Genome Rearrangements and YOU

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Genome Rearrangements and YOU!!

• Presented by
• Kevin Gaittens

2
Overview

• Bio background
• Definitions and Set-up
• Reality-Desire
• Good Components
• Bad Components
• Fin

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Biological Bakground

• Comparing entire genomes across species
• Need distance measure
• Interested in larger differences than just single
  insertions/deletions etc.
• Genome Rearrangements chromosome piece (gene)
  being moved or copied to another location or
  transferring to another chromosome altogether

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Definitions

• Block section of genome possibly containing
  more than one gene one unit
• Homologous when two blocks contain the same
  genes. Homologous blocks have the same number
  label
• Reversal reversing a series of blocks and also
  their orientations distance is measured in
  number of reversals

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Example of Reversal

• 3 4 1 2 5 3 2 1 4 5
  
• Red right orientation
• Black left orientation

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Goals

• Want shortest number of reversals to transform
  one genome to another
• Parsimony assumption assume Nature changes
  optimally
• Desire polynomial time solution
• Oriented has a poly-time solution, unoriented
  NP-hard

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Example

• 1 2 3 4 5
• 5 2 1 3 4Add circle
  if orientation changes

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One solution
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Breakpoints

• Act as a minimum
• Happens in the case of
• first/last label in original not the first/last
  label in the target
• OR 2 labels are consecutive in original, but not
  in target
• OR consecutive in original and target but duel
  orientation is different between blocks
• 5 4 and 5 4
• NOTE If a pair of labels is an exact reversal in
  the target, there is NO breakpoint
• 4 5 and 5 4 do not have a breakpoint

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Breakpoints for Last Example

• 1 2 3 4 5
• Goal reminder
• 5 2 1 3 4

1 is different than first of target
No breakpoint between 1 and 2 since exact
reversal in target
2 and 3 not consecutive in target
3 and 4 match, thus no breakpoint
4 and 5 are not consecutive in target
5 is different from last in target
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Mathy Stuff o)

• Let L be finite set of labels
• L0 U a , a for all a in L
• x -gt remove arrows
• Ex
• a a a

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Contd

• Oriented permutation over L is a
• mapping a 1..n -gt L0 such that for any a e L,
  
• there is exactly one i e 1..n with a(i) a
• Basically, permutation picks an orientation for
  each label. If a is picked, then a will not be

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Example

• n 4
• L 1, 2, 3, 4
• a ( 2, 1, 4, 3 )
• So a(3) 4

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Identity Permutation

• Special case
• Permutation I such that I(i) i for all i
  between 1 and n
• For n 3, I ( 1 2 3)

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Reversals

• Let i and j be two indices with 1 i, j n
• i,j indicates a reversal affecting elements
  a(i) through a(j)

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Example

• Given a ( 2, 3, 4, 1)
• a2,3 ( 2, 4, 3, 1)
• Note similar to boxing scheme used earlier

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More Math!

• In general
• ?i, j(k) a(i j k) if i k j
• a(k) otherwise
• a(k) means reversal of orientation of a(k)

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Sorting by Reversals

• Is the main goal
• Given 2 permutations a and ß, seek minimum number
  of reversals to transform a into ß
• ?p1p2p3pt ß where p1, p2,, pt are reversals
• t is called the reversal distance of a with
  respect to ß and denoted by dß(a)

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Sorting cont

• Look for reversals that make progress towards ß
• dß (ap) lt dß (a) or
• dß (ap) dß(a) - 1

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Breakpoints

• Add labels L and R to a to get extended version
• One example of a a is(L, 2, 3, 1, 6, 5, 4, R)
• If B is identity, then breakpoints at

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Breakpoints
L 2 3 1 6 5 4 R
L 1 2 3 4 5 6 R
2 is not the first block of ß
2 and 3 are consecutive, but the orientations are
different than what they need and are not a
complete reversal
3 and 1 are not consecutive in ß
1 and 6 are not consecutive in ß
6 and 5 are consecutive, but not a complete
reversal (orientation of 6 prevents it)
none at 5 4, reverse pair 4 5 is in ß
4 is not the final block in ß
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Breakpoints cont

• Can remove at most 2 breakpoints with each
  reversal
• Thus, b(a) b(ap) 2
• This also means that b(a)/2 d(a)
• This is a lower bound for d(a)

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Bps contd

• b(a)/2 is lower bound
• However, this is rarely achievable
• Want a better lower bound
• Look to something called reality-desire diagram

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Reality-Desire

• Happens when 2 labels are adjacent, but do not
  want to be adjacent
• Reality neighbor a certain label has in a
• Desire neighbor the label has in ß

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Diagram

• Oriented labels can be viewed as a battery
• Positive terminal at tip of arrow
• Negative at tail
• - a

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Example
Desire

• a
• ap

Reality
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Example

• Extended a L 3 2 1 4 5 R
• Replace labels by terminals reality edges
• L -3 3 2 -2 1 -1 -4 4 5 -5
  R
• Add desire edges

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Diagram

• To create diagram of reality-desire
• Arrange all terminal nodes around a circle with L
  and R at the top
• L to the left of R and all other nodes following
  a counterclockwise
• Reality edges will be along circumference
• Desire edges will be the chords

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Diagram of Reality-Desire
Happens where not breakpoint
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Interpretation

• Number of cycles in RD(a) is cß(a) and is number
  of connected parts
• cß(ß) has no breakpoints
• Notice cß(ß)n1
• Why?

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Effects of a Reversal

• Let (s,t) and (u,v) be two reality edges
  characterizing a reversal p with (s,t) preceding
  in the permutation a. Then RD(ap) differs from
  RD(a) by
• 1. Reality edges (s,t) and (u,v) are replaced
  by (s,u) and (t,v)
• 2. Desire edges remain unchanged
• 3. The section of the circle going from node t
  to node u, including these extremities, in
  counterclockwise direction, is reversed.

