Download Population genetics (III)

The neutral theory
of molecular evolution
• the neutral theory
• detecting natural selection
• exercises
Objectives
1 - learn about the neutral theory
2 - be able to detect natural selection at the molecular level
Molecular evolution and Darwin
Two theses in Darwin’s Origin of Species
1 - organisms descend with modification from common ancestors
phylogenetics - pattern
2 - mechanism for this modification is natural selection
molecular basis of adaptation - process
The field of molecular evolution has been dominated by phylogenetics and molecular
systematics. These endeavors have been extremely successful in supporting and elucidating
the dynamics of point #1 above.
Molecular evolutionists have been relatively less successful (Sharp 1997) at uncovering
evidence detailing the mechanism of descent with modification - the molecular basis of
adaptation. Recent years have seen tremendous strides in this area (Hughes 1999) but it
remains to a great extent uncharted territory.
Population genetics
• interested in genetic variation - understand generation and maintenance
• initially (until 1960s) only able to study indirectly - phenotype
• paucity of data led to controversy on the extent of genetic variation
Classic school - very little genetic
variation, cost associated with
natural selection
Muller
Balance school - lots of genetic variation
maintained by natural selection
Dobzhansky
• debate settled with advent of molecular approaches (direct) to the
study of genetic variation - electrophoresis, sequencing
• tremendous amount of genetic variation exists
• thought was that this variation was maintained by natural selection
accumulation of adaptively advantageous variants
heterozygote advantage or heterosis
A new explanation – neutralist – for the
high levels of molecular variation
Motoo Kimura (1968) Evolutionary rate at the molecular level. Nature 217: 624
• Electrophoresis studies (Lewontin & Hubby) and sequence comparisons (Pauling & Zuckerkandl)
reveal high levels of molecular variation
• Kimura reasoned that variation was too high and accumulated to rapidly to be explained by selection
This is the so-called ‘cost of selection argument’ (see appendix slide # 9)
· Conclude that these observed differences are selectively neutral – that is they do not confer any
selective advantage or disadvantage to the organisms that bear them (because they do not alter the
function of the protein that they encode)
J.L. King & T.H. Jukes (1969) Non-Darwinian evolution. Science 164: 788
• Note that many genetic changes have no effect on organismic fitness – they are neutral
• Natural selection can not alter changes that it can not perceive
• Marshall biochemical evidence in support of these assertions
e.g. synonymous substitutions, functionally equivalent cytochrome c variants, rapid evolution of
fibrinopeptides (removed from functional fibrinogen)
The neutral theory
• change too rapid to be explained by natural selection
• therefore most of the changes observed are selectively neutral no effect on fitness
Kimura reasoned that the majority of both polymorphism (allelic frequencies within
populations) and substitution (fixed differences between populations) result from
fixation of selectively neutral variants by random genetic drift
- the main role of natural selection is elimination of deleterious variants (maintenance
of the status quo) - molecular evolution is conservative
- adaptively favorable mutations fixed by natural selection are a small minority of
all nucleotide substitutions
• huge debate ensued between selectionists (believe that extensive variation
is a product of natural selection) and neutralists (believe that variation is
a product of random fixation of neutral variants)
Predictions of the neutral theory
• neutral theory makes explicit quantitative predictions about levels of
genetic variation - null hypothesis of molecular evolution
most important for our purposes:
• functionally important parts of a molecule will change more slowly than
functionally unimportant parts
“Those mutant substitutions that disrupt less the existing structure and
function of a molecule (conservative substitutions) occur more frequently
in evolution than more disruptive ones.”
Kimura and Ohta 1974
Absolutely essential concept in modern molecular biology: basis of
programs to align sequences and make functional predictions
Challenge to the Darwinian view: if selection is driving force in evolution, rate
of evolution should be most rapid where selection operates most - in the functionally
important parts of molecules (opposite to the neutral view)
Maximum evolutionary rate &
selection versus neutrality
• do relative rates of change better fit selectionist or neutralist prediction?
• overwhelming support for neutralist prediction:
1 – synonymous versus nonsynonymous subs rate (Kimura 1977, Jukes 1978)
2 – accelerated rate of psuedogene evolution (Li et al 1981)
synonymous subs - do not change encoded amino acid
nonsynoymous subs - do change encoded amino acid
GAT AAC ATC CAA GGA ATA ACT GCA ATC
GAC AAC ATC CAA GGT ATC ACG GCT ATC
Asp Asn Ile Gln Gly Ile Thr Ala Ile
· in virtually every gene ever studied synonymous sites change
at a higher rate than nonsynonymous sites
Detection of natural selection using synonymous
& non-synonymous substitution rates
Types of natural selection:
1 – purifying (negative) selection – removal of deleterious variants
2 – diversifying (positive) selection – fixation of adaptive variants
Types of substitution rates: (for protein coding genes i.e. codons)
1 – synonymous substitution rate (Ks or ds) – rate of substitution for DNA changes that do not
change the encoded amino acids
2 – non-synonymous substitution rate (Ka or dn) – rate of substitution for DNA changes that do
change the encoded amino acids
•The relative levels for these rates indicate the mode of selection for a gene
Neutral evolution (no selection):
Purifying selection:
Diversifying selection:
Ks ≈ Ka
Ks >> Ka
Ks << Ka
Exercises
Compare synonymous and nonsynonymous substitution rates for:
1 – the Drosophila alcohol dehydrogenase (Adh) gene – dros-adh.meg
2 – the human & mouse gene pair – mammal.meg
Determine the mode of selection acting on each based on these rates
1 - load alignment into DnaSP
2 - assign coding region
3 - calculate Ks and Ka and compare (which is higher)
4 - load alignment into Mega
5 - calculate ds and dn and compare (which is higher)
6 - do statistical test for difference between ds and dn
(see appendix II slide # 11)
Kimura’s derivation of the neutral theory
• noticed variation too high and change too rapid to be explained by natural selection
- based on amino acid sequence data (hemoglobin & cytochrome c) Kimura calculated an
average of 1 aa per 28 my in a 100 aa protein
- this is too high for natural selection based on Haldane’s concept of ‘the cost of selection’
- if only individuals with high fitnesses for a number of diff traits survive, only a
very small fraction of the population will remain
e.g. moth melanism - 50% mortality due to bird predation
- if simultaneous effects at 10 loci 1 / 210 (1 out of 1,024) survivors - population likely to go
extinct before all 10 alleles fixed
- Haldane calculated that a 1 new allele per 300 generations can be substituted
- Kimura noted that in fact substitutions at the molecular level occurring much more rapidly
1 - 1 sub per 28my per 100 aa
2 - mammalian genome size 4 x 109 bp
3 - 100 aa = 300 bp and 20% nucleotide subs synonymous thus 1 aa sub ≈ 1.2 bp sub
4 - time it would take for nucleotide substitution to occur in the genome is:
(28 x 106) / (4 x 109/300) / 1.2 = 1.8 years
- this is a much higher rate of substitution than 1 every 300 generations
Statistical comparison of synonymous
& non-synonymous substitution rates
Depending on the values observed one may wish to test the following hypotheses:
ds > dn ds = dn ds < dn
To do this use the normal deviate or Z test (see lecture 8 slide #6)
Z = difference between ds & dn divided by the standard error of the difference
difference – D = abs (ds – dn)
Standard error of the difference – sD =
Z = D / sD
√ (se(ds)
2
+ se(dn)2)
formula in MSExcel =abs(ds-dn)/sqrt(se(ds)^2+se(dn)^2)
Then use t-table with infinite degrees of freedom to evaluate P value