Question

In fruit flies, brown bodies are dominant to black bodies. cross two heterozygous fruit flies. determine the phenotypic and genotypic ratios. then determine how many fruit flies, if 200 are born, will have black bodies

Answer 1

By using a Punnett square, the phenotypes of offspring from two heterozygous fruit flies are 3 brown-bodied to 1 black-bodied and the genotypic ratio is 1 BB: 2 Bb: 1 bb. Out of 200 offspring, roughly 50 are expected to have black bodies.

Step-by-step explanation:

When crossing two heterozygous fruit flies, where brown bodies are dominant to black bodies, the phenotypic and genotypic ratios can be determined using a Punnett square. The genotypes of the parents are both Bb (where 'B' is brown and 'b' is black). A Punnett square shows that the offspring genotypes will be as follows: BB, Bb, Bb, and bb. The ratio of phenotypes is 3 brown-bodied to 1 black-bodied, and the genotypic ratio is 1 BB: 2 Bb: 1 bb.

If 200 fruit flies are born, you would expect approximately 25% to have black bodies (genotype bb). Hence, 200 x 0.25 = 50 fruit flies will have black bodies.