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?

Question

Rosemary Stewart

Biology

5 September, 04:29

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?

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Answers (1)

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Ximena Newman · Answer 1

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