To map the three genes, we first need to determine the order of the genes on the chromosome (i.e. which gene is in the middle). To do this is fairly straightforward. A double crossover will switch the middle gene with respect to the genes on either side of it. Double crossovers are easy to identify in the data: they will be much more rare than single crossovers, so the two phenotypic groups present in the smallest numbers will be the double crossover class. In our data, these are the flies that are echinus/yellow, and the flies that are white. Since all three of these recessive traits are found together in the non-crossover class, in the double crossover class, white has been switched relative to the other two. Therefore, the white gene must be in the middle.
To determine map distances, simply consider pairs of genes and calculate the percent of recombination between those two genes. The trick is to be able to identify recombination out of the excess amount of data. To do this, only consider the two genes you are trying to map; ignore the third for now. Compare the combinations of phenotypes (wild type or recessive for each gene) to the combinations in the parentals. If they are different, thenrecombination has occurred.
As an example, let's look at echinus and white. The parental phenotypes are wild-type phenotypes together, and recessive phenotypes (white, echinus, yellow) together. Those phenotype classes showing recombination are: the white, yellow flies (193 flies), the echinus flies (207 flies), the echinus, yellow flies (3 flies), and the white flies (3 flies).
The distance between these two genes is therefore:
(193+207+3+3)/10,000 = 406/10,000 = 0.0406
or 4.06 map units
Next, we'll determine the distance between white and yellow. In this case, the recombinants are the white, echinus flies (70 flies), the yellow flies (80 flies), the echinus, yellow flies (3 flies), and the white flies (3 flies). You'll note that the last two groups of flies have been counted twice - this is because they represent the double crossover group, with one crossover between each pair of genes.
The distance between these two genes is:
(70+80+3+3)/10,000 = 156/10,000 = 0.0156
or 1.56 map units
To calculate the distance between echinus and yellow, you can do a similar calculation to those above, or simply add the other distances together. Since white is in the middle, the distance from echinus to yellow ought to equal the distance from echinus to white plus the distance from white to yellow. This distance should therefore be 5.62 map units.
Linkage and Mapping: Summary of Key Points