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TREES. Gorilla. Human. Chimp. Human. Chimp. Gorilla. Trees. =. Gorilla. Chimp. Human. =. =. Chimp. Human. Gorilla. s1. s1. s2. s2. s3. s3. s4. s4. s5. s5. Same thing…. =. Terminology. A branch = An edge. The root. Internal nodes. Chicken. Gorilla. Human. Chimp.

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TREES


Gorilla

Human

Chimp

Human

Chimp

Gorilla

Trees

=

Gorilla

Chimp

Human

=

=

Chimp

Human

Gorilla


s1

s1

s2

s2

s3

s3

s4

s4

s5

s5

Same thing…

=


Terminology

A branch =An edge

The root

Internal nodes

Chicken

Gorilla

Human

Chimp

External node - leaf


אלו מהמשפטים הבאים נכון, בהתייחס לעץ הנ"ל?

א. האדם והגורילה יותר קרובים זה לזה מהשימפנזה והגורילה.

ב. האדם קרוב לתרנגולת ולברווז באותה מידה.

ג. התרנגולת יותר קרובה לגורילה מהאדם.

ד. א'+ב'.

ה. א'+ג'.

ו. ב'+ג'.

ז. א'+ב'+ג'.

ח. אף תשובה אינה נכונה.

תרגיל


The maximum parsimony principle.

Tree building


Tree building


Tree building

Evaluate this tree…

s2

s1

s4

s3

s5


Tree building

Gene number 1

1

0

1

s1

s4

s3

s2

s5

1

1

1

0

0


Tree building

1

0

Gene number 1, Option number 1.

1

1

s1

s4

s3

s2

s5

1

1

1

0

0


Tree building

0

1

0

1

1

1

1

0

0

Gene number 1, Option number 2.

s1

s4

s3

s2

s5

Number of changes for gene 1 (character 1) = 1


Tree building

0

1

0

1

0

1

1

0

0

Gene number 2, Option number 1.

s2

s1

s4

s3

s5


Tree building

1

1

0

1

0

1

1

0

0

Gene number 2, Option number 2.

s2

s1

s4

s3

s5


Tree building

0

0

0

0

0

1

1

0

0

Gene number 2, Option number 3.

s2

s1

s4

s3

s5

Number of changes for gene 2 (character 2) = 2


Tree building

0

0

1

0

0

0

0

1

1

Gene number 3, Option number 1.

s2

s1

s4

s3

s5


Tree building

1

0

1

0

0

0

0

1

1

Gene number 3, Option number 2.

s2

s1

s4

s3

s5

Number of changes for gene 3 (character 3) = 1


Tree building

1

1

1

1

1

1

0

0

1

Gene number 4, Option number 1.

s2

s1

s4

s3

s5


Tree building

0

0

0

1

1

1

0

0

1

Gene number 4, Option number 2.

s2

s1

s4

s3

s5

Number of changes for gene 4 (character 4) = 2


Tree building

Gene number 5 is the same as Gene number 4

Number of changes for gene 5 (character 5) = 2


Tree building

0

0

0

0

0

1

0

0

0

Gene number 6, 1 option only:

s2

s1

s4

s3

s5

Number of changes for gene 6 (character 6) = 1


Tree building

Sum of changes

Number of changes for gene 1 (character 1) = 1

Number of changes for gene 2 (character 2) = 2

Number of changes for gene 3 (character 3) = 1

Number of changes for gene 4 (character 4) = 2

Number of changes for gene 5 (character 5) = 2

Number of changes for gene 6 (character 6) = 1

Sum of changes for this tree topology = 9

Can we do better ???


Tree building

The MP (most parsimonious) tree:

s2

s1

s4

s3

s5

Sum of changes for this tree topology = 8


How to efficiently compute the MP score of a tree


The Fitch algorithm (1971):

{A,C}

U

{A,C,G}

U

{A,G}

U

{A,C}

U

C

A

C

A

G

Chicken

Duck

Gorilla

Human

Chimp

Postorder tree scan. In each node, if the intersection between the leaves is empty: we apply a union operator. Otherwise, an intersection.


