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(a) The F! are heterozygous at both loci (a+a—b+b—) and possess two genes that contribute an additional 0.6 g each to the 1-g weight of a plant with no contributing genes. Therefore, the seeds of the F1 should average 1 g + 2(0.6 g) = 2.2 g.

(b) The F2 will have the following phenotypes and proportions: V16 1 g; 4/i6 1.6 g;6/16 2.2 g; 4/i6 2.8 g; and %6 3.4 g.

2. Phenotypic variation is analyzed for milk production in a herd of dairy cattle and the following variance components are obtained.

Additive genetic variance ( VA) = .4

Dominance genetic variance (VD) = .1

Genic interaction variance (Vj) = .2

Environmental variance (VE) = .5

Genetic-environmental interaction variance (VGE) = .0

(a) What is the narrow-sense heritability of milk production?

(b) What is the broad-sense heritability of milk production?

To determine the heritabilities, we first need to calculate VP and VG.

Vp = Va + Vd + Vj + Ve + Vge = .4 + .1 + .2 + .5 + .5 = 1.2

(a) The narrow sense heritability is:

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