The linkage map of X-chromosome of fruit fly | vedclass

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(B) The distance between two genes on a chromosome is measured in map units (centimorgans),where $1$ map unit corresponds to $1\%$ recombination frequ

Vedclass · Accepted Answer

$66\%$

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(B) The distance between two genes on a chromosome is measured in map units (centimorgans),where $1$ map unit corresponds to $1\%$ recombination frequency.Since the linkage map distance between the yellow body gene $(y)$ and the bobbed hair gene $(b)$ is $66$ units,the theoretical recombination frequency is $66\%$.However,in practice,recombination frequency cannot exceed $50\%$ because multiple crossovers between distant genes lead to an observed frequency that approaches $50\%$.Given the options provided and the definition of map units,the correct representation of the map distance is $66\%$.

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

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