Sequence Evolution

Provide a brief introduction to models of sequence evolution.

From gene evolution to sequence evolution

Given a fixed species tree (representing a hypothesis about the population-level history), a standard methodology for simulating DNA sequence evolution involves a two-step model of evolution, in which gene trees are first generated (step 1), and then DNA sequence are generated for each gene tree.

There are two operant parameters in this framework:

• $m:$ the number of underlying genes in the genome
• $k:$ the length of each gene, in base-pairs.

As a simplifying assumption, we assume all genes to have the same length $k$, and the genome to consist only of our $m$ genes. As such, the total data is $mk$.

The Jukes-Cantor Model of Site Substitution

The simplest four-state model of DNA mutation is the Jukes-Cantor substitution model, which assumes four nucleotide states $A,C,G,T$ which mutate between each other at equal rates 1.

Input: A gene tree

Derired output: A length $k$ sequence of DNA for each leaf of the tree

Constructively, the Jukes-Cantor model can be thought of as a procedure that takes as input an $n$-leaf gene tree $T$ with branch lengths $(t_{e})$ and leaves labeled $1,...,n$. A probability transition matrix $P_{e}$ is associated to each edge; $P_{e}$ is a $4\times 4$ matrix which depends on the branch length $t_{e}$ and the choice of model (see below). The process is initialized by by assigning a random nucleotide from ${A,C,G,T}$ to the root according to a pre-specified root distribution $\pi$. Then, propagating away from the root, mutations occur along edges in a manner governed by the transition matrices $P_{e}$. The output is a random vector $X=(X_1,...,X_n)$ of the nucleotides at the $n$ leaves of the tree.

It is standard to assume that each site mutates independently. Hence, to generate a gene sequence of length $k$, we simply run the Jukes-Cantor process $k$ times independently, using the gene tree as input. Putting together $k$ i.i.d.~copies of $X$, we obtain DNA sequences of length $k$ as in the following figure:

Exercise 1

Suppose we are given a gene tree of the following form:

Assume that the root distribution is uniform (i.e., $P(A) = P(C) = P(G) = P(T) = \frac{1}{4}$), and that the numbers are the substitution probabilities along each edge. Using the dice provided, simulate by hand the Jukes-Cantor process 10 times using this tree as input. Record your results as a $4\times 10$ matrix.

See example output

Here is an example output

>H GCGGCGGCCC
>C GCGGCGGCCC
>G GCGGCGACCT
>O GCGGCGTCCC

Evolutionary Distance

The branch lengths $(t_{e})$ of a gene tree are measured in units of expected number of mutations per site. This is a measure of evolutionary distance, with longer branches having more time for mutations to accumulate. Let

$$\begin{aligned} p &= \text{probability of substitution along the branch} \\\ d &= \text{branch length in expected number of mutations per site.} \end{aligned}$$

Then the relationship between $d$ and $p$ is given by

$$ p = \frac{3}{4}\left(1-e^{-\frac{4}{3}d}\right) $$

or equivalently,

$$ d = -\frac{3}{4}\log\left(1-\frac{4}{3}p\right). $$

This is the Jukes-Cantor correction formula.

As an example, consider the following 30bp portion of the gene sequences of the human $\alpha$ and $\beta$ hemoglobin genes:

The $\alpha$ and $\beta$ genes diverged from a common ancestor about 400 million years ago. Since we have two samples, this corresponds to the following 2-leaf gene tree:

There are $12$ substitutions (i.e. differences) between the two genes. However, the actual number of mutations which occurred is likely greater then 12, because nucleotides may have mutated and reverted (e.g., $A\to C \to A$) or may have mutated multiple times (e.g., $A\to C\to G$).

We can estimate the number of mutations by observing that $p\approx 12/30 = .4$. By the Jukes-Cantor correction formula, the evolutionary distance for $p=.4$ is

$$ d = .57 \text{ expected mutations per site} $$

Since there are $30$ sites, this works out to about 17 mutations.

Exercise 2

1. In this example, find an estimate for the rate of mutations per site per year.

2. For the 4-leaf gene tree in the previous example, use the Jukes-Cantor correction formula to convert the labelled substitution probabilities into units of evolutionary distance, and write the gene tree as in Newick format.

See Answer

1. We estimated the path between two leaves to have length $d=.57$. This path consists of two edges, each representing 400 million years of evolution. Therefore the expected number of mutations per site per year is $\frac{.57}{8\times 10^8}=0.7\times 10^{-9}$.

2. (((H:.052,C:.052):.107,G:.167):.052,O:.233);

References and further reading

[1] Thomas H. Jukes and Charles R. Cantor, Evolution of protein molecules, Mammalian Protein Metabolism (1969) DOI: 10.1016/B978-1-4832-3211-9.50009-7

The hemoglobin example is based on Thomas H. Jukes' short article, How Many Nucleotide Substitutions Actually Took Place? Current Contents (1990).

A good introductory exposition of the Jukes-Cantor substitution model can be found at http://treethinkers.org/jukes-cantor-model-of-dna-substitution/