haplotyping algorithms and structure of human variation eecs 458 cwru fall 2004 readings: see papers...
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Haplotyping algorithms and structure of human variation
EECS 458 CWRU
Fall 2004
Readings: see papers on the course website
Roadmap
• Definition: haplotype and haplotype inference• Why infer haplotypes• Infer haplotypes from pedigree data
– Most probable haplotype configurations– Haplotype configurations with minimum recombinations
• Infer haplotypes from population data– Combinatorial: Clark’s, Perfect Phylogeny– Statistical methods: EM, Bayesian (MCMC)
• Infer haplotypes from pooled samples• Haplotype block partition• Tag SNP selection
Genotype and Haplotype
{1 2}
{1 2}
.
.
.AATGCCGCAA...GTC...
.
.
.AGTGCCGCAA...TAC...
Paternal Maternal
Typical Genotype Data
• Two alleles for each individual– Chromosome origin for
each allele is unknown
• Multiple haplotype pairs can fit observed genotype
• Molecular haplotyping is expensive
Observation:
A C
G A
T C
Marker1
Marker2
Marker3
Possible haplotypes:
A C
G A
T C
A C
G A
C T
A C
A G
T C
A C
A G
C T
Haplotypes are important!
• Phase may determine phenotype
• Phase helps exploit linkage disequilibrium
Infer state of neighboring alleles
• Phase clarifies identity-by-descent status
Common Uses of Haplotypes
• Linkage disequilibrium studies– Summarize genetic variation
• Selecting markers to genotype– Identify haplotype tag SNPs
• Candidate gene association studies– Test haplotype associations– Help interpret single marker associations
• Understanding evolution of human populations
The problem…
• Haplotypes are hard to measure directly– X-chromosome in males– Sperm typing– Other molecular techniques
• Often, statistical or combinatorial methods for reconstruction required
Haplotype Inference on population data
m=6, m’=4
{1 2}{1 1}{1 2}{1 2}{1 2}{2 2}
m
1|21|11|21|21|22|2
2|11|11|21|21|22|2
1|21|12|11|21|22|2
2|11|12|11|21|22|2
2|11|12|12|12|12|2
……
2m’
Information on Relatives
• Number of ambiguous individuals increases rapidly with number of markers
• Family information can help, but many ambiguities remain
Haplotype Inference on Pedigrees, Mendelian Law
{1 2} {1 1}
{1 1} {1 2} {2 2}
2|1 1|1
1|1 2|1 2 2
{1 2} {1 2}
{1 2} {1 2} {1 2}
{1 2} {1 *}
{1 1} {1 2} {1 2}
Haplotype inference on pooled samples
• The input contain n pools• Each pool contains k
individuals, thus 2k haplotypes and m markers
• At each marker, we are given the number of alleles for the k individuals for each pool
• The goal is to find the haplotype frequencies
• Example: n=3, k=2, m=5
2 4 3 2 2
0 2 3 1 2
1 2 2 2 3
Haptotyping pedigree data: statistical formulation
• Statistical formulation: find the most probable haplotype configuration
• Need to calculate the probability of a pedigree on every haplotype configuration
• Recall for linkage analysis, we need to calculate the probability of a pedigree, that sums over all possible haplotype configs
Haptotyping pedigree data : statistical formulation
• Thus the linkage programs like Genehunter, Allegro, Merlin could compute the most probable haplotypes
• But, it is time consuming….• In addition to exact computation, there are some
approximation algorithms, mainly based on important sampling, e.g. SimWalk.
• Still very time consuming, may consider many configurations with very small probabilities
Recombination and combinatorial formulation
{1 2}{1 2}
{1 1}{1 2}
{1 2}{1 2}
{1 2}{1 2}
1|2{1 2}
1|1{1 2}
{1 2}{1 2}
1|2{1 2}
1|2 1|2
1|11|2
1|21|2
1|2 1|2
1|2 1|2
1|11|2
1|22|1
1|2 1|2
MRHC Problem
Find a minimum recombinant haplotype configuration from a given pedigree with genotype data.
Assumptions:• Mendelian law (no mutations);• Recombination events are rare.
Well supported from real data.
