Bioinformatics Algorithms and Data Structures

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Bioinformatics Algorithms and Data Structures

• Chapter 17.4-6 Strings and Evolutionary Trees
• Lecturer Dr. Rose
• Slides by Dr. Rose
• April 17, 2003

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Ultrametric Problem Centrality

• Four related tree problems
• Ultrametric
• Additive
• Binary perfect phylogeny
• Tree compatibility
• All can be solved as ultrametric tree problems.
• Recall tree compatibility reduces to perfect
  phylogeny.
• Now we reduce additive tree (binary) perfect
  phylogeny problems to the ultrametric tree
  problem.

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Ultrametric Problem Additive Trees

• Goal reduce additive tree problem to ultrametric
  problem
• Complexity O(n2) reduction
• Approach create a matrix D? that is ultrametric
  ? D is additive.
• We will start by describing a reduction that
  involves a tree T for D and T? for D?.
• We will then describe a direct reduction of D to
  D?.

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Ultrametric Problem Additive Trees

• Assume that D is additive.
• Assume that we know of an additive tree T for D
• Assume that each of the n taxa in D labels a leaf
  of T.
• Idea label the nodes of T to create an
  ultrametric tree T?.
• Q How can we do this?

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Ultrametric Problem Additive Trees

• A we will do the following
• Select one node as the root
• Stretch the leaf edges so that they are
  equidistant from the root.
• Let v be the row of D containing the largest
  entry.
• Let mv denote the value of this entry.
• Select node v as the root of T.
• This creates a directed tree.

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Ultrametric Problem Additive Trees

• Example
• A is the row of D containing the largest entry.
• Select node A as the root of T.

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Ultrametric Problem Additive Trees

• Stretch leaf edges
• for each leaf i, add mA D(A, i) to the leaf
  edge.
• Leaf edges are now equidistant from A.

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Ultrametric Problem Additive Trees

• The resulting tree T? is
• a rooted edge-weighted tree
• distance mv from root to every leaf
• each internal node is equidistant to leaves in
  its subtree.

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Ultrametric Problem Additive Trees

• Since each internal node is equidistant to the
  leaves in its subtree
• Label each internal node by this unique distance.
• These labels can be used to define an ultrametric
  matrix D?.
• D?(i, j) is the label at the least common
  ancestor of leaves i and j in T?.
• Q How can we go directly from matrix D to matrix
  D? without involving T and T??

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Ultrametric Problem Additive Trees

• Consider leaves i j in T
• Let node w be their least common ancestor
• Let x be the distance from the root v to w.
• Let y be the distance from node w to leaf i.

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Ultrametric Problem Additive Trees

• Q What is the distance from w to i in T??
• A y mv - D(v, i) in T?.
• Q Where does mv - D(v, i) come from?
• A Recall we add mv - D(v, i) to stretch the leaf
  edges.

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Ultrametric Problem Additive Trees

• Gusfield presents the following lemma
• Without knowing T or T explicitly, we can deduce
  that D(i, j) mv (D(i, j) - D(v, i) - D(v,
  j))/2
• Q Is this equation correct?
• D(i, j) mv ((y z) - (x y) - (x z))/2 ?
• D(i, j) mv -2x/2 ?
• Should it instead be
• D(i, j) 2mv D(i, j) - D(v, i) - D(v, j)?
• i.e., D(i, j) 2mv - 2x?
• Probably, but it is not necessary for the
  reduction (slide 9)

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Ultrametric Problem Additive Trees

• This brings us to the following Theorem
• If D is an additive matrix, then D is
  ultrametric, where D(i, j) mv (D(i, j) -
  D(v, i) - D(v, j))/2
• Proof. Weve shown that
• D(i, j) y mv - D(v, i)
• y D(v, i) x
• x (D(v, i) D(v, j) - D(i, j))/2
• Putting it altogether establishes the equation in
  the theorem.
• D satisfies the ultrametric requirement.

