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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