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If the recombination frequency between genes (A) and (B) is 5.3%, what is the distance between the genes in map units on the linkage map?

a) 5.3 cM
b) 10.6 cM
c) 15.9 cM
d) 21.2 cM

1 Answer

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Final answer:

The distance between two genes on a linkage map is equivalent to the recombination frequency. Since 1% recombination equals 1 centimorgan, a recombination frequency of 5.3% indicates the genes are 5.3 cM apart.

Step-by-step explanation:

If the recombination frequency between genes (A) and (B) is 5.3%, this means that on a linkage map, the distance between the two genes is directly equivalent to the recombination frequency. In genetic mapping, 1% recombination frequency is equivalent to 1 centimorgan (cM).

Therefore, if the recombination frequency is 5.3%, the distance between the genes in map units on the linkage map is 5.3 cM. The correct answer to the question is a) 5.3 cM.

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