Feb12_Excercise

Feb12_Excercise - BINARY ADDITIVE AND NONADDITIVE...

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1 BINARY, ADDITIVE, AND NONADDITIVE CHARACTERS IB200a exercises* I NTRODUCTION : In this lab we will compare binary, multistate nonadditive, and multistate additive character coding. We will compare their effects on tree length and on optimization of internal nodes of the trees. We will begin by learning how to code complex-character-state hierarchies as additive binary and additive multistate characters. A DDITIVE B INARY C ODING (A.K.A. H IERARCHIC B INARY C ODING ): Additive binary coding allows any complex hierarchy to be translated into a series of binary characters. It can be utilized to transform any additive multistate character, or any tree structure, into variables that represent the nodes or transformations in the character. Although the easiest way to implement additive binary coding is by determining a "root," it should be noted that the root is actually arbitrary and will not change the results of an analysis using the additive character. To perform additive binary coding, first draw a character hierarchy, or character cladogram, in which each state is a terminal unit, or an internal node. You can draw this as an unrooted tree, and then root it directly to one of the character state terminals, or you can draw it in such a rooted manner from the start. For example: Take the character hierarchy, and make a matrix with the character states down the left side, as if they were taxa. Score each "monophyletic" group of character states with a group-membership character (Farris 1974). As you do this, work away from the root, and score each node and all of its descendants as a unique group (i.e., the descendents are all assigned a "1" and all others not in the group are assigned a "0"). Thus, each node defines a group of states above it as ones and all others as zeros. Also, code a group-membership character for each terminal character state in the tree, for which that terminal is scored "1" and all others "0" (i.e., an autapomorphic character). In the example above, this would result in the groups (B, C, D, E) and (C, D, E), as well as 3 variables for terminals D, E, and F, respectively: B C C D D E E D E F A 0 0 0 0 0 B 1 0 0 0 0 C 1 1 0 0 0 D 1 1 1 0 0
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This note was uploaded on 08/01/2008 for the course IB 200 taught by Professor Lindberg,mishler,will during the Spring '08 term at Berkeley.

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Feb12_Excercise - BINARY ADDITIVE AND NONADDITIVE...

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