Genome Rearrangements

1
Genome Rearrangements

• CIS 667 April 13, 2004

2
Genome Rearrangements

• We have seen how differences in genes at the
  sequence level can be used to infer evolutionary
  relations among species
• Differences in sequences in (one or more) genes
  resulted from point mutations (insert, delete,
  substitute)
• These are not the only type of changes that can
  occur in the genome

3
Genome Rearrangements

• Repair of broken chromosomes is an important
  process
• Mistakes can occur, however
• Mistakes can also occur during crossover
• These mistakes cause changes in gene order
• A large piece of chromosome can be moved or
  copied to another location
• It can also move from one chromosome to another
• We call these movements genome rearrangments

4
Crossover
5
Chromosome Repair
6
Genome Rearrangements

• These have important (usually fatal) consequences
  for the organism and its evolution
• Alignments do not capture genome rearrangments
• Two species may have nearly the same gene
  sequences, but in a different order (why would
  the two species then be different?)

7
Genome Rearrangements

• We need some other way to compare entire genomes
  (i.e. compare at a higher level)
• Rather than simple point mutations a genome is
  obtained from another by a number of a special
  kind of rearrangements Reversals
• Use the number of reversals needed to transform
  one genome into another to measure evolutionary
  distance

8
The Method

• Use combinatorial optimization techniques in an
  attempt to infer a most economical sequence of
  rearrangement operations to account for
  differences among the genomes
• Compare with character-based methods for
  phylogenetics (parsimony)

9
Reversals

• Consider the genome of a species as a sequence of
  blocks
• A block is some sequence of the genome (possibly
  containing more than one gene) transcribed as a
  unit
• Blocks are oriented since they can be transcribed
  from either strand of DNA
• Give homologous blocks the same label

10
Reversals

• Relation between chloroplast genomes of alfalfa
  and garden pea

11
Reversals

• Reversal operation for oriented blocks
• Inverts the order of affected blocks and changes
  their orientation (arrow)
• Affects a contiguous segment of blocks
• What sequence of reversal operations could have
  changed alfalfa into garden pea?
• Would like to have a polynomial time algorithm to
  find the shortest sequence

12
(No Transcript)
13
Genome Comparison vs. Gene Comparison

• In the late 1980s, J. Palmer and his colleagues
  studied the mitochondrial genomes of cabbage and
  turnips
• The gene sequences are very similar (some genes
  are 99 equal)
• Gene order, however, differs dramatically
• Genome rearrangements are now considered to be a
  common mode of molecular evolution

14
Genome Comparison vs. Gene Comparison

• Extreme conservation of genes on X chromosomes
  across mammalian species provides an opportunity
  to study the evolutionary history of X chromosome
  independently of the rest of the genomes
• According to Ohnos law, the gene content of X
  chromosome has barely changed throughout
  mammalian development in the last 125 million
  years.
• However, the order of genes on X chromosomes has
  been disrupted several times.

15
Human and Mouse X Chromosomes
16
Human and Mouse X Chromosomes
-4 -6 1 7
2 -3 5 8
3 2 7 -1
6 4 5 8
1 7 2 -3
6 4 5 8
1 2 7 -3
6 4 5 8
1
2 7 3
-6 4 5 8
1 2 -5 -4
-3 6 7 8
1 2 3 4
5 6 7 8
17
Genome Comparison vs. Gene Comparison

• The traditional molecular evolutionary technique
  is a gene comparison to construct a phylogenetic
  tree
• In the cabbage and turnip case this is hardly
  suitable, since rate of point mutations in their
  mitochondrial genes is so low that their genes
  are almost identical
• Genome comparison (i.e. comparison of gene
  orders) is the method of choice in the case of
  very slowly evolving genomes
• Another area is the case where genomes evolve
  very rapidly (genes not very similar)

18
Genome Comparison

• Only about (178?39) genome rearrangements have
  happened since human and mouse diverged 80
  million years ago
• Mouse and human genomes can be viewed as a
  collection of about 200 fragments which are
  shuffled in mice as compared to humans
• A comparative mouse-human genetic map gives the
  position of a human gene given the location of a
  related mouse gene

