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Genome Rearrangements …and YOU!!. Presented by: Kevin Gaittens. Overview. Bio background Definitions and Set-up Reality-Desire Good Components Bad Components Fin. Biological Bakground. Comparing entire genomes across species Need “distance” measure
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Genome Rearrangements …and YOU!! Presented by: Kevin Gaittens
Overview • Bio background • Definitions and Set-up • Reality-Desire • Good Components • Bad Components • Fin
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
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
Example of Reversal 3 4 1 2 53 2 1 4 5 Red – right orientation Black – left orientation
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
Example 1 2 3 4 5 5 2 1 3 4Add circle if orientation changes
1 2 3 4 5 1 2 5 4 3 1 2 5 3 4 5 2 1 3 4 One solution
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
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
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
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
Example n = 4 L = {1, 2, 3, 4} α = ( 2, 1, 4, 3 ) So α(3) = 4
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)
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)
Example Given α = ( 2, 3, 4, 1) α[2,3] = ( 2, 4, 3, 1) Note: similar to boxing scheme used earlier
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)
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)
Sorting con’t • Look for reversals that “make progress” towards β dβ (αp) < dβ (α) or dβ (αp) = dβ(α) - 1
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…
Breakpoints 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 β 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 β 4 is not the final block in β
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)
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
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
Diagram • Oriented labels can be viewed as a battery • Positive terminal at tip of arrow • Negative at tail - a +
Example α αp Desire Reality
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
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
Diagram of Reality-Desire Happens where not breakpoint
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?
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 unchanged 3. The section of the circle going from node t to node u, including these extremities, in counterclockwise direction, is reversed.
Our Example Reversing (-1,-4) and (+4, +5)
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
Walking Convergent (+3,+2) to (-1,-4) Still counterclockwise
How Reversals Affect Cycles If e and f belong to different cycles, c(αp)=c(α) -1
If e and f belong to the same cycles and converge c(αp)=c(α)
If e and f belong to the same cycles and diverge c(αp)=c(α) +1
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
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
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
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
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?
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
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
RD to Interleave Gray filled-in circles are good cycles
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
e and g • e and g diverge and produce 2 good components with 1 cycle each
e and f • e and f produce a single good component with two cycles