Number of changes

{A,C}

U

{A,C,G}

U

{A,G}

U

{A,C}

U

C

A

C

A

G

Chicken

Duck

Gorilla

Human

Chimp

Total number of changes = number of union operators => 3 in this case.


תרגיל

CAAG

GAAA

GCGA

GACA

GGGA

Chicken

Duck

Gorilla

Human

Chimp

Find minimum number of changes.


Chimpanzee

Gorilla

Human


Human

ACTAG

Chimp

ACAAC

Gorilla

AAAAT

Gorilla

Chimp

Human

U

Position 1 A A A

0

Position 2 A C C

1

Position 3 A A T

1

Position 4 A A A

0

Position 5 T C G

2

4


Human

ACTAG

Chimp

ACAAC

Gorilla

AAAAT

Gorilla

Chimp

Human

U

Position 1 A A A

0

Position 2 A C C

1

Position 3 A A T

1

Position 4 A A A

0

Position 5 T C G

2

4


Human

ACTAG

Chimp

ACAAC

Gorilla

AAAAT

Chimp

Gorilla

Human

U

Position 1 A A A

0

Position 2 C A C

1

Position 3 A A T

1

Position 4 A A A

0

Position 5 C T G

2

4


Human

ACTAG

Chimp

ACAAC

Gorilla

AAAAT

Chimp

Gorilla

Human

Gorilla

Chimp

Human

Gorilla

Chimp

Human

These 3 trees will ALWAYS get the same score


The unrooted tree represents a set of rooted trees

3

1

2

3

1

2


A general observation: the position of the root does not affect the MP score.

E

A

D

D E C A B

B

C

A B C E D

A B C E D


1

0

Intuition as to why rooting does not change the score.

1

1

s1

s4

s3

s2

s5

1

1

1

0

0

The change will always be on the same branch, no matter where the root is positioned…


Which is not a rooted version of this tree?

E

T3

תרגיל

A

D

C E D A B

B

C

T1

T2

A B D E C

A B C D E


Gorilla gorilla

(Gorilla)

Pan troglodytes (Chimpanzee)

Homo sapiens (human)

Gallus gallus (chicken)


Human

Human

Human

Chicken

Chimp

Chimp

Gorilla

Chicken

Gorilla

Chimp

Gorilla

Chicken

Evaluate all 3 possible UNROOTED trees:

MP tree


Rooting based on a priori knowledge:

Human

Chicken

Gorilla

Chimp

Chicken

Gorilla

Human

Chimp


Ingroup / Outgroup:

Chicken

Gorilla

Human

Chimp

OUTGROUP

INGROUP


Chicken

Duck

Gorilla

Human

Chimp

Subtrees

A subtree


Monophyletic groups

Chicken

Gorilla

Human

Chimp

The Gorilla+Human+Chimp are monophyletic.A clade is a monophyletic group.


Paraphyletic = Non-monophyletic groups

Whale

Chimp

Drosophila

Zebrafish

The Zebrafish+Whale are paraphyletic


When an unrooted tree is given, you cannot know which groups are monophyletic. You can only say which are not.

Human

Chicken

Rat

Gorilla

Chimp

Chicken + Rat seems to be monophyletic but they are not, since the root of the tree is between Chicken and the rest.

Human and Gorilla are not monophyletic no matter where the root is…


HOW MANY TREES


a

b

a

b

c

b

a

a

c

c

b

a

c

a

c

b

b

c

a

d

b

d

d

b

b

c

a

b

b

c

a

a

a

a

a

d

d

c

d

d

c

c

b

d

d

b

c

c

b

c

b

a

a

b

d

a

d

c

d

c

b

d

c

b

b

d

d

c

a

a

a

TR = “TREE ROOTED”

How many rooted trees

N=2, TR(2) = 1

N=3, TR(3) = 3

N=4, TR(4) = 15


a

b

c

a

b

b

a

c

d

d

b

c

a

TR = “TREE ROOTED”

How many rooted trees

c

c

c

2 branches. 3 possible places to add “c”

4 branches. 5 possible places to add “d”

6 branches. 7 possible places to add “e”

The number of branches is increased by 2 each time. The number of branches is an

arithmetic series.

