1. Genome Rearrangements, Synteny, and Comparative Mapping CSCI 4830: Algorithms for Molecular Biology Debra S. Goldberg

2. Turnip vs Cabbage Share a recent common ancestor Look and taste different

3. Turnip vs Cabbage Comparing mtDNA gene sequences yields no evolutionary information 99% similarity between genes These surprisingly identical gene sequences differed in gene order This study helped pave the way to analyzing genome rearrangements in molecular evolution

4. Turnip vs Cabbage: Different mtDNA Gene Order Gene order comparison:

5. Turnip vs Cabbage: Different mtDNA Gene Order Gene order comparison:

6. Turnip vs Cabbage: Different mtDNA Gene Order Gene order comparison:

7. Turnip vs Cabbage: Different mtDNA Gene Order Gene order comparison:

8. Turnip vs Cabbage: Gene Order Comparison Before After Evolution is manifested as the divergence in gene order

9. Transforming Cabbage into Turnip

10. Reversals 1 32 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 misread as 1, 2, 3, -8, -7, -6, -5, -4, 9, 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 Blocks represent conserved genes.

11. Types of Mutations

12. Types of Rearrangements Inversion/Reversal 1 2 3 4 5 61 2 -5 -4 -3 6

13. Types of Rearrangements Translocation 4 1 2 3 4 5 6 1 2 6 4 5 3

14. Types of Rearrangements 1 2 3 4 5 6 1 2 3 4 5 6 Fusion Fission

15. What are the similarity blocks and how to find them? What is the architecture of the ancestral genome? What is the evolutionary scenario for transforming one genome into the other? Unknown ancestor ~ 75 million years ago Mouse (X chrom.) Human (X chrom.) Genome rearrangements

16. Why do we care?

17. SKY (spectral karyotyping)

18. Robertsonian Translocation 1314

19. Robertsonian Translocation Translocation of chromosomes 13 and 14 No net gain or loss of genetic material: normal phenotype. Increased risk for an abnormal child or spontaneous pregnancy loss 1314

20. Philadelphia Chromosome 9 22

21. Philadelphia Chromosome A translocation between chromosomes 9 and 22 (part of 22 is attached to 9) Seen in about 90% of patients with Chronic myelogenous leukemia (CML) 922

22. Colon cancer

23. Colon Cancer

24. Comparative maps

25. Waardenburg’s Syndrome: Mouse Provides Insight into Human Genetic Disorder Characterized by pigmentary dysphasia Gene implicated linked to human chromosome 2 It was not clear where exactly on chromosome 2

26. Waardenburg’s syndrome and splotch mice A breed of mice (with splotch gene) had similar symptoms caused by the same type of gene as in humans Scientists identified location of gene responsible for disorder in mice Finding the gene in mice gives clues to where same gene is located in humans

27. Reversals: Example  = 1 2 3 4 5 6 7 8  1 2 5 4 3 6 7 8 

28. Reversals: Example  = 1 2 3 4 5 6 7 8  1 2 5 4 3 6 7 8  1 2 5 4 6 3 7 8

29. Reversals and Gene Orders Gene order represented by permutation   1  ------  i-1  i  i+1 ------  j-1  j  j+1 -----  n   1  ------  i-1  j  j-1 ------  i+1  i  j+1 -----  n Reversal  ( i, j ) reverses (flips) the elements from i to j in  ,j)

30. Reversal Distance Problem Goal: Given two permutations, find shortest series of reversals to transform one into another Input: Permutations  and  Output: A series of reversals  1,…  t transforming  into  such that t is minimum t - reversal distance between  and  d( ,  ) = smallest possible value of t, given  

31. Sorting By Reversals Problem Goal: Given a permutation, find a shortest series of reversals that transforms it into the identity permutation (1 2 … n ) Input: Permutation  Output: A series of reversals  1, …  t transforming  into the identity permutation such that t is minimum min t =d(  ) = reversal distance of 

32. Sorting by reversals Example: 5 steps

33. Sorting by reversals Example: 4 steps What is the reversal distance for this permutation? Can it be sorted in 3 steps?

