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 n

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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?

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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:225 fliesExplanation: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 = AABBTrue breeding ebony body and eyeless = aabbIf the two are crossed and two of the F1 offspring are mated, the phenotypic ration will be:Normal body color and eye size = 9/16Normal body color and eyeless = 3/16Ebony body and normal eye size = 3/16Ebony body and eyeless = 1/16If 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 flies225 flies will be expected to have both normal body color and normal eyes.

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