Lecture 6A – Introduction to Trees & Optimality Criteria Branches: n-taxa -> 2n-3 branches 1, 2, 4, 6, & 7 are external (leaves) 3 & 5 are internal branches (edges) Nodes A – E are terminals x, y, & z are internal (vertices)

If we break branch 3, we have two sub-trees: (A,B) and (C,(D,E)). Newick Format ((A,B),C,(D,E)). If we break branch 3, we have two sub-trees: (A,B) and (C,(D,E)).

Rooting – The tree is an unrooted tree.

Also note that there is free rotation around nodes: (1 2 3 4) (1 2 3 4) (1 2 3) (1 2)

Growth of tree space.

The Scope of the Problem Taxa Unrooted Trees 3 1 4 3 5 15 6 105 7 945 8 10,395 9 135,135 10 2.027 X 106 22 3 X 1023 50 3 X 1074 100 2 X 1082 1000 2 X 102,860 10 mil 5 X 1068,667,340

II. Optimality Criteria A. Parsimony First, the score of a tree (i.e., its length) for the entire data set is given by: li is the length of character i when optimized on tree t. wi is the weight we assign to character i.

The Fitch Algorithm (1971): state sets and accumulated lengths. (Unordered states with equal transformation costs) We erect a state set at each terminal node and assign an accumulated length of zero to terminal nodes. This is the minimum number of changes in the daughter subtree.

The Fitch Algorithm: state sets and accumulated lengths. 1 – Form the intersection of the state sets of the two daughter nodes. If the intersection is non-empty, assign the set for the internal node equal to the intersection. The accumulated length of the internal node is the sum of those of the daughter nodes. 2 – If the intersection is empty, we assign the union of the two daughter nodes to the state set for the internal node. The accumulated length is the sum of those of the daughter nodes plus one. empty Union: 0+0+1=1 non-empty Intersection: 0+0+0=0 empty Union: 1+0+1=2 So li = 2

Sankoff Algorithm – Character-state vectors and step matrices. Step Matrix – define ci,j   A C G T A -- 4 1 4 C 4 -- 4 1 G 1 4 -- 4 T 4 1 4 -- Step one: Fill in the character-state vectors for terminal nodes. Each cell is indexed by sk(i), the cost of having state i at node k.

A C G T A -- 4 1 4 C 4 -- 4 1 G 1 4 -- 4 T 4 1 4 -- Step two: Fill in vectors for other nodes, descending tree. Node 1 (k = 1): Node 2 (k = 2): s1(A) = cAG + cAA = 1 + 0 = 1, s2(A) = 4 + 4 = 8 s1(C) = cCG + cCA = 4 + 4 = 8, s2(C) = 0 + 0 = 0 s1(G) = cGG + cGA = 0 + 1 = 1, s2(G) = 4 + 4 = 8 s1(T) = cTG + cTA = 4 + 4 = 8 s2(T) = 1 + 1 = 2

For nodes below, we must calculate the cost for each possible state assignment for daughter nodes. From daughter node 1 From step matrix s3(A) = min[s1A + cAj] + min[s2A + cAj] = min[1,12,2,12] + min[8,4,9,6] = 1+4 = 5 s3(C) = min[s1C + cCj] + min[s2C + cCj] = min [5,8,5,9] + min[12,0,12,3] = 5+0 = 5 5 = min [2,12,1,12] + min[9,4,8,6] = 1+4 = 5 5 = min [5,9,5,8] + min[12,1,12,2] = 5+1 = 6 6 5 s3(G) = min[s1G + cGj] + min[s2G + cGj] A C G T A -- 4 1 4 C 4 -- 4 1 G 1 4 -- 4 T 4 1 4 -- s3(T) = min[s1T + cTj] + min[s2T + cTj] So we fill in the character-state vector for node 3:

So, li = 5 Points to note: 1) Two types of weighting are possible: weighting of transformations within characters (which we demonstrated with the step matrix) and weighting among characters, which are reflected in the weighted sum of lengths across characters (wi). 2) One can’t compare tree lengths across weighting schemes. In the first example, with all transformations having the same cost, the length of the character on this tree was 2. In the second, with a 4:1 step matrix to weight transversions, the length was 5.