Bioinformatics Algorithms and Data Structures

1
Bioinformatics Algorithms and Data Structures

• Chapter 14.6-8 Multiple Alignment
• Lecturer Dr. Rose
• Slides by Dr. Rose
• March 5, 2007

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Sum-of-Pairs

• Defn. The sum-of-pairs (SP) score of a multiple
  alignment is the sum of the score of all induced
  pairs in a global alignment.
• From the previous example
• 1 A A T - G G T T T
• 2 A A - C G T T A T
• T A T C G - A A T
• SP 4 5 4 13

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Sum-of-Pairs

• Q What theoretical justification is there for
  adopting the SP score?
• Wait for response..
• A None. Or rather none more than for any other
  multiple alignment scoring scheme.
• In practice it is a good heuristic and is
  popular.
• Q How can we compute a global alignment M using
  a minimum sum-of-pairs score?
• A Why dynamic programming of course!

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Sum-of-Pairs

• Assuming that we want to align k strings
• Q What time complexity for the DP solution?
• A Q(nk), exact SP aligment has been shown to be
  NP-complete.
• Q So what should we do?
• A Choose small a k.
• In practice, the NP-completeness of a problem
  often does not mean that the sky is falling.

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Sum-of-Pairs

• Q How will k affect the recurrence relation?
• The recurrence relation for k 3 is
• D(i, j, k) min
• D(i -1, j - 1, k - 1) ?,
• D(i -1, j - 1, k ) ?,
• D(i -1, j, k - 1) ?,
• D(i, j - 1, k - 1) ?,
• D(i -1, j , k ) ?,
• D(i, j - 1, k ) ?,
• D(i, j , k - 1) ?

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Sum-of-Pairs

• Lets consider each term of the recurrence in
  turn
• D(i -1, j - 1, k - 1) is the diagonal cell in
  all three dimensions.
• Q What should be the SP transition cost for
  D(i-1,j-1,k-1) ?D(i, j, k) ?
• Recall for k 2, if S1(i) S2(j) the cost is
  the match cost, o/w S1(i) ? S2(j) and we incur
  the mismatch cost.
• A the sum of pairwise match comparisons, i.e.,
  ij, jk, ik.

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Sum-of-Pairs

• Let m(i, j) denote the pairwise character match
  function defined as
• m(i, j) matchCost if the characters match
• m(i, j) mismatchCost if the characters mismatch
• Then the SP transition cost for D(i - 1, j - 1, k
  - 1) ?D(i, j, k) is m(i, j) m(j, k) m(i, k)
• Hence the term cost is
• D(i - 1, j - 1, k - 1) m(i, j) m(j, k)
  m(i, k)

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Sum-of-Pairs

• The next term
• D(i -1, j - 1, k ) is the diagonal cell in the
  first two dimensions.
• Q What should be the SP transition cost for
  D(i-1, j-1, k) ?D(i, j, k) ?
• We have two types of cases to consider
• The pairwise diagonal case i-1, j-1? i, j
• The two pairwise space insertion cases
• i-1, k ? i, k and j-1, k ? j, k

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Sum-of-Pairs

• The cost will be the sum of the pairwise match
  and space insertion costs.
• m(i, j) for (i-1, j-1? i, j) and
• spacecost for i-1, k ? i, k and spacecost for
  j-1, k ? j, k
• Then the SP transition cost for D(i - 1, j - 1,
  k) ?D(i, j, k) is m(i, j) 2 spacecost
• Hence the term cost is
• D(i - 1, j - 1, k) m(i, j) 2 spacecost

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Sum-of-Pairs

• Similarly, the third and fourth term costs are
• D(i - 1, j, k - 1) m(i, k) 2 spacecost,
• D(i, j - 1, k - 1) m(j, k) 2 spacecost
• Note the similarity in the fifth, sixth, and
  seventh terms
• D(i -1, j , k ) ?
• D(i, j - 1, k ) ?
• D(i, j , k - 1) ?
• Q What should be the cost for transitions from
  them?

