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Giuseppe LanciaUniversity of Udine
The phasing of The phasing of heterozygous heterozygous
traits: traits: Algorithms and ComplexityAlgorithms and Complexity
-The genomic age has allowed to look at ourselves in a detailed, comparative way
-All humans are >99% identical at genome level
-Small changes in a genome can make a big difference in how we look and who we are
What makes us different from each other?
The answer is
POLYMORPHISMSPOLYMORPHISMS
This is true for humans
as well as for other species
Polymorphisms are features existing in different“flavours”, that make us all look (and be) different
Examples can be eye-color, blood type, hair, etc…
In fact, polymorphisms in the way we look (phenotyes) are determined by polymorphisms in our genome
For a given polymorhism, say the eye-color, thepossible forms are called alleles
We all inherit two alleles (paternal and maternal)
identical HOMOZYGOUS
If they are
different HETEROZYGOUS
{
mother
father
childHomozygous
mother
father
childHomozygous
mother
father
childHeterozygous
Dominant Recessive
mother
father
childHomozygous
mother
father
childHeterozygous
mother
father
childHomozygous
Dominant Recessive
mother
father
childHomozygous
mother
father
childHeterozygous
mother
father
childHomozygous
Dominant Recessive
mother
father
child
mother
father
child
mother
father
child
??
??
??
??
??
??
mother
father
child
mother
father
child
mother
father
child
??
??
??
??
??
??
SingleSingle NucleotideNucleotidePolymorphismsPolymorphisms
At DNA level, a polymorphism is a sequence of nucleotidesvarying in a population.
The shortest possible sequence has only 1 nucleotide, hence
SSingle NNucleotide PPolymorphism (SNP)
At DNA level, a polymorphism is a sequence of nucleotidesvarying in a population.
The shortest possible sequence has only 1 nucleotide, hence
SSingle NNucleotide PPolymorphism (SNP)
atcggattagttagggcacaggacggac
atcggattagttagggcacaggacgtac
atcggcttagttagggcacaggacgtac
atcggattagttagggcacaggacggac
atcggcttagttagggcacaggacgtac
atcggcttagttagggcacaggacggac
atcggattagttagggcacaggacgtac
atcggattagttagggcacaggacgtac
atcggattagttagggcacaggacggac
atcggcttagttagggcacaggacggac
atcggattagttagggcacaggacggac
atcggcttagttagggcacaggacggac
atcggattagttagggcacaggacggac
atcggattagttagggcacaggacggac
- SNPs are predominant form of human variations
atcggattagttagggcacaggacggac
atcggattagttagggcacaggacgtac
atcggcttagttagggcacaggacgtac
atcggattagttagggcacaggacggac
atcggcttagttagggcacaggacgtac
atcggcttagttagggcacaggacggac
atcggattagttagggcacaggacgtac
atcggattagttagggcacaggacgtac
atcggattagttagggcacaggacggac
atcggcttagttagggcacaggacggac
atcggattagttagggcacaggacggac
atcggcttagttagggcacaggacggac
atcggattagttagggcacaggacggac
atcggattagttagggcacaggacggac
- Used for drug design, study disease, forensic, evolutionary...
- On average one every 1,000 bases
atcggcttagttagggcacaggacgtac
atcggattagttagggcacaggacggac
atcggcttagttagggcacaggacgtac
atcggcttagttagggcacaggacggac
atcggattagttagggcacaggacgtac
atcggattagttagggcacaggacgt
atcggattagttagggcacaggacggac
atcggcttagttagggcacaggacggac
atcggattagttagggcacaggacggac
atcggcttagttagggcacaggacggac
atcggattagttagggcacaggacggac
atcggattagttagggcacaggacggac
atcggattagttagggcacaggacggac
atcggattagttagggcacaggacgtac
- SNPs are predominant form of human variations
- Used for drug design, study disease, forensic, evolutionary...
- On average one every 1,000 bases
ag at
ct ag
ct cg
at at
ag cg
ag cg
ag ag
- SNPs are predominant form of human variations
- Used for drug design, study disease, forensic, evolutionary...
