Distance methods: p distances and the least squares (LS) approach

Distance methods: p distances and the least squares (LS) approach

Two steps:

1.Compute a distance D(i,j) between any two sequences i and j.

2.Find the tree that agrees most with the distance table.

General concept of distance based methods

SEQ1 AACAAGCGSEQ2 AACGAGCA

There are 2 differences, so the distance = 2.The problem is that now, if you have a longer pair

of sequences

SEQ3 AACAAGCGCCCTCAGTCCGCTCGCACAASEQ4 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 AACAAGCGSEQ2 AACGAGCA

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

This is called the p distance.

Simplest distance: the “p” distance

Distance estimation

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…

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).

From a distance table to a tree

Human

Chimp

Gorilla

0.30.41

0.25

d(Human,Chimp) = 0.55

d(Human,Gorilla) = 0.71

d(Chimp, Gorilla) = 0.66

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

Human

Chimp

Gorilla

XY

Z

D(Human,Chimp) = 0.3

D(Human,Gorilla) = 0.4

D(Chimp, Gorilla) = 0.5


Human

Chimp

Gorilla

XY

Z

D(Human, Chimp) = 0.3

D(Human, Gorilla) = 0.4

D(Chimp, Gorilla) = 0.5


X+Z = 0.3

X+Y = 0.4

Y+Z = 0.5

d(Human, Chimp) = X+Z

d(Human, Gorilla) = X+Y

d(Chimp, Gorilla) = Y+Z

Y-Z = 0.1

Y+Z = 0.5

Y = 0.3

Z = 0.2

X = 0.1

Human

Chimp

Gorilla

XY

Z

D(Human, Chimp) = D1

D(Human, Gorilla) = D2

D(Chimp, Gorilla) = D3

Is there always a solution?

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



d(Chimp, Gorilla) = Y+ZX+Z = D1

X+Y = D2

Y+Z = D3

Human

Chimp

Gorilla

XY

Z

D(Human, Chimp) = D1

D(Human, Gorilla) = D2

D(Chimp, Gorilla) = D3

Ex.

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(Chimp, Gorilla) = Y+ZX+Z = D1

X+Y = D2

Y+Z = D3

A

B

D

XY

Z

5 Variables,

6 Equations,

It might be that there’s no solution

Is there always a solution??

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

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

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

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

C

W

V


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).

The least square solution

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].

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

Is there always a solution???

So for the simplest of two equations with 1 parameter

a = 2

a = 3

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

Back to phylogeny

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

n

i

n

jijij dDQ

1 1

2)(

Back to phylogeny

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

n

i

n

jijij dDQ

1 1

2)(

Back to phylogeny

The general formula for LS

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

n

i

n

jijijij dDwQ

1 1

2)(

Back to phylogeny

w’s used

ijij

ijij

ij

Dw

Dw

w

1

1

1

2

Cavalli-Sforza and Edwards (1967)

Fitch Margoliash (1967)

Beyer et al (1974)

Tree search

There are the general heuristic searches.

No branch-and-bound method published so far.

Problem was shown to be NP-complete.

Minimum Evolution

The general formula for LS

n

i

n

jijijij dDwQ

1 1

2)(

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 and the Neighbor Joining algorithm

The Newick tree format is used to represent trees as strings

C

A

B

In Newick format: (A,B,C)

The Newick tree format is used to represent trees as strings

C

A

D

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

B

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

A

C

B

D

E

Red: best pair to group together

D

A

D

(C,B)A

E

((C,B),E)

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.

Neighbor-joining

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.

Neighbor-joining

http://www.ncbi.nlm.nih.gov/entrez/query.fcgi?cmd=Retrieve&db=PubMed&list_uids=3447015&dopt=Abstract

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.

OTU’s and HTU’s

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

Neighbors, we are …

BD

A C

A and B are neighbours,C and D are neighbours,But…A and C are not neighbours…

In an additive tree, the distance matrix exactly reflects the tree:

Additive trees

BD

A

XY

C

the distance between nodes A and B

= the distance between nodes A and Y

+ the distance between nodes Y and B

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

Additive trees

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.

NJ is an approximation of the Minimum evolution

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.

NJ notations:

We distinguish between L(X,Y) and D(A,B). D’s are given as input to the algorithm, L’s should be inferred…

L(x,y) notation:

BD

A

XY

C

• In each round we join as neighbours all possible pairs of leaves and evaluate the 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 which S(i,j) is minimal, where i and j denote numbers of leaves, and i<j .

• This is why NJ is approximating ME (minimum evolution).

NJ step:

How can we evaluate S(1,2) from the input (the distance matrix)?

