The RNA polymerase binds to the -35 sequence in the promoter DNA strand of E.col

Question

The RNA polymerase binds to the -35 sequence in the promoter DNA strand of E.coli. This initiates unwinding of the DNA strands at the AT - rich -10 sequence. Transcription begins within the transcription bullbe at a site five to nine base pairs beyond the -10 sequence.

I understand the mechanism, but I don't understand the "pieces". The DNA is the non-template strand? Why? Why is the mRNA the template strand? Does mRNA also grow from the 5' --> 3'? And finally, does the promoter region contain AT because of the reduced number of hydrogen bonds; thus requiring less energy to break apart rather than GC bonds?

Explanation / Answer

The RNAp will always bind to the template strand. You have to first get the wording down. There are 2 strands: coding strand and template strand. The coding strand is literally the version of your protein in the DNA form. The complement of the coding strand is the template strand. The reason you want to use the template strand is because then you can obtain the coding strand in the form of mRNA. all nucleic acids are synthesized 5' to 3'. That is the forward direction. this includes DNA and RNA. Therefore, the coding strand must be 5' to 3', meaning the template strand (the one your RNAP is reading is being read 3' to 5'). In summary, yes, mRNA grows 5' to 3'. as for the AT region, only the -35 region contains the AT rich region and what you said is for the most part correct. basically it weakens the bond so less energy is required for the promoter to leave and begin elongation

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