Eukaryotes have three RNA Polymerases that aid in differential regulation of gen

Question

Eukaryotes have three RNA Polymerases that aid in differential regulation of gene expression. Prokaryotes have only one RNA Polymerase; how do they regulate their gene expression on a broad scale?

A. They don't need to - transcription and translation occur simeltaneously

B. C and D

C. They focus on post-transcriptional regulation by actively degrading mRNA they don't need

D. They have one many different signma factors

E. They have significantly larger promoters for enhanced regulation of gene expression

Explanation / Answer

In prokaryotes the transcription and translation are coupled and regulation of transcription is carried out at the step of RNA polymerase binding to promoter . In prokaryotes, the promoter consists of two short sequences at -10 and -35 positions upstream from the transcription start site. Sigma factors not only help in enhancing RNAP binding to the promoter but also help RNAP target specific genes to transcribe. The sequence at -10 is called the Pribnow box, or the -10 element, and usually consists of the six nucleotides TATAAT. The Pribnow box is absolutely essential to start transcription in prokaryotes. The other sequence at -35 (the -35 element) usually consists of the seven nucleotides TTGACAT. Its presence allows a very high transcription rate. Both of the above consensus sequences, while conserved on average, are not found intact in most promoters. On average only 3 of the 6 base pairs in each consensus sequence is found in any given promoter.Some promoters contain a UP element (consensus sequence 5'-AAAWWTWTTTTNNNAAANNN-3'; W = A or T; N = any base) centered at -50; the presence of the -35 element appears to be unimportant for transcription from the UP element-containing promoters. It should be noted that the above promoter sequences are only recognized by the sigma-70 protein that interacts with the prokaryotic RNA polymerase. Complexes of prokaryotic RNA polymerase with other sigma factors recognize totally different core promoter sequences.

Hence, option E is correct.

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