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Sum-of-Pairs

• For D(i -1, j , k) we have two types of cases to
  consider
• The pairwise no change case j, k ? j, k
• The two pairwise space insertion cases
• i-1, j ? i, j and i-1, k ? i, k
• Then the SP transition cost for D(i - 1, j , k)
  ?D(i, j, k) is 0 2 spacecost
• Hence the term cost is
• D(i - 1, j, k) 2 spacecost

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Sum-of-Pairs

• Similarly, the sixth and seventh term costs are
• D(i - 1, j, k) 2 spacecost,
• D(i, j, k) 2 spacecost
• Hence D(i, j, k) min
• D(i -1, j - 1, k - 1) m(i, j) m(j, k) m(i,
  k),
• D(i -1, j - 1, k ) m(i, j) 2 spacecost,
• D(i -1, j, k - 1) m(i, k) 2 spacecost,
• D(i, j - 1, k - 1) m(j, k) 2 spacecost,
• D(i -1, j , k ) 2 spacecost,
• D(i, j - 1, k ) 2 spacecost,
• D(i, j , k - 1) 2 spacecost

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Sum-of-Pairs

• Q What about the boundary cells on the 3 faces
  of the table?
• D(i, j, 0),
• D(i, 0, k),
• D(0, j, k)
• Observation Each case degenerates into the
  familiar two-string alignment distance space
  costs for the empty string argument.
• Approach represent these cases in terms of
  pair-wise distance space costs.

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Sum-of-Pairs

• Let D1,2(i, j) denote the pairwise distance
  between S11..i and S21..j. D1,3(i, k) and
  D2,3(j, k) are analogously defined.
• Consider D(i, j, 0)
• D(i, j, 0) D1,2(i, j) ? spaceCost
• Q What is the space cost, i.e., how many spaces?
• A i for S1 and j for S2 hence
• D(i, j, 0) D1,2(i, j) (i j) spaceCost

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Sum-of-Pairs

• By this argument, the boundary cells are given
  by
• D(i, j, 0) D1,2(i, j) (i j) spaceCost ,
• D(i, 0, k) D1,3(i, k) (i k) spaceCost ,
• D(0, j, k) D2,3(j, k) (j k) spaceCost,
• D(0,0,0) 0

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Sum-of-Pairs Speedup

• Q How can we speedup our DP approach?
• A Use forward dynamic programming.
• Note so far we have used backward dynamic
  programming, i.e., cell (i, j, k) looks back to
  the seven cells that can influence its value.
• In contrast forward DP sends the result of cell
  (i, j, k) forward to the seven cells whose value
  it could influence.

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Sum-of-Pairs Speedup

• Q How does this speed things up?
• A it doesnt, if we always send cell (i, j, k)s
  value forward.
• The only significant way to speed up the Q(nk) is
  to avoid computing all nk cells in the DP table.
• We will use forward DP to reduce the number of
  cells that we compute in the DP table.

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Sum-of-Pairs Speedup

• Lets rethink this problem
• View the optimal alignment problem as the
  shortest path through the weighted edit distance
  graph.
• We are looking for the shortest path from (0,0,0)
  to (n,n,n).
• When node (i, j, k) is computed, we have the
  shortest path from (0,0,0) to (i, j, k).
• The value of node (i, j, k) is sent forward to
  the seven neighboring nodes that it can influence

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Sum-of-Pairs Speedup

• Let w be reached by an outgoing edge from (i, j,
  k)
• the true shortest distance from (0,0,0) to w is
  the value computed after it has been updated by
  every node with a ingoing edge to it.
• A queue is used to order the nodes for
  processing.
• The final shortest distance for the node v at the
  head of the queue is set and node v is removed.
• Every neighbor w of v is then updated, w is
  placed in the queue if it is not already there.

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Sum-of-Pairs Speedup

• At this point we borrow an A-like idea
• IF (i, j, k) is not on the shortest path from
  (0,0,0) to (n,n,n) then avoid passing its value
  forward.
• More importantly, avoid putting its neighbors,
  not already in the queue, into the queue.
• The trick is deciding (i, j, k) is not on the
  shortest path from (0,0,0) to (n,n,n).
• Q How do we pull this rabbit out of our hat?

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Sum-of-Pairs Speedup

• Define d1,2(i, j) to be the edit distance between
  suffixes S1i..n and S2j..n. Define d1,3(i, k)
  d2,3(j, k), analogously.
• Note these edit distances can be computed in
  O(n2) via DP on the reversed strings.
• Observation any shortest path from (i, j, k) to
  (n,n,n) must have distance at least d1,2(i, j)
  d1,3(i, k) d2,3(j, k)

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Sum-of-Pairs Speedup

• Suppose we have an alignment (from somewhere)
  with an SP distance score z.
• Core idea
• if D(i, j, k) d1,2(i, j) d1,3(i, k) d2,3(j,
  k) gt z, then node (i, j, k) can not be on any
  shortest path.
• Do not pass its value forward.
• Do not put its neighbors reached by outgoing
  edges onto the queue.

