Download - The Population Haplotyping problem
The PopulationHaplotyping problem
10 11
01 10
01 00
11 11
10 00
10 00
10 10
0*
**
10
1* 11
*0
*0
NOTATION: each SNP only two values in a population (bio).
Call them 0 and 1. Also, call * the fact that a site is heterozygous
HAPLOTYPE: string over 0, 1GENOTYPE: string over 0, 1, * where 0={0}, 1={1}, *={0,1}
10 11
01 10
01 00
11 00
00 10
10 10
0*
**
10
1*
**
*0
0 + 0 =--- 0
1 + 1 =--- 1
0 + 1 + 1 = 0 = --- --- * *
ALGEBRA OF HAPLOTYPES:
Homozygous sites Heterozygous (ambiguous) sites
1**0*
1110110000
1110010001
1100110100
1100010101
Phasing the alleles
For k heterozygous (ambiguous) sites, there are 2k-1 possible phasings
THE 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
*1**1
****1
11**1
INPUT: G = { 11**1, ****1, 11011, *1**1, 00011 }
The input is GENOTYPE data
1101111101
00011
0001111101
1101101101
1101111011
0001100011
11011
*1**1
****1
11**1
OUTPUT: H = { 11011, 11101, 00011, 01101}
INPUT: G = { 11**1, ****1, 11011, *1**1, 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
**0*1 00001 11011
E.g., always put 0 above and 1 below
1*0** 10000 11011
-the objectives/constraints must be “driven by biology”
4°) (parsimony): minimize |H|
1°) Clark’s inference rule
2°) Perfect Phylogeny
3°) Disease Association
OBJECTIVES
Obj: Clark’s rule
1st
1011001011 +?????????? =1**1001*1*
known haplotype h
known (ambiguos) genotype g
Inference Rule
for a compatible pair h , g
1011001011 +1101001110 =1**1001*1*
known haplotype h
known (ambiguos) genotype g
Inference Rule
for a compatible pair h , g
new (derived) haplotype h’
We write h + h’ = g
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)
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)
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
00001000**0011**
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
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
110000001000**0011**
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
1100 1111 SUCCESS
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
00001000**0011**
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
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
00001000**0011**
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
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
0100
00001000**0011**
If, at end, G is empty, SUCCESS, otherwise FAILURE
Step 3 is non-deterministic
0100 FAILURE (can’t resolve 1122 )
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
00001000**0011**
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)
OBJ: 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: Perfect Phylogeny
2nd
- 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 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 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
Theorem: 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
Theorem: 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: 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: We want the solution to admit a perfect phylogeny
(Rationale : we assume haplotypes have evolved independently along a tree)
0 1 * 0* 1 0 ** 0 * 0
We can consider each SNP as a binary feature
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 * 0* 1 0 ** 0 * 0
We can consider each SNP as a binary feature
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 0 1 0 0 00 0 1 0
NOT a perfect phylogeny solution !
0 1 * 0* 1 0 ** 0 * 0
We can consider each SNP as a binary feature
Objective: We want the solution to admit a perfect phylogeny
(Rationale : we assume haplotypes have evolved independently along a tree)
0 1 * 0 0 1 0 *0 0 0 *
We can consider each SNP as a binary feature
Objective: We want the solution to admit a perfect phylogeny
(Rationale : we assume haplotypes have evolved independently along a tree)
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
0 1 * 0 0 1 0 *0 0 0 *
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
** 00 + 11 or ** 01 + 10
- a complex visit of the graph decides how to phase free SNPs
Obj: Disease Association
3rd
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
0*011 *1**1
0**01
11**1
INPUT: GD = {11**1,*1**1,0*011}, GH = {11**1,0**01,00011}
11**1
1101111101
0110100001
1101101101
0101100011
0001100011
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
1100111111
00011
0*011 *1**1
0**01
11**1
INPUT: GD = {11**1,*1**1,0*011}, GH = {11**1,0**01,00011}
11**1
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!
