NEET Biology · NCERT Class 12

Molecular Basis of Inheritance notes

DNA & RNA, DNA Replication, Transcription, Genetic Code & Translation, Gene Regulation and Human Genome Project & DNA Fingerprinting — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 7 topics
  • 54 min read
  • 18 formulas
  • 98 PYQs
1

Chapter Overview

6 min

Molecular Basis of Inheritance explains how genetic information is stored, copied, expressed, regulated and used in biotechnology. DNA is the hereditary material in most organisms because it can replicate, mutate and express information through RNA and proteins. The chapter connects DNA structure with semi-conservative replication, transcription of RNA, genetic code, translation into proteins and regulation of gene expression using the lac operon. It also covers modern applications such as the Human Genome Project and DNA fingerprinting. For NEET, this chapter is highly conceptual and diagram-based; repeated questions come from DNA structure, enzymes of replication, RNA processing, codon features, lac operon and VNTR-based DNA profiling.

Key points5

  • The chapter moves from molecular structure to biological function: DNA structure → replication → transcription → translation → regulation.
  • NCERT emphasizes experimental evidence for DNA as genetic material: Griffith, Avery-MacLeod-McCarty and Hershey-Chase.
  • Replication, transcription and translation are directional processes, and 5' to 3' polarity is a frequent NEET trap.
  • Gene regulation in prokaryotes is best understood through Jacob and Monod's lac operon model.
  • HGP and DNA fingerprinting show how molecular genetics is applied in medicine, forensics and evolution.

Memory tricks2

Chapter Order Trick: Remember the learning sequence as DRT-GH: DNA/RNA → Replication → Transcription → Genetic code/Translation → Gene regulation → HGP/Fingerprinting.

Central Dogma: DNA is the library, RNA is the photocopy, protein is the working machine.

Examples2

Real-life Analogy

A recipe book is like DNA, a copied recipe is like mRNA, and the prepared dish is like the protein.

NEET Application

If a question mentions VNTR, autoradiography or banding pattern, immediately connect it to DNA fingerprinting.

Common mistakes2

Mixing replication and transcription

Replication copies the entire DNA molecule before cell division, while transcription copies only a gene or transcription unit into RNA.

Assuming every DNA sequence codes for protein

Only coding regions are translated. Eukaryotic genes also contain introns, regulatory sequences and non-coding DNA.

Reference tables1

Quick revision8

Formulas2

Central Dogma Flow
DNA→RNA→ ProteinDNA → RNA →\text{ Protein}

The usual flow of genetic information in living cells. DNA is transcribed into RNA, and RNA is translated into a polypeptide.

DNA
Genetic material storing hereditary information
RNA
Intermediate molecule carrying genetic message
Protein
Functional product formed by amino acid sequence
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Diagrams3

2

DNA & RNA

8 min

DNA and RNA are nucleic acids made of nucleotides, but DNA is the main hereditary material in most organisms. The proof came through classical experiments: Griffith showed transformation, Avery-MacLeod-McCarty identified DNA as the transforming principle, and Hershey-Chase confirmed DNA as genetic material in bacteriophages. Watson and Crick proposed the double helix model using Chargaff’s rules and X-ray diffraction evidence. DNA has antiparallel strands, complementary base pairing and a sugar-phosphate backbone. RNA is usually single-stranded and functions as mRNA, tRNA and rRNA. In cells, DNA is highly compacted by proteins: histones in eukaryotes and non-histone basic proteins in prokaryotes.

Key points6

  • A molecule must replicate, be stable, allow mutation and express information to act as genetic material.
  • RNA can act as genetic material in some viruses but is generally less stable and more reactive than DNA.
  • Chargaff’s rule states that in double-stranded DNA, A = T and G = C.
  • Watson-Crick DNA is a right-handed double helix with 10 base pairs per turn and 3.4 nm pitch.
  • Histone octamer plus wrapped DNA forms a nucleosome; nucleosomes form chromatin.
  • NCERT highlight: DNA is better genetic material, but RNA was likely the first genetic material because it can store information and catalyse reactions.

Memory tricks3

Purines: Pure As Gold: Purines are Adenine and Guanine.