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Sum-of-Pairs Speedup

• Benefits of being able to prune cell (i, j, k)
• We automatically prune many of its descendants.
• We dont process all nk cells in a k-string
  problem. Big win!!!!
• The computation is still exact will find the
  optimal alignment.

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Sum-of-Pairs Speedup

• The program called MSA implements the speedup we
  are discussing.
• Cold shower
• MSA can align 6 strings with n 200
• Unlikely to be able to align tens or hundreds of
  strings.
• Still, 2006 cells ( 6.4 1013 cells), otherwise
  impossible.

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Bounded-Error Approximation for SP-Alignment

• Q Where do we get z from?
• A We will use a bounded-error approximation
  method.
• Properties of the specific method we will
  discuss
• Polynomial worst-case time complexity
• The SP-score is less than twice the optimal value.

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Bounded-Error Approximation for SP-Alignment

• Idea focus on alignments consistent with a tree.
• Q What do we mean by consistent with a tree?
• Informal explanation
• A graph edge denotes a relation between two
  nodes.
• Recall that D(Si, Sj) is the optimal weighted
  distance between Si and Sj.
• We could let D(Si, Sj) be the edge relation.

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Bounded-Error Approximation for SP-Alignment

• Informal explanation
• A graph edge denotes a relation between two
  nodes.
• Recall that D(Si, Sj) is the optimal weighted
  edit distance between Si and Sj.
• We could let D(Si, Sj) be the edge relation
  between the node labeled Si and the node labeled
  Sj.

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Bounded-Error Approximation for SP-Alignment

• Informal explanation continued
• Suppose we have a multiple alignment M.
• Suppose we construct an unrooted tree from a
  subset of such edges? between nodes labeled with
  strings from M.
• We call the alignment of the strings represented
  in the tree consistent with the tree.
• ? recall D(Si, Sj) is the edge relation.

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Bounded-Error Approximation for SP-Alignment

• Example from text
• A X X _ Z
• A X _ _ Z
• A _ X _ Z
• A Y _ _ Z
• A Y X X Z

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Bounded-Error Approximation for SP-Alignment

• Defn. More formally, let
• S be a set of distinct strings.
• T be an unrooted tree comprised of nodes labeled
  with strings from set S.
• M be multiple alignment of the strings in S.
• M is consistent with T if the induced pairwise
  alignment of Si and Sj has score D(Si, Sj) for
  each pair of strings (Si, Sj) that label adjacent
  nodes in T.

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Bounded-Error Approximation for SP-Alignment

• Thm. For any set of strings S and for any tree T
  whose nodes are labeled by distinct strings from
  set S, we can efficiently find a multiple
  alignment M(T) of S that is consistent with T.
• Proof sketch construct M(T) of S one string at a
  time.
• Base case
• Pick two strings Si and Sj labeling nodes
  adjacent in T.
• Create M2(T) a two string alignment with
  distance D(Si,Sj).

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Bounded-Error Approximation for SP-Alignment

• Inductive Hypothesis Assume the theorem holds
  for 2 lt k strings, i.e., Mk(T) is consistent with
  T.
• Inductive Step show that the theorem holds for k
  1 strings.
• Pick a string Sj not in Mk(T) such that it labels
  a node adjacent to a node labeled Si already in
  Mk(T).
• Optimally align Sj with Si? (Si with spaces in
  Mk(T)).
• Add Sj? (Sj with spaces) to Mk(T) creating
  Mk1(T).
• Look at detailed proof (pg. 348) to see how the
  issue of inserted spaces is handled.

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Bounded-Error Approximation for SP-Alignment

• By construction
• Sj and Si have distance D(Si, Sj)
• Mk1(T) is consistent with T.
• By induction, M(T) of S is consistent with T and
  is efficiently computed.

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Bounded-Error Approximation for SP-Alignment

• We need some more definitions at this point
• Defn. the center string Sc ? S, a set of k
  strings, is the string that minimizes M SSj?S
  D(Sc, Sj).
• Defn. the center star is a star tree of k nodes,
  with the center node Sc and each of the k-1
  remaining nodes labeled by a distinct string in S
  Sc.

