1. CSCI2950-C Lecture 7 Molecular Evolution and Phylogeny

2. Early Evolutionary Studies
Anatomical features were the dominant criteria used to derive evolutionary relationships between species since Darwin till early 1960s
The evolutionary relationships derived from these relatively subjective observations were often inconclusive. Some of them were later proved incorrect

3. Evolutionary Trees from DNA analysis
For roughly 100 years scientists were unable to figure out which family the giant panda belongs to
Giant pandas look like bears but have features that are unusual for bears and typical for raccoons, e.g., they do not hibernate
In 1985, Steven O’Brien and colleagues solved the giant panda classification problem using DNA sequences and algorithms

4. Out of Africa Hypothesis
DNA-based reconstruction of the human evolutionary tree led to the Out of Africa Hypothesis that claims our most ancient ancestor lived in Africa roughly 200,000 years ago 1 2 3 4 5

5. Human Evolutionary Tree

6. The Origin of Humans: ”Out of Africa” vs Multiregional Hypothesis
Humans evolved in the last two million years as a single species. Independent appearance of modern traits in different areas
Humans migrated out of Africa mixing with other humanoids on the way
Out of Africa:
Humans evolved in Africa ~150,000 years ago
Humans migrated out of Africa, replacing other humanoids around the globe

7. Evolutionary Tree of Humans (mtDNA)
African population is the most diverse
(sub-populations had more time to diverge)
Evolutionary tree separates one group of Africans from a group containing all five populations.
Tree rooted on branch between groups of greatest difference.
Vigilant, Stoneking, Harpending, Hawkes, and Wilson (1991)

8. Evolutionary Tree of Humans: (microsatellites)
Neighbor joining tree for 14 human populations genotyped with 30 microsatellite loci.

9. Outline Constructing Phylogenetic Trees Distance-based Methods
Additive distances
4 Point condition
UPGMA & Neighbor joining
Parsimony-based methods
Sankoff + Fitch’s algorithms

10. Phylogenetic Trees How are these trees built from DNA sequences?1 4 3 2 5
Leaves represent existing species
Internal vertices represent ancestors
Root represents the oldest evolutionary ancestor

11. Rooted and Unrooted Trees
In the unrooted tree the position of the root (“oldest ancestor”) is unknown. Otherwise, they are like rooted trees

12. Phylogenetic Trees How are these trees built from DNA sequences?1 4 3 2 5
Methods
Distance
Parsimony
Minimum number of mutations
Likelihood
Probabilistic model of mutations

13. Distances in Trees Edges may have weights reflecting:
Number of mutations on evolutionary path from one species to another
Time estimate for evolution of one species into another
In a tree T, we often compute
dij(T) - the length of a path between leaves i and j
dij(T) – tree distance between i and j

14. Distance in Trees: an Examplej
d1,4 = = 69

15. Distance Matrix n x n distance matrix Dij for n species
Example: Dij = edit distance between a gene in species i and species j.
Dij – edit distance between i and j
Rat: ACAGTGACGCCCCAAACGT
Mouse: ACAGTGACGCTACAAACGT
Gorilla: CCTGTGACGTAACAAACGA
Chimpanzee: CCTGTGACGTAGCAAACGA
Human: CCTGTGACGTAGCAAACGA

16. Edit Distance vs. Tree Distance
n x n distance matrix Dij for n species
Example: Dij = edit distance between a gene in species i and species j, where edit distance = # of “mutations” to convert string i to string j
Dij – edit distance between i and j
Note the difference with
dij(T) – tree distance between i and j
Rat: ACAGTGACGCCCCAAACGT
Mouse: ACAGTGACGCTACAAACGT
Gorilla: CCTGTGACGTAACAAACGA
Chimpanzee: CCTGTGACGTAGCAAACGA
Human: CCTGTGACGTAGCAAACGA

17. Fitting Distance Matrix
Given n species, we can compute the n x n distance matrix Dij
Evolution of these genes is described by a tree that we don’t know.
We need an algorithm to construct a tree that best fits the distance matrix Dij
Unknown topology of tree makes evolutionary tree reconstruction hard!
# unrooted binary trees n leaves = (2n-3)! / ((n-2)! 2n-2)
n = 24: 5.74 x 1026

18. Fitting Distance Matrix
Fitting means Dij = dij(T)
Lengths of path in an (unknown) tree T
Edit distance between species (known)