RNA Types: MTR: mRNA gives Message, tRNA Transfers amino acid, rRNA makes Ribosome.

DNA Pairing: Apple in the Tree: A pairs with T. Car in the Garage: C pairs with G.

Examples2

Hershey-Chase Labelling

DNA contains phosphorus but not sulphur, so 32P labelled DNA. Protein contains sulphur, so 35S labelled protein coat.

Packaging Example

About 2.2 metres of human DNA fits into a nucleus because it is packed into nucleosomes and higher-order chromatin.

Common mistakes3

Saying RNA always cannot be genetic material

RNA is genetic material in many viruses. DNA is preferred in cells because it is more stable.

Confusing nucleotide and nucleoside

Nucleoside = sugar + base. Nucleotide = sugar + base + phosphate.

Forgetting antiparallel polarity

The two DNA strands run in opposite directions: 5' to 3' and 3' to 5'.

Reference tables3

Quick revision7

Formulas3

Chargaff’s Rule
A=TandG=CA = T and G = C

In double-stranded DNA, adenine pairs with thymine and guanine pairs with cytosine, so their quantities are equal.

A
Adenine
T
Thymine
G
Guanine
C
Cytosine
Base Pair Distance
Length of DNA = number of base pairs ×0.34nm\text{Length of DNA }=\text{ number of base pairs }× 0.34 nm

Adjacent base pairs in B-DNA are separated by 0.34 nm, useful for calculating DNA length.

number of base pairs
Total paired nucleotides in double-stranded DNA
0.34 nm
Distance between two adjacent base pairs
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Diagrams4

3

DNA Replication

8 min

DNA replication is the process by which a cell copies its DNA before cell division. It is semi-conservative: each daughter DNA molecule contains one parental strand and one newly synthesized strand. Meselson and Stahl proved this using 15N and 14N isotopes in E. coli. Replication starts at an origin and proceeds through a replication fork where helicase unwinds DNA, primase adds RNA primers and DNA polymerase extends the new strand only in the 5' to 3' direction. Because strands are antiparallel, one strand is synthesized continuously as the leading strand, while the lagging strand forms Okazaki fragments that are joined by ligase. Proofreading ensures high fidelity.

Key points6

  • Meselson-Stahl experiment used density gradient centrifugation to prove semi-conservative replication.
  • The replication fork is Y-shaped because parental strands separate and act as templates.
  • Deoxyribonucleoside triphosphates provide both building blocks and energy for polymerization.
  • Prokaryotes usually have one origin of replication, while eukaryotes have multiple origins.
  • NEET commonly asks enzyme functions: helicase unwinds, ligase joins, polymerase extends, primase makes primer.
  • Replication is fast yet accurate due to proofreading activity of DNA polymerases.

Memory tricks2

Replication Enzyme Order: Help Some Primers Polymerize Later: Helicase, SSB, Primase, Polymerase, Ligase.

Lagging Strand: Lagging = Lots of fragments. The lagging strand needs many primers and Okazaki fragments.

Examples2

PYQ Concept

If a newly made DNA molecule contains one old and one new strand after replication, the correct term is semi-conservative replication.

Real-life Example

DNA replication is like using each half of a zipper as a template to build the missing complementary half.

Common mistakes3

Thinking DNA polymerase starts synthesis alone

DNA polymerase can only extend from an existing 3'-OH; primase must make an RNA primer first.

Reversing synthesis direction

Template may be read 3' to 5', but the new DNA strand is always synthesized 5' to 3'.

Calling replication conservative

Replication is semi-conservative, not conservative or dispersive.

Reference tables2

Quick revision7

Formulas3

Direction of DNA Synthesis
New strand elongation:5′→3′\text{New strand elongation}: 5' → 3'

DNA polymerase adds each nucleotide to the 3'-OH end, so the newly synthesized strand always grows in the 5' to 3' direction.

