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General concept of distance based methods

- Two steps:
- Compute a distance D(i,j) between any two sequences i and j.
- Find the tree that agrees most with the distance table.

Simplest distance: the “p” distance

SEQ1 AACAAGCG

SEQ2 AACGAGCA

There are 2 differences, so the distance = 2.

The problem is that now, if you have a longer pair of sequences

SEQ3 AACAAGCGCCCTCAGTCCGCTCGCACAA

SEQ4 AACGAGCACCCTCAGTCCGCTCGCACAA

The distance is still 2, but in fact, the distance between 3 and 4 should be smaller than the distance between 1 and 2.

Simplest distance: the “p” distance

SEQ1 AACAAGCG

SEQ2 AACGAGCA

There are 2 differences, the length = 8, so the distance is 2/8

This is called the p distance.

There are better and more accurate methods to compute the distance D(i,j) between any two sequences i and j. For example, one can take into account different probabilities between transitions and transversions…

From a distance table to a tree

Each tree has branch lengths from which “predicted” set of distances can be computed: d(i,j) (small d, denotes the distance of the branches, unlike the observed pairwise distances D).

Human

d(Human,Chimp) = 0.55

d(Human,Gorilla) = 0.71

d(Chimp, Gorilla) = 0.66

0.3

0.41

Gorilla

0.25

Chimp

From a distance table to a tree

The question is can we find branch lengths, so that the d’s are equal to the D’s?

Human

D(Human,Chimp) = 0.3

D(Human,Gorilla) = 0.4

D(Chimp, Gorilla) = 0.5

X

Y

Gorilla

Z

Chimp

From a distance table to a tree

Human

D(Human, Chimp) = 0.3

D(Human, Gorilla) = 0.4

D(Chimp, Gorilla) = 0.5

X

Y

Gorilla

Z

Chimp

d(Human, Chimp) = X+Z

d(Human, Gorilla) = X+Y

d(Chimp, Gorilla) = Y+Z

X+Z = 0.3

X+Y = 0.4

Y+Z = 0.5

Y = 0.3

Z = 0.2

X = 0.1

YES

Y-Z = 0.1

Y+Z = 0.5

Human

D(Human, Chimp) = D1

D(Human, Gorilla) = D2

D(Chimp, Gorilla) = D3

X

Y

Gorilla

Z

Chimp

d(Human, Chimp) = X+Z

d(Human, Gorilla) = X+Y

d(Chimp, Gorilla) = Y+Z

X+Z = D1

X+Y = D2

Y+Z = D3

We get 3 equations with 3 variables: there’s always a solution!

Human

D(Human, Chimp) = D1

D(Human, Gorilla) = D2

D(Chimp, Gorilla) = D3

X

Y

Gorilla

Z

Chimp

d(Human, Chimp) = X+Z

d(Human, Gorilla) = X+Y

d(Chimp, Gorilla) = Y+Z

X+Z = D1

X+Y = D2

Y+Z = D3

Show that for a 3 taxa tree, there’s always a solution and it is given by:

Z=0.5(D1-D2+D3), Y=0.5(D2+D3-D1)

X=0.5(D1+D2-D3)

D

A

5 Variables,

6 Equations,

It might be that there’s no solution

X

Y

W

Z

V

B

C

D(A, B) = 2 D(A, D) = 3

D(A, C) = 3 D(B, C) = 3

D(B, D) = 3 D(C, D) = 4

An example of a case where there’s no solution (v=w=x=y=z=1 solves the first 5 equations)

In real life, for n>3 sequences, there is never a solution.

One might try to find the “best” solution.

The simplest case where it might be that equations have no solution: two equations with 1 parameter

a = 2

a = 3

We want to find the “best” solution which solves these equations

Putting it another way:

a – 2 = 0

a – 3 = 0

Let’s assign parameters instead of 0

a – 2 = e1

a – 3 = e2

Ideally, we want e1, and e2 to be as small as possible (e1=e2=0 could be the best).

a – 2 = e1

a – 3 = e2

We want the distance of the point (e1,e2) from (0,0) to be the smallest.

