What is semiconservative replication? How was it proved?

In semiconservative replication, each daughter DNA molecule consists of one original (parental) strand and one newly synthesised strand — the double helix unwinds and each strand serves as a template. Meselson and Stahl proved this in 1958 using heavy nitrogen (N-15) labelling in E. coli. After one generation in N-14 medium, all DNA was hybrid density; after two generations, hybrid and light DNA appeared in equal amounts — exactly as semiconservative predicts.

What is the genetic code? List its main features.

The genetic code is the set of rules by which nucleotide triplets (codons) in mRNA specify amino acids. Features: (1) Triplet — each codon is 3 nucleotides. (2) Non-overlapping — each nucleotide belongs to only one codon. (3) Degenerate — multiple codons can specify the same amino acid (e.g., UUU and UUC both code for phenylalanine). (4) Universal — same code in almost all organisms. (5) Unambiguous — each codon specifies only one amino acid. AUG is the start codon (methionine); UAA, UAG, UGA are stop codons.

Tamil Nadu Class 12 Biology — Molecular Biology and DNA Notes

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Overview: Why Molecular Biology Matters for NEET

Molecular biology contributes 5–7 questions in NEET. DNA structure, replication, transcription, translation, and the genetic code are high-priority topics. The lac operon is frequently tested as a 1-mark or short-answer question.

1. Nucleic Acid Structure

Nucleotides — The Building Blocks

Every nucleotide consists of three parts: 1. Phosphate group — negatively charged, forms the backbone 2. Pentose sugar — deoxyribose in DNA, ribose in RNA 3. Nitrogenous base — carries genetic information

Bases: Purines vs Pyrimidines

Type	Bases	Memory Aid
Purines (double ring)	Adenine (A), Guanine (G)	PURe As Gold
Pyrimidines (single ring)	Cytosine (C), Thymine (T) in DNA; Uracil (U) in RNA	CUT the Pye

DNA vs RNA — Key Differences

Feature	DNA	RNA
Sugar	Deoxyribose	Ribose
Bases	A, T, G, C	A, U, G, C
Strands	Double-stranded	Usually single-stranded
Location	Nucleus (mainly)	Nucleus and cytoplasm
Stability	Very stable	Less stable
Function	Genetic information storage	Protein synthesis intermediary

2. Watson-Crick Double Helix Model (1953)

James Watson and Francis Crick proposed the B-form double helix based on:

Chargaff's rules (A=T, G=C)
X-ray crystallography data from Rosalind Franklin and Maurice Wilkins

Key Features of the Double Helix

1. Two antiparallel strands — one runs 5'→3', the other 3'→5' 2. Right-handed helix — coils clockwise when viewed from top 3. Base pairing rules: A=T (2 hydrogen bonds), G≡C (3 hydrogen bonds) 4. Sugar-phosphate backbone on outside; bases stacked inside 5. Helix diameter: 2 nm; pitch (one full turn) = 3.4 nm; 10 base pairs per turn

Chargaff's Rules (Must Memorise)

For any double-stranded DNA:

A = T (always)
G = C (always)
A + G = T + C (total purines = total pyrimidines)
% A + % T + % G + % C = 100%

Exam application: If A = 30%, then T = 30%, leaving G + C = 40%, so G = C = 20%.

3. DNA Replication

Semiconservative Model

Each new double helix has one old strand (template) + one new strand.

Original:    5'—ATGCGTA—3'
             3'—TACGCAT—5'
                  ↓
After replication (two daughter molecules):

Daughter 1:  5'—ATGCGTA—3'  (old strand)
             3'—TACGCAT—5'  (new strand)

Daughter 2:  5'—ATGCGTA—3'  (new strand)
             3'—TACGCAT—5'  (old strand)

Meselson-Stahl Experiment (1958)

Grew E. coli in ¹⁵N (heavy nitrogen) medium — all DNA becomes heavy
Transferred to ¹⁴N (light nitrogen) medium
After 1 generation: all DNA = hybrid density (one heavy, one light strand)
After 2 generations: 50% hybrid + 50% light DNA

Only semiconservative replication predicts this pattern — proved definitively.

