Phenylketonuria (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 PKU carriers?

Question

Quizplus · Accepted Answer

In Hardy-Weinberg equilibrium, the frequency of homozygous recessive individuals (affected by PKU) can be calculated using the equation: q^2 = 1/4500, where q is the frequency of the recessive allele. 
Solving for q: 
q^2 = 1/4500 
q = sqrt(1/4500) 
q = 0.0219 
The frequency of carriers (heterozygotes) can be calculated using the equation: 2pq, where p is the frequency of the dominant allele and q is the frequency of the recessive allele. 
p + q = 1 
p = 1 - q 
p = 1 - 0.0219 
p = 0.9781 
2pq = 2(0.9781)(0.0219) 
2pq = 0.0428 or 4.28% 
Therefore, the frequency of PKU carriers is approximately 2.9%.

Phenylketonuria (PKU)is Extremely Common in Ireland, Affecting Approximately 1 in 4500