- On average one every 1,000 bases
ag at
ct ag
ct cg
at at
ag cg
ag cg
ag ag
HAPLOTYPEHAPLOTYPE: chromosome content at SNP sites
ag at
ct ag
ct cg
at at
ag cg
ag cg
ag ag
HAPLOTYPEHAPLOTYPE: chromosome content at SNP sites
GENOTYPEGENOTYPE: “union” of 2 haplotypes
{c}{g,t}
{a,c}{g,t}
{a}{g}
{a}{g,t} {a}{t}
{a,c}{g}
{a,c}{g}
ag at
ct ag
ct cg
at at
ag cg
ag cg
ag ag
{a,c}{g,t}
{a}{g,t}
{c}{g,t}
{a}{g}
{a}{t}
{a,c}{g}
{a,c}{g}
CHANGE OF SYMBOLSCHANGE OF SYMBOLS: each SNP only two values in a population (bio).
Call them 0 and 1. Also, call 2 the fact that a site is heterozygous
HAPLOTYPEHAPLOTYPE: string over 0, 1GENOTYPEGENOTYPE: string over 0, 1, 2
ag at
ct ag
ct cg
at at
ag cg
ag cg
ag ag
{a,c}{g,t}
{a}{g,t}
{c}{g,t}
{a}{g}
{a}{t}
{a,c}{g}
{a,c}{g}
CHANGE OF SYMBOLSCHANGE OF SYMBOLS: each SNP only two values in a population (bio).
Call them 0 and 1. Also, call 2 the fact that a site is heterozygous
HAPLOTYPEHAPLOTYPE: string over 0, 1GENOTYPEGENOTYPE: string over 0, 1, 2 where 0={0}, 1={1}, 2={0,1}
10 11
01 10
01 00
11 11
10 00
10 00
10 10
02
22
10
12 11
20
20
CHANGE OF SYMBOLSCHANGE OF SYMBOLS: each SNP only two values in a population (bio).
Call them 0 and 1. Also, call 2 the fact that a site is heterozygous
HAPLOTYPEHAPLOTYPE: string over 0, 1GENOTYPEGENOTYPE: string over 0, 1, 2 where 0={0}, 1={1}, 2={0,1}
10 11
01 10
01 00
11 00
00 10
10 10
02
22
10
12
22
20
0 + 0 =--- 0
1 + 1 =--- 1
0 + 1 + 1 = 0 = --- --- 2 2
ALGEBRA OF HAPLOTYPES:
Homozygous sites Heterozygous (ambiguous) sites
12202
1110110000
1110010001
1100110100
1100010101
Phasing the allelesPhasing the alleles
For k heterozygous (ambiguous) sites, there are 2k-1 possible phasings
THE PHASING (or HAPLOTYPING) PROBLEMTHE PHASING (or HAPLOTYPING) PROBLEM
Given genotypes of k individuals, determine the phasings
of all heterozygous sites.
It is too expensive to determine haplotypes directly
Much cheaper to determine genotypes, and then infer haplotypes in silico:
This yields a set H, of (at most) 2k haplotypes. H is a resolution of G.
The input is GENOTYPE data
00011
11011
21221
22221
11221
INPUT: G = { 11221, 22221, 11011, 21221, 00011 }
The input is GENOTYPE data
1101111101
00011
0001111101
1101101101
1101111011
0001100011
11011
21221
22221
11221
OUTPUT: H = { 11011, 11101, 00011, 01101}
INPUT: G = { 11221, 22221, 11011, 21221, 00011 }
Each genotype is resolved by two haplotypes
We will define some objectives for H
--without objectives/constraints, the haplotyping problem would be (mathematically)trivial
OBJECTIVES
22021 00001 11011
E.g., always put 0 above and 1 below
12022 10000 11011
--the objectives/constraints must be “driven by biology”
2°) 2°) (parsimony): minimize |H|
1°) 1°) Clark’s inference rule
3°) Perfect Phylogeny3°) Perfect Phylogeny
4°) Disease Association4°) Disease Association
OBJECTIVES
Obj: Clark’s ruleObj: Clark’s rule
1st1st
1011001011 +********** =1221001212
known haplotype h
known (ambiguos) genotype g
Inference RuleInference Rule
for a compatible pair h , g
1011001011 +1101001110 =1221001212
known haplotype h
