Plan

• We will learn to build phylogenetic trees
• We start building trees based on the distances between species
• We assume that we know the distances
• Distances are hard to calculate
• We will study that later

We know the distance between elements

For instance in a matrix \(D_1.\)

Hierarchical clustering

The idea is to group similar nodes in the same branch

The smallest distance in \(D_1\) is \(D_1 (a,b)=17\)

So, \(a\) and \(b\) are the closest elements

Now we update the other distances

We build a new matrix \(D_2\) with the average distance of each element to \((a,b)\)

\[ \begin{aligned} D_2((a,b),c)= & \frac{D_1(a,c) + D_1(b,c)}{2}=\frac{21+30}{2}=25.5\\ D_2((a,b),d)= & \frac{D_1(a,d) + D_1(b,d)}{2}=\frac{31+34}{2}=32.5\\ D_2((a,b),e)= & \frac{D_1(a,e) + D_1(b,e)}{2}=\frac{23+21}{2}=22 \end{aligned} \]

Updated matrix

The matrix \(D_2\) is

(values in bold are new, the ones in italics did not change)

Now the smallest distance is \(D_2 ((a,b),e)=22\).
We must join \((a,b)\) and \(e\)

And we update the distances

\[ \begin{aligned} D_3(((a,b),e),c)= & \frac{D_2((a,b),c) + D_2(e,c)}{2}\\ = & \frac{25.5 + 39}{2}=32.25\\ D_3(((a,b),e),d)= & \frac{D_2((a,b),d) + D_2(e,d)}{2}\\ = & \frac{32.5 + 43}{2}=37.75 \end{aligned} \]

We have a new matrix

The matrix \(D_3\) is

Now the closest elements are \(c\) and \(d\)

We calculate the distance matrix \(D_4\)

We calculate the only remaining distance

\[ \begin{aligned} D_4((c,d),((a,b),e)) & = \frac{D_3(c,((a,b),e)) + D_3(d,((a,b),e))}{2}\\ & = \frac{32.25+37.75}{2} =35 \end{aligned} \]

The new matrix is

Describing the tree

We can represent the complete tree by \[(((a,b),e),(c,d))\]

The parenthesis show how to connect every element

But we miss the distance of every element

We can write the distance to the parent after the node label

\[(((a\colon D_a, b\colon D_b)\colon D_{ab},e\colon D_e)\colon D_{abe},(c\colon D_c,d\colon D_d)\colon D_{cd});\]

This is hierarchical clustering

This is called Weighted Pair Group Method with Arithmetic Mean (WPGMA)

There are other hierarchical clustering methods, depending on how do we evaluate the distance between

• two single elements
• one element and a group
• two groups

One problem: we mix groups of different size

Node ((a,b),e) has three sequences, and (c,d) has two

“bigger nodes” should have more weight

Alternative: UPGMA

Unweighted pair group method with arithmetic mean

The distance between branch \(A\) and \(B\), each of size \({N_A}\) and \({N_B}\), is the average of all distances \(D(x,y)\) between pairs of objects in \(A\) and in \(B\)

\[D((A,B),X) = \frac{N_A \cdot D(A,X) + N_B \cdot D(B,X)}{N_A + N_B}\]

Redoing our example

\[\begin{aligned} D_2((a,b),c)& =\frac{D_1(a,c) \times 1 + D_1(b,c) \times 1)}{1+1}\\ & =\frac{21+30}{2}=25.5\\ D_2((a,b),d)& =\frac{D_1(a,d) + D_1(b,d)}{2}=\frac{31+34}{2}=32.5\\ D_2((a,b),e)& =\frac{D_1(a,e) + D_1(b,e)}{2}=\frac{23+21}{2}=22 \end{aligned}\]

The first step is the same as before

Next step is different

\[ \begin{aligned} D_3(((a,b),e),c)&=\frac{D_2((a,b),c) \times 2 + D_2(e,c) \times 1}{2+1}=\\ & =\frac{25.5 \times 2 + 39 \times 1}{3}=30\\ D_3(((a,b),e),d)&=\frac{D_2((a,b),d) \times 2 + D_2(e,d) \times 1}{2+1}=\\ & =\frac{32.5 \times 2 + 43 \times 1}{3}=36 \end{aligned} \]

Matrix \(D_3\) in UPGMA

Comparing WPGMA against UPGMA

UPGMA is much more used than WPGMA

In practice UPGMA is more realistic than WPGMA

But both have a problem:

The distances between leaves do not match the original distances

Moreover, the mutation rate may be different for different branches

Minimization problem

If we know the tree topology, we can find the branches’ lengths

We minimize the squared difference between observed distance \(D_{ij}\) and tree distance \(d_{ij}\)

\[\min_{d_{ij}} \sum_{i,j}(D_{ij}-d_{ij})^2\]

But we still need to find the tree topology, and that is a hard problem.

Neighbor joining

This is an heuristic to solve the minimization problem

Instead of joining the nearest nodes in the distance matrix,
we look into a new matrix \(Q\)

\[Q_{ij} = (n-2) D_{ij} -\sum_k D_{ik} -\sum_k D_{kj}\]

This “neighbor-joining” distance can be negative

Method

• calculate \(R_i = \sum_j D_{ij}\) for all \(i\)
• calculate \(Q_{ij} = (n-2) D_{ij} - R_i - R_j\)
• Find smallest \(Q_ij\)
• Join \(i\) and \(j\) into a new node \(u\) \[\begin{aligned} D_{iu} &= \frac{(n-2) D_{ij} + R_i -R_j}{2(n-2)}\\ D_{uk} &= \frac{1}{2}(D_{ik} + D_{jk} - D_{ij}) \end{aligned}\]
• Repeat until well done

First we calculate \(Q_1\)

For each \(i,j∈ \{a,b,c,d,e\}, i≠j,\) we have \[Q(i,j) = (n-2) D(i,j) - R_i - R_j\]

The nearest elements are \(c\) and \(d\)

New node \(u\) connects \(c\) and \(d\)

Matrix D2

For each \(k∈ \{a,b,e\}\) we have \[D(u,k) = \frac{1}{2}(D(c,k) + D(d,k) - D(c,d))\]

Matrix Q2

New node \(v\) connects \(a\) and \(u\)

Matrix D3

Matrix Q3

D4

thus \(D(w, e) = 14.25\)

Exercise

Redo all the calculation of these trees