Question

In a certain group of people, 4% are born with sickle-cell disease (homozygous recessive). If this group is in Hardy-Weinberg equilibrium, what percentage of the group is heterozygous for the sickle hemoglobin allele?

Answer 1

Answer:

percentage of the group heterozygous for the sickle hemoglobin allele = 32%

Explanation:

If

p = frequency of dominant allele

q = frequency of recessive allele

According to Hardy-Weinberg equilibrium: p² + 2pq + q² = 1 and p + q =1

where p² = frequency of homozygous dominant population

q² = frequency of homozygous recessive population

2pq = frequency of heterozygous population

Here, frequency of sickle cell disease (homozygous recessive) = 0.04

q² = 0.04

q = 0.2

p + q =1

So p = 1 - 0.2 = 0.8

Frequency of heterozygous population = 2pq

2 * 0.8 * 0.2 = 0.32

Hence percentage of people heterozygous for sickle hemoglobin allele = 32%