5'
Phosphate end of a nucleic acid strand
3'
Hydroxyl end where new nucleotide is added
Semi-Conservative Outcome
Parent DNA →2 daughter DNA molecules\text{Parent DNA }→ 2\text{ daughter DNA molecules}
each =1 old strand +1 new strand\text{each }= 1\text{ old strand }+ 1\text{ new strand}

Each daughter duplex conserves one parental strand and contains one freshly synthesized complementary strand.

old strand
Original parental template strand
new strand
Newly polymerized complementary strand
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Diagrams3

4

Transcription

7 min

Transcription is the synthesis of RNA from a DNA template. Only one strand of DNA, called the template strand, is copied; the other is the coding strand because its sequence resembles RNA except T is replaced by U. A transcription unit contains a promoter, structural gene and terminator. RNA polymerase binds the promoter, initiates RNA synthesis, elongates the RNA chain in the 5' to 3' direction and stops at the terminator. In prokaryotes, transcription and translation can be coupled because there is no nuclear membrane. In eukaryotes, primary RNA transcripts undergo processing: capping, tailing and splicing to produce mature mRNA.

Key points6

  • RNA polymerase does not require a primer, unlike DNA polymerase.
  • Promoter defines the template strand and direction of transcription.
  • In bacteria, a single RNA polymerase transcribes all RNA types.
  • In eukaryotes, RNA polymerase II transcribes hnRNA/mRNA precursor.
  • Capping adds methyl guanosine at 5' end; tailing adds adenylate residues at 3' end.
  • Splicing removes introns and joins exons to make mature mRNA.

Memory tricks2

Transcription Unit: PST: Promoter Starts Transcription, Structural gene is copied, Terminator Stops.

RNA Processing: CaTaS: Capping, Tailing, Splicing.

Examples2

Sequence Example

If template DNA is 3'-TAC-5', the RNA codon formed is 5'-AUG-3'.

Frequently Asked NEET Concept

A promoter is located upstream and provides binding site for RNA polymerase; it also determines which strand acts as template.

Common mistakes3

Confusing coding and template strands

The template strand is copied. The coding strand is not copied but matches RNA sequence except T/U difference.

Adding thymine to RNA

RNA contains uracil instead of thymine.

Assuming prokaryotic mRNA always needs processing

Extensive capping, tailing and splicing are characteristic of eukaryotic hnRNA.

Reference tables3

Quick revision7

Formulas2

RNA Complementarity
DNA template A→RNAU\text{DNA template }A → RNA U
T→AT → A
G→CG → C
C→GC → G

During transcription, RNA bases pair with the DNA template using uracil instead of thymine.

A
Adenine
U
Uracil in RNA
T
Thymine in DNA
G/C
Guanine-cytosine complementary pair
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Diagrams4

5

Genetic Code & Translation

9 min

The genetic code is the relationship between nucleotide triplets in mRNA and amino acids in a polypeptide. Since four bases are read in groups of three, there are 64 codons. AUG codes for methionine and acts as the start codon, while UAA, UAG and UGA are stop codons. The code is triplet, degenerate, unambiguous, nearly universal, non-overlapping and comma-less. Translation is protein synthesis on ribosomes. mRNA provides codons, tRNA brings amino acids using anticodons, and rRNA forms the catalytic ribosome. Translation includes activation of amino acids, initiation, elongation and termination. Proteins may later undergo folding, cleavage or chemical modifications.

Key points6

  • 64 codons include 61 sense codons and 3 stop codons.
  • The code is unambiguous: one codon specifies only one amino acid.
  • Wobble at the third codon position contributes to degeneracy.
  • Prokaryotic ribosome is 70S with 50S and 30S subunits; eukaryotic ribosome is 80S with 60S and 40S subunits.
  • Peptidyl transferase activity is performed by rRNA, making ribosome a ribozyme.
  • Translation terminates when a release factor recognizes a stop codon.

Memory tricks3

Stop Codons: U Are Annoying, U Are Gone, U Go Away: UAA, UAG, UGA.

Ribosome Sites: APE: A site Accepts, P site Peptide grows, E site Exits.

Genetic Code Features: T-DUUN-C: Triplet, Degenerate, Universal, Unambiguous, Non-overlapping, Comma-less.

Examples2

Translation Example

mRNA 5'-AUG GGC UAA-3' produces methionine-glycine and then stops at UAA.

Post-Translational Modification

A newly made protein may fold, form disulphide bonds or be cleaved to become functional, like insulin maturation.

