Haplotype phasing

1
Haplotype phasing

• Lecture 25 December 1, 2005
• Algorithms in Biosequence Analysis
• Nathan Edwards - Fall, 2005

2
Haplotypes

• 1ACGACTCAGATCACTACGTACGACT
• 1ACGACTCAGATAACTACGGACGACT
• 2ACGACTCAGATCACTACGTACGACT
• 2ACGACTCAGATCACTACGTACGACT
• 3ACGAGTCAGATCACTACGTACGACT
• 3ACGAGTCAGATAACTACGGACGACT

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Haplotypes

• 1ACGACTCAGATCACTACGTACGACT
• 1ACGACTCAGATAACTACGGACGACT
• 2ACGACTCAGATCACTACGTACGACT
• 2ACGACTCAGATCACTACGTACGACT
• 3ACGAGTCAGATCACTACGTACGACT
• 3ACGAGTCAGATAACTACGGACGACT

4
Genotypes

• ACGACCTCAGATCAACTACGTGACGACT
• ACGACCTCAGATCCACTACGTTACGACT
• ACGAGGTCAGATCAACTACGTGACGACT

5
Genotyping Technology

• Can only tell us about a particular locus, so we
  lose information about an individuals haplotypes
• 1 C, A,C, G,T
• 2 C, C, T
• 3 G, A,C, G,T

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Genotyping Technology

• Can only tell us about a particular locus, so we
  lose information about an individuals haplotypes
• 1 0, 2, 2
• 2 0, 1, 1
• 3 1, 2, 2

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Haplotype phasing

• Given n genotypes, resolve each genotype into its
  2 haplotypes.
• 1 0, 2, 2
• 2 0, 1, 1
• 3 1, 2, 2

• 1 0, 0, 0
• 1 0, 1, 1
• 2 0, 1, 1
• 2 0, 1, 1
• 3 1, 0, 0
• 3 1, 1, 1

Phasing
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Haplotype phasing

• Given n genotypes, resolve each genotype into its
  2 haplotypes.
• 1 0, 2, 2
• 2 0, 1, 1
• 3 1, 2, 2

• 1 0, 0, 1
• 1 0, 1, 0
• 2 0, 1, 1
• 2 0, 1, 1
• 3 1, 0, 1
• 3 1, 1, 0

Phasing
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Clarks Rule

• Find unambiguous individuals
• (at most 1 ambiguous locus)
• Form initial list of known haplotypes
• Resolve ambiguous individuals
• If possible, use two known haplotypes
• Otherwise, use one known haplotype and add new
  haplotype to list.
• If unphased individuals remain
• Assign phase randomly for one individual

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Clarks Rule

• Initial list of known haplotypes 0, 1, 1
• 1 0, 2, 2
• 2 0, 1, 1
• 3 1, 2, 2

• 1 0, 0, 0
• 1 0, 1, 1
• 2 0, 1, 1
• 2 0, 1, 1
• 3 1, 0, 0
• 3 1, 1, 1

Phasing
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Clarks Rule

• Heuristic Use unambiguous genotypes
• Doesnt necessarily minimize the number of
  haplotypes
• Doesnt pay any attention (as such) to the
  haplotype frequency

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Haplotype frequencies

• HWE says that the probability of phase h,h
  for genotype g
• P g h,h 2 P h P h if h ? h
• P h 2 if h h
• If we knew the haplotype frequencies, fh, we
  would assign the phase of g based on these
  frequencies. P h ph fh / 2n

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Estimating haplotype frequencies

• Wed like to extract the haplotype frequencies
  from our genotype data
• Well use an EM algorithm to find
  maximum-likelihood estimates of the haplotype
  frequencies.
• Find haplotype frequencies ph so thatP
  g1,....,gn ph is maximum.

