Competition

• In the short term, all viable organisms are alive

• In the long term, and when resources are scarce, some organisms do not survive

• For example, some organisms may be more efficient in capturing food or using energy

  • Some organisms have higher “fitness”

• If the environment changes, the “fitness” changes

  • There may be bottleneck effects

Coevolution

• Evolution is more complex for sexual organisms

• Some individuals do not pass their genes to the next generation, due to mate-selection

• Mate-selection also evolves

• We say that phenotype and peer-selection co-evolve

Coevolution between predator and prey

• “Every morning in Africa, a gazelle wakes up, it knows it must run faster than the fastest lion or it will be killed.

• “Every morning in Africa, a lion wakes up, it knows it must run faster than the slowest gazelle, or it will starve.

• “It doesn’t matter whether you’re the lion or a gazelle-when the sun comes up, you’d better be running.”

Looking at only one gene

For this class we will consider the 16S gene in bacteria

• Approx. 1500 nucleotides
• Highly conserved
  • Most mutations are lethal
  • Cell viability depends on 16S structure
• Asexual reproduction

Trees are networks

Trees are a kind of networks where every node is connected and there are no loops

There are two kinds of nodes

• internal nodes
  • they have “children”
• leaves
  • They do not have “children”

We see only the leaves, we want to find the internal nodes, and the branches

Rooted trees

To represent evolution we start with an initial node

It is the ancestor to all other nodes, and it has no ancestor

It has two arrows pointing to its “children”

The arrows start from the root and point to the leaves.

Unrooted trees

Looking only at the modern data, we cannot know which sequence existed before

That is, we cannot put an arrow between two nodes

We put a link, undirected, between nodes

These trees are called unrooted

Outgroups point to the root

Since we only see leaves, we cannot put arrows

So we cannot tell which internal node is the root

But, if we include a leave that we know is very distant from all the others, then we can find the root.

Mathematical Language

At least two groups of people have worked with networks: engineers and mathematicians

They use different words for the same objects

• Network: graph

• Node: vertex

• Link: edge

• Arrow: arc

Essence of a tree

The same tree can be drawn in several ways

The drawing is not important

The only important things are

• The tree topology. That is, who is connected to who
• The length of each arc (or edge)

Reconstructing the tree

there are basically three approaches

• Maximum parsimony
  • smallest tree that explains all mutations
• Maximum likelihood
  • most probable tree, using a probabilistic model
• Distance based
  • forget the sequences, use only their distances

In all cases the input is a multiple alignment of all sequences

Other problem with parsimony methods

• In some simulations the predicted tree may be very different from the real one

  • We only know “the real tree” when we create it

• It can be statistically inconsistent

  • That is, adding more sequences sometimes makes a worse tree

Maximum likelihood

An alternative is to find the most probable tree, given the available data

This method needs:

• A probabilistic model of evolution
• Looking at all the trees

So, again, we need an heuristic

Distance methods

Here we use the multiple alignment to calculate the distance between sequences. For example

We call it \(D_1.\) Then we forget about the sequences

Hierarchical clustering WPGMA

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

We join \(a\) and \(b\) into a new node \((a,b)\), and update distances

Updated distances

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

Updated matrix

now the smallest distance is \(D_2 ((a,b),e)=22\)

Update again

New distances

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

and again

new distance matrix

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

finally

This is hierarchical clustering

This is called average linking, or Weighted Pair Group Method with Arithmetic Mean

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 clusters \(\mathcal{A}\) and \(\mathcal{B}\), each of size \({|\mathcal{A}|}\) and \({|\mathcal{B}|}\), is the average of all distances \(d(x,y)\) between pairs of objects in \(\mathcal{A}\) and in \(\mathcal{B}\)

\[d_{(\mathcal{A} \cup \mathcal{B}),X} = \frac{|\mathcal{A}| \cdot d_{\mathcal{A},X} + |\mathcal{B}| \cdot d_{\mathcal{B},X}}{|\mathcal{A}| + |\mathcal{B}|}\]

Example

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

Then

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

Comparison

Minimization problem

If we know the topology, we can find the branch lengths

We minimize the squared difference between observed distance and tree distance

\[\min_{d_{ij}} \sum_i\sum_j(D_{ij}-d_{ij})^2\]

But we still need to find the tree topology

Neighbor joining

This is one of the most recommended methods

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 “normalized” value 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