Transcript Graphs and Graph Theory in Computational Biology

Graphs and Graph Theory in
Computational Biology
Dan Gusfield
Miami University, May 15, 2008
(four hour tutorial)
Some examples of graphs in
biology
• Taken from the web - see the citations for
details. Many other examples of graphs
more complex than trees in biology.
QuickTime™ and a
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are needed to see this picture.
From Max Delbrueck Center, Berlin
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Yeast protein
interactions
From http://www-personal.umich.edu/~mejn/networks/
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Protein-Protein Interactions
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Protein-Protein Interaction Modelling
Dr. Peter Uetz
Institut fur Toxikologie und Genetik Forschungszentrum Karlsruhe
NY Times May 5, 2008 The Diseasome
http://www.nytimes.com/interactive/2008/05/05/science/20080506_DISEASE.html
Graphs and Graph Theory
1. Numerous uses of graphs and networks to
represent biological phenomena at many
conceptual levels. Maybe several 1000s of papers
using graph representations, particularly trees, but
little graph theory.
2. A respectable number of papers that develop new
non-trivial graph theory for problems in biology.
100s of papers, maybe 1000.
3. A handful of papers exploiting or extending nontrivial classic graph theory for problems in
biology. Perhaps a few hundred.
Introduction and Conclusion
Very diverse biological applications and very
diverse graph theory. So no single grand reason
for graphs and no single graph topic in biology.
Lots of opportunity for graph theorists and graph
algorithmists to develop or apply graph theory to
biological problems. Even more opportunity for
combinatorial optimization.
What I will do in this tutorial
• Emphasis on points 2 and 3, i.e., Examples of the
development of new non-trivial graph theory, and
of the exploitation of classic graph theory. And
(my apologies) I will mostly emphasize topics I
have been involved with.
Still,
• There are some hot biological areas today where
graphs arise, and some graph topics that recur
commonly, and I should point those out even if I
will not talk in detail on those topics.
The digression
• Hot biology: Network biology -- biological
phenomena that are represented by networks -gene regulatory networks and protein interaction
networks, just to name two. These form the core
of Systems biology. Other relationships in biology
represented by graphs and networks. Ex.
diseasome.
• Recurring graph problems: graph problems in
clustering data ( ex. finding cliques or variants of
cliques); variants of graph isomorphism in
network motif or molecular pathway problems;
need for more random graph theory for
significance testing
Clique Problems
Clique problems are recurrent in clustering
applications, but true cliques are computationally
hard to find. Suggested research for graph theorist
and algorithmists: computationally tractable,
biologically meaningful alternatives to cliques. As
examples: maximum density subgraphs; extreme
sets in a graph.
Subgraph density
• Given a graph G, and a subset S of its nodes, let
G(S) be the subgraph of G induced by S, i.e, G(S)
has node set S and edge set E(S) consisting of all
edges in G both of whose ends are in S.
• A Maximum Density subgraph of G is induced by
the set of nodes S which the Maximizes |E(S)|/|S|.
• The maximum density subgraph can be found in
polynomial time. It has the flavor of a maximum
clique, but has different properties.
Extreme Sets
In an edge-weighted undirected graph G, a
subset S of nodes of G is called an extreme
set if for every subset S’ of S, the total
weight of the edges crossing from S’ to V-S’
is larger than the total weight of the edges
crossing from S to V-S.
All the extreme sets in a graph can be found in
polynomial time.
Also
There is also a great need for more sophisticated
application of random graph theory in the study of
biological networks. This is needed in order to
establish null models to use in assessing the
statistical significance of subgraphs, paths,
patterns and motifs that are found in biological
networks.
We need to be able to distinguish observed patterns
and subgraphs from those that occur with a high
probability in a random graph, under a
biologically appropriate model of randomness (an
open field).
End of digression
Start of the main tutorial: Examples
of Graph Theory in Bioinformatics
and Computational Biology
Outline
• Three Smaller examples: Euler paths and
sequencing; Tanglegrams and co-evolution;
Network Design and Multiple Alignment.
• Haplotyping by Perfect Phylogeny: Graph
Realization.
• Phylogenetic Networks: Incompatibility Graph;
Galled-Trees; Recombination Networks; The
Decomposition Theorem and sufficient conditions.
• Multi-state Perfect Phylogeny and Chordal
Graphs.
To start: Three small examples
1. Euler paths in sequencing and sequence
assembly.
2. Tanglegrams and planarity testing in the study of
co-evolution.
3. Application of Tree-Design approximations in
multiple sequence alignment. Interplay between
trees and strings.
