1. dij(T) - the length of a path between leaves i and j
Distance in Trees
dij(T) - the length of a path between leaves i and j i j
d1,4 = = 69

2. Phylogenetic Tree Reconstruction
Input:
Distance matrix D
Output:
Binary Tree T such that dij(T) = Dij

3. 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

4. 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

5. Trees with > 3 Leaves An tree with n leaves has 2n-3 edges
This means fitting a given tree to a distance matrix D requires solving a system of “n choose 2” equations with 2n-3 variables
This is not always possible to solve for n > 3

6. The Four Point Condition
Compute: 1. Dij + Dkl, 2. Dik + Djl, 3. Dil + Djk 2 3 1
2 and 3 represent the same number: the length of all edges + the middle edge (it is counted twice)
1 represents a smaller number: the length of all edges – the middle edge

7. The Four Point Condition
Four point condition: For i,j,k,l two of the sums Dij + Dkl, Dik + Djl, Dil + Djk are equal and the third sum is smaller
Definition : An n x n matrix D is additive provided there exists a tree T with D(T) = D. (Note: T is unique.)
Theorem: D is additive if and only if the four point condition holds for every quartet 1 ≤ i,j,k,l ≤ n

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

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

10. Reconstructing Additive Distances Given Tx v w x y z 10 17 16 15 14 9 T y D z a w
dvx + dwx = 2 dax + dvw 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)

11. Reconstructing Additive Distances Given Tx a x y z 11 10 9 15 14 T 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

12. 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

13. 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

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

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

16. 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!

17. 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).

18. 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).

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

20. 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.

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

22. 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

23. 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

24. 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

25. 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’ 2 C D(C, C0 )
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: Very technical, please read Durbin et al.! 1 3
0.1
0.1
0.1
0.4
0.4 2 4

26. 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