Question

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. Phenylketonuria (PKU) is a human metabolic disorder that results in intellectual disabilities if it is untreated in infancy. If the U.S. population is in Hardy-Weinberg equilibrium, approximately what percent of the population are heterozygous carriers of the recessive PKU allele?

Answer 1

Answer:

1.98% or 2%

Explanation:

In US, 1 out of 10000 people has the disease. So, the frequency of the recessive alleles is 1/10000 = 0.0001

This term is q^2, so now we need to find the q term alone. q= $\sqrt{x} 0.0001$ = 0.01

Now, as p+q=1 in HW equations, then p=1-q, in this case p=1-0.01=0.99

So, the next part is determine what percentage is the heterozygous carriers and the heterozigous are in the equation named as 2pq

2pq=> 2(0.01)*(0.99)=0.0198 If we convert it to percentage is 0.0198 * 100= 1.98% aproximate to 2%