Question
In DNA replication, arrange the following events in the correct sequence and identify the enzyme responsible for each: (a) Unwinding of the double helix (b) Synthesis of RNA primer (c) Addition of deoxyribonucleotides in 5’→3’ direction (d) Joining of Okazaki fragments (e) Removal of primer and gap-filling
Also explain why one strand is synthesized continuously while the other is synthesized in fragments.
Solution — Step by Step
Helicase breaks the hydrogen bonds between complementary base pairs, separating the two strands at the replication fork. This creates the template strands. The point where unwinding occurs is called the replication fork, and it moves bidirectionally from the origin of replication (ori).
DNA polymerase cannot start a new chain — it can only extend an existing one. So primase (an RNA polymerase) synthesizes a short RNA primer (~10 nucleotides) complementary to the template. This gives DNA Pol III a 3’-OH group to work from.
DNA Pol III (the main replication enzyme in prokaryotes) adds deoxyribonucleotides in the 5’→3’ direction only. This is the key constraint that creates the leading/lagging strand asymmetry. It reads the template 3’→5’ and synthesizes the new strand 5’→3’.
One template runs 3’→5’ toward the fork — DNA Pol III can follow the fork continuously. This is the leading strand (synthesized as one continuous piece).
The other template runs 5’→3’ toward the fork, meaning Pol III must work away from the fork. It synthesizes short fragments (150-200 nucleotides in eukaryotes) called Okazaki fragments, each needing its own primer. This is the lagging strand.
DNA Pol I removes RNA primers using its 5’→3’ exonuclease activity and fills the gaps with DNA. DNA Ligase then seals the nicks between adjacent fragments by forming phosphodiester bonds. Result: two complete, identical daughter duplexes.
Why This Works
The entire system is built around one non-negotiable rule: DNA polymerase can only add nucleotides to an existing 3’-OH group, and always in the 5’→3’ direction. Every quirk of replication — the primer, the Okazaki fragments, the need for ligase — exists because of this constraint.
Think of it like a one-way escalator. You can only move in one direction. If you need to go the other way, you have to take multiple short trips and then connect them afterward. The lagging strand is essentially doing exactly that.
The semi-conservative nature (each daughter cell gets one old strand and one new strand) was proven by the Meselson-Stahl experiment (1958) using ¹⁵N/¹⁴N labeling — a classic NEET question in itself.
Alternative Method: Remembering the Enzyme Order
For MCQ purposes, this mnemonic works well for the enzyme sequence:
Helicase → Primase → Pol III → Pol I → Ligase “Happy People Play Piano Loudly”
NEET loves asking which enzyme is responsible for which specific step. The most commonly tested trap: DNA Pol III does elongation, but DNA Pol I does primer removal and gap-filling. Students mix these up constantly. Also remember — Pol III has no 5’→3’ exonuclease activity; that’s Pol I’s job.
For eukaryotes, the enzymes differ slightly — DNA Pol α acts as primase, DNA Pol δ handles lagging strand, DNA Pol ε handles leading strand. NEET 2024 specifically asked about prokaryotic replication enzymes, so know both but prioritize prokaryotic for NEET.
Common Mistake
The most common error: saying DNA replication is “conservative” or “dispersive.” It is semi-conservative — each daughter molecule has one parental strand and one newly synthesized strand. Watson and Crick proposed this; Meselson and Stahl proved it. NEET has asked about the experimental evidence (cesium chloride density gradient) multiple times — don’t just memorize the result, understand the logic of the experiment.
A second mistake: writing that helicase “breaks covalent bonds.” Helicase breaks hydrogen bonds between bases. The phosphodiester backbone (covalent bonds) is intact throughout — only ligase and restriction enzymes deal with those.
Quick Revision Table
| Enzyme | Function | Direction |
|---|---|---|
| Helicase | Unwinds double helix | At replication fork |
| Primase | Synthesizes RNA primer | 5’→3’ |
| DNA Pol III | Main elongation enzyme | 5’→3’ only |
| DNA Pol I | Removes primer, fills gap | 5’→3’ synthesis, 5’→3’ exonuclease |
| DNA Ligase | Joins Okazaki fragments | Seals nicks |
| SSB proteins | Stabilize separated strands | Prevents re-annealing |
Final answer: The correct sequence is (a) → (b) → (c) → (d)/(e), with helicase, primase, DNA Pol III, DNA Pol I, and ligase as the respective enzymes. The lagging strand is synthesized discontinuously because DNA Pol III can only work 5’→3’, forcing it to synthesize away from the fork in short Okazaki fragments.