You are now trying to map a new gene locus (Gene X) on chromosome 9. You have determined that the recombination frequency between Gene X and Gene A is 12.5%, recombination frequency between Gene X and Gene B is 25%, and the recombination frequency between Gene X and Gene C is 31.25%. Use this information to determine the location of Gene X. Show your work and explain your reasoning.

We need to know that 1% of recombinations = 1 map unit. 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:

1% of recombination frequency = 1 map unit (MU)

A-X = 12.5% = 12.5 mu

B-X = 25% = 25mu

C-X = 31.25% = 31.25 mu

We do not know the distances between A-B, A-C, and B-C, so we can just estimate the location of the gene X by its distances from genes A, B, and C. To do this we need to think about it as additions. We know that A-X = 12.5 mu, a distance duplicated by B-X = 25 mu (12.5 + 12.5 = 25), and X is farther away from B than from A, so, this might tell us that A is located between B and X.

---B--------------A--------------X------

∫-------------25mu----------∫

∫--12.5mu----∫

Now, we know that X-C = 31.25 mu. And, assuming that these are linked genes, the maximum recombination frequency is always 50%. So the distance between two genes can not exceed 50%. According to this, we might assume that the C gene is near B, because if it was on the other side of X, the distance between B-C would sum 56.25 mu (25 + 31.25), meaning that they would not be linked. So, the correct order might be

--C------------B--------------------A-------------------X----

∫-----------------25mu-----------------∫

∫-----12.5mu-----∫∫-----12.5mu----∫

∫-----------------------31.25mu-------------------∫

∫6.25mu∫

∫------------18.75mu---------∫

batolisis batolisis · Answer 2 · 2022-01-14T15:00:48+01:00

The location of Gene X is closest to Gene A located 12.5

Given that :

The recombination frequency between Gene X and Gene A is 12.5%

The recombination frequency between Gene X and Gene B is 25%

The recombination frequency between Gene X and Gene C is 31.5%

The recombination frequency represents the map distance between the genes on the linkage map and the lesser the value of the recombination frequency the more linked are the genes.

The sequence in the linkage map is attached below

Hence we can conclude that the location of Gene X is closest to Gene A located 12.5

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