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Combinatorial Optimization in Computational Biology

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### Topic 3: Combinatorial Algorithms for Haplotype Inference

### Topic 3b: Haplotyping via Perfect Phylogeny

Some Applications of optimization methods

Dynamic Programming in sequence alignment

TSP and Euler paths in DNA sequencing

Integer Programming in Haplotype inference

Graphic Matroid Recognition in Haplotype inference

Integer Programming for Virus Identification

Weighted matching in genetic partition analysis

Apology

This talk focuses (mostly) on my own work, for simplicity of

preparation. But, many other wonderful examples exist

in the literature.

At present, there is no coherent story about the ways that

combinatorial optimization arises in Computational

Biology. Must be alert for meaningful applications.

Topic 1: Sequence Alignment

Intended to show the similarity of two sequences (strings).

Definition: An alignment of two strings S and S’ is obtained

by inserting spaces into, or at the ends of, the two strings, so

that the resulting strings (including the spaces) have the same

lengths.

Matches, Mismatches and Indels

Two aligned, identical characters in an alignment are a match.

Two aligned, unequal characters are a mismatch.

A character aligned with a space, represents an indel

(insertion/deletion).

A A C T A C T -- C C T A A C A C T -- --

-- -- C T C C T A C C T -- -- T A C T T T

the bars show mismatches

10 matches, 2 mismatches, 7 indels

Basic Algorithmic Problem

Find an alignment of the two strings that

Maximizes the # matches - # mismatches - # spaces

in the alignment of strings S and S’.

That maximum number defines the similarity of the two

strings.

Also called Optimal Global Alignment

Biologically, under the point mutation model, a mismatch

represents a mutation, while an indel represents a historical

insertion or deletion of a single character.

Solution By Dynamic Programming

Def: Let V(I, J) be the Maximimum Value obtainable in

any alignment of the first I characters of S and the first

J characters of S’.

If S has N characters and S’ has M characters, then

we want to compute V(N, M), and the associated

alignment.

Let M(I, J) = -1 if S(I) mismatches S’(J) and

M(I, J) = 1 if S(I) matches S’(J)

DP Solution

The DP recurrences are:

Base Cases

General Cases

I and J aligned

I opposite a space

J opposite a space

They can be evaluated in O(NM) operations.

Many Variations on the Theme

The most important alignment variant is Optimal Local Alignment:

Given S and S’ find a substring A of S, and a substring B of S’

such that the value of the Global Alignment of A and B is

maximized over all pairs of substrings of S and S’.

The point is to find a pair of substrings of S and S’ which are

highly similar to each other, even if S and S’ are not. Many

biological reasons for this: DNA problems, Protein Domains etc.

Naively, we would need to do O((NM)**2) global alignments,

for a time bound of O((NM)**3) operations overall.

DP solution to Local Alignment

But the optimal value is no longer V(N, M) as in global alignment.

Instead the optimal Local Alignment Value is the largest V

value computed over all the NM values. The time is

again O(NM). This is the Smith-Waterman Local Alignment

Algorithm.

Adding Operation Weights

Find an alignment of the two strings that

Maximizes

Vm x (# matches) - Cm x (# mismatches) - Cs x (# spaces)

in the alignment of strings S and S’.

The modification to the DP is immediate, but now the

modeling question is what Vm, Cm and Cs should be.

Those values greatly determine the biological utility of

the alignment.

This leads to Parametric Alignment: Find the optimal alignment

as a function of unknown values of Vm, Cm, and Cs

Parametric Sequence Alignment

As a function of two selected weight parameters Vm, Cm, Cs, Cg,

decompose the two dimensional parameter space into convex

polygons, so that in each polygon, an alignment inside the polygon

is optimal for any point in the polygon, and nowhere else.

The time for this decomposition is O(NM) per polygon.

In the case of global alignment, the number of polygons is bounded

by 0.88(N^(2/3)), where N < M.

package XPARAL to find the polygons is available on the web.

Gusfield, Naor, Stelling in several papers (see the web)

Adding Gaps to the Model

A Gap in an alignment is a contiguous run of spaces in one sequence.

has six spaces in three gaps

Example: A T T -- -- -- C C --

-- -- T C T G C C C

Now the objective function for alignment is Maximize

Vm x (#matches) - Cm x (#mis) - Cs x (#spaces) - Cg x (#gaps)

This is the affine gap model, the dominant model for studying

protein sequence similarity. The choice of Cs and Cg are critical for the

biological utility. The DP in this case can also be solved in O(NM) time.