I.e., we want to find “a” that satisfies:

Sqrt(e12+e22) is minimum.

The term: sqrt(e12+e22) reaches its minimum when the term: e12+e22 reaches its minimum.

So for:

a – 2 = e1

a – 3 = e2

we want to minimize: [(a-2)2+(a-3)2]

Min [(a-2)2+(a-3)2]=

Min[2a2-10a+13]=

Min[2a2-10a]=

Min[a2-5a].

a2-5ais a parabola that crosses the X axis at a=0, and a=5, and its minimum is at a=2.5

So for the simplest of two equations with 1 parameter

a = 2

a = 3

The “best” solution is a = 2.5 which makes sense.

We have the D’s (“observed distances”), and we want to find the d’s (branches) that minimize the expression

For each tree topology we get a different Q. The least square (LS) method searches for the tree with the lowest Q.

The general formula for LS

The w’s are weights that differ between different least square methods.

There are the general heuristic searches.

No branch-and-bound method published so far.

Problem was shown to be NP-complete.

The general formula for LS

Minimum Evolution (ME) for a given topology, it estimates the branch lengths using LS. But unlike LS, it chooses the topology that results in minimal sum of branches.

The Newick tree format is used to represent trees as strings

B

A

D

C

In Newick format: (A,C,(B,D)).

Each pair of parenthesis () enclose a monophyletic group,

and the comma separates the members of the corresponding group.

Neighbor-joining is based on Star decomposition

Red: best pair to group together

B

E

A

(C,B)

C

A

D

D

E

A

((C,B),E)

D

The Neighbour Joining method is used for re-constructing phylogenetic trees. Both the tree topology and branch lengths are estimated. In each stage, the two nearest nodes of the tree (the term "nearest nodes" will be defined in the following paragraphs) are chosen and defined as neighbours in our tree. This is done recursively until all of the nodes are paired together.

The algorithm was originally written by Saitou and Nei, 1987. In 1988 a correction for the paper was published by Studier & Keppler. The correction was related to the main theorem in the algorithm. Studier and Keppler also suggested a slight change to the algorithm which brought the efficiency down to O(n3).We will first of all describe the original algorithm, and then elaborate on the changes made by Studier & Kepler.

Reminder:

OTU’s = operational taxonomic units, or in other words – leaves of the tree.

HTU’s = hypothetical taxonomic units, or in other words – the internal nodes of the tree.

A

D

B

Neighbors, we are …

What are neighbours?Neighbours are defined as a pair of OTU's who have one internal node connecting them.

A and B are neighbours,

C and D are neighbours,

But…

A and C are not neighbours…

The NJ theorem: the NJ algorithm recovers the true tree, if the tree is additive.

NJ is an approximation of the Minimum evolution

In the original article, Saitou and Nei defined the two nearest nodes as the pair of nodes that give the minimal sum of branches when placed in a tree.

- First of all – some notations:
- D(i,j) is defined as the distance between leaves i and j (the observed distance which we have as an input from our distance matrix).
- L(x,y) is defined as the sum of branch lengths between node X and node Y. L is used as a notation for distances between internal nodes, or an internal node to a leaf.

We distinguish between L(X,Y) and D(A,B).

D’s are given as input to the algorithm,

L’s should be inferred…

C

A

Y

X

D

B

- In each round we join as neighbours all possible pairs of leaves and evaluatethe sum of branches for each resultant tree. This means we compare the sum of branches when 1 and 2 are joined as neighbours, denoted as S(1,2), to the sum of branches when 1 and 3 are joined as neighbours, S(1,3), and so on. We look for the i and j pair for whichS(i,j) is minimal, where i and j denote numbers of leaves, and i<j.
- This is why NJ is approximating ME (minimum evolution).

S(1,2) = L(1,X)+L(2,X)+L(X,Y)+L(Y,3)+L(Y,4)+L(Y,5)

3

1

X

Y

4

5

2

The problem is that we don’t know the L’s. We only know the D’s…

1

X

Y

4

5

2

Computing S(1,2)

S(1,2) = L(1,X)+L(2,X)+L(X,Y)+L(Y,3)+L(Y,4)+L(Y,5)

S(1,2) = D(1,2)+L(X,Y)+L(Y,3)+L(Y,4)+L(Y,5)

Since our tree is additive, we can replace L(1,X)+L(2,X), with D(1,2).

