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:

Let gene for body colour : A

Let gene for eye type : B

Parent 1 : normal body and normal eyes = AABB

Parent 2: ebony body and eyeless = aabb

F1 : AABB X aabb = AaBb ( all have normal body and normal eye )

F1 progeny is self crossed given that the two genes are on different chromosomes which means they show independent assortment:

AaBb X AaBb =

A_B_ = normal body, normal eyes = 9

aaB_ = ebony body, normal eyes = 3

A_bb = normal body, eyeless  = 3    

aabb = ebony body, eyeless  =   1

Hence the ratio is 9 : 3 : 3 : 1

Out of the 400 flies:

(9/16) * 400 = 225 flies have normal body colour and normal eyes

complete the square on the right side of the equation;

-5(x + 0)^2 + 15

answer: d. y = -x^2 + 15

answer; a muscle cell, another neuron and a of the above