Question

Fly Recombination- Male fruitfly spermatocytes do not undergo any crossing over. Only female oocytes are capable of recombination. Given this fact, If A and B are 42 map units apart on the same chromosome, and a male fly with AB and ab chromosomes mates with a female having AB and ab chromosomes, what will the genotypic offspring ratios be?

I understand that they must be linked because the recombination frequency is less than 50% here. -they are on the same chromosome. But i am so stumped on this question and I need to study!

Answer 1

Distance between the genes = 42 mu

# recombinant gametes = % gene distance = 42%

# parental gametes = 100 - 42 = 58%

AB/ab × AB/ab

Parental gametes (58%) = AB (29%), ab (29%)

Recombinant gametes (42%) = Ab (21%), aB (21%)

Gametes	AB (29%)	ab (29%)	aB (21%)	Ab (21%)
AB (29%)	AB/AB (29% × 29% = 8.41%)	AB/ab (8.41%)	AB/aB (6.09%)	AB/Ab (6.09%)
ab (29%)	AB/ab (8.41%)	ab/ab (8.41%)	ab/aB (6.09%)	ab/Ab (6.09%)
Ab (21%)	AB/Ab (6.09%)	Ab/ab (6.09%)	Ab/aB (4.41%)	Ab/Ab (4.41%)
aB (21%)	AB/aB (6.09%)	aB/ab (6.09%)	aB/aB (4.41%)	aB/Ab (4.41%)

Phenotypic ratio : 58.41 : 16.59 : 16.59 : 8.41

A_B_ = 8.41 + 8.41 + 6.09 + 6.09 + 8.41 + 6.09 + 6.09 + 4.41 + 4.41 = 58.41%

A_bb = 6.09 + 4.41 + 6.09 = 16.59%

aaB_ = 6.09 + 6.09 + 4.41 = 16.59%

aabb = 8.41%

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