In a fruit fly,the X-chromosome linkage map h | vedclass

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(A) The distance between two genes on a chromosome is measured in map units or centimorgans $(cM)$.One map unit corresponds to $1\%$ recombination fre

Vedclass · Accepted Answer

$\leq 50\%$

Vedclass · Answer

(A) The distance between two genes on a chromosome is measured in map units or centimorgans $(cM)$.One map unit corresponds to $1\%$ recombination frequency.However,the maximum observable recombination frequency between any two linked genes is capped at $50\%$,even if the map distance is greater than $50$ map units.This is because multiple crossovers (double or higher-order crossovers) between distant genes can result in the parental combination of alleles,effectively masking the recombination events.Since the map distance is $66$ units (which is $> 50$ units),the observed recombination frequency will be $50\%$ or less (approaching $50\%$ as the distance increases).Therefore,the correct answer is $\leq 50\%$.

In a fruit fly,the $X$-chromosome linkage map has a total length of $66$ units,with the yellow body gene $(y)$ at one end and the bobbed hair gene $(b)$ at the other end. What will be the recombination frequency between these two genes $(y$ and $b)$?

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