Question

Percentage of recombination between A and B is $9\%$, A and C $17\%$ and B and C is $26\%$. The arrangement of genes would be

• A. A-B-C
• B. A-C-B
• C. B-C-A
• D. B-A-C

Answer 1

Answer: (D)

B-A-C

1 map unit or centimorgan is equivalent to 1% recombination between the two genes. The frequency of recombination can be used to depict the arrangement of the genes.
Recombination frequency between three genes are
A-B=99%
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 cM and A-C being 17 cM and the distance between B-C being 26 cM.

So, the correct answer is 'B-A-C'.