Topics
6Chapter Overview
Overview
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.
- 1The chapter moves from molecular structure to biological function: DNA structure → replication → transcription → translation → regulation.
- 2NCERT emphasizes experimental evidence for DNA as genetic material: Griffith, Avery-MacLeod-McCarty and Hershey-Chase.
- 3Replication, transcription and translation are directional processes, and 5' to 3' polarity is a frequent NEET trap.
- 4Gene regulation in prokaryotes is best understood through Jacob and Monod's lac operon model.
- 5HGP and DNA fingerprinting show how molecular genetics is applied in medicine, forensics and evolution.
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.
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.
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.
The usual flow of genetic information in living cells. DNA is transcribed into RNA, and RNA is translated into a polypeptide.
Variables
DNA=Genetic material storing hereditary information
RNA=Intermediate molecule carrying genetic message
Protein=Functional product formed by amino acid sequence
DNA & RNA
Overview
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.
- 1A molecule must replicate, be stable, allow mutation and express information to act as genetic material.
- 2RNA can act as genetic material in some viruses but is generally less stable and more reactive than DNA.
- 3Chargaff’s rule states that in double-stranded DNA, A = T and G = C.
- 4Watson-Crick DNA is a right-handed double helix with 10 base pairs per turn and 3.4 nm pitch.
- 5Histone octamer plus wrapped DNA forms a nucleosome; nucleosomes form chromatin.
- 6NCERT highlight: DNA is better genetic material, but RNA was likely the first genetic material because it can store information and catalyse reactions.
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.
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.
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'.
In double-stranded DNA, adenine pairs with thymine and guanine pairs with cytosine, so their quantities are equal.
Variables
A=Adenine
T=Thymine
G=Guanine
C=Cytosine
Adjacent base pairs in B-DNA are separated by 0.34 nm, useful for calculating DNA length.
Variables
number of base pairs=Total paired nucleotides in double-stranded DNA
0.34 nm=Distance between two adjacent base pairs
DNA Replication
Overview
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.
- 1Meselson-Stahl experiment used density gradient centrifugation to prove semi-conservative replication.
- 2The replication fork is Y-shaped because parental strands separate and act as templates.
- 3Deoxyribonucleoside triphosphates provide both building blocks and energy for polymerization.
- 4Prokaryotes usually have one origin of replication, while eukaryotes have multiple origins.
- 5NEET commonly asks enzyme functions: helicase unwinds, ligase joins, polymerase extends, primase makes primer.
- 6Replication is fast yet accurate due to proofreading activity of DNA polymerases.
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.
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.
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.
DNA polymerase adds each nucleotide to the 3'-OH end, so the newly synthesized strand always grows in the 5' to 3' direction.
Variables
5'=Phosphate end of a nucleic acid strand
3'=Hydroxyl end where new nucleotide is added
Each daughter duplex conserves one parental strand and contains one freshly synthesized complementary strand.
Variables
old strand=Original parental template strand
new strand=Newly polymerized complementary strand
Transcription
Overview
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.
- 1RNA polymerase does not require a primer, unlike DNA polymerase.
- 2Promoter defines the template strand and direction of transcription.
- 3In bacteria, a single RNA polymerase transcribes all RNA types.
- 4In eukaryotes, RNA polymerase II transcribes hnRNA/mRNA precursor.
- 5Capping adds methyl guanosine at 5' end; tailing adds adenylate residues at 3' end.
- 6Splicing removes introns and joins exons to make mature mRNA.
Transcription Unit
PST: Promoter Starts Transcription, Structural gene is copied, Terminator Stops.
RNA Processing
CaTaS: Capping, Tailing, Splicing.
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.
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.
During transcription, RNA bases pair with the DNA template using uracil instead of thymine.
Variables
A=Adenine
U=Uracil in RNA
T=Thymine in DNA
G/C=Guanine-cytosine complementary pair
Genetic Code & Translation
Overview
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.