Obj: Max Parsimony See separate slides…
4th
Summary:
- haplotyping in-silico needed for economical reasons
- several objectives, all biologically driven
- nice combinatorial problems (mostly due to binary nature of SNPs)
- these problems are technology-dependant and may become obsolete (hopefully after we have retired)
011101
111111
011000
010001
010011
111111
022
222
012
221
011111 022211
012022
012
222
minimize |H|
2nd Objective (parsimony) :
1. 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
The input is GENOTYPE data
1101111101
0001111101
1101101101
1101111011
0001100011
OUTPUT: H = { 11011, 11101, 00011, 01101}
Each genotype is explained by two haplotypes
We will define some objectives for H
INPUT: G = { 11**1, ****1, 11011, *1**1, 00011 }
****1
11**1
OOO11
11O11
*1**1
1st Objective (open research problem):
minimize |H|
2nd Objective based on inference rule:
1st Objective (parsimony) :
minimize |H|
An easy SQRT(n) approximation: k haplotypes can explain at most k(k-1)/2 genotypes, hence, we need at least LB = SQRT(n) haplotypes.
BUT any greedy algorithm can find 2 haplotypes to explain a genotype, giving asolution of <= 2n haplotypes, i.e. <= SQRT(n) * LB
It’s difficult, but not impossible, to come up with better approximations, like constants(Lancia, Pinotti, Rizzi ’02)
2nd Objective based on inference rule:
xoxxooxoxx +********** =x??xoox?x?
known haplotype h
known (ambiguos) genotype g
Inference Rule
xoxxooxoxx +xxoxooxxxo =x??xoox?x?
known haplotype h
known (ambiguos) genotype g
new (derived) haplotype h’
Inference Rule
xoxxooxoxx +xxoxooxxxo =x??xoox?x?
known haplotype h
known (ambiguos) genotype g
new (derived) haplotype h’
We write h + h’ = g
g and h must be compatible to derive h’
Inference Rule
2nd Objective (Clark, 1990)
1. Start with H = nonambiguos genotypes2. while exists ambiguos genotype g in G3. take h in H compatible with g and let h + h’ = g4. set H = H + {h’} and G = G - {g}5. end while
2nd Objective (Clark, 1990)
1. Start with H = nonambiguos genotypes2. while exists ambiguos genotype g in G3. take h in H compatible with g and let h + h’ = g4. 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
2nd Objective (Clark, 1990)
1. Start with H = nonambiguos genotypes2. while exists ambiguos genotype g in G3. take h in H compatible with g and let h + h’ = g4. 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
ooooxooo??ooxx??
2nd Objective (Clark, 1990)
1. Start with H = nonambiguos genotypes2. while exists ambiguos genotype g in G3. take h in H compatible with g and let h + h’ = g4. 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
ooooxooo??ooxx??
xxoo
2nd Objective (Clark, 1990)
1. Start with H = nonambiguos genotypes2. while exists ambiguos genotype g in G3. take h in H compatible with g and let h + h’ = g4. 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
ooooxooo??ooxx??
xxoo xxxx SUCCESS
2nd Objective (Clark, 1990)
1. Start with H = nonambiguos genotypes2. while exists ambiguos genotype g in G3. take h in H compatible with g and let h + h’ = g4. 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
ooooxooo??ooxx??
2nd Objective (Clark, 1990)
1. Start with H = nonambiguos genotypes2. while exists ambiguos genotype g in G3. take h in H compatible with g and let h + h’ = g4. 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
ooooxooo??ooxx??
oxoo
2nd Objective (Clark, 1990)
1. Start with H = nonambiguos genotypes2. while exists ambiguos genotype g in G3. take h in H compatible with g and let h + h’ = g4. 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
ooooxooo??ooxx??
oxoo FAILURE (can’t resolve xx?? )
OBJ: find order of application rule that leaves the fewest elements in G
- Problem is APX-hard (Gusfield,00)
- Graph-Model + Integer Programming for practical solution (G.,01)
- Problem is APX-hard (Gusfield,00)
- Graph-Model + Integer Programming for practical solution (G.,01)
x??o?
