Question

A virgin Drosophila female whose thorax bristles are very short is mated with a males having normal (long) bristles. The F1 progeny are 1/3 short-bristle females, 1/3 long bristle females, and 1/3 long bristle males. A cross of the F1 long bristle females with their brothers gives only long bristle F2 progeny. A cross of short bristle females with their brothers gave 1/3 long bristle females, 1/3 short bristle females, and 1/3 long bristle males. A) Explain what is going on genetically and B) why short bristled females can not be homozygous.

Answer 1

A cross of the short-bristled female with a normal male can produce two phenotypes with regard to bristles and an abnormal sex ratio of two females: one male. In addition, all males are normal while the females are normal and short in equal numbers. When the sexes differ with respect to phenotype among the progeny, an X-linked gene is implicated. The only normal phenotype is observed in males. Therefore, the short-bristled phenotype (in females) must be heterozygous, and the allele must be a recessive lethal. So, the first cross was A/a X a/Y.

When long-bristled females (a/a) were crossed with long-bristled males (a/Y), all their progeny would be expected to be long-bristled (a/a or a/Y). Short-bristled females (A/a) were crossed with long bristled males (a/Y). The progeny expected are:

1/4 A/a short-bristled females

1/4 a/a long-bristled females

1/4 a/Y long-bristled males

1/4 A/Y nonviable