Orthologs vs. Paralogs Three events: 1. Specia4on Orthologs : genes - - PDF document

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CSCI1950‐Z Computa4onal Methods for Biology Lecture 12

Ben Raphael March 9, 2009

hFp://cs.brown.edu/courses/csci1950‐z/

Orthologs vs. Paralogs

Three events:

• 1. Specia4on

Orthologs: genes descended from a common ancestor.

• 2. Loss
• 3. Duplica4on

Paralogs: genes related by duplica4on.

Dis4nguishing orthologs from paralogs is difficult! Sequence similarity is not enough.

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Whole‐Genome Phylogeny

Rearrangements can be used for phylogene4cs.

Cell Division and Muta4on

Single nucleo4de change

Copy number Structural

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History of Chromosome X

Rat Consortium, Nature, 2004

Genome Rearrangement Models

• Sor4ng By Reversals

– Naïve Greedy Algorithm – Breakpoints and Greedy algorithm

• Breakpoint Graphs
• Mul4chromosomal Rearrangements

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Reversals: Biology

5’ ATGCCTGTACTA 3’ 3’ TACGGACATGAT 5’ 5’ ATGTACAGGCTA 3’ 3’ TACATGTCCGAT 5’ Break and Invert

Reversals: Simplified Representa4on

Blocks represent segments of the genome conserved across mul4ple species

• 1. Genes
• 2. Synteny blocks (iden4fied by comparing genome

sequences)

1 3 2 4 10 5 6 8 9 7

1, 2, 3, 4, 5, 6, 7, 8, 9, 10

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Reversals

1 3 2 4 10 5 6 8 9 7

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 incorrectly copied as 1, 2, 3, 8, 7, 6, 5, 4, 9, 10.

Reversals and Breakpoints

1 3 2 4 10 5 6 8 9 7

1, 2, 3, 8, 7, 6, 5, 4, 9, 10

The inversion/reversal introduced two breakpoints (disruptions in order).

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Reversals: Example

π = 1 2 3 4 5 6 7 8 ρ(3,5) 1 2 5 4 3 6 7 8

Reversals: Example

π = 1 2 3 4 5 6 7 8 ρ(3,5) 1 2 5 4 3 6 7 8 ρ(5,6) 1 2 5 4 6 3 7 8

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Reversals and Gene Orders

Gene order is represented by a permuta4on π: : π = π 1 ------ π i-1 π i π i+1 ------ π j-1 π j π j+1 ----- π n π 1 ------ π i-1 π j π j-1 ------ π i+1 π i π j+1 ----- πn Reversal ρ ( i, j ) reverses (flips) the elements from i to j in π ρ(ι,j)

Reversal Distance

For phylogene4c reconstruc4ons, need a distance between “genomes”: permuta4ons π and σ. Reversal distance d(π, σ) is the minimum number of

reversals ρ1,…ρt transforming π into σ.

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Reversal Distance Problem

Goal: Given two permuta4ons, find the shortest series of reversals that transforms one into another

• Input: Permuta4ons π and σ.
• Output: A series of reversals ρ1,…ρt transforming π into σ,

such that t is minimum. Note: A varia4on: Output t, reversal distance.

Sor4ng By Reversals Problem

By relabeling, can consider one permuta4on to be the iden4ty permuta4on (1 2 … n ). Input: Permuta4on π Output: A series of reversals ρ1, … ρt transforming π into the iden4ty permuta4on: ρ1 … ρt π = (1 2 … n) such that t is minimum.

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Sor4ng By Reversals: Example

• t =d(π ) ‐ reversal distance of π
• Example :

π = 3 4 2 1 5 6 7 10 9 8 4 3 2 1 5 6 7 10 9 8 4 3 2 1 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 So d(π ) = 3

Pancake Flipping Problem

• The chef is sloppy; he prepares

an unordered stack of pancakes

• f different sizes
• The waiter wants to rearrange

them (so that the smallest winds up on top, and so on, down to the largest at the boFom)

• He does it by flipping over

several from the top, repea4ng this as many 4mes as necessary

Christos Papadimitrou and Bill Gates flip pancakes

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Pancake Flipping Problem: Formula4on Goal: Given a stack of n pancakes, what is the minimum number of flips to rearrange them into perfect stack?

