Population Genetics Basics 1 Terminology review Allele Locus

Population Genetics Basics 1

Terminology review • • Allele Locus Diploid SNP 2

Single Nucleotide Polymorphisms Infinite Sites Assumption: Each site mutates at most once 00000101011 10001101001 01000101010 01000000011110000 0010110 3

What causes variation in a population? • • Mutations (may lead to SNPs) Recombinations Other genetic events (gene conversion) Structural Polymorphisms 4

Recombination 0000 1111 00011111 5

Gene Conversion • Gene Conversion versus crossover – Hard to distinguish in a population 6

Structural polymorphisms • Large scale structural changes (deletions/insertions/inversions) may occur in a population. 7

Topic 1: Basic Principles • In a ‘stable’ population, the distribution of alleles obeys certain laws – • HW Equilibrium – • Not really, and the deviations are interesting (due to mixing in a population) Linkage (dis)-equilibrium – Due to recombination 8

Hardy Weinberg equilibrium • • Consider a locus with 2 alleles, A, a p (respectively, q) is the frequency of A (resp. a) in the population 3 Genotypes: AA, Aa, aa Q: What is the frequency of each genotype If various assumptions are satisfied, (such as random mating, no natural selection), Then • PAA=p 2 • PAa=2 pq • Paa=q 2 9

Hardy Weinberg: why? • Assumptions: – – – • Diploid Sexual reproduction Random mating Bi-allelic sites Large population size, … Why? Each individual randomly picks his two chromosomes. Therefore, Prob. (Aa) = pq+qp = 2 pq, and so on. 10

Hardy Weinberg: Generalizations • Multiple alleles with frequencies – • By HW, Multiple loci? 11

Hardy Weinberg: Implications • • The allele frequency does not change from generation to generation. Why? It is observed that 1 in 10, 000 caucasians have the disease phenylketonuria. The disease mutation(s) are all recessive. What fraction of the population carries the mutation? Males are 100 times more likely to have the “red’ type of color blindness than females. Why? Conclusion: While the HW assumptions are rarely satisfied, the principle is still important as a baseline assumption, and significant deviations are interesting. 12

Recombination 0000 1111 00011111 13

What if there were no recombinations? • • • Life would be simpler Each individual sequence would have a single parent (even for higher ploidy) The relationship is expressed as a tree. 14

The Infinite Sites Assumption 0000 3 00100000 5 8 00100001 • • • 00101000 The different sites are linked. A 1 in position 8 implies 0 in position 5, and vice versa. Some phenotypes could be linked to the polymorphisms Some of the linkage is “destroyed” by recombination 15

Infinite sites assumption and Perfect Phylogeny • • Each site is mutated at most once in the history. All descendants must carry the mutated value, and all others must carry the ancestral value i 1 in position i 0 in position i 16

Perfect Phylogeny • • Assume an evolutionary model in which no recombination takes place, only mutation. The evolutionary history is explained by a tree in which every mutation is on an edge of the tree. All the species in one sub-tree contain a 0, and all species in the other contain a 1. Such a tree is called a perfect phylogeny. 17

The 4 -gamete condition • • • A column i partitions the set of species into two sets i 0, and i 1 A column is homogeneous w. r. t a set of species, if it has the same value for all species. Otherwise, it is heterogenous. EX: i is heterogenous w. r. t {A, D, E} A i 0 B C D i 1 E F i 0 0 0 1 18

4 Gamete Condition • 4 Gamete Condition – – – There exists a perfect phylogeny if and only if for all pair of columns (i, j), either j is not heterogenous w. r. t i 0, or i 1. Equivalent to There exists a perfect phylogeny if and only if for all pairs of columns (i, j), the following 4 rows do not exist (0, 0), (0, 1), (1, 0), (1, 1) 19

