A virgin Drosophila female bristles has noticed her thorax are much shorter than the normal. When mated with the normal male with long bristles the progeny of F1 is obtained to be 1/3 short-bristled females and 1/3 long-bristled males. When F1 is crossed with their brothers it gives the progeny of F2 long-bristled males.
In addition, a cross between short bristled female and normal male has resulted in an abnormal 2:1 ratio of female to male. While the males are normal the female are in the ratio of 1:1 in short and long bristles. It should be noted that the differences in the progeny and the differences in the bristles are linked together with the X-linked gene.
Since, the male progeny are normal the short-bristled phenotype should be heterozygous in nature. This is because of the existing ratio of 2:1, the allele should be recessive lethal when homozygous. Thus, the first cross was A/a X a/Y.
Thus, the cross between long-bristled females (a/a) and long-bristled males (a/Y) gives the following progeny.
¼ A/a short-bristled females
¼ a/a long-bristled females
¼ a/Y long-bristled males
¼ A/Y nonviable