In a chromosome, The distance between A to B is 15cM(CentiMorgan), B to C is 9cM , A to C is 6cM. If A to D is is 5cM, then what is recombinant percentage C to D ??? 1)1cM
2)11cM
3)6cM
4)Both 1 and 4​

The recombinant percentage C to D is 1% = 1cM  (Option 1)    

Explanation:

Two genes that are very close will have a few recombination events and are strongly bounded.

To calculate the recombination frequency, we have to know that 1% of recombinations = 1 map unit = 1 centiMorgan. 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.  

Available data:

In a chromosome, there are 4 genes under study, A, B, C, and DDistance A-B = 15 cMDistance B-C = 9 cMDistance A-C = 6cMDistance A-D = 5 cM

The recombinant percentage between C and D equals their distances in centiMorgan. So we need to see the order of the genes on the chromosome.

First, we need to identify the genes located in the extremes. These are the most distant from each other. A-B are 15 cM appart.

AB---

     ║15 cM ║

Now that we placed both genes, we can place the other genes. The second most distant are B-C = 9cM and then A-C = 6 cM

ACB---

    ║6 cM║ 9 cM║

Finally, we can place D, which is 5 cM appart from A

ADCB---

   ║---5cM--║

Now that we can visualize better the location of genes on the chromosome, we can figure out the distance between D-C.

If the distance from A to C is 6 cM, and the distance from A to D is 5 cM, We just need to perform a subtraction, and we will get the distance from D to C.

(C-D) = (A-C) - (A-D) = 6 cM - 5 cM = 1 cM

ADCB---

                   ║1cM║

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