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Computational Genomics 5a Distance Based Trees Reconstruction (cont.) Sections 7.1, 7.2, in Durbin et al. Chapter 17 in Gusfield (updated April 12, 2009) Slides by Shlomo Moran and Ydo Wexler (IIT), modified by Benny Chor Evolution Evolution of new organisms is driven by Diversity

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slide1

Computational Genomics 5a Distance Based Trees Reconstruction (cont.) Sections 7.1, 7.2, in Durbin et al.Chapter 17 in Gusfield(updated April 12, 2009)

Slides by Shlomo Moran and Ydo Wexler (IIT), modified by Benny Chor

.

evolution
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
the tree of life
The Tree of Life

Source: Alberts et al

slide4

Tree of life- a better picture

D’après Ernst Haeckel, 1891

slide5

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.

historical note
Historical Note
  • Until mid 1950’s 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
morphological vs molecular
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
slide8

Topology based on Morphology

Bonobo

Chimpanzee

Man

Gorilla

Sumatran orangutan

Bornean orangutan

Common gibbon

Barbary ape

Baboon

White-fronted capuchin

Slow loris

Tree shrew

Japanese pipistrelle

Long-tailed bat

Jamaican fruit-eating bat

Horseshoe bat

Little red flying fox

Ryukyu flying fox

Mouse

Rat

Glires

Vole

Cane-rat

Guinea pig

Squirrel

Dormouse

Rabbit

Pika

Pig

Hippopotamus

Sheep

Cow

Alpaca

Blue whale

Fin whale

Sperm whale

Donkey

Horse

Indian rhino

White rhino

Elephant

Carnivora

Aardvark

Grey seal

Harbor seal

Dog

Cat

Asiatic shrew

Insectivora

Long-clawed shrew

Small Madagascar hedgehog

Hedgehog

Gymnure

Mole

Armadillo

Xenarthra

Bandicoot

Wallaroo

Opossum

Platypus

(Based on Mc Kenna and Bell, 1997)

Archonta

Ungulata

slide9

From Sequences to a Phylogenetic Tree

Rat QEPGGLVVPPTDA

Rabbit QEPGGMVVPPTDA

Gorilla QEPGGLVVPPTDA

Cat REPGGLVVPPTEG

Different genes/proteins may lead

to different phylogenetic trees

slide10

From sequences to a phylogenetic tree

Rat QEPGGLVVPPTDA

Rabbit QEPGGMVVPPTDA

Gorilla QEPGGLVVPPTDA

Cat REPGGLVVPPTEG

There are many possible types of sequences to use (e.g. mitochondrial vs. nuclear proteins).

slide11

Perissodactyla

Donkey

Horse

Carnivora

Indian rhino

White rhino

Grey seal

Harbor seal

Dog

Cetartiodactyla

Cat

Blue whale

Fin whale

Sperm whale

Hippopotamus

Sheep

Cow

Chiroptera

Alpaca

Pig

Little red flying fox

Ryukyu flying fox

Moles+Shrews

Horseshoe bat

Japanese pipistrelle

Long-tailed bat

Afrotheria

Jamaican fruit-eating bat

Asiatic shrew

Long-clawed shrew

Mole

Small Madagascar hedgehog

Xenarthra

Aardvark

Elephant

Armadillo

Rabbit

Lagomorpha

+ Scandentia

Pika

Tree shrew

Bonobo

Chimpanzee

Man

Gorilla

Sumatran orangutan

Primates

Bornean orangutan

Common gibbon

Barbary ape

Baboon

White-fronted capuchin

Rodentia 1

Slow loris

Squirrel

Dormouse

Cane-rat

Rodentia 2

Guinea pig

Mouse

Rat

Vole

Hedgehog

Hedgehogs

Gymnure

Bandicoot

Wallaroo

Opossum

Platypus

Topology1 , based on Mitochondrial DNA

(Based on Pupko et al.,)

slide12

Chiroptera

Round Eared Bat

Eulipotyphla

Flying Fox

Hedgehog

Pholidota

Mole

Pangolin

Whale

1

Cetartiodactyla

Hippo

Cow

Carnivora

Pig

Cat

Dog

Perissodactyla

Horse

Rhino

Glires

Rat

Capybara

2

Scandentia+

Dermoptera

Rabbit

Flying Lemur

Tree Shrew

3

Human

Primate

Galago

Sloth

Xenarthra

4

Hyrax

Dugong

Elephant

Afrotheria

Aardvark

Elephant Shrew

Opossum

Kangaroo

Topology2 ,based on Nuclear DNA

(Based on Pupko et al. slide)

