CPSC 536A Notes for 01 Feb 6

Parsimony

• The parsimony approach to phylogenetic tree reconstruction is to find a tree, T, that explains the data with the fewest mutations(shown on edges in the example below).


• In the tree, nodes are labeled by sequences of character-based data, each with the same length m (we'll assume that these are DNA sequences with no gaps).
• Leaves are labeled by input seqs
• The tree score, S(T) is

Parsimony problem

• given n sequences, find tree T (topology and node labels) with n leaves labeled by n seqs, that minimizes the score S(T)

Small parsimony problem

• given tree topology and leaf labels, but not internal node labels, find best internal node labels (i.e. best score) for that tree
• thus, for positions j, up to length m, and edges u->v the tree score is given by: where I_j,u,v = {1  if u_j not equal v_j; 0 otherwise}

Example - focus on m=1 case, i.e. sequences 1 position long
score is 1 for this tree.

• It would be nice to be able to find an optimal solution to the small parsimony problem, for a given input tree, by recursively solving the problem on subtrees of the input tree. At first glance, this appears not to be possible: in the above example, the left subtree of the root could be labeled "A" and still yield an optimal score for that subtree, but this would not yield an optimal score for the whole tree. However, by keeping track of the set of optimal labels for internal nodes, an efficient recursive solution is possible. Here is an example:


• At the root, use the intersection of its child nodes, or the union of the child nodes if the intersection is empty.
• We can express this algorithm recursively

Fitch's Algorithm

• Fitch's algorithm is based on this idea. With respect to a (fixed) input tree, this algorithm takes as input a node u of the tree and outputs a pair (R,C), where R is the set of bases that can label u in an optimally scoring tree rooted at u and C is the score (or cost) of such an optimally scoring tree. Given u as input, the algorithm is as follows:
• if u is a leaf, output ({label of u}, Cost 0)

else let v, w be u's children.
call the algorithm recursively on v, w to obtain (Rv,Cv) and (Ru,Cu)
if Rv intersect Rw is not empty then output ({Rv intersect Rw}, Cost Cv+Cw)
else output  ({Rv union Rw}, Cost Cv+Cw+1)

Exercise:

use Fitch's algorithm to generate the sets and total cost of 4 as shown.

Large parsimony problem

• given sequence data, find minimum-score tree with leaves labeled by seqs. Unfortunately, this problem is NP-complete. We describe two methods for exploring the space of possible solutions, in order to find a good (if not optimal) solution.

Branch and Bound method

• In this method, all possible solution trees are examined in a systematic way (if the algorithm runs long enough). As trees are considered, the score of the best possible tree found so far is recorded. This score enables the method to bound the search, by not explicitly examining some trees whose score is larger than the current best bound found so far. On "lucky" runs, the bounding can sometimes prune the search space dramatically so that the algorithm halts with the optimal solution. More typically, however, one can expect that the algorithm will not halt in the time alloted, in which case the best solution found by the algorithm may not be optimal.

Nearest Neighbour Interchange

• This is a method for exploring part of the search space (i.e. set of all possible phylogenetic trees consistent with the data) via a neighbourhood structure on the set of possible solutions.   A neighbour of a tree in the search space may be defined as a tree obtained by swapping two subtrees of the tree. A simple implementation of nearest neighbour interchange moves through the neighbor structure by choosing a neighbour of the current tree at each step that has the lowest cost, and stops when a tree is reached whose neighbours all have higher cost. More sophisticated methods use a probabilistic approach to choice of neighbour, choosing neighbours that have poorer scores with low probability in order to avoid being trapped at a local optimum.

Maximum Likelihood

• Assume the data, D, was generated according to some probablistic model M.
• Find the tree T that best explains the data with respect to M, i.e. maximizes the probability P[ D | T,M ]

Method

• each sequence position is independent
• independence between branches i.e. what's going on down one branch of the tree doesn't affect a sibling branch.

Jukes-Cantor model

• parameter m is the expected # mutations at a site in 1 unit of time.
• for a sequence x mutating into a sequence y, positions x_i = x₁...x_m, y_i = y₁..y_m.

then probability P(y_i | x_i, t ) = ¼(1 - e ^-4mt/3); y_i  not equal x_i.
• Given a model M to define the likelihood of a tree, the tree edges are labeled with some measure of evolutionary distance.
• Sequences are the leaves


then probability of the tree is Sum over all u and all w of P(u)P( w | u,t2 )P( v | u,t1 )P( x | w,t4 )P( y | w,t3 )

....more on this next day.