Intro to Phylogenetic Trees Lecture 5

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Intro to Phylogenetic TreesLecture 5
Sections 7.1, 7.2, in Durbin et al. Chapter 17 in
Gusfield Slides by Shlomo Moran. 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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Morphological topology
(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
There are many possible types of sequences to use
(e.g. Mitochondrial vs Nuclear proteins).
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Mitochondrial topology
(Based on Pupko et al.,)
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Nuclear topology
(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 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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Type 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, or
• Character-based
• Examine each character (e.g., residue) separately

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Two Methods of Tree Construction

• Distance- A weighted tree that realizes the
  distances between the objects.
• 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)gt0
• 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
• eg, there is no tree which realizes the below
  distances.

i j k l
i 0 2 2 2
j 0 2 2
k 0 3
l 0
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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 Lgt4

• 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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Dangers of Paralogs

• 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.