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Molecular Basis of Inheritance – Class 12 Biology Chapter 5 Notes & NCERT PDF

Class 12BiologyChapter 5NCERT book: Biology - Class 12

Notes and a simple summary of Chapter 5 of the NCERT Class 12 Biology book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 5: Molecular Basis of Inheritance

Introduction

This chapter explains what a gene is made of and how it works. DNA is the genetic material in most organisms; RNA is genetic material in some viruses and otherwise acts as messenger, adapter, structural molecule and catalyst. We follow information from DNA structure through replication, transcription, the genetic code and translation, to regulation, the Human Genome Project and DNA fingerprinting.

Key Concepts

5.1 The DNA

  • DNA length is given in base pairs (bp): phage φX174 has 5386 nucleotides, phage lambda 48502 bp, E. coli 4.6 × 10⁶ bp, human haploid DNA 3.3 × 10⁹ bp.
  • Nucleotide = nitrogenous base + pentose sugar + phosphate. Purines: adenine, guanine. Pyrimidines: cytosine, uracil, thymine. Thymine occurs only in DNA; uracil replaces it in RNA.
  • Base joins the 1' carbon of sugar by an N-glycosidic linkage (forming a nucleoside); phosphate joins the 5'-OH by a phosphoester linkage (forming a nucleotide). Nucleotides link by 3'-5' phosphodiester bonds, so a strand has a free 5' phosphate end and a free 3'-OH end.
  • Friedrich Meischer (1869) first identified DNA and called it 'Nuclein'. Using X-ray diffraction data of Maurice Wilkins and Rosalind Franklin, Watson and Crick (1953) proposed the double helix. Chargaff's rule: A = T and G = C, so (A+G)/(T+C) = 1.
  • Double helix features: two antiparallel polynucleotide chains; sugar-phosphate backbone outside, bases inside; A pairs with T by two hydrogen bonds, G with C by three; a purine always faces a pyrimidine, giving uniform width; right-handed coiling; pitch 3.4 nm with about 10 bp per turn, so 0.34 nm between adjacent bp; plane of one base pair stacks over another.
  • Central dogma (Francis Crick): DNA → RNA → protein. In some viruses information flows from RNA to DNA (reverse transcription).
  • Packaging: 6.6 × 10⁹ bp × 0.34 × 10⁻⁹ m/bp ≈ 2.2 m of DNA fits in a nucleus about 10⁻⁶ m across. Prokaryotes hold DNA in a nucleoid with positively charged proteins. In eukaryotes, positively charged histones (rich in lysine and arginine) form an octamer; negatively charged DNA wraps around it to form a nucleosome (about 200 bp). Nucleosomes form 'beads-on-string', coiled further with non-histone chromosomal (NHC) proteins. Loosely packed, lightly stained euchromatin is transcriptionally active; densely packed, darkly stained heterochromatin is inactive.

    • 5.2 The Search for Genetic Material

  • Griffith (1928), transforming principle: S strain (smooth capsule, virulent) and R strain (non-virulent) of Streptococcus pneumoniae. Live S → mouse dies; live R → lives; heat-killed S → lives; heat-killed S + live R → mouse dies and live S bacteria are recovered. Something from dead S transformed R.
  • Avery, MacLeod and McCarty (1933–44): protein-digesting enzymes (proteases) and RNase did not stop transformation, but DNase did, so DNA is the transforming principle.
  • Hershey and Chase (1952): one batch of viruses grown on radioactive phosphorus (³²P labels DNA), another on radioactive sulphur (³⁵S labels protein). After infection, blending and centrifugation, radioactivity from ³²P was inside bacteria while ³⁵S stayed outside. DNA enters the cell, so DNA is the genetic material.

    • 5.3 RNA World

    Genetic material must replicate, be stable, allow slow mutation and express as Mendelian characters. RNA's 2'-OH makes it labile, while DNA (with thymine) is more stable. RNA was probably the first genetic material and can act as a catalyst; DNA evolved later for stability.

    5.4 Replication

  • Watson and Crick proposed semiconservative replication: each new molecule has one parental and one new strand.
  • Meselson and Stahl (1958): grew E. coli in ¹⁵NH₄Cl for many generations, then moved them to ¹⁴NH₄Cl. DNA was separated by CsCl density gradient centrifugation. After 20 minutes (one generation) DNA was hybrid (intermediate density); after 40 minutes, equal hybrid and light DNA. Taylor (1958) confirmed this in Vicia faba using radioactive thymidine.
  • Machinery: DNA-dependent DNA polymerase polymerises only 5'→3'. In E. coli 4.6 × 10⁶ bp are copied in about 18 minutes, roughly 2000 bp per second.
  • At the replication fork, the template with 3'→5' polarity is copied continuously (leading strand); the other is copied discontinuously in short fragments later joined by DNA ligase. Replication begins at a defined origin of replication. In eukaryotes replication occurs in the S-phase.

