Bioinformatics Algorithms and Data Structures

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Bioinformatics Algorithms and Data Structures

• Chapter 17.2-3 Strings and Evolutionary Trees
• Lecturer Dr. Rose
• Slides by Dr. Rose
• April 5, 2007

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Next Homework Due 4/19/07

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• 8
• Note there are three parts to this question.
  Only answer the first two parts, i.e.,
• show that if D is ultrametric and D(i, i)0 for
  each i, then D is also additive.
• Show the converse is not true.
• 14 (Grad students only.)

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Additive Distance Trees

• Real data is rarely ultrametric.
• A weaker constraint is that data be additive.
• Recall Additive distances are distances which
• can be fitted to an unrooted tree such that
• pairwise taxa distances are equal to the sum of
  the branch lengths connecting them.

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Additive Distance Trees

• Consider the relationship between additive and
  ultrametric trees
• Q Are all ultrametric trees additive?
• A Yes.
• Q Are all additive trees ultrametric?
• A No.

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Additive Distance Trees

• Assuming that
• D is a symmetric n by n distance matrix
• D contains only zero values on the diagonal
• D contains only positive off-diagonal values
• T is an n node tree, then
• Defn. T is an additive tree for D if, for every
  pair of labeled nodes (i, j), the path from i to
  j has total weight exactly D(i, j).

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Additive Distance Trees

• Additive tree problem, given
• symmetric matrix D
• zero entries on the diagonal
• Positive off-diagonal values
• Find additive tree T for D or determine that one
  does not exist.
• Imagine that you have a distance matrix D
  representing evolutionary distance between pairs
  of taxa.
• Q Do you expect the additive tree for D to be
  unique?

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Additive Distance Trees

• Q Is there a unique additive tree for D?
• If you think the answer is yes, why?
• If you think the answer is no, why?
• Consider what we know about D and T
• Is Ts branching pattern is consistent with D?
  (y/n)
• Are the edge lengths in T consistent with D?
  (y/n)
• Does D specifies directed edges? (y/n)
• Does D imply directed edges? (y/n)

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Additive Distance Trees
(Table and figure from http//imbs.massey.ac.nz/Re
search/MolEvol/Farside/DNA/00312.html)
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Additive Distance Trees

• Concept Additive tree problem
• Given n by n symmetrical matrix D
• with zero diagonal entries
• positive off-diagonal values
• Find additive tree or determine none exists.

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Additive Distance Trees

• Concept Compact Additive tree problem
• Given n by n symmetrical matrix D
• with zero diagonal entries
• positive off-diagonal values
• Find additive tree with exactly n nodes.
• Q What does this definition say about the
  topology of the tree?
• A For every node, there must be a corresponding
  row in D.

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Additive Distance Trees

• Consider the symmetrical matrix D above and the
  tree T.
• Q Is T an additive tree for D?
• Q Is T a compact additive tree for D?

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Additive Distance Trees

• Consider the symmetrical matrix D above.
• Q Does D have a additive tree?
• Q Does D have a compact additive tree?

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Additive Distance Trees

• Defn. Let G(D) be the n-node complete graph
  corresponding to D where nodes are labeled 1 n
  and edges have weight D(i, j).
• Thm. If there is a compact additive tree T for D,
  then T must be the unique minimum spanning tree
  of G(D).

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Additive Distance Trees

• Proof. Let
• T be a compact additive tree for D.
• e (x, y) be any edge not in T.
• We know
• The path from x to y in T is D(x, y)
• The edge weight for e is also D(x, y)
• Since e is not in T, e is strictly greater than
  any edge in the path from x to y in T.

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Additive Distance Trees

• Proof. continued
• Assume that there is some other minimum spanning
  tree T containing e.
• Removing e splits T into two sets of nodes, S
  S.
• WLOG, S contains x S contains y.
• In T there is an edge e that connects the nodes
  in S S
• Furthermore, e is on the path from x to y in T.
• Hence e lt e.

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Additive Distance Trees

• Proof. continued
• Create a new spanning tree T by removing e from
  T and adding e.
• The edge weight of T is less than that of T.
• This contradicts the assumption that T is a
  minimum spanning tree.
• T must itself be the unique minimum spanning tree
  of G(D).

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Additive Distance Trees

• How can we use this theorem to solve the compact
  additive tree problem in O(n2) time?
• Answer
• Construct G(D) from D.
• Use an O(n2) mst algorithm, such as Prims
  algorithm, that extends a single growing tree T.
• When an edge e (x, y) is added to T, and x is
  already in T.
• Compute d(i, y) d(i, x) D(x, y) for all i in
  T. This takes O(n) per iteration and O(n2) for
  all of T.
• Verify d(i, y) D(i, y)

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Parsimony

• Q What is parsimony?
• A Parsimony extreme or excessive frugality.
• Q So what does frugal mean?
• A Frugal thrifty, economical.
• In this chapter, parsimony is a character-based
  method for reconstructing evolutionary history.
• Characters are attributes, traits
• In this section we will look at highly
  constrained trees that express evolutionary
  history.

