To show very simply how weighting can affect an outcome, consider the data set
If we now weight character #2 by 2 (a formal statement that it takes 2 changes by any other character or combination of other characters to refute the information about relationships offered by a change in character 2) we get only one most parsimonious tree:one aaaa two acac three cccc four cacawhich defines two equally parsimonious trees (homoplasy in red):_____ one aaaa |_____ two acac | __ three cccc `--|__ four caca and _____ one aaaa | __ two acac |--|__ three cccc |_____ four caca
Weighting methods can be roughly divided into two classes_____ one aaaa | __ two acac |--|__ three cccc |_____ four caca
adaptive
independence
a a
c g a t g
cagtcgg tgattcc actga c
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ttcagcc actaagg tgact c
t a g c a
c a
chemistry
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For example,
a a
c g a t g
cagtcgg tgattcc actga c
||||||| ||||||| ||||| t
ttcagcc actaagg tgact c
t a g c a
c a
Knight and Mindell (1993) stated that "a level of TIs near 50% indicates that this class of change is saturated with multiple changes and is therefore not an indicator of phylogeny but is largely 'noise'" (see also Mindell and Honeycutt, 1990). In some cases (e.g., Mindell et al., 1995) all gapped regions in alignments as well as third position transitions are eliminated. And this can be determined from saturation plots (below) in which the number of transitions and transversions are plotted against total substitutions for all pairwise comparisons of taxa.

More often, though, it is argued that one should simply weight according to the inverse of the frequency.
If so, the frequency has to be determined across some tree. This can be a preliminary tree based on unweighted data (which leaves you open to the criticism of cricularly biasing your results towards that unweigted tree because ofweighting in light of the unweigted tree), or across a random set of trees.
Given the tree here for primate mtDNA (left), one can get a preliminary tree (by a search on the unweighted data) and then calulate the amount of change in 1st, 2nd and 3rd positions, which is 260:146:594 and would imply a weighting scheme for positions of about 2:4:1.

Random trees for the same data suggest a number of Ti's between 576 and 715 and a number of Tv's between 938 and 1113. So, weighting transversion somewhere between 1.5 and 2 appears to be indicated.
How is this implemnented? In PAUP, one can set up a Sankoff matrix to define these weights as follows:
BEGIN ASSUMPTIONS;
Williams and Fitch (and Fitch and Ye) took this one step further by suggesting dynamic weighting. In this, each class of trnasformation in each direction is weighted inversely to it frequency. For the primate data, the frequencies of each transformation is:
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Because transformations are weighted inversely to their frequency, this would suggest the following Sankoff matrix:
BEGIN ASSUMPTIONS;
Frequencies are inherently rate-based arguments and thus if one is going to justify weighting according to this principle, one should weight not according to the inverse of the frequency but accoring to the natural logarithm of the inverse of the frequency.Thus, a Sankoff matrix that looks like this: BEGIN ASSUMPTIONS;USERTYPE tv = 4 a c g t - 2 1 2 2 - 2 1 1 2 - 2 2 1 2 -; END; Really applies to a situation where transitions are more than 7 times as frequent as transversions.
The procedure involves
234 1238
Steps on tree 1 : 2 14
Steps on tree 2 : 1 15
The proportional difference in number of steps for character 234 is much larger than for character 1238. That is, for character 234 the proportional disagreement is:
(2 - 1)/(2*1) = 0.500Whereas for character 1238 is:
(15-14)/(15*14) = 0.005Or, character 234 is very different on the two trees (strongly supports tree #2) while character 1238 is not very different on the two trees (weakly supports tree #1).
The argument is that one should prefer tree #2.
Goloboff developed a way to weight characters without requiring a priori reference to a tree in which the implied weight for a character is:
W = K/(K+ (maxsteps - observed steps))K is an arbitrary constant and there is no a priori way to choose a value for K so it is reccoemended that one try 1, 2, 3 etc and see what happens.