Phenylketonuria (PKU) is a genetic disease caused by the inability to break down the amino acid phenylalanine. If untreated it leads to severe mental disabilities. PKU is due to a recessive allele. Assume that the US population is in Hardy-Weinberg equilibrium with respect to this gene. In the US the PKU rate is 1 out of every 10,000 babies born. What percent of the US population have no alleles for PKU

Therefore, 98% of the US population have no alleles for PKU

Explanation:

The Hardy-Weinberg equilibrium states that the amount of genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors.

The Hardy-Weinberg equilibrium is expressed quantitatively using a mathematical equation known as the Hardy-Weinberg equation. the equation is given below:

p² + 2pq + q² = 1

also, p + q = 1

Given a pair of alleles, S and s with A dominant and a recessive

where p is the frequency of the dominant allele in the population,

q is the frequency of the recessive allele in the population,

p² represents the frequency of the (SS)  dominant genotype,

q² represents the frequency of the (ss) recessive genotype,

2pq represents the frequency of the heterozygous genotype

From the given question,

q² = 1/10000 = 0.0001

q = 0.01

from p + q = 1

p = 1 - 0.01 = 0.99

p² = 0.98

Therefore, 98% of the US population have no alleles for PKU

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