Suppose 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. If this population conforms to Hardy–Weinberg expectations for this gene, what percentage of the population must be heterozygous for this trait?

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%

answer: that would be cheating

explanation:

what subject is this