Presentation is loading. Please wait.

Presentation is loading. Please wait.

Molecular evolution:   how do we explain the patterns of variation observed in DNA sequences? how do we detect selection by comparing silent site substitutions.

Similar presentations


Presentation on theme: "Molecular evolution:   how do we explain the patterns of variation observed in DNA sequences? how do we detect selection by comparing silent site substitutions."— Presentation transcript:

1 Molecular evolution: how do we explain the patterns of variation observed in DNA sequences? how do we detect selection by comparing silent site substitutions to replacement substitutions? how do we detect selection by comparing fixed differences between species to polymorphisms within species? how do we detect selection by using hitchhiking? Goal: understand the logic behind key tests.

2 Neutralist vs. selectionist view
Are most substitutions due to drift or natural selection? “Neutralist” vs. “selectionist” Agree that: Most mutations are deleterious and are removed. Some mutations are favourable and are fixed. Dispute: Are most replacement mutations that fix beneficial or neutral? Is observed polymorphism due to selection or drift? Don’t know! Different faculty here have different views.

3 Reminder: substitution vs. polymorphism
What happen after a mutation changes a nucleotide in a locus Polymorphism: mutant allele is one of several present in population Substitution: the mutant allele fixes in the population. (New mutations at other nucleotides may occur later.)

4 Substitution schematic
Individual Time 0: aaat aaat aaat aaat aaat aaat aaat Time 10: aaat aaat aaat aaat acat aaat aaat Time 20: aaat aaat acat aaat acat acat acat Time 30: acat acat acat acat acat acat acat Time 40: acat acat actt acat acat acat acat Times 10-29: polymorphism Time 30: mutation fixed -> substitution Time 40: new mutation: polymorphism

5 Reminder: substitution rates for neutral mutations
Most neutral mutations are lost Only 1 out of 2N fix Most that are lost go quickly (< 20 generations for population sizes from ) Most replacement mutations are lost since deleterious: rate of loss is faster than neutral

6 Data in favor of neutrality
Substitutions in DNA appear to be clock-like Figure 6.21 Why are these observations evidence for neutrality? it’s a bit subtle. We think silent sites don’t matter for selection – so they must be neutral. If mutation rate is relatively constant, then a steady rate. fine. But, we expect replacement substitutions to matter. Yet, we find a steady rate – suggesting that the substitutions are being driven by selection.

7 Drift model pseudocode
Population with 2N – 1 copies of allele A, 1 of allele a For each generation, draw from prior generation alleles. -> generate a random number. If less than f(A), new allele = A. Otherwise, allele = a. -> repeat until 2N alleles drawn Check to see outcome of drift ->If a is lost, start over. ->If a has fixed, note the number of years ->Otherwise, next year with the new allele frequencies Repeat 100x per population size Test populations of 100, 500, 1000, 1500, and 2000

8 Times to fix for neutral alleles (Only 1/2N fix: how long do they take
Estimated formula: fixation time = 4.07 * N – 57 Theoretical formula: fixation time = 4N

9 Puzzle for neutrality Rates of substitution are clock-like per year, not per generation. Years Substitutions Actual pattern rabbits elephants Years Substitutions Expected pattern rabbits elephants However, there are some strange patterns that don’t quite fit with neutrality. From mutation data, we expect a correlation between substitution rate and generation time. We don’t see one. This means that something more complex is going on.

10 Revised theory: the nearly – neutral theory
Figure 6.22 So, a compromise theory was born. Nearly neutral. Idea is that generation time and population size balance out. Longer generation time – slower mutation rate.

11 Can we distinguish selection from drift using sequence data?
Compare two species: infer where substitutions have occurred. Silent site substitutions should be neutral (dS) Non-synonymous substitutions are expected to be deleterious (usually) (dN) so, expect < 1 Translation: rate of non-synonymous (dN) is less than the rate of synonymous substitutions (dS) So, does this theory match the data? Take a look at two more ways to look at this. first, we can look at lots of genes. We expect in any gene to find more silent substitions than replacements.

12 and inferences about selection
< 1: replacements are deleterious = 1: replacements are neutral But, sometimes we find more replacements. How can this be? > 1: replacements are beneficial

13 What happens to fixation time with selection? Model pseudocode
Population with 2N – 1 copies of allele A, 1 of allele a WA = 1 + s; Wa = 1 For each generation, draw from prior generation alleles. -> generate a random number. If greater than f(A), new alleel = a. Otherwise, test fitness: if random < WA, new allele = A. -> repeat until 2N alleles drawn Check to see outcome of drift ->If a is lost, start over. ->If a has fixed, note the number of years ->Otherwise, next year with the new allele frequencies Repeat 100x per fitness Test populations of 100

14 Time to fix favourable allele

15 Time to fix: neutral vs. favourable
Simulation results: black – neutral mutations; red – favourable mutations

16 Time to fixation: drift is slow
Neutral: New mutations per generation: 2Neu Probability of fixing a new mutation: 1 / 2Ne Fixations per generation: = 2Neu * 1 / 2Ne = u Time to fix: 4Ne Favored by selection New mutations per generation: 2Neu (but how many favourable??) Favored mutation probability of fixing: 2|s| Fixations per generation: 2Neu * 2|s| * prob. favourable Time to fix: 2 ln (2Ne) / |s| 2 ln (2Ne) / |s| << 4Ne Key: drift is slow. Shorter time to fixation Derivations of these results are tough! See Kimura (1962) and Kimutra and Ohta (1969).

