identification of snp alleles in dna sequences giuseppe lancia università di padova e celera...

Identification of SNP Alleles Identification of SNP Alleles in DNA Sequencesin DNA Sequences

Giuseppe LanciaUniversità di Padova e Celera Genomics

PolymorphismsPolymorphismsA polymorphism is a feature

PolymorphismsPolymorphismsA polymorphism is a feature - common to everybody

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

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

PolymorphismsPolymorphisms

E.g. think of eye-coloreye-color

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

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

E.g. think of eye-coloreye-color

Or blood-typeblood-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, hence

SSingle NNucleotide PPolymorphism (SNP)

atcggattagttagggcacaggacggac

atcggattagttagggcacaggacgtac

atcggcttagttagggcacaggacgtac

atcggcttagttagggcacaggacggac

- SNPs are predominant form of human variations

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

- On average one every 1,000 bases

HOMOZYGOUSHOMOZYGOUS: same allele on both chromosomes

HETEROZYGOUSHETEROZYGOUS: different alleles

HAPLOTYPEHAPLOTYPE: chromosome content at SNP sites

atcggattagttagggcacaggacgt

GENOTYPEGENOTYPE: “union” of 2 haplotypes

OaE OaOt

CHANGE OF SYMBOLSCHANGE OF SYMBOLS: each SNP only two values in a poplulation (bio).

Call them X and O. Also, call ? the fact that a site is heterozygous

HAPLOTYPEHAPLOTYPE: string over X,OGENOTYPEGENOTYPE: string over X,O,?

CHANGE OF SYMBOLSCHANGE OF SYMBOLS: each SNP only two values in a poplulation (bio).

Call them X and O. Also, call ? the fact that a site is heterozygous

HAPLOTYPEHAPLOTYPE: string over X,OGENOTYPEGENOTYPE: string over X,O,?

THE HAPLOTYPING PROBLEMTHE HAPLOTYPING PROBLEM

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

Population 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 (Lancia, Bafna, Istrail, Lippert, Schwartz 01 & Bafna, Lancia, 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 (Lancia, Pesole 02)

The Single-IndividualThe Single-IndividualHaplotyping problemHaplotyping problem

TGAGCCTAG GATTT GCCTAG CTATCTT

ATAGATA GAGATTTCTAGAAATC ACTGA

TAGAGATTTC TCCTAAAGAT CGCATAGATA

fragmentation

sequencing

assembly

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

ACTGCAGCCTAGAGATTCTCAGATATTTCTAGGCGTATCTATCTTACTGCAGCCTAGAGATTCTCAGATATTTCTAGGCGTATCTATCTTACTGCAGCCTAGAGATTCTCAGATATTTCTAGGCGTATCTATCTT

ACTGCAGCCTAGAGATTCTCAGATATTTCTAGGCGTATCTATCTT

Sequencing errors:

ACTGCCTGGCCAATGGAACGGACAAG CTGGCCAAT CATTGGAAC AATGGAACGGA

Paralogous regions:

ACAAACCCTTTGGGACT … CTAGTAAACCCTATGGGGA AAACCCTT TAAACCCT CTATGGGA CCTATGG CTTTGGGACT ACCCTATGGG

ERROR SOURCESERROR SOURCES

Given errorserrors (sequencing errors, and/or paralogous) the data may be inconsistentinconsistent with exactly 2 haplotypes

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

Hence, assembler is unable Hence, assembler is unable to build 2 chromosomesto build 2 chromosomes

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

The data: a SNP matrix

Snips 1,..,n

1 2 3 4 5 6 7 8 9 1 - - - O X X O O - 2 - O - O X - - - X3 X X O X X - - - - 4 O O X - - - - O - 5 - - - - - - - X O6 - - - - O O O X -

Fragments 1,..,m

Snips 1,..,n

Fragments 1,..,m

Fragment conflict: can’t be on same haplotype

Snips 1,..,n

Fragments 1,..,m

Fragment conflict: can’t be on same haplotype

Fragment Conflict Graph GF(M)

We have 2 haplotypes iff GF is BIPARTITE

Snips 1,..,n

Fragments 1,..,m

PROBLEM (Fragment Removal): make GF Bipartite

Snips 1,..,n

Fragments 1,..,m

PROBLEM (Fragment Removal): make GF Bipartite

1 2 3 4 5 6 7 8 9 1 - - - O X X O O - 2 - O - O X - - - X4 O O X - - - - O -

3 X X O X X - - - -5 - - - - - - - X O

O O X O X X O O X

X X O X X - - X 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

---OXXOO---OXOOX--- gap

---OXXOOXOXOXOOX--- gapless

---OXX--XO----OX--- 2 gaps

Why gaps?