1. expand genotypes
- Problem is APX-hard (Gusfield,00)
- Graph-Model + Integer Programming for practical solution (G.,01)
x??o?
xxxox
xxxoo
xxoox
xxooo
xoxox
xooox
xoxoo
xoooo
1. expand genotypes
- Problem is APX-hard (Gusfield,00)
- Graph-Model + Integer Programming for practical solution (G.,01)
x??o?
xxxox
xxxoo
xxoox
xxooo
xoxox
xooox
xoxoo
xoooo
2. create (h, h’) if exists g s.t. h’ can bederived from g and h
1. expand genotypes 3. Largest number of nodes in forest
rooted at unambiguos genotpes = = largest number of ambiguous genotypes resolved
Hence, find largest number of nodes in forest rooted at unambiguos genotpes. Use I.P. model with vars x(ij).
This reduction is exponential. Is there a better practical approach?
3rd Objective (open research problem)Disease Detection:
oooxx
??oxx
?x??x
????x
xx??x
INPUT: G = { xx??x, ????x, ??oxx, ?x??x, oooxx }
3rd Objective (open research problem)Disease Detection:
xxoxxxxxox
oooxx
oooxxxxxox
xxoxxoxxox
xxoxxoooxx
oooxxoooxx
??oxx
?x??x
????x
xx??x
OUTPUT: H = { xxoxx, xxxox, oooxx, oxxox}
H contains H’, s.t. each diseased has one haplotype in H’ and each healty none
minimize | H|
INPUT: G = { xx??x, ????x, ??oxx, ?x??x, oooxx }
Genome Rearrangements and Evolutionary Distances
Each species has a genome (organized in pairs of chromosomes)
tcgtgatggat………………ttgatggattga
tcgattatggat………………ttttgatatcca
Genomes evolve by means of
• Insertions• Deletions• Inversions• Transpositions• Translocations
of DNA regions
deletion
deletioninsertion
deletioninsertion
translocation
deletioninsertion
translocation
inversion
deletioninsertion
translocation
inversion
transposition
Combinatorial problem: given 2 permutations P, Q and operators in a set F find ashortest sequence f1, ..fk of operators such that Q = fk(fk-1(…(f1(P))))
Very difficult problem! We focus on operators all of the same type (e.g. inversions)(…still difficult…)
Wlog we can take Q = (1 2 … n). Hence we talk of sorting by … (inversions, transpositions…)
5 6 4 8 3 2 1 9 7Example:
We focus on inversions, that are the most important in Nature
1 2 3 8 4 6 5 9 7
1 2 3 8 4 5 6 9 7
1 2 3 6 5 4 8 9 7
1 2 3 6 5 4 8 7 9
1 2 3 4 5 6 8 7 9
1 2 3 4 5 6 7 8 9
Combinatorial problem: given 2 permutations P, Q and operators in a set F find ashortest sequence f1, ..fk of operators such that Q = fk(fk-1(…(f1(P))))
Very difficult problem! We focus on operators all of the same type (e.g. inversions)(…still difficult…)
Wlog we can take Q = (1 2 … n). Hence we talk of sorting by … (inversions, transposition…)
+5 +6 -4 -8 -3 -2 -1 -9 +7Example:
We focus on inversions, that are the most important in Nature
+1 +2 +3 +8 +4 -6 -5 -9 +7
+1 +2 +3 +8 +4 +5 +6 -9 +7
+1 +2 +3 -6 -5 -4 -8 -9 +7
+1 +2 +3 -6 -5 -4 -8 -7 +9
+1 +2 +3 +4 +5 +6 -8 -7 +9
+1 +2 +3 +4 +5 +6 +7 +8 +9
There is also a SIGNED VERSION of the problem !