known (ambiguos) genotype g
Inference RuleInference Rule
for a compatible pair h , g
new (derived) haplotype h’
We write h + h’ = g
1st Objective (Clark, 1990)1st Objective (Clark, 1990)
1. Start with H = “bootstrap” haplotypes2. while Clark’s rule applies to a pair (h, g) in H x G3. apply the rule to any such (h, g) obtaining h’4. set H = H + {h’} and G = G - {g}5. end while
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
1st Objective (Clark, 1990)1st Objective (Clark, 1990)
1. Start with H = “bootstrap” haplotypes2. while Clark’s rule applies to a pair (h, g) in H x G3. apply the rule to any such (h, g) obtaining h’4. set H = H + {h’} and G = G - {g}5. end while
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
1st Objective (Clark, 1990)1st Objective (Clark, 1990)
1. Start with H = “bootstrap” haplotypes2. while Clark’s rule applies to a pair (h, g) in H x G3. apply the rule to any such (h, g) obtaining h’4. set H = H + {h’} and G = G - {g}5. end while
0000100022001122
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
1st Objective (Clark, 1990)1st Objective (Clark, 1990)
1. Start with H = “bootstrap” haplotypes2. while Clark’s rule applies to a pair (h, g) in H x G3. apply the rule to any such (h, g) obtaining h’4. set H = H + {h’} and G = G - {g}5. end while
0000100022001122
1100
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
0000100022001122
1100 1111 SUCCESS
1st Objective (Clark, 1990)1st Objective (Clark, 1990)
1. Start with H = “bootstrap” haplotypes2. while Clark’s rule applies to a pair (h, g) in H x G3. apply the rule to any such (h, g) obtaining h’4. set H = H + {h’} and G = G - {g}5. end while
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
1st Objective (Clark, 1990)1st Objective (Clark, 1990)
1. Start with H = “bootstrap” haplotypes2. while Clark’s rule applies to a pair (h, g) in H x G3. apply the rule to any such (h, g) obtaining h’4. set H = H + {h’} and G = G - {g}5. end while
0000100022001122
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
1st Objective (Clark, 1990)1st Objective (Clark, 1990)
1. Start with H = “bootstrap” haplotypes2. while Clark’s rule applies to a pair (h, g) in H x G3. apply the rule to any such (h, g) obtaining h’4. set H = H + {h’} and G = G - {g}5. end while
0000100022001122
0100
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
0000100022001122
0100 FAILURE (can’t resolve 1122 )
1st Objective (Clark, 1990)1st Objective (Clark, 1990)
1. Start with H = “bootstrap” haplotypes2. while Clark’s rule applies to a pair (h, g) in H x G3. apply the rule to any such (h, g) obtaining h’4. set H = H + {h’} and G = G - {g}5. end while
1. Start with H = “bootstrap” haplotypes2. while Clark’s rule applies to a pair (h, g) in H x G3. apply the rule to any such (h, g) obtaining h’4. set H = H + {h’} and G = G - {g}5. end while
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic: the algorithm could end without explainingall genotypes even if an explanation was possible.
The number of genotypes solved depends on order of application.
