CSE280b: Population Genetics

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CSE280b Population Genetics

• Vineet Bafna/Pavel Pevzner

www.cse.ucsd.edu/classes/sp06/cse280b
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Population Sub-structure
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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)

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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 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1
Pop. AB
p10.5 q10.5 P110.1 D0.1-0.250.15
1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0
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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.

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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

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Iterative algorithm for population sub-structure

• Define
• N number of individuals (each has a single
  chromosome)
• k number of sub-populations.
• Z ? 1..kN is a vector giving the
  sub-population.
• Zik gt 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

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Example

• Ex consider the following assignment
• P1,1,0 0.9
• P2,1,0 0.1

0 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1 0 .. 1
1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2
1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0 1 .. 0
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Goal

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

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AlgorithmStructure

• 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?

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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

0 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1 0 .. 1
1 2 2 1 1 2 1 2 1 2 1 2 2 1 1 2 1 2 2 1
1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0 1 .. 0
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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,
• N1,1,0 4
• N1,1,1 6
• p1,1,0 4/10
• p1,2,0 4/10
• Thus, we can sample P(m)

0 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1 0 .. 1
1 2 2 1 1 2 1 2 1 2 1 2 2 1 1 2 1 2 2 1
1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0 1 .. 0
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Sampling Z

• PrZ1 1 Pr01 belongs to population 1?
• We know that each position should be in linkage
  equilibrium and independent.
• Pr01 Population 1 p1,1,0 p1,2,1
  (4/10)(6/10)(0.24)
• Pr01 Population 2 p2,1,0 p2,2,1
  (6/10)(4/10)0.24
• Pr Z1 1 0.24/(0.240.24) 0.5

Assuming, HWE, and LE
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Sampling

• Suppose, during the iteration, there is a bias.
• Then, in the next step of sampling Z, we will do
  the right thing
• Pr01 pop. 1 p1,1,0 p1,2,1 0.70.7
  0.49
• Pr01 pop. 2 p2,1,0 p2,2,1 0.30.3
  0.09
• PrZ1 1 0.49/(0.490.09) 0.85
• PrZ6 1 0.49/(0.490.09) 0.85
• Eventually all 01 will become 1 population, and
  all 10 will become a second population

0 .. 1 0 .. 1 0 .. 0 1 .. 1 0 .. 1 0 .. 1 0 ..
1 0 .. 1 0 .. 1 0 .. 1
1 1 1 2 1 2 1 2 1 1 2 2 2 1 2 2 1 2 2 1
1 .. 0 1 .. 0 0 .. 0 1 .. 1 1 .. 0 1 .. 0 1 ..
0 1 .. 0 1 .. 0 1 .. 0
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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))

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Estimating Z (admixture case)

• Instead of estimating Pr(Z(i)kX,P,Q), (origin
  of individual i is k), we estimate
  Pr(Z(i,j,l)kX,P,Q)

i,1
i,2
j
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Results on admixture prediction
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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.

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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)

Oceania
Eurasia
East Asia
America
Africa
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Population sub-structureresearch 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

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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))

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Estimating Z (admixture case)

• Instead of estimating Pr(Z(i)kX,P,Q), (origin
  of individual i is k), we estimate
  Pr(Z(i,j,l)kX,P,Q)

i,1
i,2
j
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Results on admixture prediction simulated data
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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.

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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)

Oceania
Eurasia
East Asia
America
Africa
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NJ versus Structurethrush data

• Objective function is different in standard
  clustering algorithms!

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Population sub-structureresearch 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

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Admixture mapping
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Estimating Recombination Rates
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Recombination in human chromosome 22 (Mb scale)
Dawson et al. Nature 2002
Q Can we give a direct count of the number of
the recombination events?
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Recombination hot-spots (fine scale)
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Recombination rates (chimp/human)

• Fine scale recombination rates differ between
  chimp and human
• The six hot-spots seen in human are not seen in
  chimp

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Combinatorial Bounds for estimating recombination
rate

• Recall that expected recombinations ? log n
• Procedure
• Generate N random ARGs that results in the given
  sample
• Compute mean of the number of recombinations
• Alternatively, generate a summary statistic s
  from the population.
• For each ?, generate many populations, and
  compute the mean and variance of s (This only
  needs to be done once).
• Use this to select the most likely ?
• What is the correct summary statistic?
• Today, we talk about the min. number of
  recombination events as a possible summary
  statistic. It is not the most natural, but it is
  the most interesting computationally.