19. Reconstructing a 3 Leaved Tree
Tree reconstruction for any 3x3 matrix is straightforward
We have 3 leaves i, j, k and a center vertex c
Observe:
dic + djc = Dij
dic + dkc = Dik
djc + dkc = Djk

20. Reconstructing a 3 Leaved Tree (cont’d)
dic + djc = Dij
+ dic + dkc = Dik
2dic + djc + dkc = Dij + Dik
2dic + Djk = Dij + Dik
dic = (Dij + Dik – Djk)/2
Similarly,
djc = (Dij + Djk – Dik)/2
dkc = (Dki + Dkj – Dij)/2

21. Trees with > 3 Leaves A binary tree with n leaves has 2n-3 edges
Titting a given tree to a distance matrix D requires solving a system:
n(n-1)/2 equations and 2n-3 variables
This is not always possible to solve for n > 3

22. Additive Distance Matrices
Matrix D is ADDITIVE if there exists a tree T with dij(T) = Dij
NON-ADDITIVE otherwise

23. Reconstructing Additive Distances Given Tx T D y 5 4 v w x y z 10 17 16 15 14 9 3 z 3 4 7 w 6 v
If we know T and D, but do not know the length of each edge, we can reconstruct those lengths

24. Reconstructing Additive Distances Given Tx T D y v w x y z 10 17 16 15 14 9 z w v

25. Reconstructing Additive Distances Given Tx v w x y z 10 17 16 15 14 9
Find neighbors v, w
(common parent) y D z a w v a x y z 11 10 9 15 14
dax = ½ (dvx + dwx – dvw)
day = ½ (dvy + dwy – dvw) D1
daz = ½ (dvz + dwz – dvw)

26. Reconstructing Additive Distances Given Tx a x y z 11 10 9 15 14
Neighbors x, y
(common parent) y 5 4 D1 b 3 z 3 a 7 c 4 w 6
d(a, c) = 3
d(b, c) = d(a, b) – d(a, c) = 3
d(c, z) = d(a, z) – d(a, c) = 7
d(b, x) = d(a, x) – d(a, b) = 5
d(b, y) = d(a, y) – d(a, b) = 4
d(a, w) = d(z, w) – d(a, z) = 4
d(a, v) = d(z, v) – d(a, z) = 6
Correct!!! v a b z 6 10 D3 D2 a c 3

27. Distance Based Phylogeny Problem
Goal: Reconstruct an evolutionary tree from a distance matrix
Input: n x n distance matrix Dij
Output: weighted tree T with n leaves fitting D
If D is additive, this problem has a solution and there is a simple algorithm to solve it

28. Using Neighboring Leaves to Construct the Tree
Find neighboring leaves i and j with parent k
Remove the rows and columns of i and j
Add a new row and column corresponding to k, where the distance from k to any other leaf m can be computed as:
Dkm = (Dim + Djm – Dij)/2
Compress i and j into k, iterate algorithm for rest of tree

29. Finding Neighboring Leaves
To find neighboring leaves we simply select a pair of closest leaves.

30. Finding Neighboring Leaves
To find neighboring leaves we simply select a pair of closest leaves.
WRONG

31. Finding Neighboring Leaves
Closest leaves aren’t necessarily neighbors
i and j are neighbors, but (dij = 13) > (djk = 12)
Finding a pair of neighboring leaves is
a nontrivial problem!

32. Degenerate Triples
A degenerate triple is a set of three distinct elements 1≤i,j,k≤n where
Dij + Djk = Dik
Element j in a degenerate triple i,j,k lies on the evolutionary path from i to k (or is attached to this path by an edge of length 0).

33. Looking for Degenerate Triples
If distance matrix D has a degenerate triple i,j,k then j can be “removed” from D thus reducing the size of the problem.
If distance matrix D does not have a degenerate triple i,j,k, one can “create” a degenerative triple in D by shortening all hanging edges (in the tree).

34. Shortening Hanging Edges to Produce Degenerate Triples
Shorten all “hanging” edges (edges that connect leaves) until a degenerate triple is found

35. Finding Degenerate Triples
If there is no degenerate triple, all hanging edges are reduced by the same amount δ, so that all pair-wise distances in the matrix are reduced by 2δ.
Eventually this process collapses one of the leaves (when δ = length of shortest hanging edge), forming a degenerate triple i,j,k and reducing the size of the distance matrix D.
The attachment point for j can be recovered in the reverse transformations by saving Dij for each collapsed leaf.