4 -gamete condition: proof • • • Depending on which edge the mutation j occurs, either i 0, or i 1 should be homogenous. (only if) Every perfect phylogeny satisfies the 4 gamete condition (if) If the 4 -gamete condition is satisfied, does a prefect phylogeny exist? i i 0 i 1 20

An algorithm for constructing a perfect phylogeny • • We will consider the case where 0 is the ancestral state, and 1 is the mutated state. This will be fixed later. In any tree, each node (except the root) has a single parent. – • • It is sufficient to construct a parent for every node. In each step, we add a column and refine some of the nodes containing multiple children. Stop if all columns have been considered. 21

Inclusion Property • • For any pair of columns i, j – i < j if and only if i 1 j 1 Note that if i<j then the edge containing i is an ancestor of the edge containing i i j 22

Example 1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 D 0 0 1 E 1 0 0 r A B C D E Initially, there is a single clade r, and each node has r as its parent 23

Sort columns • • Sort columns according to the inclusion property (note that the columns are already sorted here). This can be achieved by considering the columns as binary representations of numbers (most significant bit in row 1) and sorting in decreasing order A B C D E 1 1 0 1 2 1 0 0 3 0 1 0 4 0 0 1 0 0 5 0 0 0 1 0 24

Add first column • In adding column i – – Check each edge and decide which side you belong. Finally add a node if you can resolve a clade A B C D E 1 2 3 4 5 1 1 0 0 0 1 1 0 0 0 1 1 0 0 r u A C E B D 25

Adding other columns • Add other columns on edges using the ordering property A B C D E 1 1 0 1 2 1 0 0 3 0 1 0 4 0 0 1 0 0 5 0 0 0 1 0 r 1 E 3 2 B 4 C 5 D A 26

Unrooted case • • Switch the values in each column, so that 0 is the majority element. Apply the algorithm for the rooted case 27

Handling recombination • • A tree is not sufficient as a sequence may have 2 parents Recombination leads to loss of correlation between columns 28

Linkage (Dis)-equilibrium (LD) • • Consider sites A &B Case 1: No recombination – Pr[A, B=0, 1] = 0. 25 • • Linkage disequilibrium Case 2: Extensive recombination – Pr[A, B=(0, 1)=0. 125 • Linkage equilibrium A 0 0 1 1 B 1 1 0 0 0 29

Handling recombination • • A tree is not sufficient as a sequence may have 2 parents Recombination leads to loss of correlation between columns 30

Recombination, and populations • • • Think of a population of N individual chromosomes. The population remains stable from generation to generation. Without recombination, each individual has exactly one parent chromosome from the previous generation. With recombinations, each individual is derived from one or two parents. We will formalize this notion later in the context of coalescent theory. 31

Linkage (Dis)-equilibrium (LD) • • • Consider sites A &B Case 1: No recombination Each new individual chromosome chooses a parent from the existing ‘haplotype’ A 0 0 1 1 B 1 1 0 0 0 1 0 32

Linkage (Dis)-equilibrium (LD) • • • Consider sites A &B Case 2: diploidy and recombination Each new individual chooses a parent from the existing alleles A 0 0 1 1 B 1 1 0 0 0 1 1 33

Linkage (Dis)-equilibrium (LD) • • • Consider sites A &B Case 1: No recombination Each new individual chooses a parent from the existing ‘haplotype’ – Pr[A, B=0, 1] = 0. 25 • Linkage disequilibrium Case 2: Extensive recombination Each new individual simply chooses and allele from either site – Pr[A, B=(0, 1)=0. 125 • Linkage equilibrium A 0 0 1 1 B 1 1 0 0 0 34

LD • In the absence of recombination, – – • Correlation between columns The joint probability Pr[A=a, B=b] is different from P(a)P(b) With extensive recombination – Pr(a, b)=P(a)P(b) 35

