Molecular Basis of InheritanceClass 12 Biology, Chapter 5

This chapter explains how DNA stores, copies and uses genetic information: its structure, the experiments proving it is the genetic material, replication, transcription, the genetic code and translation. It also covers the lac operon, the Human Genome Project and DNA fingerprinting. After revising, you can explain each process and answer typical board and NEET questions.

Follows the Class 12 Biology textbook, Chapter 5. Made with Classly and checked by our team.

Cheat sheet

Molecular Basis of Inheritance on one page, for a last look before a test.

Molecular Basis of Inheritance

Cheat sheetClass 12 · Biology

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Base pairing — $A=T$ (2 H-bonds); $G\equiv C$ (3); strands are complementary and antiparallel.

Replication — semiconservative: each daughter DNA has one old and one new strand; synthesis runs $5'\to3'$.

Gene expression — DNA $\to$ RNA $\to$ protein; in some viruses, RNA $\to$ DNA.

Code — triplet, degenerate, nearly universal; AUG starts and codes Met; UAA/UAG/UGA stop.

1. DNA: structure and packaging

Nucleotide — nitrogenous base + pentose sugar + phosphate · nucleoside — base + sugar

Bases — purines: A, G · pyrimidines: C, T, U · DNA has T; RNA has U and a 2′-OH group

Chain links — base–sugar: N-glycosidic · phosphate–sugar: phosphoester · nucleotides: 3′–5′ phosphodiester

Polarity — 5′ end has free phosphate · 3′ end has free OH · sugar–phosphate backbone; bases project inward

DNA featureKey pointRecall check
StrandsTwo, right-handed helix; antiparallelOne runs $5'\to3'$, other $3'\to5'$
Pairing$A=T$; $G\equiv C$Purine pairs with pyrimidine
Chargaff rule$A=T$, $G=C$; total purines = pyrimidinesIn double-stranded DNA
Dimensions0.34 nm per bp; 3.4 nm per turnAbout 10 bp per turn
StabilityH-bonds + stacked base pairsComplementary strands can re-form

Packaging — prokaryotic DNA: protein-held loops in nucleoid · eukaryotic DNA: wraps around histone octamers → nucleosomes (~200 bp each) → chromatin fibre → condensed chromosome

Histones — positively charged, lysine- and arginine-rich proteins · Euchromatin — loosely packed, active · Heterochromatin — tightly packed, usually inactive

2. Genetic material: evidence and properties

EvidenceExperimental patternConclusion
Griffith (1928)Heat-killed virulent S + live non-virulent R → mice die; live S recoveredR cells transformed; factor not identified
Avery, MacLeod, McCartyProtease/RNase do not stop transformation; DNase doesDNA is transforming principle
Hershey–Chase (1952)$^{32}P$ labels phage DNA; $^{35}S$ labels protein; only DNA label enters bacteriaDNA, not protein, is inherited by infected bacteria

Hershey–Chase logic — infect E. coli with labelled phages → blend off coats → centrifuge → radioactivity with bacterial pellet only for DNA-labelled phages

Genetic material must — replicate · remain sufficiently stable · allow mutation · express inherited traits

DNA vs RNA — DNA is less reactive and more stable (deoxyribose; T instead of U) → better information storage · RNA is reactive, less stable, and can act as messenger, adapter, structure or catalyst

RNA world — RNA likely preceded DNA: it could store information and catalyse reactions; more stable DNA later became the main information store

3. Replication

Semiconservative replication — each parental strand templates a complementary new strand; each daughter double helix contains one parental + one newly made strand

Meselson–Stahl — grow E. coli in $^{15}N$ → transfer to $^{14}N$ → separate DNA by CsCl density gradient: generation 1, all hybrid; generation 2, half hybrid and half light. This supports semiconservative copying.

Replication featureWhat to recallCondition / consequence
DNA polymeraseDNA-dependent; adds deoxynucleotidesSynthesises only $5'\to3'$
SubstratesDeoxyribonucleoside triphosphatesBuilding blocks and energy source
Replication forkLocal opening of long DNA helixWhole molecule does not unwind at once
Leading strandContinuous synthesisTemplate runs $3'\to5'$
Lagging strandDiscontinuous fragments; DNA ligase joins themTemplate runs $5'\to3'$
OriginSpecific DNA site where copying beginsPolymerase cannot initiate randomly

Timing — eukaryotic DNA replication occurs in S phase · replication errors can cause mutations

4. Transcription and RNA processing

Transcription — copying one DNA segment, using one strand, into RNA · RNA is made $5'\to3'$; A pairs with U

Transcription unitPosition / role
PromoterUpstream; RNA polymerase binding; sets strand choice and direction
Structural geneDNA region transcribed into RNA
TerminatorDownstream; signals transcription end

Template strand — read $3'\to5'$; RNA is complementary · Coding strand — $5'\to3'$; same sequence as RNA except DNA has T where RNA has U

