To determine the most likely order of markers within a linkage group, we could imagine using the following simple procedure: For each possible order of that group, we calculate the maximum-likelihood map (e.g. the distances between all markers given the data), and the corresponding map's likelihood. We then compare these likelihoods and choose the most likely order as the answer. This type of exhaustive analysis may be performed using MAPMAKER's "compare" command.
In practice however, this sort of "exhaustive" analysis is not practical for even medium sized groups: a group of N markers has N!/2 possible orders, a number which becomes unwieldy (for most computers) when N gets to be between 6 and 10. (In practice, one needs to order subsets of the linkage group and then overlap those subsets, mapping any remaining markers relative to those already mapped, a process we will illustrate later).
Luckily, group1, consisting of markers 1, 2, 3, 5, and 7 is small enough that we can use a fully exhaustive analysis.
To do this, we first change MAPMAKER's sequence to "{1 2 3 5 7}". Here, the set-braces indicate that the order of the markers contained within them is unknown, and thus that all possible orders need to be considered.
We then type the "compare" command, instructing MAPMAKER to compute the maximum likelihood map for each specified order of markers, and to report the orders sorted by the likelihoods of their maps. Note that while MAPMAKER examines all *** orders, only the 20 most likely ones are reported (by default).
For each of these 20 orders, MAPMAKER displays the log-likelihood of that order relative to the best likelihood found. Thus the best order:
1 3 2 5 7is indicated as having a relative log-likelihood of 0.0. The second best order:
3 1 2 5 7is significantly less likely than the best, having a relative log-likelihood of -6.0. Said a different way, the best order of this group is supported by an odds ratio of roughly 1,000,000:1 (10 to the 6th power to one), over any other order. We consider this good evidence that we have found the right order.
5> sequence {1 2 3 5 7}
sequence #2= {1 2 3 5 7}
6> compare
Best 20 orders:
1: 1 3 2 5 7 Like: 0.00
2: 3 1 2 5 7 Like: -6.00
3: 5 7 2 3 1 Like: -20.20
4: 5 7 2 1 3 Like: -26.26
5: 2 5 7 3 1 Like: -27.25
6: 2 5 7 1 3 Like: -28.39
7: 2 3 1 5 7 Like: -28.85
8: 5 2 3 1 7 Like: -32.33
9: 2 1 3 5 7 Like: -34.12
10: 5 7 1 3 2 Like: -35.55
11: 5 2 1 3 7 Like: -37.61
12: 1 3 5 2 7 Like: -37.76
13: 3 1 5 2 7 Like: -39.09
14: 5 7 3 1 2 Like: -40.38
15: 1 3 5 7 2 Like: -40.87
16: 3 1 5 7 2 Like: -41.55
17: 5 2 7 3 1 Like: -43.67
18: 5 2 7 1 3 Like: -44.78
19: 5 1 3 2 7 Like: -47.63
20: 2 5 3 1 7 Like: -52.28
order1 is set