Key Enzymes in DNA Replication

Enzyme	Function
Helicase	Unwinds double helix at replication fork
SSB proteins	Keep strands separated
Primase	Synthesises short RNA primer (needed to start DNA synthesis)
DNA Polymerase III	Main synthesis enzyme; adds nucleotides 5'→3' only
DNA Polymerase I	Removes RNA primers and fills gaps
DNA Ligase	Joins Okazaki fragments and seals nicks
Topoisomerase	Relieves torsional stress ahead of replication fork

Leading vs Lagging Strand

                    Replication fork moves →
5'————————————————————————————————> 3' (template)
         Leading strand (continuous synthesis 5'→3')
         ←————————————

         Okazaki fragment   Okazaki fragment
         ←—————           ←—————
3'————————————————————————————————> 5' (template)
                    Lagging strand (discontinuous)

Leading strand: Synthesised continuously toward the fork (5'→3')
Lagging strand: Synthesised away from fork in short segments (Okazaki fragments, ~1000–2000 nucleotides in bacteria)
DNA ligase joins Okazaki fragments after RNA primers are replaced

4. Transcription (DNA → RNA)

Key Terminology

Template strand (antisense strand): DNA strand used as template; read 3'→5' by RNA polymerase
Coding strand (sense strand): DNA strand with same sequence as the mRNA (but with T instead of U)
mRNA: Messenger RNA — carries code from nucleus to ribosomes

Steps of Transcription

1. Initiation: RNA polymerase binds to promoter sequence (TATA box in eukaryotes) upstream of gene

2. Elongation: RNA polymerase reads template 3'→5', synthesises mRNA 5'→3' using complementary base pairing (A→U, T→A, G→C, C→G)

3. Termination: RNA polymerase reaches terminator sequence → transcription stops → mRNA released

Post-transcriptional Modifications in Eukaryotes

Modification	What happens	Purpose
5' capping	7-methylguanosine added to 5' end	Protects mRNA from degradation; needed for ribosome binding
Poly-A tail	150–200 adenine nucleotides added to 3' end	Stability and export from nucleus
RNA splicing	Introns removed; exons joined	Only exons code for protein

Introns = intervening sequences (non-coding, removed) Exons = expressed sequences (coding, retained)

Splicing is carried out by spliceosome (RNA + protein complex).

Types of RNA

RNA Type	Function	Structure
mRNA	Carries genetic code from DNA to ribosome	Linear
tRNA	Carries amino acids to ribosome	Cloverleaf (L-shape in 3D)
rRNA	Structural and catalytic component of ribosome	Forms ribozyme

tRNA structure: Cloverleaf with 4 arms:

Anticodon loop: recognises mRNA codon (complementary, antiparallel)
Acceptor stem (3'-CCA end): amino acid attaches here

5. Translation (mRNA → Protein)

Ribosome Structure

Component	Prokaryotes	Eukaryotes
Small subunit	30S	40S
Large subunit	50S	60S
Complete ribosome	70S	80S

The large subunit has three sites: A (aminoacyl), P (peptidyl), E (exit)

Genetic Code

Triplet: Three nucleotides = one codon
64 codons total: 61 sense codons (specify amino acids) + 3 stop codons
Start codon: AUG → methionine (first amino acid in all proteins)
Stop codons: UAA, UAG, UGA ("UAA = U Are All done")
Degenerate: Multiple codons → same amino acid (e.g., GGU, GGC, GGA, GGG all = glycine)
Unambiguous: Each codon → only one amino acid
Universal: Same code in nearly all organisms (minor exceptions in mitochondria)
Non-overlapping: Each nucleotide is part of only one codon

Steps of Translation

1. Initiation:

Answer

Small ribosome subunit binds mRNA at 5' cap (eukaryotes) or Shine-Dalgarno sequence (prokaryotes)
tRNA carrying methionine (initiator tRNA) pairs with AUG start codon at P site
Large subunit joins → complete ribosome formed

2. Elongation (repeat for each amino acid):

Answer

Aminoacyl-tRNA enters A site → anticodon pairs with codon
Peptidyl transferase (rRNA enzyme in large subunit) forms peptide bond
Ribosome translocates 3 nucleotides toward 3' end (translocation)
Empty tRNA exits through E site

3. Termination:

Answer

Stop codon (UAA/UAG/UGA) reaches A site
Release factor (not tRNA) binds to A site
Peptide chain released; ribosome dissociates

6. Gene Regulation — Lac Operon

Background

Bacteria control gene expression based on environmental conditions. The lac operon is the model for negative inducible regulation in E. coli.

