Lecture 17 - The Comparative Method

Lecture 17 The - Lecture 17 Phylogenetic Inference con't Evaluating trees e.g if the most parsimonious tree has 500 steps and there are 500

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Lecture 17 - Phylogenetic Inference, con't - Evaluating trees e.g., if the most parsimonious tree has 500 steps and there are 500 alternative possible trees with 501 steps, you would be justified in having little confidence that the most-parsimonious tree is more likely to be the "true" tree than any of the other 500 trees. A common way to assess the robustness of a tree (or a node) is to use a statistical technique called "bootstrapping". 1. Resample the data WITH REPLACEMENT to construct a pseudo-dataset 2. Construct a tree from the pseudodata 3. Repeat a large number of times. Clades that appear in most bootstrap trees are robust (~ "likely to be true"), those that do not are not. - "Phylogenetically Informative" characters are those that can discriminate between at least two of the n possible trees by the criterion of parsimony. Example: Taxon Character A B C D 1 1 1 0 0 2 1 1 0 0 3 1 1 1 0 4 1 1 1 0 5 1 1 1 0 6 1 1 1 0 7 0 1 0 0 8 0 1 0 0 9 0 1 0 0 10 1 0 1 0 11 1 0 0 1 If we assume that "D" is the outgroup and 0 is the ancestral character state, the problem is to discriminate among the three possible three-taxon trees in the in-group. The phylogenetically informative characters are thus: 1, 2, and 10. Chars 3-6 are obviously not informative. Neither are chars 7-9, because the derived character state is unique to taxon B and there is no information that allows us to decide between the three possible trees. Similarly, char. 11 is not phylogenetically informative because, given that we "know" D is the outgroup, the derived character state in D must be homoplasy. 1
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- Phenetics Revisited (same data from the parsimony example) Taxon Number of shared character states Character A B C D A B C D 1 1 1 0 0 A - 6 7 3 2 1 1 0 0 3 1 1 1 0 B - 4 0 4 1 1 1 0 5 1 1 1 0 C - 5 6 1 1 1 0 7 0 1 0 0 D - 8 0 1 0 0 9 0 1 0 0 To construct a tree using a common phenetic algorithm, 1) Join the pair of taxa that share the most characters (A and C in this example) 2) See which pair is most similar either to the first pair or to another unjoined taxon. Similarity to interior nodes is the mean of the two comparisons. In this example the similarity of B to D is 0 and the similarity of B to the A,C pair is (6+4)/2 = 5, so the final tree is [((AC)B)D] Contrast the tree constructed by overall similarity with the tree constructed last lecture using the criteria of phylogenetic systematics (shared DERIVED characters). (Figure 4, phenetic tree), (((AC),B),D) Lecture 17.1 - The Comparative Method I. Phylogenetic inference, continued. II. The comparative method - "Comparative" biology is the practice of drawing comparisons between taxa with the goal of understanding both similarities and differences between organisms. Since Darwin, comparative biology is an explicitly evolutionary enterprise. Recall the Mission of Evolutionary Biology: Explain Adaptation
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This note was uploaded on 06/08/2011 for the course PCB 4674 taught by Professor Baer during the Fall '08 term at University of Florida.

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Lecture 17 The - Lecture 17 Phylogenetic Inference con't Evaluating trees e.g if the most parsimonious tree has 500 steps and there are 500

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