Genome Rearrangements, Synteny, and Comparative Mapping

1
Genome Rearrangements, Synteny, and Comparative
Mapping

• CSCI 4830 Algorithms for Molecular Biology
• Debra S. Goldberg

2
Turnip vs Cabbage

• Share a recent common ancestor
• Look and taste different

3
Turnip vs Cabbage

• Comparing mtDNA gene sequences yields no
  evolutionary information
• 99 similarity between genes
• These surprisingly identical gene sequences
  differed in gene order
• This study helped pave the way to analyzing
  genome rearrangements in molecular evolution

4
Turnip vs Cabbage Different mtDNA Gene Order

• Gene order comparison

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Turnip vs Cabbage Different mtDNA Gene Order

• Gene order comparison

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Turnip vs Cabbage Different mtDNA Gene Order

• Gene order comparison

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Turnip vs Cabbage Different mtDNA Gene Order

• Gene order comparison

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Turnip vs Cabbage Gene Order Comparison

• Evolution is manifested as the divergence in gene
  order

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Transforming Cabbage into Turnip
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Reversals
1
3
2
10
9
8
4
7
5
6
1, 2, 3, -8, -7, -6, -5, -4, 9, 10

• Blocks represent conserved genes.
• In the course of evolution or in a clinical
  context, blocks 1,,10 could be misread as 1, 2,
  3, -8, -7, -6, -5, -4, 9, 10.

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Types of Mutations
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Types of Rearrangements
Inversion/Reversal
1 2 3 4 5 6
1 2 -5 -4 -3 6
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Types of Rearrangements
Translocation
1 2 3 4 5 6
1 2 6 4 5 3
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Types of Rearrangements
Fusion
1 2 3 4 5 6
1 2 3 4 5 6
Fission
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Genome rearrangements
Mouse (X chrom.)
Unknown ancestor 75 million years ago
Human (X chrom.)

• What are the similarity blocks and how to find
  them?
• What is the architecture of the ancestral genome?
• What is the evolutionary scenario for
  transforming one genome into the other?

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Why do we care?
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SKY (spectral karyotyping)
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Robertsonian Translocation
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Robertsonian Translocation

• Translocation of chromosomes 13 and 14
• No net gain or loss of genetic material normal
  phenotype.
• Increased risk for an abnormal child or
  spontaneous pregnancy loss

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Philadelphia Chromosome
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Philadelphia Chromosome

• A translocation between chromosomes 9 and 22
  (part of 22 is attached to 9)
• Seen in about 90 of patients with Chronic
  myelogenous leukemia (CML)

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Colon cancer
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Colon Cancer
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Comparative maps
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Waardenburgs Syndrome Mouse Provides Insight
into Human Genetic Disorder

• Characterized by pigmentary dysphasia
• Gene implicated linked to human chromosome 2
• It was not clear where exactly on chromosome 2

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Waardenburgs syndrome and splotch mice

• A breed of mice (with splotch gene) had similar
  symptoms caused by the same type of gene as in
  humans
• Scientists identified location of gene
  responsible for disorder in mice
• Finding the gene in mice gives clues to where
  same gene is located in humans

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Reversals Example

• p 1 2 3 4 5 6 7 8
• 
  
• 1 2 5 4 3 6 7 8

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Reversals Example

• p 1 2 3 4 5 6 7 8
• 
  
• 1 2 5 4 3 6 7 8
• 1 2 5 4 6 3 7 8

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Reversals and Gene Orders

• Gene order represented by permutation p
• p p 1 ------ p i-1 p i p i1 ------ p j-1 p
  j p j1 ----- p n
• p 1 ------ p i-1 p j p j-1 ------ p i1
  p i p j1 ----- pn
• Reversal r ( i, j ) reverses (flips) the elements
  from i to j in p

r(i,j)
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Reversal Distance Problem

• Goal Given two permutations, find shortest
  series of reversals to transform one into another
• Input Permutations p and s
• Output A series of reversals r1,rt
  transforming p into s, such that t is minimum
• t - reversal distance between p and s
• d(p, s) smallest possible value of t, given p,
  s

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Sorting By Reversals Problem

• Goal Given a permutation, find a shortest series
  of reversals that transforms it into the identity
  permutation (1 2 n )
• Input Permutation p
• Output A series of reversals r1, rt
  transforming p into the identity permutation such
  that t is minimum
• min t d(p ) reversal distance of p

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Sorting by reversalsExample 5 steps
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Sorting by reversalsExample 4 steps
What is the reversal distance for this
permutation? Can it be sorted in 3 steps?
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Pancake Flipping Problem

