Illustration of the idea of a) cMAF and b) cLD.

An example to show the calculation of cLD, inspired by cMAF. a) Out of six haplotypes, there are two [1, 4] who have mutations in region A. Therefore, the cMAF P(A) for region A is 2/6 = 0.33. b) There are three haplotypes [3, 4, 5] who have mutations in region B and the cMAF P(B) for region B is 3/6 = 0.50. If one considers regions A and B together, there is one individual with mutations in both regions: [4]. Thus, the P(AB) is 1/6 = 0.17. Finally, by yielding P(A), P(B) and P(AB) into the standard formula of LD we have cLD = 0.375. (This is only an illustrating example with 6 haplotypes. In practice, the variants contributing to cMAF or cLD are rare ones with MAF < 0.5%.