# Consensus methods. Strict consensus methods

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1 Consensus methods A consensus tree is a summary of the agreement among a set of fundamental trees There are many consensus methods that differ in: 1. the kind of agreement 2. the level of agreement Consensus methods can be used with multiple trees from a single analysis or from multiple analyses Strict consensus methods Strict consensus methods require agreement across all the fundamental trees They show only those relationships that are unambiguously supported by the parsimonious interpretation of the data The commonest method (strict component consensus) focuses on clades/components/full splits This method produces a consensus tree that includes all and only those full splits found in all the fundamental trees Other relationships (those in which the fundamental trees disagree) are shown as unresolved polytomies 1

2 Strict consensus methods TWO FUNDAMENTAL TREES A B C D E F G A B C E D F G A B C D E F G STRICT COMPONENT CONSENSUS TREE Majority-rule consensus methods Majority-rule consensus methods require agreement across a majority of the fundamental trees May include relationships that are not supported by the MP tree This method produces a consensus tree that includes all and only those full splits found in a majority (>50%) of the fundamental trees Other relationships are shown as unresolved polytomies Of particular use in bootstrapping 2

3 Majority rule consensus THREE FUNDAMENTAL TREES A B C D E F G A B C E F D G A B C E D F G A B C E D F G Numbers indicate frequency of clades in the fundamental trees MAJORITY-RULE COMPONENT CONSENSUS TREE Reduced consensus methods TWO FUNDAMENTAL TREES A B C D E F G A G B C D E F A B C D E F A B C D E F G STRICT REDUCED CONSENSUS TREE Taxon G is excluded Strict component consensus completely unresolved 3

4 Parsimonious Character Optimization 1 => 0 origin and reversal (ACCTRAN) A B C D E * = 0 => 1 * = OR parallelism 2 separate origins 0 => 1 (DELTRAN) Homoplastic characters often have alternative equally parsimonious optimizations Commonly used varieties are: ACCTRAN - accelerated transformation DELTRAN - delayed transformation Consequently, branch lengths are not always fully determined PAUP reports minimum and maximum branch lengths Questions History? India Sri lanka 4

5 Questions History? India Sri lanka Missing data Missing data is ignored in tree building but can lead to alternative equally parsimonious optimizations in the absence of homoplasy single origin 0 => 1 on any one of 3 branches 1?? 0 0 A B C D E * * * Abundant missing data can lead to multiple equally parsimonious trees. This can be a serious problem with morphological data but is less likely to arise with molecular data 5

6 Maximum Likelihood Maximum Likelihood To estimate the probability that we would observe a particular dataset, given a phylogenetic tree and some notion of how the evolutionary process worked over time. Ï a b c d ) b a e f Ì c e a g Ó d c f a Probability of given (p = [ a,c,g,t] 6

7 What is the probability of observing a datum? If we flip a coin and get a head and we think the coin is unbiased, then the probability of observing this head is 0.5. If we think the coin is biased so that we expect to get a head 80% of the time, then the likelihood of observing this datum (a head) is 0.8. Therefore: The likelihood of making some observation is entirely dependent on the model that underlies our assumption. p =? Lesson: The datum has not changed, our model has. Therefore under the new model the likelihood of observing the datum has changed. What is the probability of observing a 'G' nucleotide? Model 1: frequency of G = 0.4 => likelihood(g) = 0.4 Model 2: frequency of G = 0.25 => likelihood(g) = 0.25 One rule the rule of 1. The sum of the likelihoods of all the possibilities will always equal 1. E.g. for DNA p(a)+p(c)+p(g)+p(t)=1 7

8 What about longer sequences? If we consider a gene of length 2: Gene 1: ga The the probability of observing this gene is the product of the probabilities of observing each character. E.g p(g) = 0.4; p(a)=0.15 (for instance) likelihood(ga) = 0.4 x 0.15 = 0.06 or even longer sequences? Gene 1: gactagctagacagatacgaattac Model (simple base frequency model): p(a)=0.15; p(c)=0.2; p(g)=0.4; p(t)=0.25; (the sum of all probabilities must equal 1) Like(Gene 1) =

9 Note about models You might notice that our model of base frequency is not the optimal model for our observed data. If we had used the following model: p(a)=0.4; p(c) =0.2; p(g)= 0.2; p(t) = 0.2; The likelihood of observing the gene is: Like(gene 1) = (a value that is almost 10,000 times higher) Lesson: The datum has not changed, our model has. Therefore under the new model the likelihood of observing the datum has changed. How does this relate to phylogenetic trees? Consider an alignment of two sequences: Gene 1: gaac Gene 2: gacc We assume these genes are related by a (simple) phylogenetic tree with branch lengths. 9

