This MCQ module is based on: Replication Transcription
TOPIC 21 OF 64
Replication Transcription
🎓 Class 12
Biology
CBSE
Theory
Ch 5 – Molecular Basis of Inheritance
⏱ ~14 min
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Replication Transcription
6.3 DNA Replication — Copying the Genetic Code
Watson and Crick noted that the structure of DNA — two complementary strands — immediately suggests how it copies itself. Each strand can serve as a template for synthesizing the new partner strand. They proposed semi-conservative replication: each daughter DNA contains one old (parental) strand + one new strand.
6.3.1 Meselson-Stahl Experiment (1958)
Matthew Meselson and Franklin Stahl elegantly proved semi-conservative replication using E. coli and density-gradient centrifugation:
- Grew bacteria for many generations in heavy nitrogen (¹⁵N) medium → all DNA contained heavy nitrogen → "heavy DNA" band on density gradient.
- Transferred bacteria to light nitrogen (¹⁴N) medium and let them divide once.
- After 1st generation: DNA showed an intermediate density band (¹⁵N/¹⁴N hybrid).
- After 2nd generation: two bands — half intermediate, half light. This pattern is predicted ONLY by semi-conservative replication.
| Model | 1st generation | 2nd generation |
|---|---|---|
| Conservative (parent stays intact, makes copy) | 1 heavy + 1 light | 1 heavy + 3 light |
| Semi-conservative ✓ (each strand becomes template) | All hybrid | 2 hybrid + 2 light |
| Dispersive (mixture of old + new in each strand) | All hybrid | All hybrid |
6.3.2 Mechanism of DNA Replication
Replication begins at specific sites called origins of replication. Several enzymes work together:
| Enzyme | Function |
|---|---|
| Helicase | Unwinds the double helix, breaking H-bonds |
| Topoisomerase / Gyrase | Relieves supercoiling tension ahead of the fork |
| SSB proteins | Single-strand binding — keep separated strands stable |
| Primase | Synthesizes RNA primers (DNA polymerase needs primer to start) |
| DNA polymerase III | Adds nucleotides only in 5'→3' direction |
| DNA polymerase I | Replaces RNA primers with DNA |
| DNA ligase | Joins Okazaki fragments on lagging strand |
6.3.3 Leading and Lagging Strands
Because DNA polymerase only synthesises in 5'→3' direction, the two antiparallel strands are replicated differently:
- Leading strand (3'→5' template): synthesised continuously, in the same direction as fork movement.
- Lagging strand (5'→3' template): synthesised discontinuously as short fragments called Okazaki fragments — these are joined later by DNA ligase.
6.4 Transcription — DNA → RNA
Transcription is the process of making an RNA copy from a DNA template. The enzyme that catalyses it is RNA polymerase.
6.4.1 Key Features
- Only ONE strand of DNA is used as template — called the template strand (3'→5').
- The other strand has the same sequence as RNA (with T → U) and is called the coding strand / sense strand.
- RNA is synthesised in 5'→3' direction.
- RNA differs from DNA: contains uracil (U) instead of thymine; has ribose sugar; usually single-stranded.
6.4.2 Steps of Transcription
- Initiation: RNA polymerase binds at the promoter (a specific DNA sequence upstream of the gene). With help from sigma factor (in bacteria), the helix unwinds locally.
- Elongation: RNA polymerase moves along the template, adding ribonucleotides 5'→3' according to base-pairing rules: A→U, T→A, G→C, C→G.
- Termination: RNA polymerase reaches a terminator sequence; the RNA transcript is released, and the polymerase falls off.
6.4.3 Three Major Types of RNA
| RNA type | Function | % of total RNA |
|---|---|---|
| mRNA (messenger) | Carries genetic message from DNA to ribosome; specifies amino acid sequence of protein | ~5% |
| tRNA (transfer) | Carries amino acids to the ribosome; reads mRNA codons via its anticodon | ~15% |
| rRNA (ribosomal) | Structural and catalytic component of the ribosome; performs peptide bond formation | ~80% |
In prokaryotes, a single RNA polymerase makes all three types. In eukaryotes, three different polymerases make them: RNA pol I (rRNA), pol II (mRNA), pol III (tRNA + small RNAs).
6.4.4 Post-transcriptional Processing in Eukaryotes
The newly made eukaryotic mRNA undergoes processing before leaving the nucleus:
- 5' capping: a methylated guanine cap is added to the 5' end (protects from degradation).
