ASBP Training_Alignment and Phylogeny

8/3/2019 ASBP Training_Alignment and Phylogeny

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III. Evolutionary Change in DNA Sequences


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Kinds of questions

Indentification of INDIVIDUALS does the fish in the freezer match the carcass on

the field

Detecting RELATEDNESS can kin selection (i.e. high level of relatedness)explain cooperative courtship behavior?

Assigning INDIVIDUALS to POPULATIONS do fish populations across the Bohol

Sea show sufficient differentiation to allow us to identify unknown samples to asource population with a high level of confidence?

Defining structure of POPULATIONS what forces could explain the geneticdifferentiation among populations of rabbit fish in western Philippines.

Identifying SPECIES boundaries are these two forms of rock fish a single

species or tow distinct speices

PHYLOGENETIC TREES where do whales (Cetaceans) fit in a phlogenetictreee of mammalian groups.

What is the grand arrangement of the tree of life in terms of kingdoms and phyla?


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WHY USE MOLECULAR MARKERS?

Only genetically transmitted traits are informative to phylogenyestimation

Molecular markers open the whole biological world to geneticscrutiny

Genetic markers access an almost unlimited pool of geneticvariability

Molecular data distinguishes Homology (common ancestry) fromAnalogy (convergence from different ancestors

Provides a common yardstick for measuring divergence

Facilitate Mechanistic appraisals of evolution


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PHYLOGENY and SYSTEMATICS

How are taxa arranged in thetree of life?

MORPHOLOGY

MOLECULARAPPROACHES


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0.092 0.060 0.019 0.0075

Gibbon

Orangutan

Human

Chimpanzee

Gorilla


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Terms

Nodes (terminal observed taxa; internalhypotheticalancestors)

Dichotomous or polytonous (uncertainty of relationships

or multiple simultaneous branching) Rooted vs unrooted trees

Clades and ingroups monophyly vs paraphyletic

Ingroup and outgroup


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Nucleotide Difference Between Sequences

A simple measure of the extent of sequence divergenceis p proportion of nucleotide sites at which the twosequences are different. This is estimated by:

p = nd/n

And is called the p distance. Although the overallnucleotide difference.

^


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Different types of nucleotide pairs between Xand Y

Class Nucleotide Pair

Identical nucleotidesfrequency

AA TT CC GG Total

O1 O2 O3 O4 O

Transition-type pair frequency AG GA TC CT

P11 P12 P21 P22 P

Transversion-type pairfrequency

AT TA AC CA

Q11

Q12

Q21

Q22

Q

TG GT CG GC

Q31 Q32 Q41 Q42


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Transition/ Transversion ratio

R = P / Q

R is usually 0.5 2.0 in many nuclear genes. In mtDNA it can be as highas 15.

R is subject to a large sampling error when the number of nucleotidesexamined (n) is small.

V(R) = R2 (1/nP + 1/nQ)

Assumption

P11= P12 ; P21= P22; Q11= Q12; Q 21= Q22; Q31= Q32 ;Q41= Q42

^ ^ ^

^


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Estimation of the number of substitutions

When p is large, it gives an underestimateWhy?

It does not consider backward and parallel substitutions

A number of mathematical models have beendeveloped to address this. We will discuss:

Jukes and Cantors Method

Kimuras Two-Parameter Method

Tajima and Neis MethodTamuras Method

Tamura and Neis Method


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Jukes-Cantor model

Assumes that nucleotide substitution occurs at anynucleotide site with equal frequency

Each site and nucleotide changes to one of the

remaining nucleotides with a probability of per year

Probability of change in nucleotide= rate of substitution

r = 3


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Jukes-Cantor model

A T C G

A -

T

-

C -

G -


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Consider X and Y

Let qt = proportion of identical nucleotides at time t

Let pt = 1-qt = proportion of different nucleotides

Probability that site with similar nucleotides in X and Y

at t will be remain similar by t+1:(1-r)2 or approximately 1-2r

Probability that site with different nucleotides in X and Ywill be similar by t+1:

(1-r) * 2

= 2r (1-r)/3 or approximately 2r/3


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Deriving a value for d

qt+1 = (1-2r) qt + 2r/3 (1-qt)qt+1 qt =2r/3 8r/3 qt

Using a continuous time model using dq/dtto

represent qt+1 qt

dq/dt =2r/3 8r/3 q

The solution of this equation with initial

conditions q=1 at t=0

q= 1-3/4 (1-e-8rt/3)


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Under our present model, the expected number ofnucleotide usbstituions per site (d) for the twosequences is 2rt. Therefore, d is given by:

d = -(3/4) ln [1-(4/3 p)]

where; p= 1-q is the proportion of different nucelotidesbetween X and Y. An estimate d can be obtained by

using p. The large-sample variance of d is:

V(d) = 9p(1-p)

(3-4p)2 n

^ ^


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Kimura Two Parameter model

Considers the higher rate of transitional vs trasversionalnucleotide substitution and 2

Total substitution rate per year r = + 2

C

A

T

G


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Kimura Two Parameter model

A T C G

A -

T

-

C

-

G -


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Deriving d

P = (1-2 e-4(+)t +e -8t)

Q = (1-e-8 t)

Where t is the time for transitional substitution:

d = 2rt = 2t + 4rt

= - ln (1-2P-Q)- ln (1-2Q)


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Variance of d (Kimuras model)

Variance of d is:

V(d) = 1/n [c12P + c3

2Q (c1P + c3Q)2]

Where;

c1 = 1 , c2 = 1 , and c3 = (c1 +c2)/2

1-2P-Q 1-2Q


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Notes:

In both the Kimura and Jukes Cantor models, theexpected frequencies of A,C,T and G will eventuallybecome equal to 0.25.

