Question

"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?"

Answer 1

Answer:

50% wild type and 50% black i.e. 1:1 phenotypic ratio

Explanation:

Given, b+ = wild type allele

b = recessive allele

b+b+ = dominant wild type

b+b = heterozygous wild type

bb = recessive black

First cross: female b+b (wild type) X male bb (black) = bb, bb, b+b, b+b

Since bb and b+b are in 1:1 phenotypic ratio, 52 black and 58 wild type progeny are produced.

From this progeny second cross occurs between:

female bb (black) X male (wild type) b+b = bb, bb, b+b, b+b

Again the same 1:1 phenotypic ratio is obtained. Half of the progeny will be be black (bb), other half will be wild type (b+b).