CPM '05 Sensitivity Analysis for Ungapped Markov Models of Evolution David Fernández-Baca Department of Computer Science Iowa State University (Joint work with Balaji Venkatachalam)

CPM '05 Motivation Alignment scoring schemes are often based on Markov models of evolution Optimum alignment depends on evolutionary distance Our goal: Understand how optimum alignments are affected by choice of evolutionary distance

CPM '05 Ungapped local alignments Only matches and mismatches — no gaps An ungapped local alignment of sequences X and Y is a pair of equal-length substrings of X and Y X Y

CPM '05 Ungapped local alignments P. Agarwal and D.J. States. Bayesian evolutionary distance. Journal of Computational Biology 3(1):1—17, matches 2 mismatches 34 matches 11 mismatches A:A: B:B:

CPM '05 Which alignment is better? Score =  ∙ #matches +  ∙ #mismatches In practice, scoring schemes depend on evolutionary distance score(B) // -11/9 score(A) > 0< 0

CPM '05 Log-odds scoring Let q X =  base frequency of nucleotide X m XY (t) =  Prob(X  Y mutation in t time units) A be an alignment X 1 X 2 X 3  X n Y 1 Y 2 Y 3  Y n Then, Log odds score of A =

CPM '05 Log-odds scoring Simplest model: –m XX (t) = r(t) for all X –m XY (t) = s(t) for all X  Y –q X = ¼ for all X Log-odds score of alignment:  (t) ∙ #matches +  (t) ∙ #mismatches where  (t) = 4 + log r(t)  (t) = 4 + log s(t)

CPM '05 Scores depend nonlinearly on evolutionary distance

CPM '05 This talk An efficient algorithm to compute optimum alignments for all evolutionary distances Techniques –Linearization –Geometry –Divide-and-conquer

CPM '05 Related Work Combinatorial/linear scoring schemes: –Waterman, Eggert, and Lander 1992: Problem definition –Gusfield, Balasubramanian, and Naor 1994: Bounds on number of optimality regions for pairwise alignment –F-B, Seppäläinen, and Slutzki 2004: Generalization to multiple and phylogenetic alignment Sensitivity analysis for statistical models: –P. Agarwal and D.J. States 1996 –L. Pachter and B. Sturmfels 2004a & b: connections between linear scoring and Markov models

CPM '05 A simple Markov model of evolution Sites evolve independently through mutation according to a Markov process For each site: –Transition probability matrix: M = [m ij ], i, j  {A, C, T, G} where m ij = Prob(i  j mutation in 1 time unit) –Transition matrix for t time units is M (t)

CPM '05 Jukes-Cantor transition probability matrix where

CPM '05    versus  t = +∞ t = 0  (t) = 4 + log r(t)  (t) = 4 + log s(t)

CPM '05 Linearization Allow  and  to vary arbitrarily, ignoring that they –are functions of t and –must satisfy laws of probability Result is a linear parametric problem Recall: Score(A) =  ∙ #matches +  ∙ #mismatches

CPM '05 Theorem (ii) The parameter space decomposition looks like this: Let n be the length of the shorter sequence. Then,   (i) The number of distinct optimal solutions over all values of  and  is O(n 2/3 ).

CPM '05 Re-introducing distance The  vs.  curve intersects every boundary line with slope  (-∞, +1] The optimum solutions for t = 0 to +  are found by varying  /  from -  to 1 Non-linear problem in t reduces to a linear one-parameter problem in  / 

CPM '05 An algorithm 1.Start with a simple, but highly parallel, algorithm for fixed-parameter problem 2.Lift the fixed-parameter algorithm Lifted algorithm runs simultaneously for all parameter values in linearized problem Output: A decomposition of parameter space into optimality regions 3.Construct solution to original problem by finding the optimality regions intersected by the  (t),  (t) curve

CPM '05 A naïve dynamic programming algorithm Let C be the matrix where C ij = score of opt alignment ending at X i and Y j Subdiagonals correspond to alignments Diagonals are independent of each other –Process each diagonal separately –Pick best answer over all diagonals Total time: O(nm) caatttgtcacttttt... C aattcaattcaatc... X Y

CPM '05 Divide and conquer for diagonals Split diagonal in half, solve each side recursively, and combine answers. E.g.: X Y Y (1) X (1) X (2) Y (2) Y (1) X (1) X (2) Y (2) Y (1) X (1) X (2) Y (2) T(N) = 2 T(N/2) + O(1)  T(N) = O(N) length of diagonal #subproblems

CPM '05 Lifting Run naïve DP algorithm for all parameter values by manipulating piecewise linear functions instead of numbers: –“+”  “+” for piecewise linear functions –“max”  “max” of piecewise linear functions

CPM '05 Adding piecewise linear functions f g f + g Time = O(total number of segments)

CPM '05 Computing the maximum of piecewise linear functions f g max (f,g) Time = O(total number of segments)

CPM '05 Analysis Processing a diagonal: –T(n) = 2 T(n/2) + O(n 2/3 )   T(n) = O(n) Merging score functions for diagonals: –O(n 2/3 ) line segments per function, m+n-1 diagonals –Total time:O(mn + mn 2/3 lg m) #(optimum solutions for diagonal)

CPM '05 Further Results (1): Parametric ancestral reconstruction Given a phylogeny, find most likely ancestors AATACTAGC AAT AAC Sensitive to edge lengths Result: O(n) algorithm for uniform model (all edge lengths equal)

CPM '05 Further Results (2) Bounds on number of regions for gapped alignment (indels are allowed) –Lead to algorithms, but not as efficient as ungapped case

CPM '05 Open Problems Tight bounds on size of parameter space decomposition Evolutionary trees with different branch lengths Efficient sensitivity analysis for gapped models Evaluation of sensitivity to changes in structure and parameters –Useful in branch-swapping

CPM '05 Thanks to National Science Foundation –CCR CCR –EF EF