A heterozygous fruit fly with normal wings and a gray body (VvBb) is crossed with homozygous flies with vestigial wings and black bodies (vvbb). If there are 800 F₁ progeny, the expected phenotypes would be: A) 200 normal gray, 200 normal black, 200 vestigial gray, and 200 vestigial black B) 400 normal gray and 400 vestigial black C) 800 normal gray D) 800 vestigial black E) 450 normal gray, 150 normal black, 150 vestigial gray, and 50 vestigial black

Question

Quizplus · Accepted Answer

The heterozygous fruit fly (VvBb) will produce four types of gametes in equal proportions: VB, Vb, vB, and vb. The homozygous fruit fly (vvbb) will produce only one type of gamete: vb. The cross between these two flies will produce the following F1 progeny:- 25% VVBb: normal gray- 25% Vvbb: normal black- 25% vvBb: vestigial gray- 25% vvbb: vestigial blackTherefore, the expected phenotypes would be 200 normal gray, 200 normal black, 200 vestigial gray, and 200 vestigial black.

A Heterozygous Fruit Fly with Normal Wings and a Gray