We solve the problem using the Hardy-Wineberg equation.
Let us start with the homozygous recessive percentage of sickle cell anaemia trait first,
∴ q² = 0.2
We have the frequency for sickle cell anaemia allele as q = √0.2 = 0.45
Solving for the frequency of normal allele that is not causing anaemia by using the fact that,
p + q = 1
∴ p = 1 - q = 1 - 0.45 = 0.55
Now we can calculate 2pq in p² + 2pq + q², which is the heterozygous alleles frequency.
∴ 2pq = 2 x 0.45 x 0.55 = 0.50
Hence 50% of the population is heterozygous.