    • 5.5 Transcription

  • Only one strand of a segment is copied. A transcription unit has a promoter, a structural gene and a terminator. The template strand has 3'→5' polarity; the coding strand (5'→3') has the same sequence as RNA, with T instead of U. The promoter lies towards the 5' end (upstream) of the coding strand.
  • A cistron is a DNA segment coding for one polypeptide. Eukaryotic structural genes are mostly monocistronic; bacterial ones are polycistronic. Eukaryotic genes are split into exons (expressed) and introns (intervening).
  • Bacteria have a single RNA polymerase. It binds the promoter with the σ (sigma) factor to initiate, elongates, and terminates with the ρ (rho) factor. In bacteria, transcription and translation can be coupled.
  • Eukaryotes: RNA polymerase I makes rRNAs (28S, 18S, 5.8S); II makes hnRNA (precursor of mRNA); III makes tRNA, 5S rRNA and snRNAs. hnRNA undergoes splicing (introns removed), capping (methyl guanosine triphosphate at the 5' end) and tailing (200–300 adenylate residues at the 3' end).

    • 5.6 Genetic Code

    George Gamow argued that 4 bases coding 20 amino acids needs a triplet code: 4³ = 64 codons. Khorana, Nirenberg and Severo Ochoa helped decode it. Features: triplet; 61 codons code amino acids and 3 (UAA, UAG, UGA) are stop codons; unambiguous and specific; degenerate (more than one codon per amino acid); read contiguously without punctuation; nearly universal; AUG codes methionine and also acts as the initiator codon.

  • Mutations: in sickle-cell anaemia one base change replaces glutamate with valine in β-globin. Inserting or deleting one or two bases causes a frameshift; three bases add or remove one amino acid (the RAM HAS RED CAP example).
  • tRNA, the adapter: has an anticodon loop and an amino-acid acceptor end; secondary structure is clover-leaf, 3-D structure an inverted L.

    • 5.7 Translation

  • Charging (aminoacylation): amino acids are activated with ATP and linked to their tRNA.
  • Initiation: ribosome binds mRNA at the start codon AUG with initiator tRNA.
  • Elongation: peptide bonds form; 23S rRNA in bacteria acts as a ribozyme.
  • Termination: a release factor binds the stop codon and the polypeptide is released. mRNA also carries untranslated regions (UTRs) at both ends.

    • 5.8 Regulation of Gene Expression

    In prokaryotes the key control is transcription initiation. Lac operon (Jacob and Monod): regulatory gene i codes the repressor; structural genes z (β-galactosidase), y (permease) and a (transacetylase). Without lactose the repressor binds the operator and blocks transcription. Lactose (the inducer) inactivates the repressor, so RNA polymerase transcribes. This is negative regulation.

    5.9 Human Genome Project (HGP)

    Launched 1990, completed 2003 by the US Department of Energy and NIH, with the Wellcome Trust (UK). At about US $3 per bp it cost near 9 billion dollars. Methods: Expressed Sequence Tags and sequence annotation; fragments cloned in BAC/YAC vectors and sequenced by Sanger's method. Chromosome 1 was completed last, in May 2006.

    Salient features: 3164.7 million bp; average gene 3000 bases, the largest (dystrophin) 2.4 million bases; about 30,000 genes; 99.9% of bases identical in all people; functions unknown for over 50% of genes; less than 2% codes for proteins; chromosome 1 has most genes (2968) and Y the fewest (231); about 1.4 million SNPs identified.

    5.10 DNA Fingerprinting

    Repetitive DNA forms satellite DNA (micro- and mini-satellites). A variant at a locus with a frequency above 0.01 is a polymorphism. Alec Jeffreys used VNTR (a mini-satellite, 0.1 to 20 kb) as the probe. Steps: (i) isolate DNA; (ii) digest with restriction endonucleases; (iii) separate fragments by electrophoresis; (iv) blot onto nitrocellulose or nylon; (v) hybridise with labelled VNTR probe; (vi) detect by autoradiography. Every individual has a unique pattern except identical twins. Used in forensics and paternity disputes.

    Diagrams to Know

  • Nucleosome: histone octamer wrapped by DNA; H1 linker; beads-on-string.
  • Replication fork: leading strand continuous, lagging strand in fragments joined by ligase.
  • Transcription unit: promoter – structural gene – terminator, with template and coding strands labelled.
  • Lac operon: i, p, o, z, y, a in both 'off' and 'on' states.

