Genetics: In fruit flies black body(b) and vestigial wing(v) are each recessive

Question

Genetics:

In fruit flies black body(b) and vestigial wing(v) are each recessive mutants. How would you set up a testcross to determine if these loci are linked? Show the chromosomes, b v, b+ v+ that you would have at each step. For your test cross, what would be the predicted frequency of each genotype and each phenotype resulting from your testcross if the genes are unlinked? What would be the predicted frequency of each genotype and phenotype if the recombination rate between the loci is 0.20(20%)? Show how you got your answers!

Explanation / Answer

Answer:

bv, b+ v+ X b v, bv ------------Testcross (if the genes are unlinked)

b v

bv

bv / bv –25% (black body, vestigial wing)

b+v+

b+v+ / bv—25% (normal body, normal wing)

b+ v

b+ v / bv—25% (normal body, vestigial wing)

b v+

b v+ / bv—25% (black body, normal wing)

If the genes are unliked, all the progeny would be produced in equal proportions.

If the genes are linked, the progeny are as follows.

bv, b+ v+ X b v, bv ------------Testcross (if the genes are unlinked)

b v, b+ v+ genotype produces recombinant gametes, b+ v & b v+ in 20% (each 10%) and non-recombinant gametes, b+ v+ & b v in 80% (each 40%)

b v

Bv (40%)

bv / bv –40% (black body, vestigial wing)

b+v+ (40%)

b+v+ / bv—40% (normal body, normal wing)

b+ v (10%)

b+ v / bv—10% (normal body, vestigial wing)

b v+ (10%)

b v+ / bv—10% (black body, normal wing)

b v

bv

bv / bv –25% (black body, vestigial wing)

b+v+

b+v+ / bv—25% (normal body, normal wing)

b+ v

b+ v / bv—25% (normal body, vestigial wing)

b v+

b v+ / bv—25% (black body, normal wing)

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