Components of the Lac Operon

DNA: ——[regulator]——[promoter]——[operator]——[lacZ]——[lacY]——[lacA]——

Regulator gene: codes for lac repressor protein
Promoter: RNA polymerase binding site
Operator: repressor binding site (between promoter and structural genes)
lacZ: β-galactosidase (cleaves lactose)
lacY: permease (transports lactose into cell)
lacA: transacetylase (minor function)

When Lactose is ABSENT (genes OFF)

1. Regulator gene constitutively expressed → repressor protein synthesised 2. Repressor binds to operator → blocks RNA polymerase from reaching structural genes 3. lacZ, lacY, lacA are NOT transcribed → no enzyme made

When Lactose is PRESENT (genes ON)

1. Lactose enters cell → converted to allolactose (the inducer) 2. Allolactose binds to repressor → changes repressor shape → repressor can no longer bind operator 3. RNA polymerase reads through operator → lacZ, lacY, lacA transcribed 4. Enzymes made → lactose digested 5. When lactose is used up → allolactose disappears → repressor active again → genes turn off

Why this is "negative inducible":

Negative: repressor normally turns genes OFF
Inducible: inducer (allolactose) turns genes ON

7. Human Genome Project — Key Facts

Feature	Data
Total base pairs	~3 billion
Protein-coding genes	~20,000–25,000
Non-coding DNA	~98% of genome
Completed	2003 (draft 2001)
Technology used	Automated DNA sequencing, bioinformatics

Significance:

Identifies genes linked to diseases (e.g., BRCA1/BRCA2 for breast cancer)
Enables pharmacogenomics — personalised medicine
Forensic identification using DNA fingerprinting
Evolutionary studies across species

Board Examination Questions

1-Mark Questions

Q1: Name the enzyme that seals nicks in DNA and joins Okazaki fragments. DNA Ligase

Q2: What is the anticodon of tRNA that recognises the codon AUG? The anticodon is UAC (antiparallel and complementary: A pairs with U, U pairs with A, G pairs with C).

Q3: Which strand of DNA serves as the template for transcription? The template strand (also called antisense strand or non-coding strand), read in the 3'→5' direction.

Q4: What are introns? Introns are intervening, non-coding sequences within eukaryotic genes that are transcribed into pre-mRNA but are removed by RNA splicing before translation. Only exons are retained in the mature mRNA.

Q5: Name two stop codons. UAA and UAG (also UGA — any two are accepted).

2-Mark Questions

Q1: Distinguish between template strand and coding strand.

Answer

Template Strand	Coding Strand
Read by RNA polymerase	Not used as template
Runs 3' to 5'	Runs 5' to 3'
Complementary to mRNA	Same sequence as mRNA (with T instead of U)
Also called: antisense / non-coding strand	Also called: sense / coding strand

Q2: What is the significance of RNA splicing in eukaryotes?

Answer

RNA splicing removes introns (non-coding sequences) from the pre-mRNA and joins the exons to form mature mRNA. This allows one gene to produce multiple proteins through alternative splicing (different exon combinations). It also allows parts of non-coding DNA to have regulatory functions. Unspliced mRNA containing introns cannot be properly translated.

Q3: Why does DNA polymerase require a primer?

Answer

DNA polymerase can only add nucleotides to an existing 3'-OH group — it cannot start synthesis on a bare template strand. Therefore, primase first synthesises a short RNA primer (8–12 nucleotides) that provides a free 3'-OH end. DNA polymerase III then extends from this primer. After synthesis, the RNA primer is removed by DNA Polymerase I and replaced with DNA, and the gap is sealed by DNA Ligase.

Q4: Explain why the genetic code is called degenerate.