Measures of LD • • Consider two bi-allelic sites with alleles marked with 0 and 1 Define – – • • P 00 = Pr[Allele 0 in locus 1, and 0 in locus 2] P 0* = Pr[Allele 0 in locus 1] Linkage equilibrium if P 00 = P 0* P*0 D = abs(P 00 - P 0* P*0) = abs(P 01 - P 0* P*1) = … 36

LD over time • With random mating, and fixed recombination rate r between the sites, Linkage Disequilibrium will disappear – – Let D(t) = LD at time t P(t)00 = (1 -r) P(t-1)00 + r P(t-1)0* P(t-1)*0 D(t) = P(t)00 - P(t)0* P(t)*0 = P(t)00 - P(t-1)0* P(t-1)*0 D(t) =(1 -r) D(t-1) =(1 -r)t D(0) 37

LD over distance • Assumption – – • • Recombination rate increases linearly with distance LD decays exponentially with distance. The assumption is reasonable, but recombination rates vary from region to region, adding to complexity This simple fact is the basis of disease association mapping. 38

LD and disease mapping • • • Consider a mutation that is causal for a disease. The goal of disease gene mapping is to discover which gene (locus) carries the mutation. Consider every polymorphism, and check: – – • There might be too many polymorphisms Multiple mutations (even at a single locus) that lead to the same disease Instead, consider a dense sample of polymorphisms that span the genome 39

LD can be used to map disease genes LD D N N D D N • • 0 1 1 0 0 1 LD decays with distance from the disease allele. By plotting LD, one can short list the region containing the disease gene. 40

LD and disease gene mapping problems • • • Marker density? Complex diseases Population sub-structure 41

Human Samples • • We look at data from human samples Gabriel et al. Science 2002. – 3 populations were sampled at multiple regions spanning the genome • • • 54 regions (Average size 250 Kb) SNP density 1 over 2 Kb 90 Individuals from Nigeria (Yoruban) 93 Europeans 42 Asian 50 African American 42

Population specific recombination • • D’ was used as the measure between SNP pairs were classified in one of the following – – – • Strong LD Strong evidence for recombination Others (13% of cases) This roughly favors out-ofafrica. A Coalescent simulation can help give confidence values on this. Gabriel et al. , Science 2002 43

Haplotype Blocks • A haplotype block is a region of low recombination. – • • Define a region as a block if less than 5% of the pairs show strong recombination Much of the genome is in blocks. Distribution of block sizes vary across populations. 44

Testing Out-of-Africa • • Generate simulations with and without migration. Check size of haplotype blocks. – – • Does it vary when migrations are allowed? When the ‘new’ population has a bottleneck? If there was a bottleneck that created European and Asian populations, can we say anything about frequency of alleles that are ‘African specific’? – Should they be high frequency, or low frequency in African populations? 45

Haplotype Block: implications • • The genome is mostly partitioned into haplotype blocks. Within a block, there is extensive LD. – Is this good, or bad, for association mapping? 46

Coalescent reconstruction • Reconstructing likely coalescents 47

Re-constructing history in the absence of recombination 48

An algorithm for constructing a perfect phylogeny • • We will consider the case where 0 is the ancestral state, and 1 is the mutated state. This will be fixed later. In any tree, each node (except the root) has a single parent. – • • It is sufficient to construct a parent for every node. In each step, we add a column and refine some of the nodes containing multiple children. Stop if all columns have been considered. 49

Inclusion Property • • For any pair of columns i, j – i < j if and only if i 1 j 1 Note that if i<j then the edge containing i is an ancestor of the edge containing i i j 50

Example 1 2 3 4 5 A 1 1 0 0 0 B 0 0 1 0 0 C 1 1 0 D 0 0 1 E 1 0 0 r A B C D E Initially, there is a single clade r, and each node has r as its parent 51

Sort columns • • Sort columns according to the inclusion property (note that the columns are already sorted here). This can be achieved by considering the columns as binary representations of numbers (most significant bit in row 1) and sorting in decreasing order A B C D E 1 1 0 1 2 1 0 0 3 0 1 0 4 0 0 1 0 0 5 0 0 0 1 0 52

