Population Genetics II. Bio

Transcription

1 Population Genetics II. Bio Don Conrad

2 Agenda Population Genetic Inference Mutation Selection Recombination

3 The Coalescent Process ACTT T G C G ACGT ACGT ACTT ACTT AGTT Backward in time process Discovered by JFC Kingman, F. Tajima, R. R. Hudson c DNA sequence diversity is shaped by genealogical history Genealogies are unobserved but can be estimated Conceptual framework for population genetic inference: mutation, recombination, demographic history

4 2 sample coalescent MRCA N = population size of diploid individuals n = sample size of haploid chromosomes T 2 MRCA = most recent common ancestor T 2 = coalescence time for 2 chromosomes sequence1 sequence2

5 2 sample coalescent MRCA Probability that the time of MRCA is t generations ago T 2 " P(T 2 = t) = 1 1 % $ ' # 2N & t 1 " $ # 1 2N If we blur our eyes a bit (as N gets very large) this becomes % ' & sequence1 sequence2! P(T 2 = t) = 1 $ # &e " 2N %! 1 $ # &t " 2 N %

6 2 sample coalescent MRCA E(T 2 ) = 2N Var(T 2 ) = 2N 2 T 2 If we consider t = t/2n, call that coalescent time Then E(T 2 ) = 1 sequence1 sequence2

7 n-coalescent There are possible pairs, each coalescing at rate 1/2N n 2! " # $ % & = n(n 1) 2 P(T n = t) = n 2! " # $ % & 2N! " # # # # # $ % & & & & & e n 2! " # $ % & 2 N! " # # # # # $ % & & & & & t E(T n ) = 2N n 2! " # $ % &

8 n-coalescent Mean elapsed time (*2N) T 2 1 T 3 T 4 T 5 1/3 1/6 1/10 = 2/(5*4) E(TMRCA for n chromosomes)= T 2 + T 3 + T 4 + T n = 2(1-1/n) coalescent units

9 Simon Myers N e and coalescence times in humans and other animals In humans, it is known that appropriate values for N e are surprisingly small. This is approximation is called the effective population size : N e 10,000 in Europe N e 9,500 in East Asia N e < 25,000 for all human populations, highest in Africa

10 N e and coalescence times in humans and other animals The mean coalescence time for two lineages is just E( T 2 ) = 1 in units of 2N e generations, so if we have G=22 years per generation, the average ancestry depth for 2 human chromosomes is 1 2N e G in years (20,000-50,000) 22 = 440,000-1,100,000 years N e varies widely across species (Charlesworth, Nature Reviews Genetics 2009): 25,000,000 for E.coli 2,000,000 for fruit fly D. Melanogaster <100 for Salamanders (Funk et al. 1999) Simon Myers

11 Adding mutations For neutral models, can separately model the genealogical process (the tree) and the mutation process (genetic types) -Infinite sites mutation model

12 What is expected number of mutations between 2 chroms? μ = mutation rate per bp per generation t E(t) = 2N E(π t) = 2tµ E(π ) = 2µE(t) = 4Nµ t ~ Expo (2N) π ~ Pois(2tμ) 4Nu comes up repeatedly in population genetics, often referred to as theta θ = 4Nµ

13 Number of segregating sites in a sample of size n The total time in the tree for a sample of n chromosomes is t 2 L = 4t 4 + 3t 3 + 2t 2 or in general: L = n i=2 i *t i t 3 t 4 n n 1 = ie(t i ) 1 E(L) E(S) = µe(l) = 4Nµ =θ i=2 i=1 i n # 4N & = i% ( $ i(i 1) ' i=2 n 1 = 4N 1 i i=1 Watterson Estimator for population scaled mutation rate n 1 i=1 1 i θˆ W = S n 1 1 i i=1

14 Estimators of Theta Watterson estimator is just one approach, uses summary of the data θˆ W = S n 1 1 i i=1 There is also the Tajima estimator Average # of pairwise differences ˆ θ T = n S 2 p i (1 p i ) n 1 i=1 p i = allele frequency of mutation i S = number of polymorphic sites

15 The site frequency spectrum Under standard neutral model, the site frequency spectrum is a beta distribution with parameters theta/2,theta/2 Expected proportion of sites with count x is theta/x Number of sites Derived allele count Plot of theoretical SFS for 1Mb

16 The sfs under neutrality and selection

17 Daniel MacArthur, Suganti Balasubramaniam The sfs of genic variants splice-disrupting 621 stop-gain 1,654 non-synonymous 84,358 synonymous 61,155