Computing S(1,2)

2

4

1

X Y

3

5

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

Computing S(1,2)

2

4

1

YX

3

5

S(1,2) = L(1,X)+L(2,X)+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).

Computing S(1,2)

2

4

1

YX

3

5

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

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

Computing L(X,Y) in terms of the D’s

2

4

1

YX

3

5

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

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

So L(1,X) is canceled out…

]),(2)),2(),1()(2()),2(),1(([)2(2

1),(

33

N

i

N

K

YiLXLXLNkDkDN

YXL

N denotes the number of leaves


2

4

1

YX

3

5

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

Once here

So L(3,Y) is canceled out…

L(3,Y) is counted once here

]),(2)),2(),1()(2()),2(),1(([)2(2

1),(

33

N

i

N

K

YiLXLXLNkDkDN

YXL


2

4

1

YX

3

5

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

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

]),(2)),2(),1()(2()),2(),1(([)2(2

1),(

33

N

i

N

K

YiLXLXLNkDkDN

YXL


2

4

1

YX

3

5

]),(2))2,1()(2()),2(),1(([)2(2

1

33

N

i

N

K

YiLDNkDkDN

]),(2)),2(),1()(2()),2(),1(([)2(2

1),(

33

N

i

N

K

YiLXLXLNkDkDN

YXL

We still have to replace this term by the D’s

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

2

4

1

YX

3

5

3 3

1( , ) ( , )

3

N N

i i j

L i Y D i jN

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


2

4

1

YX

3

5

3 3

( , )

1 2[ ( (1, ) (2, )) ( 2)( (1,2)) ( , )]

2( 2) 3

N N

K i j

L X Y

D k D k N D D i jN N

Back to S(1,2)

3

3 3

3

3 3

(1,2) (1,2) ( , ) ( , )

(1,2)

1 2[ ( (1, ) (2, )) ( 2)( (1,2)) ( , )]

2( 2) 3

1( , )

3

1 1 1[ ( (1, ) (2, ))] (1,2) ( , )

2( 2) 2 2

N

i

N N

K i j

N

i j

N N

K i j

S D L X Y L Y i

D

D k D k N D D i jN N

D i jN

D k D k D D i jN N

Finding the best neighbor

2

4

1

YX3

5

Let’s assume that S(1,2) is minimal in round 1…We call the new node that joins 1 and 2, X.

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.

2

4

13

5

Finding the best neighbor

2

4

1

YX

3

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: (1, ) (2, )

(12, )2

D j D jD j

12

4

3

5

Branch lengths

2

4

1

YX

3

5 Z

2

),2(

),2(2

),1(

),1( 33

N

iD

ZDN

iD

ZD

N

i

N

i

2

),1(),2()2,1(),2(

2

),2(),1()2,1(),1(

ZDZDDXL

ZDZDDXL

Only the branches in red are being computed.

Branch lengths

412

3

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).

4

12

3

5

X

Y

Branch lengths

4

12

3

5

X

Y

)]4,3(2)5,12(2)4,5()3,5()4,12()3,12([(4

1

),(

DDDDDD

YXL

Complexity of computing S(1,2)

3

3

(1,2)

1[ ( (1, ) (2, ))]

2( 2)

1(1,2)

21

( , )2

N

K

N

i j

S

D k D kN

D

D i jN

This part requires O(N2) computations

Complexity of the original NJ algorithm

Computing each S(i,j) sums up to N2

computations.

There are N2 combinations 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.

The UPGMA tree building method and the phylogeny of Carnivores

Minimum evolution

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.

Molecular clocks

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:

rtd

Molecular clocks

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.

Human MouseHuman Mouse

Molecular clocks

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.

Human MouseHuman Mouse

WITHOUT CLOCKWITH CLOCK

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.

Ultrametric

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

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!

UPGMA

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.

UPGMA

The algorithm:

1.Find the i and j that have the smallest D(i,j)2.Create a new group (ij) which has n(ij)=n(i)+n(j)3.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.

UPGMA

The algorithm:

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

),())()(

)((),()

)()(

)(()),(( kjD

jnin

jnkiD

jnin

inkijD

UPGMA

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.

Complexity

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).

An example

Distances based on immunological data of Sarich (1969).

The players

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]

http://www.cliffshade.com/dpfwiw/images/dog-beach.jpg

The players

Ursus americanus

Common name = bear.

The species = americanus.Genus = Ursus . Family = Ursidae. Order = Carnivora.

The players

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

The raccoon (דביבון)


• 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 players

Mustela nivalis

Common name = weasel.

Order = Carnivora.

In Hebrew (Samor)

The players

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 players

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.

The players

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 players

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

Weasel distribution

The players

Phoca vitulinaCommon name = Harbor seal.Order = Carnivora.