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Estimating haplotyping frequencies

• If we had haplotype frequencies, but no genotype
  frequencies
• Pg Sh,h g Ph,h
• Compute Ph,h from HW (and ph) as before
• So P genotype g with phase h,h P phase
  h,h g P genotype g Ph,h/Pg x
  1/n
• Use previous estimate of ph to compute first term

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Estimating haplotype frequencies

• Given P genotype g with phase h,h , based on
  our estimated ph, we now re-estimate fh.
• Efh Sg Sh,h g d(h,h,h)
  Pgh,hwhere d(h,h,h) is 0,1,2, times
  h in h,h
• ph Efh/2n

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Estimating haplotype frequencies

• This EM algorithm (Excoffier and Slatkin, 1995)
  will converge to the maximum likelihood estimates
  of the haplotype frequencies
• Initial guess for frequencies is not clear
• Pgh,h 1/ of phasings of g

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Estimating haplotype frequencies

• Unfortunately, this approach is exponential in
  the number of ambiguous sites (or number of
  phasings)
• Many heuristics have been proposed, lots of open
  problems to consider.
• Unambiguous or partially phased genotypes
  contribute most.

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Alternative method

• For a large enough sample, if we assign a new
  genotype to its most likely phase, the haplotype
  frequencies wont change (much).
• Choose g at random, re-assign phase based on
  haplotype frequencies of other genotypes phases.
• Iterate until stable.
• Monte Carlo Markov Chain (MCMC) approach
• Estimates maximum-likelihood haplotype
  frequencies too!
• Variations allow extra population assumptions to
  be built in.

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Combinatorial Approach

• There is good reason to use these statistical
  approaches, especially if n is large, and the
  haplotypes are not too rare.
• However, with smaller samples, or when we have
  more structure, or rarer haplotypes, we might
  consider each phasing carefully.
• These approaches make the most sense when there
  is additional structure to consider.

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Perfect Phylogeny

• Perfect phylogeny
• all-0 haplotype at the root of tree
• remaining haplotypes at leaves of tree
• minor alleles accumulated on path from root to
  leaf
• each locus is represented by exactly one edge.
  (infinite sites model)

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Perfect phylogeny

• all-0 at root
• otherwise at leaves
• minor-alleles accumulated on path from root
• each locus on exactly one edge
• Implies no evidence of recombination!

00000
1
4
3
2
10100
00010
10000
5
01011
01010
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Haplotyping by Perfect Phylogeny

• PPH Given a set of genotypes, find phasing
  haplotypes that fit a perfect phylogeny.
• 1 2, 2
• 2 0, 2
• 3 1, 0

1 1, 0 1 0, 1 2 0, 0 2 0, 1 3 1, 0 3 1, 0
Phasing
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Perfect phylogeny

• 1 1, 0
• 1 0, 1
• 2 0, 0
• 2 0, 1
• 3 1, 0
• 3 1, 0
• This phasing fits a perfect phylogeny!

00
1
2
00
10
01
01
10
10
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Haplotyping by Perfect Phylogeny

• PPH Given a set of genotypes, find phasing
  haplotypes that fit a perfect phylogeny.
• 1 2, 2
• 2 0, 2
• 3 1, 0
• No perfect phylogeny is possible!

1 0, 0 1 1, 1 2 0, 0 2 0, 1 3 1, 0 3 1, 0
Phasing
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4 Gamete Test

• Format the haplotypes in a matrix, two rows for
  each individual.
• The haplotypes fit a perfect phylogeny if and
  only if no two columns contain all four pairs
  00, 01, 10, 11.

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Combinatorial Problem

• Input matrix G (n rows) with elements from
  0,1,2
• Output matrix H (2n rows) with elements 0,1,
  each 2 of G replaced by 0 and 1, such that H
  passes 4 gamete test.
• Independently, from 2002
• Gusfield
• Eskin, Halperin, Karp
• Bafna, Gusfield, Lancia, Yooseph

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Initial Observations

• Forced Expansions
• G has two columns (loci) with these rows
• 2 0
• 0 2
• H must have
• 0 0
• 1 0
• 0 0
• 0 1
• That is, we know that 0 1 and 1 0 are present in
  the column

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Initial Observations

• Forced Expansions
• G has two columns (loci) with these rows
• 2 1
• 0 2
• H must have
• 0 1
• 1 1
• 0 0
• 0 1
• That is, we know that 0 0 and 1 1 are present in
  the column

