L5: Estimating Recombination Rates

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L5 Estimating Recombination Rates
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Review

• mM min. number of recombination events in any
  explanation of the haplotypes in M
• Last time, we covered 3 lower bounds on mM
• The only exact algorithm that is known is super
  exponential. Not even an exponential time
  algorithm is known.
• Can we get efficient upper bounds that are tight.
• Idea An Rs like method can be used to get an
  upper bound.

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Upper bounds

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

• Upper Bound
• Procedure Compute_U(M)
• if ? non-informative column
• return (Compute_U(M-s))
• else if ? redundant row
• return (Compute_U(M-h))
• else
• return(minh(f(h,M-h)Compute_U(M-h))

Number of recombinations needed to explain h
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Many approaches to estimating ?
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1. Counting methods

• Rm
• Rh
• Rs
• ARG with min number of recombinations
• These numbers correlate with ? but how do we get
  a value for ? given this number
• These numbers still have value in defining
  hot-spots of recombination (showing variance in
  local recombination rates)
• They generally underestimate the true number of
  recombinations

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2. Model based approaches

• Full likelihood approaches
• Approximate likelihood approaches

Fearnhead, Donnelly
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Approximate Likelihood approaches

• Two locus sampling
• 4 gamete violation implies recombination.
• Generalization
• Define vector n n00, n01 , n10, n11 for a
  pair of loci
• The distribution of n depends upon ?, ?
• Can we compute Pr(n ?, ?)? Then, we can iterate
  to get the Max likelihood estimator for ?.

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Two locus method

• Generate MANY random ARGs with n n00 n01 n10
  n11 leaves.
• For each ARG, generate the two trees
  corresponding to the two loci
• Drop 2 mutations at random, to get a value for n
• How can you make this more efficient?
• Given an ARG (topology), we know the edge pairs
  that would generate desired n.

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Two locus estimation
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Multi locus estimator

• For a site with multiple loci, assume each pair
  to be independent, each generating a vector ni
• Assume recombination rate (per bp) to be constant
  in the region

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Performance of the 2 locus estimator

• The composite likelihood estimator performs
  well in practice.
• Note that the values of ? can be pre-computed
  making this a fast method.
• Note that this plot does not describe the variance

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Performancs 90/10 percentile
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Research 2 locus versus other statistics

• Q1 Can we use some of the counting based methods
  as summary statistic?
• It is better than composite likelihood in that
• It does not assume independence between loci.
• There is a direct linear relationship (expected
  number of recombination events is ? log n)
• Variation might be better.
• Can we compute Pr(Rh ?, ?) efficiently? In a
  sense, it does not matter, because we can
  pre-compute the numbers.
• Incorporate distance constraints in computing
  these summary statistics. It is reasonable to
  assume that the rate is constant per bp within a
  window.

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Research Problem

• Recombination hot-spots are NOT correlated
  between humans and Chimps.
• 99 sequence identity
• Virtually no overlap between hot-spots (generated
  using pop. Genetics).
• What can cause this?
• Method
• Europeans/Africans share hot-spots
• Concordance with sperm typing
• Population sub-structure? Not (as shown by
  structure)
• Genomic factors

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Genomic factors

• Recombination is elevated in GC rich regions
• Epigenetic factors (such as acetylation,
  methylation) that affect chromatin structure
  might be key.
• Yeast is a useful model for studying
  recombination
• In yeast, recombination hotspots can be
  eliminated by insertion of transposable elements!
• Can differential insertion of Alus explain the
  differences between chimps/humans?

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Haplotype Phasing
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Genotypes and Haplotypes

• Each individual has two copies of each
  chromosome.
• At each site, each chromosome has one of two
  alleles
• Current Genotyping technology doesnt give phase

0 1 1 1 0 0 1 1 0 1 1 0 1 0 0 1 0
0
2 1 2 1 0 0 1 2 0
Genotype for the individual
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• Why is haplotype phasing important ?

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Haplotype Phasing

• Haplotype Phasing is the resolution of a genotype
  into the two haplotypes.
• Haplotypes increase the power of an association
  between marker loci and phenotypic traits
• Current approaches to Haplotyping
• Via technological innovations (expensive)
• Statistical Methods (ML, Phase,PL)
• This lecture, we will consider a combinatorial
  approach to the phasing problem
• Efficient, provable quality of solution
• Not completely generalizable (as yet)

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The Perfect Phylogeny Model

• We assume that the evolution of extant haplotypes
  can be displayed on a rooted, directed tree, with
  the all-0 haplotype at the root, where each site
  changes from 0 to 1 on exactly one edge, and each
  extant haplotype is created by accumulating the
  changes on a path from the root to a leaf, where
  that haplotype is displayed.
• In other words, the extant haplotypes evolved
  along a perfect phylogeny with all-0 root.

