Question

3. About one child in 2500 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 phenylketonuria allele?

b. What proportion of the population are carriers of the phenylketonuria allele (that is what proportion are heterozygous)?

Answer 1

ANSWER a ) = 0.02 ANSWER b ) 0.032

EXPLANATION

PHENYLKETONURIA IS AN AUTOSOMAL RECESSIVE DISORDER

SUPPOSE THE GENOTYPE FOR INDIVIDUAL WITH PHENYLKETONEURIA IS pp AND INDIVIDUAL WITH NORMAL PHENOTYPE IS PP WHILE INDIVIDUAL WHO IS NORMAL BUT CARRIER IS HAVING GENOTYPE Pp

If a population follow hardy weinberg equilibrium then f(pp)= q² , f(PP)= p² and f(Pp)= 2pq

The frequency of recessive homozygote that is individual with phenylketonuria would have q²

since according to question 1/2500 is affected by the phenylketonuria which is q²

hence q² = 1/2500 the q = √ 1/2500 = 0.02

Thus the frequency of Phenylketonuria allele = q = 0.02

p+q =1 ( according to hardy weinberg equilibrium ) p is wild type allele

then p =1-q

p = 1- 0.02

p = 0.08

THE PROPORTION OF POPULATION WHICH ARE CARRIER OF PHENYLKETONURIA IS 2pq

2pq = 2 x 0.08 x 0.02 = 0.032