Does someone think himself a BIOLOGIST? Try this one.?

Question

1) About one child in 2,500 is born with phenylketonuria ( an inability to metabolize the amino acid phenylalanine). This is known to be a recessive autosomal trait.

a. If the population is in Hardy-Weinberg equilibrium for this trait, what is the frequency of the phenlyketonuria allele?

b. What proportion of the population are carriers of the phenylketonuria allele (that is,

what proportion are heterozygous?

Answer 1

Hardy Weinberg Equilibrium

Dominant allele = p

Recessive allele = q

p+q = 1

At population equilibrium

p^2 are homozygous dominant

2pq are heterozygous

q^2 are homozygous recessive

1/2500 have PKU, so q^2 = .0004

q therefore = sqrt(.0004) = .02

p = 1 - .02

p = .98

2pq = 2(.02*.98) = .0392

So, the frequency of the PKU allele is equal to q, which is .02

The proportion that are heterozygous carriers is equal to 2pq, which is .0392.