(Unsigned) Sorting by Inversions is NP-hard (longstanding question, settled by Caprara ‘98)
Surprisingly, Signed Sorting by Inversions is Polynomial (beautiful theory, by Hannenhalli and Pevzner)
The complexity of Sorting by Transpositions, e.g., is unknown
5 7 8 2 1 4 3 6 9
The concept of breakpoint
Breakpoint at position i if | p(i) - p(i+1) | > 1
0 10
(Unsigned) Sorting by Inversions is NP-hard (longstanding question, settled by Caprara ‘98)
Surprisingly, Signed Sorting by Inversions is Polynomial (beautiful theory, by Hannenhalli and Pevzner)
The complexity of Sorting by Transpositions, e.g., is unknown
(Unsigned) Sorting by Inversions is NP-hard (longstanding question, settled by Caprara ‘98)
Surprisingly, Signed Sorting by Inversions is Polynomial (beautiful theory, by Hannenhalli and Pevzner)
The complexity of Sorting by Transpositions, e.g., is unknown
5 7 8 2 1 4 3 6 9
The concept of breakpoint
Breakpoint at position i if | p(i) - p(i+1) | > 1
0 10
d(p) = inversion distanceb(p) = # breakpoints
TRIVIAL BOUND: d(p) >= b(p) / 2
Example: d(p) >= 6 / 2 = 3
The Breakpoint Graph
5 7 8 2 1 4 3 6 9 0
10
The Breakpoint Graph
5 7 8 2 1 4 3 6 9 0
10
The Breakpoint Graph
5 7 8 2 1 4 3 6 9 0
10
The Breakpoint Graph
5 7 8 2 1 4 3 6 9 0
10
The Breakpoint Graph
5 7 8 2 1 4 3 6 9 0
10
10 64
Each node has degree...
0 2 or 4 …
hence the graph can be decomposed in cycles!
The Breakpoint Graph
5 7 8 2 1 4 3 6 9 0
10
Alternating cycle decomposition
The Breakpoint Graph
5 7 8 2 1 4 3 6 9 0
10
Alternating cycle decomposition
The Breakpoint Graph
5 7 8 2 1 4 3 6 9 0
10
Alternating cycle decomposition
c(p) = max # cycles in alternating decomposition
VERY STRONG BOUND : d (p) >= b(p) - c(p)
Example: c(p)= 2 and d (p) >= 6 - 2 = 4
The Breakpoint Graph
5 7 8 2 1 4 3 6 9 0
10
The best algorithm for this problem is based on an Integer Programmingformulation of the max cycle decomposition
A variable xC for each cycle (exponential # of vars…)
A constraint S xC = 1 for each edge e
Objective: maximize SC xC
C containing e
max S xCC
S xC = 1 for all edges eC\ni e
xC \in {0,1} for all alt. cycles C
PRIMAL
min S yee
S ye <= 1 for all alt. Cycles Ce\in C
ye \in R for all edges e
DUAL
max S xCC
S xC = 1 for all edges eC\ni e
xC \in {0,1} for all alt. cycles C
PRIMAL
min S yee
S ye <= 1 for all alt. Cycles Ce\in C
ye \in R for all edges e
DUAL
5 7 8 2 1 4 3 6 9 0
10
Pricing out the cycles for which y*(C) < 1
5 7 8 2 1 4 3 6 9 0
10
5 7 8 2 1 4 3 6 9 0
10
Split the graph in two copies
5 7 8 2 1 4 3 6 9 0
10
5 7 8 2 1 4 3 6 9 0
10
Connect twins
5 7 8 2 1 4 3 6 9 0
10
5 7 8 2 1 4 3 6 9 0
10
A perfect matching corresponds to (a set of) alternating cycles
5 7 8 2 1 4 3 6 9 0
10
5 7 8 2 1 4 3 6 9 0
10
A perfect matching corresponds to (a set of) alternating cycles
5 7 8 2 1 4 3 6 9 0
10
5 7 8 2 1 4 3 6 9 0
10
A perfect matching corresponds to (a set of) alternating cycles
5 7 8 2 1 4 3 6 9 0
10
5 7 8 2 1 4 3 6 9 0
10
A perfect matching corresponds to (a set of) alternating cycles
5 7 8 2 1 4 3 6 9 0
10
5 7 8 2 1 4 3 6 9 0
10
A perfect matching corresponds to (a set of) alternating cycles
5 7 8 2 1 4 3 6 9 0
10
5 7 8 2 1 4 3 6 9 0
10
The weight of the matching is the y*-weight of the cycles
.2
.4
.5
1
.6
0
5 7 8 2 1 4 3 6 9 0
10
5 7 8 2 1 4 3 6 9 0
10
Forcing a cycle to use a certain node
.2
.4
.5
1
.6
100000
- These cycles would not use the same node twice, but with simple trick is possible to model (OMISSIS)
BRANCH&PRICE algorithm by Caprara, Lancia, Ng (1999,2001)
BRANCH&BOUND combinatorial algorithm by Kececioglu, Sankoff (1996)
KS can solve at most n=40. Take days for n=50
CLN can solve for n=200. Takes few seconds (say 5) for n=100
NP-hard problem practically solved to optimality!
