.

Phylogenetic TreesLecture 12

Based on pages 160-176 in Durbin et al (the black text book).This class has been edited from Nir Friedman’s lecture which was available at www.cs.huji.ac.il/~nir. Pictures from Tal Pupko slides. Changes by Dan Geiger and Shlomo Moran. 2

EvolutionEvolution of new organisms

is driven byDiversity

Different individuals carry different variants of the same basic blue print

MutationsThe DNA sequence can be changed due to single base changes, deletion/insertion of DNA segments, etc.

Selection bias

3

The Tree of Life

Sour

ce: A

lber

ts e

t al

4

Tree of life- a better picture

D’après Ernst Haeckel, 1891

5

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.

6

Morphological vs. MolecularClassical phylogenetic analysis: morphological features:number of legs, lengths of legs, etc.

Modern biological methods allow to use molecular featuresGene sequencesProtein sequences

Analysis based on homologous sequences (e.g., globins) in different species

Important for many aspects of biologyClassificationUnderstanding biological mechanisms

7

Morphological topology(Based on Mc Kenna and Bell, 1997)

BonoboChimpanzee

ManGorillaSumatran orangutanBornean orangutanCommon gibbon

Barbary apeBaboonWhite-fronted capuchinSlow lorisTree shrewJapanese pipistrelleLong-tailed batJamaican fruit-eating batHorseshoe bat

Little red flying foxRyukyu flying foxMouseRatVoleCane-ratGuinea pigSquirrelDormouseRabbitPikaPigHippopotamusSheepCowAlpacaBlue whaleFin whaleSperm whaleDonkeyHorseIndian rhinoWhite rhinoElephantAardvarkGrey sealHarbor sealDogCatAsiatic shrewLong-clawed shrewSmall Madagascar hedgehog

HedgehogGymnureMoleArmadilloBandicootWallarooOpossumPlatypus

Archonta

Glires

CarnivoraInsectivoraXenarthra

Ungulata

8

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).

9

Mitochondrial topology(Based on Pupko et al.,)

DonkeyHorseIndian rhinoWhite rhinoGrey sealHarbor sealDogCatBlue whaleFin whaleSperm whaleHippopotamusSheepCowAlpacaPig

Little red flying foxRyukyu flying foxHorseshoe batJapanese pipistrelleLong-tailed batJamaican fruit-eating bat

Asiatic shrewLong-clawed shrew

MoleSmall Madagascar hedgehogAardvarkElephantArmadilloRabbitPikaTree shrewBonoboChimpanzeeManGorillaSumatran orangutanBornean orangutanCommon gibbonBarbary apeBaboon

White-fronted capuchinSlow lorisSquirrelDormouseCane-ratGuinea pigMouseRatVoleHedgehogGymnureBandicootWallarooOpossumPlatypus

Perissodactyla

CarnivoraCetartiodactyla

Rodentia 1

HedgehogsRodentia 2

Primates

ChiropteraMoles+ShrewsAfrotheria

XenarthraLagomorpha+ Scandentia

10

Nuclear topology(Based on Pupko et al. slide)(tree by Madsenl)

Round Eared Bat

Flying Fox

Hedgehog

Mole

Pangolin

Whale

Hippo

Cow

Pig

Cat

Dog

Horse

Rhino

Rat

Capybara

Rabbit

Flying Lemur

Tree Shrew

Human

Galago

Sloth

Hyrax

Dugong

Elephant

Aardvark

Elephant Shrew

Opossum

Kangaroo

1

23

4

Cetartiodactyla

Afrotheria

ChiropteraEulipotyphla

Glires

Xenarthra

CarnivoraPerissodactyla

Scandentia+Dermoptera

Pholidota

Primate

11

Theory of EvolutionBasic 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

12

Phylogenenetic trees

Aardvark Bison Chimp Dog Elephant

Leafs - current day speciesNodes - hypothetical most recent common ancestorsEdges length - “time” from one speciation to the next

13

Dangers in Molecular PhylogeniesGene and protein sequences can be homologous for various reasons:

Orthologs -- sequences diverged after a speciationevent. Indicative of a new specie.Paralogs -- sequences diverged after a duplicationevent.Xenologs -- sequences diverged after a horizontaltransfer (e.g., by virus).

14

Gene PhylogeniesPhylogenies can be constructed to describe evolution genes.

Species Phylogeny

Speciation events

Gene Duplication

1A 2A 3A 3B 2B 1B

Three species termed 1,2,3.Two paralog genes A and B.

15

Dangers of ParalogsIf we happen to consider only species 1A, 2B, and 3A, 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.

Speciation events

Gene Duplication

2B 1B3A 3B2A1A

In the sequel we assume all given sequences are orthologs.16

Types of TreesA natural model to consider is that of rooted trees

CommonAncestor

17

Types of treesUnrooted tree represents phylogeny without the root node

Depending on the model, data from current day species does not distinguish between different placements of the root.

