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Class XII 🧬 Biology Ch 5 of 13

Molecular Basis of Inheritance

CUET unit: Genetics and Evolution → Molecular Basis of Inheritance

📌 Snapshot

  • Understand DNA structure, replication, transcription, the genetic code, the lac operon and DNA fingerprinting.
  • The classic experiments of Griffith, Hershey and Chase, and Meselson and Stahl, explained step by step.
  • Replication, transcription and translation compared side by side, with every number NTA likes to test.
  • The Human Genome Project findings and DNA fingerprinting steps, plus the traps that cost marks.

📖 Detailed Notes

Written and reviewed by Anjali Sharma, Biology faculty at UniDrill (M.Sc. Botany, Banaras Hindu University · 8 years teaching NCERT Biology). Based on NCERT Biology Class XII, Chapter 5.

How do two metres of DNA fit inside one cell?

Multiply the number of base pairs in a diploid human cell (6.6 × 10⁹) by the distance between two base pairs (0.34 nm) and you get about 2.2 metres of DNA. It sits inside a nucleus only a few micrometres across. This chapter explains how that molecule is built, packed, copied, read and switched on and off, and how scientists proved it is the genetic material in the first place.

DNA and RNA: the building blocks

A nucleotide has three parts: a nitrogenous base, a pentose sugar and a phosphate group. A base joined to a sugar is a nucleoside, and a nucleoside with a phosphate (at the 5′ carbon) is a nucleotide. Nucleotides link through 3′–5′ phosphodiester bonds to make a polynucleotide chain.

The bases are of two kinds. Purines (adenine and guanine) have two rings. Pyrimidines (cytosine, thymine and uracil) have one ring. Thymine is found in DNA, and uracil replaces it in RNA.

Feature DNA RNA
Sugar Deoxyribose Ribose (has a reactive 2′ hydroxyl)
Pyrimidines Cytosine, thymine Cytosine, uracil
Strands Two, coiled as a double helix Usually one
Stability High Lower, mutates faster
Main role Genetic material Messenger, adapter, structural and catalytic roles

Sizes worth knowing. Bacteriophage φX174 has 5386 nucleotides, bacteriophage lambda has 48,502 base pairs, E. coli has 4.6 × 10⁶ base pairs, and the haploid human genome has about 3.3 × 10⁹ base pairs.

The double helix

Friedrich Miescher first isolated DNA from the nucleus in 1869 and called it nuclein. In 1953 James Watson and Francis Crick proposed the double helix, using X-ray diffraction data from Maurice Wilkins and Rosalind Franklin, and Chargaff's rule: in double stranded DNA, the amount of A equals T, and G equals C, so the ratio A+G to T+C is 1.

Feature of B-DNA Value or fact
Strands Two, antiparallel (one runs 5′ to 3′, the other 3′ to 5′)
Backbone Sugar and phosphate, on the outside
Pairing A with T by 2 hydrogen bonds; G with C by 3 hydrogen bonds
Handedness Right handed
Pitch (one full turn) 3.4 nm
Base pairs per turn 10
Distance between adjacent base pairs 0.34 nm

A purine always pairs with a pyrimidine, so the width of the helix stays constant. The stacking of base pairs also stabilises the structure.

Packaging DNA

A bacterium has no nucleus. Its negatively charged DNA is held by some positively charged proteins in a region called the nucleoid, organised in large loops.

Eukaryotes use histones, positively charged basic proteins rich in lysine and arginine. Eight histone molecules form a histone octamer, and about 200 base pairs of negatively charged DNA wrap around it to form a nucleosome. Under the electron microscope, nucleosomes look like "beads on a string", and this is chromatin. Chromatin is then folded into fibres and into the metaphase chromosome, and the higher levels of packing need non-histone chromosomal (NHC) proteins.

Euchromatin Heterochromatin
Packing Loose Dense
Staining Light Dark
Transcription Active Inactive

Beyond the textbook. For the exam: a nucleosome has about 200 bp of DNA on a histone octamer. In reality: the DNA tightly wound on the octamer core is about 147 base pairs, and the remaining stretch of linker DNA between nucleosomes brings the repeat to about 200. If a question gives 147 or 146 as an option, the stem is asking about the core, not the repeat.

Which molecule carries the genes? Three experiments

Until the 1940s many scientists believed proteins were the genetic material. Three experiments settled the question.