• Input: Permuta4on π
• Output: A series of prefix reversals ρ1, … ρt

transforming π into the iden4ty permuta4on such that t is minimum. A prefix reversal is ρ(1, k) for some k. Pancake Flipping Problem: Greedy Algorithm

• Greedy approach: 2 prefix reversals at most to

place a pancake in its right posi4on, 2n – 2 steps total

• William Gates and Christos Papadimitriou

showed in the mid‐1970s that this problem can be solved by at most 5/3 (n + 1) prefix reversals

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Sor4ng By Reversals: A Greedy Algorithm

• If sor4ng permuta4on π = 1 2 3 6 4 5, the first

three elements are already in order so it does not make any sense to break them.

• The length of the already sorted prefix of π is

denoted prefix(π)

– prefix(π) = 3

• This results in an idea for a greedy algorithm:

increase prefix(π) at every step

• Doing so, π 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 permuta4on of

length n is at most (n – 1)

Greedy
Algorithm:
An
Example


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Greedy Algorithm: Pseudocode

SimpleReversalSort(π) 1 for for i  1 to n – 1 2 j  position of element i in π (i.e., πj = i) 3 if if j ≠i 4 π  π * ρ(i, j) 5 output

• utput π

6 if if π is the identity permutation 7 return return

Analyzing SimpleReversalSort

• SimpleReversalSort does not guarantee the

smallest number of reversals and takes five steps on π = 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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• But it can be sorted in two steps:

π = 6 1 2 3 4 5 – Step 1: 5 4 3 2 1 6 – Step 2: 1 2 3 4 5 6

• So, SimpleReversalSort(π) is not op4mal
• Op4mal algorithms are unknown for many

problems; approxima4on algorithms are used

Analyzing
SimpleReversalSort
(cont’d)


π = π1π2π3…πn‐1πn

• A pair of elements π i and π i + 1 are adjacent if

πi+1 = πi + 1

• For example:

π = 1 9 3 4 7 8 2 6 5

• (3, 4) or (7, 8) and (6,5) are adjacent pairs

Adjacencies and Breakpoints

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There is a breakpoint between any adjacent element that are non‐consecu4ve: π = 1 9 3 4 7 8 2 6 5

• Pairs (1,9), (9,3), (4,7), (8,2) and (2,5) form

breakpoints of permuta4on π

• b(π) ‐ # breakpoints in permuta4on π

Breakpoints: An Example

• We put two elements π 0 =0 and π n + 1=n+1 at the

ends of π Example:

Extending with 0 and 10

Note: A new breakpoint was created after extending

Extending Permuta4ons

π = 1 9 3 4 7 8 2 6 5 π = 0 1 9 3 4 7 8 2 6 5 10

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• Each reversal eliminates at most 2 breakpoints.

π = 2 3 1 4 6 5 0 2 3 1 4 6 5 7 b(π) = 5 0 1 3 2 4 6 5 7 b(π) = 4 0 1 2 3 4 6 5 7 b(π) = 2 0 1 2 3 4 5 6 7 b(π) = 0

Reversal Distance and Breakpoints

• Each reversal eliminates at most 2 breakpoints.
• This implies:

reversal distance ≥ #breakpoints / 2 π = 2 3 1 4 6 5 0 2 3 1 4 6 5 7 b(π) = 5 0 1 3 2 4 6 5 7 b(π) = 4 0 1 2 3 4 6 5 7 b(π) = 2 0 1 2 3 4 5 6 7 b(π) = 0

Reversal Distance and Breakpoints

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Sor4ng By Reversals: A BeFer Greedy Algorithm

BreakPointReversalSort(π) 1 while while b(π) > 0 2 Among all possible reversals, choose reversal ρ minimizing b(π • ρ) 3 π  π • ρ(i, j) 4 output

• utput π

5 return return

Sor4ng By Reversals: A BeFer Greedy Algorithm

BreakPointReversalSort(π) 1 while while b(π) > 0 2 Among all possible reversals, choose reversal ρ minimizing b(π • ρ) 3 π  π • ρ(i, j) 4 output

• utput π

5 return return

Problem: Will this algorithm terminate?

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Strips

• Strip: an interval between two consecu4ve

breakpoints in a permuta4on – Decreasing strip: strip of elements in decreasing

• rder (e.g. 6 5 and 3 2 ).

– Increasing strip: strip of elements in increasing

• rder (e.g. 7 8)

0 1 9 4 3 7 8 2 5 6 10

A single‐element strip can be declared either increasing or decreasing. We will choose to declare them as decreasing with excep4on of the strips with 0 and n+1

Reducing the Number of Breakpoints Theorem 1:

If permuta4on π contains at least one decreasing strip, then there exists a reversal ρ which decreases the number of breakpoints (i.e. b(π • ρ) < b(π) )

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Proof & Example Proof & Example

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Proof & Example

0 1 2 3 8 7 5 6 4 9

b(π) = 4

0 1 4 6 5 7 8 3 2 9

b(π) = 5

Reducing the Number of Breakpoints Again

What if no decreasing strip?