Topic 2: Sequencing By Hybridization

A proposed approach to sequencing DNA

The technology allows you to determine all the K-tuples,

for a fixed K, that are contained in a DNA string S.

The goal is to try to determine the string S from the given

K-tuples.

One can cast this problem as a Hamilton Path problem on

a graph, but it is more productive to view it as an Euler

Path Problem.

This approach was introduced by Pavel Pavzner.

I will focus on the results in

Graph Traversals, Genes and Matroids: An Efficient Special Case

of the Traveling Salesman Problem

D. Gusfield, R. Karp, L. Wang, P. Stelling

Discrete Applied Math 88 special issue on

Computational Molecular Biology, 1998, p. 167-180.

Sequencing by Hybridization

Example: S =

ACTTAAACGCGCACAA

K = 3

So we learn that ACT, CTT etc. are in S, but we learn these

in no predictable order.

The Hamilton Path View

CGC

ACG

add a C on the

left

Each node represents a K-tuple, and a directed edge from one

node to another if the second K-tuple can be obtained from the

first by deleting the left character, and adding a right character:

i.e, a shift and add.

Then a Hamilton Path in this graph generates a sequence S’

that has the same K-tuples as S

The Euler Path View

The Euler Path view is more productive, leading to more

efficient algorithms both for the base problem, and for

variants of the problem.

Due to P. Pevzner

Graph G: One node for every k-1 tuple, and one

edge for every observed k-tuple

A

A

C

AA

AC

CA

A

C

A

A

GC

G

T

C

TA

A

T

CG

CT

TT

Observed triples from S: AAA, AAC, ACA, CAA, CAC, ACG, CGC,

GCA, GCC, ACT, CTT, TTA, TAA

Any Euler path in G produces a string which has the same

K-tuples as the original string S. So if there is only one

Euler path in G, then you have reconstructed S.

The problem is that typically there is more than one Euler

path. So how to choose one Euler path as a most promising

one? That is, one that is most likely to correspond to S?

Frequently there is additional information about the K-tuples,

for example the rough distances between K-tuples in S. So,

we can give a value to each pair of successive edges in G,

which roughly reflects the probability that the two edges

would appear successively in the Euler Path for S. Another

way to state this is that we evaluate the goodness of S’

according to the (k+1)-tuples that it contains.

Back to the Hamilton View

Now we are back to the Hamilton Path view, where each

node is a K-tuple of S, but now each

edge has a distance reflecting the goodness of any (K+1)-tuple

that would be generated by a Hamilton path, and we want the

Maximum distance Hamilton Path in the graph.

This sounds hard, but there is structure in this problem: the

graphs are not arbitrary.

The structure of the Hamilton Graph

The Hamilton Graph for the data, where nodes represent

K-tuples in S, and edges represent (K+1)-tuples in a

solution string S’, is the line di-graph of the Euler Graph

for the data, where nodes represent (K-1) tuples, and

edges represent K-tuples.

The Hamilton graph is the

line di-graph L(G) of the Euler Graph G.

Each edge of G becomes a node in L(G), and each edge in L(G)

represents a successive pair of edges in G.

Each red edge has the value of the successive pair of black edges.

So the problem now is to find a Maximum Value TSPath.

In the case that each node has only two in and two out edges

(which happens when using a simplified DNA alphabet),

this TSP problem has a polynomial time solution.

In fact, if L(G) is any line digraph where every node has

in and out degrees at most 2, then even for arbitrary edge

distances, the TSP problem can be solved in O(n log n) time.

Moreover, if every edge value is distinct, then the TSP solution

is unique.

The Hamilton paths for such 2-in, 2-out-degree graphs have

a matroid structure that reveals many features of the TSP solutions.

When degrees are greater than 2, we branch-and-bound down to

cases of degree 2, which we then can solve optimally.

Approximations

When the in and out-degrees are bounded by X, a greedy

algorithm is guaranteed to find a Hamilton path whose

value is within 1/X that of the optimal. So 1/3 or 1/4

for the case of DNA.

Use of this approximation can speed up the branch-and-

bound until the in and out-degree bound of 2 is reached.

True Parsimony and Perfect Phylogeny

Dan Gusfield

Genotypes and Haplotypes

Each individual has two “copies” of each chromosome.

