Question

One out of every 10,000 newborns in a country has phenylketonuria (pku), a debilitating disease (when untreated) caused by a recessive allele. The frequency of carriers of this disease is

Answer 1

The Hardy-Weinberg Equation is useful for predicting the percent of a human population that may be heterozygous carriers of recessive alleles for certain genetic diseases.

For a population in genetic equilibrium: p2 + 2pq + q2= 1

p2= frequency of AA (homozygous dominant)

2pq= frequency of Aa (heterozygous)

q2= frequency of aa (homozygous recessive)

oncalculations,

q2= 1/10,000  

q =√ 1/10,00=1/100

p = 1

2pq = 2 (1) (1/100) = 1/50

so, 1/50 individuals in the general population are carriers of PKU