The trap that costs marks in DNA replication questions: confusing the direction of synthesis with the direction of fork movement.
DNA polymerase synthesises new strands exclusively in the 5′→3′ direction. No known DNA polymerase works 3′→5′. This single constraint creates the entire leading-strand / lagging-strand distinction.
Leading strand: The template runs 3′→5′ toward the fork, so synthesis proceeds continuously 5′→3′ in the same direction the fork moves. One primer, one continuous strand.
Lagging strand: The template runs 5′→3′ toward the fork. Since polymerase cannot synthesise 3′→5′, it must work in short bursts away from the fork. Each burst is an Okazaki fragment (1000–2000 nucleotides in prokaryotes, 100–200 in eukaryotes). Each fragment needs its own RNA primer. DNA ligase seals the nicks after primer removal.
Meselson-Stahl experiment (1958): Used ¹⁵N (heavy) and ¹⁴N (light) to demonstrate semi-conservative replication in E. coli. After one generation in ¹⁴N medium: all DNA was hybrid (one heavy, one light strand) — intermediate band in CsCl gradient. After two generations: 50% hybrid, 50% light. This ruled out conservative and dispersive models.
Key enzymes at the fork: Helicase (unwinds), SSB proteins (stabilise single strands), primase (lays RNA primer), DNA pol III (main synthesis in prokaryotes), DNA pol I (primer removal + gap filling), ligase (seals nicks).
Watch-out for NEET: When a question says "the lagging strand grows 3′→5′," that describes the net extension toward the fork — each Okazaki fragment is still synthesised 5′→3′. Picking "3′→5′ synthesis" as the answer loses you marks.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
DNA replication is termed semi-conservative because:
Answer: C. Semi-conservative means each daughter duplex conserves one original (parental) strand paired with one newly synthesised strand, as demonstrated by Meselson and Stahl (NCERT Class 12 Biology Chapter 5, page 89).
Why A is wrong: A is impossible under any model: both parental strands would have to be discarded. Semi-conservative replication keeps one parental strand in each daughter molecule.
Why B is wrong: B describes the conservative model (one daughter all-old, one all-new), which Meselson-Stahl ruled out.
Why D is wrong: D describes the dispersive model where parental DNA is fragmented among daughters — also ruled out by Meselson-Stahl.
In the Meselson-Stahl experiment, after two generations of growth in ¹⁴N medium, what proportion of DNA molecules would show a hybrid (intermediate) density band?
Answer: C. After generation 1: all hybrid. After generation 2: each hybrid molecule produces one hybrid + one light daughter. So 2 hybrid + 2 light out of 4 total = 50% hybrid (NCERT Class 12 Biology Chapter 5, page 89).
Why A is wrong: A (100%) is the result after one generation, not two. By the second round each hybrid produces one hybrid and one light daughter.
Why B is wrong: B (75%) does not correspond to any generation in semi-conservative replication — it confuses the doubling pattern.
Why D is wrong: D (25%) would apply after three generations (2 hybrid out of 8), not two.
Which of the following is the correct direction of synthesis of a new DNA strand by DNA polymerase?
Answer: A. All DNA polymerases add nucleotides only to the 3′-OH end, synthesising in the 5′→3′ direction. This applies to BOTH leading and lagging strands (NCERT Class 12 Biology Chapter 5, page 90).
Why B is wrong: B (3′→5′) — no DNA polymerase synthesises in this direction. The 3′→5′ activity of some polymerases is proofreading (exonuclease), not synthesis (trap: 5′→3′ synthesis polarity).
Why C is wrong: C is wrong because polymerase direction is invariant — always 5′→3′ on both strands. The difference is continuous vs discontinuous, not direction of synthesis.
Why D is wrong: D is wrong because the lagging strand is also synthesised 5′→3′; each Okazaki fragment grows 5′→3′. The net extension toward the fork is opposite, but synthesis polarity remains 5′→3′ (trap: 5′→3′ synthesis polarity).
Okazaki fragments are formed on the:
Answer: B. The lagging strand template runs 5′→3′ toward the fork, forcing polymerase to synthesise in short 5′→3′ bursts (Okazaki fragments) away from the fork, each requiring a separate RNA primer (NCERT Class 12 Biology Chapter 5, page 90 describes the discontinuous synthesis and its joining by DNA ligase; the names Okazaki fragment and RNA primer go beyond NCERT).
Why A is wrong: A is wrong — the leading strand uses only one primer and is synthesised continuously. Multiple primers are a lagging-strand feature.
Why C is wrong: C is wrong — Okazaki fragments form exclusively on the lagging strand. The leading strand is continuous.
Why D is wrong: D is wrong — it's the lagging strand (not leading) that has its template oriented such that synthesis moves away from the fork. The leading strand's template allows continuous synthesis toward the fork.
A student claims that the lagging strand is synthesised in the 3′→5′ direction. What is the correct rebuttal?
Answer: D. Every Okazaki fragment is synthesised 5′→3′ by the same DNA polymerase. The fragments appear to extend the strand toward the fork, but each individual fragment grows 5′→3′ away from the fork. No polymerase synthesises 3′→5′ (mistake: lagging strand direction confusion).
Why A is wrong: A is wrong — both daughter duplexes have one newly synthesised strand. The lagging strand IS synthesised; it's just done in fragments.
Why B is wrong: B is wrong — no DNA polymerase in any organism synthesises 3′→5′. The 3′→5′ exonuclease activity is for proofreading, not strand extension.
