Question

1) You are a breeder of exotic dogs. Assume that you have conducted a standard set of crosses: (P1 x P2  F1  F2) and found that for the F2 population which displays a variance of V_P = 1.0, tail length in a dog breed has a narrow sense heritability of 0.4. How much of the variation in phenotype in the F2 population is due to additive genetics factors?

2) If we assume there is no epistasis (V_I) or gene x environment interaction (V_GE), and that dominance effects on the phenotype are 0.1, what is the variance in the phenotype due to environmental factors?

Answer 1

1)

Given figures :

Narrow sense heritability (h^2) = 0.4

Phenotypic variance (Vp) = 1

Narrow sense heritability (h^2) = Va /Vp

Additive genetic variance (Va) = narrow sense heritability * Vp

= 0.4 * 1 = 0.4

Hence, Additive genetic variance (Va) = 0.4

2)

Phenotypic Variance (Total Variance ) = Va + Vd + Vi + Ve + Vge

Additive genetice variance (Va) = 0.4

Dominant Variance (Vd) = 0.1

Epistatic variance (Vi) = 0

Gene-environmental interaction variance (Vge) = 0

Vp = Va +Vd + Vi + Vge +Ve

1 = 0.4 + 0.1 + 0 + 0 +Ve

1 = 0.5 + Ve

Ve (Environmental variance) = 1 - 0.5 = 0.5

Hence, Ve (environmental variance) = 0.5