Directionality of DNA chain elongation by DNA polymerases during replication

Difficulty: Easy

Correct Answer: 5′ to 3′ direction

Explanation:


Introduction / Context:
All known DNA polymerases add nucleotides to the 3′-OH of a growing strand, making synthesis intrinsically 5′→3′. Understanding this constraint explains leading and lagging strand strategies with Okazaki fragments.


Given Data / Assumptions:

  • Enzyme: DNA polymerase (replicative or repair).
  • Substrate: dNTPs; primer with free 3′-OH; template strand.
  • Direction: determined by chemistry of phosphodiester bond formation.


Concept / Approach:
The 3′-OH of the primer attacks the alpha-phosphate of an incoming dNTP, releasing pyrophosphate. This mechanism only permits 5′→3′ extension. On the lagging strand, discontinuous synthesis produces Okazaki fragments that are later joined by ligase; still, each fragment is synthesized 5′→3′.


Step-by-Step Solution:
Identify the nucleophile (primer 3′-OH) and electrophile (dNTP α-phosphate).Recognize that reversal would create unstable chemistry and impair proofreading.Conclude universal 5′→3′ chain elongation.


Verification / Alternative check:
No cellular DNA polymerase has been found that synthesizes 3′→5′; this is consistent across domains of life.


Why Other Options Are Wrong:

  • 3′→5′, 5′→5′, 3′→3′: chemically inconsistent with polymerase mechanism.
  • Either direction: not supported by known biology; both strands are synthesized 5′→3′.


Common Pitfalls:
Assuming the leading strand is 3′→5′ because it is continuous—direction refers to synthesis, not template orientation.


Final Answer:
5′ to 3′ direction

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