(tree by Madsenl)

theory of evolution
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
phylogenenetic trees

Aardvark

Bison

Chimp

Dog

Elephant

Phylogenenetic trees
  • Leafs - current day species
  • Nodes - hypothetical most recent common ancestors
  • Edges length - “time” from one speciation to the next
types of trees
Types of Trees

A natural model to consider is that of rooted trees

Common

Ancestor

types of trees16
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.

slide17

Tree a

Tree b

Rooted versus unrooted trees

Tree c

b

a

c

Represents all three rooted trees

number of different trees on n leaves
Number of Different Trees on n Leaves

For n=4, there are 3 different unrooted trees

For n=5, there are 3*5=15 different unrooted trees

For n=6, there are 3*5*7=15 different unrooted trees

For n, there are n!! = 3*5*…*(2n-5)=(2n-5)!/[2n-3 *(n-3)!]

different unrooted trees on n labeled leaves.

There are 3*5*…*(2n-3)=(2n-3)!/[2n-2 *(n-2)!]

different rooted trees on n labeled leaves (same as number of

unrooted trees on n+1 labeled leaves.

In both cases, number is too large to go over all of them!

positioning roots in unrooted trees
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

phylogenetic trees methods
Phylogenetic Trees - Methods
  • There are several methods with which we construct trees and estimate how good a tree describes the data (and thus the evolution process)
  • Distance based methods
  • Parsimony
  • character based methods
  • Likelihood
  • Whole genome/proteome methods
methods
Methods
  • 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.
tree construction different outcomes
Tree Construction: Different Outcomes
  • Distance Based- Output is a weighted tree that realizes the distances between the objects (or gets close to it).
  • Character Based – Output is 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, and consider 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.

exact solution additive sets
Exact solution: Additive sets

Given a set M of L objects with an L×Ldistance 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?

We should first clarify what realizes means.

additive sets cont
Additive sets (cont)

We say that the set of distances M on 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 just non-negative.

three objects sets is always additive

k

c

b

j

m

a

i

Three objects sets is always additive:

For L=3: There is always a (unique) tree with one internal node.

3 equations in 3 unknowns: a,b,c

Thus

how about four objects
How about four objects?

L=4: Not all sets with 4 objects are additive:

e.g., there is no tree which realizes the distances below.

the four points condition

k

i

l

j

The Four Points Condition

Theorem: A set M of L objectsis 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 (and maybe the board...)

additive distances
Additive Distances

We say that a distance metric D on L objects is additive if there is an unrooted binary tree on L leaves, with positive edge weights, that realizes the distanceD.

Namely for all i,j, D(i,j)=DT(i,j)

characterizing additive distances
Characterizing Additive Distances

Thm: Any additive distance is fully characterized by the four point condition: Any 4 points can be renamed such that 

trees from additive distances algorithm

7

C

A

Trees from Additive Distances: Algorithm
  • Verify that the distance matrix constitutes an additive metric
  • Choose a pair of objects, which results in the first path in the tree.
  • Choose a third object and establish the linear equations to let the object branch off the path.
  • Choose a pair of leaves in the tree constructed so far and compute the point a newly chosen object is inserted at.
  • 1. If the new path branches off an existing branch in the tree: Do the insertion step once more, replacing one of the two original leaves by another leaf along the branching path.
  • 2. Once the new path branches off an edge in the tree, this insertion is finished.
trees from additive distances algorithm31
Trees from Additive Distances: Algorithm
  • Verify that the distance matrix constitutes an additive metric
  • Choose a pair of objects, which results in the first path in the tree.
  • Choose a third object and establish the linear equations to let the object branch off the path.
  • Choose a pair of leaves in the tree constructed so far and compute the point a newly chosen object is inserted at.
  • 1. If the new path branches off an existing branch in the tree: Do the insertion step once more, replacing one of the two original leaves by another leaf along the branching path.
  • 2. Once the new path branches off an edge in the tree: This insertion is finished.