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Parsimony

• Can be used to deduce evolutionary trees
• Specifies branching order
• Does not specify divergence times
• Can be used as basis for a taxonomy
• This section is a limited introduction to maximum
  parsimony problems
• Binary-character problems
• Focus on perfect phylogeny problem

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Parsimony

• Defn. Let M be an n by m, binary matrix
  representing n objects with m character traits.
• Since M is binary, each character trait has two
  possible states, 1 or 0.
• Cell (p, i) of M has value 1 iff object p has
  character i.
• M has a flavor similar to the old chestnut animal
  guessing program that uses a binary tree.

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Parsimony

• Defn. a phylogenetic tree for M is a rooted tree
  T with exactly n leaves such that
• Each of the n leaves is labeled by exactly one
  object.
• Each of the m character-traits labels exactly one
  edge of T.
• For any object p, the character-traits labeling
  the edges along the path from the root to p are
  exactly those character-traits whose state is one.

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Parsimony

• Consider the matrices below
• do either M1 or M2 have a phylogenetic tree?
• If so, what does the tree look like?

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Parsimony

• Q What is the interpretation of the phylogenetic
  tree?
• A It is an estimate of the divergent
  evolutionary history of the objects. (does not
  give time)
• The root represents an ancestor with none of the
  m character-traits.
• Each character-trait transitions from 0 to 1 only
  once.
• No character-trait ever transitions from 1 to 0.

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Parsimony

• Q In what sense are phylogenetic trees
  parsimonious?
• A Each character-trait labels exactly 1 edge of
  the tree. The biological assumptions are
• The root represents an ancestor with none of the
  m character-traits.
• Each character-trait transitions from 0 to 1 only
  once.
• No character-trait ever transitions from 1 to 0.

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Parsimony

• Q What character-traits can be used?
• Morphological features
• (from http//anthro.palomar.edu/hominid/australo_
  2.htm)
• (Also see http//www.cfsan.fda.gov/frf/rfe3pc00.
  html )

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Parsimony

• Q What character-traits can be used?
• Morphological features
• Gross anatomical features
• OTU-specific esoterica
• DNA-based characters
• specific substring patterns
• Specific nucleotides in fixed positions
• See pages 460 461 for more discussion

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Parsimony

• Defn. perfect phylogeny problem given the binary
  matrix M, determine if there is a phylogenetic
  tree for M, if there is one, build it.
• We will discuss an O(nm)-time algorithm
• First we need to preprocess M.
• Consider each column as a binary number
• msb in row 1
• sort columns in decreasing order.
• Let M denote the reordered matrix M.

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Parsimony

• Example.

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Parsimony

• Defn. for any column k of M, let Ok be the set
  of objects with a one in column k.
• Obs. If Oj is a proper subset of Ok, then column
  k must be to the left of column j in M.

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Parsimony

• Thm. Matrix Mhas a phylogenetic tree iff for
  every pair of columns i, j, either Oi and Oj are
  disjoint or one contains the other.
• Proof. (Sketch starting on next slide)

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Parsimony

• Proof. ?
• Let T be the phylogenetic tree for M.
• Consider characters i, j.
• Let ej be the edge that character j transitions
  from 0 to 1.
• Let ei be the edge that character i transitions
  from 0 to 1.
• Objects with character i are below ei in T.
• Objects with character j are below ej in T.

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Parsimony

• Proof. ?
• There are 4 possible cases
• ei ej
• ei is on the path from the root to ej.
• ej is on the path from the root to ei.
• The paths diverge before reaching ei or ej.
• In case 1, Oi Oj.
• In case 2, Oj ? Oi since all objects possessing j
  possess i.
• In case 3, Oi ? Oj since all objects possessing i
  possess j.
• In case 4, Oi ? Oj ?

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Parsimony

• Proof. ?for all i, j Oi Oj are disjoint or one
  contains the other
• Consider objects p and q.
• Let k be the largest character common to both.
• All characters i lt k possessed by p are also
  possessed by q
• All characters i lt k possessed by q are also
  possessed by p
• So they have share exactly the same characters up
  till k, and none thereafter.

