Kosambi versus Haldane
Mapping Function

C =coefficient of coincidence; I = interference; I = 1 - C;

C =	obs. d.c.o.
	expected d.c.o.

Expected number of double crossovers = r_ABr_BCN.

Example:

Expected number of d.c.o. = 0.148(0.107)740 = 12.
Observed number of d.c.o. = 6.
C = 6/12 = 0.5; I = 1 - 0.5 = 0.5.

Now the expected number of double crossovers would equal the observed number of double crossovers (except for sampling) when the probability of a crossover in the G-S region was independent of the probability of a crossover in the S-L region. When the probability of a crossover in the first region is reduced due to the occurrence of a double crossover in the second region, then we have interference. The crossover in one region interfered with a crossover in the second region. If the probability of a crossover in the G-L region is independent of the probability of a crossover in the S-L region, then Haldane's mapping function would fit this biological situation.

However, in this example I = 0.5, so use Kosambi's map function. Haldane's map function assumes that the probability of a crossover in one region is independent of the probability of a crossover in the second region.
Haldane's map function r_AC = r_AB + r_BC - 2r_ABr_BC
Kosambi's map function r_AC = r_AB + r_BC - 2Cr_ABr_BC

When C=1 there is no interference. If you substitute C = 1 into Kosambi's map function, you can see that in this case the two map functions are equivalent. In Kosambi's map function C = 2r. When r_AB = r_BC = 0.5, then C = 1. This means that when the loci are located far apart, interference is absent. When the loci are close together, interference occurs and Kosambi's map function would be expected to give the best results.

Kosambi used C = 2r. When r = 0.5, then C = 2(0.5) =1 and there is no interference. In this case the two mapping functions are equivalent.