Answer:
WILD TYPE= 425;
TAN-BARE= 425;
TAN= 75;
BARE= 75
Explanation:
Fifteen map units apart implies that 15% of the offspring are products of recombination.
Out of 1000 offsprings, therefore 15% of 1000= 0.15 × 1000=150
150 offsprings are products of recombination.
100-15 =85%
Therefore, 85% of the offspring are parental.
So, if tan-bodied, ware-winged female was mated with wild-type male resulting in F1 phenotypically wild-type females were mated to tan-bodied, bare-winged males.
The expected number of offsprings that are tanned but have normal wings would be:
Wild type= 50% of 850
0.50 × 850 = 425
Tan-bare= 50% of 850
0.50 × 850 = 425
tan= 50% of 150
0.50×150= 75
Bare= 50% of 150
0.50×150=75