In Hebrew: “Kelev-Yam”

The players

Eumetopias jubatus

Common name = Steller sea lion.

Order = Carnivora.

The players

In Hebrew:Arye-Yam

The players

Felis catus

Common name = cat.

Order = Carnivora.

The players

Pan troglodytes

Common name = chimpanzee.

Order = Primates.

The distance table

dog bear raccoon weasel seal sea lion

cat chimp

dog 0 32 48 51 50 48 98 148bear 0 26 34 29 33 84 136

raccoon 0 42 44 44 92 152weasel 0 44 38 86 142

seal 0 24 89 142sea lion 0 90 142

cat 0 148chimp 0

The distance table


cat chimp

dog 0 32 48 51 50 48 98 148bear 0 26 34 29 33 84 136

raccoon 0 42 44 44 92 152weasel 0 44 38 86 142

seal 0 24 89 142sea lion 0 90 142

cat 0 148chimp 0

Starting tree

seal sea lion

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

12 12

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

ss

Removing the seal and sea-lion rows and columns,and adding the ss row and columns

dog bear raccoon weasel ss cat chimp

dog 0 32 48 51 ? 98 148bear 0 26 34 ? 84 136

raccoon 0 42 ? 92 152weasel 0 ? 86 142

ss 0 89 142cat 0 148

chimp 0

Computing dog-ss distance


cat chimp

dog 0 32 48 51 50 48 98 148

),())()(

)((),()

)()(

)(()),(( kjD

jnin

jnkiD

jnin

inkijD

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.

The new table. Starting second iteration…


dog 0 32 48 51 49 98 148bear 0 26 34 31 84 136

raccoon 0 42 44 92 152weasel 0 41 86 142

ss 0 89 142cat 0 148

chimp 0

Starting tree

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

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

seal sea lion

12 12

ss

bear raccoon

13 13

br

Computing br-ss distance


ss 49 31 44 41 0 89.5 142

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.

),())()(

)((),()

)()(

)(()),(( kjD

jnin

jnkiD

jnin

inkijD


dog br weasel ss cat chimp

dog 0 40 51 49 98 148br 0 38 37.5 88 144

weasel 0 41 86 142ss 0 89 142

cat 0 148chimp 0

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

seal sea lion

12 12

ss

bear raccoon

6.75

13

brss

br

5.75

13

Computing dog-(br-ss) distance

dog br weasel ss cat chimp

dog 0 40 51 49 98 148

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.

),())()(

)((),()

)()(

)(()),(( kjD

jnin

jnkiD

jnin

inkijD


dog Br-ss weasel cat chimp

dog 0 44.5 51 98 148br-ss 0 39.5 88.75 143

weasel 0 86 142cat 0 148

chimp 0

Starting tree

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

seal sea lion

0

ss

bear raccoon

br-ss

br 1312

19.7518.75

weasel

wbrss

Computing dog-wbrss distance

dog br-ss weasel cat chimp

dog 0 44.5 51 98 148

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

),())()(

)((),()

)()(

)(()),(( kjD

jnin

jnkiD

jnin

inkijD


dog wbrss cat chimp

dog 0 45.8 98 148wbrss 0 88.2 142.8

cat 0 148chimp 0

Starting tree

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

seal sea lion

0

ss

bear raccoon

br-ss

br 1312

22.9

18.75

weasel

dwbrss

19.75

dogl

wbrss


dwrbss cat chimp

dwrbss 0 89.833 143.66cat 0 148

chimp 0

Starting tree

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

seal sea lion

0

ss

bear raccoon

br-ss

br 1312

44.9165

18.75

weasel

cdwbrss

19.75

dog

wbrss22.9

cat

dwbrss


cdwrbss chimp

cdwrbss 0 144.2857chimp 0

Starting tree

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

seal sea lion

0

ss

bear raccoon

brss

br 1312

72.14

18.75

weasel

dwbrss

19.75

dog

wbrss22.9

cat

cdwbrss44.9165

chimp

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

Assume that this is the true tree:

13

4

10

2 2

4

B C

D

A

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

A B C D

A 17 21 27

B 12 18

C 14

D

Then, the “true” distance matrix is

Gene,

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

Networks

A network is sometimes used to represent a tree in which recombination occurred.

b c d ea

Known phylogenies

Known phylogenies

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…

Known phylogenies

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

Known phylogenies

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

Known phylogenies

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 tree output of these methods was then compared to the true tree.

Known phylogenies

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

All methods reconstructed the true tree.

Known phylogenies

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

97.3% of the ancestral states were correctly reconstructed.

Encouraging!

Known phylogenies

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.

Documents

Distance methods: p distances and the least squares (LS) approach