Common mistakes3

Calling AUG only a start signal

AUG is the start codon and also codes for methionine.

Assuming stop codons code amino acids

UAA, UAG and UGA do not code for amino acids; they recruit release factors.

Confusing codon and anticodon

Codon is on mRNA; anticodon is on tRNA.

Reference tables3

Quick revision8

Formulas3

Codon Possibilities
43=64 codons4³ = 64\text{ codons}

Four RNA bases arranged in triplets form 64 possible codons.

4
RNA bases A, U, G and C
3
Three bases per codon
64
Total possible triplet codons
Codon to Amino Acid Count
61 sense codons +3 stop codons =64 codons61\text{ sense codons }+ 3\text{ stop codons }= 64\text{ codons}

Most codons specify amino acids, while three codons terminate translation.

61
Codons that code for amino acids
3
Termination codons UAA, UAG and UGA
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Diagrams4

6

Gene Regulation

8 min

Gene regulation means controlling when, where and how much a gene is expressed. It saves energy and allows cells to respond to environmental changes. In prokaryotes, genes involved in a common function may be organized into an operon, containing structural genes, promoter, operator and regulator gene. The lac operon of E. coli is an inducible operon that controls lactose metabolism. In the absence of lactose, a repressor binds the operator and blocks transcription. In the presence of lactose, allolactose binds the repressor, inactivates it and allows transcription of lacZ, lacY and lacA. Eukaryotic regulation is more complex and occurs at chromatin, transcriptional, post-transcriptional, translational and post-translational levels.

Key points6

  • Jacob and Monod proposed the operon model for regulation of gene expression.
  • lacZ codes beta-galactosidase, lacY codes permease and lacA codes transacetylase.
  • When lactose is absent, the repressor remains active and binds the operator.
  • When lactose is present, allolactose inactivates the repressor and transcription proceeds.
  • Glucose level also affects lac operon through catabolite repression, but NCERT NEET focus is mainly inducer-repressor logic.
  • Eukaryotic genes are regulated more elaborately because of nuclear membrane, chromatin and RNA processing.

Memory tricks2

Lac Structural Genes: ZYA: Z breaks lactose, Y lets lactose enter, A assists.

Operon Switch: Operator is the ON/OFF switch; repressor is the finger pressing OFF.

Examples2

PYQ Concept

If lactose is absent, beta-galactosidase is not produced because the repressor blocks transcription.

Real-life Analogy

A lac operon works like a kitchen appliance switched on only when the ingredient lactose is available.

Common mistakes3

Calling lactose the direct repressor

Lactose/allolactose is an inducer; it inactivates the repressor rather than repressing transcription.

Confusing regulator gene and operator

Regulator gene codes for repressor protein; operator is the DNA site where repressor binds.

Assuming lac operon is repressible

Lac operon is inducible because the substrate lactose turns the operon on.

Reference tables3

Quick revision8

Formulas2

Lac Operon ON Condition
Lactose present → allolactose binds repressor → operator free → transcription ON\text{Lactose present }→\text{ allolactose binds repressor }→\text{ operator free }→\text{ transcription ON}

Lactose acts as inducer through allolactose, preventing the repressor from blocking RNA polymerase.

allolactose
Inducer molecule derived from lactose
repressor
Protein that blocks transcription by binding operator
operator
DNA control sequence near promoter
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Diagrams4

7

Human Genome Project & DNA Fingerprinting

8 min

The Human Genome Project was an international effort to sequence and map the entire human genome. It aimed to identify all human genes, determine the sequence of approximately 3.2 billion base pairs, store data in databases, improve analysis tools and address ethical, legal and social issues. NCERT highlights that only a small percentage of the genome codes for proteins, most genes have unknown functions, chromosome 1 has the most genes and chromosome Y has the fewest. DNA fingerprinting is a technique for individual identification based on variation in repetitive DNA sequences, especially VNTRs. It is used in forensics, paternity disputes, population genetics, biodiversity studies and medical diagnosis.

