Optimization Problems for Polymorphisms of Single Nucleotides

Optimization Problems for

Polymorphisms of Single Nucleotides

Polymorphisms

A polymorphism is a feature

Polymorphisms

A polymorphism is a feature - common to everybody

Polymorphisms

A polymorphism is a feature - common to everybody - not identical in everybody

Polymorphisms

A polymorphism is a feature - common to everybody - not identical in everybody- the possible variants (alleles) are just a few

Polymorphisms

E.g. think of eye-color


Polymorphisms


E.g. think of eye-color

Or blood-type for a feature not visible from outside

At DNA level, a polymorphism is a sequence of nucleotidesvarying in a population.


The shortest possible sequence has only 1 nucleotide, henceSingle Nucleotide Polymorphism (SNP)



atcggattagttagggcacaggacggacatcggattagttagggcacaggacggac









atcggattagttagggcacaggacggacatcggattagttagggcacaggacgtac

atcggcttagttagggcacaggacgtacatcggattagttagggcacaggacggac

atcggcttagttagggcacaggacgtacatcggcttagttagggcacaggacggac

atcggattagttagggcacaggacgtacatcggattagttagggcacaggacgtac

atcggattagttagggcacaggacggacatcggcttagttagggcacaggacggac



- SNPs are predominant form of human variations








- Used for drug design, study disease, forensic, evolutionary...

- On average one every 1,000 bases

- Multimillion dollar SNP consortium project








- Goal: associate SNPs (or group of SNPs) to genetic diseases

- 1st step: build maps of several thousand SNPs








HOMOZYGOUS: same allele on both chromosomes

















HETEROZYGOUS: different alleles



















HAPLOTYPE: chromosome content at SNP sites



atcggattagttagggcacaggacgtacatcggattagttagggcacaggacgt








ag at

ct ag

ct cg

at at

ag cg

ag cg

ag ag




ag at

ct ag

ct cg

at at

ag cg

ag cg

ag ag




GENOTYPE: “union” of 2 haplotypes

OcE

EE

OaOg

OaE OaOt

EOg

OgE

ag at

ct ag

ct cg

at at

ag cg

ag cg

ag ag

OcE

EE

OaOg

OaE OaOt

EOg

OgE

CHANGE OF SYMBOLS: each SNP only two values in a poplulation (bio). Call them 1 and O. Also, call * the fact that a site is heterozygous

HAPLOTYPE: string over 1,OGENOTYPE: string over 1,O,*

1o 11

o1 1o

o1 oo

11 11

1o oo

1o oo

1o 1o

o*

**

*o

1* 11

*o

*o

CHANGE OF SYMBOLS: each SNP only two values in a poplulation (bio). Call them 1 and O. Also, call * the fact that a site is heterozygous

HAPLOTYPE: string over 1,OGENOTYPE: string over 1,O,*

THE HAPLOTYPING PROBLEM

Single Individual: Given genomic data of one individual, determine 2 haplotypes (one per chromosome)

Population : Given genomic data of k individuals, determine (at most) 2k haplotypes (one per chromosome/indiv.)

For the individual problem, input is erroneous haplotype data, from sequencing

For the population problem, data is ambiguous genotype data, from screening

OBJ is lead by Occam’s razor: find minimum explanation of observed data under given hypothesis (a.k.a. parsimony principle)

Theory and Results

- Polynomial Algorithms for gapless haplotyping (L, Bafna, Istrail, Lippert, Schwartz 01 & Bafna, L, Istrail, Rizzi 02)

- Polynomial Algorithms for bounded-length gapped haplotyping (BLIR 02)

Single individual

- NP-hardness for general gapped haplotyping (LBILS 01)

- APX-hardness (Gusfield 00)

- Reduction to Graph-Theoretic model and I.P. approach (Gusfield 01)

Population

-New formulations and Disease Detection (L, Ravi, Rizzi, 02)

- Exact algorithms for min-size solution (L,Serafini 2011)

- Heuristics (Tininini, L, Bertolazzi 2010)

