Question

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?

Answer 1

16%

Step-by-step explanation:

The frequency of homozygous recessive genotype in the given population= 36% or 0.36

So, the frequency of the recessive non-taster allele in the given population would be = square root of 0.36= 0.6

Since the population is in HWE, the frequency of the dominant allele for this locus in the population= 1- frequency of the recessive allele= 1-0.6= 0.4

Therefore, the frequency of the homozygous dominant genotype in the given population = 0.4 x 0.4 = 0.16 or 16%