Intro to Phylogenetic Trees Computational Genomics Lecture 4b

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Intro to Phylogenetic TreesComputational
GenomicsLecture 4b
Sections 7.1, 7.2, in Durbin et al. Chapter 17 in
Gusfield Slides by Shlomo Moran and Ido Wexler.
Slight modifications by Benny Chor
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Evolution

• Evolution of new organisms is driven by
• Diversity
• Different individuals carry different variants of
  the same basic blue print
• Mutations
• The DNA sequence can be changed due to single
  base changes, deletion/insertion of DNA segments,
  etc.
• Selection bias

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The Tree of Life
Source Alberts et al
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Tree of life- a better picture
Daprès Ernst Haeckel, 1891
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Primate evolution
A phylogeny is a tree that describes the sequence
of speciation events that lead to the forming of
a set of current day species also called a
phylogenetic tree.
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Historical Note

• Until mid 1950s phylogenies were constructed by
  experts based on their opinion (subjective
  criteria)
• Since then, focus on objective criteria for
  constructing phylogenetic trees
• Thousands of articles in the last decades
• Important for many aspects of biology
• Classification
• Understanding biological mechanisms

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Morphological vs. Molecular

• Classical phylogenetic analysis morphological
  features number of legs, lengths of legs, etc.
• Modern biological methods allow to use molecular
  features
• Gene sequences
• Protein sequences
• Analysis based on homologous sequences (e.g.,
  globins) in different species

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Topology based on Morphology
(Based on Mc Kenna and Bell, 1997)
Archonta
Ungulata
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From Sequences to a Phylogenetic Tree
Rat QEPGGLVVPPTDA Rabbit QEPGGMVVPPTDA Gorilla QE
PGGLVVPPTDA Cat REPGGLVVPPTEG
Different genes/proteins may lead to different
phylogenetic trees
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From sequences to a phylogenetic tree
Rat QEPGGLVVPPTDA Rabbit QEPGGMVVPPTDA Gorilla QE
PGGLVVPPTDA Cat REPGGLVVPPTEG
There are many possible types of sequences to use
(e.g. mitochondrial vs. nuclear proteins).
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Topology1 , based on Mitochondrial DNA
(Based on Pupko et al.,)
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Topology2 ,based on Nuclear DNA
(Based on Pupko et al. slide)
(tree by Madsenl)
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Theory of Evolution

• Basic idea
• speciation events lead to creation of different
  species.
• Speciation caused by physical separation into
  groups where different genetic variants become
  dominant
• Any two species share a (possibly distant) common
  ancestor

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Phylogenenetic trees

• Leafs - current day species
• Nodes - hypothetical most recent common ancestors
• Edges length - time from one speciation to the
  next

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Types of Trees

• A natural model to consider is that of rooted
  trees

Common Ancestor
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Types of trees

• Unrooted tree represents the same phylogeny
  without the root node

Depending on the model, data from current day
species often does not distinguish between
different placements of the root.
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Rooted versus unrooted trees
Tree c
b
a
c
Represents all three rooted trees
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Positioning Roots in Unrooted Trees

• We can estimate the position of the root by
  introducing an outgroup
• a set of species that are definitely distant from
  all the species of interest

Proposed root
Falcon
Aardvark
Bison
Chimp
Dog
Elephant
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Dangers of Gene Duplication

• If we happen to consider genes 1A, 2B, and 3A of
  species 1,2,3, we get a wrong tree that does not
  represent the phylogeny of the host species of
  the given sequences because duplication does not
  create new species.

Gene Duplication
S
S
S
Speciation events
2B
1B
3A
3B
2A
1A
In the sequel we assume all given sequences are
orthologs.
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Types of Data

• Distance-based
• Input is a matrix of distances between species.
• Can be fraction of residue they disagree on, or
  alignment score between them, etc.
• Character-based
• Input is a multiple sequence alignment. Sequences
  consist of characters (e.g., residues) that are
  examined separately.
• Genome/Proteome based
• Input is whole genome or proteome sequences.
• No MSA or obvious distance definition.

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Tree Construction Two Popular Methods

• Distance Based- A weighted tree that realizes the
  distances between the objects (or gets close to
  it).
• Character Based A tree that optimizes an
  objective function based on all characters in
  input sequences (major methods are parsimony and
  likelihood).

We start with distance based methods, considering
the following question Given a set of species
(leaves in a supposed tree), and distances
between them construct a phylogeny which best
fits the distances.
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Exact solution Additive sets

• Given a set M of L objects with an LL distance
  matrix
• d(i,i)0, and for i?j, d(i,j)0
• d(i,j)d(j,i).
• For all i,j,k it holds that d(i,k)
  d(i,j)d(j,k).
• Can we construct a weighted tree which realizes
  these distances?