{1 2}{1 2}{1 2} …
{1 2}{1 2}{2 2} …
{1 1}{1 2}{2 2} ...
{1 2}{1 2}{1 2} ...
{1 2}{2 2}{2 2} …
{1 1}{1 2}{2 2} …
{1 2}{1 2}{1 2} ...
{1 1}{1 2}{2 2} ...
Input
MRHC Problem (cont’d)• PS: parental source of the two
alleles at the locus (i.e. phase)
• GS: grandparental source of an allele
• Haplotype configuration = assignment of PS and GS values.
Output
PS=0
PS=1
GS2=1
1|21|22|1…
1|22|12|2 …
1|11|22|2 ...
1|21|22|1 ...
1|22|22|2 …
1|12|12|2…
1|21|22|1 ...
1|11|22|2 ...
A
B
GS2=1GS2=0
Previous Results• Genotype elimination (O’Connell’00).
– For data requiring no recombinant, exhaustive elimination.
• Genetic algorithm (Tapadar et al.’00).– Time consuming.
• MRH (Qian & Beckmann’02).– Six step rule-based algorithm.– Locus by locus at every step, extremely slow for biallelic (e.g.
SNP) markers.
Thm. MRHC is NP-Hard.
Idea: Reduction from a variant of set cover.
First complexity result.
Remains hard for two loci.
Remains hard when no loops.
Li & Jiang’03, Doi, Li & Jiang’03
Block-Extension Algorithm
Iterative, heuristic, five steps. Rules are derived from Mendelian law, MR principle, block concept and some greedy ideas based on the following observations:
• Block structures are common in haplotypes.• Double recombination events are rare.• Common haplotype blocks shared in siblings.• …
Advantages/Disadvantages Time complexity (BE: O(dmn) / MRH: O(2dm3n2))
Li & Jiang’03
Block-Extension Algorithm
1 3
4 2
2 4
3 2
1 2
3 2
2 1
3 4
3 3
3 2
2 4
3 41 *
4 2
2 4
3 2
* *
* *
* *
* *
1 2
3 4 5
6
1 1
1 2
2 3
3 4
1 3
4 2
2 4
3 2
1 2
3 2
2 1
3 4
2 3
* *
* *
* *
1 2
3 4 5
6
1 1
1 2
2 3
3 4
3 3
3 2
2 4
3 41 3
4 2
2 4
3 2
1 3
4 2
2 4
3 2
1 2
3 2
2 1
3 4
2 3
3 4
1 4
2 *
1 2
3 4 5
6
1 1
1 2
2 3
3 4
3 3
3 2
2 4
3 41 3
4 2
2 4
3 2
Block-Extension Algorithm
1|1
1|2
2 3
3 4
1|2(-1,0)
2|3(1,-1)
2|1(-1,-1)
3 4
1|3(-1,1)
2|4(1,-1)
2|4(-1,-1)
3|2(-1,-1)
1|3
3 2
2 4
3 41 3
4|2(1,-1)
2 4
2|3(1,-1)
2|3
3 4
1 4
2 *
1 2
3 4 5
6
1|1
1|2
2 3
3 4
1|2(-1,0)
2|3(1,-1)
2|1(-1,-1)
3 4
1|3(-1,1)
2|4(1,-1)
2|4(1,-1)
3|2(-1,-1)
1|3
3 2
2 4
3 43|1(1,0)
4|2(1,-1)
4|2(1,-1)
2|3(1,-1)
2|3
3 4
1 4
2 *
1 2
3 4 5
6
Dynamic Programming Algorithms• Locus-based dynamic programming algorithm
– Linear time in the number of the members– Applicable to only tree pedigrees
• Member-based dynamic programming algorithm– Linear time in the number of the loci– Applicable to general pedigrees with small sizes
Doi, Li & Jiang’03
root
21
5
43
6
7 8
Locus-Based Dynamic Programming
21
5
43
6
7
8
Constraint-Finding Algorithm
• Assumptions:– No missing alleles, no errors. – Zero recombinants.
• Idea: finding all feasible (i.e. 0-recombinant) haplotype configurations is equivalent to reducing the degree of freedom in PS/GS assignment.
