The bacterial insertion sequence IS10 uses antisense RNA to regul...

Question

The bacterial insertion sequence IS10 uses antisense RNA to regulate translation of the mRNA that produces the enzyme transposase, which is required for insertion sequence transposition. Transcription of the antisense RNA gene is controlled by POUT, which is more than 10 times more efficient at transcription than the PIN promoter, which controls transposase gene transcription.

If a mutation reduced the transcriptional efficiency of so as to be equal to that of , what is the likely effect on the transposition of IS10?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem together. T N three is a transposon that can move within a bacterial genome. And its transposition is facilitated by the transposes enzyme encoded by the T N P A gene. The T N P A gene has two promoters. The P one promoter produces a full length M R N A transcript. While the P two promoter produces a shorter M R N E transcript that encodes a nonfunctional transposes enzyme. The P one promoter is active in the presence of a transposon while the P two promoter is active in its absence. If a mutation decreases the activity of P two promoter sequence and increases the activity of P one promoter sequence, what will be the likely outcome? The answer options are a increase in the frequency of transposition B decrease in the frequency of transposition C increased number of truncated M R N A and D premature stock codon in the M R N A sequence. OK. So right away, we can eliminate option D. The mutation in the question does not have a specific relationship with a stop codon and there's nothing of a stop codon mentioned. Otherwise, this answer option is irrelevant to our question. So we are eliminating option D. Now to solve this problem, let's organize the information from the question. For the transposes enzyme encoded by the gene T N P A. There are two promoters whose activity is regulated by a transposon. So basically, there are two scenarios for this question. So in our first scenario, it would be if P one is active in the presence of the transposon and this would produce a normal transposes. Now, the second scenario is if P two was active in the absence of a transposon and this produces a nonfunctional transpose. So if we have our mutation, which is decreasing P two activity, that means we are decreasing the amount of the nonfunctional transposes. In addition, the mutation increases P one and therefore it is going to increase the amount of normal transposes. So overall, you have more production of the normal transposes. Now, what does the transposes do again? Remember the question said that it facilitates transposition. Therefore, the correct answer is option a increase in the frequency of transposition. All right, everyone. I hope you found this helpful and I'll see you soon for the next practice problem.

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