Add first column • In adding column i – – Check each edge and decide which side you belong. Finally add a node if you can resolve a clade A B C D E 1 2 3 4 5 1 1 0 0 0 1 1 0 0 0 1 1 0 0 r u A C E B D 53

Adding other columns • Add other columns on edges using the ordering property A B C D E 1 1 0 1 2 1 0 0 3 0 1 0 4 0 0 1 0 0 5 0 0 0 1 0 r 1 E 3 2 B 4 C 5 D A 54

Unrooted case • • Important point is that the perfect phylogeny condition does not change when you interchange 1 s and 0 s at a column. Switch the values in each column, so that 0 is the majority element. Apply the algorithm for the rooted case. Homework: show that this is a correct algorithm 55

Population Sub-structure 56

Population sub-structure can increase LD • • • Consider two populations that were isolated and evolving independently. They might have different allele frequencies in some regions. Pick two regions that are far apart (LD is very low, close to 0) 0 0 0 1 0 0 0 . . . . 1 1 0 1 1 1 . . . . 0 0 0 1 0 0 0 Pop. A p 1=0. 1 q 1=0. 9 P 11=0. 1 D=0. 01 Pop. B p 1=0. 9 q 1=0. 1 P 11=0. 1 D=0. 01 57

Recent ad-mixing of population • • If the populations came together recently (Ex: African and European population), artificial LD might be created. D = 0. 15 (instead of 0. 01), increases 10 -fold This spurious LD might lead to false associations Other genetic events can cause LD to arise, and one needs to be careful 0 0 0 1 0 0 0 . . . . 1 1 0 1 1 1 . . . . 0 0 0 1 0 0 0 Pop. A+B p 1=0. 5 q 1=0. 5 P 11=0. 1 D=0. 1 -0. 25=0. 15 58

Determining population sub-structure • • Given a mix of people, can you sub-divide them into ethnic populations. Turn the ‘problem’ of spurious LD into a clue. – – – Find markers that are too far apart to show LD If they do show LD (correlation), that shows the existence of multiple populations. Sub-divide them into populations so that LD disappears. 59

Determining Population sub-structure • • • Same example as before: The two markers are too similar to show any LD, yet they do show LD. However, if you split them so that all 0. . 1 are in one population and all 1. . 0 are in another, LD disappears 0 0 0 1 0 0 0 . . . . 1 1 0 1 1 1 . . . . 0 0 0 1 0 0 0 60

Iterative algorithm for population substructure • • Define N = number of individuals (each has a single chromosome) k = number of sub-populations. Z {1. . k}N is a vector giving the sub-population. – • • Zi=k’ => individual i is assigned to population k’ Xi, j = allelic value for individual i in position j Pk, j, l = frequency of allele l at position j in population k 61

Example • • • Ex: consider the following assignment P 1, 1, 0 = 0. 9 P 2, 1, 0 = 0. 1 1 1 0 0 0 . . 1 1 0 1 1 1 1 2 2 2 2 2 1 1 0 1 1 1 1 . . 0 0 0 1 0 0 0 62

Goal • • X is known. P, Z are unknown. The goal is to estimate Pr(P, Z|X) Various learning techniques can be employed. – – – • max. P, Z Pr(X|P, Z) (Max likelihood estimate) max. P, Z Pr(X|P, Z) Pr(P, Z) (MAP) Sample P, Z from Pr(P, Z|X) Here a Bayesian (MCMC) scheme is employed to sample from Pr(P, Z|X). We will only consider a simplified version 63

Algorithm: Structure • Iteratively estimate – • • (Z(0), P(0)), (Z(1), P(1)), . . , (Z(m), P(m)) After ‘convergence’, Z(m) is the answer. Iteration – – Guess Z(0) For m = 1, 2, . . • • • Sample P(m) from Pr(P | X, Z(m-1)) Sample Z(m) from Pr(Z | X, P(m)) How is this sampling done? 64

