The linkage map of X -chromosome of fruitfly 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 ) should be:

Step-by-step solution

The yellow body gene (y) and bobbed hair (B) gene are present 66 map unit apart which means that there is < 50 % chances of recombination between them (recombination frequency). Related Theory One map unit is equal to 1 % recombination frequency. This linear relationship holds true for lower values only; as the recombination frequency increases beyond 50 % , the linear relationship does not hold true owing to double and multiple cross overs and recombination frequency is always less than map distance and never exce