Topic I: Eulerian paths in
sequencing problems
The general situation is that we have a (DNA say)
molecule S whose sequence is unknown, but
we know all the k-mers that occur in S, for some
fixed k. Given those k-mers, we want to determine
S, if possible, or determine whatever is possible to
determine about S. Note that k is not related to the
alphabet size.
A very useful approach to problems of this type is to
build an Eulerian digraph, based on the (k-1)mers.
Euler graph for general k
For general k, there is one node for each (k-1)-mer contained in
an observed k-mer. Then there is a directed edge from the node for (k-1)mer A
to the node for (k-1)mer B, if the
(k-2) suffix of A matches the (k-2) prefix of B, so that A and B
can be overlapped to form the observed k-mer.
Example: k = 5 and we observe the 5-mer XXYZW.
Then there will be a node for XXYZ and a node for XYZW
and a directed edge from the first node to the second node. Those
two nodes and the directed edge between them represent the
5-mer XXYZW. In some applications, there will be one such edge for each
observation of that 5-mer.
Ex. k = 3. The graph will have one node for each of the 2-mers
in the observed 3-mers. Then there is a directed edge from
the node for the 2-mer XY to the node for the 2-mer YZ, for any X, Z.
The Euler graph derived from the sequence
ACACGCAACTTAAA
If a triple is observed more than once, there should be
One directed edge for each observation of the triple.
The point: Every Eulerian path in the graph specifies a
sequence whose k-mers match the given data, and conversely
every sequence whose k-mers match the data specifies an
Eulerian path in the graph. So the set of Eulerian paths specifies
the set of candidate sequences for the unknown original sequence.
Algorithms exist for efficiently finding Eulerian paths, for
counting their number, for determining uniqueness etc. so
we can use this representation to study the set of candidate
sequences.
Compare this approach to earlier efforts to represent the set of
candidates by a graph with a Hamilton path: each node represents
an observed k-mer, not a (k-1)-mer.
Making finer distinctions in Euler paths
In general there may be many Eulerian paths in the graph,
and we want some additional criteria to distinguish the goodness
of one Eulierian path compared to another.
Different biological considerations translate into having a value
for each subpath of length two. Then the value of an Eulerian path
P with n edges is the sum of the n-1 values of the n-1 length-two
subpaths in P.
The problem is to find an Eulerian path with maximum value.
We have some reasonable approximations for that, but a simpler
case can be solved optimally in polynomial time.
The case of a binary alphabet, but
arbitrary k
Since the alphabet size is two, each node in
the graph has at most two incoming edges
and two outgoing edges. Assume exactly two
each. 001
110
Ex. k = 4
101
011
110
The case of a binary alphabet, but
arbitrary k
At any node, there are two possible ways for
an Euler path to pass through the node.
001
110
turning
Ex. k = 4
101
011
110
The case of a binary alphabet, but
arbitrary k
At any node, there are two possible ways for
an Euler path to pass through the node.
001
Ex. k = 4
101
110
011
crossing
110
So in terms of subpaths of length two, we have two choices at each
node.
Restating the optimal Euler path
problem
We are given an Eulerian graph where the in
and out degrees are at most two at each
node, and at each node there is a given
value for the turning pair, and a value for
the crossing pair. Then choose the turning or
the crossing pairs at the nodes to maximize
the total value of the choices, subject to the
requirement that the choices create an Euler
path in the graph.
Main Result
• The problem can be solved in polynomial
time.
• The set of choices that give Euler paths has
a matroidal structure, which allows a
matroid-greedy algorithm to find the
optimal Euler path.
• A more direct algorithm based on Minimum
Spanning Trees also solves the problem.
The Matroid Structure
• At every node v, the edge pair (crossing or
turning) which has the lowest value is called the
low pair, and the other pair is the high pair. The
difference in values is called the loss at v.
• A subset S of nodes is called independent if there
is an Euler path in the graph where at every node
in S, the low pair is chosen.
• As defined, the family of independent sets form a
matroid, and so we can find, by a greedy
algorithm, an independent set which minimizes
the loss - and this gives the optimal Euler path.
Topic II: Tanglegrams
• A Tanglegram is a pair of trees drawn in the plane
with no crossing edges, with the same labeled leaf
set. The leaves of one tree are displayed on a line,
and the leaves of the other tree are displayed on a
parallel line.
• A straight line connect each leaf in one tree to the
leaf with the same label in the other tree.
• The number of crossing lines is a measure of the
similarity of the trees.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
Topic III: Multiple Sequence
Alignment
Interplay between sequences and trees.