Statistical view of evolution
• Genome evolve by random inversions
• It’s like a random walk on a huge graph with an edge for
each permutation an edge for each inversion
• It is not clear why the shortest solution should be the
one followed by Nature (in fact, often it isn’t)
• We want to find the most likely number of inversions
that lead from (1 2 … n ) to p
• We use the expected number of breakpoints after k
inversions as a way to guess the # of inversions
Let B(k) be the (r.v.) number of breakpoint after k random inversions from (1..n)
Given a p obtained by h random inversions from (1 … n ) we want to estimate h
The inversion distance is only a lower bound: h >= d(p) but the gap could be big
We estimate E[B(k)]. Then, faced with some p, we pick h such that E[B(h)] is as close as possible to b(p) (maximum likelihood). CL ,2000, have shown:
Question: estimate E[D(k)], the (r.v.) inversion distance after k random inversions
E[B(k)] = ( n - 1 ) ( 1 - ( ) )
n - 3n - 1
k
Example: n = 200, k (u.a.r. in 1…n) inversions
8 8 8 1619 19 19 3468 67 67 9869 73 68 10473 79 73 10985 91 83 12086 85 83 11587 90 84 119118 117 109 138184 184 135 168
k k’ d(p) b
Protein Structure Alignments: the Maximum Contact Map Overlap
Problem
A Protein is a complex molecule with a primary, linear structure (a sequence of aminoacids) and a3-Dimensional structure (the protein fold).
Protein STRUCTURE determines its FUNCTION
For instance, the Drug Design problemcalls for constructing peptides with a 3Dshape complementary to a protein, so asto dock onto it.
Motivation:Structure Alignment is Important for:
- Discovery of Protein Function (shape determines function)
- Search in 3D data bases
- Protein Classification and Evolutionary Studies
- ...
Problem: Align two 3D protein structures
Contact Maps
Unfolded protein
CONTACT MAPS
Unfolded protein
Folded protein = contacts
CONTACT MAPS
Unfolded protein
Folded protein = contacts
Contact map = graph
CONTACT MAPS
CONTACT MAPS
Unfolded protein
Folded protein = contacts
Contact map = graph
OBJECTIVE: align 3d folds of proteins = align contact maps
Contact Map Alignments
Non-crossing Alignments
Protein 1
Protein 2
non-crossing map of residues in protein 1 and protein 2
The value of an alignment
The value of an alignment
The value of an alignment
Value = 3
The value of an alignment
Value = 3We want to maximize the value
The value of an alignment
NP-Hard
The value of an alignment
Integer Programming Formulation
Integer Programming Formulation
0-1 VARIABLES
yef for e and f contacts
e
f
yef
Integer Programming Formulation
0-1 VARIABLES
yef + ye’f’ <= 1
yef for e and f contacts
e
f
yef
CONSTRAINTS
e
f
e’
f’
Integer Programming Formulation