The Single-IndividualHaplotyping problem

TGAGCCTAG GATTT GCCTAG CTATCTTATAGATA GAGATTTCTAGAAATC ACTGATAGAGATTTC TCCTAAAGAT CGCATAGATA

fragmentation

sequencing

assembly

Shotgun Assembly of a Chromosome [ Webber and Myers, 1997]

ACTGCAGCCTAGAGATTCTCAGATATTTCTAGGCGTATCTATCTTACTGCAGCCTAGAGATTCTCAGATATTTCTAGGCGTATCTATCTTACTGCAGCCTAGAGATTCTCAGATATTTCTAGGCGTATCTATCTT

ACTGCAGCCTAGAGATTCTCAGATATTTCTAGGCGTATCTATCTT

-Sequencing errors:

ACTGCCTGGCCAATGGAACGGACAAG CTGGCCAAT CATTGGAAC AATGGAACGGA

-Contaminants

MAIN ERROR SOURCES

Given errors, the data may be inconsistent with exactly 2 haplotypes

PROBLEM: Find and remove the errors so that the data becomes consistent with exactly 2 haplotypes

Hence, assembler is unable to build 2 chromosomes

ACTGAAAGCGA ACTAGAGACAGCATGACTGATAGC GTAGAGTCAACTG TCGACTAGA CATGACTGA CGATCCATCG TCAGCACTGAAA ATCGATC AGCATGACTGAAAGCGA ACTAGAGACAGCATGACTGATAGC GTAGAGTCAACTG TCGACTAGA CATGACTGA CGATCCATCG TCAGCACTGAAA ATCGATC AGCATG 1 1 O O O 1 1 1 1 1 O

The data: a SNP matrix

Snips 1,..,n

Fragments 1,..,m

1 2 3 4 5 6 7 8 9 1 - - - O 1 1 O O - 2 - O - O 1 - - - 13 1 1 O 1 1 - - - - 4 O O 1 - - - - O - 5 - - - - - - - 1 O6 - - - - O O O 1 -

Snips 1,..,n

Fragments 1,..,m

Fragment conflict: can’t be on same haplotype

1 2 3 4 5 6 7 8 9 1 - - - O 1 1 O O - 2 - O - O 1 - - - 13 1 1 O 1 1 - - - - 4 O O 1 - - - - O - 5 - - - - - - - 1 O6 - - - - O O O 1 -

Snips 1,..,n

Fragments 1,..,m

Fragment conflict: can’t be on same haplotype

1

62

3

45

Fragment Conflict Graph GF(M)

We have 2 haplotypes iff GF is BIPARTITE

1 2 3 4 5 6 7 8 9 1 - - - O 1 1 O O - 2 - O - O 1 - - - 13 1 1 O 1 1 - - - - 4 O O 1 - - - - O - 5 - - - - - - - 1 O6 - - - - O O O 1 -

Snips 1,..,n

Fragments 1,..,m

1

62

3

45

PROBLEM (Fragment Removal): make GF Bipartite

1 2 3 4 5 6 7 8 9 1 - - - O 1 1 O O - 2 - O - O 1 - - - 13 1 1 O 1 1 - - - - 4 O O 1 - - - - O - 5 - - - - - - - 1 O6 - - - - O O O 1 -

Snips 1,..,n

1 2 3 4 5 6 7 8 9 1 - - - O 1 1 O O - 2 - O - O 1 - - - 13 1 1 O 1 1 - - - - 4 O O 1 - - - - O - 5 - - - - - - - 1 O6 - - - - O O O 1 -

Fragments 1,..,m

PROBLEM (Fragment Removal): make GF Bipartite

1

62

3

45

1 2 3 4 5 6 7 8 9 1 - - - O 1 1 O O - 2 - O - O 1 - - - 14 O O 1 - - - - O -

3 1 1 O 1 1 - - - -5 - - - - - - - 1 O

O O 1 O 1 1 O O 1

1 1 O 1 1 - - 1 O

Removing fewest fragments is equivalent to maximum induced bipartite subgraph

NP-complete [Yannakakis, 1978a, 1978b; Lewis, 1978] O(|V|(log log |V|/log |V|)2)-approximable [Halldórsson, 1999] not O(|V|)-approximable for some [Lund and Yannakakis, 1993]

Are there cases of M for which GF(M) is easier?