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Ultrametric Problem Additive Trees

• Q What is the value of y?
• A y D(v, i) - x.
• Q What is the value of x in terms of values in
  D?
• A x (D(v, i) D(v, j) - D(i, j))/2

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Ultrametric Problem Additive Trees

• So D additive ? D ultrametric
• By contraposition ?D ultrametric ? ?D additive
• Q does D ultrametric ? D additive?
• A Theorem D ultrametric ? D additive
• Proof. (constructive)
• Let T be the ultrametric tree for D
• Assign weights to edges of T
• Note the sum of edges from a leaf to an
  ancestor must match the ancestors label.
• For each edge (p, q), assign the weight p-q

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Ultrametric Problem Additive Trees

• Assign weights to edges of T continued
• Note the path distance between leaves (i, j) is
  twice the value labeling the least common
  ancestor
• Hence, 2D(i, j) 2mv D(i, j) - D(v, i) - D(v,
  j)
• Now shrink the edge into each leaf i by mv - D(v,
  i)
• The path from leaf i to leaf j is now D(i, j)
• The result is an additive tree for matrix D from
  Ds ultrametric tree.
• Putting all of this together results in a method
  for contructing and additive tree for an additive
  matrix.

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Ultrametric Problem Additive Trees

• Additive Tree Algorithm
• Create matrix D from D.
• Create ultrametric tree T from D
• Create T from T
• Label edge (p, q) with the value p-q
• For each leaf i, shrink the leaf edge by mv -
  D(v, i)
• Note no step takes more than O(n2) time.
• Thm. An additive tree for an additive matrix can
  be constructed in O(n2) time.

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Ultrametric Problem Additive Trees

• Example Given D, first find D
• Recall D(i, j) mv (D(i, j) - D(v, i) - D(v,
  j))/2

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Ultrametric Problem Additive Trees

• Example From D find T
• Recall label edge inner edges (p, q) by p-q

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Ultrametric Problem Additive Trees

• Example From T find T
• Recall shrink leaf edge i by mv - D(v, i)

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Ultrametric Problem Additive Trees

• Example Finally compare the derived T with the
  original tree as a sanity check.

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Ultrametric Problem Perfect Phylogeny

• We now recast perfect phylogeny in terms of an
  ultrametric tree problem.
• Defn. DM the n by n matrix of shared characters
• More formally
• Given the n by m character matrix M, define the n
  by n matrix DM for each pair of objects, set
  DM(p, q) to be the number of characters that p
  and q both possess.

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Ultrametric Problem Perfect Phylogeny

• Lemma If M has a perfect phylogeny, then DM is a
  min-ultrametric matrix.
• Proof convert Ms perfect phylogeny T to a
  min-ultrametric tree for DM
• Let T be the perfect phylogeny for M.
• Label Ts root be zero.
• Traverse T from top to bottom, for each node v
• Let pv be the number labeling node vs parent.
• Let ev be the of characters labeling the edge
  into v.
• Label node v with the sum pv ev

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Ultrametric Problem Perfect Phylogeny

• The label of node v is the number of characters
  common to all leaves in the subtree rooted at v.
• if v is the immediate parent of leaves p and q,
  then the label of v is DM(p, q)
• The numbers labeling nodes on any path from the
  root are strictly increasing.
• The result is an ultrametric tree for matrix DM.

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Ultrametric Problem Perfect Phylogeny

• Algorithm perfect phylogeny via ultrametrics
• Create matrix DM from M.
• Attempt to create a min-ultrametric tree T from
  DM. If not possible, then M has no perfect
  phylogeny.
• If T was successfully created in step 2
• Attempt to label its edges with the m characters
  of M.
• If not possible, then M has no perfect phylogeny.
• O/w the modified T is the perfect phylogeny T.
• Note T may be min-ultrametric but M may not
  have a perfect phylogeny, hence the check in step
  3

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Ultrametric Problem Perfect Phylogeny

• Final notes on the centrality ultrametric
  problem.
• We can see that the following problems
• perfect phylogeny
• tree compatibility
• can be cast as ultrametric problems.
• This is not an efficient way to address these
  problems.