1st Objective (Clark, 1990)1st Objective (Clark, 1990)
OBJ: find order of application rule that leaves the fewest elements in GOBJ: find order of application rule that leaves the fewest elements in G
The problem was studied by Gusfield(ISMB 2000, and Journal of Comp. Biol., 2001)
- problem is APX-hard
- it corresponds to finding largest forest in a graph with haplotypes as nodes and arcs for possible derivations
-solved via ILP of exponential-size (practical for small real instances)
Obj: Max ParsimonyObj: Max Parsimony
2nd2nd
- Clark conjectured solution (when found) uses min # of haplotypes
- this is clearly false
- solution with few haplotypes is biologically relevant (as we all descend from a small set of ancestors)
011101
111111
011000
010001
010011
111111
011101
111111
011000
010001
010011
111111
022
222
012
221
011111 022211
012022
012
222
minimize |H|
2nd Objective (parsimony)2nd Objective (parsimony) :
1. The problem is APX-Hard1. The problem is APX-Hard
Reduction from VERTEX-COVER
A
B
C
D E
A
B
C
D E
A B C D E *
A
B
C
D E
A B C D E *
AB BC AE DE AD
A
B
C
D E
A B C D E *
AB BC AE DE AD
A B C D E
A
B
C
D E
A B C D E *
AB 2 2BC 2 2AE 2 2DE 2 2AD 2 2
ABCDE
A
B
C
D E
A B C D E *
AB 2 2BC 2 2AE 2 2DE 2 2AD 2 2
A 0B 0C 0D 0E 0
A
B
C
D E
A B C D E *
AB 2 2 2 BC 2 2 2 AE 2 2 2 DE 2 2 2 AD 2 2 2
A 0 0 B 0 0C 0 0 D 0 0 E 0 0
A
B
C
D E
A B C D E *
AB 2 2 1 1 1 2BC 1 2 2 1 1 2AE 2 1 1 1 2 2DE 1 1 1 2 2 2 AD 2 1 1 2 1 2
A 0 1 1 1 1 0 B 1 0 1 1 1 0C 1 1 0 1 1 0 D 1 1 1 0 1 0 E 1 1 1 1 0 0
A
B
C
D E
A B C D E *
AB 2 2 1 1 1 2BC 1 2 2 1 1 2AE 2 1 1 1 2 2DE 1 1 1 2 2 2 AD 2 1 1 2 1 2
A 0 1 1 1 1 0 B 1 0 1 1 1 0C 1 1 0 1 1 0 D 1 1 1 0 1 0 E 1 1 1 1 0 0
G = (V,E) has a node cover X of size k there is a set H of |V | + k haplotypes that explain all genotypes
A
B
C
D E
A B C D E *
AB 2 2 1 1 1 2BC 1 2 2 1 1 2AE 2 1 1 1 2 2DE 1 1 1 2 2 2 AD 2 1 1 2 1 2
A 0 1 1 1 1 0 B 1 0 1 1 1 0C 1 1 0 1 1 0 D 1 1 1 0 1 0 E 1 1 1 1 0 0
G = (V,E) has a node cover X of size k there is a set H of |V | + k haplotypes that explain all genotypes
A
B
C
D E
A B C D E *
AB 2 2 1 1 1 2BC 1 2 2 1 1 2AE 2 1 1 1 2 2DE 1 1 1 2 2 2 AD 2 1 1 2 1 2
A 0 1 1 1 1 0 B 1 0 1 1 1 0C 1 1 0 1 1 0 D 1 1 1 0 1 0 E 1 1 1 1 0 0 A’ 0 1 1 1 1 1B’ 1 0 1 1 1 1E’ 1 1 1 1 0 1
G = (V,E) has a node cover X of size k there is a set H of |V | + k haplotypes that explain all genotypes
A basic ILP formulation
Expand your input G in all possible ways
220 120 022
A basic ILP formulation
Expand your input G in all possible ways
010 + 100, 000 + 110100 + 110 000 + 011, 001 + 010
220 120 022
A basic ILP formulation
hx
21,hh
hx
yhh 21 ,
Expand your input G in all possible ways
010 + 100, 000 + 110100 + 110 000 + 011, 001 + 010
220 120 022
A basic ILP formulation
The resulting Integer Program (IP1):
Other ILP formulation are possible. E.g. POLY-SIZE ILP formulations
Obj: Perfect PhylogenyObj: Perfect Phylogeny
3rd3rd
- Parsimony does not take into account mutations/evolution of haplotypes
- parsimony is very relialable on “small” haplotype blocks
- when haplotypes are large (span several SNPs, we should consider evolutionionary events and recombination)
- the cleanest model for evolution is the perfect phylogeny
- A phylogeny expalains set of binary features (e.g. flies, has fur…) with a tree
- Leaf nodes are labeled with species
- Each feature labels an edge leading to a subtree that possesses it
3rd objective is based on perfect phylogenyperfect phylogeny
- A phylogeny expalains set of binary features (e.g. flies, has fur…) with a tree
- Leaf nodes are labeled with species
- Each feature labels an edge leading to a subtree that possesses it
has 2 legs
3rd objective is based on perfect phylogenyperfect phylogeny
has tailflies
- A phylogeny expalains set of binary features (e.g. flies, has fur…) with a tree
- Leaf nodes are labeled with species
- Each feature labels an edge leading to a subtree that possesses it
has 2 legs
But…a new species may come along so that noPerfect phylogeny is possible…
has tailflies
TheoremTheorem: such matrix has p.p. iff there is not a 00 4x2 minor 10 01 11
Human 1 0 0
Mouse 0 1 0
Spider 0 0 0
Eagle 1 0 1
two legs
tail
flies
TheoremTheorem: such matrix has p.p. iff there is not a 00 4x2 minor 10 01 11
Human 1 0 0
Mouse 0 1 0
Spider 0 0 0
Eagle 1 0 1
Mickey mouse 1 1 0
two legs
tail
flies
We can consider each SNP as a binary feature
Objective:Objective: We want the solution to admit a perfect phylogeny
(Rationale : we assume haplotypes have evolved independently along a tree)
We can consider each SNP as a binary feature
Objective:Objective: We want the solution to admit a perfect phylogeny
(Rationale : we assume haplotypes have evolved independently along a tree)
0 1 2 02 1 0 22 0 2 0
We can consider each SNP as a binary feature
Objective:Objective: We want the solution to admit a perfect phylogeny
(Rationale : we assume haplotypes have evolved independently along a tree)
0 1 0 00 1 1 01 1 0 10 1 0 01 0 0 00 0 1 0
0 1 2 02 1 0 22 0 2 0
We can consider each SNP as a binary feature
Objective:Objective: We want the solution to admit a perfect phylogeny
(Rationale : we assume haplotypes have evolved independently along a tree)
0 1 2 02 1 0 22 0 2 0
0 1 0 00 1 1 01 1 0 10 1 0 0 1 0 0 00 0 1 0
NOT a perfect phylogeny solution !
We can consider each SNP as a binary feature
Objective:Objective: We want the solution to admit a perfect phylogeny
(Rationale : we assume haplotypes have evolved independently along a tree)
0 1 2 0 0 1 0 20 0 0 2
We can consider each SNP as a binary feature
Objective:Objective: We want the solution to admit a perfect phylogeny
(Rationale : we assume haplotypes have evolved independently along a tree)
0 1 2 0 0 1 0 20 0 0 2
0 1 0 0 0 1 1 00 1 0 0
1 1 0 1 0 0 0 00 0 0 1
A perfect phylogeny
Theorem: The Perfect Phylogeny Haplotyping problem is polynomial
Theorem: The Perfect Phylogeny Haplotyping problem is polynomial
Algorithms are of combinatorial nature
- There is a graph for which SNPs are columns and edges are of two types (forced and free)
- forced edges connect pairs of SNPs that must be phased in the same way
22 00 + 11 or 22 01 + 10
- a complex visit of the graph decides how to phase free SNPs
Obj: Disease AssociationObj: Disease Association
4th4th
Some diseases may be due to a gene which has “faulty” configurations
RECESSIVE DISEASE (e.g. cystic fibrosis, sickle cell anemia): to be diseased one must have both copies faulty. With one copy one is a carrier of the disease
DOMINANT DISEASE (e.g. Huntington’s disease, Marfan’s syndrome): to be diseased it is enough to have one faulty copy
Two individuals of which one is healthy and the other diseased may have the same genotype.
The explanation of the disease lies in a difference in their haplotypes
00011
02011 21221
02201
11221
INPUT: GD = {11221,21221,02011}, GH = {11221,02201,00011}
11221
1101111101
00011
0110100001
1101101101
0101100011
0001100011
02011 21221
02201
11221
OUTPUT: H = { 11011,01011,00001,11111,11101,00011,01101}
H contains HD, s.t. each diseased has >=1 haplotype in HD and each healty none
INPUT: GD = {11221,21221,02011}, GH = {11221,02201,00011}
1100111111
11221
Theorem 1 is proved via a reduction from 3 SAT
Theorem 2 has a mathematical proof (coloring argument) with little relation to biology:There is R (depending on input) s.t. a haplotype is healthy if the sum of its bits is congruent to R modulo 3
This means the model must be refined!
Summary:
- haplotyping in-silico needed for economical reasons
- several objectives, all biologically driven
- nice combinatorial problems (mostly from binary nature of SNPs)
- these problems are technology-dependant and may become obsolete (hopefully after we have retired)
ThanksThanks