We can also use this method to map three genes with respect to each other on a single chromosome. The three genes we will use as an example are three genes found on the X chromosome in Drosophila. Recessive mutations in these genes result in the phenotypes white eyes (w), echinus eyes (ec), and yellow body (y).

The mapping cross is set up just as it was for the two-point cross, although it has the added complication of sex linkage, which means males will always be hemizygous. First, we need to produce heterozygotes, so we can observe the effects of crossing over in the subsequent generation. These heterozygotes will necessarily be female. This is done as follows:

All of the offspring of this cross are either wild-type females or mutant males. Because there are three genes involved in this mapping exercise, the order of the genes becomes important, i.e. which gene is in the middle? We don't know at this point the order of the genes on the chromosome, so we just draw them in any order. To do the mapping cross, we'll focus on the heterozygous female, because no crossovers will occur between the X and Y chromosomes in the male. We need to cross the female heterozygote with a mutant male, which in this case will be a hemizygous recessive (the F1 male is conveniently what we need).

In the offspring of this cross (as in the two-point cross), we want to look for recombinants - those that don't conform exactly to the parental phenotypes. In addition to single crossovers between genes, there is a small chance of two crossovers occurring between two crossovers. These must be considered as well (because they are less likely to happen, they will be the smallest phenotypic class). Assessing the offspring, we get the following numbers (for the phenotypes, if a particular characteristic is not mentioned, you can assume it is wild type):