Phenylketonuria (PKU) is an autosomal recessive genetic disorder. The frequency of homozygotes with this disorder is 0.0001. What percentage of the population is heterozygous?

Answer: The percentage of heterozygous population is 99%

Explanation:

According to the Hardy-Weinberg Equilibrium equation, the homozygous recessive individuals are represented by the q^2 term

Thus, q^2 = 0.0001

q = √0.0001 = 0.01 or 1%

Now, to calculate the percentage of heterozygous individuals in the population: subtract percentage of homozygotes from 100%

i. e 100% - 1% = 99%

Therefore, the percentage of heterozygous population is 99%. Also, this means 1 out of 100 people in the population suffer from Phenylketonuria (PKU).