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Parsimony

• Proof. ?for all i, j Oi Oj are disjoint or one
  contains the other
• Label each p with the string that is the
  concatenation of the column numbers for which it
  has nonzero entries. Likewise for q.
• Append to the string so that no string is a
  prefix of any other.
• p q have a common prefix but diverge after k
• The keyword tree (sans failure links) for the n
  objects in M specifies a perfect phylogeny for
  M.

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Parsimony

• O(nm) alg. for the perfect phylogeny problem
• Reorder columns of M in descending order using
  radix sort.
• Let M be the resulting matrix.
• Label each column by its column position in M.
• Q Why do you think we are using radix sort?
• A radix sort is O(nm). Also it can be applied to
  a number with an arbitrary number of digits.

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Parsimony

• For each row p of M, construct the string
  consisting of the characters, in sorted
  (increasing) order, that p possesses.
• Recall that in step 1 we labeled each character
  by its column position.
• The string for a given row will be the
  concatenation of the column labels for which the
  row has the value one.

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Parsimony

• Build the keyword tree T for the n strings from
  step 2.
• Recall that the keyword tree for set P is a
  rooted directed tree K satisfying
• Each edge is labeled with one character
• Any two edges out of the same node have distinct
  labels.
• Every pattern Pi in P maps to some node v of K
  s.t. the path from the root to v spells out Pi
• Every leaf in K is mapped by some pattern in P.

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Keyword Trees

• Example From textbook P potato, poetry,
  pottery, science, school

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Parsimony

• Test whether T is a perfect phylogeny for M.
• Verify that T has exactly n leaves such that
• Each of the n leaves is labeled by exactly one
  object.
• Each of the m character-traits labels exactly one
  edge of T.
• For any object p, the character-traits labeling
  the edges along the path from the root to p are
  exactly those character-traits whose state is one.

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Tree Compatibility

• Suppose you have two different phylogenetic
  trees.
• Note even for the same set of taxa we can derive
  different trees by basing the comparison on
  different proteins.
• Q How can we determine if they describe a
  consistent evolutionary history?
• Q How can we combine them into a single tree?
• This section addresses these questions.

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Tree Compatibility

• Defn. Phylogenetic tree refinement
• A phylogenetic tree T? is a refinement of T if T
  can be obtained by a series of contractions of
  edges of T?.
• Nutshell T? agrees with T, but expresses
  additional evolutionary history.

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Tree Compatibility

• Tree refinement T1 T2? T1 T3? T1 T4? Etc?

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Tree Compatibility

• Defn. Phylogenetic tree compatibility
• Trees T1 and T2 are compatible if there exists a
  phylogenetic tree T3 refining both T1 and T2.

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Tree Compatibility

• Tree compatibility problem
• Given two trees, T1 and T2
• determine if they are compatible.
• if so, return the refinement tree T3.
• We will consider a matrix method for finding T3.

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Tree Compatibility

• Consider a binary matrix representation of a
  phylogenetic tree
• There is one row for each object (OTU)
• There is one column for each internal node
• Entry (i, j) is one iff the leaf for object i is
  in the subtree rooted at j.
• Q Would an example help?
• A Ok, then suggest a simple phylogenetic tree.

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Tree Compatibility

• Let M1 be the matrix representation of T1 and
  similarly M2 for T2.
• Let M3 be the matrix formed by taking the union
  of the columns of M1 and M2.
• Q What is meant by taking the union of columns?
• A M3 will contain
• all columns found only in M1
• all columns found only in M2
• One copy of all columns appearing in both M1 and
  M2
• Obviously, columns will have a different order

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Tree Compatibility

• Q What should M3 look like? What about T3?

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Tree Compatibility

• Q Do you agree?

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Tree Compatibility

• Note In refining T3 to produce T4, in M4 there
  is no impact wrt to the preceding columns in M3

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Tree Compatibility

• Theorem
• T1 and T2 are compatible iff there is a
  phylogenetic tree for M3. A phylogenetic tree T3
  for M3 is a refinement of both T1 and T2.

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Generalized Perfect Phylogeny

• Generalization of perfect phylogeny
• Allow multiple states (gt2) for character-traits
• Label edges with triple (c x y) where
• c is the character trait
• x is the value of the state before the edge
• y is the value of the state after the edge
• The starting state for each character is
  specified at the root

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Generalized Perfect Phylogeny

• Generalization of perfect phylogeny continued
• The path from the root to the leaf labeled p
  describes the character traits of the object p.
• The ending states along this path specify ps
  traits
• A combination of trait and ending state can
  appear only once in the tree
• Example character c, ending state y
• There can only be one edge labeled (c ? y)
• Where ? Matches any state of c.

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Generalized Perfect Phylogeny

• Time complexity for generalized perfect
  phylogeny
• polynomial in n and m for fixed the number of
  states
• NP-complete otherwise