The cross over frequency between linked genes C and D is 40%; between A and D is 35%; between C and A is 5%; between A and E is 25%; and between E and D is 10%. Construct a linkage map of these 4 gense and show the distances between the adjacent genes.

The first line shows the Correct order of genes in the chromosome and the second and third lines show the distances between genes.

C A E D (Correct order)

/5MU--//25 MU//10MU/  (Distances)

/40 MU /   (Total distance)

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 every 100 meiotic products, one 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)

C-D = 40% = 40 MU

A-D = 35% = 35 MU

C-A = 5% = 5 MU

A-E = 25% = 25 MU

E-D = 10% = 10 MU

Now that we know the distances between pairs of genes, we need to figure out which is the correct order in the chromosome.

C and D are the genes with the biggest distance between them (40MU), which probably means they are in the extremes. The rest of the genes are in the middle.

C (40MU) D

A is 5 MU apart from C and 35 MU apart from D, which means it is closer to C.

C (5MU)A(35 MU)D

E is 25 MU apart from A and only 10 MU apart from D, which means it is between A and D, and closer to D

CA (25 MU)E(10MU)D

The sum of C-A distance + A-E distance + E-D distance equals the C-D distance. This is:

C-A 5MU + A-E 25MU + E-D 10MU = C-D 40 MU

5 + 25 + 10 = 40

In conclusion, the correct order of the genes in the chromosome is

C A E D

/5MU--//25 MU//10MU/

/40 MU /

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