The linkage map of X chromosome of fruit fly has 66 units , with yellow body gene (y) at one end and bobbed hair gene (b) at the other end. This recombination frequency between these two genes (y and b) should be(a) 66% (b) > 50 % (c)

The actual distance between two genes is said to be equivalent to the percentage of crossing over between these genes i.e. 66%. Crossing over chances between y and b genes suggest that these are to be placed on the chromosome at a distance of 66 units.