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Initial Observations

• 2 2 can be phased in two ways
• 0 0, 1 1 and 0 1, 1 0.
• If two columns contain 0 0 and 1 1 already
  (forced or unambiguous), then 2 2 must be phased
  0 0, 1 1.
• These columns are forced in-phase
• If two columns contain 0 1 and 1 0 already,
  then 2 2 must be phased 0 1, 1 0
• These columns are forced out-of-phase

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Immediate Failure

• Forced Expansions
• G has two columns (loci) with these rows
• 2 1
• 2 0
• H must have
• 0 1
• 1 1
• 0 0
• 1 0
• 4 Gamete condition violated!

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Combinatorial Algorithm

• Consider all columns in turn.
• - Do all forced expansions
• - Find all forced in-phase or out-of- phase
  column pair relationships
• - Use invariant column phase pairs
• to fix unknown phase pairs
• - Find a set of column pairs whose
• phase can be arbitrarily set
• - Force the remaining phase pairs

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Running Example
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Companion Graph

• Node column of G
• Edge 2 cols with a row of 2s.

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Companion Graph

• Red Edge
• forced in-phase
• Blue Edge
• forced out-of-phase

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Forced Graph

• Drop dashed edges and define connected components
  
• 3 connected components
• Fill in each component in turn by coloring dashed
  edges

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Phase parity lemma

• There is a PP solution for G if and only if the
  dashed edges can be colored so that
• For every triangle in the companion graph with
  at least one dashed edge, either 0 or 2 edges
  are colored blue.

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Weak Triangulation Rule

• If any dashed edges have ends are in the same
  component of the forced graph, at least one must
  be in a triangle with two colored edges.
• Color this edge appropriately

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Weak Triangulation Rule
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Corollary

• For each connected component of the forced graph,
  all the edge relationships are uniquely
  determined.
• Phase relationships of all columns in a connected
  component of the forced graph are invariant for
  all solutions that satisfy PP

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Phase Parity Lemma

• Consider
• 2 x
• y 2
• 2 2

• If x ? 2 and y ? 2
• 2 0
• 1 2
• 2 2
• then the columns are forced in, or out-of phase.

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Phase parity lemma

• Lemma If a triangle contains a dashed edge, then
  a PP solution exists only if there are 0 or 2
  blue edges in coloring.
• A B C
• 2 2 y
• x 2 2
• 2 z 2

• Must have x2, y2 or z2 or no dashed edge.
• Row of all 2s must have even of blue edges.

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Weak Triangulation Rule

• Theorem
• If any dashed edges have ends are in the same
  component of the forced graph, at least one must
  be in a triangle with two colored edges.

• In forced graph component, can find path from any
  node to any other, using forced edges.
• Find cycle of one unforced (dashed) edge and
  forced edges of length gt 3

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Weak Triangulation Rule

• Let (J,J) be dashed edge connecting path of gt 2
  forced edges J, K, ..., K, J
• ...K J J K...
• 2 2 x
• y 2 2
• 2 2

• If x 2, then (K,J) is an edge
• If y 2, then (J,K) is an edge
• If x ? 2 and y ? 2, then (J,J) is forced.

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Finishing up

• We still need to color the dashed edges between
  components of the forced graph.
• If there are k components, select (k-1) edges
  that form a spanning tree on the component graph
• Color these edges arbitrarily. Remaining dashed
  edges will be forced by weak triangulation rule.

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Corollary

• If companion graph has r connected components,
  and forced graph has t connected components,
  there are exactly 2(t-r) perfect phylogeny
  phasings for input matrix G.
• PP solution is unique iff each component of
  companion graph is connected in forced graph.

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Summary

• If no (evidence) of recombination with the
  infinite sites model, haplotypes will fit a
  perfect phylogeny
• O(ns2) to test for PP, determine if solution is
  unique, represent all possible solutions, and
  generate one solution
• Others are pushing towards O(ns)
• Can we relax the no recombination requirement?