Li & Jiang’03
Four Levels of Constraints
PS=0
GS2=1
1|21|22|1…
1|22|12|2 …
1|11|22|2 ...
1|21|22|1 ...
1|22|22|2 …
1|12|12|2…
1|21|22|1 ...
1|11|22|2 ...
A
B
Based on 0-recombinant (for a pair of loci): Level 3: Haplotype constraint Level 4: Grouping constraint
Based on Mendelian law (on single locus) :
Level 1: GS constraint Level 2: PS constraint
Level 3 and Level 4 Constraints
{1 2}
{1 2}
{1 2}
{1 2}
{1 2}
{1 2}
{1 2}
{1 2}
{1 1}
{1 1}
{1 2}
{1 2}
1 2
3 4 5
6 1 2
2 1
1 2
1 2
1 2
1 2
2 1
2 1
4 5
6
1 2
2 1
1 2
2 1
4 5
6
{1 2}
{1 2}
{1 2}
{1 2}
4 5
6{1 2}
{1 2}
Level 3 and Level 4 ConstraintsLevel 3 and Level 4 Constraints
The variables represent PS values and the equations are over Z2
Analysis of Constraint-Finding Algorithm
Thm. Every solution consistent with the constraint equations is a feasible solution and vice versa.
• Steps: – find all constraints, in the form of linear equations over Z2
– solve the equations by Gaussian elimination – enumerate all feasible haplotype configurations
• Exact polynomial time (O(n3m3); genotype elimination: exponential)
Integer Linear Programming
• Combines missing data imputation and haplotype inference.• Regardless of the pedigree structure, number of recombinants,
number of variables are linear of problem size. • Implicitly checks the Mendelian consistency for pedigree
genotype data with missing alleles, which is also an NPC problem.
• Could find all possible optimal solutions.• Solved by a branch-and-bound algorithm.• Effective for practical size problems in terms of time efficiency.• Accurate in terms of missing alleles imputation and haplotype
inference.
Li & Jiang’04a
ILP for MRHC with Missing Data
1. Define variables . 2. Define linear constraints.3. Define a linear objective function of the variables. 4. Preprocess constraints.5. Apply branch-and-bound strategy to find solutions. (a
partial order relationship and some other special relationships).
6. Estimate bounds.7. Apply a maximum likelihood approach to multiple
optimal solutions.
FormulationMj:={mk} set of all possible alleles at marker locusj and let tj = |Mj|. M1 = {1, 2} , M2 = {1,2}
{1 2}{1 2}
{1 1}{1 2}
{1 2}{1 0}
{1 0}{1 2}
1 2
3 4
Individual 4:
11,4f 1
2,4f 11,4m 1
2,4m
21,4f 2
2,4f 21,4m 2
2,4m
1 12,4
11,4 ff
…
Define tj f vars for each paternal allele and tj m vars
for each maternal allele at locus j of individual i: )1( , ,, j
jki
jki tkmf
kjki mf is allele paternal iff 1 ,
Formulation: Variables Define 2 g vars for each paternal allele and
maternal allele at locus j for individual iji
ji gg 2,1, ,
Var g1 = 0 (or 1) iff paternal allele is copied from father’s paternal (or maternal) allele. Var g2 defined similarly.
Define r vars:
iff 1
)11( , 1
1,1,1,
2,1,
j
ij
ij
i
ji
ji
ggr
mjrr
Formulation: Objective Function
) (Founders-Non
1
12,1,
m
j
ji
ji rr
Subject to Genotype constraints:
} 1{},{
} 1{},{
} 1, 1, 1{}0,{
} 1, 1{}0,0{
,,,,,,,,
,,
1,
1,,,
1,
1,
jsi
jsi
jri
jri
jsi
jri
jsi
jri
js
jr
jri
jri
jr
jr
t
k
jki
t
k
jki
jri
jri
jr
t
k
jki
t
k
jki
mfmfmmffmm
mfmm
mfmfm
mf
jj
jj
Objective function:
Formulation: Constraints
Mendelian law of inheritance constraints (a child i and its father f ):
1
0
1,,,
1,,,
j
ij
kfjki
ji
jkf
jki
gmf
gff
Constraints for r vars:
0
0
2
0
1,,,
1,,,
1,,,
1,,,
jli
jli
jli
jli
jli
jli
jli
jli
jli
jli
jli
jli
ggr
ggr
ggr
ggr
A Partial Order Relationship
Denote:
0 1
1
y
yy
Inequalities with 2 variables:
ji yy
1
2
3
45
8
69
11
107
1
2
3’
8
9
11
10
Forced Variables
• Rule 1:
• Rule 2:
• Rule 3:
ncyInconsiste, 10 Syy
1)()(
0)()(1
1
jjiji
ijiji
yyyyy
yyyyy
01 iii yyy
Lower and Upper Bounds
• Lower bounds– Linear relaxation.– Summation of the number of recombinants in each
nuclear family.– Effective for data with large number of recombinants.
