Difficulty: Easy
Correct Answer: They are removed by DNA polymerase I
Explanation:
Introduction / Context:
Lagging-strand synthesis requires repeated priming. Understanding primer removal and fragment maturation clarifies roles of individual enzymes at the replication fork and during post-replicative processing in bacteria.
Given Data / Assumptions:
Concept / Approach:
DNA polymerase I has a unique 5′→3′ exonuclease activity that removes RNA primers and simultaneously fills the resulting gaps with DNA (nick translation). DNA ligase then seals the remaining nicks between adjacent DNA fragments. Helicase unwinds DNA ahead of the fork but does not remove primers, and the leading strand does not form Okazaki fragments.
Step-by-Step Solution:
Recognize that RNA primers are necessary for Pol III to start synthesis.After extension, primers must be removed: Pol I performs 5′→3′ exonucleolytic removal.Pol I replaces RNA with DNA to create a continuous DNA stretch.DNA ligase seals the final phosphodiester bond between fragments.
Verification / Alternative check:
Mutations in Pol I impair primer removal and nick translation; ligase-deficient mutants accumulate nicks, confirming the division of labor.
Why Other Options Are Wrong:
They result in Okazaki fragments on the leading strand: Okazaki fragments occur on the lagging strand only.They are joined together by DNA ligase: ligase seals DNA–DNA nicks, not RNA primers.They are removed by helicase + ATP: helicase unwinds duplex DNA; it does not excise RNA primers.
Common Pitfalls:
Confusing the roles of Pol I and ligase; ligase cannot replace RNA with DNA or remove primers.
Final Answer:
They are removed by DNA polymerase I
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