A female Drosophila with singed bristles is crossed to a male with sepia (dark brown) eyes.In the F₁,all females are wild type,and all males have singed bristles.F₁ males and females are crossed together,and the following F₂ phenotypic ratios are obtained.

748 wild-type females
753 wild-type males
742 singed females
747 singed males
249 sepia females
250 sepia males
244 singed, sepia females
246 singed, sepia males

If sn and se represent the mutant alleles of the singed and sepia gene,respectively,and sn⁺ and se⁺ are their wild-type counterparts,what are the most likely genotypes of the original parents?

Question

Quizplus · Accepted Answer

The singed bristle phenotype is recessive, so the female parent must be homozygous recessive sn/sn. All offspring males have singed bristles, which means that the male parent must carry the singed allele as well. Since all females are wild-type, the male parent must have given a dominant allele to all female offspring. Therefore, the male parent is X/Y;sn/sn⁺. The sepia phenotype is also recessive, so both parents must have at least one dominant allele. Based on the phenotypic ratios of the offspring, the female parent must be heterozygous for the sepia gene, se⁺. The male parent either has two dominant alleles or is heterozygous, so the most likely genotype of the male parent is se⁺/se⁺ or se⁺/se. The only answer choice that includes these two possibilities is option D.

A Female Drosophila with Singed Bristles Is Crossed to a Male