Perform the Hardy Weinburg Equation AA201 Aa35 aa21 A AA 2

Perform the Hardy Weinburg Equation :

AA(201) Aa(35) aa(21)

A = (AA) 2 + Aa

a = (Aa) + 2(aa)

Solution

If there are two alleles, A frequency of which is determined by P and a, frequency of which is determined by q.

Frequency of allele A

f(A) = P = (2 × AA) + Aa/ 2 × total number of individuals

f(A) = P = (2 × 201) + 35 ÷( 2 ×257)

= 402 + 35 ÷ 514

=437 ÷ 514= 0.85.

Therefore frequency of allele A = P = 0.85.

Frequency of allele a

f(a) = q = (2 × aa) + Aa ÷ 2 × 257

= (2 × 21) + 35 ÷ 514

= 42 + 35 ÷ 514

= 87 ÷ 514 = 0.16.

Frequency of allele a = q = 0.16.