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

Kallie Stewart

Biology

24 August, 02:01

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?

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Answers (1)

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Ean Underwood · 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).