• Upper bound– Obtained by block-extension algorithm.– Effective for data with small number of recombinants.
Statistical Assessment
• E-M algorithm to estimate haplotype frequencies for data that consist of multiple pedigrees.
N
hhf i
i 2
)()(ˆ
11
i
imifii
ii hhhPhfhffHGPfounder -non
)()(founder
21 )|()(ˆ)(ˆ)ˆ|,(
PedPhase software• Simulated data were generated to compare our
algorithms, as well as MRH in terms of efficiency, accuracy.
• Three different pedigree structures.• Multiallelic and biallelic data.• Numbers of loci: 10, 25 and 50.• Number of recombinants: 0-4.• 100 runs per data set.
Pedigree Structures
Accuracy Results of BE Algorithm
Efficiency Results
More Results from ILP
Real Data Analysis Data set (Gabriel et al.’02)
93 members, 12 pedigrees (each with 7-8 members); chromosome 3, 4 regions, each region 1-4 blocks.
Common Haplotypes
&Frequencies
Results From ILP on the Whole Dataset
3.82 4.00 0.45 0.034
What if there are no relatives?
• Rely on linkage disequilibrium
• Assume that population consists of small number of distinct haplotypes
• Haplotypes tend to be similar
Clark’s Haplotyping Algorithm
• Clark (1990) Mol Biol Evol 7:111-122
• One of the first haplotyping algorithms– Computationally efficient– Very fast
• Today, more accurate alternatives are often available
Clark’s Haplotyping Algorithm
• Find homozygous individuals– Initialize a list of known haplotypes
• Resolve ambiguous individuals– If possible, use two haplotypes from list– Otherwise, use one known haplotype and augment
list
• If unphased individuals remain– Assign phase randomly to one individual– Augment haplotype list and continue from previous
step
Haplotyping via Perfect Phylogeny - Model,
Algorithms, Empirical studies
Dan Gusfield, Ren Hua Chung
U.C. Davis
Cocoon 2003
00000
1
2
4
3
510100
1000001011
00010
01010
12345sitesAncestral haplotype
Extant haplotypes at the leaves
Site mutations on edges
The Perfect Phylogeny Model of Haplotype Evolution
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.
Perfect Phylogeny Haplotype (PPH)
Given a set of genotypes S, find an explaining set of haplotypes that fits a perfect phylogeny.
1 2
a 2 2
b 0 2
c 1 0
sitesA haplotype pair explains agenotype if the merge of thehaplotypes creates thegenotype. Example: Themerge of 0 1 and 1 0 explains 2 2.
Genotype matrix
S
The PPH Problem
Given a set of genotypes, find an explaining set of haplotypes that fits a perfect phylogeny
1 2
a 2 2
b 0 2
c 1 0
1 2
a 1 0
a 0 1
b 0 0
b 0 1
c 1 0
c 1 0
The Haplotype Phylogeny Problem
Given a set of genotypes, find an explaining set of haplotypes that fits a perfect phylogeny
1 2
a 2 2
b 0 2
c 1 0
1 2
a 1 0
a 0 1
b 0 0
b 0 1
c 1 0
c 1 0
1
c c a a
b
b
2
10 10 10 01 01
00
00
The Alternative Explanation
1 2
a 2 2
b 0 2
c 1 0
1 2
a 1 1
a 0 0
b 0 0
b 0 1
c 1 0
c 1 0
No treepossiblefor thisexplanation
Efficient Solutions to the PPH problem - n genotypes, m sites
• Reduction to a graph realization problem (GPPH) - build on Bixby-Wagner or Fushishige solution to graph realization O(nm alpha(nm)) time.
