28. In swine, two loci determine color such that WW= White (no expression of col

Question

28. In swine, two loci determine color such that WW= White (no expression of color at B locus) BB = Black if also ww Ww = White (no expression of color at B locus) Bb = Black if also ww ww = Expression of color at B locus (black or red) bb = Red if also ww An autosomal recessive gene (s) causes porcine stress syndrome, where homozygous recessive (ss) hogs may die when stressed, such as when loaded or handled by humans. A boar (male) that is heterozygous for the stress gene, with the color genotype Ww Bb is mated to a sow (female) also heterozygous for the stress gene, with color genotype Ww bb. If the mating results in a litter of eight pigs, calculate the probability that:

Explanation / Answer

The genotype of boar (male) heterozygous to the stress gene, with the colour genotype BbWw is, SsWwBb. The genotype of sow (female) heterozygous to the stress gene, with the colour genotype Wwbb is, SsWwbb. Cross between these two (SsWwBb* SsWwbb) will have the offspring with 18 different phenotypes (from punnett square analysis).

(a). The genotype of pig born is neither affected by, nor is it a carrier for stress condition is, SS____

(Since stress condition is inherited by autosomal recessive pattern, presence of two recessive genes causes the disease, and one recessive gene makes it a carrier).

The chances of having the first offspring born is neither affected by, nor is it a carrier for stress condition is, 16/ 64 = 1/4.

(b). The chances of having 5 boars (males) is = 1/2* 1/2* 1/2* 1/2* 1/2 = 1/32; the chances of having 3 gilts (females) = 1/2* 1/2* 1/2 = 1/8. Thus, the chances of having 5 boars and 3 gilts in any order = 1/32 + 1/8 = 5/32.

(c). The genotype of red glit should be __wwbb.

The chances of having first pig is a red pig = 8/64. The chances of having red glit = 8/64* 1/2 = 1/8* 1/2 = 1/16

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