19
Man-Mouse Comparative Physical Map
20
Definitions

• A signed permutation a over the set of labels L
  1, 2, , n is a permutation such that a(i) a
  or a, where a Î L
• Example 3, 2, 1 is a signed permutation over
  L 1, 2, 3
• Note that no label may appear twice in the
  permutation
• A reversal i,j is an operation that transforms
  one signed permutation into another, reversing
  the order or a contiguous portion and flipping
  the signs

21
Definitions

• a ai,j a(1), , a(i 1), a(j), , a(i),
  a(j 1), , a(n)
• We are interested in the problem of sorting by
  reversals Given two signed permutations a and b,
  find the minimum number of reversals r1, , rt
  that will transform a into b - a r1rt b
• The reversal distance db(a) t

22
Definitions

• Note that the reversal operation does not
  directly correspond to the biological operations
  (inversion, translocation, fission, fusion)
• Given a and b, can we always transform a into b
  using only the reversal operation? If so, how
  many reversals are required in the worst case?

23
Breakpoints

• A breakpoint is a point between consecutive
  labels in the initial permutation that must
  necessarily be separated by at least one reversal
  to reach the target permutation
• The two consecutive labels are not consecutive in
  the target, or their orientations are not the
  same in a relative sense

24
Breakpoints

• To formalize the idea of breakpoint, we introduce
  the extended version of a
• Let a a(1), , a(n)
• Then the extended version of a is (L, a(1), ,
  a(n), R)
• For example let extended a be (L, 2, 3, 1, 6,
  5, 4, R) and let extended b be (L, 1, 2, 3,
  4, 5, 6, R)
• The breakpoints are (L,2), (2,3), (3,1),
  (1,6), (6,5), (4,R)

25
Breakpoints

• The number of breakpoints of a permutation a is
  denoted by b(a)
• In the example, 6
• Can you characterize the situations where L is
  involved in a breakpoint? When R is involved in a
  breakpoint?

26
A Lower Bound

• A reversal can remove at most two breakpoints
• Cuts the permutation in exactly two places
• So, if ar1 rt b then
• b(a) b(ar1) 2
• b(ar1) b(ar1r2) 2
• b(ar1rt-1) b(ar1rt) 2
• So b(a) 2t. If t d(a), b(a)/2 d(a)

27
Reality and Desire Diagram

• The lower bound found is not very tight
• We can derive a better l.b. based on a structure
  called the reality-desire diagram of a
  permutation with respect to another
• To draw the diagram, we will represent a with
  the tuple (-a a) and -a with the tuple (a -a)
• The orientation is given by the rightmost member
  of the tuple

28
Reality and Desire Diagram

• A permutation is a sequence of adjacent tuples
• a 3, 2, 1, 4, 5 can be represented as
  L---(3 3)---(2 2)---(1 1)---(4 4)---(5
  5)---R
• b L---(1 1)---(2 2)---(3 3)---(4
  4)---(5 5)---R

29
Reality and Desire Diagram

• Now we will draw a graph to represent a (L, 3,
  -2, -1, 4, -5, R)
• The reality diagram

30
Reality and Desire Diagram

• Suppose that b is the identity (L, 1, 2, 3,
  4, 5)
• We will add desire edges to the previous graph to
  represent b

L -3 3 2 -2 1 -1
-4 4 5 -5 R
31
Reality and Desire Diagram

• a is the reality
• b is what is desired
• The diagram (a multigraph) shows both reality and
  desire
• Call it RD(a)
• We can rearrange the nodes of the graph to make
  it easier to understand

32
Reality and Desire Diagram
L
R
Desire
Reality
33
Properties of RD(a)

• Each vertex has degree 2
• Each node is incident to one edge from A, the set
  of reality edges, and B, the set of desire edges
• The connected components of the graph are
  alternating cycles (edges alternate between
  reality - blue - and desire - red)
• Each cycle has an even number of edges, half
  reality and half desire