– There may be no reversal ρ that reduces the number of breakpoints; i.e. b(π • ρ) ≥ b(π) for any reversal ρ. – In this case, reverse an increasing strip:

• # of breakpoints remains unchanged
• Creates a decreasing strip at the next step.
• Number of breakpoints will be reduced in the next step

(Theorem 1).

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Example

• There are no decreasing strips in π, for:

π = 0 1 2 5 6 7 3 4 8 b(π) = 3 π • ρ(6,7) = 0 1 2 5 6 7 4 3 8 b(π) = 3  ρ(6,7) does not change the # of breakpoints  ρ(6,7) creates a decreasing strip thus guaranteeing that the next step will decrease the # of breakpoints.

ImprovedBreakpointReversalSort

ImprovedBreakpointReversalSort(π) 1 while while b(π) > 0 2 if if π has a decreasing strip 3 Among all possible reversals, choose reversal ρ that minimizes b(π • ρ) 4 else else 5 Choose a reversal ρ that flips an increasing strip in π 6 π  π • ρ 7 output

• utput π

8 return return

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• ImprovedBreakPointReversalSort is an approxima4on

algorithm

• Claim: Performance guarantee of at most 4

– Eliminates at least one breakpoint in every two steps  at most 2b(π) steps – Approxima4on ra4o: 2b(π) / d(π) – Op4mal algorithm eliminates at most 2 breakpoints in every step: d(π) ≥ b(π) / 2 – Performance guarantee:

• ( 2b(π) / d(π) ) ≤ [ 2b(π) / (b(π) / 2) ] = 4

ImprovedBreakpointReversalSort: Performance Guarantee

Breakpoint Graph

1) Represent the elements of the permuta4on π = 2 3 1 4 6 5 as ver4ces in a graph (ordered along a line)

0 2 3 1 4 6 5 7

2) Connect ver4ces in order given by π with black edges (black path) 3) Connect ver4ces in order given by 1 2 3 4 5 6 with grey edges (grey path) 4) Superimpose black and gray paths

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Two Equivalent Representa4ons of the Breakpoint Graph

0 2 3 1 4 6 5 7 0 1 2 3 4 5 6 7

Showing gray path instead of black path

• n a horizontal line, gives identical graph

What is the Effect of the Reversal ?

0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7

• The gray paths stayed the same for both graphs
• There is a change in the graph at this point
• There is another change at this point

How does a reversal change the breakpoint graph? Before: 0 2 3 1 4 6 5 7 After: 0 2 3 5 6 4 1 7

• Other black edges are unaffected by the reversal so they remain the

same for both graphs

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A reversal affects 4 edges in the breakpoint graph

0 1 2 3 4 5 6 7

• A reversal removes 2 edges (red) and replaces them with 2

new edges (blue)

• Gray edges unchanged

Maximum Cycle Decomposi4on

• Since the identity permutation of size n contains the maximum cycle

decomposition of n+1, c(identity) = n+1

• Breakpoint graph can be decomposed into edge-disjoint alternating (gray-

black) cycles.

• Let c(π) =number of alternating cycles in maximal decomposition

0 2 3 1 4 6 5 7 0 1 2 3 4 5 6 7

c(π) = ? 4

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Effects of Reversals

Case 1: Both edges belong to the same cycle

• Remove the center black edges and replace them with new black

edges (there are two ways to replace them)

• (a) After this replacement, there now exists 2 cycles instead of 1 cycle

c(πρ) – c(π) = 1

This is called a proper reversal since there’s a cycle increase after the reversal.

