Question

Bar eye (B) is a dominant and sex-linked trait of Drosophila. Curved wings (cu) and black body (bl) are recessive traits, both found on chromosome 2, separated by 24 map units. A homozygous Bar-eyed, black-bodied female is crossed to a true-breeding male expressing curved wings. The heterozygous F1 female is then crossed to a male expressing all three traits. At what frequency among the F2 would you expect to find a completely wild type male?

How often would you find a Bar eyed, curved wing female?

Answer 1

Cross 1: (Black bodied, Bar eyed female crossed to Male with Curved wings)
bl c+/bl c+; X^B/X^Bx bl+ c/bl+ c; X/Y

F1 --> bl c+/bl+ c ; X^B/X (F1 Heterozygous Female)

Cross 2 (F1 Heterozygous female crossed to male expressing all three traits):

bl c+/bl+ c ; X^B/X x bl c/bl c ; X^B/Y

The female parent of Cross 2 produces the following gametes:

Gametes and the proportion of gametes

Gamete	Gamete Type	Proportion	Rationale
bl c+; XB	Parental	19%	Since the two loci are 24 map units apart, recombination would occur in 24% of all meioses. Therefore, 76% of all gametes would be parental, with each class of parental gametes represented equally
bl c+; X	Parental	19%
bl+ c ; XB	Parental	19%
bl+ c ; X	Parental	19%
bl+ c+; XB	Recombinant	6%	24% of all meioses would be recombinant, and all recombinant gametes would be represented equally.
bl+ c+; X	Recombinant	6%
bl c ; XB	Recombinant	6%
bl c ; X	Recombinant	6%

Now, the tester male (Male used in Cross 2) would produce half gametes with the X chromosome and half with the Y chromosome.

Therefore, the progeny proportions are:

A completely normal male is 3% of the F2 progeny.

A Bar-eyed, Curved Wing Female would occur in 19% of female progeny.