- 164 codons include 61 sense codons and 3 stop codons.
- 2The code is unambiguous: one codon specifies only one amino acid.
- 3Wobble at the third codon position contributes to degeneracy.
- 4Prokaryotic ribosome is 70S with 50S and 30S subunits; eukaryotic ribosome is 80S with 60S and 40S subunits.
- 5Peptidyl transferase activity is performed by rRNA, making ribosome a ribozyme.
- 6Translation terminates when a release factor recognizes a stop codon.
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.
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.
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.
Four RNA bases arranged in triplets form 64 possible codons.
Variables
4=RNA bases A, U, G and C
3=Three bases per codon
64=Total possible triplet codons
Most codons specify amino acids, while three codons terminate translation.
Variables
61=Codons that code for amino acids
3=Termination codons UAA, UAG and UGA
Gene Regulation
Overview
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.
- 1Jacob and Monod proposed the operon model for regulation of gene expression.
- 2lacZ codes beta-galactosidase, lacY codes permease and lacA codes transacetylase.
- 3When lactose is absent, the repressor remains active and binds the operator.
- 4When lactose is present, allolactose inactivates the repressor and transcription proceeds.
- 5Glucose level also affects lac operon through catabolite repression, but NCERT NEET focus is mainly inducer-repressor logic.
- 6Eukaryotic genes are regulated more elaborately because of nuclear membrane, chromatin and RNA processing.
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.
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.
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.
Lactose acts as inducer through allolactose, preventing the repressor from blocking RNA polymerase.
Variables
allolactose=Inducer molecule derived from lactose
repressor=Protein that blocks transcription by binding operator
operator=DNA control sequence near promoter
Human Genome Project & DNA Fingerprinting
Overview
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.
- 1HGP used sequencing, genetic mapping, physical mapping and bioinformatics.
- 2A major goal was to transfer related technologies to industry and improve biological research.
- 3Chromosome 1 has maximum genes and Y chromosome has minimum genes according to NCERT.
- 4DNA fingerprinting was developed by Alec Jeffreys; in India, Dr. Lalji Singh is associated with its development.
- 5VNTRs are highly polymorphic because repeat number varies greatly among individuals.
- 6DNA fingerprints are unique except in identical twins.
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.
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.
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.
Humans share most of their DNA sequence, but the small variable fraction is enough for identification and disease association studies.
Variables
99.9%=Approximate DNA sequence similarity among humans
0.1%=Approximate variable DNA fraction
Different individuals may have different numbers of tandem repeat units, producing different DNA fragment lengths.
Variables
repeat unit length=Length of one repeated DNA sequence
number of repeats=How many times the repeat occurs at a locus
Formula Sheet
10The usual flow of genetic information in living cells. DNA is transcribed into RNA, and RNA is translated into a polypeptide.
Variables
DNA=Genetic material storing hereditary information
RNA=Intermediate molecule carrying genetic message
Protein=Functional product formed by amino acid sequence
Four nucleotide bases taken three at a time produce 64 possible codons, enough to code for 20 amino acids and stop signals.
Variables
4=Four nitrogen bases in RNA: A, U, G and C
3=Number of bases in one codon
64=Total possible codons
In double-stranded DNA, adenine pairs with thymine and guanine pairs with cytosine, so their quantities are equal.
Variables
A=Adenine
T=Thymine
G=Guanine
C=Cytosine
Adjacent base pairs in B-DNA are separated by 0.34 nm, useful for calculating DNA length.
Variables
number of base pairs=Total paired nucleotides in double-stranded DNA
0.34 nm=Distance between two adjacent base pairs
B-DNA has approximately 10 base pairs in one complete helical turn.
Variables
base pairs=Number of paired nucleotide units
10=Base pairs per turn in Watson-Crick B-DNA
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NEET PYQs — Molecular Basis of Inheritance
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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 :
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:
In the lac operon, the z gene codes for:
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