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  1. TREES

  2. Gorilla Human Chimp Human Chimp Gorilla Trees = Gorilla Chimp Human = = Chimp Human Gorilla

  3. s1 s1 s2 s2 s3 s3 s4 s4 s5 s5 Same thing… =

  4. Terminology A branch =An edge The root Internal nodes Chicken Gorilla Human Chimp External node - leaf

  5. אלו מהמשפטים הבאים נכון, בהתייחס לעץ הנ"ל? א. האדם והגורילה יותר קרובים זה לזה מהשימפנזה והגורילה. ב. האדם קרוב לתרנגולת ולברווז באותה מידה. ג. התרנגולת יותר קרובה לגורילה מהאדם. ד. א'+ב'. ה. א'+ג'. ו. ב'+ג'. ז. א'+ב'+ג'. ח. אף תשובה אינה נכונה. תרגיל

  6. The maximum parsimony principle. Tree building

  7. Tree building

  8. Tree building Evaluate this tree… s2 s1 s4 s3 s5

  9. Tree building Gene number 1 1 0 1 s1 s4 s3 s2 s5 1 1 1 0 0

  10. Tree building 1 0 Gene number 1, Option number 1. 1 1 s1 s4 s3 s2 s5 1 1 1 0 0

  11. Tree building 0 1 0 1 1 1 1 0 0 Gene number 1, Option number 2. s1 s4 s3 s2 s5 Number of changes for gene 1 (character 1) = 1

  12. Tree building 0 1 0 1 0 1 1 0 0 Gene number 2, Option number 1. s2 s1 s4 s3 s5

  13. Tree building 1 1 0 1 0 1 1 0 0 Gene number 2, Option number 2. s2 s1 s4 s3 s5

  14. Tree building 0 0 0 0 0 1 1 0 0 Gene number 2, Option number 3. s2 s1 s4 s3 s5 Number of changes for gene 2 (character 2) = 2

  15. Tree building 0 0 1 0 0 0 0 1 1 Gene number 3, Option number 1. s2 s1 s4 s3 s5

  16. Tree building 1 0 1 0 0 0 0 1 1 Gene number 3, Option number 2. s2 s1 s4 s3 s5 Number of changes for gene 3 (character 3) = 1

  17. Tree building 1 1 1 1 1 1 0 0 1 Gene number 4, Option number 1. s2 s1 s4 s3 s5

  18. Tree building 0 0 0 1 1 1 0 0 1 Gene number 4, Option number 2. s2 s1 s4 s3 s5 Number of changes for gene 4 (character 4) = 2

  19. Tree building Gene number 5 is the same as Gene number 4 Number of changes for gene 5 (character 5) = 2

  20. Tree building 0 0 0 0 0 1 0 0 0 Gene number 6, 1 option only: s2 s1 s4 s3 s5 Number of changes for gene 6 (character 6) = 1

  21. Tree building Sum of changes Number of changes for gene 1 (character 1) = 1 Number of changes for gene 2 (character 2) = 2 Number of changes for gene 3 (character 3) = 1 Number of changes for gene 4 (character 4) = 2 Number of changes for gene 5 (character 5) = 2 Number of changes for gene 6 (character 6) = 1 Sum of changes for this tree topology = 9 Can we do better ???

  22. Tree building The MP (most parsimonious) tree: s2 s1 s4 s3 s5 Sum of changes for this tree topology = 8

  23. How to efficiently compute the MP score of a tree

  24. The Fitch algorithm (1971): {A,C} U {A,C,G} U {A,G} U {A,C} U C A C A G Chicken Duck Gorilla Human Chimp Postorder tree scan. In each node, if the intersection between the leaves is empty: we apply a union operator. Otherwise, an intersection.

  25. Number of changes {A,C} U {A,C,G} U {A,G} U {A,C} U C A C A G Chicken Duck Gorilla Human Chimp Total number of changes = number of union operators => 3 in this case.

  26. תרגיל CAAG GAAA GCGA GACA GGGA Chicken Duck Gorilla Human Chimp Find minimum number of changes.