Exploitation of network design approximation.
Intro to Hours 2 and 3: Two
“Post-HGP” Topics
Two topics in Population Genomics
• SNP Haplotyping in populations
• Reconstructing a history of recombination
These topics in Population Genomics illustrate
current challenges in biology, and illustrate the use
of graph theory, combinatorial algorithms and
discrete mathematics in biology.
What is population genomics?
• The Human genome “sequence” is done.
• Now we want to sequence many individuals
in a population to correlate similarities and
differences in their sequences with genetic
traits (e.g. disease or disease susceptibility).
• Presently, we can’t sequence large numbers
of individuals, but we can sample the
sequences at SNP sites.
SNP Data
• A SNP is a Single Nucleotide Polymorphism - a site in the
genome where two different nucleotides appear with
sufficient frequency in the population (say each with 5%
frequency or more).
• SNP maps have been compiled with a density of about 1
site per 1000.
• SNP data is what is mostly collected in populations - it is
much cheaper to collect than full sequence data, and
focuses on variation in the population, which is what is of
interest.
Haplotype Map Project:
HAPMAP
• NIH lead project ($100M) to find common SNP
haplotypes (“SNP sequences”) in the Human population.
• Association mapping: HAPMAP used to try to associate
genetic-influenced diseases with specific SNP haplotypes,
to either find causal haplotypes, or to find the region near
causal mutations.
• The key to the logic of Association mapping is historical
recombination in populations. Nature has done the
experiments, now we try to make sense of the results.
Topic IV: Perfect Phylogeny
Haplotyping via Graph
Realization
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 haplotype pairs that
generated the n genotypes. This is hopeless without a
genetic model.
The Perfect Phylogeny Model for
SNP sequences
Only one mutation per site
allowed.
sites 12345
Ancestral sequence 00000
1
4
Site mutations on edges
3
The tree derives the set M:
2
10100
10100
5
10000
01011
01010
00010
10000
00010
01010
01011
Extant sequences at the leaves
When can a set of sequences be
derived on a perfect phylogeny?
Classic NASC: Arrange the sequences in a
matrix. Then (with no duplicate columns),
the sequences can be generated on a unique
perfect phylogeny if and only if no two
columns (sites) contain all four pairs:
0,0 and 0,1 and 1,0 and 1,1
This is the 4-Gamete Test
So, in the case of binary characters, if each pair of columns
allows a tree, then the entire set of columns allows a tree.
For M of dimension n by m, the existence of a perfect phylogeny
for M can be tested in O(nm) time and a
tree built in that time, if there is one. Gusfield, Networks 91
We will use the classic theorem in two more modern
and more genetic applications.
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.
Justification: Haplotype Blocks, rare recombination, base problem whose
solution to be modified to incorporate more biological complexity.
Perfect Phylogeny Haplotype (PPH)
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
a
S
b
haplotypes creates the
c
genotype. Example: The
Genotype matrix merge of 0 1 and 1 0 explains
2 2.
1
2
0
1
2
2
2
0
The PPH Problem
Given a set of genotypes, find an explaining set
of haplotypes that fits a perfect phylogeny
a
b
c
1
2
0
1
2
2
2
0
a
a
b
b
c
c
1
1
0
0
0
1
1
2
0
1
0
1
0
0
The Haplotype Phylogeny Problem
Given a set of genotypes, find an explaining set
of haplotypes that fits a perfect phylogeny
00
a
b
c
1
2
0
1
2
2
2
0
a
a
b
b
c
c
1
1
0
0
0
1
1
2
0
1
0
1
0
0
1
2
b
00
cc
10 10
a
10
a
01
b
01
The Alternative Explanation
a
b
c
1
2
0
1
2
2
2
0
a
a
b
b
c
c
1
1
0
0
0
1
1
2
1
0
0
1
0
0
No tree
possible
for this
explanation
Efficient Solutions to the PPH
problem - n genotypes, m sites
• Reduction to a graph realization problem (GPPH) - build on BixbyWagner or Fushishige solution to graph realization O(nm alpha(nm))
time. Gusfield, Recomb 02
• Reduction to graph realization - build on Tutte’s graph realization
method O(nm^2) time. Chung, Gusfield 03
• Direct, from scratch combinatorial approach -O(nm^2) Bafna, Gusfield
et al JCB 03
• Berkeley (EHK) approach - specialize the Tutte solution to the PPH
problem - O(nm^2) time.
• Linear-time solutions - Recomb 2005, Ding, Filkov, Gusfield and a
different linear time solution.
The Reduction Approach
This is the original polynomial time method.