At each site, each chromosome has one of two alleles (states) denoted by 0 and 1 (motivated by

SNPs)

0 1 1 1 0 0 1 1 0

1 1 0 1 0 0 1 0 0

Two haplotypes per individual

Merge the haplotypes

2 1 2 1 0 0 1 2 0

Genotype for the individual

Haplotyping Problem

- Biological Problem: For disease association studies, haplotype data is more valuable than genotype data, but haplotype data is hard to collect. Genotype data is easy to collect.
- Computational Problem: Given a set of n genotypes, determine the original set of n haplotypepairs that generated the n genotypes. This is hopeless without a genetic model.

We consider two models in this talk

- Pure (True) Parsimony Model, Gusfield unpublished (Note added 3/18/03 technical report is available on the web, and will appear in the 2003 CPM conference)
- Perfect Phylogeny Model, Gusfield RECOMB Proceedings April 2002

Topic 3a: The Pure Parsimony Objective

For a set of genotypes, find a Smallest set H of

haplotypes, such that each genotype can be

explained by a pair of haplotypes in H.

Example of Parsimony

00100

01110

02120

3 distinct haplotypes

set S has size 3

01110

10110

22110

00100

10110

20120

Pure Parsimony is NP-hard

Earl Hubbel (Affymetrix) showed that Pure Parsimony

is NP-hard.

However, for a range of parameters of current interest

(number of sites and genotypes) a Pure Parsimony

solution can be computed efficiently, using Integer

Linear Programming, and one speed-up trick.

For larger parameters (100 sites and 50 genotypes)

A near-parsimony solution can be found efficiently.

The Conceptual Integer Programming Formulation

For each genotype (individual) j, create one integer

programming variable Yij for each pair of haplotypes

whose merge creates genotype j. If j has k 2’s, then

This creates 2^(k-1) Y variables.

Create one integer programming variable Xq for

Each distinct haplotype q that appears in one of the

pairs for a Y variable.

Conceptual IP

For each genotype, create an equality that says that

exactly one of its Y variables must be set to 1.

For each variable Yij, whose two haplotypes are

given variables Xq and Xq’, include an inequality

that says that if variable Yij is set to 1, then both

variables Xq and Xq’ must be set to 1.

Then the objective function is to Minimize the

sum of the X variables.

Example

02120 Creates a Y variable Y1 for pair 00100 X1

01110 X2

and a Y variable Y2 for pair 01100 X3

00110 X4

Include the following (in)equalities into the IP

The objective function will

include the subexpression

X1 + X2 + X3 + X4

But any X variable is included

exactly once no matter how many

Y variables it is associated with.

Y1 + Y2 = 1

Y1 - X1 <= 0

Y1 - X2 <= 0

Y2 - X3 <= 0

Y2 - X4 <= 0

Efficiency Trick

Ignore any Y variable and its two X variables if those X

variables are associated with no other Y variable. The

Resulting IP is much smaller, and can be used to find

the optimal to the conceptual IP.

Also, we need not first enumerate all X pairs for a given

genotype, but can efficiently recognize the pairs we

need. Another simple trick.

How Fast? How Good?

Depends on the level of recombination in the underlying

data. Pure Parsimony can be computed in seconds to

minutes for most cases with 50 genotypes and up to 60

sites, faster as the level of recombination increases.

As the level of recombination increases, the accuracy

of the Pure Parsimony Solution falls, but remains within

10% of the quality of PHASE (for comparison).

Conceptual Framework and Efficient (almost linear-time) Solutions

Dan Gusfield

U.C. Davis

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.

The Perfect Phylogeny Model

sites

12345

Ancestral haplotype

00000

1

4

Site mutations on edges

3

00010

2

10100

5

10000

01010

01011

Extant haplotypes at the leaves

Justification for Perfect Phylogeny Model

- In the absence of recombination each haplotype of any individual has a single parent, so tracing back the history of the haplotypes in a population gives a tree.
- Recent strong evidence for long regions of DNA with no recombination. Key to the NIH haplotype mapping project. (See NYT October 30, 2002)
- Mutations are rare at selected sites, so are assumed non-recurrent.
- Connection with coalescent models.

The Haplotype PhylogenyProblem

Given a set of genotypes S, find an explaining set

of haplotypes that fits a perfect phylogeny.

sites

A haplotype pair explains a

genotype if the merge of the

haplotypes creates the

genotype. Example: The

merge of 0 1 and 1 0 explains

2 2.

