Molecular Basis of Inheritance Class 12 Notes - CBSE Biology Chapter 6

Chapter summary

Molecular Basis of Inheritance traces how genetic information is stored in DNA, copied, and expressed: the Watson-Crick double helix, the experiments that proved DNA is the genetic material, semiconservative replication, transcription, the genetic code, and translation, ending with gene regulation via the lac operon and genomics. It is one of the highest-weightage chapters in NEET Biology, so mastering its processes and key numbers is essential for scoring marks.

Chapter notes
🃏 Flash Cards: Molecular Basis of Inheritance

Class 12 Biology · Chapter 6 – swipe through all 10 cards to understand the whole chapter.

🧬Start here1/10

DNA Structure (Watson-Crick)

DNA is an antiparallel, right-handed double helix of deoxyribonucleotides.

A=T (2 H-bonds) · G≡C (3 H-bonds)

Purines A,G (double ring) pair with pyrimidines T,C (single ring).

  • Strands antiparallel: one 5’→3′, other 3’→5′
  • Backbone joined by 3′-5′ phosphodiester bonds
  • Right-handed helix, base + sugar = nucleoside; + phosphate = nucleotide
📏Key numbers2/10

Helix Dimensions & Chargaff

The B-DNA numbers NEET asks directly in numericals.

0.34 nm/bp · 3.4 nm/turn · 10 bp/turn

Chargaff’s rule: A = T and G = C, so A+G = T+C.

  • Length = (number of bp) × 0.34 nm
  • Turns = (number of bp) ÷ 10
  • If C = 22% then G = 22%, so A = T = (100 − 44)/2 = 28%
📦Packaging3/10

Nucleosome & Chromatin

2.2 m of DNA fits in a nucleus by wrapping on histone spools.

Nucleosome = ~200 bp DNA + histone octamer (2× H2A,H2B,H3,H4)

H1 is a linker histone, NOT part of the octamer. Human cell ~6.6×109 bp.

  • Negative DNA wraps positive histones (rich in lysine & arginine)
  • Beads-on-string → solenoid → chromatin fibre → chromosome
  • Euchromatin = loose/active; Heterochromatin = dense/inactive
🔬Classic proof4/10

Search for Genetic Material

Three experiments proved DNA, not protein, is the genetic material.

Hershey-Chase: 32P → DNA · 35S → protein

Only 32P (DNA) entered bacteria; 35S coat stayed outside.

  • Griffith (1928): transforming principle in Streptococcus pneumoniae
  • Avery-MacLeod-McCarty: transforming principle is DNA (DNase destroys it)
  • Hershey-Chase (1952) used bacteriophages → DNA is genetic material
🪜Core process5/10

DNA Replication

Replication is semiconservative: each daughter keeps one old strand.

DNA polymerase adds dNTPs only 5’→3′ · needs RNA primer

Meselson-Stahl (1958): 15N→14N in E. coli, hybrid DNA after 1 generation (CsCl).

  • Leading strand continuous; lagging strand = Okazaki fragments joined by DNA ligase
  • Starts at origin (ori); occurs in S phase in eukaryotes
  • dNTPs supply both bases and energy; polymerase proofreads
📝Key process6/10

Transcription

RNA polymerase copies one DNA strand into RNA, 5’→3′, no primer.

RNA = coding strand sequence with T → U

Template strand read 3’→5′; coding (sense) strand NOT copied.

  • Transcription unit = promoter + structural gene + terminator
  • Eukaryotes: Pol I → rRNA, Pol II → mRNA (hnRNA), Pol III → tRNA
  • hnRNA processing: splicing (remove introns) + 5′ cap + 3′ poly-A tail
🔑Code rules7/10

The Genetic Code

Codons are mRNA base triplets that specify amino acids.

64 codons · AUG = start (Met) · UAA, UAG, UGA = stop

Degenerate (many codons→1 aa) but unambiguous, universal, comma-less.

  • Triplet, non-overlapping, read continuously
  • 3 stop codons code no amino acid
  • Amino acids = (coding nucleotides ÷ 3) − 1 (for the stop codon)
🏭Protein build8/10

Translation

Ribosomes read mRNA codon-by-codon to build a polypeptide.

tRNA = adaptor (anticodon ↔ codon) · rRNA = ribozyme

UTRs flank the coding region but are not translated.

  • Initiation: ribosome assembles at start codon AUG
  • Elongation: peptide bonds join amino acids as ribosome moves
  • Termination: at a stop codon, polypeptide is released
🔀Regulation9/10

lac Operon (Jacob & Monod)

Bacteria make lactose-digesting enzymes only when lactose is present.

Lactose = inducer → inactivates repressor → genes transcribed

Inducible system, negative regulation; lactose is substrate AND switch.

  • Structural genes z (β-galactosidase), y (permease), a (transacetylase)
  • i gene makes repressor that binds operator and blocks transcription
  • Add lactose → repressor off → RNA polymerase transcribes the operon
🧾Genomics10/10

HGP & DNA Fingerprinting

Reading the human genome and identifying individuals by repeat DNA.

Human genome ~3×109 bp · ~2% codes protein · ~30,000 genes

DNA fingerprinting (Alec Jeffreys) compares VNTRs (minisatellites).