Conceptually simplest at a high level (but
not at the implementation level) and most
extendable to other problems; nearly lineartime but not linear-time.
The case of the 1’s
1) 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.
2) 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”.
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.
1 2 3 4 5 6 7
S
a
b
c
d
1
0
1
2
0
1
2
2
1
0
0
0
0
1
0
0
0
0
2
0
0
0
0
2
0
0
2
0
Count 5 4 2 2 1 1 1
Simple Conclusions
sites
1234567
i:0 1 0 1 2 2 2
Subtree for row i data
Root
2
4
5
The order is
known for the red
mutations
together with the
leftmost blue(?)
mutation.
But what to do with the
remaining blue entries (2’s) in a
row?
More Simple Tools
3) 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
sites
1234567
Subtree for row i data
Root
2
4
i:0 1 0 1 2 2 2
5
i
i
Edges 5, 6 and 7
must be on the blue path,
and 5 is already known to
follow 4, but we don’t
where to put 6 and 7.
The Graph Theoretic Problem
Given a genotype matrix S with n sites, and a
red-blue subgraph for each row i,
create a directed tree T where each
integer from 1 to n labels exactly one
edge, so that each subgraph is
i
i contained in T.
Powerful Tool: Tree and 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.
• More generally, each set specifies a fundamental cycle in the unknown
graph.
Tree Realization Example
P1: 1, 5, 8
P2: 2, 4
P3: 1, 2, 5, 6
P4: 3, 6, 8
P5: 1, 5, 6, 7
5
1
2
8
4
6
7
3
Realizing Tree T
More generally, think of each path set as specifying a fundamental
cycle containing the edges in the specified path.
Graph Realization
Polynomial time (almost linear-time) algorithms exist for
the graph realization problem, given the family of
fundamental cycles the unknown graph should contain –
Whitney, Tutte, Cunningham, Edmonds, Bixby, Wagner,
Gavril, Tamari, Fushishige, Lofgren 1930’s - 1980’s
Most of the literature on this problem is in the context of
determining if a binary matroid is graphic.
The algorithms are not simple; none implemented before
2002.
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 the needed subgraph
for each row, and glue them together at the root.
From Row Data to Tree
Constraints
sites
1234567
Subtree for row i data
Root
2
4
i:0 1 0 1 2 2 2
5
i
i
Edges 5, 6 and 7
must be on the blue path,
and 5 is already known to
follow 4.
Encoding a Red-Blue directed
path
2
4
5
P1: U, 2
P2: U, 2, 4
P3: 2, 4
forced
P4: 2, 4, 5 In T
P5: 4, 5
U
2
4
5
U is a glue edge used to glue together the directed
paths from the different rows.
Now add a path set for the blues
in row i.
sites
1234567
Root
2
4
i:0 1 0 1 2 2 2
5
i
i
P: 5, 6, 7
That’s the Reduction
The resulting path-sets encode everything that is
known about row i in the input.
The family of path-sets are input to the graphrealization problem, and every solution to the
that graph-realization problem specifies a solution to
the PPH problem, and conversely.
Whitney (1933?) characterized the set of all solutions to graph
realization (based on the three-connected components of a graph)
and Tarjan et al showed how to find these in linear time.
An implicit representation of all
solutions
Whitney (1930) proved that a graph realization problem has a
unique solution if and only if the graph is three-connected. That
is, at least three nodes must be removed in order to disconnect
the graph (assuming it is connected).
Whitney (1931) proved that if the solution is not unique, then
there is a semi-unique decomposition of the graph into threeconnected components, so that the graph realizations are in oneone correspondence with all the ways that these components can
be ``twisted” relative to each other. So the number of solutions
is 2^(number of three connected comps. -1).
Tree Realization Example
P1: 1, 5, 8
P2: 2, 4
P3: 1, 2, 5, 6
P4: 3, 6, 8
P5: 1, 5, 6, 7
5
1
2
8
4
6
7
3
Realizing Tree T with
edges added to create a
fundamental cycle for each
path
Topic V: Phylogenetic Networks
with Recombination
When can a set of sequences be
derived on a perfect phylogeny?
Classic NASC: Arrange the sequences in a
matrix. Then (with no duplicate columns),
the sequences can be generated on a unique
perfect phylogeny if and only if no two
columns (sites) contain all four pairs:
0,0 and 0,1 and 1,0 and 1,1
This is the 4-Gamete Test
Incompatible Sites
A pair of sites (columns) of M that fail the
4-gametes test are said to be incompatible.
A site that is not in such a pair is compatible.
