Loading in 5 sec....

Phylogenetic Trees Lecture 1PowerPoint Presentation

Phylogenetic Trees Lecture 1

- 289 Views
- Updated On :
- Presentation posted in: Pets / Animals

Phylogenetic Trees Lecture 1. Credits: N. Friedman, D. Geiger , S. Moran, . Evolution. Evolution of new organisms is driven by Diversity Different individuals carry different variants of the same basic blue print Mutations

Phylogenetic Trees Lecture 1

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Phylogenetic TreesLecture 1

.

Credits: N. Friedman, D. Geiger , S. Moran,

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

Source: Alberts et al

Tree of life- a better picture

D’après Ernst Haeckel, 1891

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.

- 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

- 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

Morphological topology

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

From sequences to a phylogenetic tree

RatQEPGGLVVPPTDA

RabbitQEPGGMVVPPTDA

GorillaQEPGGLVVPPTDA

CatREPGGLVVPPTEG

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

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

Mitochondrial topology

(Based on Pupko et al.,)

Nuclear topology

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

(Based on Pupko et al. slide)

(tree by Madsenl)

- 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

Basic Assumptions

- Closer related organisms have more similar genomes.
- Highly similar genes are homologous (have the same ancestor).
- A universal ancestor exists for all life forms.
- Molecular difference in homologous genes (or protein sequences) are positively correlated with evolution time.
- Phylogenetic relation can be expressed by a dendrogram (a “tree”) .

.

Aardvark

Bison

Chimp

Dog

Elephant

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

- We have to emphasize that gene/protein sequence can be homologous for several different reasons:
- Orthologs -- sequences diverged after a speciation event
- Paralogs -- sequences diverged after a duplication event
- Xenologs -- sequences diverged after a horizontal transfer (e.g., by virus)

Gene Duplication

Speciation events

2B

1B

3A

3B

2A

1A

Species Phylogeny

Phylogenies can be constructed to describe evolution genes.

Three species termed 1,2,3.

Two paralog genes A and B.

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.

A natural model to consider is that of rooted trees

Common

Ancestor

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.

Tree A

Tree B

Rooted versus unrooted trees

Tree C

b

a

c

Represents the three rooted 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

- 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

- Distance- A tree that recursively combines two nodes of the smallest distance.
- Parsimony – A tree with a total minimum number of character changes between nodes.
- Maximum likelihood - Finding the best Bayesian network of a tree shape. The method of choice nowadays. Most known and useful software called phylip uses this method.

Input: distance matrix between species

For two sequences si and sj, perform a pairwise (global)

alignment. Let f = the fraction of sites with different residues. Then

Outline:

- Cluster species together
- Initially clusters are singletons
- At each iteration combine two “closest” clusters to get a new one

(Jukes-Cantor Model)

Unweighted Pair Group Method using

Arithmetic Averages (UPGMA)

- UPGMA is a type of Distance-Basedalgorithm.
- Despite its formidable acronym, the method is simple and intuitively appealing.
- It works by clustering the sequences, at each stage amalgamating two clusters and, at the same time, creating a new node on the tree.
- Thus, the tree can be imagined as being assembled upwards, each node being added above the others, and the edge lengths being determined by the difference in the heights of the nodes at the top and bottom of an edge.

An example showing how UPGMA produces

a rooted phylogenetic tree

An example showing how UPGMA produces

a rooted phylogenetic tree

An example showing how UPGMA produces

a rooted phylogenetic tree

An example showing how UPGMA produces

a rooted phylogenetic tree

An example showing how UPGMA produces

a rooted phylogenetic tree

- Let Ci and Cj be clusters, define distance between them to be
- When we combine two cluster, Ci and Cj, to form a new cluster Ck, then
- Define a node K and place its children nodes at depth
d(Ci, Cj)/2

UPGMA construction on five objects.

The length of an edge = its (vertical) height.

9

8

d(7,8) / 2

6

7

d(2,3) / 2

2

3

4

5

1

This phylogenetic tree has all leaves in the same level. When this property holds, the phylogenetic tree is said to satisfy a molecular clock. Namely, the time from a speciation event to the formation of current species is identical for all paths (wrong assumption in reality).

3

2

2

3

4

1

1

4

UPGMA constructs trees that satisfy a molecular clock, even if the true tree does not satisfy a molecular clock.

UPGMA

Proof idea: Move a horizontal line from the bottom of the T to the top. Whenever an internal node is formed, the algorithm will create it.

Proposition: If the distance function is derived by adding edge distances in a tree T with a molecular clock, then UPGMA will reconstruct T.

k

c

b

j

a

i

Molecular clock defines additive distances, namely,

distances between objects can be realized by a tree:

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

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?

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.

k

c

b

j

m

a

i

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

Thus

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

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

k

i

l

j

Theorem: A set M of L objectsis additive iffany 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 4P Condition: By the figure...

Induction on the number of objects, L.

For L≤ 3 the condition is empty and tree exists.

Consider L=4.

B = d(i, k) +d(j, l) = d(i, l) +d(j, k) ≥ d(i, j) + d(k, l) = A

k

c

f

l

Let y = (B – A)/2 ≥ 0. Then the tree should look as follows:

We have to find the distances

a,b, c and f.

n

y

b

a

m

i

j

- 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), c+f = d(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

l

k

f

c

n

y

b

j

m

a

i

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.

B = d(i, k) +d(j, l) = d(i, l) +d(j, k) ≥ d(i, j) + d(k, l) = A,

y = (B – A)/2 ≥ 0.

L

cij

bij

j

aij

mij

i

- 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:

L

cij

bij

j

aij

mij

T’

i

- 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).

L

cij

k

bij

j

mij

n

aij

T’

i

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}.

L

cij

k

bij

j

n

mij

aij

T’

i

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 hypothesis, and

d(L,j) = dT(L,j) by the construction.

Hence d(L,k) = dT(L,k). QED

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}.

- The four point condition doesn’t provides an algorithm to construct a tree from distance matrix, or to decide whether there is such a tree.
- The first methods for constructing trees for additive sets used neighbor joining methods:

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

j

c

b

m

k

a

i

Thus

- Let i, jbe neighboring leaves in a tree, let k be their parent, and let m be any other vertex.
- The formula
- shows that we can compute the distances of k to all other leaves. This suggest the following method to construct tree from a distance matrix:
- Find neighboring leaves i, j in the tree,
- Replace i, j by their parent kand recursively construct a tree T for the smaller set.
- Add i, j as children of kinT.

A

B

C

D

How can we find from distances alone a pair of nodes which are neighboring leaves?

Closest nodes aren’t necessarily neighboring leaves.

Next we show one way to find neighbors from distances.

T1

T2

m

l

k

i

j

Theorem [Saitou & Nei] Assume all edge weights are positive. If D(i, j) is minimal (among all pairs of leaves), then i and j are neighboring leaves in the tree.

The proof is rather involved!

m

k

i

j

- Set L to contain all leaves
Iteration:

- Choose i, j such that D(i, j) is minimal
- Create new node k, and set
- Remove i, j from L, and add k
Termination Condition:when |L| =2 , connect two remaining nodes

1

•

•

a

d

c

e

b

•

g

f

2

•

4

3

5

•

Let (i, j) = d(i, j) – (ri + rj)

“ L-2 ” is crucial!

D12 = (a+c+d) – (1/3)(a+b + a+c+d + a+c+e+f

+ a+c+e+g + d+c+a + d+c+b + d+e+f + d+e+g)

D13 = (a+b) – (1/3)(a+b + a+c+d + a+c+e+f

+ a+c+e+f + b+a + b+c+d + b+c+e+f + b+c+e+g)

Hence D12 - D13 = (4/3) c

B

A

e2

F

E

e3

e1

C

D

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=(p , q):

Ni(e) = number of elements in the set: {k : e is on p(i, k), k is a leave}.

e.g.

ND(e1) = 3, ND(e2) = 2, ND(e3) = 1

NC(e1) = 1

Rest of T

l

k

i

j

T1

T2

l

k

i

j

Proof of Theorem:Assume for contradiction that D(i, j) is minimized for i, j which are not neighboring leaves.

Let (i, l, ..., k, j) be the path from i to j. let T1 and T2 be the subtrees rooted at k and l which do not contain edges from P(i,j) (see figure).

Notation: |T| = #(leaves in T).

T2

m

l

k

i

j

Case 1:i or j has a neighboring leaf. WLOG j has a neighbor leaf m.

A. D(i,j) - D(m,j)=(L-2)(d(i,j) - d(j,m)) – (ri+rj) +(rm+ rj)

=(L-2)(d(i,k)-d(k,m))+rm-ri

B.rm-ri ≥ (L-2)(d(k,m)-d(i,l)) + (4-L)d(k,l)

(since for each edge eP(k,l), Nm(e) ≥ 2 and Ni(e) L-2)

Substituting B in A:

D(i,j) - D(m,j) ≥

(L-2)(d(i,k)-d(i,l)) + (4-L)d(k,l)

= 2d(k,l)> 0,

contradicting the minimality assumption.

T1

m

n

p

T2

k

l

i

j

Case 2: Not case 1. Then both T1andT2contain 2 neighboring leaves.

WLOG |T2|≥|T1|. Let n,m be neighboring leaves in T1. We shall prove that D(m,n) < D(i,j), which will again contradict the minimality assumption.

A. 0 ≤ D(m,n) - D(i,j)= (L-2)(d(m,n) - d(i,j) ) + (ri+rj) – (rm+rn)

B. rj-rm< (L-2)(d(j,k) – d(m,p)) + (|T1|-|T2|)d(k,p)

C. ri-rn <(L-2)(d(i,k) – d(n,p)) + (|T1|-|T2|)d(l,p)

Adding B and C, noting that d(l,p)>d(k,p):

D. (ri+rj) – (rm+rn) < (L-2)(d(i,j)-d(n,m)) +

2(|T1|-|T2|)d(l,p)

T1

m

n

p

T2

k

Substituting D in the right hand side of A:

D(m,n ) - D(i,j)< 2(|T1|-|T2|)d(l,p) ≤ 0,

as claimed. QED

l

i

j

Select an arbitrary leave r.

- For each pair of labeled nodes (i, j)let C(i, j) be defined by the following 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

m

k

i

j

- Set M to contain all leaves, and select a leave r. |M|=L
- If L =2, return tree of two vertices
Iteration:

- Choose i, j such that C(i, j) is maximal
- Create new vertex k, and set
- remove i, j, and add k to M
- Recursively construct a tree on the smaller set, then add i, j as children on k, at distances d(i,k) and d(j,k).

m

k

i

j

Naive Implementation:

Initialization:Θ(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).