The map distance between genes A and B is 3 u | vedclass

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(D) Given distances are:Distance between $A$ and $B = 3$ units.Distance between $B$ and $C = 10$ units.Distance between $C$ and $A = 7$ units.Since th

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$B, A, C$

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(D) Given distances are:Distance between $A$ and $B = 3$ units.Distance between $B$ and $C = 10$ units.Distance between $C$ and $A = 7$ units.Since the distance between $B$ and $C$ is the sum of the distances between $B$ and $A$ ($3$ units) and $A$ and $C$ ($7$ units),i.e.,$3 + 7 = 10$ units,it implies that gene $A$ lies between genes $B$ and $C$.Therefore,the linear order of the genes on the chromosome is $B-A-C$.

The map distance between genes $A$ and $B$ is $3$ units,between $B$ and $C$ $10$ units,and between $C$ and $A$ $7$ units. The order of the genes in a linkage map constructed on the above data would be:

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