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Question

The recombination frequency between three genes X, Y and Z are as follows
X-Y = 5.5%, Y-Z = 3.5% and X-Z = 1.5%
Which of the following shows the correct order of the arrangement of these genes on the chromosome?

A
Z-X-Y
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B
X-Y-Z
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C
Y-X-Z
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D
X-Z-Y
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Solution

The correct option is D X-Z-Y
Linkage basically means association. When genes are close together on the same chromosome, they are said to be linked. When they are far apart or on the same chromosome or on a different chromosome and they assort independently are said to be unlinked. If the genes are linked and do not independently assort, then the strength of the linkage can be measured using recombination frequency. Using the the recombination frequencies, the order and the distance of the genes can be determined.
Linkage is inversely proportional to the crossover and recombination. Crossing over and recombination breaks the strength of the linkage. Therefore, the lesser the distance, the less will be the recombination and the higher will be the linkage strength and vice versa. The genes remain linked together in case of the lowest recombination frequency.
Here, gene X-Z has the lowest recombination frequency i.e. 1.5%; so they will have the highest linkage between them. Y-Z has a recombination frequency of 3.5%, so gene Y will be situated next to gene Z and finally the highest recombination frequency gene i.e. X-Y 5% will have the lowest linkage among the three genes. Therefore they will be situated away from each other.
So the correct order of the gene is X-Z-Y.

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