AP Biology· difficulty 4/5
A heterozygous fruit fly with linked genes (AB/ab) is testcrossed. Of 1000 offspring, 410 are AB, 405 are ab, 90 are Ab, 95 are aB. The recombination frequency between A and B is approximately
18.5 cM
9 cM
37 cM
50 cM (genes are unlinked)
Explanation
Recombinants = Ab + aB = 90 + 95 = 185. RF = 185/1000 = 0.185 = 18.5 cM. RF < 50% → linked.
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