Question

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 offspring from several litters are as follows: 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 1

The genes for coat color and pattern in the guinea pigs are linked because the observed offspring ratios differ from the expected 1:1:1:1 ratio of unlinked genes. The black-spotted and brown solid offspring are recombinants, and their frequencies indicate the genes are 25 map units apart.

Step-by-step explanation:

In guinea pigs, when determining if genes for coat color and pattern are linked or unlinked, we need to examine the offspring's phenotypic ratios. In a mating between a heterozygous black, solid-colored guinea pig (Bsbs) and a brown, spotted guinea pig (bbss), the expected offspring ratio for unlinked genes would follow a 1:1:1:1 ratio because of independent assortment.

However, in the scenario provided, the offspring from several litters are as follows: black solid (16), black spotted (5), brown solid (5), and brown spotted (14).

This does not fit the 1:1:1:1 expected ratio; thus, it suggests that the genes are indeed linked. To calculate the map units apart, we can use the recombination frequency formula, which is (number of recombinant offspring) / (total number of offspring) x 100. In this case, recombinant offspring are black-spotted and brown solid, which sum to 10. The total number of offspring is 40. Thus, the genes are (10/40) x 100 = 25 map units apart.