A richer model
M
10100
10000
01011
01010
00010
10101 added
Pair 4, 5 fails the four
gamete-test. The sites 4, 5
are incompatible.
12345
00000
1
4
3
10100
2
00010
5
10000
0101101010
Real sequence histories often involve recombination.
Sequence Recombination
01011
10100
S
P
5
Single crossover recombination
10101
A recombination of P and S at recombination point 5.
The first 4 sites come from P (Prefix) and the sites
from 5 onward come from S (Suffix).
Network with Recombination:
ARG
M
10100
10000
01011
01010
00010
10101 new
12345
00000
1
4
3
2
10100
The previous tree with one
recombination event now derives
all the sequences.
P
00010
5
10000
5
10101
S
0101101010
A Min ARG for Kreitman’s data
ARG created by
SHRUB
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
An illustration of why we are
interested in recombination:
Association Mapping of
Complex Diseases Using
ARGs
Association Mapping
• A major strategy being practiced to find genes
influencing disease from haplotypes of a subset of
SNPs.
– Disease mutations: unobserved.
• A simple example to explain association mapping
and why ARGs are useful, assuming the true ARG is
known.
Disease mutation site
0
1
0
SNPs
0
1
Very Simplistic Mapping the Unobserved
Mutation of Mendelian Diseases with ARGs
The single disease
00000
mutation occurs
4
00010
between
a:00010 sites 2 and
3
What part of 01100 3!10010
1
d, e, f inherit?
00100
1 2 3 4 5
2
b:10010
d:
01100
P
S
e:
c:00100
P
3
f:
? ?
Where is the
disease mutation?
d:10100
Assumption (for now):
A sequence is diseased
iff it carries the single
disease mutation
5
00101
S
4
01101
g:00101
f:01101
e:01100
Diseased
Mapping Disease Gene with
Inferred ARGs
• “..the best information that we could possibly
get about association is to know the full
coalescent genealogy…” – Zollner and
Pritchard, 2005
• But we do not know the true ARG!
• Goal: infer ARGs from SNP data for
association mapping
– Not easy and often approximation (e.g. Zollner and
Pritchard)
– Improved results to do the inference Y. Wu (RECOMB 2007)
Results on Reconstructing the
Evolution of SNP Sequences
• Part I: Clean mathematical and algorithmic results: Galled-Trees, nearuniqueness, graph-theory lower bound, and the Decomposition
theorem
• Part II: Practical computation of Lower and Upper bounds on the
number of recombinations needed. Construction of (optimal)
phylogenetic networks; uniform sampling; haplotyping with ARGs;
LD mapping …
• Part III: Varied Biological Applications
• Part IV: Extension to Gene Conversion
• Part V: The Minimum Mosaic Model of Recombination
This talk will discuss topics in Parts I
Problem: If not a tree, then what?
If the set of sequences M cannot be derived on
a perfect phylogeny (true tree) how much
deviation from a tree is required?
We want a network for M that uses a small
number of recombinations, and we want the
resulting network to be as ``tree-like” as
possible.
A tree-like network
for the same
sequences generated
by the prior network.
4
3
1
p s
a: 00010
2
b: 10010
c: 00100
d: 10100
2
p
e: 01100
5
4
s
f: 01101
g: 00101
Recombination Cycles
• In a Phylogenetic Network, with a
recombination node x, if we trace two paths
backwards from x, then the paths will
eventually meet.
• The cycle specified by those two paths is
called a ``recombination cycle”.
Galled-Trees
• A phylogenetic network where no
recombination cycles share an edge is called
a galled tree.
• A cycle in a galled-tree is called a gall.
• Question: if M cannot be generated on a
true tree, can it be generated on a galledtree?
Results about galled-trees
• Theorem: Efficient (provably polynomial-time) algorithm to determine
whether or not any sequence set M can be derived on a galled-tree.
• Theorem: A galled-tree (if one exists) produced by the algorithm
minimizes the number of recombinations used over all possible
phylogenetic-networks.
• Theorem: If M can be derived on a galled tree, then the Galled-Tree is
``nearly unique”. This is important for biological conclusions derived
from the galled-tree.
Papers from 2003-2007.
Elaboration on Near Uniqueness
Theorem: The number of arrangements (permutations) of the
sites on any gall is
at most three, and this happens only if the gall has two
sites.
If the gall has more than two sites, then the number of
arrangements is at most two.
If the gall has four or more sites, with at least two sites
on each side of the recombination point (not the side of
the gall) then the arrangement is forced and unique.
Theorem: All other features of the galled-trees for M are invariant.