YES: the gapless M

---O11OO---O1OO1--- gap

---O11OO1O1O1OO1--- gapless

---O11--1O----O1--- 2 gaps

Why gaps?

Sequencing errors (don’t call with low confidence)

---OO11?11--- ===> ---OO11-11---

Why gaps?

Sequencing errors (don’t call with low confidence)

---OO11?11--- ===> ---OO11-11---

Celera’s mate pairs

attcgttgtagtggtagcctaaatgtcggtagaccttga

attcgttgtagtggtagcctaaatgtcggtagaccttga

THEOREM

For a gapless M, the Min Fragment RemovalProblem is Polynomial

NOTE: Does not need to be gapless. Enough if it can be sorted to become such (Consecutive Ones Property, Booth and Lueker, 1976)

An O(nm + n ) D.P. algo3

1 - O O 1 1 O O - -2 - - 1 O 1 1 O - -3 - - - 1 1 O - - - 4 - - - - O O 1 O - 5 - - - - - 1 O 1 O


LFT(i) RGT(i)

sort according to LFT

1 - O O 1 1 O O - -2 - - 1 O 1 1 O - -3 - - - 1 1 O - - - 4 - - - - O O 1 O - 5 - - - - - 1 O 1 O


1 - O O 1 1 O O - -2 - - 1 O 1 1 O - -3 - - - 1 1 O - - - 4 - - - - O O 1 O - 5 - - - - - 1 O 1 O

LFT(i) RGT(i)

D(i;h,k) := min cost to solve up to row i, with k, h not removed and put in different haplotypes, and maximizing RGT(k), RGT(h)

sort according to LFT

D(i; h,k) =

D(i-1; h,k) if i, k compatible and RGT(i) <= RGT(k) or i, h compatible and RGT(i) <= RGT(h)

1 + D(i-1; h, k) otherwise{

OPT is min h,k D( n; h, k ) and can be found in time O(nm + n^3)

Th: NP-Hard if 2 gaps per fragment

proof: (simple) use fact that for every G there is M s.t. G = GF(M) and reduce from Max Bip. Induced Subgraph on 3-regular graphs (in each row, max 3 non-bit, hence max 2 gaps)

WITH GAPS…..

Th : NP-Hard if even 1 gap per fragment proof: technical. reduction from MAX2SAT

WITH GAPS…..



Th : NP-Hard if even 1 gap per fragment proof: technical. reduction from MAX2SAT

WITH GAPS…..

But, gaps must be long for problem to be difficult.

We have O( 2 mn + 2 n ) D.P.

for MFR on matrix with total gaps length L

2L 3L 3



What for MFR with gaps? Why not ILP...

min∑𝑓𝑥 𝑓

∑𝑓 ∈𝐶

𝑥 𝑓 ≥1 for all odd cycles𝐶𝑥∈ {0,1 }𝑛


1

5 2

34

1/2

1/3

1/41/2

0

min∑𝑓𝑥 𝑓

∑𝑓 ∈𝐶



1

5 2

34

1/2

1/3

1/41/2

01

5 2

34

1

5 2

34

min∑𝑓𝑥 𝑓

∑𝑓 ∈𝐶



1

5 2

34

1/2

1/3

1/41/2

01

5 2

34

1

5 2

34

5/12 5/12

min∑𝑓𝑥 𝑓

∑𝑓 ∈𝐶



1

5 2

34

1/2

1/3

1/41/2

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5 2

34

1

5 2

34

5/12 5/12

min∑𝑓𝑥 𝑓

∑𝑓 ∈𝐶



1

5 2

34

1/2

1/3

1/41/2

01

5 2

34

1

5 2

34

5/12 5/12

min∑𝑓𝑥 𝑓

∑𝑓 ∈𝐶



1

5 2

34

1/2

1/3

1/41/2

01

5 2

34

1

5 2

34

5/12 5/12

Randomized rounding heuristic: round and repeat. Worked well at Celera

min∑𝑓𝑥 𝑓

∑𝑓 ∈𝐶


The fragment removal is good to get rid of contaminants.

