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.

answers are:

lysogenic     viral dna merges with host cell dna

lysogenic     no symptoms

lytic               viral dna takes over

lysogenic     virus does not destroy host cell

lytic               virus destroys host cell

lytic               induces symptoms of viral infection

lysogenic       cell reproduces and spreads provirus