1. Comparison of large sequences
18/09/2018
First part:
Alignment of large sequences

2. Dynamic programming What about genomes?
18/09/2018
accaccacaccacaacgagcata … acctgagcgatat a c . t
acc agt
| | | |xx
acc a--
Quadratic cost of space and time.
Quadratic cost of space and time.
Short sequences (up to bps) can be aligned using dynamic programming
As you have seen this morning ....
What about genomes?

3. (1 minute becomes 2 years)
Genomic sequences
18/09/2018
Genomic sequences have millions of base pairs.
The length of sequences is 1000 times longer.
The running time is times higher !
(1 second becomes 11 days)
(1 minute becomes 2 years)
Dealing with genomes increase the length of sequences 100 times …
then the cost is inceased at least one million times
…makes unpractical this algorithm
In which case Dynamic Programming can be applied?

4. First assumption ………………………….………………...…………...…. Genome A
……………………………………………………………….
………………………….………………...…………...….
Genome B
Genome A
……………………………………
…………………………….

5. Realistic assumption? Unrealistic assumption!
……………………………………………………………….
………………………….………………...…………...….
Genome B
Genome A
Unrealistic assumption!
………………………………………………………………….
………………………………………………...…………...….
Genome A
Genome B
More realistic assumption
…………………
………………
Genome A
Genome B

6. Realistic assumptions?
……………………………………………………………….
………………………….………………...…………...….
Genome B
Genome A
Unrealistic assumption!
…………………………………………………………………
………………………………………………...…………...….
Genome A
Genome B
More realistic assumption
But, now is it a real case?
…………………
………………
Genome A
Genome B

7. Preview in a real case    
Chlamidia muridarum: bps Chlamidia Thrachomatis: bps      

8. Preview in a real case Pyrococcus abyssis: 1.790.334 bps
Pyrococcus horikoshu: bps    

9. Methodology of an alignment
Make a preview:
……………………..….
…………………...….
1st:
2nd:
3th:
Identify the portions that can be aligned.
Make the alignment:
…..…………
……………….
(Linear cost)
(Linear cost)

10. Methodology of an alignment
Make a preview:
……………………..….
…………………...…. ?
1st:
2nd:
3th:
Identify the portions that can be aligned.
Make the alignment:
…..…………
……………….
(Linear cost)

11. Preview-Revisited MUM Maximal Unique Matching Connect to MALGEN
… a a t g….c t g...
… c g t g….c c c ...
Maximal
Unique
Matching
MUM
Connect to MALGEN

12. Methodology of an alignment
Make a preview:
……………………..….
…………………...….
1st:
2nd:
3th:
Linear cost
Suffix trees
with
How can MUMs be found?
Identify the portions that can be aligned.
Make the alignment:
…..…………
……………….
How can these portions be determined?
With CLUSTALW, TCOFFEE,…

13. Bioinformatics PhD. Course
18/09/2018
Second part:
Introducing Suffix trees

14. Suffix trees Given string ababaas: Suffixes: What kind of queries?

15. Applications of Suffix trees
1. Exact string matching
Does the sequence ababaas contain any ocurrence of patterns abab, aab, and ab?
………………………… a ba
baas,1
as,3
baas,2
as,4
s,6
as,5
s,7

16. Quadratic insertion algorithm
Invariant Properties:
Given the string …………………………...... 
and the suffix-tree
…...
P1: the leaves of suffixes from  have been inserted

17. Quadratic insertion algorithm
Given the string ababaabbs
ababaabbs,1

18. Quadratic insertion algorithm
Given the string ababaabbs
ababaabbs,1
babaabbs,2

19. Quadratic insertion algorithm
Given the string ababaabbs
aba
baabbs,1
ababaabbs,1
babaabbs,2

20. Quadratic insertion algorithm
Given the string ababaabbs
abbs,3
aba
baabbs,1
babaabbs,2

21. Quadratic insertion algorithm
Given the string ababaabbs
abbs,3
aba
baabbs,1 ba
baabbs,2
babaabbs,2

22. Quadratic insertion algorithm
Given the string ababaabbs
abbs,3
aba
baabbs,1 ba
baabbs,2
abbs,4

23. Quadratic insertion algorithm
Given the string ababaabbs
abbs,3
aba
baabbs,1
abbs,3 ba a
baabbs,1 ba
abbs,4
abbs,4
baabbs,2