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Our Example
Reversing (-1,-4) and (4, 5)
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Definitions

• Let e and f be two reality edges belonging to the
  same cycle in RD(a)
• If orientations induced by e and f coincide, they
  are convergent
• Walk counterclockwise from start of e (passing
  through desire edges) until you reach the
  beginning of f. If the end of f is still
  counterclockwise, then converge
• Divergent otherwise

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Walking Convergent
(3,2) to (-1,-4)
Still counterclockwise
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How Reversals Affect Cycles

• If e and f belong to different cycles, c(ap)c(a)
  -1

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If e and f belong to the same cycles and
converge c(ap)c(a)
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If e and f belong to the same cycles and
diverge c(ap)c(a) 1
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Summary

• If e and f
• belong to different cycles, c(ap)c(a) -1
• belong to same cycle converge, c(ap)c(a)
• belong to same cycle diverge, c(ap)c(a)1

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Lower Bound

• Since number of cycles changes by at most 1 per
  reversal, can get a new lower bound for reversals
• Suppose ap1p2..ptß
• --cß(ap1p2...pt)cß(ß)n1
• cß(ap1) cß(a) 1
• cß(ap1p2) cß(ap1) 1
• cß(ap1...pt) cß(ap1...pt-1) 1

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Lower Bound

• Add to get
• n1 cß(a) t
• If p1,p2,...,pt is an optimal sorting, then
  tdß(a)
• n1 cß(a) dß(a)
• Very good lower bound

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Good/Bad Cycles

• A cycle is good if it has two divergent reality
  edges
• If not, it is considered bad
• Good cycles have at least two desire edges that
  cross
• Not all cycles that have crossing edges are good
• Call cycles proper if they have at least four
  edges

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Good/Bad contd

• If we only have good cycles, lower bound d(a)
  n1 c(a) is an equality
• How could it be possible for it to be an equality
  if there are a few bad cycles mixed in to start?

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Interleave

• Twisting another cycle while breaking another is
  only possible if the two cycles are such that
  some desire edge from one of the cycles crosses
  some desire edge from the other
• These two cycles interleave in this case

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Interleave
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Interleaving Graph

• Important to verify which cycles interleave with
  which other cycles
• Take as nodes the proper cycles of RD(a)
• Two nodes adjacent iff the cycles interleave
• Connected components are classified as good or
  bad
• If a component contains all bad cycles, it is
  bad. Otherwise, it is said to be good

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RD to Interleave
Gray filled-in circles are good cycles
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Choosing a Reversal
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Choosing a Reversal

• C is the only good cycle
• Let e (L, 3), f(-3,-4), g(-1,2)
• f g converge, so not a good choice

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e and g

• e and g diverge and produce 2 good components
  with 1 cycle each

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e and f

• e and f produce a single good component with two
  cycles

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Reversal Choosing contd

• A reversal characterized by two divergent edges
  of the same cycle is a sorting reversal iff its
  application does not lead to the creation of bad
  components
• So reversing e f or e g are both acceptable

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Bad Components

• Good components can be sorted as in previous
  slide
• First step in dealing with bad components is to
  classify them
• Component Y separates components X and Z if all
  chords in RD(a) that link a terminal in X to one
  in Z cross a desire edge of Y

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• E separates F and D
• What are some other separations?

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Definitions

• Hurdle bad component that does not separate two
  bad components
• Nonhurdle bad component that separates two bad
  components

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Definitions contd

• X protects nonhurdle Y if removal of X would
  cause Y to become a hurdle
• If anytime Y separates 2 bad components, X is one
  of them
• Superhurdle hurdle that protects a nonhurdle
• Simple hurdle does not protect a nonhurdle

F protects E
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Classification
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Formula for Reversal Distance

• d(a) n 1 c(a) h(a) f(a)
• h(a) number of hurdles
• f(a) 0 or 1
• 1 if a is a fortress
• A nonhurdle will become a hurdle at some point

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Fortress

• A fortress is a permutation where there are an
  odd number of hurdles and all of them are super
  hurdles. They require an extra reversal since a
  nonhurdle will become a hurdle at some point

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Definitions

• X and Y are opposite hurdles when we find the
  same number of hurdles when walking around the
  circle counterclockwise from X to Y as we do
  clockwise.

Note only wheneven number hurdles
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Hurdle Cutting

• Reverse edges in same component
• Used only with simple hurdles

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Final Algorithm

• While a not B
• If there is a good component in RD(a) then pick
  two divergent edges in this component ensuring
  that it does not create a bad component
• Else
• if h(a) is even then
• return merging of two opposite hurdles
• else
• if there is a simple hurdle
• return a reversal cutting this hurdle
• else //fortress
• return merging of any two hurdles

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Fortress Handling
A
C
B
Fortress, so choose any 2 hurdles and merge C is
good
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Complexity

• Construction RD(a) takes linear
• Finding the cycles is O(n)
• For each cycle, determine good/bad
• This is O(n) per cycle, so O(n2) total
• Determining interleaving can be done in O(n2)
• Counting hurdles etc. can be done linearly with
  the other knowledge

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Complexity contd

• Figuring out a Sorting Reversal for good
  components is the worst since need ensure we
  dont create bad components
• Since reversal is identified with a pair of
  edges, O(n2) reversals.
• For each one, O(n2) time checking the resulting
  permutation. O(n4) total
• We need to do this dß(a) times so O(n5) all
  together

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Final Slide, Huzzah!

• Found accurate distance measure for genome
  movements
• Found a poly-time solution for solving the
  problem
• Played with fun graphs