Percentage of recombination between A and B is 9%, A and C is 17%, B and C is 26%, then the arrangement of genes is
ACB
BCA
ABC
BAC
A
ABC
B
BAC
C
BCA
D
ACB
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Solution
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The correct option is D BAC
1 map unit or centimorgan is equivalent to 1% recombination between two genes. The frequency of recombination can be used to depict the arrangement of the genes. Recombination frequency between three genes is A−B=9%
A−C=17% and B−C=26%.
By manipulating the three possibilities of their arrangements A-B-C, A-C-B and B-A-C, it was found that the three genes must be arranged in the order B-A-C with the distance between B-A being 9 and A-C being 17 and the distance between B-C being 26.
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