    • Key-Facts Sheet

  • Meselson–Stahl: ¹⁵N → ¹⁴N, CsCl, 20 min per generation.
  • Stop codons UAA, UAG, UGA; start AUG.

    • Board Exam Focus

  • If cytosine is 20% of a double-stranded DNA, G = 20%, so A + T = 60% and A = 30%.
  • Write the complementary strand or the mRNA from a coding strand (copy it, T → U).
  • Explain Hershey–Chase, Meselson–Stahl and the lac operon with diagrams.
  • Differentiate template and coding strand, mRNA and tRNA, repetitive and satellite DNA.
  • Explain the steps of DNA fingerprinting and the goals of HGP.

    • Key Terms

  • Nucleosome (न्यूक्लियोसोम)
  • Replication (प्रतिकृतियन)
  • Transcription (अनुलेखन)
  • Translation (स्थानांतरण)
  • Codon (कोडोन / प्रकूट)
  • Exon and intron (व्यक्तेक और अव्यक्तेक)
  • Splicing (समबंधन)
  • Operon (ऑपेरॉन)
  • Repressor (दमनकारी)
  • Inducer (प्रेरक)
  • Polymorphism (बहुरूपता)
  • Semiconservative (अर्धसंरक्षी)
  • Transforming principle (रूपांतरण सिद्धांत)

    • Common Mistakes

  • Saying DNA polymerase can work 3'→5'; it polymerises only 5'→3'.
  • Writing mRNA as the complement of the coding strand; mRNA matches the coding strand with U for T.
  • Confusing which label marks what: ³²P for DNA, ³⁵S for protein.
  • Saying lactose binds the operator; it inactivates the repressor.
  • Treating 'degenerate' and 'ambiguous' as the same; the code is degenerate but unambiguous.

    • 💡 Key Learning Points

    • ✓DNA is a right-handed antiparallel double helix (pitch 3.4 nm, 0.34 nm per bp) held by A=T and G≡C hydrogen bonds and packed with histones into nucleosomes.
    • ✓Griffith, Avery–MacLeod–McCarty and Hershey–Chase experiments proved that DNA, not protein, is the genetic material.
    • ✓Replication is semiconservative (Meselson–Stahl), runs 5'→3', continuous on one strand and discontinuous on the other with fragments joined by DNA ligase.
    • ✓Transcription copies the template strand into RNA; eukaryotic hnRNA is spliced, capped and tailed; the triplet, degenerate, nearly universal code is translated on ribosomes using tRNA adapters.
    • ✓Gene expression is regulated (lac operon), the human genome has about 30,000 genes (HGP), and VNTR polymorphism underlies DNA fingerprinting.

    👨‍🏫 Teaching Tips

    • →Build a paper double helix with colour-coded bases so students count 10 bp per turn and see antiparallel ends.
    • →Tell the Griffith and Hershey–Chase experiments as detective stories: list suspects (protein, RNA, DNA) and eliminate them step by step.
    • →Draw Meselson–Stahl tubes for generations 0, 1 and 2 on the board and ask students to predict band positions before revealing them.
    • →Practise base-pairing drills: give a coding strand and have students write template strand, mRNA and amino acids using a codon chart.
    • →Use the RAM HAS RED CAP sentence to demonstrate frameshift versus three-letter insertions live on the board.
    • →Model the lac operon as a light switch: repressor as a lock on the operator, lactose as the key that removes the lock.
    • →Show a real DNA fingerprinting gel image and ask students to solve a mock paternity case.
    • →End each section with one board-style numerical or differentiation question from the exercises.

    📋 Assessment Questions

    1. Can the student describe the double helix model with correct dimensions and base-pairing rules?
    2. Can the student explain the Hershey–Chase and Meselson–Stahl experiments and their conclusions?
    3. Can the student write the mRNA and polypeptide from a given DNA coding strand?
    4. Can the student explain the lac operon in the presence and absence of lactose?
    5. Can the student list the steps of DNA fingerprinting and key findings of HGP?

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    Keep learning

    All chapters in Biology - Class 12

    1. 1. Sexual Reproduction in Flowering Plants
    2. 2. Human Reproduction
    3. 3. Reproductive Health
    4. 4. Principles of Inheritance and Variation
    5. 5. Molecular Basis of Inheritance
    6. 6. Evolution
    7. 7. Human Health and Disease
    8. 8. Microbes in Human Welfare
    9. 9. Biotechnology: Principles and Processes
    10. 10. Biotechnology and its Applications
    11. 11. Organisms and Populations
    12. 12. Ecosystem
    13. 13. Biodiversity and Conservation