1

X

Y

4

5

2

Computing L(X,Y) in terms of the D’s

L(1,X) is counted here N-2 times

Here, -L(1,X) is counted N-2 times

So L(1,X) is canceled out…

N denotes the number of leaves

1

X

Y

4

5

2

Computing L(X,Y) in terms of the D’s

L(3,Y) is counted once here

Once here

Here, -L(3,Y) is counted 2 times

So L(3,Y) is canceled out…

1

X

Y

4

5

2

Computing L(X,Y) in terms of the D’s

L(X,Y) is counted N-2 times here

N-2 here

So L(X,Y) is counted altogether 2(N-2) times. Dividing by 2(N-2) we get L(X,Y)

1

X

Y

4

5

2

Computing L(X,Y) in terms of the Ds

L(3,Y) is counted here N-3 times: once in D(3,4), once in D(3,5), till D(3,N).

1

4

3

1

X

Y

4

5

2

5

2

Finding the best neighbor

So, we compute S(1,2), S(1,3), … , S(4,5) and join the two leaves i and j for which S(i,j) is minimal.

Let’s assume that S(1,2) is minimal in round 1…

We call the new node that joins 1 and 2, X.

1

X

Y

4

5

2

Finding the best neighbor

3

4

12

5

For the next step of the algorithm, we need to create a distance table of (N-1)x(N-1). Let 12 denote the new node that joins 1 and 2. We define:

3

3

4

12

4

Y

5

X

12

5

If now (12) and (5) are joined, it is equivalent to joining (3) and (4). So we can already compute the branch lengths L((12),X),L(5,X), L(3,Y) and L(4,Y).

Complexity of computing S(1,2)

This part requires O(N2) computations

Complexity of the original NJ algorithm

Computing each S(i,j) sums up to N2computations.

There are N2combinations of S(i,j),

and N joining steps.

Altogether, the algorithm is O(N5).

More things to know about the NJ algorithm

- Studier and Keppler introduced a way to reduce the complexity of the algorithm from O(N5) to O(N3).
- The NJ-theorems were not presented.
- BioNJ is a close relative to NJ, but with a slightly better performance.
- NJ constraints.

In minimum evolution branch lengths are computed by the LS method for each possible tree topology.

However, the criterion to choose among tree topologies is not the lowest sum-of-squares, but rather the minimum sum of branch lengths.

Branch lengths measure average number of replacements per position. It is, thus, equal to

the number of replacements per position per year, multiplied by year.

Putting it another way:

Human

Human

Mouse

Mouse

Clearly, the time t, from the root to the tips is the same for all sequences. However, the rate r, can differ, and might depend on factors such as the DNA repair mechanisms, generation time, and much more.

WITH CLOCK

WITHOUTCLOCK

Human

Human

Mouse

Mouse

A molecular clock is the assumption that the rate of all species is approximately the same. Clearly, this is not the general case, but it might be true, for example when comparing very close species of ants. If the rate is the same, the branch lengths should be the same too.

Two kinds of tree search methods

Methods like least-squares, maximum parsimony, minimum evolution and maximum likelihood have an explicit criterion which they try to maximize or minimize.

There are some other methods (UPGMA, WPGMA, NJ) that apply some direct algorithm that result in a tree. These methods are usually very fast, but their statistical justification is unclear. These methods are usually some kind of a clustering algorithm.

Trees which satisfy a molecular clock are called ultrametric.

When trees are ultrametric it is very easy to estimate the LS branch lengths (Farris 1969a).

UPGMA is one such direct method, receiving as input a distance matrix and giving as output an ultrametric tree.

It was suggested by Sokal and Michener (1958).

NOT TO BE USED, UNLESS YOU NEED A VERY FAST METHOD, AND YOU ARE SURE THE TREE IS ULTRAMETRIC!

