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EQUILIBRIA IN POPULATIONS

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Population data

• Recall that we often study a population in the form of a SNP matrix– Rows correspond to individuals (or individual chromosomes), columns

correspond to SNPs– The matrix is binary (why?)– The underlying genealogy is hidden. If the span is large, the genealogy is

not a tree any more. Why?

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Basic Principles

• In a ‘stable’ population, the distribution of alleles obeys certain laws– Not really, and the deviations are interesting

• HW Equilibrium – (due to mixing in a population)

• Linkage (dis)-equilibrium– Due to recombination

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Hardy Weinberg equilibrium

• Assume a population of diploid individuals;• Consider a locus with 2 alleles, A, a• Define p (respectively, q): the frequency of A (resp. a)

in the population• 3 Genotypes: AA, Aa, aa• Q: What is the frequency of each genotypes

If various assumptions are satisfied, (such as random mating, no natural selection), Then• PAA=p2

• PAa=2pq• Paa=q2

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Hardy Weinberg: why?

• Assumptions:– Diploid– Sexual reproduction– Random mating– Bi-allelic sites– Large population size, …

• Why? Each individual randomly picks his two parental chromosomes. Therefore, Prob. (Aa) = pq+qp = 2pq, and so on.

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01001000

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Hardy Weinberg: Generalizations

• Multiple alleles with frequencies– By HW,

• Multiple loci?

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Hardy Weinberg: Implications

• The allele frequency does not change from generation to generation. True or false?

• 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?

• Individuals homozygous for S have the sickle-cell disease. In an experiment, the ratios A/A:A/S: S/S were 9365:2993:29. Is HWE violated? Is there a reason for this violation?

• A group of individuals was chosen in NYC. Would you be surprised if HWE was violated?

• Conclusion: While the HW assumptions are rarely satisfied, the principle is still important as a baseline assumption, and significant deviations are interesting.

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A modern example of HW application• SNP-chips can give us the allelic

value at each polymorphic site based on hybridization.

• Typically, the sites sampled have common SNPs– Minor allele frequency > 10%

• We simply plot the 3 genotypes at each locus on 3 separate horizontal lines.

• What is peculiar in the picture?• What is your conclusion?

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Linkage Equilibrium

• HW equilibrium is the equilibrium in allele frequencies at a single locus.

• Linkage equilibrium refers to the equilibrium between allele occurrences at two loci.– LE is due to recombination

events– First, we will consider the case

where no recombination occurs.

HWE

Perfect phylogeny

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What if there were no recombinations?

• Life would be simpler• Each individual sequence would have a single parent

(even for higher ploidy)• True for mtDNA (maternal) , and y-chromosome

(paternal)• The genealogical relationship is expressed as a tree.

This principle can be used to track ancestry of an individual

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Phylogeny

• Recall that we often study a population in the form of a SNP matrix– Rows correspond to individuals (or individual chromosomes), columns

correspond to SNPs– The matrix is binary (why?)– The underlying genealogy is hidden. If the span is large, the genealogy is

not a tree any more. Why?

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Reconstructing Perfect Phylogeny

• Input: SNP matrix M (n rows, m columns/sites/mutations)

• Output: a tree with the following properties– Rows correspond to leaf nodes– We add mutations to edges– each edge labeled with i splits

the individuals into two subsets. • Individuals with a 1 in column i• Individuals with a 0 in column i

i

1 in position i0 in position i

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Reconstructing perfect phylogeny

• Each mutation can be labeled by the column number

• Goal is to reconstruct the phylogeny• genographic atlas

1

3 4

2

85

67

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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.

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Columns

• Define i0: taxa (individuals) with a 0 at the i’th column

• Define i1: taxa (individuals) with a 1 at the i’th column

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Inclusion Property

• For any pair of columns i,j, one of the folowing holds– i1 j1

– j1 i1

– i1 j1 = • For any pair of columns i,j

– i < j if and only if i1 j1 • Note that if i<j then the edge

containing i is an ancestor of the edge containing j

i

j

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Perfect Phylogeny via Example

1 2 3 4 5A 1 1 0 0 0B 0 0 1 0 0C 1 1 0 1 0D 0 0 1 0 1E 1 0 0 0 0

r

A B C D E

Initially, there is a single clade r, and each node has r as its parent

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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

1 2 3 4 5A 1 1 0 0 0B 0 0 1 0 0C 1 1 0 1 0D 0 0 1 0 1E 1 0 0 0 0

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Add first column• In adding column i– Check each individual

and decide which side you belong.

– Finally add a node if you can resolve a clade

r

A BC DE

1 2 3 4 5A 1 1 0 0 0B 0 0 1 0 0C 1 1 0 1 0D 0 0 1 0 1E 1 0 0 0 0

u

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Adding other columns• Add other

columns on edges using the ordering property r

E B

C

D

A

1 2 3 4 5A 1 1 0 0 0B 0 0 1 0 0C 1 1 0 1 0D 0 0 1 0 1E 1 0 0 0 0

1

2

4

3

5

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Unrooted case

• Important point is that the perfect phylogeny condition does not change when you interchange 1s and 0s at a column.

• Alg (Unrooted)– Switch the values in each column, so that 0 is the majority

element. – Apply the algorithm for the rooted case.– Relabel columns and individuals.

