1. Phylogeny

2. Tree construction methods
Character based
Parsimony
Fitch
Sankoff
Probabilistic
Maximum likelihood
Distance based
UPGMA

3. Maximum Likelihood Method
Input:
𝑛 strings of length 𝑚 (multiple alignment)
Substitution matrix
Character frequency
Output: A tree topology with the input strings at the leaves

4. Maximum Likelihood Method
Input:
𝑛 strings of length 𝑚 (multiple alignment)
Substitution matrix
Character frequency
for each possible tree topology with leaf labeling 𝑇:
for each position 𝑖 from 1 to 𝑚:
𝐿 𝑖 = 𝑃(𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑖𝑛𝑛𝑒𝑟 𝑛𝑜𝑑𝑒 𝑙𝑎𝑏𝑒𝑙𝑖𝑛𝑔)
𝐿 𝑇 = 𝐿 1 ∗ 𝐿 2 ∗…∗ 𝐿 𝑚
Pick the tree with the Maximal Likelihood

5. Maximum Likelihood Computation for a specific tree
Given a tree topology and a leaf labeling
Every possible inner node labeling should be considered
How many different trees with inner nodes labeling exist for the given tree? (DNA alphabet)
4 4 =256 T G C A

6. Maximum Likelihood Computation for a specific tree
Compute the likelihood for the given tree: A T G C T G C A
0.1
0.2 1
*All other options for inner node labeling were already computed, and their sum is 0.2
𝑃 𝐴 =0.3, 𝑃 𝑇 =0.3
𝑃 𝐶 =0.2, 𝑃 𝐺 =0.2

7. Maximum Likelihood Computation for a specific tree
Compute the likelihood for the given tree: A T G C T G C A
0.1
0.2 1
*All other options for inner node labeling were already computed, and their sum is 0.2
𝑃 𝐴 =0.3, 𝑃 𝑇 =0.3
𝑃 𝐶 =0.2, 𝑃 𝐺 =0.2
𝐿(𝑇)=0.2+𝑃 𝐴 ∗𝑃 𝐴→𝑇 ∗𝑃 𝑇→𝐶 ∗𝑃 𝐶→𝑇 ∗𝑃 𝐶→𝐺 ∗𝑃 𝑇→𝐶 ∗𝑃 𝐴→𝐺 ∗𝑃 𝐺→𝐴 ∗𝑃 𝐺→𝐺

8. Maximum Likelihood Computation for a specific tree
Compute the likelihood for the given tree: A T G C T G C A
0.1
0.2 1
*All other options for inner node labeling were already computed, and their sum is 0.2
𝑃 𝐴 =0.3, 𝑃 𝑇 =0.3
𝑃 𝐶 =0.2, 𝑃 𝐺 =0.2
𝐿(𝑇)=0.2+𝑃 𝐴 ∗𝑃 𝐴→𝑇 ∗𝑃 𝑇→𝐶 ∗𝑃 𝐶→𝑇 ∗𝑃 𝐶→𝐺 ∗𝑃 𝑇→𝐶 ∗𝑃 𝐴→𝐺 ∗𝑃 𝐺→𝐴 ∗𝑃 𝐺→𝐺 =0.3∗0.1∗0.2∗0.2∗0.1∗0.2∗0.2∗0.2∗1= ∗ 10 −7 =

9. UPGMA
UPGMA is a greedy algorithm that constructs a phylogenetic tree, given 𝑛 species and a table 𝐷[𝑛×𝑛] of distances between each 2 species.

10. Some definitions
Additive distance matrix: A distance matrix is called additive if there exists a tree in which the distances between the leaves correspond to the matrix’s distances. Another definition is the “4 point criterion”, which is easier to verify.

