Question

Linkage map of X-chromosome of fruitfly has $66$ map units with yellow body gene(y) at one end and bobbed hair(b) at the other. The recombination frequency between y and b gene would be

• A. $66\%$
• B. $>50\%$
• C. $\le 50\%$
• D. $100\%$

Answer 1

The correct option is C $\le 50\%$

Recombination frequencies are directly proportional to the distance between the genes and therefore are used in linkage map preparation. The distance between the two genes is measured in the map unit. One map unit is equal to a 1% recombination frequency and is also referred to as centimorgan.

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%. In the question, 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).

So, the correct answer is '≤50%'.