Bacterial RNA polymerase composition: Which subunit of prokaryotic RNA polymerase is the removable factor that confers promoter recognition?

Difficulty: Easy

Correct Answer: Sigma subunit

Explanation:

Introduction / Context: Bacterial RNA polymerase core enzyme requires an additional specificity factor to recognize promoter sequences efficiently. This highlights how transcription initiation is regulated at the level of polymerase–DNA recognition.

Given Data / Assumptions:

  • The prokaryotic holoenzyme equals core (α2ββ'ω) plus σ.
  • Only one subunit is transiently associated and exchangeable to alter promoter specificity.

Concept / Approach: The sigma (σ) factor binds core polymerase to form the holoenzyme, enabling recognition of −35/−10 elements. Different σ factors direct transcription of distinct regulons (e.g., heat shock, sporulation).

Step-by-Step Solution: Identify removable specificity factor → σ. Confirm role: promoter recognition and closed-complex formation. Choose “Sigma subunit”.

Verification / Alternative check: In vitro, core polymerase binds DNA weakly and initiates poorly; adding σ restores accurate start site selection.

Why Other Options Are Wrong: α participates in assembly and regulation; β and β' form catalytic center; ω assists assembly/stability; δ is not a standard E. coli subunit.

Common Pitfalls: Thinking α or β control specificity; confusing σ with transcription factors that bind DNA directly.

Final Answer: Sigma subunit.

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