- 3' polyadenylation: a tail of ~200 adenines (poly-A tail) added to 3' end (stability + transport).
- Splicing: non-coding introns are removed; exons (coding parts) are joined together to form mature mRNA.
🧬 Interactive: DNA → mRNA Transcriber
Enter a DNA template strand (3'→5') and see the transcribed mRNA:
📐 Activity 6.2 — Predict Replication Outcome
Setup: A bacterial culture is grown for 4 generations in heavy ¹⁵N medium (all DNA heavy), then transferred to ¹⁴N medium and allowed to divide TWICE.
Predict: What proportion of DNA will be (a) hybrid (¹⁵N/¹⁴N), (b) fully light (¹⁴N/¹⁴N), (c) fully heavy (¹⁵N/¹⁵N)?
After 1st division: 2 hybrid molecules (each has one ¹⁵N old strand + one ¹⁴N new strand).
After 2nd division: Each hybrid → 1 hybrid (the ¹⁵N old strand is template + new ¹⁴N) + 1 light (the ¹⁴N strand is template + new ¹⁴N).
Total = 4 molecules. (a) Hybrid: 2/4 = 50%; (b) Light: 2/4 = 50%; (c) Heavy: 0/4 = 0%.
This pattern (hybrid + light, no heavy) directly proves semi-conservative replication.
Worked Examples
Worked Example 1: mRNA from Coding Strand
The coding strand of a gene reads: 5'-ATGGCATTAGCC-3'. Write the mRNA produced and identify the template strand.
Step 1: The coding strand has the same sequence as mRNA, except T → U.
Step 2: Coding strand: 5'-ATGGCATTAGCC-3'
Template strand (complementary, antiparallel): 3'-TACCGTAATCGG-5'
mRNA (read off template, 5'→3'): 5'-AUGGCAUUAGCC-3'
Note: mRNA = coding strand with U instead of T. This shortcut works for any prokaryotic gene without introns.
Step 2: Coding strand: 5'-ATGGCATTAGCC-3'
Template strand (complementary, antiparallel): 3'-TACCGTAATCGG-5'
mRNA (read off template, 5'→3'): 5'-AUGGCAUUAGCC-3'
Note: mRNA = coding strand with U instead of T. This shortcut works for any prokaryotic gene without introns.
Worked Example 2: Replication Time
The E. coli genome is ~4.6 million base pairs. If DNA polymerase synthesises ~1000 bases/second, how long does it take to replicate (assuming a single replication fork)?
Time = base pairs / rate = 4,600,000 / 1000 = 4600 seconds ≈ 76 minutes.
But E. coli divides every ~20 minutes! How? Bidirectional replication — replication starts at one origin (oriC) and proceeds in both directions, halving the time. Plus, multiple origins in fast-growing cells. So effective replication is ~38 minutes — fits the doubling time.
Eukaryotic genomes (~3 billion bp in humans) need thousands of origins to replicate in S phase (~6–8 hours).
But E. coli divides every ~20 minutes! How? Bidirectional replication — replication starts at one origin (oriC) and proceeds in both directions, halving the time. Plus, multiple origins in fast-growing cells. So effective replication is ~38 minutes — fits the doubling time.
Eukaryotic genomes (~3 billion bp in humans) need thousands of origins to replicate in S phase (~6–8 hours).
🎯 Competency-Based Questions
Q1. The Meselson-Stahl experiment proved that DNA replication is:L1 Remember
Answer: (b) Semi-conservative. After 1 generation in light nitrogen, all DNA was hybrid; after 2 generations, half was hybrid and half was light. Only the semi-conservative model predicts this exact pattern.
Q2. Fill in the blank: The lagging strand is synthesised in short fragments called _____. L2 Understand
Answer: Okazaki fragments. Because DNA polymerase only adds nucleotides 5'→3', and the lagging strand template runs the wrong way, cells synthesise it as short fragments which are later joined by DNA ligase.
Q3. If a DNA template reads 3'-TACGGTACG-5', what is the corresponding mRNA sequence? L3 Apply
Answer: Pair each base: T→A, A→U, C→G, G→C, G→C, T→A, A→U, C→G, G→C. Reading 5'→3': 5'-AUGCCAUGC-3'. Note that mRNA always has U where DNA has T.
