Answer:
C) gray-bodied, normal-winged flies PLUS black-bodied, vestigial-winged flies = 17% of the total.
Step-by-step explanation:
Parent 1 : Black body, normal wings
Parent 2: Gray body, vestigial wings
F1 progeny has all gray body and normal wings.
when F1 is crossed with a fly recessive for both traits (test cross), both parental types and recombinants are produced.
Distance in mu = recombination frequency so 17 mu means that there were 17% recombinants.
In the progeny from test cross:
Black body, normal wings = Parental
Gray body, vestigial wings = Parental
Black body, vestigial wings = Recombinants
Gray body, normal wings = Recombinants
The two types of recombinants will make 17% of flies. Hence, gray-bodied, normal-winged flies PLUS black-bodied, vestigial-winged flies = 17% of the total.