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Linkage mapping example problem

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1.) In guinea pigs, black (B) is dominant to brown (b), and dolid color (S) is dominant to spotted (s). A heterozygous black, solid-colored pig is mated with a brown, spotted pig. The total offspring for several litters are black solid=16, black spotted=5, brown solid=5, and brown spotted=14.

Are these genes linked or unlinked? If they are linked, how many map units are they apart?

2.) A woman is a carrier for a sex-linked lethal gene that causes an embryo with the gene to spontaneously abort. She has nine children. How many of these do you expect to be boys?

3.) A dominant sex-linked gene B produces white bars on black chickens, as seen in the Barred Plymouth Rock breed. A clutch of chicks has equal numbers of black and barred chicks. (Sex is determined by the Z-W system in birds: ZZ are males, ZW are females.)
a.) If only the females are found to be black, what were the genotypes of the parents?
b.) If males and females are evenly represented in the black and barred chicks, what were the genotypes of the parents?

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Genetics Problems

1.) In guinea pigs, Black (B) is dominant to Brown (b), and solid color (S) is dominant to spotted (s). A heterozygous black, solid-colored pig is mated with a brown, spotted pig. The total offspring for several litters are black solid=16, black spotted=5, brown solid=5, and brown spotted=14. Are these genes linked or unlinked? If they are linked, how many map units are they apart?

Answer - If the genes were independent(unlinked), using standard genetic techniques you would predict that the offspring would have the phenotypic ratio: 1 Black solid : 1 Black spotted : 1 Brown solid : 1 Brown spotted.

Given the 40 offspring (16+14+5+5), that would result in 10 of each of the above phenotypic classes. Since this isn't the case in our cross, these genes are linked.

Parents (1) heterozygous black, solid - BbSs
(2) brown, spotted - bbss

The only parent that matters for figuring things out in this ...

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