Question

In Drosophila, the genes for body coloration and eye size are on different chromosomes. Normal-colored bodies are dominant to ebony-colored (very dark) bodies, and normal-sized eyes are dominant to being eyeless. Line A is true breeding for normal body and normal eye, whereas line B is true breeding for ebony bodies and eyeless. Individuals from lines A and B are crossed. From a dihybrid cross between the F1 generation, 400 flies are scored. How many of these F2 flies are expected to have both normal body color and normal eyes?

Answer 1

Let’s say that the gene for body coloration is M and the gene for the eye size is N.Than:

Normal-colored bodies are with genotype: Mm (heterozygous) or MM (dominant homozygous)

And ebony-colored bodies are genotype: mm (recessive homozygous)

normal-sized eyes are: NN or Nn

eyeless are: nn

Line A: MMNN

Line B:mmnn

If we cross A and B:

F1 phenotypes: 9:3:3:1

400/(9+3+3+1)=25

9*25=225 normal body and normal eye

3*25=75 normal body, eyeless

3*25=75 ebony-colored bodies normal eye

1*25=25 ebony-colored bodies, eyeless

Answer 2

Answer:

225 flies

Explanation:

Since the two genes are on different chromosomes, they will assort independently and produce phenotypes in 9:3:3:1 ratio according to Mendel.

Let body color allele be A and eye size allele be B.

True breeding normal colored body and normal-sized eyes = AABB

True breeding ebony body and eyeless = aabb

If the two are crossed and two of the F1 offspring are mated, the phenotypic ration will be:

Normal body color and eye size = 9/16

Normal body color and eyeless = 3/16

Ebony body and normal eye size = 3/16

Ebony body and eyeless = 1/16

If there are 400 flies in F2, then the number of flies with normal body color and normal eyes will be:

         9/16 x 400 = 225 flies

225 flies will be expected to have both normal body color and normal eyes.