Sect 19 Molecular Evolution

8/11/2019 Sect 19 Molecular Evolution

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MOLECULAR EVOLUTION

Focus is on long-term evolution leading to differences between species.

Differences between species are determined by the same factors that determine

differences between individuals within a species: mutation, selection, and drift. But thereare some critical differences:1. Time period

Population genetics: !4Negenerations (!time since most recent common ancestor of

all copies of a gene in a species)

Evolutionary genetics: > 4Negenerations

2. Selection causes hitchhiking within species; we only discussed this in terms of theAdhF/S difference.

3. Method of study

Rates of sequence evolution

Molecular evolution is generally studied by following procedure:

(1)Sequence one copy of a gene from each of a number of different species oforganisms

(2)Align sequences.(3)Calculate the proportion of sites that differ, for each pair of species.

d = pairwise sequence difference in differences per bp


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(4) Correct for multiple hits, especially if d !0.10

sequence divergence = K = -(3/4) ln[1-(4/3)d] in substitutions per

bp


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(5) Calculate rate of sequence divergence if desired

E = K/2T in substitutions per bp per year


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A species is a population of organisms.During speciation, one species splits into two populations that evolve

independently of each other.To study evolution, we usually sequence one gene from each species.

Many mutations occur in the subsequent evolution of both species, but mostare eliminated. The sequences only show differences that were fixed in oneor the other species.

K = frequency of base pair substitutionsthat occurred along both evolutionarypaths; i.e. the number of mutations that occurred AND were fixedin one or theother species instead of being lost.


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We can state this with a remarkably simple equation:

E = MF

rate of molecular Evolution = total Mutation rate X Fixation probability

M = 2Nu

whereu = mutations per site (bp) per gameteF = mutations fixed/total mutations (substitutions per mutation)2N = number of gametes in population or number of copies of the gene inpopulation

K = number (per site) of mutations that occurred AND were fixedin one or theother species instead of being lost.


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E = 2NuF

If we want to express this in units of time, then we have to incorporate

generation time by assigning units:2N = gametes per year

2N u F gametes X mutations X substitutions year site X gamete mutation

e.g.

u = 5 !10-9mutations per site per gamete2N = 106 gametesF = 10-7substitutions per mutationE = 5 !10-10 substitutions per site per year

Haploids, organelles, asexuals: E = NuF


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The rate of neutral substitution equals the mutation rate.

Neutral mutations:

Fn= 1/2N for a new mutation

En= 2Nu(1/2N) = u

E = u !!

This remarkably simple result is also remarkably important. It means that

The mutation rate can be estimated from the substitution rate for neutralmutations.

The mutation rate equals the pseudogene substitution rate becausepseudogene substitutions are neutral.Synonymous substitution rates in functional protein-coding genes are about

equal to pseudogene rates in eukaryotes, which shows that they are alsoneutral or effectively neutral (average Ne|s|


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Directional selection reduces F and thus E

F(new mutation in diploids)= (1 e-2Nes/N) / (1 e-4Nes)

The three classes of mutations with different levels of polymorphism also havedifferent rates of substitution:

(1) neutral s = 0 H !4Neu F = 1/2N E = u

(2) detrimental s < 0 H < 4Neu F < 1/2N E < u

(3) advantageous s > 0 H < 4Neu F > 1/2N E > u

Note that we are ignoring those subject to balancing selection, as they are rare.

The great majority of mutations are either neutral or detrimental, so on average,F < 1/2N and E < u.


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Emphasize: this is average over a large number of sites.

This is actually what we observe:

For large sample of genes in mammals:

Synonymous rate 3.51 X 10-9

substitutions per site per year

Nonsynonymous rate 0.74 X 10-9

e.g. comparisons of globin genes in cow and goatK (mean std. error)

!-globin pseudogenes 9.1 0.9!- and "-globin exon synonymous 8.6 2.5!- and "-globin intron 8.1 0.7!- and "-globin 5-flanking 5.3 1.2

Note that this is evidence of natural selection which is predominantlypurifying,eliminating detrimental mutations.


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MEASURING THE STRENGTH OF PURIFYING SELECTION

Calculate ratio nonsynonymous substitutions/synonymous substitutions =

Kn/Ksor Ka/Ks.

0.74 X 10-9

= 0.21

3.51 X 10-9

neutral mutations + detrimental mutations Kn/Ks < 1neutral mutations only Kn/Ks = 1

neutral mutations + advantageous mutations Kn/Ks > 1

DETECTING POSITIVE SELECTION (FOR ADVANTAGEOUS MUTATIONS)

Selection for advantageous mutations: Kn/Ks > 1

e can use this to detect positive selection for advantageous mutations. Use

computer to isolate specific sites and calculate Kn/Ksfor each site. Then find if

find some sites have Kn/Ks > 1, these probably had one or more advantageousmutations fixed in fairly recent time.


