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Andrew [12]
2 years ago
6

A number of experiments have demonstrated that areas of the genome that are relatively inert transcriptionally are resistant to

DNase I digestion; however, those areas that are transcriptionally active are DNase I sensitive. Describe how DNase I resistance or sensitivity might indicate transcriptional activity.
Biology
1 answer:
algol [13]2 years ago
6 0

Answer:

The DNA strands are not free in the nucleus, but forming a compact structure called chromatin along special proteins, known as histones. The chromatin structure has an important role in gene expression, as the level of compaction and the histone modification act as signals for the transcription machinery

In order to be transcribed, different areas in the genome need to unfold from the histone proteins, so the RNA polymerase can access the DNA.

That means, <u>that DNA sequences that transcribe at high rate, are more often unfold from the histones, and more exposed to other proteins to bind them. Particularly, those areas are more sensitive to the attack of endonucleases, such as DNAse I.</u>

As a consequence, an assay of DNAse I digestion can be used to identify highly sensitive or resistant to DNAse cleavage areas in the genome, and therefore serve as an insight to which sequences are more and less transcriptionally active.

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A calculated risk is intellectual in nature. <br> a. True <br> b. False
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Suppose that Gene B occurs in a sea turtle gene pool where the frequency of the B1 allele is 0.5 and the frequency of the B2 all
disa [49]

Answer:

Frequency of the B3 allele = 0.1

Explanation:

The allele frequencies in a given population remain constant which means that every generation will have the same gene pool across the generations. As per Hardy Weinberg’s first equilibrium equation, the sum of frequency of all the alleles at a given locus is equal to one  

Thus, if "p" represents the frequency of B1 allele and "q" represents the frequency of B2 allele and "r" represents the allele for B3 allele. Then, sum or "p", "q" and "r" is equal to one.

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0.5 + 0.4 + r = 1\\0.9 + r = 1\\r = 1-0.9\\r = 0.1

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Hope that answers the question, have a great day!

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2 years ago
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