1. Gene: A sequence of nucleotides coding for protein Gene Prediction Problem: Determine the beginning and end positions of genes in a genome Gene Prediction: Computational Challenge

2. Central Dogma and Splicing exon1 exon2exon3 intron1intron2 transcription translation splicing exon = coding intron = non-coding Batzoglou

3. Splice site detection 5’ 3’ Donor site Position % From lectures by Serafim Batzoglou (Stanford)

4. Six Frames in a DNA Sequence stop codons – TAA, TAG, TGA start codons - ATG GACGTCTGCTTTGGAGAACTACATCAACCGGACTGTGGCTGTTATTACTTCTGATGGCAGAATGATTGTG CTGCAGACGAAACCTCTTGATGTAGTTGGCCTGACACCGACAATAATGAAGACTACCGTCTTACTAACAC GACGTCTGCTTTGGAGAACTACATCAACCGGACTGTGGCTGTTATTACTTCTGATGGCAGAATGATTGTG CTGCAGACGAAACCTCTTGATGTAGTTGGCCTGACACCGACAATAATGAAGACTACCGTCTTACTAACAC

5. Codon Usage in Human Genome

6. Promoter Structure in Prokaryotes (E.Coli) Transcription starts at offset 0. Pribnow Box (-10) Gilbert Box (-30) Ribosomal Binding Site (+10)

7. TestCode Statistics Define a window size no less than 200 bp, slide the window the sequence down 3 bases. In each window: Calculate for each base {A, T, G, C} max (n 3k+1, n 3k+2, n 3k ) / min ( n 3k+1, n 3k+2, n 3k ) Use these values to obtain a probability from a lookup table (which was a previously defined and determined experimentally with known coding and noncoding sequences

8. Distribution of Each Base

9. Position Parameter

10. TestCode Sample Output Coding No opinion Non-coding

11. Popular Gene Prediction Algorithms GENSCAN: uses Hidden Markov Models (HMMs) TWINSCAN Uses both HMM and similarity (e.g., between human and mouse genomes)

12. Statistics

13. Weight

14. TestCode Method Compute A,C,G,T position and content parameters Look up from probability of coding value and get p 1, p 2, …p 8 Get corresponding weights w 1, w 2, …w 8 Compute p 1 w 1 + p 2 w 2 +…+p 8 w 8 This is the indicator of coding function

15. Comparing Genes in Two Genomes Small islands of similarity corresponding to similarities between exons

16. Finding Local Alignments Use local alignments to find all islands of similarity Human Genome Frog Genes (known)

17. Exon Chaining Problem: Graph Representation This problem can be solved with dynamic programming in O(n) time.

18. Infeasible Chains Red local similarities form two non -overlapping intervals but do not form a valid global alignment Human Genome Frog Genes (known)

19. Genome Rearrangements

21. Simple Rearrangements

22. Phylogenetic Reconstruction

23. Rearrangement Phylogeny

25. Turnip vs Cabbage: Look and Taste Different Although cabbages and turnips share a recent common ancestor, they look and taste different

26. Turnip vs Cabbage: Comparing Gene Sequences Yields No Evolutionary Information

27. Turnip vs Cabbage: Almost Identical mtDNA gene sequences In 1980s Jeffrey Palmer studied evolution of plant organelles by comparing mitochondrial genomes of the cabbage and turnip 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

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

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

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

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

32. Turnip vs Cabbage: Different mtDNA Gene Order Gene order comparison: Befo re After Evolution is manifested as the divergence in gene order

33. Transforming Cabbage into Turnip

34. 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

35. History of Chromosome X Rat Consortium, Nature, 2004

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

37. 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.

38. Reversals and Breakpoints 1 32 4 10 5 6 8 9 7 1, 2, 3, -8, -7, -6, -5, -4, 9, 10 The reversion introduced two breakpoints (disruptions in order).

39. Reversals: Example 5’ ATGCCTGTACTA 3’ 3’ TACGGACATGAT 5’ 5’ ATGTACAGGCTA 3’ 3’ TACATGTCCGAT 5’ Break and Invert

40. Types of Rearrangements Reversal 1 2 3 4 5 61 2 -5 -4 -3 6 Translocation 4 1 2 3 4 5 6 1 2 6 4 5 3 1 2 3 4 5 6 1 2 3 4 5 6 Fusion Fission

41. Comparative Genomic Architectures: Mouse vs Human Genome Humans and mice have similar genomes, but their genes are ordered differently ~245 rearrangements Reversals Fusions Fissions Translocation

42. Waardenburg ’ s Syndrome: Mouse Provides Insight into Human Genetic Disorder Waardenburg’s syndrome is characterized by pigmentary dysphasia Gene implicated in the disease was linked to human chromosome 2 but it was not clear where exactly it is located on chromosome 2