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Bounded-Error Approximation for SP-Alignment

• Defn. the multiple alignment Mc of strings in S
  is the multiple alignment consistent with the
  center star.
• Defn. let d(Si, Sj) denote the score of the
  pairwise alignment of strings Sj and Si induced
  by Mc.
• Defn. let d(M) denote the score of the alignment
  M.
• Observations
• d(Si, Sj) ? D(Si, Sj)
• d(Mc) Siltjd(Si, Sj).

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Bounded-Error Approximation for SP-Alignment

• Defn. the triangle inequality wrt a scoring
  scheme is defined as the relation s(x, z) ? s(x,
  y) s(y, z) for any three characters x, y, and
  z.
• We can extend the triangle inequality from the
  scoring scheme for characters to string alignment.

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Bounded-Error Approximation for SP-Alignment

• Lemma. If a 2-string scoring scheme that
  satisfies the triangle inequality is used, then
  for any Si Sj
• d(Si, Sj) ? d(Si, Sc) d(Sc, Sj) D(Si, Sc)
  D(Sc, Sj)
• Proof sketch Notice that for each column we
  have
• s(x, z) ? s(x, y) s(y, z)
• The inequality in the lemma follows immediately.
• The equality holds since all strings are
  optimally aligned with Sc.

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Bounded-Error Approximation for SP-Alignment

• We can now establish the bounded-error
  approximation
• Defn. Let M denote the optimal alignment of the
  k string of S.
• Defn. Let d(Si, Sj) denote the pairwise
  alignment score of the strings Si and Sj induced
  by M.

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Bounded-Error Approximation for SP-Alignment

• Thm. d(Mc)/d(M) ? 2(k 1)/k lt 2
• See proof on page 350 for details. (basically
  depends on the previous lemma)
• Corollary
• kM ? SiltjD(Si, Sj) ? d(M) ? d(Mc) ? 2(k
  1)/k SiltjD(Si, Sj)
• Recall that M SSj?SD(Sc, Sj)
• The alignment score D(Si, Sj) is not based on Mc
  or M
• Observation d(Mc)/SiltjD(Si, Sj) gives a measure
  of the goodness of Mc and is guaranteed to be
  less than 2.

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Consensus Objective Functions

• First fact of consensus representations
• There is no consensus as to how to define
  consensus.
• Consequently, we will look at several
  definitions.
• Steiner consensus strings
• Defn. Given a set of string S and a string S,
  the consensus error of S relative to S is E(S)
  SSj?SD(S, Sj).
• S is not required to be a member of S.

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Consensus Objective Functions

• Defn. Given a set of strings S, an optimal
  Steiner string S for S minimizes the consensus
  error E(S).
• S is not required to be a member of S.
• Observations
• in S we are trying to capture the essential
  common features in S.
• Computing E(S) appears to be a hard problem.

42
Consensus Objective Functions

• No known efficient method for finding S.
• We will consider an approximate method.
• Lemma Assume that S contains k strings and that
  the scoring scheme satisfies the triangle
  inequality. There exists a string S? S such that
  E(S)/E(S) ? 2.
• Q What does this lemma say?
• (Proof sketch next slide)

43
Consensus Objective Functions

• Proof sketch
• For any i, D(S, Si) ? D(S, S) D(S, Si) so,
• E(S) SSj?S D(S, Sj) and
• SSj?S D(S, Sj) ? SSj?S D(S, S) D(S, Sj)
• But SSj?S D(S, S) D(S, Sj) (k-2) D(S,
  S) E(S)
• Therefore E(S) ? (k-2) D(S, S) E(S)

44
Consensus Objective Functions

• QWhere do we find a good candidate for S?
• A Sc, the center string.
• Recall Sc minimizes SSj?S D(Sc, Sj).
• Thm. E(Sc)/E(S) ? 2 - 2/k, assuming the scoring
  scheme satisfies the triangle inequality.
• Proof. Follows immediately from the previous
  lemma and the observation that E(Sc) ? E(S)

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Consensus Objective Functions

• Consensus strings from multiple alignment
• Defn. Let M be a multiple alignment of strings S,
  the consensus character of column i of M is the
  character that minimizes the summed distance to
  all the characters in column i.
• Note
• the summed distance depends on the pairwise
  scoring scheme.
• The plurality character is the consensus
  character for some scoring schemes.