S

Genotype matrix

The Haplotype PhylogenyProblem

Given a set of genotypes, find an explaining set

of haplotypes that fits a perfect phylogeny

The Haplotype PhylogenyProblem

Given a set of genotypes, find an explaining set

of haplotypes that fits a perfect phylogeny

00

1

2

b

00

a

a

b

c

c

01

01

10

10

10

When does a set of haplotypes to fit a perfect phylogeny?

Classic NASC: 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

Solving the Haplotype Phylogeny Problem (PPH)

- Simple Tools based on classical Perfect Phylogeny Problem.
- Complex Tools based on Graph Realization
Problem (graphic matroid realization).

Simple Tools – treeing the 1’s

- For any row i in S, the set of 1 entries in row i specify the exact set of mutations on the path from the root to the least common ancestor of the two leaves labeled i, in every perfect phylogeny for S.
- The order of those 1 entries on the path is also the same in every perfect phylogeny for S, and is easy to determine by “leaf counting”.

In any column c, count two for each 1, and

count one for each 2. The total is the number

of leaves below mutation c, in every perfect

phylogeny for S. So if we know the set of

mutations on a path from the root, we know

their order as well.

Leaf CountingS

Count 4 4 2 2 1 1 1

The Subtree of 1’s is Forced

The columns of S with at least one 1 entry

specify a unique upper subtree that must be

in every perfect phylogeny for S.

2

1

4

3

S

a

a

b

b

4 4 2 2 1 1 1

Corollary: A Useful Sufficient Condition

If every column of S has at least one 1, then there is a unique perfect phylogeny for S, if there is any.

Build the forced tree based on the 1-entries, then every mutation (site) is in the tree, and any 2-entries dictate where the remaining leaves must be attached to the tree.

What to do when the Corollary does not apply?

- Build the forced tree of 1-entries.
- 2-entries in columns with 1-entries are done.
- Remaining problem: 2-entries in columns without any 1-entries

Uniqueness without 1’s

There are 8 haplotype solutions

that can explain the data, but

only one that fits a perfect

phylogeny, i.e. only one solution

to the PPH problem.

More Simple Tools

- For any row i in S, and any column c, if S(i,c) is 2, then in every perfect phylogeny for S, the path between the two leaves labeled i, must contain the edge with mutation c.
Further, every mutation c on the path between the two i leaves must be from such a column c.

From Row Data to Tree Constraints

Subtree for row i data

sites

Root

1 2 3 4 5 6 7

i:0 1 0 2 212

The order is known for the red mutations

2

6

4,5,7

order

unknown

i

i

The Graph Theoretic Problem

Given a genotype matrix S with n sites, and a red-blue fork for each row i,

create a directed tree T where each

integer from 1 to n labels exactly one

edge, so that each fork is

contained in T.

i

i

Complex Tool: Graph Realization

- Let Rn be the integers 1 to n, and let P be an unordered subset of Rn. P is called a path set.
- A tree T with n edges, where each is labeled with a unique integer of Rn, realizes P if there is a contiguous path in T labeled with the integers of P and no others.
- Given a family P1, P2, P3…Pk of path sets, tree T realizes the family if it realizes each Pi.
- The graph realization problem generalizes the consecutive ones problem, where T is a path.

Graph Realization Example

5

P1: 1, 5, 8

P2: 2, 4

P3: 1, 2, 5, 6

P4: 3, 6, 8

P5: 1, 5, 6, 7

1

8

6

2

4

3

7

Realizing Tree T

Graph Realization

Polynomial time (almost linear-time) algorithms exist for the graph realization problem – Whitney, Tutte, Cunningham, Edmonds, Bixby, Wagner, Gavril, Tamari, Fushishige.

The algorithms are not simple; no known implementations of the near linear-time methods.

But an implementation from UCD CS of a slightly

slower method is available on the web.

Recognizing graphic Matroids

- The graph realization problem is the same problem as determining if a binary matroid is graphic, and the algorithms come from that literature.
- The fastest algorithm is due to Bixby and Wagner. For n rows and m columns the time is O(nm x alpha(nm).
- Representation methods due to Cunningham et al.

Reducing PPH to graph realization

We solve any instance of the PPH problem by creating appropriate path sets, so that a solution to the resulting graph realization problem leads to a solution to the PPH problem instance.

The key issue: How to encode a fork by path sets.