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Additive sets (cont)

• We say that the set M with L objects is additive
  if there is a tree T, L of its nodes correspond
  to the L objects, with positive weights on the
  edges, such that for all i,j, d(i,j) dT(i,j),
  the length of the path from i to j in T.
• Note Sometimes the tree is required to be
  binary, and then the edge weights are required to
  be non-negative.

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Three objects sets always additive

• For L3 There is always a (unique) tree with one
  internal node.

Thus
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How about four objects?

• L4 Not all sets with 4 objects are additive
• e.g., there is no tree which realizes the
  distances below.

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The Four Points Condition

• Theorem A set M of L objects is additive iff any
  subset of four objects can be labeled i,j,k,l so
  that
• d(i,k) d(j,l) d(i,l) d(k,j) d(i,j)
  d(k,l)
• We call i,j,k,l the split of i,j,k,l.

Proof Additivity ?4 Points Condition By the
figure...
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4P Condition ?Additivity

• Induction on the number of objects, L.
• For L 3 the condition is empty and tree
  exists.
• Consider L4.
• B d(i,k) d(j,l) d(i,l) d(j,k) d(i,j)
  d(k,l) A

Let y (B A)/2 0. Then the tree should look
as follows We have to find the distances a,b, c
and f.
k
c
l
f
n
y
b
a
m
i
j
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Tree construction for L4

• Construct the tree by the given distances as
  follows
• Construct a tree for i, j,k, with internal
  vertex m
• Add vertex n ,d(m,n) y
• Add edge (n,l), cfd(k,l)

l
k
f
f
f
f
c
Remains to prove d(i,l) dT(i,l) d(j,l)
dT(j,l)
n
n
n
n
y
b
j
m
a
i
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Proof for L4
By the 4 points condition and the definition of
y d(i,l) d(i,j) d(k,l) 2y - d(k,j) a y
f dT(i,l) (the middle equality holds since
d(i,j), d(k,l) and d(k,j) are realized by the
tree) d(j,l) dT(j,l) is proved similarly.
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Induction step for L4

• Remove Object L from the set
• By induction, there is a tree, T, for
  1,2,,L-1.
• For each pair of labeled nodes (i,j) in T, let
  aij, bij, cij be defined by the following figure

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Induction step

• Pick i and j that minimize cij.
• T is constructed by adding L (and possibly mij)
  to T, as in the figure. Then d(i,L) dT(i,L)
  and d(j,L) dT(j,L)
• Remains to prove For each k ? i,j d(k,L)
  dT(k,L).

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Induction step (cont.)

• Let k ?i,j be an arbitrary node in T, and let n
  be the branching point of k in the path from i to
  j.
• By the minimality of cij , i,j,k,L is not a
  split of i,j,k,L. So assume WLOG that
  i,L,j,k is a
• split of i,j, k,L.

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Induction step (end)

• Since i,L,j,k is a split, by the 4 points
  condition
• d(L,k) d(i,k) d(L,j) - d(i,j)
• d(i,k) dT(i,k) and d(i,j) dT(i,j) by
  induction, and
• d(L,j) dT(L,j) by the construction.
• Hence d(L,k) dT(L,k).
• QED

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From Additive Distance to a Tree
By following the proof, the four point condition
can be used to construct a tree from a distance
matrix, or to decide that there is no such tree
(namely that the distance is not additive). But
this algorithm will go over all quartets,
resulting in O(L4) many steps for L species (too
sllllllllllllow). The most popular method for
constructing trees for additive sets uses the
neighbor joining approach.
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Constructing additive treesThe neighbor joining
problem

• Let i, j be sisters (neighboring leaves) in a
  tree, let k be
• their father, and let m be any other vertex.
• Using eq.
• we can compute the distances from k to all
  other leaves.
• This suggest the following method to construct
  tree from an
• additive distance matrix
• Find sisters i,j in the tree,
• Replace i,j by their father, k, and recursively
  construct a tree T for the smaller set.
• Add i,j as children of k in T.

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Neighbor Finding

• How can we find from distances alone a pair of
  sisters (neighboring leaves)?
• Closest nodes are not necessarily neighboring
  leaves.

