Answer:
Step-by-step explanation:
In this case, we can use Punnett square to determine the genotypic and phenotypic ratios of the offspring:
B b
B BB (brown) Bb (brown)
b Bb (brown) bb (black)
So, when we cross two heterozygous fruit flies (Bb x Bb), we get the following genotypic ratios:
25% BB (brown)
50% Bb (brown)
25% bb (black)
And the following phenotypic ratios:
75% brown bodies
25% black bodies
Therefore, if 200 fruit flies are born, we can estimate that 25% of them will have black bodies, which is:
0.25 x 200 = 50 fruit flies