The algorithm:

Input: a distance matrix D which is symmetric, i.e., D(i,j)=D(j,i).

Variables: for each group of species we give a number which indicates how many species are in this group. N(i) will indicate the number of species in group i. Intially, all sequences have n=1.

- The algorithm:
- Find the i and j that have the smallest D(i,j)
- Create a new group (ij) which has n(ij)=n(i)+n(j)
- Connect i and j to a new node (which corresponds to the new group (ij)). Give the two branches connecting i to (ij) and j to (ij) each length of D(i,j)/2.

- The algorithm:
- 4. Compute the distance between the new group and all other groups (except for i and j) by using:

- The algorithm:
- 5. Delete the columns and rows of the data (modified input) matrix that correspond to groups i and j, and add a column and row for group (ij).
- 6. Go to step 1, unless there is only 1 item left in the data matrix.

O(n3), because it takes O(n2) to find the minimum D(i,j) in a matrix and you have n iterations of that.

However, we can keep a record of the smallest number in each row, and then finding the minimum goes down to O(n).

Thus, the overall time-complexity is O(n2).

Distances based on immunological data of Sarich (1969).

Canis familiaris

Common name = Dog.

The species = familiaris.

Genus = Canis. [First letter always in capital]

Family = Canidae. [First letter always in capital]

Order = Carnivora. [First letter always in capital]

Class = Mammalia. [First letter always in capital]

Phylum = Chordata. [First letter always in capital]

Kingdom = Metazoa [=Multi-cellular organism. First letter always in capital]

Ursus americanus

Common name = bear.

The species = americanus.

Genus = Ursus .

Family = Ursidae.

Order = Carnivora.

Procyon lotor

Common name = raccoon.

The species = lotor.

Genus = Procyon.

Family = Procyonidae.

Order = Carnivora.

- Reddish-brown above and black or greyish below.
- Bushy tail with 4-6 black or brown rings
- Black mask outlined in white
- Small ears
- The feet and forepaws are dexterous

- Native to the southern part of the Canadian provinces and most of the United States
- Most common along stream edges, open forests and coastal marshes

- Inhabit hollow trees and logs and often use the ground burrows of other animals for raising their young or for sleeping during the coldest part of the winter months.
- An average of 4-5 young are born in April-May; the mother at first carries them by the nape of the neck like a cat; they are weaned by late summer.
- Omnivorous, it feeds on grapes, nuts, grubs, crickets, small mammals, birds' eggs and nestlings.
- Often seen washing their food, the raccoon is actually feeling for matter that should be rejected as wetting the paws enhances its sense of feel.
- Winter is the raccoons’ greatest enemy when food is scarce.

HEBREW: Nape = “OREF” ; Grub = “Zachal” ; Nestling = “Gozal”

The color of the weasel is chocolate brown on its back side and white with brown spots on its underparts. The summer coat is about 1 cm in length. The winter coat, which is about 1.5 cm in length, turns to all white in northern populations and remains brown in the southern populations.

The body of the least weasel is long and slender, with a long neck; a flat, narrow head; short limbs. This animal has large black eyes and large, round ears. The weasel's feet have five fingers with sharp claws. Breeding can occur throughout the year, but most of the breeding occurs in the spring and late summer. Gestation in the least weasel lasts from 34 - 37 days. Litters may range from 1 - 7.

A higher number of offsprings per litter can be found in northern populations. Newborns weigh from 1.1 g to 1.7 g and are wrinkled, pink, naked, blind, and deaf. After 49 - 56 days, they have reached their adult length. By week 6, the males are larger than the females. In 9 - 12 weeks family groups begin to break up, and in 12 - 15 weeks the weasels reach their adult mass.

The young spend their time play fighting and play mating. Weasels watch the movement of their prey before they attack. When they kill, they go for the neck of the victim.

Distribution:

Europe, northern Africa, Asia, North America; introduced to New Zealand

Diet:

Rodents, birds

Distance between these two taxa was 24, so each branch has a length of 12.

ss

12

12

seal

sea lion

We call the father node of seal and sea lion “ss”.