Griffith, 1928. Griffith worked with Streptococcus pneumoniae and mice. Smooth (S) strain bacteria have a capsule and cause pneumonia. Rough (R) strain bacteria have none and are harmless.

Injection Result
Live S Mouse dies
Live R Mouse lives
Heat killed S Mouse lives
Heat killed S plus live R Mouse dies, and live S bacteria are recovered

Something from the dead S cells had transformed the living R cells into virulent S cells. Griffith called it the transforming principle but could not say what it was.

Avery, MacLeod and McCarty (1933 to 1944). They purified the biochemicals in the heat killed S cells and tested each. Only DNA caused transformation. Protein digesting enzymes (proteases) and RNA digesting enzymes (RNases) did not stop it, but a DNA digesting enzyme (DNase) did. They concluded DNA was the transforming material, though many scientists were still unconvinced.

Hershey and Chase, 1952. The experiment was designed around a chemical difference: DNA contains phosphorus but no sulphur, and proteins contain sulphur but no phosphorus. They grew T2 bacteriophages in two media, one with radioactive 32P (labelling DNA) and the other with radioactive 35S (labelling protein). They let the phages infect E. coli, then blended the mixture to shake the empty phage coats off the bacteria and centrifuged it so the heavy bacteria settled to the bottom.

The bacteria that received 32P phages carried radioactivity inside. In the 35S experiment, the radioactivity stayed in the supernatant with the coats. So DNA entered the cell, and the protein did not. DNA is the genetic material.

Pause and predict: in the Hershey and Chase experiment, which labelled molecule would you find in the pellet of infected bacteria if the phage's protein had been the genetic material?

Check your answer: 35S, the protein label. The pellet carries only what entered the bacteria, so the label that is found there shows which molecule carries the genetic message. Because 32P was found in the pellet, it is DNA.

Why DNA and not RNA?

A genetic material must be able to replicate, be chemically and structurally stable, allow slow changes (mutation) for evolution, and be able to express itself as characters. DNA scores well on stability. RNA's 2′ hydroxyl group makes it reactive, and DNA's thymine (rather than uracil) and double strands add further stability. RNA, however, can both carry information and act as a catalyst, so it mutates faster and is the genetic material of many viruses.

Beyond the textbook. For the exam: the RNA world idea says RNA was the first genetic material, and DNA evolved from it later because it is more stable. In reality: this is a leading hypothesis, not an observed fact. It rests on the discovery that some RNAs, called ribozymes, are catalysts. The bacterial 23S rRNA forms the peptide bonds during translation, for instance.

Replication: copying the code

Watson and Crick predicted that DNA replicates by a semi-conservative mechanism: the two strands separate, and each acts as a template for a new complementary strand. Each daughter DNA therefore has one old strand and one new strand.

Meselson and Stahl, 1958. They grew E. coli for many generations in a medium containing the heavy isotope 15N (as 15NH4Cl), so all its DNA became heavy. They then shifted the bacteria to a medium with ordinary 14N, and took samples every 20 minutes (one generation). The DNA was separated by density gradient centrifugation in caesium chloride.

Time after shift DNA density found Meaning
0 minutes All heavy Starting DNA, both strands 15N
20 minutes (1 generation) All hybrid Each molecule has one old, one new strand
40 minutes (2 generations) Equal hybrid and light Half the molecules have two new strands

Had replication been conservative, the first generation would have shown separate heavy and light bands, not a single hybrid band. Taylor and colleagues (1958) showed the same semi-conservative pattern in the chromosomes of Vicia faba (broad bean) using radioactive thymidine.

Pause and predict: after three generations in 14N, what will be the ratio of hybrid to light DNA molecules?

Check your answer: 1 hybrid to 3 light. Eight molecules result from one original, and the two original heavy strands are always found in two hybrid molecules. The other six are fully light.

E. coli copies 4.6 × 10⁶ base pairs in about 18 minutes, which works out to roughly 2000 base pairs per second. This speed means that the copying and the cell division cycle need to be tightly coordinated, because a failure leads to unequal chromosome distribution.

How the cell does it.