34
Properties of RD(a)

• The number of cycles of RD(a) is denoted by cb(a)
  
• Note that cb(b) n 1 since b has no
  breakpoints
• All cycles are two parallel edges between the
  same pair of nodes
• We have 2n 2 nodes, so n 1 cycles
• This is the only permutation for which cb(a) 1

35
Properties of RD(a)

• So transforming a into b can be seen as
  transforming RD(a) into a graph with as many
  cycles as possible - n 1
• Now we need to see how a reversal affects the
  cycles of RD(a)
• Note that a reversal is characterized by the two
  points where it cuts the current permutation,
  which each correspond to a reality edge

36
Reversals and RD(a)

• Let r be a reversal defined by two reality edges
  (s,t) and (u,v), then RD(ar) differs from RD(a)
  as follows
• Reality edges (s,t) and (u,v) are replaced by
  (s,u) and (t,v)
• Vertices u, , t are reversed
• Desire edges remain unchanged
• See example on following slide

37
Example
Some nodes/edges omitted
L
L
R
R
38
Orientation of Cycles

• How many cycles are affected by a reversal?
• First we define convergent and divergent edges
• Two reality edges on the same cycle converge if
  they are traversed in the same direction
  (clockwise or counterclockwise on the circle in
  the diagram) on the cycle
• Otherwise they diverge

39
Orientation of Cycles
L
Convergent (3,2) (-1,-4) Divergent (L,-3)
(3,2)
R
40
Reversals and Cycles

• Let r be a reversal acting on two reality edges e
  and f
• If e and f belong to different cycles, c(ar)
  c(a) 1
• If e and f belong to the same cycle and converge,
  c(ar) c(a)
• If e and f belong to the same cycle and diverge,
  c(ar) c(a) 1

41
First Case

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

42
Second Case

• If e and f belong to the same cycle and converge,
  c(ar) c(a)

43
Third Case

• If e and f belong to the same cycle and diverge,
  c(ar) c(a) 1

44
Reversals and Cycles

• Note that the number of cycles changes by at most
  one with each reversal
• Use that to find another lower bound for reversal
  distance
• Suppose we have ar1r2rt b we know that c(b)
  n 1 and we have
• c(ar1) - c(a) ? 1
• c(ar1r2) - c(ar1) ? 1
• c(ar1r2rt) - c(ar1r2rt-1) ? 1
• Adding and cancelling terms we get
• n 1 - c(a) ? t
• If r1r2rt is optimal then t d(a), n 1 - c(a)
  ? d(a)

45
Interleaving Graph

• This new lower bound is better than the old one -
  b(a)/2
• For most signed permutations, it is close to the
  actual distance, however it does not always work
  (we cant always choose two divergent edges)
• We can classify the cycles of RD(a) as good or
  bad
• A cycle is good if it has two divergent reality
  edges
• Otherwise it is bad

46
Interleaving Graph

• The classification only applies to proper cycles
  (those with at least four edges)
• Those with three edges dont need to be touched
  since reality desire
• If we have only good cycles in a permutation,
  then the lower bound previously given is an
  equality
• We sort, increasing the number of cycles by one
  per reversal

47
Interleaving Graph

• If a desire edge from one cycle crosses some
  desire edge from another cycle we say that the
  two cycles interleave
• Interleaved cycles allow us to change a bad cycle
  into a good one while breaking another cycle
• This good cycle can then broken in the next step
• To find interleaving cycles, we construct an
  interleaving graph

48
Interleaving Graph
49
Interleaving Graph

• Nodes in the interleaving graph are cycles
• Edge between two nodes if the cycles interleave
• The connected components of the graph are called
  bad components if they consist entirely of bad
  cycles
• Component otherwise is a good component

50
Interleaving Graph

• What is the interleaving graph of the previous
  example?
• Suppose that F and C are good cycles.
• Which components of the interleaving graph are
  good and which are bad?