Key points6

  • HGP used sequencing, genetic mapping, physical mapping and bioinformatics.
  • A major goal was to transfer related technologies to industry and improve biological research.
  • Chromosome 1 has maximum genes and Y chromosome has minimum genes according to NCERT.
  • DNA fingerprinting was developed by Alec Jeffreys; in India, Dr. Lalji Singh is associated with its development.
  • VNTRs are highly polymorphic because repeat number varies greatly among individuals.
  • DNA fingerprints are unique except in identical twins.

Memory tricks3

HGP Goals: GIS-TE: Genes identify, Information store, Sequence genome, Tools improve, ELSI address.

VNTR: VNTR = Variable Number of Tandem Repeats; the number varies, the repeated sequence stays tandem.

Fingerprinting Steps: I Dig Gel, Blot Probe Detect: Isolation, Digestion, Gel, Blotting, Probe, Detection.

Examples3

Forensic Case Study

If the band pattern of DNA from a crime scene matches a suspect's DNA profile, it provides strong evidence linking the suspect to the sample.

Paternity Testing

A child’s DNA bands must be explainable by bands inherited from the mother and the alleged father.

Medical Genomics

HGP data helps identify disease-associated genes and supports personalized medicine.

Common mistakes3

Thinking DNA fingerprinting uses coding genes only

It mainly uses highly variable repetitive non-coding DNA such as VNTRs.

Assuming all humans have very different genomes

Humans are about 99.9% similar; DNA fingerprinting detects small but useful differences.

Forgetting identical twins

DNA fingerprints are generally unique, but identical twins have nearly identical DNA profiles.

Reference tables3

Quick revision8

Formulas3

Human Similarity and Variation
Genetic similarity ≈99.9\text{Genetic similarity }≈ 99.9%;\text{ variation }≈ 0.1%

Humans share most of their DNA sequence, but the small variable fraction is enough for identification and disease association studies.

99.9%
Approximate DNA sequence similarity among humans
0.1%
Approximate variable DNA fraction
VNTR Length Concept
VNTR length = repeat unit length × number of repeats\text{VNTR length }=\text{ repeat unit length }×\text{ number of repeats}

Different individuals may have different numbers of tandem repeat units, producing different DNA fragment lengths.

repeat unit length
Length of one repeated DNA sequence
number of repeats
How many times the repeat occurs at a locus
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Diagrams4

Videos for Molecular Basis of Inheritance

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NEET PYQs from Molecular Basis of Inheritance

The latest 3 of 98 previous-year questions from this chapter.

  1. NEET 2026Set 11MediumQ1

    Which of the following statements are correct with reference to packaging of DNA helix? A. Histones are organized to form a unit of eight molecules called histone octamer. B. Histones are negatively charged basic proteins. C. Histones are rich in the basic amino acid residues - lysine and arginine. D. The positively charged DNA is wrapped around the histone octamer to form nucleosome. E. The packaging of chromatin at higher levels requires an additional set of proteins called non-histone chromosomal proteins. Choose the correct answer from the options given below :

    • AA, C and E only
    • BB, D and E only
    • CC, D and E only
    • DA, B and D only
  2. NEET 2026Set 11MediumQ2

    Arrange the following steps of DNA fingerprinting in a correct sequence. A. Isolation of DNA and its digestion by restriction endonucleases. B. Hybridisation using a labelled VNTR probe. C. Transferring of separated DNA fragments to synthetic membranes. D. Detection of hybridised DNA fragments by autoradiography. E. Separation of DNA fragments by electrophoresis. Choose the correct answer from the options given below :

    • AA, B, D, C, E
    • BA, D, B, E, C
    • CA, E, C, B, D
    • DA, E, B, C, D
  3. NEET 2026Set 11MediumQ3

    Which of the following statements are correct with reference to a transcription unit? A. A transcription unit in DNA is defined primarily by three regions : promoter, structural gene and terminator. B. The promoter is said to be located towards the 5'-end of the structural gene. C. The promoter is a DNA sequence that provides binding site for RNA polymerase. D. The promoter defines the template and coding strands. E. The terminator is located towards the 3'-end of the coding strand and it defines the end of the process of transcription. Choose the correct answer from the options given below:

    • AA, B, C and D only
    • BA, C, D and E only
    • CB, C, D and E only
    • DA, B, C, D and E

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