0-1 VARIABLES
yef + ye’f’ <= 1
yef for e and f contacts
e
f
yef
CONSTRAINTS
e
f
e’
f’
OBJECTIVE max SeSf yef
Independent Set ProblemIt’s just a huge max independent set problem in Gy:
• a node for each sharing • an edge for each pair of incompatible sharings
e
f
e’
f’f’’
e’’
ef
e’f’
e’’f’’
Independent Set ProblemIt’s just a huge max independent set problem in Gy:
• a node for each sharing • an edge for each pair of incompatible sharings
e
f
e’
f’f’’
e’’
ef
e’f’
e’’f’’
|Gy|=|E1|*|E2| (approximately 5000 for two proteins with 50 residues and 75 contacts each)
The best exact algorithm for independent set can solve for at most a few hundred nodes
Node to Node VariablesNew variables x provide an easy check for the non-crossing conditions
NEW VARIABLES
xij for i and j residues
e
f
yef
i
jxij
Node to Node VariablesNew variables x provide an easy check for the non-crossing conditions
NEW VARIABLES
xij for i and j residues
e
f
yef
NEW CONSTRAINTS
i
j
i’
j’
xij + xi’j’ <= 1
i
jxij
Node to Node VariablesNew variables x provide an easy check for the non-crossing conditions
NEW VARIABLES
y(ip)(jq) <= xij and y(ip)(jq) <= xpq
xij for i and j residues
e
f
yef
NEW CONSTRAINTS
i
j
i’
j’
xij + xi’j’ <= 1
i
jxij
i
j
p
q
Clique ConstraintsVariables x define a graph Gx:
• A node for each line• An edge between each pair of crossing lines
i
j
i’
j’
ij
i’j’
Clique ConstraintsVariables x define a graph Gx:
• Gx is much smaller than Gy
• Gx has nice proprieties (it’s a perfect graph)• It’s easier to find large independent sets in Gx
• A node for each line• An edge between each pair of crossing lines
i
j
i’
j’
ij
i’j’
Clique ConstraintsNon-crossing constraints can be extended to
CLIQUE CONSTRAINTS
S xij <= 1[i,j] in M
For all sets M of mutually incompatible (i.e. crossing) lines
All clique constraints satisfied (and Gx perfect) imply a strong bound!
Structure of Maximal cliques in Gx
1. Pick two subsets of same size
Structure of Maximal cliques in Gx
2. Connect them in a zig-zag fashion
Structure of Maximal cliques in Gx
Structure of Maximal cliques in Gx
Structure of Maximal cliques in Gx
Structure of Maximal cliques in Gx
Structure of Maximal cliques in Gx
Structure of Maximal cliques in Gx
Structure of Maximal cliques in Gx
Structure of Maximal cliques in Gx
Structure of Maximal cliques in Gx
Structure of Maximal cliques in Gx
3. Throw in all lines included in a zig or a zag
Structure of Maximal cliques in Gx
3. Throw in all lines included in a zig or a zag
Structure of Maximal cliques in Gx
The result is a maximal clique in Gx
Separation of Clique Inequalities
Separation of Clique InequalitiesPROBLEM
There exist exponentially many such cliques (O(22n) inequalities).
