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%
