Question

phenylketonuria is an autosomal recessive disorder of man.if the frequency of affected newborn infants is about 1 in 14,000 assuming random matting , what is the frequency of heterozygotes?

Answer 1

Phenylketonuria is an autosomal recessive genetic disorder in which the amino acid, phenylalanine starts to accumulate in the body. Therfore, it is also, characterized as an autosomal recessive metabolic genetic disorder.

Now, according to the Hardy-Weinberg Principle:

$p^{2} + 2pq + q^{2} =1$

Where,

$p^{2}$ = frequency of AA (homozygous dominant)

2pq = frequency of Aa (heterozygotes)

$q^{2}$ = frequency of aa (homozygous recessive)

Given:

$p^{2}$ = 1/14000 = 0.000071428

Therefore q= $\sqrt{0.000071428}$ =0.0085

Thus, q = 0.0085

Also, p + q = 1

Therefore,

p = 1 - 0.0085

p = 0.9915

Now, Frequency of heterozygotes (2pq) = 2 *0.9915 * 0.0085 = 0.017

Therefore, the answer is 2pq = 0.017

Thanks

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