- 282 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'rRNA structure is conserved in evolution.' - ivanbritt

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

a good example where molecular sequences have led to a big improvement of our understanding of evolution.

Cao et al. (2000) Gene

rRNA structure is conserved in evolution.The sequence changes slowly. Therefore it can be used to tell us about the earliest branches in the tree of life.

69 Mammals with complete motochondrial genomes.

Used two models simulatneously

Total of 3571 sites

= 1637 single sites

+ 967 pairs

Hudelot et al. 2003

- Distances
- Clustering Methods
- Likelihood Methods
- Parsimony

Recommended books:

P. G. Higgs and T. K. Attwood – Bioinformatics and Molecular Evolution

R. Page and A. Holmes – Molecular evolution: a phylogenetic approach

W. H. Li – Molecular evolution

Part of sequence alignment of Mitochondrial Small Sub-Unit rRNA

Full gene is length ~950

11 Primate species with mouse as outgroup

G

A

A

C

A

2 substitutions happened - only 1 is visible

2 substitutions happened - nothing visible

From alignment construct pairwise distances.

Species 1: AAGTCTTAGCGCGAT

Species 2: ACGTCGTATCGCGAT

* * * D = 3/15 = 0.2

D = fraction of differences between sequences

BUT - D is not an additive distance,

D does not increase linearly with time.

rijis the rate of substitution from state i to state j

States label bases A,C,G & T

i

Pij(t) = probability of being in state j at time t

given that ancestor was in state i at time 0.

t

j

d

D = 3/4

D

Jukes - Cantor Model

All substitution rates = a

All base frequencies are 1/4

t = 2t

Mean number of substitutions per site:

d increases linearly with time

The HKY model has a more general substitution rate matrix

to

from

The frequencies of the four bases are

kis the transition-transversion rate parameter

* means minus the sum of elements on the row

Part of the Jukes-Cantor Distance Matrix for the Primates example

Baboon Gibbon Orang Gorilla PygmyCh. Chimp Human

Baboon 0.00000 0.18463 0.19997 0.18485 0.17872 0.18213 0.17651

Gibbon 0.18463 0.00000 0.13232 0.11614 0.11901 0.11368 0.11478

Orang 0.19997 0.13232 0.00000 0.09647 0.09767 0.09974 0.09615

Gorilla 0.18485 0.11614 0.09647 0.00000 0.04124 0.04669 0.04111

PygmyChimp 0.17872 0.11901 0.09767 0.04124 0.00000 0.01703 0.03226

Chimp 0.18213 0.11368 0.09974 0.04669 0.01703 0.00000 0.03545

Human 0.17651 0.11478 0.09615 0.04111 0.03226 0.03545 0.00000

Mouse-Primates ~ 0.3

Use as input to clustering methods

Follow a clustering procedure on the distance matrix:

1. Join closest 2 clusters

2. Recalculate distances between all the clusters

3. Repeat 1 and 2 until all species are connected in a single cluster.

Initially each species is a cluster on its own. The clusters get bigger during the process, until they are all connected to a single tree.

Neighbour Joining method is commonly used clustering method

n

n

j

k

but ...

Neighbour-Joining method

Take two neighbouring nodes i and j and replace by a single new node n.

din + dnk = dik ; djn + dnk = djk ; din + djn = dij ;

therefore dnk = (dik + djk - dij)/2 ; applies for every k

define

Let din = (dij + ri - rj)/2 ; djn = (dij + rj - ri)/2 .

Rule: choose i and j for which Dijis smallest, where Dij = dij - ri - rj .

Gorilla

PygmyChimp

Gibbon

Chimp

Human

Marmoset

The tree has been rooted using the Mouse as outgroup

Baboon

SakiMonkey

Mouse

Lemur

47

0.1

Tarsier

Mouse

SakiMonkey

100

Tarsier

Marmoset

Lemur

100

Baboon

100

Gibbon

100

Orang

99

Gorilla

100

Human

84

PygmyChimp

0.1

100

Chimp

NJ method produces an Unrooted, Additive Tree

Additive means distance between species = distance summed along internal branches

x

t1

t1

t2

t2

A

y

G

z

The Maximum Likelihood Criterion

Calculate the likelihood of observing the data on a given the tree.

