Question

The linkage map of X-chromosome of fruit fly has 66 units with yellow body gene (y) at one end and bobbed hair (b) gene at the other end. The recombination frequency between these two genes (y and b) would be

• A. 100 %
• B. 66 %
• C. 50 %
• D. 5.50 %

Answer 1

The correct option is B 66 %

Recombination frequency can be defined as the measure of genetic linkage in a gene map, a centimorgan (cM) is a unit that describes a recombination frequency of 1%, in this way the genetic distance between two loci can be measured based on their recombination frequency.

So, the correct answer is "66 %".