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answer; the above answer to the above statement is (true)

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%.