genome rearrangements …and you!! presented by: kevin gaittens
TRANSCRIPT
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Genome Rearrangements Genome Rearrangements …and YOU!!…and YOU!!
Genome Rearrangements Genome Rearrangements …and YOU!!…and YOU!!
Presented by:Presented by:
Kevin GaittensKevin Gaittens
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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 orientationBlack – 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 4
Add circle if orientation changes
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One solution
1 2 3 4 5
1 2 5 4 3
1 2 5 3 4
5 2 1 3 4
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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
5 is different from last in target
4 and 5 are not consecutive 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 | -> remove arrows
Ex:| a | = | a | = a
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Cont’d
Oriented permutation over L is a mapping α: [1..n] -> L0 such that for any a ε L,
there is exactly one i ε [1..n] with |α(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}
α = ( 2, 1, 4, 3 )
So α(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 α(i) through α(j)
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Example
Given α = ( 2, 3, 4, 1)
α[2,3] = ( 2, 4, 3, 1)
Note: similar to boxing scheme used earlier
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More Math!
In general:
Α[i, j](k) = α(i + j – k) if i ≤ k ≤ j
α(k) otherwise
α(k) means reversal of orientation of α(k)
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Sorting by Reversals
• Is the main goal
• Given 2 permutations α and β, seek minimum number of reversals to transform α into β
• Αp1p2p3…pt = β where p1, p2,…, pt are reversals
• t is called the reversal distance of α with respect to β and denoted by dβ(α)
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Sorting con’t
• Look for reversals that “make progress” towards β
dβ (αp) < dβ (α) or
dβ (αp) = dβ(α) - 1
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Breakpoints
• Add labels L and R to α to get “extended version”
• One example of a α is:
(L, 2, 3, 1, 6, 5, 4, R)
• If B is identity, then breakpoints at…
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Breakpoints
none at 5 4, reverse pair 4 5 is in β
L 2 3 1 6 5 4 R
L 1 2 3 4 5 6 R
2 is not the first block of β
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 β
1 and 6 are not consecutive in β6 and 5 are consecutive, but not a complete reversal (orientation of 6 prevents it)
4 is not the final block in β
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Breakpoints con’t
• Can remove at most 2 breakpoints with each reversal
• Thus, b(α) – b(αp) ≤ 2
• This also means that b(α)/2 ≤ d(α)
• This is a lower bound for d(α)
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Bps cont’d
• b(α)/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 α
• Desire – neighbor the label has in β
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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
α
αp
Desire
Reality
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Example
Extended α: 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
α 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(α) is cβ(α) and is number of connected parts
• cβ(β) has no breakpoints
• Notice cβ(β)=n+1– 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 α. Then RD(αp) differs from RD(α) by:1. Reality edges (s,t) and (u,v) are replaced by (s,u) and (t,v)2. Desire edges remain unchanged3. 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(α)
• 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
Still counterclockwise
(+3,+2) to (-1,-4)
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How Reversals Affect Cycles
If e and f belong to different cycles, c(αp)=c(α) -1
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If e and f belong to the same cycles and converge
c(αp)=c(α)
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If e and f belong to the same cycles and diverge
c(αp)=c(α) +1
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Summary
If e and f:• belong to different cycles, c(αp)=c(α) -1• belong to same cycle & converge, c(αp)=c(α)• belong to same cycle & diverge, c(αp)=c(α)+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 αp1p2..pt=β--cβ(αp1p2...pt)=cβ(β)=n+1
cβ(αp1) – cβ(α) ≤ 1
cβ(αp1p2) – cβ(αp1) ≤ 1…
cβ(αp1...pt) – cβ(αp1...pt-1) ≤ 1
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Lower Bound
• Add to get n+1 – cβ(α) ≤ t
• If p1,p2,...,pt is an optimal sorting, then t=dβ(α)
n+1 – cβ(α) ≤ dβ(α)
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 cont’d
• If we only have good cycles, lower bound d(α) ≥ n+1 – c(α) 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(α)• 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 cont’d
• 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(α) 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 cont’d
• 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(α) = n + 1 – c(α) + h(α) + f(α)
h(α) = number of hurdlesf(α) = 0 or 1
1 if α is a fortressA 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 AlgorithmWhile α not B:
If there is a good component in RD(α) then pick two divergent edges in this component ensuring that it does not create a bad component
Else if h(α) is even then
return merging of two opposite hurdleselse
if there is a simple hurdlereturn a reversal cutting this hurdle
else //fortressreturn merging of any two hurdles
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Fortress Handling
Fortress, so choose any 2 hurdles and merge
C is good
C
A
B
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Complexity
• Construction RD(α) 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 cont’d
• Figuring out a Sorting Reversal for good components is the worst since need ensure we don’t 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 dβ(α) 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