Suppose 64% of a remote 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?
A) 16%
B) 32%
C) 40%
D) 48%

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ajeigbeibraheem ajeigbeibraheem · Answer 1 · 2020-02-17T05:39:30+01:00

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%