we solve the problem using the hardy-wineberg equation.
let us start with the homozygous recessive proportion of individuals with sickle cell anaemia,
we have the proportion as q² = 0.25.
∴ q = 0.5 (frequency of hbs in population)
further,
solving for the frequency of normal cell by using the fact that,
p + q = 1
∴ p = 1 - 0.5 = 1 - 0.5 = 0.5
now we can calculate 2pq in p² + 2pq + q², which is the frequency for heterozygous carriers of sickle cell anaemia,
∴ 2pq = 2 x 0.5 x 0.5 = 0.5
and the homozygous normal cell individuals proportion is given by the expression,
∴ p² = 0.5² = 0.25
hence, the carriers for sickle cell anaemia are in proportion, 0.5, or 50%.