We need to generate in polynomial time a clique inequality when needed,i.e., when violated by the current LP solution x*
S x*ij > 1[i,j] in M
THEOREM
We can find the most violated clique inequality in time O(n2)
Separation of Clique InequalitiesPROOF (sketch)
1) Clique = zigzag path
Separation of Clique InequalitiesPROOF (sketch)
1) Clique = zigzag path
1 2 3 4 5 6 7 8
Separation of Clique InequalitiesPROOF (sketch)
1) Clique = zigzag path 2) Flip one graph: zigzag leftright
1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1
Separation of Clique InequalitiesPROOF (sketch)
1) Clique = zigzag path 2) Flip one graph: zigzag leftright
1 2 3 4 5 6 7 8 8 7 6 5 4 3 2 1
3) Define a grid with lengths for arcs so that length(P) = x*(clique(P)). Use Dyn. Progr.to find longest path in grid, time O(n^2)
Separation of cliques
n2
1n11 2
2
i
u
Create n1 x n2 gridOrient all edges and give weights
Separation of cliques
n2
1n11 2
2
i
u
Create n1 x n2 gridOrient all edges and give weights
x*iu
x*iu
Separation of cliques
Create n1 x n2 gridOrient all edges and give weightsThere is violated clique iff longest A,B path has length > 1
A=(1,n2)
B=(n1,1)
Gx is a Perfect Graph
We show why polynomial separation is possible:
Gx is weakly triangulated (no chordless cycles >= 5 in Gx or Gx)
=> Gx is perfect (Hayward, 1985)
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5
PROOF (Sketch, for Gx)
L1 and L3 don’t cross. Wlog RIGHT(L3, L1)
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5L1 L3
L1 and L3 don’t cross. Wlog RIGHT(L3, L1)
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5L1 L3
For i=4,5,… Li crosses Li-1 but not L1
=> RIGHT (Li, L1)
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5L1 L3
For i=4,5,… Li crosses Li-1 but not L1
=> RIGHT (Li, L1)
L4
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5
For i=4,5,… Li crosses Li-1 but not L1
=> RIGHT (Li, L1)
L1
L4
L5
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5
For i=4,5,… Li crosses Li-1 but not L1
=> RIGHT (Li, L1)
L1 L5L6
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5L1
We get LEFT(L1, {L3, L4, L5, L6})
L3, L4, L5 L6
L6
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5L1
A symmetric argument started at L6, with LEFT(L1, L6) implies LEFT(Li, L6) for i=2,3,4,5
L3, L4, L5 L6
L6
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5L1
A symmetric argument started at L6, with LEFT(L1, L6) implies LEFT(Li, L6) for i=2,3,4,5
L3, L4, L5 L6
L6
L2, L3, L4 L5
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5L1
Then {L3, L4, L5} are between L1 and L6
L3, L4, L5 L6
L6
L2, L3, L4 L5
Gx is a Perfect Graph
L1
L2
L3
L4
L7
L6
L5L1
Then {L3, L4, L5} are between L1 and L6
L3, L4, L5 L6
L6
L2, L3, L4 L5
But L7 crosses L1 and L6, and so should cross them all !
L7
The approach just seen is due to Lancia, Carr, Istrail, Walenz (2001)It can be applied to small or moderate proteins (up to 80 residues/150 contacts)
In 2002, a new approach, by Caprara and Lancia, based on LAGRANGIANRELAXATION. Approach borrowed from Quadratic Assignment. With newapproach we can solve important proteins (up to 150 residues/300 contacts)
What about Heuristics?E.g., genetic algorithms…
Genetic Algorithm Overview
• A Population of candidate solutions thatevolve (improve) over time
• Recombination creates new candidate solutions viacrossover and mutation
Populationat time t
Populationat time t+1
Recombinationoperators
Evaluationfunction
Crossover
• Crossover selects pieces from both parents and creates two offspring solutions
Blue Parent
Offspring
Red Parent
Crossover
• Crossover selects pieces from both parents and creates two offspring solutions– Select a set of edges in one parent to copy to the child
Crossover
• Crossover selects pieces from both parents and creates two offspring solutions– Select a set of edges in one parent to copy to the child
Crossover
• Crossover selects pieces from both parents and creates two offspring solutions– Select a set of edges in one parent to copy to the child– Copy as many edges as possible from the other parent
Crossover
• Crossover selects pieces from both parents and creates two offspring solutions– Select a set of edges in one parent to copy to the child– Copy as many edges as possible from the other parent