A

1

6

C

X

1

B

trees from additive distances algorithm32
Trees from Additive Distances: Algorithm
  • Verify that the distance matrix constitutes an additive metric
  • Choose a pair of objects, which results in the first path in the tree.
  • Choose a third object and establish the linear equations to let the object branch off the path.
  • Choose a pair of leaves in the tree constructed so far and compute the point a newly chosen object is inserted at.
  • 1. If the new path branches off an existing branch in the tree: Do the insertion step once more, replacing one of the two original leaves by another leaf along the branching path.
  • 2. Once the new path branches off an edge in the tree: This insertion is finished.

d(A,B)=d(A,X)+d(X,B)

d(A,C)=d(A,X)+d(X,C)

d(B,C)=d(B,X)+d(X,C)

trees from additive distances algorithm33
Trees from Additive Distances: Algorithm
  • Verify that the distance matrix constitutes an additive metric
  • Choose a pair of objects, which results in the first path in the tree.
  • Choose a third object and establish the linear equations to let the object branch off the path.
  • Choose a pair of leaves in the tree constructed so far and compute the point a newly chosen object is inserted at.
  • 1. If the new path branches off an existing branch in the tree: Do the insertion step once more, replacing one of the two original leaves by another leaf along the branching path.
  • 2. Once the new path branches off an edge in the tree: This insertion is finished.

A

1

6

C

X

1

B

trees from additive distances algorithm34
Trees from Additive Distances: Algorithm
  • Verify that the distance matrix constitutes an additive metric
  • Choose a pair of objects, which results in the first path in the tree.
  • Choose a third object and establish the linear equations to let the object branch off the path.
  • Choose a pair of leaves in the tree constructed so far and compute the point a newly chosen object is inserted at.
  • 1. If the new path branches off an existing branch in the tree: Do the insertion step once more, replacing one of the two original leaves by another leaf along the branching path.
  • 2. Once the new path branches off an edge in the tree: This insertion is finished.

C

5

A

1

1

1

2

B

D

(we now add E on the A-to-D path)

trees from additive distances algorithm35
Trees from Additive Distances: Algorithm
  • Verify that the distance matrix constitutes an additive metric
  • Choose a pair of objects, which results in the first path in the tree.
  • Choose a third object and establish the linear equations to let the object branch off the path.
  • Choose a pair of leaves in the tree constructed so far and compute the point a newly chosen object is inserted at.
  • 1. If the new path branches off an existing branch in the tree: Do the insertion step once more, replacing one of the two original leaves by another leaf along the branching path.
  • 2. Once the new path branches off an edge in the tree: This insertion is finished.

C

5

A

1

1

E

5

1

2

B

NO!

D

trees from additive distances algorithm36
Trees from Additive Distances: Algorithm
  • Verify that the distance matrix constitutes an additive metric
  • Choose a pair of objects, which results in the first path in the tree.
  • Choose a third object and establish the linear equations to let the object branch off the path.
  • Choose a pair of leaves in the tree constructed so far and compute the point a newly chosen object is inserted at.
  • 1. If the new path branches off an existing branch in the tree: Do the insertion step once more, replacing one of the two original leaves by another leaf along the branching path.
  • 2. Once the new path branches off an edge in the tree: This insertion is finished.

E

3

A

2

1

1

3

C

1

2

B

D

trees from additive distances algorithm37
Trees from Additive Distances: Algorithm

is this necessary?

  • Verify that the distance matrix constitutes an additive metric
  • Choose a pair of objects, which results in the first path in the tree.
  • Choose a third object and establish the linear equations to let the object branch off the path.
  • Choose a pair of leaves in the tree constructed so far and compute the point a newly chosen object is inserted at.
  • 1. If the new path branches off an existing branch in the tree: Do the insertion step once more, replacing one of the two original leaves by another leaf along the branching path.
  • 2. Once the new path branches off an edge in the tree: This insertion is finished.

