1. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Phylogenetic Reconstruction: Parsimony Anders Gorm Pedersen gorm@cbs.dtu.dk

2. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Trees: terminology

3. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Trees: terminology

4. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Trees: terminology “Reptilia” is a non-monophyletic group

5. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Trees: representations Three different representations of the same tree-topology

6. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Trees: rooted vs. unrooted A rooted tree has a single node (the root) that represents a point in time that is earlier than any other node in the tree. A rooted tree has directionality (nodes can be ordered in terms of “earlier” or “later”). In the rooted tree, distance between two nodes is represented along the time-axis only (the second axis just helps spread out the leafs) EarlyLate

7. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Trees: rooted vs. unrooted EarlyLate A rooted tree has a single node (the root) that represents a point in time that is earlier than any other node in the tree. A rooted tree has directionality (nodes can be ordered in terms of “earlier” or “later”). In the rooted tree, distance between two nodes is represented along the time-axis only (the second axis just helps spread out the leafs)

8. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Trees: rooted vs. unrooted EarlyLate A rooted tree has a single node (the root) that represents a point in time that is earlier than any other node in the tree. A rooted tree has directionality (nodes can be ordered in terms of “earlier” or “later”). In the rooted tree, distance between two nodes is represented along the time-axis only (the second axis just helps spread out the leafs)

9. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Trees: rooted vs. unrooted In unrooted trees there is no directionality: we do not know if a node is earlier or later than another nodeIn unrooted trees there is no directionality: we do not know if a node is earlier or later than another node Distance along branches directly represents node distanceDistance along branches directly represents node distance

10. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Trees: rooted vs. unrooted In unrooted trees there is no directionality: we do not know if a node is earlier or later than another nodeIn unrooted trees there is no directionality: we do not know if a node is earlier or later than another node Distance along branches directly represents node distanceDistance along branches directly represents node distance

11. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Reconstructing a tree using non- contemporaneous data

12. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Cladistics: group organisms based on shared, derived characters (“synapomorphies”)

13. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Homology: limb structure Homology: any similarity between characters that is due to their shared ancestry

14. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Homology vs. Homoplasy Homology: similar traits inherited from a common ancestor Homoplasy: similar traits are not directly caused by common ancestry (convergent evolution). X X XX

15. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Homoplasy: wings

16. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Molecular phylogeny A A G C G T T G G G C A A B A G C G T T T G G C A A C A G C T T T G T G C A A D A G C T T T T T G C A A 1 2 3 1 2 3 DNA and protein sequencesDNA and protein sequences Homologous characters inferred from alignment.Homologous characters inferred from alignment. Other molecular data: absence/presence of restriction sites, DNA hybridization data, antibody cross-reactivity, etc. (but losing importance due to cheap, efficient sequencing).Other molecular data: absence/presence of restriction sites, DNA hybridization data, antibody cross-reactivity, etc. (but losing importance due to cheap, efficient sequencing).

17. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Morphology vs. molecular data African white-backed vulture (old world vulture) Andean condor (new world vulture) New and old world vultures seem to be closely related based on morphology. Molecular data indicates that old world vultures are related to birds of prey (falcons, hawks, etc.) while new world vultures are more closely related to storks Similar features presumably the result of convergent evolution

18. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Phylogenetic reconstruction Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

19. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Phylogenetic reconstruction Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

20. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Parsimony criterion: choose simplest hypothesis Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

21. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Parsimonious reconstruction A G.. B G.. C T.. D T.. Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

22. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Parsimonious reconstruction A G.. B G.. C T.. D T.. G.. Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

23. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Parsimonious reconstruction A G.. B G.. C T.. D T.. G.. Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

24. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Alternative tree: homoplasy A G.. B G.. C T.. D T.. A G.. B G.. C T.. D T.. Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT T.. G..

25. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Alternative tree: homoplasy A G.. B G.. C T.. D T.. A G.. B G.. C T.. D T.. Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT T.. G..

26. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Alternative tree: homoplasy A G.. B G.. C T.. D T.. A G.. B G.. D T.. C T.. Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT T.. G.. T..

27. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS One character: Assumption of no homoplasy is equivalent to finding shortest tree A G... B G... C T... D T... A G.. B G.. C T.. D T.. Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT T.. G.. T..

28. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Phylogenetic reconstruction A..G B..G C..T D..T..G Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

29. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Phylogenetic reconstruction A G.G B G.G C T.T D T.T G.G Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

30. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Phylogenetic reconstruction: conflicts ABCD A.G. C.G. B.T. D.T..G. Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

31. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Phylogenetic reconstruction: conflicts A.G.A.G. B.T. C.G.C.G. D.T. ACBD Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

32. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Phylogenetic reconstruction: conflicts ABCD AG.GAG.G C T.T BG.GBG.G D T.T Taxon Nucleotide position 12 3 AGGG BGTG CTGT DTTT

33. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Several characters: choose shortest tree (equivalent to fewer assumptions of homoplasy) AGGGAGGG B GTG CTGTCTGT D TTT AGGGAGGG BGTGBGTG C TGT D TTT GTG TTT TGT Total length of tree: 4 Total length of tree: 5

34. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Maximum Parsimony Maximum parsimony: the best tree is the shortest tree (the tree requiring the smallest number of mutational events)Maximum parsimony: the best tree is the shortest tree (the tree requiring the smallest number of mutational events) This corresponds to the tree that implies the least amount of homoplasy (convergent evolution, reversals)This corresponds to the tree that implies the least amount of homoplasy (convergent evolution, reversals) How do we find the best tree for a given data set?How do we find the best tree for a given data set?

35. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Maximum Parsimony: first approach 1.Construct list of all possible trees for data set 2.For each tree: determine length, add to list of lengths 3.When finished: select shortest tree from list 4.If several trees have the same length, then they are equally good (equally parsimonious)

36. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Maximum Parsimony: problems We need algorithm for constructing list of all possible treesWe need algorithm for constructing list of all possible trees We need algorithm for determining length of given treeWe need algorithm for determining length of given tree Should all mutational events have same cost?Should all mutational events have same cost?

37. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Constructing list of all possible unrooted trees 1.Construct unrooted tree from first three taxa. There is only one way of doing this 2.Starting from (1), construct the three possible derived trees by adding taxon 4 to each internal branch 3.From each of the trees constructed in step (2), construct the five possible derived trees by adding taxon 5 to each internal branch. 4.Continue until all taxa have been added in all possible locations

38. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Maximum Parsimony: problems We need algorithm for constructing list of all possible trees We need algorithm for constructing list of all possible trees  We need algorithm for determining length of given treeWe need algorithm for determining length of given tree Should all mutational events have same cost?Should all mutational events have same cost?

39. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch What is the length of this tree? (How many mutational steps are required?) C A C A G

40. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Root the tree at an arbitrary internal node (or internal branch)Root the tree at an arbitrary internal node (or internal branch) Visit an internal node x for which no state set has been defined, but where the state sets of x’s immediate descendants (y,z) have been defined.Visit an internal node x for which no state set has been defined, but where the state sets of x’s immediate descendants (y,z) have been defined. If the state sets of y,z have common states, then assign these to x.If the state sets of y,z have common states, then assign these to x. If there are no common states, then assign the union of y,z to x, and increase tree length by one. Repeat until all internal nodes have been visited. Note length of current tree.Repeat until all internal nodes have been visited. Note length of current tree. Algorithm for determining length of given tree: Fitch

41. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch C A C A G

42. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch C A C A G 

43. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch C A C A G 

44. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch CACAG

45. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch CACAG Length so far = 0

46. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch Length so far = 1 {C, A} CACAG

47. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch Length so far = 2 {C, A} {A, G} CACAG

48. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch Length so far = 3 {A, C} {A, G} {A, C, G} CACAG

49. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch {A, C} {A, G} {A, C, G} {A, C} Length so far = 3 CACAG

50. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch Length of tree = 3 {A, C} {A, G} {A, C, G} {A, C} CACAG

51. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Fitch CACAG Length of tree = 3 A A A A One possible reconstruction (several others exist)

52. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Maximum Parsimony: problems We need algorithm for constructing list of all possible trees We need algorithm for constructing list of all possible trees  We need algorithm for determining length of given tree We need algorithm for determining length of given tree  Should all mutational events have same cost?Should all mutational events have same cost?

53. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Mutational events need not have the same cost A C G T A 0 3 1 3 C 3 0 3 1 G 1 3 0 3 T 3 1 3 0 CACAG Sankoff algorithm

54. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Maximum Parsimony: problems We need algorithm for constructing list of all possible trees We need algorithm for constructing list of all possible trees  We need algorithm for determining length of given tree We need algorithm for determining length of given tree  Should all mutational events have same cost? Should all mutational events have same cost? 

55. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS How many branches are there on an unrooted tree with x tips? There is only one way of con- structing the first tree. This tree has 3 tips and 3 branchesThere is only one way of con- structing the first tree. This tree has 3 tips and 3 branches Each time an extra taxon is added, two branches are created.Each time an extra taxon is added, two branches are created. A tree with x tips will therefore have the following number of branches:A tree with x tips will therefore have the following number of branches: n branches = 3+(x-3)*2 = 3+2x-6 = 2x-3 A B C A B C D

56. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS How many unrooted trees are there? A tree with x tips has 2x-3 branchesA tree with x tips has 2x-3 branches For each tree with x=n-1 tips, we can therefore construct 2(n-1)-3 derived trees (with n tips).For each tree with x=n-1 tips, we can therefore construct 2(n-1)-3 derived trees (with n tips). The number of unrooted trees with n tips is therefore:The number of unrooted trees with n tips is therefore:  (2i-3) = 1 x 3 x 5 x 7 x... i=2 n-1

57. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Exhaustive search impossible for large data sets No. taxa No. trees 31 43 515 6105 7945 810,395 9135,135 102,027,025 1134,459,425 12654,729,075 1313,749,310,575 14316,234,143,225 157,905,853,580,625

58. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Branch and bound: shortcut to perfection Step 1: Construction of initial tree by sequential addition

59. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Branch and bound: shortcut to perfection Step 2: search treespace, discard suboptimal parts

60. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Heuristic search 1.Construct initial tree (e.g., sequential addition); determine length 2.Construct set of “neighboring trees” by making small rearrangements of initial tree; determine lengths 3.If any of the neighboring trees are better than the initial tree, then select it/them and use as starting point for new round of rearrangements. (Possibly several neighbors are equally good) 4.Repeat steps 2+3 until you have found a tree that is better than all of its neighbors. 5.This tree is a “local optimum” (not necessarily a global optimum!)

61. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Heuristic search: hill-climbing

62. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Heuristic search: local vs. global optimum

63. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Types of rearrangement I: nearest neighbor interchange (NNI) Original tree Two neighbors per internal branch: tree with n tips has 2(n-3) neighbors (For example, a tree with 20 tips has 34 neighbbors)

64. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Types of rearrangement II: subtree pruning and regrafting (SPR)

65. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Types of rearrangement III: tree bisection and reconnection (TBR)

66. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Algorithm for determining length of given tree: Sankoff A CGT CACAG

67. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: length of subtrees starting at terminal node A 0GT CACAG

68. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: length of subtrees starting at terminal node  0 CACAG

69. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: length of subtrees starting at terminal node  0 C0 ACAG

70. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: length of subtrees starting at terminal node  0 C0 A0 CAG

71. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: length of subtrees starting at terminal node  0 C0 A0 C0 AG

72. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: length of subtrees starting at terminal node  0 C0 A0 C0 A0 G

73. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees starting at internal node?  0 C0 A0 C0 A0 G????

74. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node  0 C0 AS S A = min i [cost Ai + S left, i ] + min j [cost Aj + S right, j ] cost AA = 0, cost AC = cost AG = cost AT = 1

75. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node  0 C? 0 Nt on left branch CostA0 C G T S A = min i [cost Ai + S left, i ] + min j [cost Aj + S right, j ]

76. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node  0 C? 0 Nt on left branch CostA 0 +  C G T S A = min i [cost Ai + S left, i ] + min j [cost Aj + S right, j ]

77. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node  0 C? 0 Nt on left branch CostA 0 +  =  C G T S A = min i [cost Ai + S left, i ] + min j [cost Aj + S right, j ]

78. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node  0 C? Nt on left branch CostA 0 +  =  C 1 + 0 = 1 G T 1 S A = min i [cost Ai + S left, i ] + min j [cost Aj + S right, j ]

79. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node  0 C? Nt on left branch CostA 0 +  =  C 1 + 0 = 1 G 1 +  =  T 1 S A = min i [cost Ai + S left, i ] + min j [cost Aj + S right, j ]

80. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node  0 C? Nt on left branch CostA 0 +  =  C 1 + 0 = 1 G 1 +  =  T 1 S A = min i [cost Ai + S left, i ] + min j [cost Aj + S right, j ]

81. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node  0 C? Nt on left branch CostA 0 +  =  C 1 + 0 = 1 G 1 +  =  T 1 S A = min i [cost Ai + S left, i ] + min j [cost Aj + S right, j ]

82. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node Nt on left branch Cost A 0 + 0 = 0 C G T S A = 1 + min j [cost Aj + S right, j ]0 AS 0

83. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node Nt on left branch Cost A 0 + 0 = 0 C 1 +  =  G T S A = 1 + min j [cost Aj + S right, j ]0 AS 1

84. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node Nt on left branch Cost A 0 + 0 = 0 C 1 +  =  G T S A = 1 + min j [cost Aj + S right, j ]0 AS 1

85. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node Nt on left branch Cost A 0 + 0 = 0 C 1 +  =  G T S A = 1 + min j [cost Aj + S right, j ]0 AS 1

86. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node Nt on left branch Cost A 0 + 0 = 0 C 1 +  =  G T S A = 1 + min j [cost Aj + S right, j ]0 AS 0

87. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node S A = min i [cost Ai + S left, i ] + min j [cost Aj + S right, j ] S A = 1 + 0 = 1  0 C0 A1 1 0

88. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “A” at internal node  0 C0 A0 C0 A0 G1

89. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “C” at internal node  0 C0 A11 0 1 S C = 0 + 1 = 1

90. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “G” at internal node  0 C0 A112 1 1 S G = 1 + 1 = 2

91. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees having nucleotide “T” at internal node  0 C0 A1122 1 1 S T = 1 + 1 = 2

92. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of all possible subtrees starting at internal node  0 C0 A0 C0 A0 G1122

93. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: minimal length of possible subtrees starting at all internal nodes  0 C0 A0 C0 A0 G1122 1212 2223 3345

94. CENTER FOR BIOLOGICAL SEQUENCE ANALYSIS Sankoff: smallest possible length of tree = 3  0 C0 A0 C0 A0 G1122 1212 2223 3345