A recessive allele causes phenylketonuria (PKU) when homozygote. If the frequency of the recessive allele is 0.05 in the population, what is the heterozygote frequency in the population? (Assume that people mate randomly with respect to PKU, and that the mutant allele does not have any effects on fitness.)

0.095

Explanation:

Phenylkentonuria is a disease caused by a recessive allele.

The frequency of the recessive allele + the frequency of the dominant allele equals 1.

The frequency of the recessive allele is q = 0.05

The frequency of the dominant allele then is p = 1 - q = 0.95

If people mate randomly, the frequency of the homozygous dominant genotype will be p², the frequency of the heterozygous genotype will be 2pq and the frequency of the homozygous recessive genotype will be q².

2pq=2 * 0.05 * 0.95

2pq=0.095

The heterozygote frequency in the population is 0.095