Next, we show a way to find neighbors from
distances.
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Neighbor Finding Seitou Nei method
Theorem (SaitouNei) Assume d is additive, with
all tree edge weights positive. If D(i,j) is
minimal (among all pairs of leaves), then i and j
are sister taxa in the tree.
The proof is rather involved, and will be skipped
(no tears pls).
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Saitou Nei proof (to be skipped)
Notations used in the proof p(i,j) the path
from vertex i to vertex j P(D,C) (e1,e2,e3)
(D,E,F,C)
For a vertex i, and an edge e(i,j) Ni(e)
k e is on p(i,k). ND(e1) 3, ND(e2) 2,
ND(e3) 1 NC(e1) 1
E
F
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A simpler neighbor finding method

• Select an arbitrary (fixed) node r.
• For each pair of labeled nodes (i,j) let C(i,j)
  be defined by the following expression (also see
  figure)

r
C(i,j)
j
Claim Let i, j be such that C(i,j) is
maximized. Then i and j are neighboring leaves.
i
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Sisters Identification Example

• Select arbitrarily rA.
• C(B,C)(1525-30)/25
• C(B,D)(1534-31)/28
• C(C,D)(2534-49)/25

5
4
6
20
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Claim Let i, j be such that C(i,j) is
maximized. Then i and j are neighboring leaves.
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Neighbor Joining Algorithm

• Set M to contain all leaves, and select a root r.
  ML
• If L 2, return a tree of two vertices
• Iteration
• Choose i,j such that C(i,j) is maximal
• Create a new vertex k, and update distances
• remove i,j, and add k to M
• Recursively construct a tree on the smaller set.
• When done, add i,j as children on k, at distances
  d(i,k) and d(j,k).

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Complexity of Neighbor Joining Algorithm

• Naive Implementation
• Initialization ?(L2) to compute the C(i,j)s.
• Each Iteration
• O(L) to update C(i,k)i? L for the new node k.
• O(L2) to find the maximal C(i,j).
• Total of O(L3).

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Complexity of Neighbor Joining Algorithm

• Using a Heap to store the C(i,j)s
• Initialization ?(L2) to compute and heapify the
  C(i,j)s.
• Each Iteration
• O(1) to find the maximal C(i,j).
• O(L log L) to delete C(m,i), C(m,j) and add
  C(m,k) for all vertices m.
• Total of O(L2 log L).
• (implementation details are omitted)

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Reconstructing Trees from Additive Matrices
Given a distance matrix constituting an additive
metric, the topology of the corresponding
additive tree is unique.
Q Do we have to test additivity before running
NJ? A This would be bad news, as this takes
O(L4) time!
E
3
A
2
1
1
3
C
1
2
B
D
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Reconstructing Trees from Additive Matrices
Q Do we have to test additivity before running
NJ? A By Seito-Nei, if matrix is additive, NJ
will construct the correct tree. Algorithm does
not care about awareness and need not know
anything about the matrix!
E
3
A
2
1
1
3
C
1
2
B
D
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NJ Algorithm Example

• Identify i,j as neighbours if their divergence
  is minimal.
• Combine i,j into a new node u.
• update the distance matrix.
• If only 3 nodes are left finish.

Let ri be the sum of distances from i to every
other node
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Distance Matrix
A
48
Distance Matrix
Y
C
A
49
Distance Matrix
Z
Y
D
C
A
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Reconstructing Trees from non Additive Matrices
Q What if the distance matrix is not additive?
A We could still run NJ! Q But can anything
be said about the resulting tree? A Not
really. Resulting tree topology could even vary
according to way ties are resolved on the way.
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Almost Additive Matrix
A distance matrix d is almost additive if
there exists an additive matrix d such that
Atteson If d is almost additive with respect to
a tree T, then the output of NJ is a tree T with
the same topology as T
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Distance Matrix
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Unrooted Tree - NJ
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Output - NJ
Branch length is proportional to distance
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N-J Method produces an Unrooted, Additive tree
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0. Distance Matrix
Neighbor-Joining Method An Example
What is required for the Neighbour joining method?
Distance matrix
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1. First Step
PAM distance 3.3 (Human - Monkey) is the minimum.
So we'll join Human and Monkey to MonHum and
we'll calculate the new distances.
Mon-Hum
Monkey
Human
Spinach
Mosquito
Rice
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2. Calculation of New Distances
After we have joined two species in a subtree we
have to compute the distances from every other
node to the new subtree. We do this with a simple
average of distances DistSpinach, MonHum
(DistSpinach, Monkey DistSpinach, Human)/2
(90.8 86.3)/2 88.55
Mon-Hum
Monkey
Human
Spinach
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3. Next Cycle
Mos-(Mon-Hum)
Mon-Hum
Human
Mosquito
Monkey
Spinach
Rice
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4. Penultimate Cycle
Mos-(Mon-Hum)
Spin-Rice
Mon-Hum
Human
Mosquito
Monkey
Spinach
Rice
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5. Last Joining
(Spin-Rice)-(Mos-(Mon-Hum))
Mos-(Mon-Hum)
Spin-Rice
Mon-Hum
Human
Mosquito
Monkey
Spinach
Rice
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The resultUnrooted Neighbor-Joining Tree
Human
Spinach
Monkey
Mosquito
Rice