Question

Can someone please help. I don't even know where to begin with this question.

Brown eyes and curly wings are recessive traits in fruit flies. Two red-eyed, normal-winged fruit flies produced red-eyed, normal-winged progeny and 28 offspring with brown eyes and normal wings. If two of the brown-eyed F1 flies were crossed, their progeny would be:

Answer 1

Let us suppose that in the given problem R is the allele for red eyes and r for brown eyes. The allele for normal wings is N and curly wings is n. Since all the progeny regardless of eye color have normal wings therefore we know that both the parents are heterozygous for red eye color and have homozygous normal wings. Thus, the genotype of parents will be:

RrNN X RrNN

Let us take an example of the progeny from this cross. The progeny will be:

	RN	rN
RN	RRNN	RrNN
rN	RrNN	rrNN

Now from the above cross the brown-eyed F1 flies will have the genotype rrNN.

Crossing two of the F1 flies will be:

rrNN X rrNN

The resulting offspring will have the genotype rrNN. Thus, the progeny will have brown eyes and normal wings.