How to Solve Linkage Map Problems

Transcription

1 Page 1 of 6 Examples to Accompany How to Solve Linkage Map Problems Note that these examples are invented. Real numbers would be much messier than these. Determining Linkage/Independence Suppose you want to map the distance between genes "N" and "P." Each has two alleles (N and n, P and p). Both pairs of alleles have a complete dominance relationship; we will identify the phenotypes simply by the letters of the alleles. Your mating will be: NnPp x nnpp If your genes are independent, then your first parent's four possible gametes (NP, Np, np, and np) will be produced in equal numbers, and thus you will get four offspring classes, also with equal numbers, like so: NP offspring: 100 Np offspring: 100 np offspring: 100 np offspring: 100 These equal numbers are the evidence that these two genes are independent. However, if you get numbers like: NP offspring: 150 Np offspring: 50

2 Page 2 of 6 np offspring: 50 np offspring: 150 then you have established that the genes are not independent. Moreover, you know that in your original heterozygous parent, one chromosome had N and P on it and the other had n and p on it (as opposed to N and p on one and n and P on the other). In other words, your parental linkage is N linked with P and n linked with p. In this example, the parental offspring are the NP and np offspring; the recombinanant offspring are the Np and np offspring. Two-point Cross Using the example above, the linkage map distance between the N and P gene loci is calculated by calculating the percentage of recombination between the two genes. This is done by adding the number of recombinant offspring, dividing by the total number of offspring, and multiplying by 100 (to convert to a percentage. In this example, we have a total of 100 recombinant offspring (the Np and np offspring added together), and an overall total of 400 offspring. So... (100/400) x 100 = 25% 25% of our offspring are recombinants, thus the linkage map distance between these two genes is 25 LMU. Three-point Cross This time we are looking at the A, B, and C genes (alleles A and a, B and b, C and c). Again,

3 Page 3 of 6 complete dominance for all. Here's our mating: AaBbCc x aabbcc If the genes are independent, we should get eight classes of offspring, all pretty much equal in number. However, here's what we get: ABC: 95 abc: 50 ABc: 5 abc: 700 AbC: 700 abc: 5 Abc: 50 abc: 95 Obviously, not all phenotypic classes are equal, so we definitely have linkage. Determining Gene Order To determine gene order, you must first identify the parental classes and the double crossover classes. Don't fall into the trap of assuming that the parentals must always be ABC and abc. Parental classes are always the largest ones, since the connection between the genes will tend to preserve parental linkages above the expected level of independence. In our example, the AbC and abc classes are the parental classes. Notice that these are reciprocals of each other--they have "opposite" phenotypes. This should make sense. The double crossover classes will always be the smallest classes, since they require a crossover between A and B and a crossover between B and C. So our double crossover classes are ABc and abc. Again, note that they are reciprocals. Now we compare the parentals with the double crossovers. The only gene which has recombined in a double crossover will be the middle one, because the chromatids will have crossed over between the first two genes, then crossed back between the second and third genes. Compare each double crossover with the parental which is most like it. For double crossover ABc, the most similar parental is abc. (It's the same in two of the genes; the other parental is the same in only one gene.) Comparing these, we see that the only gene that is different is the A/a gene. If we compare the other double

4 Page 4 of 6 crossover to the other parental, we get the same result--only the A/a gene is difrerent. So A must be the gene in the middle. Our actual gene order is B-A-C. To keep from getting confused, it's an excellent idea to rewrite the offspring chart above with the genes in the correct order, like this: BAC: 95 BaC: 50 BAc: 5 Bac: 700 bac: 700 bac: 5 bac: 50 bac: 95 Map Distance To figure out the distances, we must identify our single crossover classes. We've got four of them: BAC, bac, BaC and bac. Note that this is two pairs of reciprocals (BAC and bac are reciprocals and bac and BaC are also reciprocals). Again, we want to compare these to the parentals which are most like them to determine just where the crossover occurred to produce them. Comparing single crossover BAC to parental bac, we see that it has recombined between the B and the A loci. Same for its reciprocal. Comparing the BaC single crossover to the Bac parental, we see that it has recombined between the A and the C genes, as has its reciprocal. Note also that the numbers of offspring give you hints as to which crossover classes go together. Now we are ready to calculate distances. We have four offspring classes which have recombined between the B and the A genes. These are the single crossovers BAC (95) and bac (95) as well as the double crossovers Bac (5) and bac (5). This gives us a total of 200 offspring who have crossed over between these two genes. There are a total of 1700 offspring, so we'd calculate this distance as (200/1700) x 100, which equals So we've calculated a distance of 11.8 LMU between genes B and A. There are also four offspring classes which have recombined between the A and C genes. bac (50), BaC (50) and again, our double crossover classes BAc (5) and bac (5), for a total of 110. Calculating: (110/1700) x 100 = 6.5 LMU between genes A and C.

5 Page 5 of 6 To calculate the distance between B and C, just add 11.8 and 6.5, to get 18.3 LMU. 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 A gene 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 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.

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