  1. Replication begins at a specific site, the origin of replication, where the strands open into a Y shaped replication fork.
  2. The main enzyme is a DNA dependent DNA polymerase. It uses deoxynucleoside triphosphates, which are both the raw material and the energy source.
  3. DNA polymerase works only in the 5′ to 3′ direction. On the template read 3′ to 5′, the new strand is built continuously. On the other template, the new strand is built in short pieces (Okazaki fragments), discontinuously, and the pieces are joined by DNA ligase.
  4. In eukaryotes, replication takes place in the S phase of the cell cycle.

Transcription: DNA to RNA

Transcription copies one strand of DNA into RNA, following the same base pairing rules except that adenine pairs with uracil.

A transcription unit has three regions: a promoter, a structural gene and a terminator. The promoter lies at the 5′ end of the coding strand, upstream of the gene, and is where RNA polymerase binds. The terminator lies at the 3′ end, downstream.

Template strand Coding strand
Polarity 3′ to 5′ 5′ to 3′
Role Read by RNA polymerase Not transcribed
Sequence Complementary to RNA Same as RNA, with T in place of U

Why only one strand? If both strands were transcribed, two complementary RNAs would form that could pair with each other into double stranded RNA, and the message would not be translated. The cell would also make two different proteins from one gene.

Pause and predict: a template strand reads 3′ TACGGA 5′. What is the mRNA?

Check your answer: 5′ AUGCCU 3′. The RNA is complementary and antiparallel to the template, with U in place of T. Notice it matches the coding strand, 5′ ATGCCT 3′, except for U.

A cistron is a segment of DNA that codes for one polypeptide.

Prokaryotes Eukaryotes
Structural genes Polycistronic (many cistrons on one mRNA), continuous Monocistronic, split into exons and introns
RNA polymerase One enzyme makes all RNAs Three nuclear polymerases
Initiation and termination Sigma factor and rho factor Other protein factors
mRNA processing None Splicing, capping, tailing
Place Transcription and translation in the same compartment Transcription in nucleus, translation in cytoplasm

Because bacterial mRNA needs no processing, translation can begin before transcription has finished.

Eukaryotic polymerase Makes
RNA polymerase I rRNAs (28S, 18S, 5.8S)
RNA polymerase II hnRNA, the precursor of mRNA
RNA polymerase III tRNA, 5S rRNA and snRNAs

Processing in eukaryotes. The primary transcript is heterogeneous nuclear RNA (hnRNA). It contains exons (the sequences that appear in mature RNA) and introns (intervening sequences that do not). Three steps turn it into mRNA. Splicing removes the introns and joins the exons. Capping adds methyl guanosine triphosphate to the 5′ end. Tailing adds 200 to 300 adenylate residues to the 3′ end, without a template.

The genetic code

Gamow argued that 4 bases must be read in groups of 3, because 4³ gives 64 codons, enough for 20 amino acids. Har Gobind Khorana synthesised RNAs of known sequence chemically. Marshall Nirenberg developed a cell free protein synthesis system, and Severo Ochoa's enzyme polynucleotide phosphorylase helped make RNA polymers. Together these work broke the code.

Property Meaning
Triplet Three bases per codon, 64 codons
Degenerate Most amino acids have more than one codon (61 codons code for 20 amino acids)
Unambiguous One codon codes for only one amino acid
Contiguous Read continuously with no punctuation
Nearly universal The same across organisms, with exceptions in some mitochondria and protozoa
Start and stop AUG is the start codon and codes for methionine; UAA, UAG and UGA are stop codons and code for no amino acid

Mutations and the code. A point mutation changes one base, as in sickle cell anaemia, where one base change replaces glutamic acid with valine in the beta globin chain. If bases are inserted or deleted in numbers that are not multiples of three, the reading frame shifts and every codon downstream is changed. This is a frameshift mutation. Insertions or deletions of exactly three bases (or multiples) leave the frame intact and add or remove just one amino acid.

Translation: RNA to protein

tRNA, the adapter. Crick predicted an adapter molecule that reads the code. A tRNA has an anticodon loop that pairs with the mRNA codon and an amino acid acceptor end that carries the amino acid. There is a specific tRNA for every amino acid, plus a separate initiator tRNA, and no tRNA for stop codons. On paper the tRNA is drawn as a clover leaf, but in three dimensions it is an inverted L.

The mRNA also has untranslated regions (UTRs) at both ends, which are needed for efficient translation. A ribosome is made of rRNA and about 80 proteins, and it exists as separate large and small subunits when inactive.