24. Quadratic insertion algorithm
Given the string ababaabbs
abbs,5
abbs,3 ba a
baabbs,1 ba
abbs,4
abbs,4
baabbs,2

25. Quadratic insertion algorithm
Given the string ababaabbs
abbs,5
abbs,3 ba a
baabbs,1 ba
abbs,4
abbs,4
baabbs,2

26. Quadratic insertion algorithm
Given the string ababaabbs
abbs,5 a b a
abbs,3
baabbs,1 ba ba
abbs,4
abbs,4
baabbs,2

27. Quadratic insertion algorithm
Given the string ababaabbs
abbs,5 a
bs,6 b a
abbs,3
baabbs,1 ba
abbs,4
abbs,4
baabbs,2

28. Quadratic insertion algorithm
Given the string ababaabbs
abbs,5 a
bs,6 b a
abbs,3
baabbs,1 ba
abbs,4
abbs,4
baabbs,2

29. Quadratic insertion algorithm
Given the string ababaabbs
abbs,5 a
bs,6 b a
abbs,3
baabbs,1 a
baabbs,2 b
abbs,4
bs,7

30. Quadratic insertion algorithm
Given the string ababaabbs
abbs,5 a
bs,6 b a
abbs,3
baabbs,1 a
baabbs,2 b
abbs,4
bs,7
s,8

31. Quadratic insertion algorithm
Given the string ababaabbs
abbs,5 a
bs,6 b a
abbs,3
baabbs,1
s,9 a
baabbs,2 b
abbs,4
bs,7
s,7

32. Generalizad suffix tree
The suffix tree of many strings …
is called the generalized suffix tree …
and it is the suffix tree of the concatenation
of strings.
For instance,
the generalized suffix tree of ababaabb and aabaat …
is the suffix tree of ababaabαaabaatβ, :

33. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
Given the suffix tree of ababaabα :
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9 a
baabbα,2 b
abbα,4
bα,7
α,8

34. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9 a
baabbα,2 b
abbα,4
bα,7
α,8

35. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
aaβ,1 ab a
bα,5
bα,6 b a
abbα,3
baabbα,1
α,9 a
baabbα,2 b
abbα,4
bα,7
α,8

36. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
aaβ,1 ab a
bα,5
bα,6 b a
abbα,3
baabbα,1
α,9 a
baabbα,2 b
abbα,4
bα,7
α,8

37. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
aaβ,1 ab a
bα,5
β,2
bα,6 b a
α,9
bbα,3 a a
baabbα,2 b
abbα,4
bα,7
baabbα,1
α,8

38. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
aaβ,1 ab a
bα,5
β,2
bα,6 b a
α,9
bbα,3 a a
baabbα,2 b
abbα,4
bα,7
baabbα,1
α,8

39. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
aaβ,1 ab a
bα,5
β,2
bα,6 b a
α,9
bbα,3 a b
bα,7
baabbα,1
α,8 a
β,3 a
bbα,4
baabbα,2

40. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
aaβ,1 ab a
bα,5
β,2
bα,6 b a
α,9
bbα,3 a b
bα,7
baabbα,1
α,8 a
β,3 a
bbα,4
baabbα,2

41. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
β,4
aaβ,1 a b a
bα,5
β,2
bα,6 b a
α,9
bbα,3 a b
bα,7
baabbα,1
α,8 a
β,3 a
bbα,4
baabbα,2

42. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
β,4
aaβ,1 a b a
bα,5
β,2
bα,6 b a
α,9
bbα,3 a b
bα,7
baabbα,1
α,8 a
β,3 a
bbα,4
baabbα,2

43. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
β,5
β,4
aaβ,1 a b a
bα,5
β,2
bα,6 b a
α,9
bbα,3 a b
bα,7
baabbα,1
α,8 a
β,3 a
bbα,4
baabbα,2

44. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
β,5
β,4
aaβ,1 a b a
bα,5
β,2
bα,6 b a
α,9
bbα,3 a b
bα,7
baabbα,1
α,8 a
β,3 a
bbα,4
baabbα,2

45. Generalizad suffix tree
Construction of the suffix tree of ababaabbαaabaaβ :
β,5
β,4
β,6
aaβ,1 a b a
bα,5
β,2
bα,6 b a
α,9
bbα,3 a b
bα,7
baabbα,1
α,8 a
β,3 a
bbα,4
baabbα,2

46. Generalizad suffix tree
Generalized suffix tree of ababaabbαaabaaβ : a
bα,5 b
bbα,3
baabbα,1
bα,6
baabbα,2
bbα,4
bα,7
α,8
α,9
aaβ,1
β,2
β,3
β,4
β,5
β,6

47. Applications of Suffix trees
1. Exact string matching
Does the sequence ababaas contain any ocurrence of patterns abab, aab, and ab?
………………………… a ba
baas,1
as,3
baas,2
as,4
s,6
as,5
s,7

48. Applications of Suffix trees
2. The substring problem for a database of strings DB
Does the DB contain any ocurrence of patterns abab, aab, and ab? a
bα,5 b
bbα,3
baabbα,1
bα,6
baabbα,2
bbα,4
bα,7
α,8
α,9
aaβ,1
β,2
β,3
β,4
β,5
β,6

49. Applications of Suffix trees
3. The longest common substring of two strings a
bα,5 b
bbα,3
baabbα,1
bα,6
baabbα,2
bbα,4
bα,7
α,8
α,9
aaβ,1
β,2
β,3
β,4
β,5
β,6

50. Applications of Suffix trees
5. Finding MUMs. a
bα,5 b
bbα,3
baabbα,1
bα,6
baabbα,2
bbα,4
bα,7
α,8
α,9
aaβ,1
β,2
β,3
β,4
β,5
β,6

51. Bioinformatics PhD. Course
18/09/2018
Third part:
Suffix links

52. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8

53. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 ? α,9 bα,7 a
α,8

54. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8 ?

55. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8 ?

56. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8 ?

57. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8 ?

58. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8 ?

59. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8 ?

60. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8 ?

61. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8

62. Suffix links a  abbα,5 a bα,6 b a abbα,3 baabbα,1 α,9 bα,7 a
α,8

63. Traversal using Suffix links
Given S2 = a a b a a
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

64. Traversal using Suffix links
Given S2 = a a b a a
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

65. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a
aa in S2 [1]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

66. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a
aa in S2 [1]
aab in S2 [1] =
S1[5..6-7] in S2 [1]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

67. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a
S1[5..6-7] in S2 [1]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

68. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a
S1[5..6-7] in S2 [1]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

69. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a b b a
S1[5..6-7] in S2 [1]
S1[3..6-…] in S2 [2]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

70. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a b b a
S1[5..6-7] in S2 [1]
S1[3..6-…] in S2 [2]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

71. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a b b a
S1[5..6-7] in S2 [1]
S1[3..6-…] in S2 [2]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

72. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a b b a
S1[5..6-7] in S2 [1]
S1[3..6-…] in S2 [2]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

73. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a b b a
S1[5..6-7] in S2 [1]
S1[3..6-8] in S2 [2]
S1[4..6-8] in S2 [3]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

74. Traversal using Suffix links
Unique matchings
Given S2 = a a b a a b b a
S1[5..8] in S2 [4]
S1[3..6-8] in S2 [2]
S1[4..6-8] in S2 [3]
S1[6..8] in S2 [5]
S1[7..8] in S2 [6]
abbα,5 a
bα,6 b a
abbα,3
baabbα,1
α,9
bα,7 a
baabbα,2 b
abbα,4
α,8

75. From UMs to MUMs Unique matchings Given S2 = a a b a a b b a
S1[5..8] in S2 [4]
and S1 = a b a b a a b b α
S1[3..6-8] in S2 [2]
S1[4..6-8] in S2 [3]
Array of UMs
S1[6..8] in S2 [5] 1 2
5 8
6 8
7 8 8 9
S1[7..8] in S2 [6]
MUM: S1[3..6-8] in S2[2]

76. Bioinformatics PhD. Course
18/09/2018
Third part:
Linear insertion algorithm

77. Quadratic insertion algorithm
Invariant Properties:
Given the string …………………………...... 
and the suffix-tree
…...
P1: the leaves of suffixes from  have been inserted

78. Linear insertion algorithm
Invariant Properties: 
Given the string …………………………...... 
and the suffix-tree 
…...
P1: the leaves of suffixes from  have been inserted
P2: the string  is the longest string that can be
spelt through the tree.

79. Linear insertion algorithm: example  a 
Given the string ababaababb... ba
baababb...,2 a
ababb...,5
ababb...,3
baababb...,1
ababb...,4

80. Linear insertion algorithm: example 
Given the string ababaababb...
6 7 8 ba
baababb...,2 a
ababb...,5
ababb...,3
baababb...,1
ababb...,4

81. Linear insertion algorithm: example
Given the string ababaababb...
6 7 8  ba
baababb...,2 a
ababb...,5
ababb...,3
baababb...,1
ababb...,4

82. Linear insertion algorithm: example
Given the string ababaababb...
6 7 89  ba
baababb...,2 a
ababb...,5
ababb...,3
baababb...,1
ababb...,4

83. Linear insertion algorithm: example
Given the string ababaababb...  
6 7 89 a
ababb...,5 ba
ababb...,3
baababb...,1
baababb...,2
ababb...,4
baababb...,1 b
b...,6
aababb...,1

84. Linear insertion algorithm: example
Given the string ababaababb...  
7 89 a
ababb...,5 ba
ababb...,3
baababb...,2
ababb...,4 b
b...,6
aababb...,1

85. Linear insertion algorithm: example
Given the string ababaababb...  
7 89 a
ababb...,5 ba
ababb...,3
baababb...,2
ababb...,4 b
b...,6
aababb...,1

86. Linear insertion algorithm: example
Given the string ababaababb...  
7 89 a
ababb...,5
ababb...,3 ba b
b...,6
aababb...,1 ba
ababb...,4
baababb...,2