• Chef prepares unordered stack of pancakes of
  different sizes
• The waiter wants to sort (rearrange) them,
  smallest on top, largest at bottom
• He does it by flipping over several from the top,
  repeating this as many times as necessary

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Sorting By Reversals A Greedy Algorithm

• Unsigned permutations
• Example permutation p 1 2 3 6 4 5
• First three elements are already in order
• prefix(p) length of already sorted prefix
• prefix(p) 3
• Idea increase prefix(p) at every step

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Greedy Algorithm An Example

• Doing so, p can be sorted
• 1 2 3 6 4 5
• 1 2 3 4 6 5
• 1 2 3 4 5 6
• Number of steps to sort permutation of length n
  is at most (n 1)

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Greedy Algorithm Pseudocode

• SimpleReversalSort(p)
• 1 for i ? 1 to n 1
• 2 j ? position of element i in p (i.e., pj
  i)
• 3 if j ?i
• 4 p ? p r(i, j)
• 5 output p
• 6 if p is the identity permutation
• 7 return

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Analyzing SimpleReversalSort

• SimpleReversalSort does not guarantee the
  smallest number of reversals and takes five steps
  on p 6 1 2 3 4 5
• Step 1 1 6 2 3 4 5
• Step 2 1 2 6 3 4 5
• Step 3 1 2 3 6 4 5
• Step 4 1 2 3 4 6 5
• Step 5 1 2 3 4 5 6

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Analyzing SimpleReversalSort (contd)

• But it can be sorted in two steps
• p 6 1 2 3 4 5
• Step 1 5 4 3 2 1 6
• Step 2 1 2 3 4 5 6
• So, SimpleReversalSort(p) is not optimal
• Optimal algorithms are unknown for many problems
  approximation algorithms used

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Approximation Algorithms

• These algorithms find approximate solutions
  rather than optimal solutions
• The approximation ratio of an algorithm A on
  input p is
• A(p) / OPT(p)
• where
• A(p) -solution produced by algorithm A
  OPT(p) - optimal solution of the problem

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Approximation Ratio / Performance Guarantee

• Approximation ratio (performance guarantee) of
  algorithm A max approximation ratio of all
  inputs of size n
• For algorithm A that minimizes objective function
  (minimization algorithm)
• maxp n A(p) / OPT(p)
• For maximization algorithm
• minp n A(p) / OPT(p)

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Adjacencies and Breakpoints

• p p1p2p3pn-1pn
• A pair of elements p i and p i 1 are adjacent
  if
• pi1 pi 1
• For example
• p 1 9 3 4 7 8 2 6 5
• (3, 4) or (7, 8) and (6,5) are adjacent pairs

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Breakpoints An Example

• There is a breakpoint between any adjacent
  element that are non-consecutive
• p 1 9 3 4 7 8 2 6 5
• Pairs (1,9), (9,3), (4,7), (8,2) and (2,5) form
  breakpoints of permutation p
• b(p) - breakpoints in permutation p

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Adjacency Breakpoints

• An adjacency consecutive
• A breakpoint not consecutive

Extend p with p0 0 and pn1 n1
adjacencies
p 5 6 2 1 3 4
0 5 6 2 1 3 4 7
breakpoints
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Extending Permutations

• Add p 0 0 and p n 1n1 at ends of p
• Example

p 1 9 3 4 7 8 2 6 5
Extending with 0 and 10
p 0 1 9 3 4 7 8 2 6 5 10
Note A new breakpoint was created after extending
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Reversal Distance and Breakpoints

• Each reversal eliminates at most 2 breakpoints.
• p 2 3 1 4 6 5
• 0 2 3 1 4 6 5 7 b(p) 5
• 0 1 3 2 4 6 5 7 b(p) 4
• 0 1 2 3 4 6 5 7 b(p) 2
• 0 1 2 3 4 5 6 7 b(p) 0

This implies reversal distance breakpoints /
2
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Sorting By Reversals A Better Greedy Algorithm

• BreakPointReversalSort(p)
• 1 while b(p) gt 0
• 2 Among all possible reversals, choose
  reversal r minimizing b(p r)
• 3 p ? p r(i, j)
• 4 output p
• 5 return

Problem this algorithm may work forever
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Strips

• Strip an interval between two consecutive
  breakpoints in a permutation
• Decreasing strip elements in decreasing order
  (e.g. 6 5)
• Increasing strip elements in increasing order
  (e.g. 7 8)
• 0 1 9 4 3 7 8 2 5 6 10
• Consider single-element strips decreasing except
  strips 0 and n1 are increasing

