A fruit fly of genotype A/a.B/b (parent 1) is crossed to another fruit fly of ge
ID: 313670 • Letter: A
Question
A fruit fly of genotype A/a.B/b (parent 1) is crossed to another fruit fly of genotype a/a.b/b (parent 2). The progeny of this cross were: Genotype Number of individuals A/a.B/b 32 a/a.b/b 33 A/a.b/b 17 a/a.B/b 18
a. What is the recombination frequency (RF)?
b. If one sets up a selfing for parent 1 (A/b.B/b X A/a.B/b), what fraction of the offpring will have the genotype (a/a.b/b)?if one sets up a selfing for parent 1 (A/b.B/b X A/a.B/b), what fraction of the offpring will have the genotype (a/a.b/b)? Hint: use the RF you calculated
Explanation / Answer
A fruit fly of genotype A/a.B/b (parent 1) is crossed to another fruit fly of genotype a/a.b/b (parent 2). The progeny of this cross were:
Genotype Number of individuals
A/a.B/b 32
a/a.b/b 33
A/a.b/b 17
a/a.B/b 18
total = 100
a. What is the recombination frequency (RF)?
The two gene pairs must be linked; the %R (RF) = 17 + 18 = 35%-------answer
b. If one sets up a selfing for parent 1 (A/b.B/b X A/a.B/b),
what fraction of the offpring will have the genotype (a/a.b/b)?
A/a.B/b X A/a.B/b
AB
A b
aB
ab
AB
AB AB
A b AB
aB AB
ab AB
A b
AB A b
A b A b
aB A b
ab A b
aB
AB aB
A b aB
aB aB
ab aB
ab
AB ab
A b ab
aB ab
ab ab
Answer = genotype (a/a.b/b)=1/16------------answer
AB
A b
aB
ab
AB
AB AB
A b AB
aB AB
ab AB
A b
AB A b
A b A b
aB A b
ab A b
aB
AB aB
A b aB
aB aB
ab aB
ab
AB ab
A b ab
aB ab
ab ab
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