The linkage map of a fruit fly has a total ma | vedclass

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(C) The recombination frequency is directly proportional to the map distance between two genes,provided the distance is small.However,as the map dista

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$\leq 50\%$

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(C) The recombination frequency is directly proportional to the map distance between two genes,provided the distance is small.However,as the map distance increases,the probability of double crossovers increases,which leads to an underestimation of the recombination frequency.Recombination frequency can never exceed $50\%$ because independent assortment occurs at $50\%$ recombination.Since the total map distance is $66$ units (which is greater than $50$ map units),the observed recombination frequency will be capped at $50\%$ due to the occurrence of multiple crossovers.Therefore,the recombination frequency between genes at a distance of $66$ map units will be $\leq 50\%$.

The linkage map of a fruit fly has a total map distance of $66$ units. The gene for yellow body $(a)$ is at one end and the gene for short hair $(y)$ is at the other end. What will be the recombination frequency between these two genes $(a$ and $y)$?

Similar Questions

There are three genes $a, b,$ and $c$. The percentage of crossing over between $a$ and $b$ is $20\%$,between $b$ and $c$ is $28\%$,and between $a$ and $c$ is $8\%$. What is the sequence of these genes on the chromosome?

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$(A) \quad\quad (B)$

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