From Row Data to Tree Constraints

Subtree for row i data

sites

Root

1 2 3 4 5 6 7

i:0 1 0 2 212

The order is known for the red mutations

2

6

4,5,7

order

unknown

i

i

Encoding a Red path, named g

2

P1: U, 2

P2: U, 2, 5

P3: 2, 5

P4: 2, 5, 7

P5: 5, 7

P6: 5, 7, g

P7: 7, g

U

5

2

7

5

forced

In T

7

g

U is a universal glue edge, used for all red paths.

g is used for all red paths identical to this one – i.e.

different forks with the identical red path.

Encoding and gluing a blue path

For a specific blue path, simply specify a single

path set containing

the integers in the blue path. However, we need to

force that path to be glued to the end of a specific

red path g, or to the root of T.

In the first case, just add g to the path set. In the

second case, add U to the path set.

After the Reduction

After all the paths are given to the graph realization algorithm, and a realizing tree T is obtained (assuming one is), contract all the glue edges to obtain a perfect phylogeny for the original haplotyping problem (PPH) instance.

Conversely, any solution can be obtained in

this way. The reduction takes time O(nm) for n rows and m columns.

Uniqueness

In 1932, Whitney established the NASC for the uniqueness of a solution to a graph realization problem:

Let T be the tree realizing the family of required path sets. For each path set Pi in the family, add an edge in T between the ends of the path realizing Pi. Call the result graph G(T).

T is the unique realizing tree, if and only if

G(T) is 3-vertex connected. e.g., G(T) remains connected after the removal of any two vertices.

Uniqueness of PPH Solution

- Apply Whitney’s theorem, using all the paths implied by the reduction of the PPH problem to graph realization.
- (Minor point) Do not allow the removal of the endpoints of the universal glue edge U.

Multiple Solutions

- In 1933, Whitney showed how multiple solutions to the graph realization problem are related.
- Partition the edges of G into two, connected graphs, each with at least two edges, such that the two graphs have exactly two nodes in common. Then twist one graph around those nodes.
- Any solution can be transformed to another via a series of such twists.

Twisting Example

2

2

3

1

3

1

x

y

x

y

8

8

4

4

7

7

5

5

6

6

All the cycle sets are preserved by twisting

Representing all the solutions

All the solutions to the PPH problem can be

implicitly represented in a data structure which

can be built in linear time, once one solution is

known. Then each solution can be generated in

linear time per solution. Method is a small

modification of ones developed by Cunningham

and Edwards, and by Hopcroft and Tarjan.

See Gusfield’s web site for errata.

The biological meaning

Each subgraph in the representation identifies a set of

mutations (columns) whose 0/1 settings are invariant

over all PPH solutions. However, by switching all

0/1 values that had been 2, in all of those mutations,

the relative 0/1

settings between columns in different subgraphs can

be set arbitrarilly, and all PPH solutions can be

generated in this way.

Starting from a PPH

Solution, if all

shaded cells in a

block switch value,

then the result is

also a PPH solution,

and any PPH

solution can be

obtained in this

way, i.e. by

choosing in each

block whether to

switch or not.

a

a

b

b

c

c

d

d

e

e

Secondary information and optimization

- The partition shows explicitly what added information is useful and what is redundant. Information about the relationship of a pair of columns is redundant if and only if they are in the same block of the column partition. Apply this successively as additional information is obtained.
- Problem: Minimize the number of haplotype pairs (individuals) that need be laboratory determined in order to find the correct tree.
- Minimize the number of (individual, site1, site2) triples whose phase relationship needs to be determined, in order to find the correct tree.

The implicit representation of all solutions provides

a framework for solving these secondary problems,

as well as other problems involving the use of

additional information, and specific tree-selection

criteria.

A Phase-Transition

Problem, as the ratio of sites to genotypes changes,

how does the probability that the PPH solution is

unique change?

For greatest utility, we want genotype data where the

PPH solution is unique.

Intuitively, as the ratio of genotypes to sites increases,

the probability of uniqueness increases.

Frequency of a unique solution with 50 and 100 sites, 5% rule and 2500 datasets per entry

# geno. Frequency of

unique solution

An alternative solution rule and 2500 datasets per entry

Recently, we developed an algorithm that is not based

on graph realization, and which is much easier to understand.

However, it runs in O(nm^2) time. See Gusfield’s website.

Implementations rule and 2500 datasets per entry

See wwwcsif.cs.ucdavis.edu/~gusfield/

for papers and software.

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