These edges conflict with existingedges and are not copied
Crossover
• Crossover selects pieces from both parents and creates two offspring solutions– Select a set of edges in one parent to copy to the child– Copy as many edges as possible from the other parent– Add random edges to fill any remaining space
Crossover
• Crossover selects pieces from both parents and creates two offspring solutions– Select a set of edges in one parent to copy to the child– Copy as many edges as possible from the other parent– Add random edges to fill any remaining space
Mutation
• Mutation introduces small changes to existing solutions by shifting edge endpoints
Mutation
• Mutation introduces small changes to existing solutions by shifting edge endpoints– Select a set of endpoints to shift
Mutation
• Mutation introduces small changes to existing solutions by shifting edge endpoints– Select a set of endpoints to shift
Mutation
• Mutation introduces small changes to existing solutions by shifting edge endpoints– Select a set of endpoints to shift
This edge “fell off” theend of the contact map
and is removed
Mutation
• Mutation introduces small changes to existing solutions by shifting edge endpoints– Select a set of endpoints to shift– Randomly add new edges
Mutation
• Mutation introduces small changes to existing solutions by shifting edge endpoints– Select a set of endpoints to shift– Randomly add new edges
Computational Results
Computational Results
• 269 proteins– 70 -100 residues– 80 to 140 contacts
• Picked 10,000 pairs of proteins out of 36046 possible• Took a weekend on PC• 500 were solved to optimality• 2500 had a gap <= 10 contacts
Skolnick Clustering Test
Skolnick Results• Four Families
1 Flavodoxin-like fold Che-Y related2 Plastocyanin3 TIM Barrel4 Ferratin
• alpha-beta• 8 structures• up to 124 residues• 15-30% sequence similarity• < 3Å RMSD
Skolnick Results• Four Families
1 Flavodoxin-like fold Che-Y related2 Plastocyanin3 TIM Barrel4 Ferratin
• beta• 8 structures• up to 99 residues• 35-90% sequence similarity• < 2Å RMSD
Skolnick Results• Four Families
1 Flavodoxin-like fold Che-Y related2 Plastocyanin3 TIM Barrel4 Ferratin
• alpha-beta• 11 structures• up to 250 residues• 30-90% sequence similarity• < 2Å RMSD
Skolnick Results• Four Families
1 Flavodoxin-like fold Che-Y related2 Plastocyanin3 TIM Barrel4 Ferratin
• alpha• 6 structures• up to 170 residues• 7-70% sequence similarity• < 4Å RMSD
Skolnick Results
Family Style Residues Seq. Sim. RMSD Proteins1 alpha-beta 124 15-30% < 3A 1b00, 1dbw, 1nat, 1ntr,
1qmp, 1rnl, 3cah, 4tmy2 beta 99 35-90% < 2A 1baw, 1byo, 1kdi, 1nin,
1pla, 3b3i, 2pcy, 2plt3 alpha-beta 250 30-90% < 2A 1amk, 1aw2, 1b9b, 1btm,
1hti, 1tmh, 1tre, 1tri,1ydv, 3ypi, 8tim
4 170 7-70% < 4A 1b71, 1bcf, 1dps, 1fha,1ier, 1rcd
• Four Families1 Flavodoxin-like fold Che-Y related2 Plastocyanin3 TIM Barrel4 Ferratin
Clustering
Define score(P1, P2) as
0 <= # shared contacts
Min # of contacts of P1,P2
<= 1
Put P1, P2 in same family if score(P1, P2) >= threshold
Clustering
Define score(P1, P2) as
0 <= # shared contacts
Min # of contacts of P1,P2
<= 1
Put P1, P2 in same family if score(P1, P2) >= threshold
If P1, P2 too big, use G.A. and local search to compute score
L.P. gives then bounds:
HEUR score <= OPT score <= LP bound
and we know how far off OPT we are
Clustering validation
We got some known families from biologists, PDB.
Experiment: Take a family F of proteins and align them against each other and against the remaining.
Clustering validation
We got some known families from biologists, PDB.
0.05 MISMATCH0.1 MISMATCH0.15 MISMATCH0.2 MISMATCH0.25 MISMATCH0.3 MISMATCH0.35 MATCH…… ……1.0 MATCH
score proteins were…
Experiment: Take a family F of proteins and align them against each other and against the remaining.
TYPICAL BEHAVIOUR
Skolnick Results
• Performance– 528 alignments– 1.3% false negative– 0.0% false positive
Clustering
Computed, for 1st time, provably optimal alignments for 150 pairs(inter-family)
Used the CMO value to cluster: retrieves the clusters.
Set S(i,j) = 1 if CMO >= a, S(i,j) = 0 otherwise
Use TSP to find a block diagonal structure for S
Clustering
Last Open Problem
? ?