The ability to taste phenylthiocarbamide (PTC) is a trait controlled by 2 alleles allele? (PTC taster and PTC non-taster). Suppose 36% of a remote mountain village cannot taste PTC and must, therefore, be homozygous recessive (aa) for the PTC non-taster allele. If this population conforms to Hardy-Weinberg expectations for this gene, what percentage of the population must be homozygous (AA) for the PTC taster allele?

a. 32%
b. 16%
c. 48%
d. 60%
e. 40%

The correct answer is b: 16%

Explanation:

According to Hardy-Weinberg the allelic frequencies in a locus are represented as p and q, referring to the allelic dominant or recessive forms respectively. The genotypic frequencies after one generation are p² (Homozygous dominant), 2pq (Heterozygous), q² (Homozygous recessive). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example, the genotypic frequency q² or aa is 36% which equals 0,36.

If q²= 0.36, then q = √0.36 = 0.6

The allelic frequency for a is 0.6.

This means that the allelic frequency for p is 0.4, which we can deduce by clearing the equation p + q = 1

                          p + 0.6 = 1

                          p = 1 - 0.6

                          p = 0.4

Then, we can calculate the genotypic frequency for AA or p², that is: 0.4²= 0.16

Then, the percentage of the population that must be homozygous (AA) for the PTC taster allele is 16%.

well

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