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You have mated a phenotypically wild-type female of the genotype Xy+w+Xyw to a male with the genotype XywY. You counted 2205 total F1 progeny and found 17 flies with gray bodies and white eyes and 12 flies that had yellow body and red eyes. What is the distance between these genes controlling body color and eye color on the X chromosome?

Respuesta :

Answer:

1.31 cM

Explanation:

Total offspring = 2205

Since two genes are involved, F1 progeny should have four types of combination. Out of them two are 17 and 12 which definitely means they are in lesser number. Since recombinants are always less than parental progeny in linkage, the given two types are recombinants.

Recombination frequency = (Number of recombinants / Total progeny) * 100

= [ ( 17 + 12 ) / 2205 ] * 100

= ( 29 / 2205 ) * 100

= 1.31 %

Map distance = Recombination frequency

Hence, distance between two genes = 1.31 cM

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