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Maximum Parsimony

• Maximum parsimony
• Perfect phylogeny is a special instance
• Can be viewed as a Steiner tree problem on a
  hypercube
• Presentation Approach
• Introduce Steiner trees
• Hypercube graphs
• Maximum parsimony as a Steiner tree problem

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Maximum Parsimony

• Definitions
• Let N be a set of nodes
• Let E be a set undirected edges with non-negative
  weight
• Let G (N, E) be an undirected graph
• Let X ? N be a subset of nodes.
• A Steiner tree ST for X is any connected subtree
  of G that contains all nodes of X and possibly
  nodes in N-X.
• Weighted Steiner Tree Problem Given G and X,
  find the Steiner tree of minimum total weight.

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Maximum Parsimony

• More Definitions
• A hypercube of dimension d is an undirected graph
  with 2d nodes, labeled 0..2d-1. Adjacent nodes
  differ in only one label bit position.
• The weighted Steiner tree problem on hypercubes
  G must be a hypercube.

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Maximum Parsimony

• More Definitions
• Maximum Parsimony Occams razor applied to
  phylogenetic reconstruction. A preference for
  trees requiring fewer evolutionary events to
  explain data.
• Gusfields definition
• The Maximum Parsimony problem is the unweighted
  Steiner tree problem on a d-dimensional hypercube.

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Maximum Parsimony

• More about the hypercube formulation of MP
• The X input taxa are described as d-length binary
  vectors.
• Recall adjacent nodes differ in only one label
  bit position.
• Correspondingly, taxa that differ by a single
  mutation will be adjacent.
• ?? Steiner tree of X nodes and l edges iff ? a
  corresponding phylogenetic tree that entails l
  character-state mutations.

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Steiner interpretation of Perfect Phylogeny

• Define a nontrivial binary character to be a
  character contained by some taxa but not all.
• Consider an MP dataset of d nontrivial binary
  characters
• Q what is the minimal number of mutations in the
  MP tree?
• A at least d.

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Steiner interpretation of Perfect Phylogeny

• Q What is the relation to binary perfect
  phylogeny?
• A the binary perfect phylogeny problem is
  equivalent to asking if there is an MP solution
  with a cost of exactly d.
• Q What about generalized perfect phylogeny?
• A Its similar. The lower bound must reflect
• the number of character states in the input taxa.
• a character having r states in the input taxa is
  allowed only r-1 transitions.

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Steiner interpretation of Perfect Phylogeny

• Complexity
• No known efficient solution for Steiner tree
  problem on unweighted graphs.
• Polynomial time solution for generalized perfect
  phylogeny problem when r is fixed.
• ? this particular Steiner tree problem can be
  answer in polynomial time.

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Steiner interpretation of Perfect Phylogeny

• MP approximations
• The weighted Steiner tree problem on hypercubes
  is NP-hard.
• There is an approximate method with an error
  bound of a factor of 11/6.
• Also MST can be used to find a Steiner tree with
  weight less than twice the optimal Steiner tree.

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Phylogenetic Alignment

• Recall
• phylogenetic alignment was discussed in section
  14.8
• The focus was on deriving a multiple alignment
  enlightened by evolutionary history.
• The tree focused emphasis on specific alignment
  groupings
• Internal node sequences were a secondary artifact

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Phylogenetic Alignment

• Phylogenetic alignment as a parsimony problem
• In contrast
• we are now interested in the internal sequences
• These sequences are waypoints in the evoutionary
  trajectory leading to the extant taxa
• phylogenetic alignment is thus a parsimony problem

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Phylogenetic Alignment

• Hypothesis optimal phylogenetic alignment
  describes evolutionary history.
• Assumptions
• Edit distance realistically models evolutionary
  distance
• Globally optimal phylogenetic alignment captures
  essence of the evolutionary process
• We will look at minimum mutation, a variant of
  phylogenetic alignment

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Fitch-Hartigan minimum mutation problem

• Defn. minimum mutation problem variant of
  phylogenetic alignment problem.
• Input comprised of
• Tree
• Strings labeling the leaves
• A multiple alignment of those strings

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Fitch-Hartigan minimum mutation problem

• Q If you are given the tree and the multiple
  alignment, what is left to compute?
• A the mutations that accounts for the input
  data.
• These mutations should be
• minimum sequence of site mutations that is
• compatible with the given tree and
• the given multiple alignment.