Why C is wrong: C is wrong — the lagging strand is NOT continuous. Discontinuous synthesis (Okazaki fragments) is its defining characteristic.
Which enzyme seals the gaps between Okazaki fragments after RNA primers are removed and replaced with DNA?
Answer: A. DNA ligase catalyses the formation of phosphodiester bonds between the 3′-OH of one fragment and the 5′-phosphate of the next, sealing nicks in the sugar-phosphate backbone (NCERT Class 12 Biology Chapter 5, page 90).
Why B is wrong: B — Helicase unwinds the double helix at the fork; it has no ligating activity.
Why C is wrong: C — DNA pol III is the main replicative polymerase that synthesises new DNA but does not seal nicks between fragments.
Why D is wrong: D — Primase synthesises RNA primers to initiate each Okazaki fragment; it does not join fragments.
In the Meselson-Stahl experiment, if replication were conservative instead of semi-conservative, what would the CsCl gradient show after one generation in ¹⁴N medium?
Answer: D. Conservative replication keeps both parental strands together (heavy band) and produces one entirely new duplex (light band). No intermediate band would appear. Meselson-Stahl observed ONLY an intermediate band after one generation, ruling out conservative replication (NCERT Class 12 Biology Chapter 5, page 89).
Why A is wrong: A (intermediate only) is what semi-conservative replication produces after one generation — the actual experimental result that confirmed semi-conservative.
Why B is wrong: B (three bands) does not correspond to any single-generation prediction from conservative, semi-conservative, or dispersive models.
Why C is wrong: C (light only) would mean all parental heavy DNA disappeared in one generation — not possible under any proposed model.
During DNA replication, the enzyme that unwinds the double helix at the replication fork is:
Answer: B. Helicase breaks hydrogen bonds between complementary base pairs, separating the two strands at the replication fork. Topoisomerase relieves torsional strain ahead of the fork but does not unwind the helix directly (helicase and topoisomerase go beyond NCERT, whose account in NCERT Class 12 Biology Chapter 5, page 90 names only DNA polymerase and DNA ligase).
Why A is wrong: A — Topoisomerase relieves supercoiling (torsional stress) ahead of the fork by cutting and re-joining the backbone, but it does not separate the two strands.
Why C is wrong: C — DNA pol I removes RNA primers and fills gaps with DNA; it does not unwind the helix.
Why D is wrong: D — Ligase seals nicks between Okazaki fragments; it has no unwinding function.
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Given
• Starting: 1 DNA molecule, both strands heavy (¹⁵N-¹⁵N).• Grown in ¹⁴N for 3 generations.
Required
• Total number of DNA molecules after 3 generations.• Number (and fraction) showing hybrid density.
Concept
Semi-conservative replication: each round doubles the molecule count. Each parental strand is conserved in one daughter molecule forever. After n generations: 2ⁿ total molecules, exactly 2 contain a parental (heavy) strand (one from each original strand), producing hybrid-density molecules.
Formula
• Total molecules = 2ⁿ (n = number of generations)• Hybrid molecules = 2 (constant for n ≥ 1, since only 2 original parental strands exist)• Light molecules = 2ⁿ − 2
Substitution
• Total = 2³ = 8• Hybrid = 2• Light = 8 − 2 = 6
Calculation
Fraction hybrid = 2/8 = 1/4 = 25%
Note: The number 2 (parental strands) and 3 (generations) are exact counting integers and do not limit significant figures.
Final answer
After 3 generations: 8 DNA molecules total, of which 2 are hybrid (¹⁵N-¹⁴N, intermediate band) and 6 are light (¹⁴N-¹⁴N). Fraction hybrid = 25%.
Common trap
Students sometimes think the number of hybrid molecules increases with each generation (e.g., "half are always hybrid"). In reality, only 2 parental strands exist — once separated in the first round, they persist in exactly 2 hybrid duplexes forever, regardless of further divisions. The hybrid fraction decreases each generation: 100% → 50% → 25% → 12.5%...
Similar NEET-style question
"After four generations of E. coli grown in ¹⁴N medium (originally ¹⁵N-labelled), how many molecules would appear in the light band of a CsCl gradient?" (Answer: 2⁴ − 2 = 14 light molecules.)
---
Semi-conservative (Meselson-Stahl 1958, ¹⁵N → ¹⁴N density gradient). DNA polymerase III. Leading strand continuous, lagging discontinuous (Okazaki fragments). Bidirectional from origin. RNA primer + ligase + helicase + topoisomerase.
-- NCERT Class 12 Biology, Ch. 5, p. 89These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Sign Convention
DNA polymerase synthesises only 5'→3'. Leading strand: continuous, same direction as fork. Lagging: discontinuous (Okazaki), opposite to fork.
Question on Okazaki, leading vs lagging, primer direction.
Reading template 3'→5'; synthesising 5'→3'. Lagging strand needs short fragments because it can't run continuously against fork direction.
Root cause: concept gap
ALL DNA pol synthesises 5'→3'. Lagging strand uses Okazaki fragments to APPEAR to extend toward fork while each fragment grows 5'→3'.
More in Genetics and Evolution: 19 exam traps and mistakes · 4 formulas · 1 question pattern from its other lessons.
4 questions from NEET 2021, 2022, 2023, 2024. Answers verified against NTA official keys.
Which of the following statement is correct regarding the process of replication in E.coli?
Among eukaryotes, replication of DNA takes place in :
Match List-I with List-II. Choose the correct answer from the options given below. (a) (b) (c) (d)
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →