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Phylogenetic Trees Lecture 11

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Phylogenetic Trees Lecture 11

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Phylogenetic TreesLecture 11

Sections 7.1, 7.2, in Durbin et al.

.

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

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

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 weighted tree that realizes the distances between the objects.
- Parsimony – A tree with a total minimum number of character changes between nodes.

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.

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 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:

eg, there is no tree which realizes the below distances.

k

i

l

j

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 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

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

f

l

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.

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, and

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

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

QED