E

3

A

2

1

1

3

C

1

2

B

D

reconstructing a tree from an additive distance
Reconstructing a Tree from an Additive Distance

By algorithm, given a distance matrix constituting an additive metric, the topology of the corresponding additive tree is unique.

Q.: Given an additive metric on n leaves, what is the run time of the algorithm?

A.: Number of phases is n. Work per phase is O(n).

So total is O(n2).

E

3

A

2

1

1

3

C

1

2

B

D

approximating additive metrices
Approximating Additive Metrices

In practice, the distance matrix between molecular sequences will not be additive. In such case we want to find a tree T whose distance matrix is “close” to the given one.

The methods for exact tree reconstruction provide an inventory for heuristics for tree construction based on approximating additive metrics.

Heuristics give exact results when operating on additive metrics, but the performance of solutions gets unclear when non additive metrics are handled.

neighbor finding

A

B

C

D

Neighbor Finding

How can we find from distances alone a pair of sisters (neighboring leaves)?

Important observation: Closest nodes are not necessarily neighboring leaves.

Next, we show a way to find neighbors from distances. The method

is simple, but unfortunately not very intuitive 

neighbour joining algorithm outline
Neighbour Joining Algorithm: Outline
  • Identifya pair of leaves u,v asneighbors.
  • Combineu,v into a new node, w.
  • Update the distance matrix: Calculate w’s distance from
  • any other node x of the tree using
  • Notice that all 3 quantities on rhs are known.
  • When only 3 nodes are left – compute 3 distances & finish.
neighbour joining algorithm

i

m

0.1

0.1

0.1

k

l

0.4

0.4

j

n

Neighbour Joining Algorithm
  • Identify a pair of neighborsi,j among n leaves.
  • Combine i,j into a new node u.
  • Update the distance matrix.
  • When only 3 nodes are left – finish.

Let ri be the sum of distances

from i to every other node

The measure between i and j we use

in the algorithm is

neighbour joining algorithm43

i

m

0.1

0.1

0.1

k

l

0.4

0.4

j

n

Neighbour Joining Algorithm

Let ri be the sum of distances

from i to all other nodes

The measure between i and j we use in the

algorithm is

neighbor finding seitou nei method

T1

T2

m

l

k

i

j

Neighbor Finding: Seitou & Nei method

Theorem (Saitou&Nei)Assume D is additive, and all tree edge weights are positive. If XD(i,j) is minimal (among all pairs of leaves), then iandj are sister

taxa in the tree.

The proof is rather involved, and will be skipped (no tears, pls).

complexity of neighbor joining algorithm

m

k

i

j

Complexity of Neighbor Joining Algorithm

Naive Implementation:

Initialization:θ(L2) to compute the XD(i,j)’s.

Each Iteration:

  • O(L) to update {XD(i,k):i L} for the new node k.
  • O(L2) to find the minimalXD(i,j).

Total of O(L3).

  • This can be improved using better data structures

(e.g. heap)

reconstructing trees from additive matrices
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 constructthe 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

running nj example on 4 leaves

U

B

Running NJ: Example on 4 Leaves

A

Remark: The XD values imply that

the distances are not additive (why?).

updated distance matrix choosing a b as neighbors

U

B

Updated Distance Matrix,Choosing A,B as Neighbors

V

d(U,x)=[d(A,x)+d(B,x)-d(A,B)]/2

D

A

Notice that now we have only one

Choice: The neighbors are U and D.

final distance matrix

U

B

Final Distance Matrix

V

C

D

A

Remark: Resulting tree is unrooted.

reconstructing trees from non additive matrices
Reconstructing Trees from non Additive Matrices

Q: What if the distance matrix is not additive?

A: We could still run NJ (like we just did)!

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.

Remark: This indeed was the case with last example.

output nj tree
Output - NJ Tree

Branch length

is proportional

to distance

almost additive matrix
Almost Additive Matrix

A distance matrix d’ is “almost additive” if there exists an additive matrix D such that

Thm (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