A whiff of the ideas behind the
results
Incompatible Sites
A pair of sites (columns) of M that fail the
4-gametes test are said to be incompatible.
A site that is not in such a pair is compatible.
M
12345
a 00010
b 10010
c 00100
d 10100
e 01100
f 01101
g 00101
Incompatibility Graph G(M)
4
1
3
2
5
Two nodes are connected iff the pair
of sites are incompatible, i.e, fail the
4-gamete test.
THE MAIN TOOL: We represent the pairwise
incompatibilities in a incompatibility graph.
The connected components of
G(M) are very informative
• Theorem: The number of non-trivial connected components is a lowerbound on the number of recombinations needed in any network.
• Theorem: When M can be derived on a galled-tree, all the
incompatible sites in a gall must come from a single connected
component C, and that gall must contain all the sites from C.
Compatible sites need not be inside any blob.
•
In a galled-tree the number of recombinations is exactly the number of
connected components in G(M), and hence is minimum over all
possible phylogenetic networks for M.
4
Incompatibility Graph
4
3
1
1
3
2
5
p s
a: 00010
2
b: 10010
c: 00100
d: 10100
2
p
e: 01100
5
4
s
f: 01101
g: 00101
A Graph Theoretic Necessary
Condition for a Galled-Tree
If M can be generated on a galled-tree, then
the incompatibility graph must be a bipartite
bi-convex graph. Other structural
properties
of the conflict graph can be deduced and
exploited.
Galled-Tree Haplotyping
Problem: Given genotype matrix G, if there
is no PPH solution for G, is there a
haplotyping H for G such that H can be
derived on a Galled-Tree?
A different Neccessary Condition
for a one-gall tree
1. There exists a set of sequences S such that
for every pair of incompatible sites p,q, a
single p,q state-pair appears in all sequences
in S, and does not appear in any sequence
outside S.
2. There must be a number x such that
p < x < q, for each incompatible pair p,q.
Example
4
3
H
1
a: 000100
p s
2
b: 100100
6
a
b
c
d
e
f
g
000100
100100
001000
101000
110001
011000
001010
c: 001000
d: 101000 2
5
e:1010001
S = {e,d} the sequences
below the recombination node.
f: 011000
g: 001010
Surprising Result - Yun Song
The necessary condition is also sufficient.
Yun S. Song in TCBB 2006
Coming full circle - back to
genotypes
When can a set of genotypes be explained by a set of
haplotypes derived on a galled-tree, rather than on a
perfect phylogeny?
The Song NASC can be translated into an ILP,
using the part of the
MinIncompat ILP that identifies which site pairs are
incompatibile.
For the one gall problem, the ILP formulation
solves very efficiently (200 rows x 40 sites
in seconds to minutes). So far, the 2-gall
case does not solve well (ongoing work).
(Dan Brown, Gusfield 2006).
Change of Scope: Minimizing
Recombinations in unconstrained
networks
• Problem: given a set of sequences M, find a phylogenetic
network generating M, minimizing the number of
recombinations used to generate M, allowing only one
mutation per site. This has biological meaning in
appropriate contexts.
• We can solve this problem in poly-time for the special case
of Galled-Trees.
• The minimization problem is NP-hard in general.
Minimization is an NP-hard
Problem
What we have done:
1. Solve small data-sets optimally with exponential-time methods
or with algorithms that work well in practice;
2. Efficiently compute lower and upper bounds on the number of
needed recombinations.
3. Apply these methods to address specific
biological and bio-tech questions.
The Decomposition Theorem
Since the minimization problem is NP-hard
we want to break up a problem into
subproblems that can be solved separately
and combined.
M
12345
a 00010
b 10010
c 00100
d 10100
e 01100
f 01101
g 00101
Incompatibility Graph G(M)
4
1
3
2
5
Two nodes are connected iff the pair
of sites are incompatible, i.e, fail the
4-gamete test.
THE MAIN TOOL: We represent the pairwise
incompatibilities in a incompatibility graph.
The connected components of
G(M) are very informative
For example we have the Theorem:
The number of non-trivial connected components is a lower-bound on the
number of recombinations needed in any network.
Recombination Cycles
• In a Phylogenetic Network, with a
recombination node x, if we trace two paths
backwards from x, then the paths will
eventually meet.
• The cycle specified by those two paths is
called a ``recombination cycle”.
A maximal set of intersecting
cycles forms a Blob
00010
10010
00000
4
3
1
00100
2
P
S
3
01100
p
5
00101
S
4
01101
If directions on the edges are removed, a blob is
a bi-connected component of the network.
