In a herd of hereford cattle the recessive autosomal gene fo

In a herd of hereford cattle, the recessive autosomal gene for \"lineback\" trait is expressed in 10/1000 animals. If the lineback cows and bulls are not allowed to reproduce (W22=0) How many lineback cattle will there be in the next generation of 1000 cattle? (Assume W11 & W12 both equal 1). Allelic frequency is 0.1. Please show work not just final answer!

Solution

Let

p: frequency of allele A1 in the current generation

q: frequency of allele A2 (or “lineback” trait) in the current generation

p’: frequency of allele A1 in the next generation

q’: frequency of allele A2 (or “lineback” trait) in the next generation

W₁₁, W₁₂, W₂₂: relative fitnesses of the genotypes A1A1 , A1A2 , A2A2.

: Population mean (total) fitness ( = p²W₁₁ + 2pqW₁₂ + q²W₂₂)

Formula used

p = 1 – q

= p²W₁₁ + 2pqW₁₂ + q²W₂₂ = p₁ + q₂

p’ = p²W₁₁ / + pqW₁₂/

Given

W₁₁ = W₁₂ = 1

W₂₂ = 0

q = 0.1 => p = 1 - 0.1 = 0.9

Distribution of lineback cattle in current generation = 10/1000 = 0.01 = q²

Average fitness W = p²W₁₁ + 2pqW₁₂ + q²W₂₂           

W₁ = 1

W₂ = (1*pq + 0*q²)/(pq+q²) = p

W = p(1) + q(p) = p(1+q) = p(2-p)

p’ = p[W₁/W] = p[1/p(2-p)] = 1/(2-p)

=> p’ = 1/(2-0.9) = 1/1.1 = 0.909

q’ = 1 – p’ = 1 - 0.909 = 0.091

Next generation genotype freq. for linback = q’² = 0.091² = 0.008281

Number of lineback cattle (with A2A2 gene) in next generation = 0.008281 * 1000 ~=8