  27. Chimpanzee Gorilla Human

  28. Human ACTAG Chimp ACAAC Gorilla AAAAT Gorilla Chimp Human U Position 1 A A A 0 Position 2 A C C 1 Position 3 A A T 1 Position 4 A A A 0 Position 5 T C G 2 4

  29. Human ACTAG Chimp ACAAC Gorilla AAAAT Gorilla Chimp Human U Position 1 A A A 0 Position 2 A C C 1 Position 3 A A T 1 Position 4 A A A 0 Position 5 T C G 2 4

  30. Human ACTAG Chimp ACAAC Gorilla AAAAT Chimp Gorilla Human U Position 1 A A A 0 Position 2 C A C 1 Position 3 A A T 1 Position 4 A A A 0 Position 5 C T G 2 4

  31. Human ACTAG Chimp ACAAC Gorilla AAAAT Chimp Gorilla Human Gorilla Chimp Human Gorilla Chimp Human These 3 trees will ALWAYS get the same score

  32. The unrooted tree represents a set of rooted trees 3 1 2 3 1 2

  33. A general observation: the position of the root does not affect the MP score. E A D D E C A B B C A B C E D A B C E D

  34. 1 0 Intuition as to why rooting does not change the score. 1 1 s1 s4 s3 s2 s5 1 1 1 0 0 The change will always be on the same branch, no matter where the root is positioned…

  35. Which is not a rooted version of this tree? E T3 תרגיל A D C E D A B B C T1 T2 A B D E C A B C D E

  36. Gorilla gorilla (Gorilla) Pan troglodytes (Chimpanzee) Homo sapiens (human) Gallus gallus (chicken)

  37. Human Human Human Chicken Chimp Chimp Gorilla Chicken Gorilla Chimp Gorilla Chicken Evaluate all 3 possible UNROOTED trees: MP tree

  38. Rooting based on a priori knowledge: Human Chicken Gorilla Chimp Chicken Gorilla Human Chimp

  39. Ingroup / Outgroup: Chicken Gorilla Human Chimp OUTGROUP INGROUP

  40. Chicken Duck Gorilla Human Chimp Subtrees A subtree

  41. Monophyletic groups Chicken Gorilla Human Chimp The Gorilla+Human+Chimp are monophyletic.A clade is a monophyletic group.

  42. Paraphyletic = Non-monophyletic groups Whale Chimp Drosophila Zebrafish The Zebrafish+Whale are paraphyletic

  43. When an unrooted tree is given, you cannot know which groups are monophyletic. You can only say which are not. Human Chicken Rat Gorilla Chimp Chicken + Rat seems to be monophyletic but they are not, since the root of the tree is between Chicken and the rest. Human and Gorilla are not monophyletic no matter where the root is…

  44. HOW MANY TREES

  45. a b a b c b a a c c b a c a c b b c a d b d d b b c a b b c a a a a a d d c d d c c b d d b c c b c b a a b d a d c d c b d c b b d d c a a a TR = “TREE ROOTED” How many rooted trees N=2, TR(2) = 1 N=3, TR(3) = 3 N=4, TR(4) = 15

  46. a b c a b b a c d d b c a TR = “TREE ROOTED” How many rooted trees c c c 2 branches. 3 possible places to add “c” 4 branches. 5 possible places to add “d” 6 branches. 7 possible places to add “e” The number of branches is increased by 2 each time. The number of branches is an arithmetic series. 0,2,4,6,8,…. A(n) = A(1)+(n-1)d. A(1) = 0; d=2. => A(n) = (n-1)*2 = 2n-2

  47. a b TR = “TREE ROOTED” How many rooted trees The number of branches is increased by 2 each time. The number of branches is an arithmetic series. 0,2,4,6,8,…. A(n) = A(1)+(n-1)d. A(1) = 0; d=2. => A(n) = (n-1)*2 = 2n-2 c c c 2 branches. 3 possible places to add “c” Each time we can add a new branch in Br(n)+1 places. [Br(n)=number of branches] [Tr(n)=number of trees with n sequences] TR(n+1) = TR(n)*(BR(n)+1)=TR(n)*(2n-1) TR(5) = TR(4)*7=TR(3)*5*7=TR(2)*3*5*7=1*3*5*7 … TR(n) = 1*3*5*7*…..*(2n-3)

  48. TR = “TREE ROOTED” How many rooted trees n!=1*2*3*4*5*6…..*n = n factorial. TR(n) = 1*3*5*7*…..*(2n-3) = 1*2*3*4*5*6*7*…*(2n-3) = 2*4*6*8*….*(2n-4) 1*2*3*4*5*6*7*…*(2n-3) = (2*1)*(2*2)*(2*3)*(2*4)*….*(2*(n-2)) (2n-3)! = (2(n-2))*(1*2*3*4*….(n-2)) (2n-3)! = (2(n-2))*(n-2)!

  49. TR = “TREE ROOTED” How many rooted trees TR(n) = 1*3*5*7*…..*(2n-3) = =(2n-3)!! (2n-3)! = (2(N-2))*(n-2)!