• Reduction to graph realization - build on Tutte’s graph realization method O(nm^2) time.
• Direct, from scratch combinatorial approach -O(nm^2) Bafna et al.
• Berkeley (EHK) approach - specialize the Tutte solution to the PPH problem - O(nm^2) time.
The DPPH Method
• Bafna et al. O(nm^2) time
• Based on deeper combinatorial observations about the PPH problem.
• A matrix-centric approach (rather than tree-centric), although a graph is used in the algorithm.
First, we need to understand why some sets of haplotypeshave a perfect phylogeny, and some do not.
When does a set of haplotypes fit a 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 three pairs:
0,1 and 1,0 and 1,1
This is the 3-Gamete Test
The Alternative Explanation
1 2
a 2 2
b 0 2
c 1 0
1 2
a 1 1
a 0 0
b 0 0
b 0 1
c 1 0
c 1 0
No treepossiblefor thisexplanation
1 2
a 2 2
b 0 2
c 1 0
1 2
a 1 0
a 0 1
b 0 0
b 0 1
c 1 0
c 1 0
1
c c a a
b
b
2
0 0
0 1 0 1
0 0
The Tree Explanation Again
PPH: The Combinatorial Problem
Input: A ternary matrix (0,1,2) M with 2N rowspartitioned into N pairs of rows, where thetwo rows in each pair are identical.
Def: If a pair of rows (r,r’) in the partition have entry values of 2 in a column j then positions (r,j) and (r’,j) are called Mates.
Output: A binary matrix M’ created from Mby replacing each 2 in M with either 0 or 1,such that
a) A position is assigned 0 if and only if its Mate is assigned 1.
b) M’ passes the 3-Gamete Test, i.e., does not contain a 3x2 submatrix (after row and column permutations) with all three combinations 0,1; 1,0; and 1,1
Initial Observations
If two columns of M contain the following rows 2 0 2 0 mates then M’ will contain a row with 1 0 and a row with 0 1 in
those columns. This is a forced expansion.
Initial Observations
Similarly, if two columns of M contain the mates 2 1 2 1 then M’ will contain a row with 1 1 and a row with
0 1 in those columns.
This is a forced expansion.
If a forced expansion of two columns creates 0 1 in those columns, then any 2 2 1 0 2 2 in those columns must be set to be0 11 0 We say that two columns are forced out-of-phase.
If a forced expansion of two columns creates 1 1 in those columns, then any 2 2 2 2 in those columns must be set to be1 10 0 We say that two columns are forced in-phase.
1 2 2
1 2 2
2
0 2
2 0 2
1 2 2
1 2 2
1 2 2
1 2 2
2 2 0
2 2 0
a
a
b
b
cc
d
ed
e
1 2 3
Columns 1 and 2, and 1 and3 are forced in-phase.Columns 2 and 3 are forced out-of-phase.
Example:
1 0
1 1
1 0
0 0
1 3
a
ae
e
1 0
1 1
0 0
1 0
1 2
a
ab
b
0 0
0 1
1 0
0 0
2 3
b
be
e
Overview of Bafna et al. algorithm
First, represent the forced phase relationships, andthe needed decisions, in a graph G.
1 7
2
5
3
4
6Each node representsa column in M, and eachedge indicates that thepair of columns hasa row with 2’sin both columns.
The algorithm builds thisgraph, and then checkswhether any pair of nodesis forced in or out of phase.
Graph G
1 7
2
5
3
4
6Each Red edge indicatesthat the columns areforced in-phase.
Each Blue edge indicatesthat the columns areforced out-of-phase.
Let Gf be the subgraph of Gcdefined by the red and blueedges.
Graph Gc
1 7
2
5
3
4
6
Graph Gf has threeconnected components.
The Central Theorem
There is a solution to the PPH problem for M ifand only if there is a coloring of the dashed edges of Gc with the following property:
For any triangle (i,j,k) in Gc, where there is one row containing 2’s in all three columns i,j and k (any triangle containing at least one dashed edge will be of this type), the coloring makes either 0 or 2 of the edges blue (out-of-phase). Nice, but how do we find such a coloring?