Sequencing errors (don’t call with low confidence)

---OOXX?XX--- ===> ---OOXX-XX---

Celera’s mate pairs

attcgttgtagtggtagcctaaatgtcggtagaccttga

THEOREM

For a gapless M, the Min Fragment RemovalProblem is Polynomial

NOTENOTE: 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 X X O O - -2 - - X O X X O - -3 - - - X X O - - - 4 - - - - O O X O - 5 - - - - - X O X O

LFT(i) RGT(i)

sort according to LFT

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. InducedSubgraph on 3-regular graphs

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

WITH GAPS…..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

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.

- - - O X X O O - - O X O X - - - XX X O X X - - - - O O X - - - O O - - - - - - - X X O- - - - O O O X -

SNP conflicts

CONFLICT !

SNP conflicts

CONFLICT !

SNP conflicts

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

1 2 3 4 5 6 7 8 9 - - - O X X O O - - O X O X - - - XX X O X X - - - - O O X - - - O O - - - - - - - X X O- - - - O O O X -

SNP conflicts

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

--OOXXOO-------------OOXOOXOXXO-----------XXOXOXXX-----XXOOXOXXO-----------XOOOX-----------XXXXXO-------XXOXXOXOO------

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

Take an odd cycle in GF

--O?X???-------------O????????O-----------??O??X??-----??????X??-----------???O?-----------????X?-------X???????O------

There is a generic structure of hor-vert cycle

--O?X???-------------O????????O-----------??O??X??-----??????X??-----------???O?-----------????X?-------X???????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?X???-------------O????????O-----------??O??X??-----??????X??-----------???O?-----------????X?-------X???????O------

Must be X

--O?X???-------------O?????X??O-----------??O??X??-----??????X??-----------???O?-----------????X?-------X???????O------

Must be X

Merge the rightmost lines

--O?X???-------------O?????X--------------??O----------??????X-------------???O------------????X--------X???????O------

Still a counterexample!

Merge the rightmost lines

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

Note: Theorem not true if there are gaps

GF(M) GS(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 - X O O ? X O X - - O X O ? X X X -

Equal:conflicts with i

Different:conflicts with k

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).

We have better, D.P., algorithms, O(mn + m^2)

What if gaps ?

The PopulationThe PopulationHaplotyping problemHaplotyping problem

The input is GENOTYPE data

INPUT: G = { xx??x, ????x, xxoxx, ?x??x, oooxx }

The input is GENOTYPE data

xxoxxxxxox

oooxxxxxox

xxoxxoxxox

xxoxxxxoxx

oooxxoooxx

OUTPUT: H = { xxoxx, xxxox, oooxx, oxxox}

INPUT: G = { xx??x, ????x, xxoxx, ?x??x, oooxx }

Each genotype is explained by two haplotypes

We will define some objectives for H

1st Objective1st Objective (open research problem):

minimize |H|

2nd Objective2nd Objective based on inference rule:

xoxxooxoxx +********** =x??xoox?x?

known haplotype h

known (ambiguos) genotype g

Inference RuleInference Rule

xoxxooxoxx +xxoxooxxxo =x??xoox?x?

known haplotype h

new (derived) haplotype h’

xoxxooxoxx +xxoxooxxxo =x??xoox?x?

known haplotype h

new (derived) haplotype h’

We write h + h’ = g

g and h must be compatible to derive h’

2nd Objective (Clark, 1990)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

ooooxooo??ooxx??

oxoo FAILURE (can’t resolve xx?? )

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

- Problem is APX-hard (Gusfield,00)

- Graph-Model + Integer Programming for practical solution (G.,01)

1. expand genotypes

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 Objective3rd Objective (open research problem)Disease Detection:

INPUT: G = { xx??x, ????x, ??oxx, ?x??x, oooxx }

3rd Objective3rd Objective (open research problem)Disease Detection:

xxoxxxxxox

oooxxxxxox

xxoxxoxxox

xxoxxoooxx

oooxxoooxx

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 }

identification of snp alleles in dna sequences giuseppe lancia università di padova e celera...

fewpolymorphismsa polymorphism

colora polymorphism

feature common

sequence of nucleotidesvarying

shortest possible sequence

possible variants alleles

outsideat dna level

think of eye

Documents

lancia - acdac · pdf filelancia (lancia automobiles s.p.a....

far maco vigi lancia

19.2 multiple alleles

lancia musa owner handbook english.pdf

package alfa/fiat/lancia*

july 2021 lancia 2000 berlina sedan parts cars-lancia... ·...

lancia - fcagroup

celera networks on cloud computing

lancia parts from lancia auto ®: custodians of the proud...

02. celera assembler -...

celera and the virus celera and the virus a review of...

skemaer for -...

anticorpos anti-hla no transplante cardíaco encontro... ·...

lancia am 81. internationalen autosalon genf filelancia am...

icohtec newsletter...icohtec newsletter no 159 january 2019...

uputstvo lancia y

celera cracks genetic code - iese · celera cracks genetic...

blood genetics multiple alleles - more than 2 alleles, but...

lancia motors

lethal recessive alleles