A dominant gene b⁺ is responsible for the wild-type body color of Drosophila;its recessive allele b produces black body color.A testcross of a heterozygous b⁺/b female by a black b/b male gave 52 black and 58 wild-type progeny.If a black female from these progeny were crossed with a wild-type brother,what phenotypic ratios would be expected in their offspring? A)All males will be wild type,and all females will be black. B)All progeny will be black. C)All progeny will be wild type. D)75% will be wild type,25% will be black. E)50% will be wild type,50% will be black.

Question

Quizplus · Accepted Answer

The black female from the testcross is homozygous recessive for the b allele since she gave only black progeny when crossed with the black male. The wild-type brother is homozygous dominant for the b⁺ allele. Therefore, the cross between them would be a simple heterozygous x homozygous dominant cross, and the expected phenotypic ratio would be 1:1 for the two possible genotypes (b⁺/b x b⁺/b) - 50% wild type and 50% black.

A Dominant Gene B+ Is Responsible for the Wild-Type Body

A Dominant Gene B⁺ Is Responsible for the Wild-Type Body