Steps of translation.

  1. Charging. Each amino acid is activated using ATP and attached to its own tRNA (aminoacylation).
  2. Initiation. The small ribosomal subunit binds the mRNA at the start codon AUG, and then the large subunit joins.
  3. Elongation. The ribosome reads one codon at a time. Each incoming charged tRNA pairs through its anticodon, and the ribosome forms a peptide bond between the amino acids. This reaction is catalysed by 23S rRNA in bacteria, which acts as a ribozyme. The ribosome then moves along the mRNA.
  4. Termination. At a stop codon no tRNA can bind. A release factor binds instead, and the completed polypeptide is released.

Regulation of gene expression: the lac operon

A cell does not need every protein all the time. Eukaryotes control gene expression at several levels: transcription, processing, transport of mRNA to the cytoplasm, and translation. In bacteria, the main control point is the start of transcription. A cluster of genes under common control is an operon.

The lac operon, worked out by François Jacob and Jacques Monod, controls the use of lactose in E. coli. Lactose is a disaccharide, and the cell only needs the enzymes to digest it when lactose is present.

Gene or element Function
i gene (regulator) Makes the repressor; i stands for inhibitor
Promoter (p) RNA polymerase binds here
Operator (o) The repressor binds here
z gene Beta galactosidase, which splits lactose into glucose and galactose
y gene Permease, which lets lactose into the cell
a gene Transacetylase

Without lactose, the repressor binds the operator and blocks RNA polymerase, so z, y and a are silent. With lactose, the lactose (or its isomer allolactose) acts as the inducer. It binds the repressor and inactivates it, the operator is free, and RNA polymerase transcribes all three genes as one polycistronic mRNA. This is negative regulation, because the regulator acts by turning genes off. Glucose and galactose are not inducers.

Why it makes sense. The cell saves energy by making the lactose enzymes only when their substrate is present. The substrate itself throws the switch.

Beyond the textbook. For the exam: lactose is the inducer, and the repressor is always made. In reality: if glucose is also available, E. coli still prefers glucose, and the operon stays low even when lactose is present. This second layer of control (catabolite repression) is positive regulation.

The Human Genome Project (HGP)

The HGP was a 13 year project, launched in 1990 and completed in 2003. It was coordinated by the US Department of Energy and the National Institutes of Health, with major contributions from the Wellcome Trust (UK) and partners in Japan, France, Germany and China. It created a need for storing and analysing huge amounts of data, and so helped to build bioinformatics. At the start, its cost was estimated at around 9 billion US dollars, at 3 US dollars per base pair for 3 × 10⁹ base pairs.

Goals included identifying all human genes, sequencing the 3 × 10⁹ bases, storing the data in databases, improving data analysis tools, transferring the technology and addressing the ethical, legal and social issues (ELSI).

Methods. One approach labelled the genes that are expressed as RNA, using Expressed Sequence Tags (ESTs). The other, sequence annotation, sequenced the whole genome and assigned functions later. The DNA was broken into fragments, cloned in bacteria or yeast using BAC and YAC (bacterial and yeast artificial chromosomes), and sequenced by automated machines based on Frederick Sanger's method. Chromosome 1 was the last to be completed, in May 2006.

Finding Value
Total bases 3164.7 million base pairs
Genes About 30,000 (earlier estimates were far higher)
Average gene About 3000 bases
Largest known gene Dystrophin, 2.4 million bases
Genome shared by any two people 99.9 percent
Protein coding sequence Less than 2 percent of the genome
Genes with unknown function More than 50 percent
Most genes Chromosome 1 (2968)
Fewest genes Y chromosome (231)
SNPs identified About 1.4 million (single nucleotide polymorphisms, "snips")

Beyond the textbook. For the exam: about 30,000 genes. In reality: current counts put protein coding genes at roughly 20,000, and the "complete" 2003 sequence actually left gaps in repetitive regions, which were filled in only in 2022.

DNA fingerprinting

Everyone's DNA is 99.9 percent the same, so fingerprinting looks at the part that differs. About 0.1 percent of base differences make each person unique, and the technique finds those differences in repetitive DNA. Alec Jeffreys developed the method.

