Question

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

• A. ABC
• B. ACB
• C. BCA
• D. BAC

Answer 1

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 $26$.