A scientist performed the gene mapping experiments in maize. He mapped the genes on chromosomes on the basis of % crossing over between different genes. One map unit corresponds to one % crossing over or recombination. The genes showing more than 50% recombination were not supposed to be linked on same chromosome. In crossing over studies on maize, scientist observed the following % crossing over between genes A, B, C, D: between A and D 10%, between A and C 3%, between genes C and D 7%, between genes A and B 5%, and between genes C and B 8%. On the basis of above observation, find out the correct sequence of genes A, B, C and D on chromosomes.

The correct option is A BACD
Since, recombination frequency is directly proportional to distance between genes, the values are used to locate genes on a chromosome. Here, recombination frequency for A and D (X) = 10%, that for A and C (Y) = 3% and for C and D (Z) = 7%. As we can see that X = Y + Z or 7 + 3 = 10% which means that genes A and D are present at extremes and C is in the middle. Further, recombination frequency for A and B (p) = 5%, that for C and B (q) = 8% and that for A and C (Y) = 3% which means that q = p+Y; thus the gene A is present in middle of genes C and B. This gives the whole sequence of genes as BACD. Correct answer is C. For gene sequence BCDA, X = Y+Z; but it does not hold true for given values of recombination frequency which makes option A wrong. For gene sequence ABCD, X = p+q which does not hold true for given values of recombination frequency and makes option B wrong. For sequence DACB, p = X+q; but it does not hold true for given values of recombination frequency which makes option D wrong.