  • HGP 1990–2003; Chromosome 1 most genes, Y fewest
  • Methods: ESTs (expressed genes) and sequence annotation (whole genome)
  • Fingerprinting steps: isolation → digestion → electrophoresis → blotting → probe → autoradiography
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📝 Practice Molecular Basis of Inheritance — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2021
What is the role of RNA polymerase-III in the process of transcription in eukaryotes?
Correct answer: B. RNA polymerase III transcribes tRNA, 5S rRNA and snRNAs. Pol I makes 28S/18S/5.8S rRNA; Pol II makes hnRNA (mRNA precursor).
🔎 See the full step-by-step solution in the app →
Q2NEET 2021
Statement I: The codon ‘AUG’ codes for methionine and phenylalanine. Statement II: ‘AAA’ and ‘AAG’ both codons code for the amino acid lysine. Choose the correct option.
Correct answer: D. Statement I is false: AUG codes only for methionine (and acts as the start codon), not phenylalanine (which is UUU/UUC). Statement II is true: AAA and AAG both code for lysine (degeneracy).
🔎 See the full step-by-step solution in the app →
Q3NEET 2021
DNA fingerprinting involves identifying differences in some specific regions in the DNA sequence, called:
Correct answer: B. DNA fingerprinting compares highly variable repetitive DNA (satellite/VNTR sequences) that do not code for protein and are repeated thousands of times, differing between individuals.
🔎 See the full step-by-step solution in the app →
Q4NEET 2019
Purines found in both DNA and RNA are:
Correct answer: A. Adenine and guanine are the two purines present in both DNA and RNA. Cytosine and thymine are pyrimidines (in DNA); in RNA thymine is replaced by uracil, so the purines stay the same.
🔎 See the full step-by-step solution in the app →
Q5NEET 2019
Which of the following nucleic acids is present in an organism having 70S ribosomes only?
Correct answer: B. Organisms with only 70S ribosomes are prokaryotes. Their genetic material is a double-stranded, circular, ‘naked’ DNA (no histones, no nuclear membrane).
🔎 See the full step-by-step solution in the app →
Q6NEET 2019
What will be the sequence of mRNA produced from a template strand 3′-ATGCATGCATGCATG-5′?
Correct answer: B. mRNA is complementary and antiparallel to the template. Template 3′-ATG…-5′ gives mRNA 5′-UAC…-3′ (A->U, T->A, G->C, C->G), i.e. 5′-UACGUACGUACGUAC-3′.
🔎 See the full step-by-step solution in the app →
Q7NEET 2018
The experimental proof for the semiconservative replication of DNA was first shown in a:
Correct answer: B. Meselson and Stahl (1958) first demonstrated semiconservative replication in the bacterium Escherichia coli using 15N/14N density-gradient experiments.
🔎 See the full step-by-step solution in the app →
Q8NEET 2017
The association of histone H1 with a nucleosome indicates that:
Correct answer: C. The linker histone H1 sits on the linker DNA between nucleosomes and seals the DNA, promoting higher-order folding of the ‘beads-on-string’ into a condensed 30 nm chromatin fibre. Its presence indicates the DNA is in the condensed (less accessible) state, not actively being read.
🔎 See the full step-by-step solution in the app →
Q9NEET 2017
DNA replication in bacteria occurs:
Correct answer: C. Bacteria have no nucleus and no distinct S phase. DNA replication occurs during the C period of the bacterial cell cycle, just prior to binary fission (cell division).
🔎 See the full step-by-step solution in the app →
Q10NEET 2015
Which one of the following is NOT applicable to RNA?
Correct answer: D. Chargaff’s rule (A = T, G = C) describes double-stranded DNA. RNA is usually single-stranded, so its base ratios are not fixed by Chargaff’s rule. The other features (base pairing where folded, 5’/3′ ends, heterocyclic bases) apply to RNA.
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Frequently Asked Questions

What is the molecular basis of inheritance about?

It is the study of DNA as the genetic material, including its double-helix structure, how it replicates, and how the information in genes is expressed into proteins through transcription and translation.

What does semiconservative replication mean and who proved it?

Semiconservative replication means each daughter DNA molecule keeps one old parental strand and one newly made strand. Meselson and Stahl proved this in 1958 using heavy nitrogen (15N) and CsCl density-gradient centrifugation in E. coli, getting hybrid DNA after one generation.

How important is this chapter for NEET?

It is part of the NEET syllabus from Class 12 Biology and is one of the most heavily weighted chapters, frequently giving 4 to 6 questions, including numericals on DNA length, base pairing (Chargaff), and experiments like Hershey-Chase.

What is the difference between transcription and translation?

Transcription is copying one DNA strand into RNA by RNA polymerase, happening in the nucleus in eukaryotes. Translation is reading the mRNA codons on ribosomes to build a polypeptide, happening in the cytoplasm.

What is the genetic code and why is it called degenerate?

The genetic code is the set of 64 mRNA triplet codons that specify amino acids, with AUG as start and UAA, UAG, UGA as stop codons. It is called degenerate because most amino acids are coded by more than one codon, yet each codon is unambiguous (codes only one amino acid).

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