Question

Assuming hardy-weinberg equilibrium, if 23 geese in a population of 142 are white, and 119 are blue, how many of the blue geese would you expect to be carriers of the b allele (i.e., bb heterozygotes)

Answer 1

We solve the problem using the Hardy-Wineberg equation.

Let us start with the homozygous recessive number of white trait,

∴ $q^{2} = \frac{23}{142} = 0.16$

We have the frequency for white trait as q = 0.4

Solving for the frequency of blue trait by using the fact that,

p + q = 1

∴ p = 1 - q = 1 - 0.4 = 0.6

Now we can calculate 2pq in p² + 2pq + q², which is the frequency for heterozygous blue trait,

∴ 2pq = 2 x 0.6 x 0.4 = 0.48

And the homozygous blue trait is given by the expression,

∴ p² = 0.6² = 0.36

Hence, the carriers for the trait of blue are, 0.48, or 48% of 142.

∴ 48% of 142 = $\frac{48 \times 142}{100}$ = 68.

Answer is 68.