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:

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.

AB---

     ║15 cM ║

ACB---

    ║6 cM║ 9 cM║

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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answer; i believe that the correct answer is(d) nondisjunction during meiosis;