43. 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 succeeded in identifying location of gene responsible for disorder in mice Finding the gene in mice gives clues to where the same gene is located in humans

44. Comparative Genomic Architecture of Human and Mouse Genomes To locate where corresponding gene is in humans, we have to analyze the relative architecture of human and mouse genomes

45. Reversals: Example  = 1 2 3 4 5 6 7 8  (3,5) 1 2 5 4 3 6 7 8 

46. Reversals: Example  = 1 2 3 4 5 6 7 8  (3,5) 1 2 5 4 3 6 7 8  (5,6) 1 2 5 4 6 3 7 8

47. Reversals and Gene Orders Gene order is represented by a 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)

48. Reversal Distance Problem Goal: Given two permutations, find the shortest series of reversals that transforms 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  and 

49. 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

50. Sorting By Reversals: Example t =d(  ) - reversal distance of  Example :  = 3 4 2 1 5 6 7 10 9 8 4 3 2 1 5 6 7 10 9 8 4 3 2 1 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 So d(  ) = 3

51. Sorting by reversals: 5 steps

52. Sorting by reversals: 4 steps

53. What is the reversal distance for this permutation? Can it be sorted in 3 steps?

54.  =    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

55. 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

56. Adjacency & Breakpoints An adjacency - a pair of adjacent elements that are consecutive A breakpoint - a pair of adjacent elements that are not consecutive π = 5 6 2 1 3 4 0 5 6 2 1 3 4 7 adjacencies breakpoints Extend π with π 0 = 0 and π 7 = 7

57. We put two elements  0 =0 and  n + 1 =n+1 at the 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

58.  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 7 b(  ) = 2 0 1 2 3 4 5 6 7 b(  ) = 0 Reversal Distance and Breakpoints

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

60. Breakpoint Graph 1)Represent the elements of the permutation π = 2 3 1 4 6 5 as vertices in a graph (ordered along a line) 0 2 3 1 4 6 5 7 2)Connect vertices in order given by π with black edges (black path) 3)Connect vertices in order given by 1 2 3 4 5 6 with grey edges (grey path) 4) Superimpose black and grey paths

61. Two Equivalent Representations of the Breakpoint Graph 0 2 3 1 4 6 5 7 0 1 2 3 4 5 6 7 Consider the following Breakpoint Graph If we line up the gray path (instead of black path) on a horizontal line, then we would get the following graph Although they may look different, these two graphs are the same

62. What is the Effect of the Reversal ? 0 1 2 3 4 5 6 7 The gray paths stayed the same for both graphs There is a change in the graph at this point There is another change at this point How does a reversal change the breakpoint graph? Before: 0 2 3 1 4 6 5 7 After: 0 2 3 5 6 4 1 7 The black edges are unaffected by the reversal so they remain the same for both graphs

63. A reversal affects 4 edges in the breakpoint graph 0 1 2 3 4 5 6 7 A reversal removes 2 edges (red) and replaces them with 2 new edges (blue)

64. Effects of Reversals (Continued) Case 2: Both edges belong to different cycles Remove the center black edges and replace them with new black edges After the replacement, there now exists 1 cycle instead of 2 cycles c(πρ) – c(π) = -1 Therefore, for every permutation π and reversal ρ, c(πρ) – c(π) ≤ 1

65. Reversal Distance and Maximum Cycle Decomposition Since the identity permutation of size n contains the maximum cycle decomposition of n+1, c(identity) = n+1 c(identity) – c(π) equals the number of cycles that need to be “added” to c(π) while transforming π into the identity Based on the previous theorem, at best after each reversal, the cycle decomposition could increased by one, then: d(π) = c(identity) – c(π) = n+1 – c(π) Yet, not every reversal can increase the cycle decomposition Therefore, d(π) ≥ n+1 – c(π)

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

67. Signed Permutation Genes are directed fragments of DNA and we represent a genome by a signed permutation If genes are in the same position but there orientations are different, they do not have the equivalent gene order For example, these two permutations have the same order, but each gene’s orientation is the reverse; therefore, they are not equivalent gene sequences 1 2 3 4 5 -1 2 -3 -4 -5

68. From Signed to Unsigned Permutation 0 +3 -5 +8 -6 +4 -7 +9 +2 +1 +10 -11 12 Begin by constructing a normal signed breakpoint graph Redefine each vertex x with the following rules:  If vertex x is positive, replace vertex x with vertex 2x-1 and vertex 2x in that order  If vertex x is negative, replace vertex x with vertex 2x and vertex 2x-1 in that order  The extension vertices x = 0 and x = n+1 are kept as it was before 0 3a 3b 5a 5b 8a 8b 6a 6b 4a 4b 7a 7b 9a 9b 2a 2b 1a 1b 10a 10b 11a 11b 23 0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23 +3 -5 +8 -6 +4 -7 +9 +2 +1 +10 -11