34. Pancake Flipping Problem Chef prepares unordered stack of pancakes of different sizes The waiter wants to sort (rearrange) them, smallest on top, largest at bottom He does it by flipping over several from the top, repeating this as many times as necessary Christos Papadimitrou and Bill Gates flip pancakes

35. Sorting By Reversals: A Greedy Algorithm Unsigned permutations Example: permutation  = 1 2 3 6 4 5 First three elements are already in order prefix(  ) = length of already sorted prefix – prefix(  ) = 3 Idea: increase prefix(  ) at every step

36. Doing so,  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 permutation of length n is at most (n – 1) Greedy Algorithm: An Example

37. Greedy Algorithm: Pseudocode SimpleReversalSort(  ) 1 for i  1 to n – 1 2 j  position of element i in  (i.e.,  j = i) 3 if j ≠i 4    *  (i, j) 5 output  6 if  is the identity permutation 7 return

38. Analyzing SimpleReversalSort SimpleReversalSort does not guarantee the smallest number of reversals and takes five steps on  = 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

39. But it can be sorted in two steps:  = 6 1 2 3 4 5 –Step 1: 5 4 3 2 1 6 –Step 2: 1 2 3 4 5 6 So, SimpleReversalSort(  ) is not optimal Optimal algorithms are unknown for many problems; approximation algorithms used Analyzing SimpleReversalSort (cont ’ d)

40. Approximation Algorithms These algorithms find approximate solutions rather than optimal solutions The approximation ratio of an algorithm A on input  is: A(  ) / OPT(  ) where A(  ) -solution produced by algorithm A OPT(  ) - optimal solution of the problem

41. Approximation Ratio / Performance Guarantee Approximation ratio (performance guarantee) of algorithm A: max approximation ratio of all inputs of size n –For algorithm A that minimizes objective function (minimization algorithm): max |  | = n A(  ) / OPT(  ) –For maximization algorithm: min |  | = n A(  ) / OPT(  )

42.  =    2  3 …  n-1  n A pair of elements  i and  i + 1 are adjacent if  i+1 =  i + 1 For example:  = 1 9 3 4 7 8 2 6 5 (3, 4) or (7, 8) and (6,5) are adjacent pairs Adjacencies and Breakpoints

43. There is a breakpoint between any adjacent element that are non-consecutive:  = 1 9 3 4 7 8 2 6 5 Pairs (1,9), (9,3), (4,7), (8,2) and (2,5) form breakpoints of permutation  b(  ) - # breakpoints in permutation   Breakpoints: An Example

44. Adjacency & Breakpoints An adjacency – consecutive A breakpoint – not consecutive π = 5 6 2 1 3 40 5 6 2 1 3 4 7 adjacencies breakpoints Extend π with π 0 = 0 and π n+1 = n+1

45. Add  0 =0 and  n + 1 =n+1 at ends of  Example: Extending with 0 and 10 Note: A new breakpoint was created after extending Extending Permutations  = 1 9 3 4 7 8 2 6 5  = 0 1 9 3 4 7 8 2 6 5 10

46.  Each reversal eliminates at most 2 breakpoints.  = 2 3 1 4 6 5 0 2 3 1 4 6 5 7 b(  ) = 5 0 1 3 2 4 6 5 7 b(  ) = 4 0 1 2 3 4 6 5 7b(  ) = 2 0 1 2 3 4 5 6 7 b(  ) = 0 Reversal Distance and Breakpoints This implies: reversal distance ≥ #breakpoints / 2

47. Sorting By Reversals: A Better Greedy Algorithm BreakPointReversalSort(  ) 1 while b(  ) > 0 2 Among all possible reversals, choose reversal  minimizing b(   ) 3     (i, j) 4 output  5 return Problem: this algorithm may work forever

48. Strips Strip: an interval between two consecutive breakpoints in a permutation –Decreasing strip: elements in decreasing order (e.g. 6 5) –Increasing strip: elements in increasing order (e.g. 7 8) 0 1 9 4 3 7 8 2 5 6 10 –Consider single-element strips decreasing except strips 0 and n+1 are increasing

49. Reducing the Number of Breakpoints Theorem 1: If permutation  contains at least one decreasing strip, then there exists a reversal  which decreases the number of breakpoints (i.e. b(   ) < b(  ) )

50. Things To Consider For  = 1 4 6 5 7 8 3 2 0 1 4 6 5 7 8 3 2 9 b(  ) = 5 –Choose decreasing strip with the smallest element k in  (k = 2 in this case)