46
Consensus Objective Functions

• Defn. Let d(i) denote the minimum sum in column
  i.
• Defn. The consensus string SM derived from
  alignment M is the concatenation of consensus
  characters for each column of M.
• Q How can we evaluate the goodness of SM ?
• A One possibility is Goodness(SM ) SiD(SM,
  Si), i.e., see how good of a Steiner string SM
  is.
• Consider a different approach..

47
Consensus Objective Functions

• Defn. The alignment error of SM, a consensus
  string containing q characters, is Sqi1d(i).
• Defn. The alignment error of M is defined as the
  -alignment error of SM, its consensus string.

• Example
• 1 A A T - G - T T T
• 2 A A - C G T T A T
• T A T C G - A A T

A A T C G - T A T Consensus (alignment
error of ?)
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Consensus Objective Functions

• Defn. The optimal consensus multiple alignment is
  a multiple alignment M whose consensus string SM
  has the smallest alignment error over all
  possible multiple alignments of S.

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Consensus Objective Functions

• The 3 notions of consensus we have discussed are
• The Steiner string S defined from S.
• The consensus string SM derived from M, with
  goodness related to its function as a Steiner
  string.
• The consensus string SM derived from M, with
  goodness related to is ability to reflect the
  column-wise properties of M.
• Surprisingly (or not) they lead to the same
  multiple alignment.

50
Consensus Objective Functions

• Lets investigate the assertion these concepts
  result in the same multiple alignment.
• Let S be a set of k strings.
• Let T be the star tree with Steiner string S at
  the root and each of the k strings of S at
  distinct leave of T, then
• Defn. the multiple alignment consistent with S
  is the multiple alignment of S ? S consistent
  with T.

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Consensus Objective Functions

• Thm. Let S denote the consensus string of the
  optimal consensus multiple alignment.
• Removing the spaces from S results in the optimal
  Steiner string S.
• Removal of S from the multiple alignment
  consistent with S results in the optimal
  consensus multiple alignment of S.
• Proof on page 353.

52
Consensus Objective Functions

• Q Why should we care about this theorem?
• A The theorem stating E(Sc)/E(S) ? 2 - 2/k
  plus this theorem can be used to approximate the
  optimal consensus alignment
• Find the center string Sc. Recall the center
  string Sc ? S, a set of k strings, is the string
  that minimizes M SSj?S D(Sc, Sj).
• Place Sc at the center of a k node star.

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Consensus Objective Functions

• Label each leaf with a string from S.
• Construct the multiple alignment M consistent
  with this tree T.
• Recall M is consistent with T if the induced
  pairwise alignment of Si and Sj has score D(Si,
  Sj) for each pair of strings (Si, Sj) that label
  adjacent nodes in T.

54
Consensus Objective Functions

• Revelation The multiple alignment M is the same
  as Mc used to approximate the SP objective
  function.
• Thm. The multiple alignment Mc created by the
  center star method has
• An SP score ? (2-2/k) score of the optimal SP
  alignment.
• A consensus alignment error ? (2-2/k) the
  alignment error of the optimal consensus multiple
  alignment.

55
Phylogenetic trees Multiple alignment

• Phylogenetic tree a depiction of the
  evolutionary history of set of taxa. The leaves
  of the tree are labeled by taxa names.
• Convention
• Each edge (u,v) denotes an ancestor-descendant
  relation.
• This relation may be on the basis of
  morphological attributes or sequence similarity.
• The internal nodes represent extinct taxa.
• The leafs represent currently existing taxa.

56
Phylogenetic trees Multiple alignment

• Two related problems
• Problem find a multiple alignment for a tree
• Given a phylogenetic tree, deduce sequences for
  the internal nodes to optimize some objective
  function.
• Find the multiple alignment consistent with the
  tree.
• Delete the deduced sequences (internal node
  labels)
• Find a tree from a set of leaf sequences.(Chapter
  17)

57
Phylogenetic trees Multiple alignment

• Let T be a tree with leaf nodes labeled with
  distinct strings from a set S.
• Defn. a phylogenetic alignment for T is an
  assignment of one string to each internal node.
• Note strings labeling internal nodes need not
  come from S.

58
Phylogenetic trees Multiple alignment

• Recall that D(S1, S2) denotes the edit distance
  between strings S1 and S2.
• Defn. The edge distance of edge (i, j) is D(Si,
  Sj) where Si and Sj are the strings labeling
  nodes i and j, respectively.
• Defn. Path distance is the sum of edge distances
  along the path.
• Defn. Phylogenetic alignment distance is the sum
  of all edge distances in the tree.