Transcription in bacteria — RNA polymerase binds promoter (initiation) → extends RNA (elongation) → reaches terminator and releases RNA (termination) · $\sigma$ helps initiation; $\rho$ helps termination

Eukaryotic polymeraseMain transcripts
RNA polymerase I28S, 18S, 5.8S rRNA
RNA polymerase IIhnRNA, the mRNA precursor
RNA polymerase IIItRNA, 5S rRNA, snRNA

hnRNA processing — splice out introns and join exons · add 5′ methyl-guanosine cap · add 3′ poly-A tail (about 200–300 A residues) → mature mRNA leaves nucleus

RNA roles — mRNA carries codons · tRNA brings amino acids and reads codons · rRNA forms ribosome structure and catalyses peptide-bond formation

Gene terms — Cistron: DNA segment coding for a polypeptide · Exons: retained in mature RNA · Introns: removed during processing · bacterial genes often polycistronic; eukaryotic structural genes usually monocistronic and split

5. Genetic code and translation

Genetic code — triplet (3 bases/codon) · 64 codons total: 61 amino-acid codons + 3 stops · degenerate (several codons may specify one amino acid) · contiguous, no punctuation · nearly universal

Start / stop — AUG: methionine and initiation · UAA, UAG, UGA: termination; no tRNA reads stop codons

Mutation pattern — single-base substitution can change an amino acid (β-globin Glu → Val in sickle-cell anaemia) · insertion/deletion of 1 or 2 bases shifts reading frame · insertion/deletion of 3 or a multiple changes whole codons but preserves downstream frame

Translation componentRoleDistinguish it by
mRNACodon templateCodons are read in triplets
tRNAAdapter; carries amino acidAnticodon pairs with codon; amino-acid acceptor end
RibosomeBinds mRNA; positions tRNAsrRNA catalyses peptide-bond formation; bacterial 23S rRNA is a ribozyme
Release factorEnds translationRecognises stop codon; releases polypeptide

Translation sequence — charge each tRNA with its amino acid using ATP → small ribosomal subunit binds mRNA at AUG with initiator tRNA → matching charged tRNAs enter as ribosome advances codon by codon → peptide bonds form → release factor recognises stop codon and frees polypeptide

Translational unit — mRNA sequence from start to stop codon · UTRs — untranslated regions before AUG and after stop; support efficient translation

6. Regulation: lac operon

Gene expression can be regulated — transcription · RNA processing · mRNA transport · translation (eukaryotes)

Operon — coordinated bacterial gene unit: shared promoter and regulatory control for related structural genes · transcription initiation is the main prokaryotic control point

Lactose conditionRepressor / operatorOutcome
No lactoseActive repressor binds operatorTranscription blocked
Lactose or allolactose presentInducer binds and inactivates repressorRNA polymerase can transcribe operon
Lactose used upRepressor becomes active againOperon switches off

lac operon parts — i: makes repressor · z: β-galactosidase, splits lactose · y: permease, increases β-galactoside entry · a: transacetylase

Inducer — lactose/allolactose, not glucose or galactose · repressor control is negative regulation; low baseline expression allows lactose entry

7. Human Genome Project

HGP — international genome-sequencing project launched in 1990; completed in 2003 · enabled by DNA sequencing, cloning and bioinformatics

Goals — identify human genes · sequence ~3 billion base pairs · store and analyse data · develop tools and transfer technology · address ethical, legal and social issues (ELSI)

Two approaches — EST: identify genes expressed as RNA · sequence annotation: sequence the whole genome, then assign functions to regions

Workflow — isolate DNA → make random fragments → clone/amplify in host using BAC or YAC vectors → sequence fragments → align overlaps computationally → annotate and assign to chromosomes

Genome finding (chapter figures)Recall value
Genome size3164.7 million bp
Protein-coding portionLess than 2%
Human similarity99.9% of bases shared among people
Repeated DNALarge fraction; often non-coding, useful for studying genome structure and variation
Gene countAbout 30,000 in the chapter’s account; over half had unknown function
SNPsAbout 1.4 million identified variable single-base sites

SNP — single-nucleotide polymorphism · Bioinformatics — computational storage, retrieval and analysis of biological data

8. DNA fingerprinting

DNA polymorphism — inherited sequence variation in a population; conventionally, a variant at a locus occurs above 1% frequency

Basis — highly variable repetitive DNA (satellite DNA, including minisatellite VNTRs) differs in repeat number → DNA fragments differ in length → individual banding patterns

VNTR — tandem repeats with copy number varying among chromosomes; chapter range: 0.1–20 kb · pattern generally differs between people, except identical twins

Method — isolate DNA → cut with restriction enzymes → separate fragments by electrophoresis → blot onto nylon/nitrocellulose → hybridise labelled VNTR probe → detect bands by autoradiography

Uses — forensic identification · paternity disputes · population and genetic diversity studies · DNA from different tissues of one individual shares the same polymorphism pattern

9. Traps

  • DNA polymerase works in either direction → DNA synthesis is only $5'\to3'$.
  • Both DNA strands are transcribed → one strand per transcription unit; two RNAs could be complementary and encode different products.
  • Every codon has a matching tRNA → stop codons have no tRNA; release factors terminate translation.
  • All mutations shift the reading frame → only insertions/deletions not in multiples of three shift it.
  • Lactose permanently keeps lac operon on → lactose depletion restores active repressor and shuts transcription down.