• (b) Or after this replacement, there still exists 1 cycle

c(πρ) – c(π) = 0 Therefore, after the reversal c(πρ) – c(π) = 0 or 1

Effects of Reversals (Con4nued)

Case 2: Both edges belong to different cycles

• Remove the center black edges and replace them with new black edges
• After the replacement, there now exists 1 cycle instead of 2 cycles

c(πρ) – c(π) = -1 Therefore, for every permutation π and reversal ρ, | c(πρ) – c(π) | ≤ 1

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Reversal Distance and Maximum Cycle Decomposi4on

• Since the identity permutation of size n contains the maximum cycle

decomposition of n+1, c(identity) = n+1

• c(identity) – c(π) equals the number of cycles that need to be “added”

to c(π) while transforming π into the identity

• Based on the previous theorem, at best after each reversal, the cycle

decomposition could increased by one, then: d(π) = c(identity) – c(π) = n+1 – c(π)

• Reversal distance problem is NP-hard (Caprara 1997)

Therefore, d(π) ≥ n+1 – c(π)

• Yet, not every reversal can increase the cycle decomposition

Signed Permuta4ons

• Up to this point, all permuta4ons to sort were

unsigned

• But genes (and DNA) have direc4ons… so we

should consider signed permuta4ons

5’ 3’

π = 1 -2 ‐ 3 4 -5

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Reversals and Breakpoints

1 3 2 4 10 5 6 8 9 7

1, 2, 3, -8, -7, -6, -5, -4, 9, 10

The inversion/reversal introduced two breakpoints (disruptions in order).

Sor4ng by reversals: 5 steps

Step 0: π 2 -4 -3 5 -8 -7 -6 1 Step 1: 2 3 4 5 -8 -7 -6 1 Step 2: 2 3 4 5 6 7 8 1 Step 3: 2 3 4 5 6 7 8 -1 Step 4:

• 8 -7 -6 -5 -4 -3 -2 -1

Step 5: γ 1 2 3 4 5 6 7 8

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Sor4ng by reversals: 4 steps

Step 0: π 2 -4 -3 5 -8 -7 -6 1 Step 1: 2 3 4 5 -8 -7 -6 1 Step 2:

• 5 -4 -3 -2 -8 -7 -6

1 Step 3:

• 5 -4 -3 -2 -1

6 7 8 Step 4: γ 1 2 3 4 5 6 7 8

Sor4ng by reversals: 4 steps

Step 0: π 2 -4 -3 5 -8 -7 -6 1 Step 1: 2 3 4 5 -8 -7 -6 1 Step 2:

• 5 -4 -3 -2 -8 -7 -6

1 Step 3:

• 5 -4 -3 -2 -1

6 7 8 Step 4: γ 1 2 3 4 5 6 7 8

What is the reversal distance for this permutation? Can it be sorted in 3 steps?

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Sor4ng Permuta4ons by Reversals

(Sankoff et al.1990)

Polynomial 4me algorithms

O(n4) : Hannenhalli and Pevzner, 1995. O(n2) : Kaplan, Shamir, Tarjan, 1997. O(n) [distance t] : Bader, Moret, and Yan, 2001. O(n3) : Bergeron, 2001.

Reversal ρ(i,j) [inversion] π = π1π2…πn signed permuta4on Problem: Given π, find a sequence of reversals ρ1, …, ρt with such that: π . ρ1 . ρ2 … ρt = (1, 2, …, n) and t is minimal.

π1…πi‐1 ‐πj ... ‐πi πj+1…πn SoluDon: Analysis of breakpoint graph

From Signed to Unsigned Permuta4on

0 +3 -5 +8 -6 +4 -7 +9 +2 +1 +10 -11 12

• Begin by constructing a normal signed breakpoint graph
• Redefine each vertex x with the following rules:
• If vertex x is positive, replace vertex x with

vertex 2x-1 and vertex 2x in that order

• If vertex x is negative, replace vertex x with

vertex 2x and vertex 2x-1 in that order

• The extension vertices x = 0 and x = n+1 are

kept as it was before

0 3a 3b 5a 5b 8a 8b 6a 6b 4a 4b 7a 7b 9a 9b 2a 2b 1a 1b 10a 10b 11a 11b 23 0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23

+3 -5 +8 -6 +4 -7 +9 +2 +1 +10 -11

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From Signed to Unsigned Permuta4on (Con4nued)

0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23

• Construct the breakpoint graph as usual
• Notice the alternating cycles in the graph between every other vertex

pair

• Since these cycles came from the same signed vertex, we will not be

performing any reversal on both pairs at the same time; therefore, these cycles can be removed from the graph

Increasing Number of Cycles

Proper reversal acts on black edges: c(ρ π) – c (π) = 1

x y

x+1 y+1

x y

x+1 y+1 ρ

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Breakpoint graph ⇒ rearrangement scenario

Oriented and Unoriented Cycles

• Oriented Cycles

E

• Unoriented Cycles

No proper reversal acting on an unoriented cycle These are “impediments” in sorting by reversals. Proper reversal acts on black edges: c(ρ π) – c (π) = 1

x y

x+1 y+1

x y

x+1 y+1 ρ

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Reversal Distance with Hurdles

• Hurdles are obstacles in the genome rearrangement problem
• They cause a higher number of required reversals for a permutation

to transform into the identity permutation

• 1
• 2

3 2 3 1 2 1 3

• 3
• 2

1

c(π) = 2, n +1 – c(π) =2

Every hurdle can be transformed into oriented cycles by reversal

• n arbitrary cycle in hurdle.

h(π) = 1

Safe Reversals

Let Δc = c(ρ π) – c (π) Δh(ρ π) – h(π) A reversal p is safe if Δc – Δh = 1.

