Answer:
The percentage of the population that is heterozygous for this trait is 48%
Explanation:
They are two alleles, the phenylthiocarbamide tasters (PTC) and the non phenylthiocarbamide tasters (non PTC). PTC testers are dominant and non PTC tasters are recessive.
let the frequency of the dominant allele(A) be p
and the frequency of the recessive allele(a) be q
We are told that 64 percent of people living in a remote, isolated mountain village can taste phenylthiocarbamide (PTC) and must, therefore, have at least one copy of the dominant PTC taster allele (that is AA and Aa)
Frequency of AA = p², Frequency of Aa = 2pq and Frequency of aa = q²
Therefore p² + 2pq = 64% = 0.64
According to Hardy–Weinberg:
p² + 2pq + q² = 1 and
p + q = 1
Since p² + 2pq = 0.64
∴ 0.64 + q² = 1
q² = 1 - 0.64 = 0.36
q = √0.36 = 0.6
Since p + q = 1
p = 1 - q = 1 - 0.6 = 0.4
The frequency of heterozygous = 2pq = 2 × 0.4 × 0.6 = 0.48
Therefore the percentage of the population that is heterozygous for this trait is 48%
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This mechanism is known as histone acetylation and it is a part of gene expression. The lysine residues inside the N-terminal tail from the histone core of the nucleosome are the substrate for enzymes-histone acetyltransferase which transfer an acetyl functional group from another molecule. Acetylation removes the positive charge on the histones and thus decrease the interaction negatively charged phosphate groups of DNA. As a result, chromatin becomes more relaxed structure and transcription is promoted.
Answer: cell —> tissue —> organ —> organ system
aka. D