A fruit fly of genotype AaBb parent 1 is crossed to another

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

Solution

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

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 in
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 in

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