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) should be

• A. 60%
• B. $>$ 50%
• C. $\le$ 50%
• D. 100%

Answer 1

Answer: (C)

$\le$ 50%

Recombination frequencies are directly proportional to the distance between the genes and therefore are used in linkage map preparation. Distance between two genes is measured in map unit. One map unit is equal to 1% recombination frequency and is also referred 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). Correct option is C.