Choose tree for which the likelihood is the highest.

Can calculate total likelihood for the site recursively.

Likelihood is a function of tree topology, branch lengths, and parameters in the substitution rate matrix. All of these can be optimized.

Lemur

47

Tarsier

SakiMonkey

100

Marmoset

100

Baboon

100

Gibbon

100

Orang

99

Gorilla

100

Human

84

PygmyChimp

0.1

100

Chimp

Using ML to rank the alternative trees for the primates example

The NJ Tree has:

(Gorilla,(Human,(Chimp,PygmyCh)))

((Lemur,Tarsier),Other Primates))

Tree log L difference S.E. Significantly worse

1 -5783.03 0.39 7.33 no NJ tree

2 -5782.64 0.00 <-------------best tree (Human,(Gorilla,(Chimp,PygmyCh)))

3 -5787.99 5.35 5.74 no ((Human,Gorilla),(Chimp,PygmyCh))

4 -5783.80 1.16 9.68 no (Lemur,(Tarsier,Other Primates))

5 -5784.76 2.12 9.75 no (Tarsier,(Lemur,Other Primates))

????????????????????????????????????????????????????????????????????

6 -5812.96 30.32 15.22 yes (Gorilla,(Chimp,(Human,PygmyCh)))

7 -5805.21 22.56 13.08 no Orang and Gorilla form clade

8 -5804.98 22.34 13.15 no Gorilla branches earlier than Orang

9 -5812.97 30.33 14.17 yes Gibbon and Baboon form a clade

2

+

3

+

*

+

+

A(0) C(1) B(0) D(1)

A(0) C(1) B(0) D(1)

The Parsimony Criterion

Try to explain the data in the simplest possible way with the fewest arbitrary assumptions

Used initially with morphological characters.

Suppose C and D possess a character (1) that is absent in A and B (0)

A(0) B(0) C(1) D(1)

In 1, the character evolves only once (+)

In 2, the character evolves once (+) and is lost once (*)

In 3, the character evolves twice independently

The first is the simplest explanation, therefore tree 1 is to be preferred by the parsimony criterion.

C(T)

A(T)

A(T)

*

*

*

B(T)

B(T)

D(G)

D(G)

C(T)

E(G)

A(T)

E(G)

C(T)

A(T)

*

*

B(T)

D(T)

B(T)

D(T)

C(T)

E(G)

This site is non-informative. Whatever the arrangement of species, only one mutation is required. To be informative, a site must have at least two bases present at least twice.

The best tree is the one that minimizes the overall number of mutations at all sites.

Parsimony with molecular data

1. Requires one mutation 2. Requires two mutations

By parsimony, 1 is to be preferred to 2.

Subtree pruning and regrafting

Searching Tree Space

Require a way of generating trees to be tested by Maximum Likelihood, or Parsimony.

No. of distinct tree topologies

N Unrooted (UN) Rooted (RN)

3 1 3

4 3 15

5 15 105

6 105 945

7 945 10395

UN = (2N-5)UN-1 RN = (2N-3)RN-1

Conclusion: there are huge numbers of trees, even for relatively small numbers of species. Therefore you cannot look at them all.

Dating from phylogeniesThe molecular clock

known to be 100 m.y.

must be approx 50 m.y.

must be approx 150 m.y.

Can lead to controversies because the clock does not always go at a constant rate

Mammalian orders (primates, rodents, carnivores, bats ....)

Molecular dates tend to be earlier (100 m.y.) than those coming from the fossil record (65 m.y.)

Animal phyla - Molecular phylogenies are resolving the relationships between phyla that were not understood from morphology.

Still uncertainty about dates - Molecular dates suggest about 1 b.y.Earliest fossil evidence is 560 m.y.

Old morphological phylogeny New molecular phylogeny

Molecular dating of key events in early evolution

Puts LUCA just prior to 4, and origin of eukaryotes at 2.7.

Origin of cyanobacteria is at 2.5, whereas they are claimed to be present in the fossil record at 3.5.

Download Presentation

Connecting to Server..