Question

in 4500 live births. Assuming that the population is in Hardy-Weinberg equilibrium, what is the frequency of homozygous wild-type individuals (i.e., people without the recessive mutation)? Phenylketonuria (PKU) is a disease caused by a recessive mutation. PKU is extremely common in Ireland, affecting approximately 1 in 4500 live births. Assuming that the population is in Hardy-Weinberg equilibrium, what is the frequency of homozygous wild-type individuals (i.e., people without the recessive mutation)? 98.5% 2.96% 97% 1.5% 0.022%

Answer 1

Answer:

97%

Explanation:

PKU is caused by a recessive mutation, which means that affected individuals are homozygous and recessive (tt).

The frequency of genotypes according to Hardy-Weinberg equilibrium is:

$TT= p^2\\Tt=2pq\\tt=q^2$

where p is the frequency of the T allele and q is the frequency of the t allele.

Affected individuals are 1 in 4500, which means the frequency of tt genotype is 1/4500.

If the population is in Hardy-Weinberg equilibrium,

$q^2=1/4500\\q=\sqrt{1/4500} \\\\q=0.015$

The sum of the allele frequencies adds up to 1, therefore:

p + q = 1

p= 1 - q

p= 1 - 0.015

p= 0.985

The frequency of the TT genotype (homozygous wild-type individuals) will be:

$p^2=0.985^2\\p^2=0.97$

97% of the population will have the homozygous wild-type TT genotype.