• Show that this is a correct algorithm.

Unrooted perfect phylogeny

• We transform matrix M to a 0-major matrix M0.

• if M0 has a directed perfect phylogeny, M has a perfect phylogeny.

• If M has a perfect phylogeny, does M0 have a directed perfect phylogeny?

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Unrooted case

• Theorem: If M has a perfect phylogeny, there exists a relabeling, and a perfect phylogeny s.t.– Root is all 0s– For any SNP (column), #1s <= #0s– All edges are mutated 01

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Proof• Consider the perfect phylogeny of M.• Find the center: none of the clades has

greater than n/2 nodes.– Is this always possible?

• Root at one of the 3 edges of the center, and direct all mutations from 01 away from the root. QED

• If the theorem is correct, then simply relabeling all columns so that the majority element is 0 is sufficient.

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‘Homework’ Problems

• What if there is missing data? (An entry that can be 0 or 1)?

• What if there are recurrent mutations?

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Linkage Disequilibrium

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Quiz• Recall that a SNP data-set is a ‘binary’ matrix.– Rows are individual (chromosomes)– Columns are alleles at a specific locus

• Suppose you have 2 SNP datasets of a contiguous genomic region but no other information – One from an African population, and one from a

European Population.– Can you tell which is which?– How long does the genomic region have to be?

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Linkage (Dis)-equilibrium (LD)

• Consider sites A &B• Case 1: No recombination• Each new individual

chromosome chooses a parent from the existing ‘haplotype’

A B0 10 10 00 01 01 01 01 0

1 0

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Linkage (Dis)-equilibrium (LD)

• Consider sites A &B• Case 2: diploidy and

recombination• Each new individual chooses a

parent from the existing alleles

A B0 10 10 00 01 01 01 01 0

1 1

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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 B0 10 10 00 01 01 01 01 0

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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)

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Measures of LD

• Consider two bi-allelic sites with alleles marked with 0 and 1

• Define– P00 = Pr[Allele 0 in locus 1, and 0 in locus 2]

– P0* = Pr[Allele 0 in locus 1]

• Linkage equilibrium if P00 = P0* P*0

• D = (P00 - P0* P*0) = -(P01 - P0* P*1) = …

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Other measures of LD• D’ is obtained by dividing D by the largest possible

value– Suppose D = (P00 - P0* P*0) >0.

– Then the maximum value of Dmax= min{P0* P*1, P1* P*0}

– If D<0, then maximum value is max{-P0* P*0, -P1* P*1}

– D’ = D/ Dmax

Site 10

Site 2

1

10

D -D

-D D

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Other measures of LD• D’ is obtained by dividing D by the largest possible value

– Ex: D’ = abs(P11- P1* P*1)/ Dmax

• = D/(P1* P0* P*1 P*0)1/2

• Let N be the number of individuals• Show that 2N is the 2 statistic between the two sites

Site 10

Site 2

1

10

P00N P0*N

Digression: The χ2 test0 1

1

0 O1

O3 O4

O2

• The statistic behaves like a χ2 distribution (sum of squares of normal variables).

• A p-value can be computed directly

Observed and expected

Site 10

1

P00N

10 10

P01N

P10N P11N

P0*P*0N

P1*P*0N P1*P*1N

P0*P*1N

• = D/(P1* P0* P*1 P*0)1/2

• Verify that 2N is the 2 statistic between the two sites

LD over time and distance

• The number of recombination events between two sites, can be assumed to be Poisson distributed.

• Let r denote the recombination rate between two adjacent sites

• r = # crossovers per bp per generation• The recombination rate between two sites l

apart is rl

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LD over time• Decay in LD– Let D(t) = LD at time t between two sites– r’=lr– 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 (Why?)– D(t) =(1-r’) D(t-1) =(1-r’)t D(0)

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LD over distance

• Assumption– Recombination rate increases linearly with

distance and time– LD decays exponentially.

• 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.

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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

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LD can be used to map disease genes

• LD decays with distance from the disease allele.• By plotting LD, one can short list the region containing the

disease gene.

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LD

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• 269 individuals– 90 Yorubans– 90 Europeans (CEPH)– 44 Japanese– 45 Chinese

• ~1M SNPs

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Haplotype blocks

• It was found that recombination rates vary across the genome– How can the recombination rate be measured?

• In regions with low recombination, you expect to see long haplotypes that are conserved. Why?

• Typically, haplotype blocks do not span recombination hot-spots

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Long haplotypes• Chr 2 region with high r2 value (implies little/no recombination)• History/Genealogy can be explained by a tree ( a perfect phylogeny)• Large haplotypes with high frequency

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19q13

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LD variation across populations• LD is maintained upto

60kb in swedish population, 6kb in Yoruban population

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Reich et al.Nature 411, 199-204(10 May 2001)

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Understanding Population Data• We described various population genetic concepts (HW,

LD), and their applicability• The values of these parameters depend critically upon the

population assumptions.– What if we do not have infinite populations– No random mating (Ex: geographic isolation)– Sudden growth– Bottlenecks– Ad-mixture

• It would be nice to have a simulation of such a population to test various ideas. How would you do this simulation?