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Reducing the Number of Breakpoints

• Theorem 1
• If permutation p contains at least one
  decreasing strip, then there exists a reversal r
  which decreases the number of breakpoints (i.e.
  b(p r) lt b(p) )

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Things To Consider

• For p 1 4 6 5 7 8 3 2
• 0 1 4 6 5 7 8 3 2 9 b(p)
  5
• Choose decreasing strip with the smallest element
  k in p (k 2 in this case)

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Things To Consider (contd)

• For p 1 4 6 5 7 8 3 2
• 0 1 4 6 5 7 8 3 2 9 b(p)
  5
• Choose decreasing strip with the smallest element
  k in p (k 2 in this case)

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Things To Consider (contd)

• For p 1 4 6 5 7 8 3 2
• 0 1 4 6 5 7 8 3 2 9 b(p)
  5
• Choose decreasing strip with the smallest element
  k in p (k 2 in this case)
• Find k 1 in the permutation

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Things To Consider (contd)

• For p 1 4 6 5 7 8 3 2
• 0 1 4 6 5 7 8 3 2 9 b(p)
  5
• Choose decreasing strip with the smallest element
  k in p (k 2 in this case)
• Find k 1 in the permutation
• Reverse segment between k and k-1 0 1
  2 3 8 7 5 6 4 9 b(p) 4

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Reducing the Number of Breakpoints Again

• If there is no decreasing strip, there may be no
  reversal r that reduces the number of
  breakpoints (i.e. b(p r) b(p) for any
  reversal r).
• By reversing an increasing strip ( of
  breakpoints stay unchanged ), we will create a
  decreasing strip at the next step. Then the
  number of breakpoints will be reduced in the next
  step (theorem 1).

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Things To Consider (contd)

• There are no decreasing strips in p, for
• p 0 1 2 5 6 7 3 4 8 b(p) 3
• p r(3,4) 0 1 2 5 6 7 4 3 8 b(p)
  3
• r(3,4) does not change the of breakpoints
• r(3,4) creates a decreasing strip thus
  guaranteeing that the next step will decrease the
  of breakpoints.

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ImprovedBreakpointReversalSort

• ImprovedBreakpointReversalSort(p)
• 1 while b(p) gt 0
• 2 if p has a decreasing strip
• Among all possible reversals, choose reversal
  r
• that minimizes b(p
  r)
• 4 else
• 5 Choose a reversal r that flips an
  increasing strip in p
• 6 p ? p r
• 7 output p
• 8 return

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Performance Guarantee

• ImprovedBreakPointReversalSort is an
  approximation algorithm with a performance
  guarantee of at most 4
• It eliminates at least one breakpoint in every
  two steps at most 2b(p) steps
• Approximation ratio 2b(p) / d(p)
• Optimal algorithm eliminates at most 2
  breakpoints in every step d(p) ? b(p) / 2
• Performance guarantee
• ( 2b(p) / d(p) ) ? 2b(p) / (b(p) / 2) 4

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Signed Permutations

• Up to this point, reversal sort algorithms sorted
  unsigned permutations
• But genes have directions so we should consider
  signed permutations

p 1 -2 - 3 4 -5
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Signed Permutation

• Genes are directed fragments of DNA

• Genes in the same position but different
  orientations do not have same gene order

• These two permutations are not equivalent gene
  sequences

1 2 3 4 5
-1 2 -3 -4 -5
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Signed permutations are easier!

• Polynomial time (optimal) algorithm is known

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Genome rearrangements
Mouse (X chrom.)
Unknown ancestor 75 million years ago
Human (X chrom.)

• What are the similarity blocks and how to find
  them?
• What is the architecture of the ancestral genome?
• What is the evolutionary scenario for
  transforming one genome into the other?

62
Genome rearrangements
Mouse (X chrom.)
Unknown ancestor 75 million years ago
Human (X chrom.)

• What are the similarity blocks and how to find
  them?
• What is the architecture of the ancestral genome?
• What is the evolutionary scenario for
  transforming one genome into the other?

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Comparative maps
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A brief history

• Chromosome comparisons
• no information about genes
• 1920s Sturtevant, Weinstein
• Today many organisms, many uses
• Humans
• primates, mouse, cat, dog, zebrafish, ...
• Alzheimer, cancers, diabetes, obesity, ...

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Why construct comparative maps?