10 Increase in model sophistication It is no longer possible to simply invoke a model that encompasses base composition, we must also include the mechanism of sequence change and stasis. There are two parts to this model - the tree and the process (the latter is confusingly referred to as the model, although both parts really compose the model). Note: We will stay with the confusing notation - to avoid further confusion. The model The two parts of the model are the tree and the process (the model). The model is composed of the composition and the substitution process -rate of change from one character state to another character state. Model = + Ï a b c d b a e f Ì c e a g Ó d c f a [ ] p = a,c,g,t 10

11 Simple time-reversible model A simple model is that the rate of change from a to c or vice versa is 0.4, the composition of a is 0.25 and the composition of c is 0.25 (a simplified version of the Jukes and Cantor 1969 model) Ï Ì.... Ó.... [ ] P = p = Probability of the third nucleotide position in our current alignment p(a) =0.25; p(c) = 0.25; p a Æc = 0.4 Starting with a, the likelihood of the nucleotide is 0.25 and the likelihood of the substitution (branch) is 0.4. So the likelihood of observing these data is: *Likelihood(D M) = 0.25 x 0.4 =0.01 Note: you will get the same result if you start with c, since this model is reversible *The likelihood of the data, given the model. 11

12 Substitution matrix For nucleotide sequences, there are 16 possible ways to describe substitutions - a 4x4 matrix. Ï a b c d e f g h P = Ì i j k l Ó m n o p Convention dictates that the order of the nucleotides is a,c,g,t Note: for amino acids, the matrix is a 20 x 20 matrix and for codon-based models, the matrix is 61 x 61 Substitution matrix - an example Ï P = Ì Ó In this matrix, the probability of an a changing to a c is 0.01 and the probability of a c remaining the same is 0.979, etc. Note: The rows of this matrix sum to 1 - meaning that for every nucleotide, we have covered all the possibilities of what might happen to it. The columns do not sum to anything in particular. 12

13 To calculate the likelihood of the entire dataset, given a substitution matrix, base composition and a branch length of one "certain evolutionary distance" or "ced" Gene 1: ccat Likelihood of Gene 2: ccgt given Ï P = Ì Ó π=[0.1,0.4,0.2,0.3] Likelihood of a two-sequence alignment. ccat ccgt p c P c-> c p c P c ->c p a P a-> g p t P t-> t =0.4x0.983x0.4x0.983x0.1x0.007x0.3x0.979 = Likelihood of going from the first to the second sequence is

14 Different Branch Lengths For very short branch lengths, the probability of a character staying the same is high and the probability of it changing is low (for our particular matrix). For longer branch lengths, the probability of character change becomes higher and the probability of staying the same is lower. The previous calculations are based on the assumption that the branch length describes one Certain Evolutionary Distance or CED. If we want to consider a branch length that is twice as long (2 CED), then we can multiply the substitution matrix by itself (matrix 2 ). 2 CED model Ï P = Ì Ó = X Ï P = Ì Ó È Í Í Í Í Í Î Which gives a likelihood of Note the higher likelihood 14

15 For 3 CED È Í Í P 3 = Í Í Í Î This gives a likelihood of Note that as the branch lengths increase, the values on diagonal decrease and the values on the off-diagonals increase. For higher values of CED units L i k e l i h o o d Branch Length 15

16 Likelihood of the alignment at various branch lengths ccat ccgt The maximum likelihood value is at a branch length of

17 The evolutionary revolution Organisms share a common ancestry and our classification should reflect these histories (Darwin) Philosophy and methodology for reconstructing evolutionary history - cladistics (Hennig) Philosophical nature of natural groups (Ghiselin, Hull) A nomenclatural system adapted to phylogenetic systematics (Ghiselin, Griffiths, de Queiroz & Gauthier, etc) Content Ancestry 17

18 Bryant & Cantino (2002) claim that Traditional taxonomists tend to conceptualize taxa in terms of content. Proponents of the phylocode tend to conceptualize taxa in terms of ancestry. Definitions or Fixing the reference Name Name A B C A B C A B Name C x node stem apomorphy Node based: Name refers to the least inclusive clade comprising B and C Stem based: Name refers to the most inclusive clade comprising B and C, but not A. Apomorphy based: Name refers to all taxa descending from the first ancestor possessing apomorphy x. 18

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