Linkage genes in Drosophila

Where you have 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 female is XvXv and the male X+Y. The female produces all Xv gametes and the males either X+ or Y. F1 females will be all X+Xv and the males will all be XvY.

For brown, b females are ++ and males bb. F1 of both sexes will be +b.

Now 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 XvY and are normal at the brown locus, but are all vermillion, so will have vermillion eyes.

Your F1 cross:

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

Female F2 are X+Xv and XvXv. 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 1/2 X+Xv x 1/4 ++ = 1/8 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.

You will have to tidy up these results and then repeat the same methods. The results will be the same for brown in b). The vermillion cross is X+X+ x XvY.

For c the crosses are XvXvbb x X+Y++, which you can split into two crosses and repeat using the same method.

2. a) The cross is XvXvbb++ x X+Y++ss. Use what results you can from earlier crosses and then combine with sepia.

b) the cross is X+X+++ss x XvYbb++

That is to get you going, but it is up to you to complete these problems which are time consuming.

Can someone please post the answer.

Thank you, very helpful.