We solve the problem using the Hardy-Wineberg equation.
Let us start with the homozygous recessive number of white trait,
∴ q² = [tex]\frac{23}{142}[/tex] = 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 + 0.36 = 0.84, or 84% of 142.
∴ 84% of 142 = [tex] \frac{84 \times 142}{100} = 119 [/tex]
Answer is 119.
