a single human body, the thread would stretch to Pluto and back twice! How does this 2-meter chemical helix replicate and decode into living proteins? The Central Dogma of Molecular Biology.
Why This Chapter Matters
In Class 12 Biology, "Molecular Basis of Inheritance" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.
Before You Begin (Prerequisites)
DNA structure from Chapter 4 and Chemistry.
Nitrogenous bases.
Enzymes.
What You Will Learn (Core Objectives)
Describe the Double Helix model of B-DNA (Watson and Crick, Chargaff's rule $A+G = T+C$).
Explain DNA packaging in Eukaryotes: Histone Octamer, Nucleosome ($200\text{ bp}$ DNA), Euchromatin vs Heterochromatin.
Evaluate Transforming Principle experiments: Griffith (1928), Avery-MacLeod-McCarty (1944), and Hershey-Chase Bacteriophage experiment ($^{32}\text{P}$ vs $^{35}\text{S}$).
Explain Semi-Conservative DNA Replication (Meselson and Stahl experiment with $^{15}\text{N}$).
Explain Central Dogma: Transcription, Genetic Code properties (degenerate, universal, non-overlapping), and Translation.
Explain regulation of Gene Expression: The Lac Operon model (Jacob and Monod).
Chapter Roadmap & Progression
11. Hershey-Chase & Meselson-Stahl P...
22. Transcription & The Genetic Code
33. The Lac Operon: Genetic Circuitr...
Complete Concept Guide (100% Curriculum Coverage)
1. Hershey-Chase & Meselson-Stahl Proofs
Hershey-Chase (1952): Grew bacteriophages on radioactive $^{32}\text{P}$ (labels DNA) and $^{35}\text{S}$ (labels protein coat). Only $^{32}\text{P}$ entered bacteria, proving definitively that DNA is the genetic material!
Meselson-Stahl (1958): Grew E. coli in heavy $^{15}\text{NH}_4\text{Cl}$ then shifted to $^{14}\text{N}$. After one generation (20 min), CsCl centrifugation showed intermediate density ($^{15}\text{N}-^{14}\text{N}$), proving Semi-Conservative Replication!
2. Transcription & The Genetic Code
RNA polymerase transcribes DNA template ($3' \to 5'$) into mRNA ($5' \to 3'$). Post-transcriptional processing in eukaryotes: Capping (methylguanosine triphosphate), Polyadenylation tailing, and Splicing (removing non-coding introns!). Genetic Code (64 Codons): Triplet; Degenerate (61 sense codons code 20 amino acids); Unambiguous; Universal; AUG is Dual Function (initiator codon + codes for Methionine); Stop codons: UAA, UAG, UGA.
3. The Lac Operon: Genetic Circuitry
François Jacob and Jacques Monod discovered the Lac Operon in E. coli: • Absence of Lactose: Regulatory gene $i$ produces Repressor protein that binds operator ($O$), blocking RNA polymerase. (Operon is switched OFF!). • Presence of Lactose (Inducer): Lactose binds repressor, inactivating it; RNA polymerase transcribes structural genes: $z$ ($\beta$-galactosidase), $y$ (permease), $a$ (transacetylase) to metabolize lactose!
Molecular Basis of Inheritance - Key Biological & Molecular Architecture Model
Chapter Summary & 10 Key Takeaways
Takeaway 1
Hershey-Chase Experiment: Radioisotope P-32 vs S-35 proving DNA is the genetic code carrier.
Takeaway 2
Meselson-Stahl Density Centrifugation: Density gradient proving semi-conservative strand replication.
Takeaway 3
Degenerate Genetic Code: Multiple redundant codons translating the same amino acid.
Takeaway 4
Dual-Function AUG: Standard initiator codon simultaneously translating methionine.
Check Your Understanding (Diagnostic Practice Questions)
Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.
1
Describe the Hershey-Chase experiment that proved conclusively that DNA is the genetic material.
Reveal Answer & Explanation
Answer: They cultured bacteriophages on media containing radioactive phosphorus ($^{32}\text{P}$, labeling DNA) and radioactive sulfur ($^{35}\text{S}$, labeling protein capsids). Phages infected E. coli bacteria. After blending and centrifugation, radioactivity was detected inside bacterial cells for $^{32}\text{P}$, whereas $^{35}\text{S}$ remained in the supernatant liquid, proving that DNA, not protein, enters bacteria as genetic material. Radioactive P-32 entered bacteria, proving DNA is genetic material.
2
How did Meselson and Stahl prove that DNA replication is semi-conservative?
Reveal Answer & Explanation
Answer: They grew E. coli in heavy isotope $^{15}\text{NH}_4\text{Cl}$ for generations, then switched to normal $^{14}\text{N}$ medium. After one generation (20 minutes), CsCl density centrifugation yielded a hybrid density band ($^{15}\text{N}-^{14}\text{N}$); after two generations, equal amounts of hybrid and light ($^{14}\text{N}-^{14}\text{N}$) bands appeared, proving that each daughter DNA molecule retains one parental strand. CsCl density bands showed hybrid 15N-14N DNA after generation 1.
3
State any four salient features of the Genetic Code.
Reveal Answer & Explanation
Answer: (1) The code is a triplet: 61 codons code for amino acids and 3 codons (UAA, UAG, UGA) are stop signals, (2) Unambiguous and specific: one codon codes for only one amino acid, (3) Degenerate: some amino acids are coded by more than one codon, (4) Nearly universal: from bacteria to human, UUU codes for phenylalanine. Triplet, unambiguous, degenerate, and universal.
4
Explain the working of the Lac Operon when Lactose is present in the growth medium of E. coli.
Reveal Answer & Explanation
Answer: Lactose acts as an inducer; it binds to the repressor protein synthesized by the $i$ gene, inactivating the repressor. The inactive repressor can no longer bind to the operator region. RNA polymerase binds to the promoter and transcribes the structural genes $z, y, a$, synthesizing enzymes to metabolize lactose. Lactose inactivates repressor, allowing RNA polymerase to transcribe z, y, a genes.
5
What are the post-transcriptional modifications that take place in eukaryotic pre-mRNA before it leaves the nucleus?
Reveal Answer & Explanation
Answer: (1) Splicing: non-coding introns are removed and coding exons are joined together, (2) Capping: an unusual nucleotide (methyl guanosine triphosphate) is added to the 5'-end, (3) Tailing: adenylate residues (200-300 poly-A) are added to the 3'-end. Splicing of introns, 5'-capping, and 3'-polyadenylation tailing.
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