A, B, C and D are genes that are linked on the same chromosome. Given the following recombination frequencies, what is the gene order? A-B 15%, B-C 26%, C-D 30%, A-C 11%, A-D 19%, and B-D 4%. (Hint: what are recombination frequencies equivalent to?)

Answer and Explanation:

We need to know that 1% of recombination frequency = 1 map unit = 1cm. And that the maximum recombination frequency is always 50%.

The map unit is the distance between the pair of genes for which one of the 100 meiotic products results in a recombinant one.

The recombination frequencies between two genes determine their distance in the chromosome, measured in map units. So, if we know the recombination frequencies, we can calculate distances between the four genes in the problem and we can figure the genes order out. This is:

Recombination frequencies:

We know that 15 plus 4 equals 19. So we can infer that we have the gen sequence A--B--D, because the distance between A-B= 15, and the distance between B-D=4, and finally the distance between A-D=19.

ABD

  15↔4

      19

We also know that 11 + 19 = 30. So we can assume that C is next to A at a distance of 11 MU.

CAB---D

   11   15 ↔ 4

     11 ↔19    

         30                

The gene order is C, A, B, D

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