​The initiation of transcription is a complex process involving promoter recognition, conversion of the initiation complex from closed to open form, abortive initiation events, and finally promoter escape. The following statements are made regarding these steps in transcription initiation:
A. Promoter escape in bacteria is usually accompanied by the release of the sigma factor from the RNA polymerase holoenzyme complex.
B. Abortive initiation events in prokaryotes result in the formation of short transcripts ~10 nucleotides in length while such events in eukaryotes result in formation of transcripts ~75 nucleotides in length.
C. Promoter escape in eukaryotes is accompanied by the phosphorylation of the RNA polymerase large subunit on its C-terminal domain (CTD).
D. Promoter recognition in bacteria is governed by the sigma factor which binds to the -10 and -35 regions of the promoter followed by recruitment of the RNA Pol II core enzyme to form the holoenzyme.
Which one of the following options represents the combination of all correct statements?​

Correct option is A

Explanation-

Statement A - Promoter escape in bacteria is usually accompanied by the release of the sigma factor from the RNA polymerase holoenzyme complex.
This statement is correct. This is well-established. The sigma factor directs RNA polymerase to the promoter. Upon promoter clearance, the sigma factor dissociates.
Statement C - Promoter escape in eukaryotes is accompanied by the phosphorylation of the RNA polymerase large subunit on its C-terminal domain (CTD).
This statement is correct. Phosphorylation of the CTD of RNA polymerase II (specifically at Serine 5) by TFIIH is a key step for transitioning from initiation to elongation, which aligns with promoter escape.

Incorrect options-
Statement B - Abortive initiation events in prokaryotes result in the formation of short transcripts ~10 nucleotides in length while such events in eukaryotes result in formation of transcripts ~75 nucleotides in length.

This partially incorrect due to the false claim that abortive transcripts in eukaryotes are ~75 nt long — they are usually much shorter.
Statement D - Promoter recognition in bacteria is governed by the sigma factor which binds to the -10 and -35 regions of the promoter followed by recruitment of the RNA Pol II core enzyme to form the holoenzyme.

This is incorrect because it erroneously mentions RNA Pol II, which is eukaryotic, in the context of bacterial transcription.

So, the correct answer is option a – A and C only
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