18 Sample-based estimators of Θ using the sfs Estimator Sensitivity Source θ W = 1 n 11 i=1 $ θ π = & n' ) % 2( i=1 i 1 n 1 i( n i)ξ i i=1 θ ξ e = ξ e = ξ 1 # θ H = % n& ( $ 2' n 1 ξ i 1 n 1 i 2 ξ i i=1 θ L = 1 n 1 iξ i n 1 i=1 low Watterson (1975) intermediate Tajima (1989) singleton Fu and Li (1993) high Fay and Wu (2000) high Zeng et al. (2006) Sensitivity = the frequency of observed polymorphisms that makes estimates using a given estimator large relative to the others. Eckert, UCDavis

19 Tajima s D D t = 0 neutral evolution D t = θ π θ W C D t > 0 balancing selection, more intermediate variants D t < 0 positive selection

20 Bamshad & Wooding 2001

21 LCT D Nielsen, et al 2007

22 A typical population genomics study design for detecting positive selection. Akey J M Genome Res. 2009;19: by Cold Spring Harbor Laboratory Press

23 Are humans still experiencing adaptive evolution?

24 ihs(x), a homozygosity-based statistic function of test allele frequency summarizes extent of haplotype homozygosity of derived allele chromosomes compared to ancestral built-in control for local recombination rate PLoS Biology 2006

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26 Conflicting evidence of population-specific selection (A) Genic SNPs are more likely than non-genic SNPs to have extreme allele frequency differences Coop, et al PLoS 2009

27 Conflicting evidence of population-specific selection F st (B) Maximum allele frequency difference is well explained by average genetic distance between populations Coop, et al PLoS 2009

28 Polygenic model of adaptation Pritchard, et al Curr Biology, 2010

29 Linkage Disequilibrium nonrandom associations between alleles Compare to HWE: Under HWE, gametes unite at random. Pr(A,a) = pr(a) * pr(a) where A and a are alleles at same locus LD statistics measure to what extent Pr(A,B) = pr(a) * pr(b) when A and B are alleles at different loci Example applications: mapping genes and measuring recombination

30 Linkage Disequilibrium A/a B/b Frequencies: P A P B P a P b The 4 allele frequencies P AB P Ab P ab P ab A B A b a B a b The 4 haplotype frequencies Define D = P AB P A P B = P AB P ab P Ab P ab If D!= 0 then we have LD r 2 = D/(P A P B P a P b )

31 Where does LD come from? Final Aligned Data Set:

32 LD-based Case-Control Association Study l l locate disease locus Unlikely to be among our genotyped markers è Detect indirectly using available markers Cases (affected) --A C A----G---X----T---C---A T C A----G---X----C---C---A A G G----G---X----C---C---A A C A----G---X----T---C---A T C A----G---X----T---C---A T C A----T---X----T---A---A---- Controls (unaffected) --A C A----T---X----T---A---A A G G----G---X----C---C---A A G G----T---X----C---A---A A C A----G---X----T---C---A T C A---T---X----T---A---A T C G----T---X----A---A---A----

33 Where does array content come from? Publicly Funded Genomics Projects Human Genome Project Phase I HapMap Project CNV Project Phase II HapMap 1000 Genomes Project 10k 500k 1M Number of SNPs on an Affymetrix Chip

34 The International HapMap Project -HapMap project ( ) cost $120 Million USD -Measured variation at 4 million SNPs in 4 populations (90 Europeans, 90 Nigerians, 45 Chinese, 45 Japanese) Results: - Over 2.8 M SNP with > 5% allele frequency -80% of variants within each population can be captured with 30% of the most informative SNPs, tagsnps -Nigerians require most tag SNPs, followed by Europeans and then Asians

35 HapMap tagsnps are useful for other populations Conrad, et al. (2006) genotyped 3000 SNPs in 52 populations across the globe (the Human Genome Diversity Panel or HGDP) C+J: Asian, CEU: European, YRI: Nigerian

36 The serial bottleneck model

37 The recombination rate Can vary hugely along a sequence Determines association between loci in the population Is hard to measure directly, because recombination occurs on average only ~1 in 100,000,000 meioses between any pair of successive nucleotides in the genome. Can be measured indirectly, by parametric analysis of variation data Researchers in Oxford, and elsewhere, have developed such parametric approaches (Li and Stephens, 2003; Ptak et al. 2005; Hudson 2001, McVean 2002, McVean et al. 2004) Now, we are considering ρ = 4Nr

38 Marginal Trees Time Position A Position B Position C Physical position

39 Calculated a composite likelihood for a sample of haplotypes: L(4Nr) = L(n i, n j, n ij 4Nr ij ) -sum over pairs -lookup table -RJMCMC ij

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41 Nature Genet 2008

42 Science 2009

43 Nat Genet 2010