Blobed Trees
• Contracting each blob in a network results in a directed,
rooted tree, otherwise one of the “blobs” was not maximal.
Simple, but key insight.
• So every phylogenetic network can be viewed as a directed
tree of blobs - a blobbed-tree.
The blobs are the non-tree-like parts of the network.
Every network is a
tree of blobs.
Ugly tangled
network inside
the blob.
A network
where every blob is a
single cycle is a
Galled-Tree.
A Simple Observation
In any network N for M, all sites from the
same connected component of G(M) must
appear together in a single blob in N.
The Decomposition Theorem
Theorem: For any set of sequences M, there is a
phylogenetic network that derives M, where each blob
contains all and only the sites in one non-trivial connected
component of G(M). The compatible sites can always be
put on edges outside of any blob. This
“fully-decomposed” network is the finest decomposition
possible.
Example: Network for input M
with one blob
00010
a:00010
10010
00000
4
3
1
00100
2
b:10010
P
c:00100
S
3
01100
p
d:10100
5
00101
S
4
01101
f:01101
e:01100
g:00101
The fullydecomposed
network for M
4
Incompatibility Graph
4
3
1
1
3
2
5
p s
a: 00010
2
b: 10010
c: 00100
d: 10100
2
p
e: 01100
5
4
s
f: 01101
g: 00101
Moreover, the backbone tree is invariant
over all the fully-decomposed networks
for M, and can be determined in
polynomial-time.
So, we can find a network for M by solving
the recombination minimization problem for
each connected component of G(M)
separately, and then connect those
subnetworks in an invariant way.
Algorithmically
• Finding the tree part of the blobbed-tree is easy.
• Determining the sequences labeling the exterior nodes on any blob is
easy.
• Determining a “good” structure inside a blob B is the problem of
generating the sequences of the exterior nodes of B.
• It is easy to test whether the exterior sequences on B can be generated
with only a single recombination. The original galled-tree problem is
now just the problem of testing whether one single-crossover
recombination is sufficient for each blob.
• That can be solved by successively removing each exterior sequence
and testing if the remaining sequences can be generated on a perfect
phylogeny of the correct form.
However …
While fully-decomposed networks always exist, they
do not necessarily minimize the number of
recombination nodes, over all possible networks.
That is, sometimes it pays to put sites from different
connected components together on the same blob.
Sufficient Conditions
But we can prove several useful sufficient conditions
for when there is a fully-decomposed network that minimizes the
number of recombinations, over all possible networks.
The deepest result:
Theorem: Let N be a phylogenetic network for input M, let L be the
set of sequences that label the nodes of N, and let G(L) be the
incompatibility graph for L. If G(L) and G(M) have the same number
of connected components, then there is a fully-decomposed network for
M with the same number of recombinations as in N.
JCB December 2007
Corollary
A fully-decomposed network exists that
minimizes the number of recombinations,
unless every optimal network uses some
recombination node(s) labeled by sequence(s)
not in M, and the addition of those sequences
to M creates an incompatibility between sites
in different components of G(M).
000000
4
3
5
Sequences in M are in black.
Sequence 100010 is not in M.
1
G(M) has two
components. Each
requires two recs, but
this combined network
needs only three.
4
s
p 6
2
100010
p
001000
3 s 5 s
p
000100
0011010
010010 100001
100101
G(L) has one component. The addition of
sequence 100010
reduces the number of components from 2 to 1.
A Practical Sufficient Condition
If M can be derived on a network N in which
every edge contains at most
one site, and every node is labeled with a
sequence in M, then there is a
fully-decomposed network for M which
minimizes the number of recombinations
over all possible networks for M.
Another Practical Sufficient
Condition
If M can be derived on a network N where
the number of recombinations equals the
(poly-computable) Haplotype Lower Bound,
then there is a fully decomposed network
for M which minimizes the number of
recombinations over all possible networks.
Topic VI: Perfect Phylogeny
Extension to non-binary
characters
We detail the case of three allowed states per
character.
What is a Perfect Phylogeny for
non-binary characters?
• Input consists of n sequences M with m sites (characters) each, where
each site can take one of k states.
• In a Perfect Phylogeny T for M, each node of T is labeled with an mlength sequence where each site has a value from 1 to k.
• T has n leaves, one for each sequence in M, labeled by that sequence.
