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The linkage map of X-chromosome of fruit...

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 he other end. The recombination frequency between these two genes Y and B should be

A

1

B

0.66

C

0.5

D

0.055

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The correct Answer is:
B
(b) 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 .
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