59
Phylogenetic trees Multiple alignment

• Phylogenetic alignment problem for T
• Find an assignment of strings to internal nodes
  of T that minimizes the distance of the alignment.

60
Phylogenetic trees Multiple alignment

• Phylogenetic alignment problem for T
• The general problem is too hard (NP-complete).
• We will consider a heuristic approximate
  solution.
• The solution is within twice the minimal
  distance.
• The approach has polynomial time complexity.

61
Phylogenetic trees Multiple alignment

• Defn. A lifted alignment is a phylogenetic
  alignment in which the string assigned to each
  internal node is also assigned to one of its
  children.
• Example

62
Phylogenetic trees Multiple alignment

• Lifted Alignment Observation
• Each internal node v is labeled by a leaf label
  appearing in the subtree rooted at v.

63
Phylogenetic trees Multiple alignment

• Plan
• Construct a lifted alignment TL.
• Initial approach conceptually transform the
  optimal phylogenetic alignment.
• Q Why do we say conceptually?
• A Because we dont have T, the optimal
  phylogenetic alignment.
• Demonstrate property of TL total distance lt
  twice optimal phylogenetic alignment distance.
• Next show how to compute TL efficiently using DP.

64
Phylogenetic trees Multiple alignment

• Creating TL
• Start with input tree T, with leafs labeled by
  distinct strings.
• Let T denote the optimal phylogenetic alignment
  for T. (This is the assignment of strings to
  internal nodes of T that minimizes the total of
  all edge distances.)
• Successively lift each internal node.
• An internal node can only be lifted if all of its
  children have been lifted.
• Leaf nodes are defined to be lifted.

65
Phylogenetic trees Multiple alignment

• Q How do we lift a node?
• Let Sv denote the label of node v in T.
• Assume that vs children have been lifted.
• WLOG let the labels of vs children be S1,
  S2,..,Sk from S.

66
Phylogenetic trees Multiple alignment

• Find the string Sj among the children that is
  closest to Sv, i.e., the string Sj such that
  D(Sv, Sj) ? D(Sv, Si) for all i from1 to k.
• Replace Sv,with Sj.

67
Phylogenetic trees Multiple alignment

• Claim The lifted alignment TL has total distance
  less or equal to twice that of the optimal
  phylogenetic alingment T of T.
• Sketch of proof
• Suppose e(v, w) (v the parent of w) is a
  nonzero-length edge in TL.
• Suppose v is labeled Sj ? S, and w is labeled Si
  ? S.
• If Sj ? Si then the distance of e in TL is D(Sj,
  Si) ? D(Sj, Sv) D(Sv, Si).
• But D(Sj, Sv) D(Sv, Si) ? 2 D(Sv, Si)
• Q Why is this true?
• A because D(Sj, Sv) ? D(Sv, Si)

68
Phylogenetic trees Multiple alignment

• Sketch of proof (continued)
• What about paths?
• Let Pe denote the path from v to the leaf labeled
  Si in T. The distance is at most the sum of the
  edge distances.
• In TL, if e is a nonzero-length edge, then this
  path has distance at most twice Pe.

69
Phylogenetic trees Multiple alignment

• The lifted alignment can be computed with DP.
• Let Tv be the subtree of T rooted at node v.
• Defn. d(v, S) denotes the distance of the best
  lifted alignment of Tv where v is labeled with S.
• Obviously, S must be the label of a leaf in Tv.

70
Phylogenetic trees Multiple alignment

• d(v, S) is computed from the leaves up.
• The leaves are already considered lifted.
• d(v, S) for a parent of leaves is computed by
  d(v, S) SS D(S, S) where S is the label of a
  child of v.
• The general recurrence for an internal node is
  d(v, S) Sv minS D(S, S) d(v, S) ,
  where v is a child of v and S labels a leaf in
  Tv.

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Phylogenetic trees Multiple alignment

• Time analysis
• Assume that T has k leaves.
• Assume that all pairwise distances have been
  computed.

Q How long does this take? A O(N2) where N is
the total length of all the k strings. Why is
this true? How can we explain it?
72
Phylogenetic trees Multiple alignment
Time analysis The processing at an internal node
is O(k2). Why is this true? Then the total time
is O(N2 k3). Why O(N2 k3) and not O(N2
k2)? Bottom line we can compute the optimal
lifted alignment in time that is polynomial in
the length of the strings and size of the tree.