The linkage map of x-chromosome of fruitfly has 66 units, with body colour gene (y) at one end and bobbed hair (b) gene at other end. The recombination frequency between these two genes (y & b) should be

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66%

100%

$\leq 50 %$

> 50%

answer is C.

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Detailed Solution

One map unit equals to 1% of recombination frequency. According to this concept,the given map units, the recombination % should be 66%.But that is not possible.

Reason: Since recombination takes place between two chromatids out of four chromatids in a pair of homologous chromosomes, there is always a maximum of 505 recombination but not more than that.

If the linkage map of x-chromosome of fruitfly has 66 units, with body colour gene (y) at one end and bobbed hair (b) gene at other end. The recombination frequency between these two genes (y & b) should be

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