However, we may want to keep all fragments andcorrect errors otherwise

A dual point of view is to disregard some SNPs and keepthe largest subset sufficient to reconstruct the haplotypes

All fragments get assigned to one of the two haplotypes.We describe the min SNP removal problem: remove the fewest number of columns from M so that the fragmentgraph becomes bipartite.

SNP conflicts

- - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

SNP conflicts

OK

- - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

SNP conflicts

OK

- - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

SNP conflicts

OK

- - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

SNP conflicts

CONFLICT !

- - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

SNP conflicts

CONFLICT !

- - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

- - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

SNP conflicts

SNP conflict graph GS(M)1 node for each SNP (column)edge between conflicting SNPs

SNP conflicts

1 2 3 4 5 6 7 8 9 - - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

SNP conflicts

1

6

2

3

4

5

8

9

7

1 2 3 4 5 6 7 8 9 - - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

1 2 3 4 5 6 7 8 9 - - - O 1 1 O O - - O 1 O 1 - - - 11 1 O 1 1 - - - - O O 1 - - - O O - - - - - - - 1 1 O- - - - O O O 1 -

SNP conflicts

1

6

2

3

4

5

8

9

7

THEOREM 1

For a gapless M, GF(M) is bipartiteif and only if GS(M) is an independent set

THEOREM 2

For a gapless M, GS(M) is a perfect graph

COROLLARY

For a gapless M, the min SNP removalproblem is polynomial

THEOREM 1For a gapless M, GF(M) is bipartite if and only if GS(M) is an independent set

PROOF (sketch): by minimal counterexample

--OO11OO-------------OO1OO1O11O-----------11O1O111-----11OO1O11O-----------1OOO1-----------11111O-------11O11O1OO------

Assume M gapless, GS(M) an independent set, but GF(M)not bipartite.

Take an odd cycle in GF



--O?1???-------------O????????O-----------??O??1??-----??????1??-----------???O?-----------????1?-------1???????O------

There is a generic structure of hor-vert cycle



--O?1???-------------O????????O-----------??O??1??-----??????1??-----------???O?-----------????1?-------1???????O------

“vertical lines”

There cannot be only one vertical line in odd cycle

We merge rightmost and next to reduce them by 1

Hence, there cannot be a minimal (in n. of vertical lines) counterexample



--O?1???-------------O????????O-----------??O??1??-----??????1??-----------???O?-----------????1?-------1???????O------


Must be 1



--O?1???-------------O?????1??O-----------??O??1??-----??????1??-----------???O?-----------????1?-------1???????O------


Must be 1

Merge the rightmost lines



--O?1???-------------O?????1--------------??O----------??????1-------------???O------------????1--------1???????O------


Still a counterexample!

Merge the rightmost lines

1 2 31 O - O 2 - O 1 3 1 1 -

Note: Theorem not true if there are gaps

1

2 3

1

2 3

GF(M) GS(M)

M

THEOREM 2For a gapless M, GS(M) is a perfect graph

PROOF: GS(M) is the complement of a comparability graph A

Comparability graphs are perfect

Comparability Graphs: unoriented that can be oriented to become a partial order

LEMMA: If i<j<k and (i,k) is a SNP conflict then either (i,k) or (j,k) is also a SNP conflict

i j k - 1 O O ? 1 O 1 - - O 1 O ? 1 1 1 -

Equal:conflicts with i

OO

Different:conflicts with k

O1

i kj

I.e. if (i,j) is not a conflict and (j,k) is not a conflict, also (i,k) is not a conflict

So (u,v) with u < v and u not a conflict with v is a comparability graph Aand GS is A complement

NOTE: ind set on perfect graph is in P (Lovasz, Schrijvers, Groetschel, 84)

THEOREM: The min SNP removal is NP-hard if there can be gaps (Reduction from MAXCUT)

Again, gaps must be long for problem to be difficult.

We have O(mn + n ) D.P.

for MSR on matrix with total gaps length L

2L + 1 2L + 2

Hence gapless MSR is polynomial (max stable set on perfect graph).

There are better, D.P., algorithms, O(mn + m^2)

What if gaps ?

Documents

Optimization Problems for Polymorphisms of Single Nucleotides