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The Infinite Sites Assumption the 4 gamete
condition
0 0 0 0 0 0 0 0
3
0 0 1 0 0 0 0 0
5
8
0 0 1 0 1 0 0 0
0 0 1 0 0 0 0 1

• Consider a history without recombination. No pair
  of sites shows all four gametes 00,01,10,11.
• A pair of sites with all 4 gametes implies a
  recombination event

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Hudson Kaplan

• Any pair of sites (i,j) containing 4 gametes must
  admit a recombination event.
• Disjoint (non-overlapping) sites must contain
  distinct recombination events, which can be
  summed! This gives a lower bound on the number of
  recombination events.
• Based on simulations, this bound is not tight.

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Myers and Griffiths03 Idea 1

• Let B(i,j) be a lower bound on the number of
  recombinations between sites i and j.

1i1 i2 i3 i4 i5 i6
ikn

• Can we compute maxP R(P) efficiently?

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The Rm bound
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Improved lower bounds

• The Rm bound also gives a general technique for
  combining local lower bounds into an overall
  lower bound.
• In the example, Rm2, but we cannot give any ARG
  with 2 recombination events.
• Can we improve upon Hudson and Kaplan to get
  better local lower bounds?

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
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Myers Griffiths Idea 2

• Consider the history of individuals. Let Ht
  denote the number of distinct haplotypes at time
  t
• One of three things might happen at time t
• Mutation Ht increase by at most 1
• Recombination Ht increase by at most 1
• Coalescence Ht does not increase

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The RH bound
0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1
Ex Rgt 8-3-14
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RH bound

• In general, RH can be quite weak
• consider the case when SgtH
• However, it can be improved
• Partitioning idea sum RH over disjoint intervals
• Apply to any subset of columns. Ex Apply RH to
  the yellow columns

000000000000000 000000000000001 000000010000000 00
0000010000001 100000000000000 100000000000001 1000
00010000000 111111111111111
(BB05)
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The Rs bound

• Compute the minimum number of recombination
  events R in any ARG. Note that, we do not
  explicitly construct the ARG.
• Consider a matrix with M with H rows and S
  columns.
• The rows correspond to haplotypes.
• Columns correspond to sites.

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Rs bound Observation I
s

• Non-informative column If a site contains at
  most one 1, or one 0, then in any history, it can
  be obtained by adding a mutation to a branch.
• EX if a is the haplotype containing a 1, It can
  simply be added to the branch without increasing
  number of recombination events
• R(M) R(M-s)

0 0 0 1
a
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Rs bound Observation 2

• Redundant rows If two rows h1 and h2 are
  identical, then
• R(M) R(M-h1)

c
r1
r2
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Rs bound Observation 3

• Suppose M has no non-informative columns, or
  redundant rows.
• Then, at least one of the haplotypes is a
  recombinant.
• There exists h s.t.
• R(M) R(M-h)1
• Which h should you choose?

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Rs bound (Procedural)

• Procedure Compute_Rs(M)
• If ? non-informative column s
• return (Compute_Rs(M-s))
• Else if ? redundant row h
• return (Compute_Rs(M-h))
• Else
• return (1 minh(Compute_Rs(M-h))

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Results
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Additional results/problems

• Using dynamic programming, Rs can be computed in
  2n poly(mn) time.
• Also, Rs can be augmented to handle
  intermediates.
• Are there poly. time lower bounds?
• The number of connected components in the
  conflict graph is a lower bound (BB04).
• Fast algorithms for computing ARGs with minimum
  recombination.
• Poly. Time to get ARG with 0 recombination
• Poly. Time to get ARGs that are galled trees
  (Gusfield03)

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Underperforming lower bounds

• Sometimes, Rs can be quite weak
• An RI lower bound that uses intermediates can
  help (BB05)

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LPL data set

• 71 individuals, 9.7Kbp genomic sequence
• Rm22, Rh70

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