0,2,4,6,8,…. A(n) = A(1)+(n-1)d. A(1) = 0; d=2. => A(n) = (n-1)*2 = 2n-2


a

b

TR = “TREE ROOTED”

How many rooted trees

The number of branches is increased by 2 each time. The number of branches is an

arithmetic series.

0,2,4,6,8,…. A(n) = A(1)+(n-1)d. A(1) = 0; d=2. => A(n) = (n-1)*2 = 2n-2

c

c

c

2 branches. 3 possible places to add “c”

Each time we can add a new branch in Br(n)+1 places. [Br(n)=number of branches]

[Tr(n)=number of trees with n sequences]

TR(n+1) = TR(n)*(BR(n)+1)=TR(n)*(2n-1)

TR(5) = TR(4)*7=TR(3)*5*7=TR(2)*3*5*7=1*3*5*7

TR(n) = 1*3*5*7*…..*(2n-3)


TR = “TREE ROOTED”

How many rooted trees

n!=1*2*3*4*5*6…..*n = n factorial.

TR(n) = 1*3*5*7*…..*(2n-3) =

1*2*3*4*5*6*7*…*(2n-3)

=

2*4*6*8*….*(2n-4)

1*2*3*4*5*6*7*…*(2n-3)

=

(2*1)*(2*2)*(2*3)*(2*4)*….*(2*(n-2))

(2n-3)!

=

(2(n-2))*(1*2*3*4*….(n-2))

(2n-3)!

=

(2(n-2))*(n-2)!


TR = “TREE ROOTED”

How many rooted trees

TR(n) = 1*3*5*7*…..*(2n-3) =

=(2n-3)!!

(2n-3)!

=

(2(N-2))*(n-2)!


HEURISTIC SEARCH


There are many trees..,

We cannot go over all the trees. We will try to find a way to find the best tree.

These are approximate solutions…


Finding the maximum is the same thing as finding the minimum

Say we have a computer procedure that given a function, it finds its minimum, and

we want to find the maximum of a function f(x). We can just find the minimum of

-f(x) and this is minus the maximum of f(x).

Example.

f(0) = 3; f(1) = 7; f(2) = -5; f(3) = 0; max f(x) = 7. argmax f(x) = 1;

-f(0)=-3; -f(1) = -7; -f(2) = 5; -f(3) =0; min(-f(x)) = -7. argmax –(f(x) = 1;


Score = 1700


Score = 1825

Score = 1700

Score = 1710

Score = 1695

Score = 1410


Score = 1828

Score = 1825

Score = 1910

Score = 1800


Max score = 2900


Problem number 1: local maximum

Score = 3100

Global max

Score = 2900

Local max

Score = 2100


This algorithm is “greedy” – it seizes the first improvement encountered.

One way to avoid local maxima is to start from many random starting points


Option 1

Several options to define a neighbor.

Option 2


B

C

D

D

C

B

B

C

A

A

A

D

Nearest-neighbor interchange

Each internal branch

defines two neighbors


How many neighbors do we check each time?

B

C

Internal branches

A

NNI is possible only

in internal branches

D

External branches

E

For unrooted trees of n taxa, we have 2n-3 branches. However, only internal branches are interesting, thus we have n-3. Each defines two neighbors, thus the total number of neighbors in each NNI cycle is 2n-6.


I am greedy


Greedy variants

  • Most greedy: Start searching your neighbors. If you find something better – move there, and start the search again.

  • Just greedy: Check ALL your neighbors. Move to the one that is the highest.

  • Smart greedy: Try all NNI of trees that are tied for the best score.

There are many other variants of the greedy search

that would not be discussed in this course.


likelihood


  • Parsimony has many shortcomings. To name a few:

  • All changes are counted the same, which is not true for biological systems (Leu->Ile is much more likely than Leu->His).

