Question

A female fruit fly with the recessive mutant phenotype of white eyes and miniature wings is mated with a male possessing the wild-type phenotype of red eyes and normal wings. Among the F1s, all the females are wild-type while all the males exhibit the mutant phenotype. The F2s resulting from a test cross exhibit predominantly (63%) parental phenotypes. How do you explain these results?

Answer 1

These results occur when the two alleles are linked to the X chromosome.

In fruit flies, males are hemizygous for the X chromosome(have only 1 copy of the X chromosome), and therefore male F1 flies express both mutant phenotypes, it inherited from the Parental female fly. However, since the F1 females have two X chromosomes, one carrying Wild-type alleles from the parental Male, they express wild-type phneotypes for both traits.

When the F1 is test-crossed to produce the F2 progeny if the two loci were assorting independently, the expectation is that each allele would assort independently of the other and would produce 1:1:1:1 :: White-eyes, Normal Wings: White-eyes, Miniature Wings: Red-eyes, Normal Wings: Red eyes, Miniature Wings F2 progeny.

However, if the two loci are 'linked' and the genes coding for these phenotypes are present on the same X chromosome, non-parental phenotypes would only occur when a crossing over or recombination event occurs that exchanges genetic material between non-sister chromatids.