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)? 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%

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:

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,

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:

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

the correct answer is d

if you add everything up correctly you should get 69 i got this because 50 plus 19 is it