1 7
2
5
3
4
6
Theorem 1: If there are anydashed edges whose ends are in the same connected component of Gf, atleast one edge is in a trianglewhere the other edges arenot dashed, and in every PPHsolution, it must be coloredso that the triangle has aneven number of Blue (out ofPhase) edges. This is an “inferred” coloring.
Graph Gf
Triangle Rule
1 7
2
5
3
4
6
1 7
2
5
3
4
6
1 7
2
5
3
4
6
Corollary
Inside any connected component of Gf, ALL the phaserelationships on edges (columns of M) are uniquely determined, either as forced relationships based onpairwise column comparisons, or by triangle-based inferred colorings.
Hence, the phase relationships of all the columns in a connected component of Gf are INVARIANT over allthe solutions to the PPH problem.
Comparing the programs - R.H. Chung
• All three are fast and practical (under one second) on problem instances of size 50 x 30.
• DPPH is the fastest, followed by HPPH and GPPH.
• HPPH encounters memory problems with large input.
30 50 0.65 0.0206 0.0215
300 150 9.3 3.0 4.49
500 250 36 11.5 21.5
2000 1000 2331 640 1866
sites individ GPPH DPPH HPPH
times shown are in seconds on an 800 Mhz machine.
A Phase-Transition
Problem, as the ratio of sites to genotypes changes,how does the probability that the PPH solution isunique change?
For greatest utility, we want genotype data where thePPH solution is unique.
Intuitively, as the ratio of genotypes to sites increases,the probability of uniqueness increases.
Extension
• With recombination
• The papers: See wwwcsif.cs.ucdavis.edu/~gusfield
The E-M Haplotyping Algorithm
• Excoffier and Slatkin (1995) Mol Biol Evol 12:921-927
• Provide a clear outline of how the algorithm can be applied to genetic data
• Combination of two strategies– E-M statistical algorithm for missing data– Counting algorithm for allele frequencies
E-M Algorithm For Haplotyping
1. “Guesstimate” haplotype frequencies
2. Use current frequency estimates to replace ambiguous genotypes with fractional counts of phased genotypes
3. Estimate frequency of each haplotype by counting
4. Repeat steps 2 and 3 until frequencies are stable
E-M Algorithm for Haplotyping
• Cost grows rapidly with number of markers
• Typically appropriate for < 25 SNPs– Fewer microsatellites
• More accurate than Clark’s method
• Fully or partially phased individuals contribute most of the information
Enhancements to E-M
• List only haplotypes present in sample– Gradually expand subset of markers under
consideration, eliminating haplotypes with low estimated frequency from consideration at each stage
• SNPHAP [Clayton (2001)]
• HAPLOTYPER [Qin et al. (2002)]
Divide-And-Conquer Approximation
• No. of potential haplotypes increases exponentially– Actual no. of haplotypes doesn’t
• Approximation– Successively divide marker set– Run E-M assuming segments associate randomly– Proceed, ignoring composites of segments with zero
frequency
• Order: ~ m log m• Exact E-M is order ~ 2m
Other Recent Developments …
• Newer methods try to further improve haplotype estimation by favoring sets of similar haplotypes
• Stephens et al. (2001) Am J Hum Genet 68:978-89
• Genealogical approach, which implies haplotypes are similar to each other…
Method based on Gibbs sampler
• MCMC method– Stochastic, random procedure– Improves solution gradually
• Given initial set of haplotypes
• Sample haplotypes for one individual at a time, assuming other haplotypes are true
• Repeat a few million times…
Update Procedure I
• Pick individual U to update at random
• Calculate haplotype frequencies F in all other individuals– Since everyone is “phased”, this is done by
counting
• Sample new haplotypes for U from conditional distribution of U’s haplotypes given F
Update Procedure I
• This procedure would produce an estimate of haplotype frequencies that equivalent to the E-M algorithm…
• Stephens et al (2001) suggested an alternative estimate of F…
Update Procedure II
• Estimate F from the other individuals
• Construct F* to include haplotypes in F and also other similar (possibly differing at a few sites, due to mutations)
• Update U’s haplotypes conditional on F*