Removing the seal and sea-lion rows and columns,

and adding the ss row and columns

Here, i=seal, j=sea lion, k = dog.

n(i)=n(j)=1.

D(ss,dog) = 0.5D(sea lion,dog) + 0.5D(seal,dog) = 49.

12

12

seal

sea lion

Starting tree

Distance between bear and raccoon was 26, so each branch has a length of 13.

br

13

13

bear

raccoon

We call the father node of seal and sea lion “ss”.

Here, i=raccoon, j=bear, k = ss.

n(i)=n(j)=1. D(br,ss) = 0.5D(bear,ss)+0.5D(raccoon,ss)=37.5.

12

12

seal

sea lion

Starting tree

Distance between br and ss was 37.5, so each branch has a length of 18.75. But this is the distance from br-ss to the leaves. The distance br-ss to ss is 18.75-12=6.75. The distance between br-ss to br is 18.75-13=5.75

brss

6.75

5.75

br

13

13

bear

raccoon

Computing dog-(br-ss) distance

Here, i = br, j = ss, k = dog.

n(i)=n(j)=2. D( brss , dog ) = 0.5D( br , dog ) + 0.5D( ss , dog )=44.5.

Distance between br-ss and w was 39.5, so wbrss is mapped to the line 19.75. The distance to br-ss, is thus, 1

wbrss

br-ss

19.75

18.75

br

13

ss

12

0

weasel

seal

sea lion

bear

raccoon

Here, i = br-ss, j = weasel, k = dog.

n(i)=4, n(j)=1. D( wbrss , dog ) = 0.8D( br-ss , dog ) + 0.2D( weasel , dog )=

44.5*8/10+51*2/10 = (356+102)/10=45.8

Distance between wbrss and dog was 45.8, so dwbrss is mapped to the line 22.9 The distance to wbrss, is thus, 3.15

dwbrss

22.9

wbrss

br-ss

19.75

18.75

br

13

ss

12

0

weasel

seal

sea lion

bear

raccoon

dogl

Distance between dwbrss and cat was 89.833, so cdwbrss is mapped to the line 44.9165The distance to dwbrss, is thus, 22.0165

cdwbrss

44.9165

dwbrss

22.9

wbrss

br-ss

19.75

18.75

br

13

ss

12

0

cat

weasel

seal

sea lion

bear

raccoon

dog

72.14

Distance between cdwbrss and chimp was 144.2857, so THE ROOT is mapped to the line 72.14285The distance to dwbrss, is thus, 27.22635

cdwbrss

44.9165

dwbrss

22.9

wbrss

brss

19.75

18.75

br

13

ss

12

0

cat

weasel

seal

sea lion

bear

raccoon

dog

chimp

Problems with UPGMA, when the data is not clock-like

Assume that this is the true tree:

A

Then, the “true” distance matrix is

D

13

B

C

10

4

4

2

2

In this case, B and C will be clustered first – wrong!

Volume 397, Issues 1-2, 1 August 2007, Pages 76-83

The best way to test different methods of phylogenetic reconstruction is by using trees that are known to be true from other sources…

Problem: known phylogenies are very rare.

Known phylogeny: laboratory animals, crop plants (and even those are often suspicious). Also, their evolutionary rate is very slow…

David Hillis and colleagues have created “experimental” phylogenies in the lab.

The first paper (1992) analyzed phylogeny reconstruction based on restriction sites analysis.

Later bacteriophage T7 was used. It was subdivided into cultures in the presence of a mutagen. Then they sequenced the final cultures and gave the sequences as input to a few phylogenetic reconstruction methods. The treeoutput of these methods was then compared to the true tree.

In fact, they used restriction sites to infer the phylogeny, using MP, NJ, UPGMA and others.

All methods reconstructed the true tree.

They also compared outputs of ancestral sequence reconstruction, using MP.

97.3% of the ancestral states were correctly reconstructed.

Encouraging!

Criticism:

(1) The true tree was very easy to infer, because it was well balanced, and all the nodes are accompanied by numerous changes.

(2) Mutating using a single mutagen doesn’t reflect reality.

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