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In Drosophila melanogaster, vestigial wings are caused by a recessive allele of a gene that is linked to a gene with a recessive allele that causes black body color. Morgan crossed black-bodied, normal-winged females and gray-bodied, vestigial-winged males. The F1 were all gray bodied, normal winged. The F1 females were crossed to homozygous recessive males to produce testcross progeny. Morgan calculated the map distance to be 17 map units. Which of the following is correct about the testcross progeny?

A) black-bodied, normal-winged flies = 17% of the total.
B) black-bodied, normal-winged flies PLUS gray-bodied, vestigial-winged flies = 17% of the total.
C) gray-bodied, normal-winged flies PLUS black-bodied, vestigial-winged flies = 17% of the total.
D) black-bodied, vestigial-winged flies = 17% of the total.

2 Answers

5 votes

Answer:

C

Step-by-step explanation:

User Petrus
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1 vote

Answer:

C) gray-bodied, normal-winged flies PLUS black-bodied, vestigial-winged flies = 17% of the total.

Step-by-step explanation:

Parent 1 : Black body, normal wings

Parent 2: Gray body, vestigial wings

F1 progeny has all gray body and normal wings.

when F1 is crossed with a fly recessive for both traits (test cross), both parental types and recombinants are produced.

Distance in mu = recombination frequency so 17 mu means that there were 17% recombinants.

In the progeny from test cross:

Black body, normal wings = Parental

Gray body, vestigial wings = Parental

Black body, vestigial wings = Recombinants

Gray body, normal wings = Recombinants

The two types of recombinants will make 17% of flies. Hence, gray-bodied, normal-winged flies PLUS black-bodied, vestigial-winged flies = 17% of the total.

User Zyoma
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