Question

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3. The RNA polymerase from bacteriophage T7 recognizes specific promoter sequence and melts open the DNA to form a transcription bubble. No other transcription factors are necessary for T7 RNA polymerase to initiate transcription. Dissociation constants (Ka) were measured for the interaction between the polymerase and DNA segments containing the promoter sequence. In some cases, the DNA contained a bulge, caused by a mismatch of one, four, or eight bases, to mimic the intermediates in the formation of a transcription bubble. The results are shown in the table. To which DNA segment does the polymerase bind most tightly? Explain in terms of the DNA structure. DNA promoter segment Fully base paired One-base bulge Four-base bulge Eight-base bulge K. (M) 315 0.52 0.0025 0.0013

Answer 1

Binding of polymerase depends upon the binding affinity which is the strength of the interaction between the biomolecule to its ligand. It is measured in terms of dissociation constant, Kd. The smallee the Kd, the greater is the affinity of the biomolecule for its target, and the more tightly it will bind.

T7 polymerase will bind to the promoter having big bulge and lesser Kd value, which is the DNA promoter having 8 base bulge. Because if the transcription bubble is big it will provide sufficient space for the promoter to bind tightly to the promoter region.

If the bulged base has smaller number the transcription bubble will not be able to provide sufficient binding site and the polymerase would require to use its helicase activity to make single stranded transcription bubble for transcription initiation.

So the answer will be the DNA promoter segment having 8 base bulge with 0.0013nM Kd.