• Oriented Cycles
• Unoriented Cycles

Proper reversal acts on black edges: c(ρ π) – c (π) = 1

x y

x+1 y+1

x y

x+1 y+1 ρ

• 1
• 2

3 2 1 3

c(π) = 2, h(π) = 1 c(π) = 2, h(π) = 0

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Interleaving Edges

0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23

• Interleaving edges are grey edges that cross each other

These 2 grey edges interleave

Example: Edges (0,1) and (18, 19) are interleaving

• Cycles are interleaving if they have an interleaving edge

Interleaving Graphs

0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23

An Interleaving Graph is defined on the set of cycles in the

Breakpoint graph and are connected by edges where cycles are interleaved A B C E F

0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23

A B C E F D D A B C E F

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Interleaving Graphs

0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23

Label oriented cycles. Component oriented if contains oriented cycle. A B C E F

0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23

A B C E F D D A B C E F

Interleaving Graphs

Remove oriented components from interleaving graph. A E A E F F B C B C D A E F C D B

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Hurdles

Particular type of oriented component A E A E A E

h(π) = 1

Hurdles

Define: Containment partial order on unoriented components. U < W if [Umin, Umax] contained in [Wmin, Wmax] Hurdle: Minimal or maximal element under partial order: Red and Green unoriented components

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Reversal Distance with Hurdles

• Hurdles are obstacles in the genome rearrangement problem
• They cause a higher number of required reversals for a permutation

to transform into the identity permutation

• Taking into account of hurdles, the following formula gives a

tighter bound on reversal distance:

d(π) ≥ n+1 – c(π) + h(π)

• Let h(π) be the number of hurdles in permutation π

Every hurdle can be transformed into oriented cycles by reversal

• n arbitrary cycle in hurdle.

** Doing so, might cause problems with overlapping hurdles

Superhurdles

• “Protect” non‐hurdles

– Dele4on of superhurdles creates another hurdle

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Superhurdles

• “Protect” non‐hurdles

– Dele4on of superhurdles creates another hurdle

Superhurdle Superhurdle

Superhurdles

• “Protect” non‐hurdles

– Dele4on of superhurdles creates another hurdle

Hurdle Hurdle

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Fortresses

• A permuta4on π with an odd number of

hurdles, all of which are superhurdles

Theorem (Hannenhalli-Pevzner 1995): d(π) = n + 1 – c(π) + h(π) + f(π) where c = # cycles; h = # hurdles f = 1 if π is fortress.

Complexity of reversal distance

Approximate solutions: Kececioglu & Sankoff (1992) (CPM '93) Bafna & Pevzner (FOCS '93) Exact solutions: Compute distance Exhibit steps Hannenhalli & Pevzner (STOC '95) O (n 2) O (n 4) Berman & Hannenhalli (CPM '96) O (n α(n )) O (n 2α(n )) Kaplan, Shamir, Tarjan (SODA '97) O (n α(n )+d_ n ) O (n 2) Bader, Moret, Yan (WADS '01) O (n ) N/A n = # of blocks α(n ) is the "Inverse Ackerman function" 1.5-approximation algorithm and introduced breakpoint graph 2-approximation algorithm

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Mul4chromosomal Sor4ng

• Concatenate chromosomes
• Transloca4ons modeled by reversals in

concatenate

• Minimal sequence in polynomial 4me (Hannenhalli

& Pevzner 1996, Tesler 2003, Ozery‐Flato and Shamir, 2003.)

B2 A1 B2 A1 ‐A2 ‐B1 A2 B1 A2 A1 B2 B1 A2 A1 ‐B2 ‐B1 concatena4on reversal concatena4on transloca4on

Addi4onal Problems

• 1. Other rearrangement opera4ons

Duplica4ons

• 2. Rearrangements and Phylogeny

MulDple Genomic Distance Problem: Given permuta4ons π1, …, πk find a permuta4on σ such that ∑k=1, k d(π1, σ) is minimal.