Question

4. The antibiotic cordycenin syntheon 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 Ka (NM) 315 0.52 0.0025 0.0013

Answer 1

solution:

DNA segment having 8 base bulge with 0.0013nM Kd does the polymerase bind most tightly

explanation:

• the binding of polymerase depends upon the binding affinity which is the strength of the interaction between the biomolecule and to its ligand.
• It is measured in terms of Kd. The smaller Kd has the greater affinity of the biomolecule for its target, and it will bind more tightly.
• T7 polymers will bind the to the promoter which having big bulge and lesser kd value(which is the DNA promoter having 8 base bulge). Because if the trascription bubble is big then it will provide sufficient space for the promoter to bond tightly to the promotor site.
• If the bulged base is smaller number then the transcription bubble will not be able to provide suffient binding site and then the polymers need to use its helicase activity to make single standard transcription bubble for trancription initiation.