Answer

The genetic code is degenerate because multiple codons can specify the same amino acid. There are 64 possible triplet codons but only 20 amino acids. Most amino acids have between 2 and 6 synonymous codons (e.g., leucine has 6 codons: UUA, UUG, CUU, CUC, CUA, CUG). Degeneracy is especially common in the third position of the codon (wobble position). This redundancy provides partial protection against point mutations in the third base — many such mutations still specify the correct amino acid.

5-Mark Questions

Q1: Describe the process of DNA replication with a focus on the leading and lagging strands.

Answer

DNA replication follows the semiconservative model, where each daughter molecule retains one parental strand.

Initiation: Replication begins at specific sequences called origins of replication. In E. coli, there is one origin (oriC). Initiator proteins recognise the origin, and helicase unwinds the double helix creating a replication fork. SSB proteins keep strands separated. Topoisomerase relieves tension ahead of the fork.

Primer synthesis: Primase synthesises short RNA primers (8–12 nucleotides) on both template strands, providing a 3'-OH group for DNA polymerase.

Leading strand synthesis: The leading strand template runs 3'→5'. DNA Polymerase III synthesises the leading strand continuously in the 5'→3' direction toward the replication fork.

Lagging strand synthesis: The lagging strand template runs 5'→3' (opposite to the fork direction). DNA Polymerase III can only work 5'→3', so it synthesises short fragments (Okazaki fragments, each preceded by an RNA primer) moving away from the fork. In E. coli, each fragment is 1000–2000 nucleotides long.

Primer removal and gap filling: DNA Polymerase I removes RNA primers using its 5'→3' exonuclease activity and fills the gaps with DNA.

Ligation: DNA Ligase seals the remaining nicks (single-strand breaks) between Okazaki fragments and completes the lagging strand.

ResultTwo identical daughter DNA molecules, each with one old and one new strand (semiconservative).

Q2: Describe the lac operon model of gene regulation. What happens when lactose is present vs absent?

Answer

The lac operon in E. coli is a classic model of negative inducible gene regulation. It controls the expression of three genes needed to metabolise lactose.

Structure: The operon consists of a regulator gene (separate from operon), a promoter (RNA polymerase binding site), an operator (repressor binding site), and three structural genes: lacZ (β-galactosidase), lacY (permease), lacA (transacetylase).

When lactose is absent: The regulator gene is continuously expressed, producing lac repressor protein. The repressor has a high affinity for the operator sequence. It binds the operator and physically blocks RNA polymerase from progressing to the structural genes. As a result, lacZ, lacY, and lacA are not transcribed — no enzymes are produced. This conserves energy when lactose is unavailable.

When lactose is present: Lactose enters the cell and a small amount is converted by existing β-galactosidase to allolactose (the true inducer). Allolactose binds to the lac repressor, causing a conformational change in the repressor's shape. The altered repressor loses affinity for the operator and dissociates from it. RNA polymerase can now transcribe lacZ, lacY, and lacA. The resulting enzymes allow the cell to import and digest lactose. Once all lactose is consumed, allolactose disappears, the repressor reverts to its active shape, binds the operator again, and the genes are turned off.

Why this is adaptive: It is energetically wasteful to produce lactose-digesting enzymes when there is no lactose. The lac operon ensures these enzymes are made only when needed.

NEET Focus Points

1. Chargaff's rules: If %A is given, immediately find %T (equal), then split remainder equally between G and C 2. Replication enzymes: Helicase unwinds, primase makes primer, DNA Pol III synthesises, DNA Pol I removes primer, Ligase joins fragments 3. Codons: AUG = start/methionine; UAA, UAG, UGA = stop. Know these by heart 4. Lac operon: Allolactose is the inducer (not lactose directly); repressor binds operator; negative regulation 5. tRNA anticodon: Complement of codon and antiparallel — practice working these out

Frequently Asked Questions

Chargaff's rules state that in any double-stranded DNA: (1) the amount of adenine equals thymine (A = T), and (2) the amount of guanine equals cytosine (G = C). This means purines always equal pyrimidines (A+G = T+C). These rules were key evidence for the Watson-Crick double helix model.