In this example there are seven possible ways to place a root.18

Rooted versus unrooted treesTree bTree a

ab

c

Represents the three rooted trees

Tree c

Slide by Tal Pupko

19

Positioning Roots in Unrooted TreesWe 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

Falcon

Proposed root

Aardvark Bison Chimp Dog Elephant

20

Type of Data Distance-based

Input is a matrix of distances between speciesCan be fraction of residue they disagree on, or alignment score between them, or …

Character-basedExamine each character (e.g., residue) separately

21

Three Methods of Tree Construction

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.http://evolution.genetics.washington.edu/phylip.html

22

Distance-Based (1st type Method)Input: distance matrix between speciesOutline:

Cluster species togetherInitially clusters are singletonsAt each iteration combine two “closest” clusters to get a new one

23

UPGMA ClusteringLet 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 daughter nodes at depth d(Ci,Cj)/2

ääÍ Í

=i jCp Cqji

ji qpdCC

1CCd ),(||||

),(

||||),(||),(||

),(ji

ljjliilk CC

CCdCCCdCCCd

+

+=

24

Example

UPGMA construction on five objects.The length of an edge = its (vertical) height.Claim (exercise): A node is never placed below its children.

2 3

98

0.5d(7,8)0.5d(2,3)

4 5 1

6 7

25

Molecular clock

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).

26

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

2 3 4 1

1

23

4

UPGMA

27

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

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. 28

AdditivityMolecular clock defines additive distances, namely,

distances between objects can be realized by a tree:

ab

c

i

j

k

cbkjdcakidbajid

+=

+=

+=

),(),(),(

29

Basic property of AdditivitySuppose input distances are additiveFor any three leaves

Thus

m

cbmjdcamidbajid

+=+=+=

),(),(),(

a

cb j

k

i

)),(),(),((21),( jidmjdmidmkd -+=

30

Constructing additive trees:The neighbor finding problem

Can we use this fact to construct trees assuming only additivity (but not a molecular clock)?Yes. The formula shows that if we knew that i and j are neighboring leaves, then we can construct their parent node k and compute the distances of k to all other leaves m.

We remove nodes i,j and add k.

)),(),(),((21),( jidmjdmidmkd -+=

31

Neighbor FindingHow can we find from distances alone that a pair of nodes i,j

are neighboring leaves? Closest nodes aren’t necessarily neighbors.

A B

CD

Next we show one way to find neighbors from additive distances.32

Neighbor Finding

is a leafFor a leaf , let ( , ).i

mi r d i m= ä

Definition: Let , be leaves Then( , ) ( 2) ( , ) ( )

where is the number of leaves ini j

i jD i j L d i j r r

L T= - - +

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

ij

kl

m

T1T2

33

Neighbor Joining AlgorithmSet L to contain all leaves

Iteration:Choose i,j such that D(i,j) is minimalCreate new node k, and set

remove i,j from L, and add kTerminate:

when |L| =2, connect two remaining nodes

1( , ) ( ( , ) ( , ) ( , )) (for some )2

( , ) ( , ) ( , )1for each node , ( , ) ( ( , ) ( , ) ( , ))2

d i k d i j d i m d j m m

d j k d i j d i k

m d k m d i m d j m d i j

= + -

= -

= + -

ij

k

m

34

Neighbor FindingNotations 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) = 1NC(e1) = 1

A B

C D

e1e3

e2EF

35

Neighbor FindingNotation: For e=(i,m), we denote d(i,m) by d(e).

ij

kl

Rest of T is a leaf

Observe that ( , ) ( ) ( ),i im e E

r d i m d e N eÍ

= =ä ä

[ ]=-=- äÍ ),(

)()()(jipe

jiji eNeNedrr

Lemma: For leaves i,j connected by a path (i,l,…,k,j),

[ ] [ ] [ ])1(1),(1)1(),()()()(),(

--+--+-= äÍ

LjkdLlideNeNedklpe

ji

[ ] [ ]äÍ

-+--=-),(

)()()(),(),()2(klpe

jiji eNeNedjkdlidLrr36

Neighbor FindingProof of Theorem: Assume by contradiction that D(i,j) is minimal 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 l and k.

Let |T| denote the number of leaves in T.

ij

kl

T1T2

37

Neighbor FindingCase 1: i or j has a neighboring leaf. WLOG j and m are such leaves.A. D(i,j) - D(m,j)=(L-2)(d(i,j) - d(j,m) ) – (ri+rj) + rm+ rj {Definition}

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

i j

kl

mT2

B. rm-ri (L-2)(d(k,m)-d(i,l)) + (4-L)d(k,l) {Lemma+Figure}

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

so Nm(e)- Ni(e ) 4-L )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.

38

Neighbor FindingCase 2: Not case 1. Then both T1 and T2 contain 2 neighboring leaves.We show that if D(i,j) is minimal, then we must have both |T1| > |T2| and |T2| > |T1| - which is a contradiction, hence D(i,j) is not minimal.

i j

kl

mn

p

T1

T2We prove that |T1| > |T2| by assuming that |T1| |T2| and reaching a contradiction.The proof that |T2| > |T1| is similar.Let n,m be neighboring leaves in T1.

39

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

i j

kl

mn

p

T1

T2

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) and using the assumption |T1| - |T2| 0:D. (ri+rj) – (rm+rn) < (L-2)(d(i,j)-d(n,m)) +

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

Substituting D in the right hand side of A:0 D(m,n) - D(i,j)< 2(|T1|-|T2|)d(k,p),hence |T1|-|T2| > 0, a contradiction.

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