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Making Phylogenetic Trees

If we have DNA sequences from the genes of three or more species, we can use

them to recover their evolutionary history by making a phylogenetic tree.Can do same using AA sequences of protein encoded by gene.

Saw at beginning of course in the tree of life.Need one gene that is present in all organisms: usually use gene for smallsubunit of ribosomal RNA.Not good for some organisms, so usually use protein-coding genes foreukaryotes.

Done by computer.Aligns all sequences.Calculates pairwise d.Corrects for multiple hits to get pairwise K..Uses pairwise K values to make trees.


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Give e.g. of three species from our bdelloid rotifers plus an outgroup(monogonont rotifer).Makes tree, using any one of a number of different algorithms to make tree thatis simplest or most probable given the data.

Ari1/2 WPr1/1 Flt2/1Ari1/2 -WPr 0.19394 -FlT2/1 0.17876 0.11301 -B.quadrid 0.56295 0.52794 0.53827 -

Old way: use morphological differences/similarities.Sequence data have some advantages:Use neutral or nearly neutral sites, avoids problems due to convergent or

parallel evolution.Hard to know what morphological traits are informative in many organisms.

Can make phylogenetic trees of genes even without seeing the organisms!

Ari WPr FlT Bq


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Uses:

Reconstruct evolutionary historye.g. tree of life shown at beginning of course

Potential problems.

If I sequence a gene from a cow, a langur monkey, and a rhesus monkey, whichtree will I get?

langurmonkeycow

rhesusmonkey

langurmonkeycow

rhesusmonkey


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Detect horizontal transfer: e.g. origin of chloroplasts from cyanobacterialendosymbionts


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Detect horizontal transfer: e.g. origin of chloroplasts from cyanobacterialendosymbionts


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GENOME EVOLUTION: THE ORIGIN OF NEW GENES WITH NEW

FUNCTIONS

Have focused on changes in bp sequence of genes. This can actually lead tochanges in gene function if the bp sequence changes are nonsynonymous andresult in changes in amino acids.

But in course of evolution, more complex organisms have arisen with increasednumberof genes. Genes with old functions have been retained (e.g. genes coding

for ribosomal RNA), while additional genes with new functions have arisen.Happens as result of gene duplication.

Several ways of making a copy of a gene and inserting it in a new location.

Fates of duplicate genes:1.Nonfunctionalization !pseudogene!loss by deletion

2.

Subfunctionalization: activity of both copies is reduced so still make sametotal amount of product.

3.Neofunctionalization: one copy acquires mutations that give it a new function.


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e.g. human globins: example of clustered multigene family.

myoglobin muscle (stores oxygen)! alpha major adult" beta major adult# gamma fetal$ delta infant, minor adult% epsilon embryonic& zeta embryonic' psi pseudogene

chromosome 16 !family &2 &1 'a1 !2 !1

chromosome 11 "family 'b2 % G# A# 'b1 $ "

Entire gene family arose from a single ancestral gene by duplication.Shows nonfunctionalization and neofunctionalization.Phylogenetic tree:


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SEXUAL REPRODUCTION

Sexual reproduction is widespread but most often alternates with asexualreproduction.

Mammals, birds: obligate sexOther eukaryotes include some obligately sexual; some obligately asexual;

most reproduce asexually most of the time with occasional sex

Asexual reproduction has advantages, e.g.: fas terone individual can colonizein plants and animals with separate sexes, parthenogenetic mutation has 2-

fold advantage because all offspring are parthenogenetic mutants

What are advantages of sex?

Sex in diploids makes some mutations homozygous so recessive mutations canbe eliminated more easily.

Sex with outcrossing makes natural selection more effective:

2 loci, A and B advantageous alleles, a and b detrimental mutations.

AB AB AB Ab aB

asexual Selection for B tends to fix a; selection for A tends to fix b

sexual Ab X aB --> AB and ab Selection on ab eliminates a and b


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Result: compared to sexual population, asexual population1.accumulates detrimental mutations (Mullers ratchet)!reduces fitness

of individuals and population!early extinction

2.has trouble fixing advantageous mutations !less able to adapt tochanging conditions!early extinction andless speciation

We are testing this in bdelloid rotifers, which have been parthenogenetic

(asexual) for !60 million years.Ka/Ksratio is same as for their close sexual relatives.

Why???

Join my lab next year and help us find out.


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Documents

Sect 19 Molecular Evolution