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?

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.

the right side of the picture going down is:

polymerase

protein

transport vesicle

modified protein

the left side of the picture going down is:

dna helix

single-stranded rna

endoplasmic reticulum

ribosome

golgi apparatus

protein released from vesicle

click to let others know, how is it

read more on -

the answer is :