Q4. Compare: List 3 differences between leading and lagging strand synthesis. L4 Analyse
| Feature | Leading strand | Lagging strand |
|---|---|---|
| Direction of synthesis | Same as fork movement | Opposite to fork movement |
| Continuity | Continuous (one long piece) | Discontinuous (Okazaki fragments) |
| Number of primers needed | Just one (start) | Many (one per fragment) |
| Need for DNA ligase | Not needed (1 piece) | Required (joins fragments) |
Q5. HOT (Create): A scientist invents a drug that inhibits topoisomerase. Predict the immediate effects on bacterial replication and explain why such drugs could be used as antibiotics. L6 Create
Effects on replication:
- Topoisomerase relieves supercoiling tension ahead of the replication fork. Without it, the DNA ahead of the fork becomes increasingly twisted.
- The replication fork stalls; the DNA snaps under stress.
- The bacterium cannot replicate its genome → cannot divide → eventually dies.
- Bacterial topoisomerase (DNA gyrase) is structurally different from human topoisomerase II.
- Drugs like fluoroquinolones (ciprofloxacin, levofloxacin) selectively inhibit bacterial gyrase — sparing human cells.
- This makes them effective broad-spectrum antibiotics for urinary tract infections, pneumonia, etc.
- Resistance develops via point mutations in the gyrase gene — illustrating the urgency of antibiotic stewardship.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: DNA replication produces two daughter molecules each containing one parental and one new strand.
R: DNA replication is semi-conservative as proven by Meselson and Stahl.
Answer: (A). Both true; R explains A. The 1958 Meselson-Stahl experiment is a landmark proof of semi-conservative replication.
A: RNA polymerase requires a primer to start transcription.
R: RNA polymerase reads the template strand 3'→5' and synthesises RNA 5'→3'.
Answer: (D). A is FALSE — RNA polymerase does NOT need a primer (unlike DNA polymerase). R is TRUE — that is the direction of transcription. The independence from primer is one key difference from DNA replication.
A: mRNA in eukaryotes is shorter than the original DNA gene.
R: Introns (non-coding sequences) are removed during splicing.
Answer: (A). Both true; R explains A. The pre-mRNA contains both exons and introns; introns are excised during splicing, leaving only exons in the mature mRNA — which is significantly shorter.
Frequently Asked Questions - Replication Transcription
What is the main concept covered in Replication Transcription?
In NCERT Class 12 Biology Chapter on Molecular Basis of Inheritance, "Replication Transcription" covers the core biological structures, processes, and pathways students need for board exam success. The MyAiSchool lesson explains the topic with definitions, labelled diagrams, comparison tables, and interactive simulations. Scientific terminology and physiological/genetic significance are highlighted throughout to build conceptual depth aligned with CBSE 2025-26 syllabus.
How is Replication Transcription useful in real-life or applied biology?
Real-life applications of "Replication Transcription" from NCERT Class 12 Biology Molecular Basis of Inheritance include medical diagnostics, agriculture, biotechnology, public health, evolutionary insights, and ecological monitoring. The MyAiSchool lesson links every biological concept to a tangible application so students see biology as a problem-solving framework for living systems and real-world challenges.
What are the key terms students should memorize for Replication Transcription?
Key terms in "Replication Transcription" (NCERT Class 12 Biology Molecular Basis of Inheritance) are tabulated in the MyAiSchool key-terms grid. Students should memorize each term with its precise definition, function, and example. Terminology is high-yield in CBSE board exams — 1-mark MCQs and 2-mark short answers test definitions directly. The Summary section provides a printable quick-reference card.
How does this part connect to other parts of the chapter?
NCERT Class 12 Biology Molecular Basis of Inheritance is structured so each part builds biological understanding sequentially. "Replication Transcription" connects to neighbouring parts via shared mechanisms, structural hierarchies, and physiological processes. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected biological story rather than disconnected fragments.
What types of CBSE board questions come from Replication Transcription?
CBSE board questions from "Replication Transcription" typically include: (1) 1-mark MCQs on definitions and processes, (2) 2-mark short-answer differences/comparisons, (3) 3-mark labelled-diagram questions, (4) 5-mark long-answer essays combining mechanism + diagram + significance. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Replication Transcription" lesson allows students to explore biological processes, classifications, or pathways using selectors and sliders, with live visual feedback. To use it effectively: (1) explore each option/state, (2) compare with textbook diagrams, (3) note the function/outcome changes, (4) try the integrated practice quiz. The simulation reinforces visual-spatial understanding that pure text-based study cannot.
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