Additive:


Linkage map units don't actually correspond to a specific length of chromosome. The exact distance calculated between two genes depends upon how much infomation you have about what's going on in the region between the genes.For example, consider the problem above. If we'd been following only the two end genes (B and C), and disregarding the A gene, of the offspring in our example, only the single crossovers would have counted as recombinants. The only reason we know that the double crossovers recombined (twice) between B and C is because we see that the Agene is switched; the B and C genes are still in the parental formation. If we are ignoring A, they become parentals. Thus, if we calculate the B-C distance without regard to the information about the A gene, we get:Recombinants: BC {B(A)C [95] + B(a)C [50]} and bc {b(a)c [95] + b(A)c [50]} for a total of 290.All the rest would score as parentals. We we'd calculate the distance between B and C as: (290/1700) x 100 - 17.1 LMU. Compare this to the 18.3 LMU we calculated between the same genes when we were paying attention to the A locus.Keep in mind that when you are calculating linkage map distance, there will virtually always be many genes between each pair that you are considering--genes that you are ignoring in your calculations. So linkage map units are often not strictly additive, and you will often get different precise distances, depending upon the specifics of the calculation method.