The linkage map of $\mathrm{X}$-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:

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 exceeds than $50 \%$.