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?

Question

Biology

Kayleigh Mack

11 June, 03:49

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?

+3

Answers (1)

Know the Answer?

Pamela Orr · Answer 1

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).