12345
00000
1
4
3
00010
2
10100
5
10000
01010
01011
Extant Haplotypes
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Haplotyping via Perfect Phylogeny
PPH Given a set of genotypes, find an explaining
set of haplotypes that fits a perfect phylogeny
00
1 2
a 1 0
a 0 1
b 0 0
b 0 1
c 1 0
c 1 0
1 2
a 2 2
b 0 2
c 1 0
1
2
b
00
a
a
b
c
c
01
01

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10
10
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The Alternative Explanation
1 2
a 1 1
a 0 0
b 0 0
b 0 1
c 1 0
c 1 0
No tree possible for this explanation
1 2
a 2 2
b 0 2
c 1 0
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The 4 Gamete Test for Perfect Phylogeny

• Arrange the haplotypes in a matrix, two
  haplotypes for each individual.
• Then (with no duplicate columns), the haplotypes
  fit a unique perfect phylogeny if and only if no
  two columns contain all four pairs (Buneman)
• 0,0 and 0,1 and 1,0 and 1,1

00
10
01
11
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The Alternative Explanation
1 2
a 1 1
a 0 0
b 0 0
b 0 1
c 1 0
c 1 0
No tree possible for this explanation
1 2
a 2 2
b 0 2
c 1 0
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The Tree Explanation Again
0 0
1 2
a 1 0
a 0 1
b 0 0
b 0 1
c 1 0
c 1 0
1 2
a 2 2
b 0 2
c 1 0
1
2
b
0 0
a
b
a
c
c
0 1
0 1
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The Combinatorial Problem

• Input A ternary matrix (0,1,2) M with N rows
• Output A binary matrix M created from M by
  replacing each 2 in M with a 0 and 1, such that
  M passes the 4 gamete test
• Gusfield (Recomb2002) proposed a solution which
  used a reduction to Matroids.
• We present a (slightly inefficient) solution
  using elementary techniques
• Independently by (Eskin, Halperin, Karp02)

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Initial Observations

• Forced Expansions
• EX 1 If two columns(sites) of M contain the
  following rows
• 2 0
• 0 2
• Then M will contain a row with 1 0 and a row
  with 0 1 in those columns.
• EX 2 Similarly, if two columns of M contain the
  rows
• 2 1
• 2 0
• Then M will contain rows with 1 1 and 0 0 in
  those columns

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Initial Observations
If a forced expansion of two columns creates rows
0 1, and 1 0 in those columns, then any 2 2 in
those columns must be set to be 0 1 1 0
We say that two columns are forced out-of-phase.
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If a forced expansion of two columns creates 1 1,
and 0 0 in those columns, then any 2 2 in those
columns must be set to be 1 1 0 0 We
say that two columns are forced in-phase.
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Immediate Failure
It can happen that the forced expansion of
cells creates a 4x2 submatrix that fails the
4-Gamete Test. In that case, there is no PPH
solution for M.
20 12 02
Example
Will fail the 4-Gamete Test
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An O(ns2)-time Algorithm

• Find all the forced phase relationships by
  considering columns in pairs.
• Find all the inferred, invariant, phase
  relationships.
• Find a set of column pairs whose phase
  relationship can be arbitrarily set, so that all
  the remaining phase relationships can be
  inferred.
• Result An implicit representation of all
  solutions to the PPH problem.

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A Running Example
1 2 3 4 5 6 7
1 2 2 2 0 0 0
2 0 2 0 0 0 2
1 2 2 2 0 2 0
1 2 2 0 2 0 0
2 2 0 0 0 2 0
0 0 0 0 0 0 0
A B C D E F
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Companion Graph G_c
1
1 2 3 4 5 6 7
1 2 2 2 0 0 0
2 0 2 0 0 0 2
1 2 2 2 0 2 0
1 2 2 0 2 0 0
2 2 0 0 0 2 0
0 0 0 0 0 0 0
A B C D E F

• Each node represents a column in M, and each
  edge indicates that the pair of columns has a row
  with 2s in both columns.
• The algorithm builds this graph, and then checks
  whether any pair of nodes is forced in or out of
  phase.

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Phasing Edges in G_c
1

• Each Red edge indicates that the columns are
  forced in-phase.
• Each Blue edge indicates that the columns are
  forced out-of-phase.

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4
2
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Let G_f be the sub-graph of G_c defined by the
red and blue edges.
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Connected Components in G_f
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• Graph G_f has three connected components

.
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Phase-parity Lemma

• Lemma 1 There is a solution to the PPH problem
  for M if and only if there is a coloring of the
  black edges of G_c with the following property
• For any triangle in G_c containing at least
  one black edge, the coloring makes either 0 or 2
  of the edges
• blue (i.e., out of phase)

Thats nice, but how do we assign the colors?
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A Weak Triangulation Rule
1

• Theorem 1 If there are any black edges whose
  ends are in the same connected component of G_f,
  at least one edge is in a triangle where the
  other edges are not black
• In every PPH solution, it must be colored so
  that the triangle has an even number of Blue (out
  of Phase) edges.
• This an inferred coloring.