• Identify isolate genes
• Crops drought resistance, yield, nutrition...
• Human disease genes, drug response,
• Infer ancestral relationships
• Discover principles of evolution
• Chromosome
• Gene family
• key to understanding the human genome

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Map construction
Go from this
to this
Maize 1 (target), Rice (base) Wilson et al.
Genetics 1999
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Why automate?

• Time consuming, laborious
• Needs to be redone frequently
• Codify a common set of principles
• Nadeau and Sankoff warn of arbitrary nature of
  comparative map construction

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Input/Output

• Input
• genetic maps of 2 species
• marker/gene correspondences (homologs)
• Output
• a comparative map
• homeologies identified

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A natural model?
Maize 1 (target), Rice (base) Wilson et al.
Genetics 1999
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Scoring
10L
3L
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Assumptions

• Accept published marker order
• All linkage groups of base are unique
• Simplistic homeology criteria
• At least one homeologous region

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A natural model?
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Dynamic programming

• li location of homolog to marker i
• Si,a penalty (score) for an optimal labeling
  of the submap from marker i to the end, when
  labeling begins with label a

a 1 ... i ... n
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Recurrence relation

• Sn,a m ?(a, ln)Si,a m ?(a, li) min
  (Si1,b s ?(a,b) )

a ... n ... ln
b?L
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Problem with linear model

• s 2

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The stack model
d
c
c
f
e
b
b
b
a

• Segment at top of the stack can be
• pushed (remembered), later popped
• replaced
• Push and replace cost s -- pop is free.

77
Scoring
uaz265a (7L) isu136 (2L)
isu151 (7L) rz509b (7L) cdo59c
(7L) rz698c (9L) bcd1087a (9L)
rz206b (9L) bcd1088c (9L)
csu40 (3S) cdo786a (9L) csu154
(7L) isu113a (7L) csu17 (7L)
cdo337 (3L) rz530a (7L)
7L
9L
7L
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Dynamic programming

• Si,j,a score for an optimal labeling of
• submap from marker i to marker j
• when labeling begins with label a -- i.e.,
  marker i is labeled a

a 1 ... i ... j ... n
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Recurrence relation

• Si,i,a m ?(a, li)
• Si,j,a min
• m ?(a, li) min (Si1,j,b s ?(a,b) )
• min Si,k,a Sk1,j,a

b?L
iltkltj
a a 1 ... i ... k1 ... j ... n
80
Advantage output similar to experts

• Maize 6 (target),
• Rice (base)

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Advantage proposes testable hypotheses

• New relations predictedgreater resolution maps
  confirm

Maize 7 (target), Rice (base)
Ahn-Tanksley 93 Ahn-Tanksley data Wilson et.
al. 99
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Advantage infers evolutionary events
Wilson et al.
Maize 1 (target) Rice (base)
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Problem Incomplete input

• Gene order not always fully resolved.
• Co-located genes can be ordered to give most
  parsimonious labeling.

33.0 33.0 33.0 33.0 33.0 33.0 33.0 33.0 33.0 33.0
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The reordering algorithm

• Uses a compression scheme
• Within a megalocus, group genes by location of
  related gene.
• Order these groups
• First, last groups interact with nearby genes
• Any ordering of internal groups is equally
  parsimonious

85
The reordering algorithm
86
The reordering algorithm
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Definitions

• ? extended to distance to a set A of labels
• 0 if a ? A,
• 1 otherwise
• li set of labels matching markers in
  megalocus i
• S set of megalocus start nodes

?(a, A)
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Definitions

• p(i,a,b) gives the total mismatched marker and
  segment boundary penalties attributed to hidden
  markers
• i is index for megalocus
• a and b are labels for megalocus ends
• Do any markers in megalocus match a, b?
• No dont penalize in recurrence and p(i,a,b)

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Recurrence relation

• Si,i,a m ?(a, li)

Si,j,a min m ?(a, li) min (Si1,j,b s
?(a,b) p(i,a,b)) min Si,k,a Sk1,j,a

b?L
iltkltj k ? S
90
Results Fewer mismatches

• stack reordering

Mouse 5 (target) Human (base)
91
Results Mismatches placed between segments

• stack reordering

Mouse 8 (target) Human (base)
92
Results Detects new segments

• stack reordering

Mouse 13 (target) Human (base)
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Summary

• Global view
• Finds optimal comparative map
• Arranges markers in most parsimonious way
• Biologically meaningful results
• Robust
• not species-specific
• high/low resolution, genetic/physical maps
• stable to errors in marker order