11. Additive matrix
The “4 points criterion”: A matrix is said to be additive if every 4 objects (species) can be labeled as 𝑥,𝑦,𝑧,𝑤 so that: z x c a x y b d w
𝑎+𝑏 + 𝑐+𝑑 ≤ 𝑎+𝑥+𝑐 + 𝑏+𝑥+𝑑 = 𝑎+𝑥+𝑑 +(𝑏+𝑥+𝑐)

12. Additive matrix
The “4 points criterion”: A matrix is said to be additive if every 4 objects (species) can be labeled as 𝑥,𝑦,𝑧,𝑤 so that: z x c a x y b d w
𝑎+𝑏 + 𝑐+𝑑 ≤ 𝑎+𝑥+𝑐 + 𝑏+𝑥+𝑑 = 𝑎+𝑥+𝑑 +(𝑏+𝑥+𝑐)

13. Additive matrix
The “4 points criterion”: A matrix is said to be additive if every 4 objects (species) can be labeled as 𝑥,𝑦,𝑧,𝑤 so that: z x c a x y b d w
𝑎+𝑏 + 𝑐+𝑑 ≤ 𝑎+𝑥+𝑐 + 𝑏+𝑥+𝑑 = 𝑎+𝑥+𝑑 +(𝑏+𝑥+𝑐)

14. Additive matrix
The “4 points criterion”: A matrix is said to be additive if every 4 objects (species) can be labeled as 𝑥,𝑦,𝑧,𝑤 so that: z x c a x y b d w
𝑎+𝑏 + 𝑐+𝑑 ≤ 𝑎+𝑥+𝑐 + 𝑏+𝑥+𝑑 = 𝑎+𝑥+𝑑 +(𝑏+𝑥+𝑐)

15. Additive matrix
𝑑 𝐴,𝐵 +𝑑 𝐶,𝐷 =12+6=18 𝑑 𝐴,𝐶 +𝑑 𝐵,𝐷 =14+12=26 𝑑 𝐴,𝐷 +𝑑(𝐵,𝐶)=14+12=26 𝑑 𝐴,𝐵 +𝑑 𝐶,𝐷 ≤𝑑 𝐴,𝐶 +𝑑 𝐵,𝐷 =𝑑 𝐴,𝐷 +𝑑(𝐵,𝐶) 18 26 26

16. Non-Additive matrix
𝑑 𝐴,𝐵 +𝑑 𝐶,𝐷 =2+2 𝑑 𝐴,𝐶 +𝑑 𝐵,𝐷 =2+2 𝑑 𝐴,𝐷 +𝑑 𝐵,𝐶 =2+3 A B C D 2 3

17. Ultrametric Distance Matrix
A distance matrix is called ultrametric if there exists a tree corresponding to the matrix’s distances, in which all leaves have equal distance from the root. Notice that by definition, ultrametric is a special case of additive.

18. Some definitions
The “3 point criterion”: Like the additive case, ultrametric has another definition: If all 3 taxa can be relabeled as 𝑥,𝑦,𝑧 so that:

19. Some definitions
Ultrametric distance: For example, this is an ultrametric tree:

20. UPGMA algorithm
UPGMA - Unweighted Pair Group Method with Arithmatic Mean
Input – a distance matrix D. Each cell [𝑖,𝑗] represents the distance 𝑑(𝑖,𝑗) between species 𝑖 and species 𝑗.
Output – an ultrametric phylogenetic tree T, with leaf labeling

21. UPGMA algorithm Input: 𝐷[𝑛×𝑛] – distance matrix
Initialize: 𝑇={ 𝐶 1 ,…, 𝐶 𝑛 }
While 𝑇 >1 cluster taxa:
Pick shortest distance 𝑑(𝑖,𝑗)
C← C 𝑖 , C j
Define node at height 𝑑 𝐶 𝑖 , 𝐶 𝑗 2
T← T \ { C i , C j }
T←T U {C}
Update D: ∀ 𝐶 𝑘 ∈𝑇, 𝐶 𝑘 ≠ 𝐶 𝑖 , 𝐶 𝑗 𝑑 𝐶, 𝐶 𝑘 = 𝑑 𝐶 𝑖 , 𝐶 𝑘 | 𝐶 𝑖 |+𝑑 𝐶 𝑗 , 𝐶 𝑘 | 𝐶 𝑗 | 𝐶 𝑖 +| 𝐶 𝑗 |

22. UPGMA Example
Given the distance matrix below, build a phylogenetic tree using UPGMA A B C D E 2 4 6 F 8

23. Example A B C D E 2 4 6 F 8
We begin by choosing a minimal distance, and clustering the nodes chosen.

24. UPGMA Example
Then we calculate the distances between our new cluster and all the rest of the nodes, to create an updated distance matrix D.
The distances not including our cluster’s nodes remain exactly the same.

