Question

In Drosophila, the X-linked recessive mutation vermillion (v) causes bright red eyes, in contrast to the brick-red eyes caused by the wild-type allele. A separate autosomal recessive mutation, brown (bw), causes flies to have brown eyes. Flies carrying both mutations have no eye pigmentation and are white-eyed. Determine the F1 and F2 phenotypic ratios for each sex from the following crosses:

(a) brown females X vermillion males

(b) white-eyed females X wild-type males

Answer 1

In the presence of one X-linked and one autosomal it is easier to treat the two genes separately and combine the results at the end

(a) For vermillion, v the male is Xv Y and the female X+ X+. The female produces all X+ gametes and the males produces either Xv or Y.

F1 females will be all X+Xv and the males will all be X+Y.

For brown, b females are bb and males are ++.

F1 of both sexes will be +b.

Now when we combine. Females are all +b X+Xv and have one wild type allele (+) for both genes, so will have normal wild type red eyes.

Males are all +b X+Y and are normal at the brown locus,

Your F1 cross:

+b X+Xv X +b X+Y

Vermillion X+Xv x X+Y. Females produce 1/2 X+ and 1/2 Xv gametes, males 1/2 X+ and 1/2 Y.

Female F2 are X+Xv and X+X+. Males are X+Y and XvY.

Brown +b x +b. Both produce + and b gametes which combine to produce ++ +b b+ and bb, the first

Now combine. F2 females will be 1/4 X+Xv x 1/4 ++ = 1/2 X+Xv++ normal wild type eyes
and 1/2 X+Xv x 1/2 +b = 1/4 X+Xv+b normal wild type eyes. Finally 1/2 X+Xv x 1/4 bb = 1/8 brown eyes.

Then 1/2 XvXv x 1/4 ++ = 1/8 XvXv ++ vermillion eyes; 1/2 XvXv x 1/2 +b = 1/4 XvXv +b vermillion eyes and 1/2 XvXv x 1/4 bb = 1/8 V=XvXvbb white eyes.

F2 males 1/2 X+Y x 3/4 ++ or +b = 3/8 X+Y++/X+Yb+ brown eyes, 1/2 X+Y x 1/4 bb = 1/8 X+Ybb brown eyes.

1/2 XvY x 3/4 ++/+b = 3/8 XvY++ or +b vermillion eyes and 1/2XvY x 1/4 bb = 1/8 XvYbb white eyes.