51
Sorting Good Components

• We need to choose two divergent edges in the same
  cycle to define a reversal that increases the
  number of cycles
• Example
• A reversal characterized by two divergent edges
  of the same cycle is a sorting reversal if and
  only if it does not lead to the creation of bad
  components

52
Bad Components

• Using this criterion to sort all of the good
  components, we must now sort the bad ones
• Give a hierarchy of bad components
• We say a component B separates components A and C
  if all chords in RD(a) that link a terminal in A
  to a terminal in C cross a desire edge of B

53
Diagram with no Good Components
54
Bad Components

• Reversal through reality edges in different
  components A and C will result in every component
  B that separates A and C being twisted
• A bad component becomes good when twisted
• A good component can stay good or become bad when
  twisted
• So twist only when no good components

55
Hierarchy of Bad Components

• A hurdle is a bad component that does not
  separate any other two bad components
• If a bad component separates others, then it is a
  nonhurdle
• A hurdle A protects a nonhurdle B when removal of
  A would cause B to become a hurdle
• B is protected by A when every time B separates
  two bad components, A is one of them

56
Hierarchy of Bad Components

• A hurdle A is called a superhurdle if it protects
  some other nonhurdle B
• Otherwise it is called a simple hurdle

Bad Components
Nonhurdles
Hurdles
Simple hurdles
Super hurdles
57
Fortress

• A signed permutation a is called a fortress iff
  RD(a) has an odd number of hurdles and all of
  them are super hurdles

58
Reversal Distance

• The reversal distance of oriented permutations is
  given by
• d(a) n 1 - c(a) h(a) f(a)
• c(a) - number of cycles (proper and non)
• h(a) - number of hurdles
• f(a) - a a fortress? (1 else 0)
• n 1 - c(a) good components and bad components
  which become good during sort
• h(a) - bad components require extra reversal
• f(a) - extra reversal for fortress

59
Algorithm

• If we dont have a good cycle we must use either
  a reversal on two convergent edges or a reversal
  on edges in different cycles
• In first case, number of cycles is constant
• In second case, number of cycles decreases by one
• Choose case one on a hurdle
• Transforms bad component into good
• Number of cycles remains constant

60
Algorithm

• Getting rid of a non-hurdle doesnt change the
  number of hurdles or fortress status, so distance
  remains the same
• If we reverse a superhurdle, the nonhurdle it
  protects becomes a hurdle so h remains constant
• Call reversal on some cycle in a hurdle hurdle
  cutting

61
Algorithm

• In order not to increase f(a), use hurdle cutting
  only when h(a) is odd
• Using reversal on edges in two different cycles
  increases c(a)
• However d(a) will decrease if we can decrease
  h(a) by two
• Choose edges from two different hurdles - this is
  called hurdle merging
• The two hurdles as well as any nonhurdle
  separating them become good components

62
Algorithm

• We have to be careful that hurdle merging doesnt
  transform a nonhurdle into a hurdle
• A and B are called opposite hurdles when we find
  the same number of hurdles walking the circle
  clockwise from A to B as we do walking
  counterclockwise
• This can only happen if h(a) is even
• Choosing opposite hurdles, we dont turn a
  nonhurdle into a hurdle

63
Algorithm

• To avoid creating a fortress where we dont have
  one, we choose the opposite hurdles when they
  exist
• If h(a) is odd and we have a simple hurdle, do
  hurdle cutting to avoid fortress
• If neither case if possible, we already have a
  fortress so f(a) doesnt increase with any hurdle
  merging

64
Algorithm
Algorithm Sorting Reversal input distinct
permutations a and b output a sorting reversal
for a with target b if there is a good component
in RDb(a) then pick 2 divergent edges e and f
in this component, making sure the
corresponding reversal does not create any bad
components return the reversal characterized by
e and f else if h(a) is even then return
merging of two opposite hurdles else if
h(a) is odd and there is a simple hurdle
then return a reversal cutting this hurdle
else // fortress return merging of
any two hurdles