69. From Signed to Unsigned Permutation (Continued) 0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23 Construct the breakpoint graph as usual Notice the alternating cycles in the graph between every other vertex pair Since these cycles came from the same signed vertex, we will not be performing any reversal on both pairs at the same time; therefore, these cycles can be removed from the graph

70. Interleaving Edges 0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23 Interleaving edges are grey edges that cross each other These 2 grey edges interleave Example: Edges (0,1) and (18, 19) are interleaving Cycles are interleaving if they have an interleaving edge

71. Interleaving Graphs 0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23 An Interleaving Graph is defined on the set of cycles in the Breakpoint graph and are connected by edges where cycles are interleaved A B C E F 0 5 6 10 9 15 16 12 11 7 8 14 13 17 18 3 4 1 2 19 20 22 21 23 A B C E F D DAB C EF

72. Interleaving Graphs (Continued) AB C DEF Oriented cycles are cycles that have the following form F C Unoriented cycles are cycles that have the following form Mark them on the interleave graph E In our example, A, B, D, E are unoriented cycles while C, F are oriented cycles

73. Hurdles Remove the oriented components from the interleaving graph AB C DEF The following is the breakpoint graph with these oriented components removed Hurdles are connected components that do not contain any other connected components within it A BD E Hurdle

74. Reversal Distance with Hurdles Hurdles are obstacles in the genome rearrangement problem They cause a higher number of required reversals for a permutation to transform into the identity permutation Taking into account of hurdles, the following formula gives a tighter bound on reversal distance: d(π) ≥ n+1 – c(π) + h(π) Let h(π) be the number of hurdles in permutation π

75. Median Problem Goal: find M so that D AM +D BM +D CM is minimized NP hard for most metric distances

76. Branch-and-Bound Algorithm Enumerate the solution genomes gene by gene (Genome Enumeration) After enumerated a gene, compute an upper bound based on the partial solution genome Bound: check whether the upper bound of the partial solution is less than a criteria Branch If it is true, the partial genome is discarded, enumerate another gene Otherwise update the criteria and continue enumeration

77. Genome Enumeration for Multichromosome Genomes Genome Enumeration For genomes on gene {1,2,3} 2 -2 2 2

78. Genome Enumeration for Multichromosome Genomes Genome Enumeration For genomes on gene {1,2,3} Red line: end fixing Black line: edge fixing 2 -2 2 2

80. Rearrangement Phylogeny

81. Compute A Given Tree (Start)

82. Compute A Given Tree (First Median)

83. Compute A Given Tree (Second Median)

84. Compute A Given Tree (Third Median)

85. Compute A Given Tree (After 1 st Iteration)

86. Statistical Approaches Combinatorial approaches are the focus of genome rearrangement research Only one MCMC method exists Maximum Likelihood methods have been very popular in sequence phylogenetic analysis Bootstrapping (data resampling) is a popular method to assess quality of obtained trees Hard to directly apply ML and bootstrapping to gene order

87. Sequence ML Example Aacgcaa Bacataa Catgtca Dgcgtta ABCDACBDADCB

88. All Possible Evolutionary Paths (Column 1) aaag a c g t

89. Likelihood for One Path aaag ag t

90. Sum of All Paths (Column 1) aaag a c g t

91. Whole Sequence ABCD

92. MLBE Convert the gene-orders into binary sequences based on adjacencies Convert the binary sequences into protein or DNA sequence Use RAxML to compute a ML tree on the sequences Binary encoding was used before for parsimony analysis, with reasonable results

93. Binary Encoding

94. MLBE Sequences

95. Experimental Results (Equal Content) 80% inversion, 20% transposition

96. Experimental Results (Unequal) 90% inversion, 10% of del/ins/dup, 5- 30 genes per segment

97. Post Analysis Bootstrapping has been widely used to assess the quality of sequence phylogeny The same procedure is impossible for gene order data since there is only one character We tested the procedure of jackknifing through simulated data to obtain Is jackknifing useful The best jackknifing rate What is the threshold of the support values

98. 98 DNA bootstrapping

99. Bootstrapping Results

100. Jackknifing Procedure Generate a new dataset by removing half of the genes from the original genomes (orders are preserved) Compute a tree on the new dataset Repeat K times and obtain K replicates Obtain a consensus tree with support values

101. An Example—New Genomes 1 2 3 4 5 6 7 8 9 10 1 -4 5 2 8 10 9 -7 -6 3 … 1 3 5 7 9 1 5 9 -7 3 …

102. Jackknifing Rate

103. Support Value Threshold - FP Up to 90% FP can be identified with 85% as the threshold

104. Support Value Threshold - FN

105. Low Support Branches

106. Jackknife Properties Jackknifing is necessary and useful for gene order phylogeny, and a large number of errors can be identified 40% jackknifing rate is reasonable 85% is a conservative threshold, 75% can also be used Low support branches should be examined in detail