Comp. Genomics Recitation 8 Phylogeny

Outline Phylogeny: Distance based Probabilistic Parsimony

Exercise Show that in UPGMA, for some new cluster k The distances d kl are given by: for any cluster l

Solution Since the members of k are the members of i and j, the sum of distances between members of k and l can be written as: This is equal to:

Solution By the definition of distance between clusters, we divide the latter sum by |C k |·|C l |: Which can also be written as: ·

Exercise Show that every parent in a tree constructed by UPGMA is never lower than its daughter nodes

Exercise k ji h k =d ij /2 n h n =d kl /2 l Can n be lower than k?

Solution Since h n =d kl /2, we will show that for every k and l d kl ≥d ij and therefore node n is higher than node k According to the previous exercise: 

Solution Since i and j were merged and not i and l or j and l, we can conclude that

Exercise Show an example in which the parent node height is equal to the child node height (UPGMA).

Solution Suppose 3 pairs of sequences have the same distance d. We choose to merge leafs 1 and 2 and produce node 4, with height d/2. The new distance, d 43, is exactly d So when we merge node 4 and leaf 3, we create a new node 5 of height d/2

Solution height=d/2 5

Solution

Exercise The famous paleontologist R. Geller argued to his sister that the last common ancestor of birds and dinosaurs lived 100 million years ago. His sister claimed that the ancestor lived 200 million years ago. The evidence are 1000nt long homologous genes with 350 differences (its not contamination this time…)

Exercise Both accept the Jukes-Cantor model Both accept the assumption of a molecular clock If mutations occur independently, with rate mutations per year, whose theory is more likely to be correct?

Solution According to Jukes-Cantor, the probability of a nucleotide remaining unchanged over t time units is: The probability for a specific change:

Solution BirdDinosaur Ancestor tt Molecular clock – both species evolve at the same rate Tree T

Solution The likelihood of the tree at site i is: Likelihood of a tree Jukes-Cantor Reversibility property Jukes-Cantor Additivity Less work to do

Solution Since the distance between the species is 2t, the probability of every site in which there is a match is: For a mismatch, the probability is:

Solution So the likelihood of the tree T is

Solution The log likelihood of the trees suggested by Dr. Geller and his sister is: 3α=10 -9  α=1/3*10 -9

Solution Yay!

Exercise Assume that the substitution cost for a weighted parsimony algorithm is a metric, i.e. it satisfied S(a,a)=0, S(a,b)=S(b,a) and S(a,c)≤S(a,b)+S(b,c). Show the tree with minimal cost is independent of the position of the root.

Solution We have a set of species and we are given a minimal weight tree for it. Denote the root in this tree by k k ij lm We will show that deleting k and moving it to this edge does not change the cost of the tree

Solution What is the cost of the tree before translocation of the root? k ij lm For a specific choice of character c at the root: The minimal choice is the cost of this tree:

And the minimal cost of the tree is: Solution Due to the triangle inequality, S(a,b)≤S(a,c)+S(c,b) k ij lm If we set c to a (or equivalently to b), we get:

Solution Now we move the root: k ij lm k Call this tree T’

Solution Denote the character at l as d k li jm The new cost is: where the S’ is due to the change in subtree

Solution k li jm k ij lm

k li jm k ij lm

We proved that when moving the root to an adjacent position does not change the minimal cost. Why is the case of moving the root to a non-adjacent position easier to prove?

Question Does every symmetric distance with 0 on the diagonal have a tree?

Answer No! Example: If d(a,d) = 0.25 and d(b,d)=0.25, then it must be that d(a,b) ≤ 0.5 abcd a b101 c110 d 0