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Fitch-Hartigan minimum mutation problem

• Q How is the input data used to determine the
  minimum sequence of mutations?
• The multiple alignment associates each amino acid
  with a specific position.
• The evolutionary history of the sequences is then
  treated as a combined but independent
  evolutionary history of each position.
• The tree guides the order of mutations for each
  position.

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Fitch-Hartigan minimum mutation problem

• Assumptions
• Each column of the alignment can be solved
  separately
• The strings labeling inner nodes adhere to the
  same alignment
• The problem reduces to a computation at a single
  position.

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Fitch-Hartigan minimum mutation problem

• Minimum mutation for a single position
• Input
• rooted tree with n nodes
• Each leaf is labeled by a single character
• Output
• Each interior node is labeled by a single
  character
• The labeling minimizes the number of edges
  between nodes with different labels.

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Fitch-Hartigan minimum mutation problem

• Algorithmic approach Dynamic Programming
• Let Tv denote the subtree rooted at node v
• Let C(v) be the cost of the optimal solution for
  Tv
• Let C(v, x) be the cost when v must be labeled by
  x
• Let vi denote the ith child of node v
• Base case for each leaf specify C(v) C(v, x)
  ?x ? S.
• C(v) 0 C(v, x) 0 if leaf v is labeled by x.
• C(v, x) ? if leaf v is not labeled by x.

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Fitch-Hartigan minimum mutation problem

• When v is an internal node

• The recurrence relations start from the base
  cases.
• Bottom up from leaves
• Backtracking is used to after all C(v,x) computed
  to extract the solution.

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Fitch-Hartigan minimum mutation problem

• Backtracking process
• The root is labeled by the character x s.t. C(r)
  C(r,x)
• The traversal is then top-down
• If v is labeled x, then vi is labeled
• character x if C(vi) 1 gt C(vi,x)
• o/w character y such that C(vi) C(vi,y)

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Fitch-Hartigan minimum mutation problem

• Lets evaluate an example
• C(v) 0 C(v, x) 0 if leaf v is labeled by x,
  o/w C(v, x) ? if leaf v is not labeled by x.

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Fitch-Hartigan minimum mutation problem

• Time complexity
• Bottom-up portion
• Let s S
• Each node is evaluate wrt each x ? S.
• For n nodes this gives O(ns)
• The backtracking portion is O(n)
• Overall O(ns)

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Maximum Parsimony

• Most widely used tree building algorithm
• Differs from distance-based algorithms
• Does not actually build trees from distances
• Parsimony is used to compute the cost of a tree
• A search strategy is used to search through all
  topologies
• Goal find the tree topology with the overall
  minimum cost

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Traditional Parsimony

• Algorithm Traditional parsimony Fitch 1971
• Goal count the number of substitutions at a
  site.
• Method recursion, keeping track of
• C, the current cost
• Rk, the residues at k, the current node

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Traditional Parsimony

• Algorithm Traditional parsimony Fitch 1971
• C 0, k root / initialize the cost and
• TP(k)
• If k is a leaf then return xk
• Rleft TP( k.left)
• Rright TP(k.right)
• if Rleft ? Rright ? ? return Rleft ? Rright
• else
• C C 1
• return Rleft ? Rright

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Traditional Parsimony

• Lets evaluate an example
• if Rleft ? Rright ? ? return Rleft ? Rright
• else C C 1, return Rleft ? Rright

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Traditional Parsimony

• There is a traceback procedure for finding
  ancestral assignments.
• Q How do you think the traceback works?
• A Start from the root
• Pick a residue
• Pick the same residue for each child set if
  possible
• If a child set does not contain the parents
  residue, randomly select a residue from its set.

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Traditional Parsimony

• Lets perform the traceback on our example