• For each character-state pair (C,s), the nodes of T that are labeled with
state s for character C, form a connected subtree of T. It follows that
the subtrees for any C are node-disjoint
Example: A perfect phylogeny
for input M
(3,2,1)
(2,3,2)
A B
3
2 2
3 3
4 1
5 1
1
(3,2,3)
(3,2,3)
(1,2,3)
2
3
2
1
2
M
(1,1,3)
(1,2,3)
n=5
m=3
k=3
C
1
2
3
3
3
Example
(3,2,1)
(2,3,2)
A B
3
2 2
3 3
4 1
5 1
1
(3,2,3)
(3,2,3)
The tree for
State 2 of
Character B
(1,2,3)
2
3
2
1
2
M
(1,1,3)
(1,2,3)
n=5
m=3
k=3
C
1
2
3
3
3
Perfect Phylogeny Problem
Given M, is there a Perfect Phylogeny for M?
Chordal Graphs
Basic Definition: A graph G is called Chordal if every cycle of
length four or more contains a chord.
More useful result: A graph G is chordal if and only if every
minimal vertex separator in G is a clique.
Chordal graphs have a large number of applications, more based
on the separator result than on the basic definition. For example,
a chordal graph on n nodes can have at most n maximal cliques
and n-1 minimal vertex separators.
Another Classic Chordal Graph
Theorem
A graph G is chordal if and only if it is the intersection graph of
a set S of subtrees of a tree T. Each node of G is a member of S.
{b,c}
c
b
{a,e,g}
g
d
a
{c,d,e,g}
e
T
f
{b,c,d}
G
{a,e}
{e,f,g}
Relation to Perfect Phylogeny
In a perfect phylogeny T for a table E, for any character C
and any state X of character C, the sub-forest of T
induced by the nodes labeled (C,X) form a single, connected
subtree of T.
So, there is a natural set of subtrees of T induced by E.
Chordal Completion Approach to
Perfect
Phylogeny
A B C
1
2
3
4
5
3
2
3
1
1
2
3
2
1
2
1
2
3
3
3
Table E
A
B
C
1
1
1
2
2
2
3
3
3
G(E)
Each row of table E induces a clique in G(E).
Graph G(E) has
one node for each
character-state pair
in E, and an edge
between two nodes
if and only if there
is a row in E with
both those
character-state
pairs.
Classic Theorem
Note that if table E has K columns, then G(E) is a
K-partite graph.
Theorem (Buneman 196?)
There is a perfect phylogeny for table E if and only if
edges can be added to graph G(E) to make it a chordal,
K-partite graph. If there is such a chordal graph, denote it
by G’(E).
Deeper Result: If G’(E) exists
• Let C(E) be the graph derived from graph G’(E) as
follows: create a node in C(E) for each maximal
clique in G’(E), and create an edge (u,v) in C(E)
iff the cliques for u and v in G’(E) share a node.
Weight edge (u,v) by the number of shared nodes.
Note that C(E) can be created from G’(E) in
polynomial time.
• Any Maximum Spanning Tree T in C(E) is a
perfect phylogeny for E. Actually, T can be found
more directly in linear time from G’(E).
Perfect Phylogeny Results
The perfect phylogeny problem was open for about 20 years,
but solved by Dress, Steel, Warnow and Kannan, Agarwalla and
Fernandez-Baca.
For any fixed bound on the number of states per character,
the Perfect Phylogeny Problem can be solved in polynomial time.
However, if the number of states per character is not bounded,
then the problem is NP-Complete.
Also, for any fixed number of characters, the problem can be
solved in polynomial time.
Dress-Steel solution for 3-state
Perfect phylogeny given
complete data (1991)
• Recode each site M(i) of M as three binary sites
M’(i,1), M’(i,2), M’(i,3) each indicating the taxa
that have state 1, 2, or 3.
• Theorem (DS) There is a 3-state perfect
phylogeny for M, if and only if there is a binarycharacter perfect phylogeny for some subset of M’
consisting of exactly two of the columns
M’(i,1), M’(i,2), M’(i,3), for each column i of M.
Example
M
A,1 A,2 A,3 B,1 B,2 B,3 C,1 C,2 C,3
A B C
3
2 2
3 3
4 1
5 1
1
2
3
2
1
2
M’
1
2
3
3
3
1
0 0 1 0 1 0 1 0 0
2
0 1 0 0 0 1 0 1 0
3 0 0 1 0 1 0 0 0 1
4
1 0 0 1 0 0 0 0 1
5 1 0 0 0 1 0 0 0 1
Compatible subset
Solved in Poly-Time by 2-SAT
As stated, the problem still seems like it would take exponential
time to solve, but in fact it is easy to code the problem as a
2-SAT problem (Y. Wu) and hence is solvable in polynomial
time. The Dress-Steel paper gave an independent poly-time
solution.