1. Phylogenetic Trees - Parsimony Tutorial #12
Next semester:
Project in advanced algorithms for
phylogenetic reconstruction
(236512)
Initial details in:
- Come to me for more details - .

2. Phylogenetic Reconstruction
We’d like to study the evolutionary history of species
Distance-based approach:
Calculate (ML) pairwise (evolutionary) distances between species
Find the edge-weighted tree best describing this metric
Major drawback:
Lose of information when reducing data to pairwise distances
Character-based approach:
Consider the character vector of each specie:
morphological characters
bio-molecular characters
Optimization criteria:
parsimony
likelihood / posterior-probability .

3. Most Parsimonious Tree
Parsimony-score:
Number of character-changes (mutations) along the evolutionary tree
(tree containing labels on internal vertices)
Example:
Score = 4
Score = 3
AGA
AAA
AAG
GGA
AAA 1 2 1
AGA
AAA
AAG
GGA
Most parsimonious tree:
 Tree with minimal parsimony score
Minimal Evolution Principle

4. Small vs. Large Parsimony
We break the problem into two:
Small parsimony: Given the topology find the best assignment to internal nodes
Large parsimony: Find the topology which gives best score
Large parsimony is NP-hard
We’ll show solution to small parsimony (Fitch and Sankoff’s algorithms)
Input to small parsimony:
tree with character-state assignments to leaves
Example:
A: CAGGTA
B: CAGACA
C: CGGGTA
D: TGCACT
E: TGCGTA
Aardvark
Bison
Chimp
Dog
Elephant

5. Fitch’s Algorithm Execute independently for each character:
Bottom-up phase: Determine set of possible states for each internal node
Top-down phase: Pick states for each internal node
Dynamic Programming framework 1 2
Aardvark
Bison
Chimp
Dog
Elephant
CAGGTA
CGGGTA
TGCGTA
CAGACA
TGCACT

6. Fitch’s Algorithm Bottom-up phase
Determine set of possible states for each internal node
Initialization: Ri = {si}
Do a post-order (from leaves to root) traversal of tree
Determine Ri of internal node i with children j, k: T T
Parsimony-score =
# union operations
AGT CT GT
score = 3 C T G T A T

7. Fitch’s Algorithm Top-down phase
Pick states for each internal node
Pick arbitrary state in Rroot for the root
Do pre-order (from root to leaves) traversal of tree
Determine sj of internal node j with parent i: T
Complexity: O(mnk)
#characters
#taxa/nodes
#states T
AGT CT GT
score = 3 C T G T A T

8. Weighted Parsimony Sankoff’s algorithm
Each mutation a↔b costs differently - S(a,b).
Bottom-up phase: Determine Ri(s) – cost of optimal state-assignment for subtree of i, when it is assigned state s.
Top-down phase: Pick optimal states for each internal node
Fitch’s algorithm as special case:
Ri – set of states which yield minimal-cost subtree of i
Same as algorithm for
optimal lifted tree alignment
(Tutorial #4)

9. Sankoff’s Algorithm Bottom-up phase
Determine Ri(s) for each internal node
Initialization:
Do a post-order (from leaves to root) traversal of tree
Determine Ri of internal node i with children j, k:
Natural generalization
For non-binary trees
Remember pointers
ss’ C T G T A T

10. Sankoff’s Algorithm Top-down phase
Pick states for each internal node
Select minimal cost character for root (s minimizing Rroot(s))
Do pre-order (from root to leaves) traversal of tree:
- For internal node j, with parent i, select state that produced
minimal cost at i (use pointers kept in 1st stage)
Complexity: O(mnk2)
#characters
#taxa/nodes
#states C T G T A T

11. Fitch’s Algorithm as special case of Sankoff’s algorithm
Unweighted parsimony:
Sankoff’s algorithm:
Ri(s) - cost of optimal subtree of i, when it is assigned state s
Fitch’s algorithm:
Score(i) - cost of optimal state-assignment for subtree of i
Ri set of optimal state-assignment for subtree of i
We need to show that:
Optimal tree assigns node i with state from Ri.
Fitch’s bottom-up recursive formula for Ri. is correct:
Check for yourselves

12. Fitch’s Algorithm as special case of Sankoff’s algorithm
Unweighted parsimony:
Score(i) - cost of optimal state-assignment for subtree of i
Ri set of optimal state-assignment for subtree of i
We need to show that:
Optimal tree assigns node i with state from Ri.
Trivially true for the root
Assume (to the contrary) that in an optimal assignment, some node – j is assigned sj∉Rj
root i j
Parsimony-score is integer
Why is this not
the case for the
weighted version?
sj∉Rj  Rj(sj) ≥ Score(j)+1 
By switching from sj to some s∊Rj
we do not raise the parsimony-score

13. Exploring the Space of Trees
We saw how to find optimal state-assignment for a given tree topology
We need to explore space of topologies
Given n sequences there are (2n-3)!! possible rooted trees
and (2n-5)!! possible unrooted trees
taxa (n) # rooted trees # unrooted trees
, ,395
,459,425 2,027,025

14. Exploring the Space of Trees
Possible solutions:
Heuristic solutions for “traveling” through “topology-space”
Find (basic) topology using distance-based methods (NJ)
Notice another problem:
We obtain state-assignments to taxa using multiple alignment
We obtain optimal MA using topology of phylogenetic tree
(e.g. CLUSTAL)
Solution:
Again, use some initial topology (via NJ) A - T G C
C1,C2 , … , Cm