17 Time to fixation: favourable and neutral

18 dN / dS data: BRCA1 > 1 < 1 Figure 6.21

19 Molecular evidence of selection II: McDonald-Kreitman Test
is very conservative: many selective events may be missed. Example: immunoglobins. = 0.37 overall We suspect selection favoring new combinations at key sites. Antigen recognition sites: > 3.0

20 Evidence of selection II: McDonald-Kreitman test

21 McDonald-Kreitman test III
If evolution of protein is neutral, the percentage of mutations that alter amino acids should be the same along any branch If all mutations are neutral, all should have the same probability of persisting So: dN / dS among polymorphisms should be the same as within fixed differences

22 McDonald-Kreitman logic
Silent sites - always neutral - fix slowly - contribute to polymorphism Replacement sites mainly unfavourable if neutral, fix at same rate as silent and contribute to polymorphism proportion of replacement mutations that are neutral determines dN / dS for polymorphism if favourable, fix quickly and do not contribute to polymorphism: higher dN / dS for fixed differences, lower rate for polymorphism

23 Time to fixation: favourable and neutral

24 Polymorphism and fixation
Neutral Deleterious Silent Replacement 1 / 2N neutral mutations fix

25 Polymorphism and fixation
Neutral Deleterious Favourable Silent Replacement 1 / 2N neutral mutations fix - slow 2|s| fix -fast Neutral Favourable

26 dN / dS for neutral and favourable
Polymorphism dN dN dS dS Fixation dN dN dS dS = < poly fixed poly fixed

27 McDonald-Kreitman hypotheses
H0: All mutations are neutral. Then, dN / dS for polymorphic sites should equal dN / dS for fixed differences H1: replacements are favoured. Favoured mutations fix rapidly, so dN / dS for polymorphic < dN / dS fixed

28 Example of MK test: ADH in Drosophilia
Compare sequences of D. simulans and D. yakuba for ADH (alcohol dehydrogenase) Fixed differences Polymorphic sites Replacement 7 2 Silent 17 42 % fixed 7 / 24 = 29% 2 / 44 = 5% Significance? Use χ2 test for independence

29 Evidence of selection III: selective sweeps
Imagine a new mutation that is strongly favored (e.g. insecticide resistance in mosquitoes)

30 Detecting selection using linkage: G6PD in humans
Natural history: Located on X chromosome encodes glucose-6-phosphate dehydrogenase Red blood cells lack mitochondria Glycolysis only NADPH only via pentose-phosphate shunt –requires G6PD NADPH needed for glutathione, which protects against oxidation No mt: : no citric acid cycle, no electron transport chain

31 G6PD and malaria Malaria (Plasmodium falciparum) infects red blood cells Has limited G6PD function typically (but can produce the enzyme) Uses NADPH from red blood cell In G6PD deficient individuals?

32 G6PD mutants Different mutants result in different levels of enzymatic activity Severe mutants result in destruction of red blood cells and anemia Most common mutant: G6PD-202A Usually mild effects: may increase risk of miscarriage Prediction: G6PD and malaria?

33 Frequency of G6PD deficiency

34 Has G6PD-202A been selected?
14 markers up to 413,000 bp from G6PD LD? Long distance LD implies strong, recent selection

35 Has G6PD-202A been selected?
Fig 7.14 Linkage disquilibrium kb from core region

36 Alternative hypothesis: drift caused linkage disequilibrium
Three possibilities: a new allele will have high LD but rare if common, means old – so low LD or, disappears So, if high frequency, high LD – selection only. G6PD-202A Allele frequency Figure 7.14b

37 Detecting selection II: CCR532

38 Detecting selection II: CCR5Δ32
Stephens (1998) found strong disequilibrium between CCR5-Δ32 and nearby markers Implies recent origin (< 2000 years): recombination breaks down linkage Implies selected

39 Detecting selection II: CCR5Δ32
But: new data – November 2005. Better map:

40 Detecting selection: summary
Several approaches to detecting selection dN / dS McDonald-Kreitman test using hitchhiking Challenges of each method?

41 Other uses of molecular data: the coalescent
Any two alleles in a population share a common ancestor in the last generation 1 / 2Ne Therefore, going backwards in time, the expected time to find the common ancestor is 1 / (1 / 2Ne) = 2Ne

42 Coalescent II

43 Coalescent and sequences
Imagine that you have two sequences at a locus. They shared a common ancestor 2Ne generations ago. They accumulate mutations at rate u per generation per basepair. 2Ne generations / lineage * 2 lineages * u = 4Neu differences per basepair between the two sequences.

44 Coalescent example We sequence 1000 base pairs from two sequences, and find 16 base pair differences, how large is the population/ Assume u = 2 x 10-8. 4Neu * 1000 = 16; 8 x 10-5 * Ne = 16; Ne * 10-5 = 2; Ne = 200,000

45 Neutral theory as a null model

46 Additional readings Eyre-Walker (2006) The genomic rate of adaptive evolution. Trends in Ecology and Evolution 29: (Well-written review) Gillespie (2004). Population genetics: a concise guide. John Hopkins: Baltimore, MD. (Very short, clear, but dense!) Graur and Li (2000) Fundamentals of molecular evolution. Sinauer: Sunderland, MA. (Very clear) Kimura (1962) On the probability of fixation of mutant genes in populations. Genetics 47: (If you really want the derivation) Kimura and Ohta (1969) The average number of generations until fixation of a mutant gene in a finite population. Genetics 61: (If you really want the derivation) Sabeti et al (2006) The case for selection at CCR5-32. PLoS Biology 3: Questions: 1. Explain why clock-like rates of substitutions per year did not fit with the neutral theory. See posted molecular evolution practice questions: highly recommended!


Download ppt "Molecular evolution:   how do we explain the patterns of variation observed in DNA sequences? how do we detect selection by comparing silent site substitutions."

Similar presentations


Ads by Google