  • Cannot take biological context into account (secondary structures, dependencies among sites, evolutionary distances between the analyzed organisms, etc).

  • Statistical basis questionable.


Alternative:

MAXIMUM-LIKELIHOOD METHOD.


Maximum likelihood uses a probabilistic model of evolution

Each amino acid has a certain probability to change and this probability depends on the evolutionary distances.

Evolutionary distances are inferred from the entire set of sequences.


Evolutionary distances

Positions can be conserved because of two reasons. Either because of functional constraints, or because of short evolutionary time.

5 replacements in 10 positions between 2 chimps, is considered very variable. 5 replacements between human, and cucumber, is not considered that variable…

Maximum likelihood takes this information into account.


The likelihood computations

X

t2

t1

Y

t4

t3

Z

t6

t5

K

M

A

C

We can infer the phylogenetic tree using maximum likelihood. This is more accurate than maximum parsimony.


Maximum likelihood tree reconstruction

This is incredibly difficult (and challenging) from the computational point of view, but efficient algorithms to find approximate solutions were developed.


HIV evolution – an example of using phylogeny tools


Human Immunodeficiency Virus (HIV)

The virus = HIV

The disease = AIDS (Aquired Immunodeficiency Syndrome)

First recognized clinically in 1981

By 1992, it had become the major cause of death in individuals 25-44 years of age in the States.


HIV Statistics

  • Till Dec 2007: 25 million people died of AIDS (20 million in 2002)

  • People living with HIV/AIDS in 2007 33.2 million

  • Africa has 12 million AIDS orphans (2007). 1 out of 3 children in some areas lost at least one of his/her parents


HIV is a lentivirus

Species = HIV

Genus = Lentiviruses

Family = Retroviridae

Lentiviruses have long incubation time, and are thus called “slow viruses”.


HIV-1 and HIV-2

In 1986, a distinct type of HIV prevalent in certain regions of West Africa was discovered and was termed HIV type 2.

Individuals infected with type 2 also had AIDS, but had longer incubation time and lower morbidity (# of cases/population size).


HIV subtypes


HIV subtypes

published by the International AIDS Vaccine Initiative


Five lines of evidence have been used to substantiate zoonotic transmission of primate lentivirus:

1. Similarities in viral genome organization;

2. Phylogenetic relatedness;

3. Prevalence in the natural host;

4. Geographic coincidence;

5. Plausible routes of transmission.


For HIV-2, a virus (SIVsm) that is genomically indistinguishable and closely related phylogenetically was found in substantial numbers of wild-living sooty mangabeys whose natural habitat coincides with the epicenter of the HIV-2 epidemic


מנגבי, קוף ארוך זנב מסוג סרקוסבוס מצוי באזורי היערות של אפריקה


Close contact between sooty mangabeys and humans is common because these monkey are hunted for food and kept as pets.

No fewer than six independent transmissions of SIVsm to humans have been proposed.

The origin of HIV-1 is much less certain.


HIV and SIV tree based on maximum parsimony

1990


Virus A

Primate A

Primate B

Virus B

Primate C

Virus C

Host-pathogen co-evolution in other SIV

This tree can be explained by co-evolution of virus and host.


Phylogenetic tree

1999

There are at least two different HIV-1 clades, and two different SIVcpz clades


2006. Nature


The origin of HIV-O

“We tested 378 chimpanzees and 213 gorilla fecal samples from remote forest regions in Cameroon for HIV-1 cross-reactive antibodies”

“Surprisingly, 6 of 213 fecal samples from wild-living gorillas also gave a positive HIV-1 signal”


Bayesian analysis

HIV-1 O is a sister clade of SIV from Gorilla!


The origin of HIV-O

It seems that chimpanzee transmitted SIV to gorilla and gorilla to human type O, or

Chimpanzee transmitted to both gorilla and to human type O

Note: gorilla and chimps rarely interact + gorilla are herbivores

?


Thank You…

Thanks

תודה


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