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7
6
3
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2
5
Graph G_f
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7
6
3
4
2
5
Graph G_f
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7
6
3
4
2
5
Graph G_f
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Corollary

• Inside any connected component of G_f, ALL the
  phase relationships on edges (columns of M) are
  uniquely determined, either as forced
  relationships based on pair-wise column
  comparisons, or by triangle-based inferred
  colorings.
• Hence, the phase relationships of all the columns
  in a connected component of G_f are INVARIANT
  over all the solutions to the PPH problem.
• The black edges in G_f can be ordered so that the
  inferred colorings can be done in linear time.
  Modification of DFS.

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Phase Parity Lemma Proof
2 X
Y 2
2 2
If X ? 2, and Y ? 2, Then the two columns are
forced
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Phase Parity Lemma proof
A B C

• Lemma If a triangle contains a black edge, then
  a PPH solution exists only if there are 0 or 2
  blue edges in the final coloring.
• Proof
• No black edge unless x2, or y2 or z2
  (previous lemma)
• If there is a row with all 2s, then there must be
  an even number of blue edges

2 2 y
x 2 2
2 z 2
B
A
C
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Proof of Weak Triangulation Theorem
A

• Arbitrary chordless cycles are possible in the
  graph, with forced edges.
• See example. The pattern 0,2 2,0 and 2,2
  implies a blue (out of phase) edge
• A single unforced edge changes the picture

E
B
D
C
A B C D E
2 2 0 0 0
0 2 2 0 0
0 0 2 2 0
0 0 0 2 2
2 0 0 0 2
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Proof of Weak Triangulation Theorem
K
K

• Let (J,J) be a black edge connecting a long
  path J,K,K,J of forced edges
• In the Matrix, x ? 2, otherwise there is a chord.
  Likewise y?2
• By previous lemma, (J,J) is forced

J
J
K J J K
2 2 x
y 2 2
2 2
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Finishing the Solution

• Problem A connected component C of G may
  contain several connected components of G_f, so
  any edge crossing two components of G_f will
  still be black. How should they be colored?

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1

• How should we color the remaining black edges in
  a connected component C of G_c?

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Answer

• For a connected component C of G with k
  connected components of Gf, select any subset S
  of k-1 black edges in C, so that S together with
  the red and blue edges span all the nodes of C.
• Arbitrarily, color each edge in S either red or
  blue.
• Infer the color of any remaining black edges by
• successive use of the triangle rule.

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2
5
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Theorem 2

• Any selected S works (allows the triangle rule to
  work) and any coloring of the edges in S
  determines the colors of any remaining black
  edges.
• Different colorings of S determine different
  colorings of the remaining black edges.
• Each different coloring of S determines a
  different solution to the PPH problem.
• All PPH solutions can be obtained in this way,
  i.e. using just one selected S set, but coloring
  it in all 2(k-1) ways.

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Corollary

• In a single connected component C of G with k
  connected components in Gf, there are exactly
  2(k-1) different solutions to the PPH problem in
  the columns of M represented by C.
• If G_c has r connected components and t connected
  components of G_f, then there are exactly 2(t-r)
  solutions to the PPH problem.
• There is one unique PPH solution if and only if
  each connected component in G is a connected
  component in G_f.

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Algorithm

• Build Graph G and find its connected components.
  Solve each connected component C of G separately.
• Find the forced (red or blue) edges. Let Gf be
  the subgraph of C containing colored edges.
• Find each connected component of Gf and make the
  inferred edge colorings (phase decisions).
• Find a spanning tree of uncolored edges in C, and
  color those edges arbitrarily, and follow the
  inferred edge colorings

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Conclusion

• In the special case of blocks with no
  recombination, and no recurrent mutations, the
  haplotypes satisfy a perfect phylogeny
• Given a set of genotypes, there is an efficient
  (O(ns2)) algorithm for representing all possible
  haplotype solutions that satisfy a prefect
  phylogeny
• Efficiency
• Input is size O(ns),
• All operations except building the graph are
  O(nss2)
• Valid PPH only if s O(n). Is O(ns) possible?
• Current best solution is O(nsn(1-e) s2) using
  Matrix Multiplication idea
• Future work involves combining this with some
  heuristics to deal with general cases (lo
  recombination/hi recombination)

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Simulated Data

• Coalescent model (Hudson)
• No Recombination
• 400 chromosomes, 100 sites
• Infinite sites
• Recombination
• 100 chromosomes
• Infinite sites
• R4.0 2501
• Pr(Recombination) 410(-9) between adjacent
  bases

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Error Measurement

• Discrepancy 1 (Num Haplotypes incorrectly
  predicted)
• Switch Error 2

01010 00101 01010 10101
02222 22222
00101 01010 00000 11111
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No Recombination
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No Recombination
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Choosing between solutions
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Choosing between solutions
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Choosing between solutions
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Conclusion

• Extremely low error rates (lt 1 discrepancy) if
  no recombination
• Randomly choosing between equivalent solutions is
  sufficient
• Other measures (Parsimony, Likelihood, Entropy)
  do not improve the quality of solution

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With Recombination