Flashcards

10 cards on Molecular Basis of Inheritance. Flip them here, or read them all below.

Molecular Basis of Inheritance flashcards1 of 10

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All 10 cards

  1. 1

    In DNA, which bases pair, and how many hydrogen bonds join each pair?

    A pairs with T through 2 hydrogen bonds; G pairs with C through 3.

  2. 2

    What did the Hershey–Chase experiment show about which phage component enters bacterial cells?

    DNA entered the bacteria; protein did not.

  3. 3

    What does semiconservative DNA replication produce in each daughter molecule?

    Each daughter DNA molecule contains one parental strand and one newly made strand.

  4. 4

    In which direction do DNA polymerases extend a new DNA strand?

    5′ to 3′

  5. 5

    What is transcription?

    RNA is made by copying the information of one DNA strand, the template.

  6. 6

    What is the genetic code’s basic unit size, and how many codons are sense versus stop?

    Triplet; 61 codons specify amino acids and 3 are stop codons.

  7. 7

    What is the adapter role of tRNA during protein synthesis?

    It matches mRNA codons with their corresponding amino acids.

  8. 8

    How does lactose switch on transcription of the lac operon?

    Lactose, or its derivative allolactose, binds to and inactivates the repressor.

  9. 9

    About how large is the human genome, as found by the Human Genome Project?

    About 3.2 billion base pairs (3164.7 million).

  10. 10

    Which highly variable repeated DNA sequences were used as probes in early DNA fingerprinting?

    Variable Number of Tandem Repeats (VNTRs)

Quiz

5 questions, each with the reason behind its answer.

1 of 5

A DNA strand has the sequence 5′-AGTC-3′. Which sequence is complementary to it, written in the opposite direction?

Answer key
  1. 1

    A DNA strand has the sequence 5′-AGTC-3′. Which sequence is complementary to it, written in the opposite direction?

    Answer: B. 3′-TCAG-5′

    A pairs with T and G pairs with C. The complementary strand runs antiparallel, so its sequence is 3′-TCAG-5′.

  2. 2

    In the Hershey–Chase experiment, bacteria infected by phages labelled with radioactive phosphorus became radioactive after infection. What does this result show?

    Answer: A. Phage DNA entered the bacteria.

    Phosphorus labelled the phage DNA, not its protein coat. Radioactivity inside the bacteria therefore showed that phage DNA entered the bacterial cells.

  3. 3

    DNA polymerase builds a new strand only in the 5′ to 3′ direction. At a replication fork, which template strand is copied continuously?

    Answer: C. The one read 3′ to 5′ in the direction the fork moves

    A strand built 5′ to 3′ needs a template read 3′ to 5′. The template that runs 3′ to 5′ along the fork gives the continuous leading strand; the other template is copied in fragments, the lagging strand.

  4. 4

    A eukaryotic primary RNA transcript contains both exons and introns. Which change specifically occurs during splicing?

    Answer: D. Introns are removed and exons are joined.

    Splicing removes introns and joins the exons in order, producing a processed RNA transcript.

  5. 5

    When lactose is available to a bacterium, how does it help switch on the lac operon?

    Answer: B. It binds to the repressor, preventing the repressor from blocking the operator.

    Lactose (or allolactose) binds to the repressor and inactivates it. The repressor can no longer block the operator, allowing transcription to proceed.

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Questions about Molecular Basis of Inheritance

What are the most important topics in Class 12 Biology Molecular Basis of Inheritance for NEET?

Focus on DNA structure and base pairing, the Griffith and Hershey-Chase experiments, semiconservative replication, transcription and RNA processing, the genetic code, translation and the lac operon. Human Genome Project facts and DNA fingerprinting also appear often. The cheat sheet on this page covers all of them in order.

What is the difference between transcription and translation?

Transcription makes an RNA copy of one DNA strand and happens in the 5′ to 3′ direction using RNA polymerase. Translation reads the codons of mRNA on a ribosome, with tRNA bringing amino acids, to build a polypeptide. So transcription goes from DNA to RNA, and translation goes from RNA to protein.

Why is the genetic code called degenerate?

There are 64 codons but only about 20 amino acids, so several codons can specify the same amino acid. Of the 64, 61 code for amino acids and 3 (UAA, UAG, UGA) are stop codons. Degenerate does not mean ambiguous: each codon still gives only one amino acid.

Why does DNA replication give two daughter molecules that each carry an old strand?

Replication is semiconservative: each parental strand acts as a template for a new complementary strand. Every daughter DNA therefore has one old and one new strand, as the Meselson-Stahl experiment showed. You can also make your own flashcards or quiz for any chapter in Classly.