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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

1. 100%
2. 66%
3. > 50%
4. < 50%

Correct answer: 66%

Solution

A linkage map distance of 66 units means the two genes are 66 map units apart, so the expected recombination frequency is 66%. This is possible because map units are based on observed crossovers along the chromosome.

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