Answer:
0.7
Explanation:
In solving Hardy Weinberg related question; we use the formula:
p + q = 1 and/or
p² + 2pq + q² = 1
where:
p = frequency of the dominant allele
q = frequency of the recessive allele
p² = frequency of individuals with the homozygous dominant genotype
2pq = frequency of individuals with the heterozygous genotype
q² = frequency of individuals with the homozygous recessive genotype.
So, if p + q = 1
and allele frequency of HbS = 0.3;
Let the frequency of allele of HbS be (q) = 0.3 (since HbS is a recessive sickle cell traits)
Let the frequency of allele of HbA be (p) = ??? ( since HbA is a normal hemoglobin)
Then; to determine the allele of HbA; we have
p+q = 1
p + 0.3 = 1
p= 1 - 0.3
p = 0.7
∴ the frequency of the HbA allele = 0.7
