Answer:
D. 48%
Explanation:
Using the Hardy-Weinberg equation;
p + q = 1
p² + 2pq + q² = 1
From the question,
64% of a remote mountain village can taste phenylthiocarbamide (PTC) = i.e p² = 0.64
and must, therefore, have at least one copy of the dominant PTC taster allele.
i.e 2pq
So. if p² + 2pq = 0.64
Then
p² + 2pq + q² = 1
0.64 + q² = 1
q² = 1 - 0.64
q² = 0.36
q = √0.36
q = 0.6
To find p; we have:
p + q = 1
p + 0.6 =1
p = 1 - 0.6
p = 0.4
SO for the percentage of the population that must be heterozygous for this trait (i.e 2pq)
we have:
= 2 × 0.6 × 0.4
= 0.48
= 48%
∴ the percentage of the population that must be heterozygous for this trait = 48%