36. Reconstructing Trees for Additive Distance Matrices
Trim(D, δ)
for all 1 ≤ i ≠ j ≤ n
Dij = Dij - 2δ

37. AdditivePhylogeny Algorithm
AdditivePhylogeny(D)
if D is a 2 x 2 matrix
T = tree of a single edge of length D1,2
return T
if D is non-degenerate
Compute trimming parameter δ
Trim(D, δ)
Find a triple i, j, k in D such that Dij + Djk = Dik
x = Dij
Remove jth row and jth column from D
T = AdditivePhylogeny(D)
Traceback

38. AdditivePhylogeny (cont’d)
Traceback
Add a new vertex v to T at distance x from i to k
Add j back to T by creating an edge (v,j) of length 0
for every leaf l in T
if distance from l to v in the tree ≠ Dl,j
output “matrix is not additive”
return
Extend all “hanging” edges by length δ
return T
Question: How to compute δ?

39. Additive Distance How to tell if D is additive?
AdditivePhylogeny provides a way to check if distance matrix D is additive
An even more efficient additivity check is the “four-point condition”
Let 1 ≤ i,j,k,l ≤ n be four distinct leaves in a tree

40. The Four Point Condition (cont’d)
Compute: 1. Dij + Dkl, 2. Dik + Djl, 3. Dil + Djk 2 3 1
2 and 3 represent the same number: (length of all edges) + 2 * (length middle edge)
1 represents a smaller number: (length of all edges) – (length middle edge)

41. The Four Point Condition: Theorem
The four point condition: For quartet i,j,k,l:
Dij + Dkl ≤ Dik + Djl = Dil + Djk
Theorem : An n x n matrix D is additive if and only if the four point condition holds for every quartet 1 ≤ i,j,k,l ≤ n

42. Least Squares Distance Phylogeny Problem
If the distance matrix D is NOT additive, then we look for a tree T that approximates D the best:
Squared Error : ∑i,j (dij(T) – Dij)2
Squared Error is a measure of the quality of the fit between distance matrix and the tree: we want to minimize it.
Least Squares Distance Phylogeny Problem:
Find approximation tree T with minimum squared error for a non-additive matrix D.
(NP-hard)

43. UPGMA Tree construction as clustering1 4 3 2 5

44. UPGMA Unweighted Pair Group Method with Averages
Computes the distance between clusters using average pairwise distance
Assigns height to every vertex in the tree, effectively dating every vertex (Assumes molecular clock) 1 4 3 2 5

45. Clustering in UPGMA Given two disjoint clusters Ci, Cj of sequences,1
dij = ––––––––– {p Ci, q Cj}dpq
|Ci|  |Cj|

46. UPGMA Algorithm Initialization: Assign each xi to its own cluster Ci
Define one leaf per sequence, each at height 0
Iteration:
Find two clusters Ci and Cj such that dij is min
Let Ck = Ci  Cj
Add a vertex connecting Ci, Cj and place it at height dij /2
Delete Ci and Cj
Termination:
When a single cluster remains 1 4 3 2 5

47. The molecular clock results in ultrametric distances
UPGMA Weakness
UPGMA produces an ultrametric tree;
distance from the root to any leaf is the same
The Molecular Clock:
The evolutionary distance between species x and y is 2 the Earth time to reach the nearest common ancestor
That is, the molecular clock has constant rate in all species
The molecular clock results in ultrametric distances
years 1 4 2 3 5

48. UPGMA’s Weakness: Example2 3 4 1
Correct tree
UPGMA

49. Neighbor Joining Algorithm
In 1987 Naruya Saitou and Masatoshi Nei developed a neighbor joining algorithm for phylogenetic tree reconstruction
Finds a pair of leaves that are close to each other but far from other leaves: implicitly finds a pair of neighboring leaves
Advantages: works well for additive and other non-additive matrices, it does not have the flawed molecular clock assumption

50. Neighbor-Joining
Guaranteed to produce the correct tree if distance is additive
May produce a good tree even when distance is not additive
Let C = current clusters.
Step 1: Finding neighboring clusters
Define: u(C) =1/(|C|-2) C’ D(C, C’)
u(C) measures separation of C from other clusters
Want to minimize D(C1, C2) and maximize u(C1) + u(C2)
Magic trick: Choose C1 and C2 that minimize
D(C1, C2) - (u(C1) + u(C2) )
Claim: Above ensures that Dij is minimal iff i, j are neighbors
Proof: Technical, please read Durbin et al.! 1 3
0.1
0.1
0.1
0.4
0.4 2 4