Most of the genome's bulk DNA forms one main peak on a density gradient. Smaller peaks of repetitive, non coding satellite DNA appear beside it. Satellite DNA is divided by base composition, length of segment and number of repeats into micro satellites, mini satellites and so on. It does not code for protein, but the number of repeats varies greatly between individuals. The probe used is a VNTR (Variable Number of Tandem Repeats), a mini satellite whose size varies from 0.1 to 20 kb. Because you inherit one allele from each parent, the repeat pattern is passed on following Mendel's rules.

The steps: isolate DNA → digest it with restriction enzymes → separate fragments by gel electrophoresis → transfer them to a nitrocellulose or nylon membrane (Southern blotting) → hybridise with a labelled VNTR probe → detect by autoradiography. The bands then form a pattern. PCR can be used to amplify small DNA samples.

The pattern is the same in every tissue of one person (blood, hair root, semen) and different between any two individuals, except identical twins. Its main uses are in forensic science, paternity disputes and studying genetic diversity of populations.

Exam traps

The trap Why it's tempting The truth
"A=T has 3 hydrogen bonds." Both pairs sound similar A with T has 2; G with C has 3
"The distance between base pairs is 3.4 nm." You remember 3.4 3.4 nm is the pitch; adjacent base pairs are 0.34 nm apart
"The coding strand is the one transcribed." The word "coding" sounds active Template strand is read (3′ to 5′); the coding strand matches RNA but is not transcribed
"The i gene is the inducer." It starts with i i is the inhibitor (regulator) gene; the inducer is lactose or allolactose
"Glucose and galactose induce the lac operon." They are the products of lactose digestion Only lactose (allolactose) is the inducer
"RNA polymerase I makes mRNA." I comes first Pol II makes hnRNA. Pol I makes rRNA, Pol III makes tRNA
"Hershey and Chase used 14N and 15N." Both involve isotopes They used 32P and 35S. 15N is for Meselson and Stahl
"Chromosome X has the fewest genes." X is the other sex chromosome Y has the fewest (231) and chromosome 1 the most (2968)

60-second revision

Topic Lock this in
Nucleotide Base + sugar + phosphate · nucleoside = base + sugar
Helix Antiparallel · A=T (2 bonds), G≡C (3) · pitch 3.4 nm · 10 bp/turn · 0.34 nm/bp
Nucleosome Histone octamer + about 200 bp DNA
Griffith 1928 Transforming principle (heat killed S + live R → mouse dies)
Avery et al. DNase stops transformation → DNA
Hershey and Chase 1952 32P (DNA) enters · 35S (protein) stays outside
Meselson and Stahl 1958 15N → 14N · hybrid after 20 min · hybrid + light after 40 min
Replication 5′ to 3′ only · leading continuous · lagging in fragments + ligase · about 2000 bp/s in E. coli
Transcription Promoter + gene + terminator · template 3′ to 5′ · Pol I rRNA, II hnRNA, III tRNA
hnRNA processing Splicing · 5′ capping · 3′ poly A tail (200 to 300 A)
Genetic code Triplet · degenerate · unambiguous · AUG start · UAA UAG UGA stop
lac operon i repressor · z beta galactosidase · y permease · a transacetylase · lactose inducer
HGP 1990 to 2003 · 3164.7 million bp · about 30,000 genes · chr 1 most (2968), Y fewest (231)
DNA fingerprinting Jeffreys · VNTR probe · Southern blot · autoradiography

Where this chapter connects

🎯 Practice MCQs

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Q1. In the B form of the DNA double helix, the distance between two adjacent base pairs is approximately:

▸ Show answer & explanation

Answer: B

One full turn (pitch) is 3.4 nm and contains about 10 base pairs, so each base pair adds 0.34 nm. Option A is the pitch, not the spacing between neighbouring base pairs.

Q2. A diploid human cell has about 6.6 × 10⁹ base pairs, and adjacent base pairs are 0.34 nm apart. The approximate length of its DNA is:

▸ Show answer & explanation

Answer: A

6.6 × 10⁹ multiplied by 0.34 × 10⁻⁹ metres gives about 2.2 metres. The other options are off by factors of 10 or 1000.

Q3. Which statement about a eukaryotic nucleosome is correct?

▸ Show answer & explanation

Answer: B

DNA is negatively charged because of its phosphates, and histones are rich in basic lysine and arginine, so they bind. Option A reverses the charges, C is wrong because bacteria use a nucleoid, and D is wrong because euchromatin is loosely packed and active.

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