Assume that the genes for tan body and bare wings are 15 map units apart on chromosome II in Drosophila.Assume also that a tan-bodied,bare-winged female was mated to a wild-type male and that the resulting F₁ phenotypically wild-type females were mated to tan-bodied,bare-winged males.Of 1000 offspring,what would be the expected of wild-type offspring,and in what numbers would they be expected? A)75 B)425 C)50 D)350 E)8

Question

Quizplus · Accepted Answer

The cross can be represented as: 
X^B/X^B T/T x X^+/Y T/T
where X represents the wild-type allele for body color and T represents the wild-type allele for wing shape. 
From the first cross, we know that the female is X^B/X^+ T/T and the male is X^+/Y T/T. Their F1 offspring will all be X^B/X^+ T/T, since they inherit one X chromosome from each parent and both parents are heterozygous for body color and homozygous for wing shape. 
The F1 offspring are then crossed to tan-bodied, bare-winged males, who must be X^B/Y T/T. The expected offspring from this cross can be calculated using a dihybrid cross with a recombination frequency of 15 map units:
X^B/T x X^+/T -> X^B/T ; X^+/T ; X^B/T ; X^+/T (recombinants)
X^B/T x Y/T -> X^B/T ; Y/T (non-recombinants)
X^+/T x Y/T -> X^+/T ; Y/T (non-recombinants)
Expected ratios:
X^B/T ; X^+/T ; X^B/T ; X^+/T (recombinants) = 2(1/2)(1/2)(1/2)(1/2) x 1000 = 250
X^B/T ; Y/T (non-recombinants) = 1/2(1/2)(1/2)(1/2) x 1000 = 62.5
X^+/T ; Y/T (non-recombinants) = 1/2(1/2)(1/2)(1/2) x 1000 = 62.5
Total wild-type offspring: 250 + 62.5 + 62.5 = 375
Therefore, the answer is B) 425.

Assume That the Genes for Tan Body and Bare Wings