51. Algorithm: Neighbor-joining
Initialization:
For n clusters, one for each leaf node
Define T to be the set of leaf nodes, one per sequence
Iteration:
Pick Ci, Cj s.t. D(Ci, Cj) – (u(C1) + u(C2)) is minimal
Merge C1 and C2 into new cluster with |C1| + |C2| elements
Add a new vertex C to T and connect to vertices C1 and C2
Assign length 1/2 (D(C1, C2) + (u(C1) - u(C2) ) to edge (C1, C)
Assign length 1/2 (D(C1, C2) + (u(C2) - u(C1) ) to edge (C2, C)
Remove rows and columns from D corresponding to C1 and C2;
Add row and column to D for new cluster C
Termination:
When only one cluster

52. Alignment Matrix vs. Distance Matrix
Sequence a gene of length m nucleotides in n species to generate an…
n x m alignment matrix
CANNOT be transformed back into alignment matrix because information was lost on the forward transformation
Gorilla: CCTGTGACGTAACAAACGA
Chimpanzee: CCTGTGACGTAGCAAACGA
Human: CCTGTGACGTAGCAAACGA
Transform into…
n x n distance matrix

53. Character-Based Tree Reconstruction
Better technique:
Use the n x m alignment matrix
(n = # species, m = #characters)
directly instead of using distance matrix.
GOAL: determine what character strings at internal nodes would best explain the character strings for the n observed species

54. Character-Based Tree Reconstruction
Characters may be nucleotides, where A, G, C, T are states of this character.
Characters may be morphological features
# of eyes or legs or the shape of a beak or a fin.
If set length of an edge in the tree to the Hamming distance, then define the parsimony score of the tree as the sum of the lengths (weights) of the edges

55. Character-Based Tree Reconstruction

56. Parsimony Approach to Evolutionary Tree Reconstruction
Applies Occam’s razor principle to identify the simplest explanation for the data
Assumes observed character differences resulted from the fewest possible mutations
Seeks the tree that yields lowest possible parsimony score - sum of cost of all mutations found in the tree

57. Parsimony and Tree Reconstruction

58. Parsimony – What if we don’t have distances
One of the most popular methods:
GIVEN multiple alignment
FIND tree & history of substitutions explaining alignment
Idea:
Find the tree that explains the observed sequences with a minimal number of substitutions
Two computational subproblems:
Find the parsimony cost of a given tree.
Small parsimony problem (easy)
Search through all tree topologies.
Large parsimony problem (hard)

59. Small Parsimony Problem
Input: Tree T with each leaf labeled by an m-character string.
Output: Labeling of internal vertices of the tree T minimizing the parsimony score.
We can assume that every leaf is labeled by a single character, because the characters in the string are independent.

60. Weighted Small Parsimony Problem
A more general version of Small Parsimony Problem
Input includes a k x k scoring matrix describing the cost of transformation of each of k states into another one
For Small Parsimony problem, the scoring matrix is based on Hamming distance
dH(v, w) = 0 if v=w
dH(v, w) = 1 otherwise

61. Scoring Matrices A T G C 1 A T G C 3 4 9 2
Weighted Small Parsimony Problem
Small Parsimony Problem A T G C 1 A T G C 3 4 9 2

62. Unweighted vs. Weighted
Small Parsimony Scoring Matrix: A T G C 1
Small Parsimony Score: 5

63. Unweighted vs. Weighted
Weighted Parsimony Scoring Matrix: A T G C 3 4 9 2
Weighted Parsimony Score: 22

64. Weighted Small Parsimony Problem: Formulation
Input: Tree T with each leaf labeled by elements of a k-letter alphabet and a k x k scoring matrix (ij)
Output: Labeling of internal vertices of the tree T minimizing the weighted parsimony score

65. Sankoff Algorithm Dynamic Programming
Calculate and keep track of a score for every possible label at each vertex
st(v) = minimum parsimony score of the subtree rooted at vertex v if v has character t
The score at each vertex is based on scores of its children:
st(parent) = mini {si( left child ) + i, t} +
minj {sj( right child ) + j, t}

66. Sources
Serafim Batzoglou (Sequencing slides)
(Additive Trees and Parsimony)