51. Things To Consider (cont’d) For  = 1 4 6 5 7 8 3 2 0 1 4 6 5 7 8 3 2 9 b(  ) = 5 –Choose decreasing strip with the smallest element k in  (k = 2 in this case)

52. Things To Consider (cont’d) For  = 1 4 6 5 7 8 3 2 0 1 4 6 5 7 8 3 2 9 b(  ) = 5 –Choose decreasing strip with the smallest element k in  (k = 2 in this case) –Find k – 1 in the permutation

53. Things To Consider (cont’d) For  = 1 4 6 5 7 8 3 2 0 1 4 6 5 7 8 3 2 9 b(  ) = 5 –Choose decreasing strip with the smallest element k in  (k = 2 in this case) –Find k – 1 in the permutation –Reverse segment between k and k-1: 0 1 2 3 8 7 5 6 4 9 b(  ) = 4

54. Reducing the Number of Breakpoints Again –If there is no decreasing strip, there may be no reversal  that reduces the number of breakpoints (i.e. b(   ) ≥ b(  ) for any reversal  ). –By reversing an increasing strip ( # of breakpoints stay unchanged ), we will create a decreasing strip at the next step. Then the number of breakpoints will be reduced in the next step (theorem 1).

55. Things To Consider (cont’d) There are no decreasing strips in , for:  = 0 1 2 5 6 7 3 4 8 b(  ) = 3   (3,4)= 0 1 2 5 6 7 4 3 8 b(  ) = 3  (3,4) does not change the # of breakpoints  (3,4) creates a decreasing strip thus guaranteeing that the next step will decrease the # of breakpoints.

56. ImprovedBreakpointReversalSort ImprovedBreakpointReversalSort(  ) 1 while b(  ) > 0 2 if  has a decreasing strip 3 Among all possible reversals, choose reversal  that minimizes b(   ) 4 else 5 Choose a reversal  that flips an increasing strip in  6     7 output  8 return

57. ImprovedBreakPointReversalSort is an approximation algorithm with a performance guarantee of at most 4 –It eliminates at least one breakpoint in every two steps; at most 2b(  ) steps –Approximation ratio: 2b(  ) / d(  ) –Optimal algorithm eliminates at most 2 breakpoints in every step: d(  )  b(  ) / 2 –Performance guarantee: ( 2b(  ) / d(  ) )  [ 2b(  ) / (b(  ) / 2) ] = 4 Performance Guarantee

58. Signed Permutations Up to this point, reversal sort algorithms sorted unsigned permutations But genes have directions… so we should consider signed permutations 5’5’ 3’3’  = 1 -2 - 3 4 -5

59. Signed Permutation Genes are directed fragments of DNA Genes in the same position but different orientations do not have same gene order These two permutations are not equivalent gene sequences 1 2 3 4 5 -1 2 -3 -4 -5

60. Signed permutations are easier! Polynomial time (optimal) algorithm is known

61. What are the similarity blocks and how to find them? What is the architecture of the ancestral genome? What is the evolutionary scenario for transforming one genome into the other? Unknown ancestor ~ 75 million years ago Mouse (X chrom.) Human (X chrom.) Genome rearrangements

62. What are the similarity blocks and how to find them? What is the architecture of the ancestral genome? What is the evolutionary scenario for transforming one genome into the other? Unknown ancestor ~ 75 million years ago Mouse (X chrom.) Human (X chrom.) Genome rearrangements

63. Comparative maps

64. A brief history Chromosome comparisons –no information about genes 1920’s: Sturtevant, Weinstein Today: many organisms, many uses Humans: –primates, mouse, cat, dog, zebrafish,... –Alzheimer, cancers, diabetes, obesity,...