Example • • • Choose Z at random, so each individual is assigned to be in one of 2 populations. See example. Now, we need to sample P(1) from Pr(P | X, Z(0)) Simply count Nk, j, l = number of people in pouplation k which have allele l in position j pk, j, l = Nk, j, l / N 1 2 2 1 1 2 1 2 0 0 0 1 0 0 0 . . 1 1 0 1 1 1 1 2 2 1 1 1 0 1 1 1 1 . . 0 0 0 1 0 0 0 65

Example • • Nk, j, l = number of people in population k which have allele l in position j pk, j, l = Nk, j, l / Nk, j, * N 1, 1, 0 = 4 N 1, 1, 1 = 6 p 1, 1, 0 = 4/10 p 1, 2, 0 = 4/10 Thus, we can sample P(m) 1 2 2 1 1 2 1 2 0 0 0 1 0 0 0 . . 1 1 0 1 1 1 1 2 2 1 1 1 0 1 1 1 1 . . 0 0 0 1 0 0 0 66

Sampling Z • • • Pr[Z 1 = 1] = Pr[” 01” belongs to population 1]? We know that each position should be in linkage equilibrium and independent. Pr[” 01” |Population 1] = p 1, 1, 0 * p 1, 2, 1 =(4/10)*(6/10)=(0. 24) Pr[” 01” |Population 2] = p 2, 1, 0 * p 2, 2, 1 = (6/10)*(4/10)=0. 24 Pr [Z 1 = 1] = 0. 24/(0. 24+0. 24) = 0. 5 Assuming, HWE, and LE 67

Sampling • • Suppose, during the iteration, there is a bias. Then, in the next step of sampling Z, we will do the right thing Pr[“ 01”| pop. 1] = p 1, 1, 0 * p 1, 2, 1 = 0. 7*0. 7 = 0. 49 Pr[“ 01”| pop. 2] = p 2, 1, 0 * p 2, 2, 1 =0. 3*0. 3 = 0. 09 Pr[Z 1 = 1] = 0. 49/(0. 49+0. 09) = 0. 85 Pr[Z 6 = 1] = 0. 49/(0. 49+0. 09) = 0. 85 Eventually all “ 01” will become 1 population, and all “ 10” will become a second population 1 1 1 2 1 2 1 1 0 0 0 . . 1 1 0 1 1 1 1 2 2 2 1 1 1 0 1 1 1 1 . . 0 0 0 1 0 0 0 68

Allowing for admixture • • Define qi, k as the fraction of individual i that originated from population k. Iteration – – Guess Z(0) For m = 1, 2, . . • • Sample P(m), Q(m) from Pr(P, Q | X, Z(m-1)) Sample Z(m) from Pr(Z | X, P(m), Q(m)) 69

Estimating Z (admixture case) • Instead of estimating Pr(Z(i)=k|X, P, Q), (origin of individual i is k), we estimate Pr(Z(i, j, l)=k|X, P, Q) i, 1 i, 2 j 70

Results on admixture prediction 71

Results: Thrush data • • For each individual, q(i) is plotted as the distance to the opposite side of the triangle. The assignment is reliable, and there is evidence of admixture. 72

Population Structure • 377 locations (loci) were sampled in 1000 people from 52 populations. 6 genetic clusters were obtained, which corresponded to 5 geographic regions (Rosenberg et al. Science 2003) Africa Eurasia East Asia Oceania • America 73

Population sub-structure: research problem • • Systematically explore the effect of admixture. Can admixture be predicted for a locus, or for an individual The sampling approach may or may not be appropriate. Formulate as an optimization/learning problem: – – (w/out admixture). Assign individuals to sub-populations so as to maximize linkage equilibrium, and hardy weinberg equilibrium in each of the sub-populations (w/ admixture) Assign (individuals, loci) to sub-populations 74