25. Example The updated distance matrix: A B C D E 2 4 6 F 8 AB C D E 4 6
𝑑 𝐶, 𝐶 𝑘 =
𝑑 𝐶 𝑖 , 𝐶 𝑘 | 𝐶 𝑖 |+𝑑 𝐶 𝑗 , 𝐶 𝑘 | 𝐶 𝑗 | 𝐶 𝑖 +| 𝐶 𝑗 | A B C D E 2 4 6 F 8
𝑑 𝐴𝐵, 𝐶 𝑘 = 𝑑 𝐴, 𝐶 𝑘 𝐴 +𝑑 𝐵, 𝐶 𝑘 𝐵 𝐴 +|𝐵|
𝑑 𝐴𝐵, 𝐶 = 4 𝐴 +4 𝐵 𝐴 +|𝐵| =4 AB C D E 4 6 F 8
The updated distance matrix:

26. Example AB C D E 4 6 F 8
Now, we carry on doing the exact same procedure, until we are left with only one cluster.

27. Example AB C D E 4 6 F 8
Now, we carry on doing the exact same procedure, until we are left with only one cluster.

28. Example
𝑑 𝐶, 𝐶 𝑘 =
𝑑 𝐶 𝑖 , 𝐶 𝑘 | 𝐶 𝑖 |+𝑑 𝐶 𝑗 , 𝐶 𝑘 | 𝐶 𝑗 | 𝐶 𝑖 +| 𝐶 𝑗 | AB C D E 4 6 F 8 AB C DE 4 6 F 8
𝑑 𝐷𝐸, 𝐴𝐵 = 𝑑 𝐷, 𝐴𝐵 𝐷 +𝑑 𝐸, 𝐴𝐵 𝐸 𝐸 +|𝐷| = =6

29. Example AB C DE 4 6 F 8

30. Example AB C DE 4 6 F 8

31. Example
𝑑 𝐶, 𝐶 𝑘 =
𝑑 𝐶 𝑖 , 𝐶 𝑘 | 𝐶 𝑖 |+𝑑 𝐶 𝑗 , 𝐶 𝑘 | 𝐶 𝑗 | 𝐶 𝑖 +| 𝐶 𝑗 | AB C DE 4 6 F 8
AB,C DE 6 F 8
𝑑 𝐴𝐵𝐶, 𝐷𝐸 = 𝑑 𝐴𝐵, 𝐷𝐸 𝐴𝐵 +𝑑 𝐶, 𝐷𝐸 𝐶 𝐴𝐵 +|𝐶| = 6∗2+6∗1 3 =6
𝑑 𝐴𝐵𝐶, 𝐹 = 𝑑 𝐴𝐵, 𝐹 𝐴𝐵 +𝑑 𝐶, 𝐹 𝐶 𝐴𝐵 +|𝐶| = 8∗2+8∗1 3 =8

32. Example AB,C DE 6 F 8 (AB,C),DE F 8
AB,C DE 6 F 8
(AB,C),DE F 8
𝑑 𝐴𝐵𝐶𝐷𝐸, 𝐹 = 𝑑 𝐴𝐵𝐶, 𝐹 𝐴𝐵𝐶 +𝑑 𝐷𝐸, 𝐹 𝐷𝐸 𝐴𝐵𝐶 +|𝐷𝐸| = 8∗3+8∗2 5 =8

33. UPGMA Example Our output tree! Lovely, isn’t it?
(AB,C),DE F 8
Our output tree!
Lovely, isn’t it?
Can there be more the one tree?

34. UPGMA Example A B C D E 2 4 6 F 8
What can be said about the distance matrix? Is it additive? Ultrametric?

35. UPGMA downfalls UPGMA will always return an ultrametric tree.
It assumes all species mutate at the same rate (molecular clock).
What will happen if we will try and reconstruct a tree such as this one?

36. UPGMA downfalls
This tree corresponds to the following distance matrix: A B C D E 5 4 7 10 6 9 F 8 11

37. UPGMA downfalls
If we run UPGMA on the matrix shown, will get this output: Compared to the original tree:

38. UPGMA downfalls
UPGMA returns the right tree if the distance matrix is ultrametric. Even then, we can’t be certain the original tree was also ultrametric. If the distance matrix D is not additive, UPGMA will generate a heuristic solution that does not fit D