65. Why construct comparative maps? Identify & isolate genes –Crops: drought resistance, yield, nutrition... –Human: disease genes, drug response,… Infer ancestral relationships Discover principles of evolution –Chromosome –Gene family “key to understanding the human genome”

66. Map construction 3S 8L 10L 3L Maize 1 (target), Rice (base) Wilson et al. Genetics 1999 Go from this to this

67. Why automate? Time consuming, laborious –Needs to be redone frequently Codify a common set of principles Nadeau and Sankoff: warn of “arbitrary nature of comparative map construction”

68. Input/Output Input: –genetic maps of 2 species –marker/gene correspondences (homologs) Output: –a comparative map homeologies identified

69. A natural model? Maize 1 (target), Rice (base) Wilson et al. Genetics 1999 Maize 1Rice 3S 8L 10L 3L

70. Scoring 10L 3L s m

71. Assumptions Accept published marker order All linkage groups of base are unique Simplistic homeology criteria At least one homeologous region

72. A natural model?

73. Dynamic programming l i = location of homolog to marker i S[i,a] = penalty (score) for an optimal labeling of the submap from marker i to the end, when labeling begins with label a a 1...i...n

74. Recurrence relation S[n,a] = m  (a, l n ) S[i,a] = m  (a, l i ) + min (S[i+1,b] + s  (a,b) ) bLbL ab...ii+1...n l i l i+1 l n a...n... l n

75. Problem with linear model s = 2 a-b-c motif: abc score: 2s = 4 aaabbbccc a-b-a motif: a score: 3m = 3 aaabbbaaa

76. The stack model Segment at top of the stack can be: –pushed (remembered), later popped –replaced Push and replace cost s -- pop is free. bbb fe d c a c

77. Scoring s 9L 7L “free” pop m m m uaz265a (7L) isu136 (2L) isu151 (7L) rz509b (7L) cdo59c (7L) rz698c (9L) bcd1087a (9L) rz206b (9L) bcd1088c (9L) csu40 (3S) cdo786a (9L) csu154 (7L) isu113a (7L) csu17 (7L) cdo337 (3L) rz530a (7L)

78. Dynamic programming S[i,j,a] = score for an optimal labeling of: –submap from marker i to marker j –when labeling begins with label a -- i.e., marker i is labeled a a 1...i... j...n

79. Recurrence relation S[i,i,a] = m  (a, l i ) S[i,j,a] = min: m  (a, l i ) + min (S[i+1,j,b] + s  (a,b) ) min S[i,k,a] + S[k+1,j,a] i<k<j bLbL a 1...i...k +1...j...n a 1...ii+1...n ab 1...ii+1...n

80. Advantage: output similar to experts’ Maize 6 (target), Rice (base)

81. Advantage: proposes testable hypotheses New relations predicted; greater resolution maps confirm Ahn-Tanksley ‘93 Ahn-Tanksley dataWilson et. al. ‘99 Maize 7 (target), Rice (base)

82. Advantage: infers evolutionary events Maize 1 (target) Rice (base) Wilson et al. Stack

83. Problem: Incomplete input Gene order not always fully resolved. Co-located genes can be ordered to give most parsimonious labeling. 8p 19p = 8p 19p 33.0

84. The reordering algorithm Uses a compression scheme –Within a megalocus, group genes by location of related gene. –Order these groups –First, last groups interact with nearby genes –Any ordering of internal groups is equally parsimonious

85. The reordering algorithm

87. Definitions  extended to distance to a set A of labels 0 if a  A, 1 otherwise l i = set of labels matching markers in megalocus i S = set of megalocus start nodes  (a, A) =

88. Definitions p(i,a,b) gives the total mismatched marker and segment boundary penalties attributed to “hidden markers” –i is index for megalocus –a and b are labels for megalocus ends –Do any markers in megalocus match a, b ? No: don’t penalize in recurrence and p(i,a,b)

89. Recurrence relation S[i,i,a] = m  (a, l i ) S[i,j,a] = min: m  (a, l i ) + min (S[i+1,j,b] + s  (a,b) + p(i,a,b)) min S[i,k,a] + S[k+1,j,a] i<k<j k  S bLbL a 1...i...k +1...j...n a 1...ii+1...n ab 1...ii+1...n

90. Results: Fewer mismatches stackreordering Mouse 5 (target) Human (base)

91. Results: Mismatches placed between segments stackreordering Mouse 8 (target) Human (base)

92. Results: Detects new segments stackreordering Mouse 13 (target) Human (base)

93. Summary Global view Finds optimal comparative map –Arranges markers in most parsimonious way Biologically meaningful results Robust –not species-specific –high/low resolution, genetic/physical maps –stable to errors in marker order