Both models make no assumption about the initialfrequencies. This property makes the two modelsapplicable to a wider condition than may other models.

There is no need to assume the stationarity ofnucleotide frequencies for estimating d.


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Tajima-Nei (Equal-input) model

A T C G

A - gT gC gG

T gA - gC gG

C gA gT - gG

G gA gT gC -


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Tajima-Nei (Equal-input) model

Similar model was proposed independently byFelsenstein (1981) and Tajima and Nei (1982)

It is necessary to assume stationarity of nucleotide

frequencies for estimating the number of nucleotidesubstitutions:

d = -b ln (1-p/b)

where,

b = [ 1- gi2 +p2/c]i=1

4


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And c is given by:

c = xij2

2gigj

Where xij (i


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Tamura model

In Kimuras model, the four nucleotides eventuallybecome 0.25. In real data, however, nucleotidefrequencies are rarely equal and the GC content isoften quite different from 0.5. (Drosophila for example

= 0.1) Tamuras (1992) model was developed as an extension

of Kimuras modelto the case of low or high GC content.

d = -h ln (1-P/ h-Q) () (1-h) ln (1-2Q)

Where h = 2 (1-), and is the GC content


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Tamura model

A T C G

A - 2 1 1

T 2 - 1 1

C 2 2 - 1

G 2 2 1 -

1 = gG + gC2 = gA + gT


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Tamura-Nei model

Hasegawa et al (1985) maximum likelihood method.This is a hybrid of Kimuras model, equal input model

and considers both the transition/ transversion and GC

content biases mentioned earlier. The formula for d isquite complicated but similar to Tamura and Neismodel of which it is a special case.

d = - 2gAgG ln [ 1- gR P1 1 Q]gR 2gAgG 2gR


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d = - 2gAgG ln [ 1- gR P1 1 Q]

gR 2gAgG 2gR

- 2gTgC ln [ 1- gY P2 1 Q]gY 2gTgC 2gY

- 2 [gRgY gAgGgR gTgCgR] ln [ 1 - 1 Q]

gR gY 2gRgY


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Gamma Distances

For our list of distances, the rate of nucleotidesubstitution is assumed to be the same for allnucleotide sites. In reality, this assumption rarely holds,and the rate varies from site to site.

Statistical analyses of rate substion at differentnucleotide sties suggested that the rate variationapproximately follows a gamma distribution


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Comparison of DifferentDistance Measures

0

0.5

1

1.5

2

0

0

.3

0.75

1

.2

1

.5

1

.8

Expected number of substitutions per s ite

Estimatednu

mberof

substitutions

persite

Tamura-NeiTamuraKimuraJukes-Cantorp


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Alignment of NucleotideSequences

ATGCGTCGTT

ATCCGCGAT

ATGCGTCGTTATCCG_CGAT

ATGC_GTCGTT

AT_CCG_CGAT


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Methods

Similarity index - Needleman and Wunsch (1970)

Alignment distance Sellers (1974)

E = Min w1 +w2

w1 andw2 are penalties for a mismatch and a gap (e.g 1and 4). The gap penalty is a function of the gap length.Similarly, mismatches are can be divided into

transitional and transversional mismatches and differentpenalities are given to them.


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Alignment of Multiple Sequences

Customary to use progressive alignment algorithm -pairs of sequences with small distances are first alignedand the alignment of more distantly related sequencesis done progressively for larger and larger groups.

Pairs of sequences are aligned using the progressivealignment algorithm. Groups of sequences are alignedwith each other using a profile alignment algorithm


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Handling sequence gaps in estimation ofevolutionary distances

Complete deletion delete all sites with gaps from thedata analysis. Generally desirable because differentregions of DNA sequences oftern evolve differently.

Pairwise-deletion if the number of nucloties invovledin the gap is small and gaps are distributed more orless at random, distances may be computed from pairsof sequences ignoring only those gaps that in the two

sequences compared


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ExampleA-AC-GGAT-AGGA-ATAAA

AT-CC?GATAA?GAAAC-A

ATTCC-GA/TACGATA-AGA

Differences/Comparison

Option Sequence (1,2) (1,3) (2,3)

Complete- deletion1 A C GA A GA A A A 1/10 0/10 1/10

2 A C GA A GA A C A

3 A C GA A GA A A A

Pairwise-deletion1 A-AC-GGAT-AGGA-ATAAA 2/12 3/12 3/14

2 AT-CC?GATAA?GAAAAC-A

3 ATTCC-GA?TACGATA-AGA


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Assignment:

Reading assignments

ClustalX-

http://inn-prot.weizmann.ac.i./software/ClustalX.html

http://www.biozentrum.unibas.ch/`biphit/slustal/ClustalX_help.html

Mega 2

http://www.megasoftware.net/
http://inn-prot.weizmann.ac.i./software/ClustalX.htmlhttp://www.biozentrum.unibas.ch/%60biphit/slustal/ClustalX_help.htmlhttp://www.megasoftware.net/http://www.megasoftware.net/http://www.biozentrum.unibas.ch/%60biphit/slustal/ClustalX_help.htmlhttp://inn-prot.weizmann.ac.i./software/ClustalX.htmlhttp://inn-prot.weizmann.ac.i./software/ClustalX.htmlhttp://inn-prot.weizmann.ac.i./software/ClustalX.html

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ASBP Training_Alignment and Phylogeny

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