87. Linear insertion algorithm: example
Given the string ababaababb...  
7 89 a
ababb...,5
ababb...,3 ba b
b...,6
aababb...,1 ba
ababb...,4 b
aababb...,2
baababb...,2
baababb...,2

88. Linear insertion algorithm: example
Given the string ababaababb...
7 8…  a
ababb...,5
ababb...,3 ba b
b...,6
aababb...,1 ba
ababb...,4
baababb...,2 b
b...,7
aababb...,2
baababb...,2

89. Linear insertion algorithm: example
Given the string ababaababb... 89  a
ababb...,5
ababb...,3 ba b
b...,6
aababb...,1 ba
ababb...,4 b
b...,7
aababb...,2

90. Linear insertion algorithm: example
Given the string ababaababb... 89  a
ababb...,5
ababb...,3 ba b
b...,6
aababb...,1 ba
ababb...,4 b
b...,7
aababb...,2

91. Linear insertion algorithm: example
Given the string ababaababb... 89  a
ababb...,5
ababb...,3 ba b
b...,6
aababb...,1 ba
ababb...,4 b
b...,7
aababb...,2

92. Linear insertion algorithm: example
Given the string ababaababb... 89  a
ababb...,5
ababb...,3 ba b
b...,6
aababb...,1 ba
ababb...,4 b
b...,7
aababb...,2

93. Linear insertion algorithm: example
Given the string ababaababb... 89  a
ababb...,5
ababb...,3 b
b...,6
aababb...,1 a b ba
ababb...,4 b
aababb...,2
b...,7

94. Linear insertion algorithm: example
Given the string ababaababb... 89  a
ababb...,5
ababb...,3 b
b...,6
aababb...,1 a b
b...,8 ba
ababb...,4 b
aababb...,2
b...,7

95. Linear insertion algorithm: example
Given the string ababaababb... 9  a
ababb...,5
ababb...,3 b
b...,6
aababb...,1 a b
b...,8 ba
ababb...,4 b
aababb...,2
b...,7

96. Linear insertion algorithm: example
Given the string ababaababb... 9  a
ababb...,5
ababb...,3 b
b...,6
aababb...,1 a b
b...,8 ba
ababb...,4 b
aababb...,2
b...,7

97. Linear insertion algorithm: example
Given the string ababaababb... 9  a
ababb...,5
ababb...,3 b
b...,6
aababb...,1 a b
b...,8 a b
ababb...,4 b
aababb...,2
b...,7

98. Linear insertion algorithm: example
Given the string ababaababb... 9  a
ababb...,5
ababb...,3 b
b...,6
aababb...,1 a b
b...,8 a b
ababb...,4
b...,9 b
aababb...,2
b...,7

99. Linear insertion algorithm: example
Given the string ababaababb... 9  a
ababb...,5
ababb...,3 b
b...,6
ababb...,1 a b
b...,8 a b
ababb...,4
b...,9 b
aababb...,2
b...,7

100. Linear insertion algorithm: example
Given the string ababaababb... 9  a
ababb...,5
ababb...,3 b
b...,6
ababb...,1 a b
b...,8 a b
ababb...,4
b...,9 b
aababb...,2
b...,7

101. Linear insertion algorithm: example
Given the string ababaababb... 9  a
ababb...,5
ababb...,3 b
b...,6
ababb...,1 a b
b...,8 a b
ababb...,4
b...,9 b
aababb...,2
b...,7

102. Index
Suffix arrays
Suffix-arrays: a new method for on-line string searches,
G. Myers, U. Manber

103. Suffix arrays Given string ababaa#: Suffixes:
… but lexicographically sorted
2: babaa# 1 2 3 4 5 6 7
1: #
3: abaa#
6: a#
4: baa#
5: aa#
5: aa#
6: a#
7: #
3: abaa#
1: ababaa#
4: baa#
2: babaa#
Which is the cost?
O(n log(n))

104. Applications of suffix arrays
1. Exact string matching
Does the sequence ababaas contain any ocurrence of patterns abab, aab, and ab?
1: ababaa#
2: babaa#
3: abaa#
4: baa#
5: aa#
6: a#
1: # 1 2 3 4 5 6 7
Binary search
… which is the cost?
O(log(n) |P|)
Can it be improved to …
O(log(n)+|P|) ?

105. Fast search with cost O(log(n)+|P|)1 2
… … n
Suffix array
Query:
Invariant Properties:
P1: α < query ≤ β α β
P2: matches pref( query)

106. Fast search with cost O(log(n)+|P|)1 2
… … n
Suffix array
P2: matches pref( query)
Query:
Invariant Properties:
P1: α < query ≤ β α β γ
If suff(γ)<suff(query) then α = γ
else β = γ
Algorithm: