Population Genetics

Definition: Evolution is the process of change in the gene pool.

Question: What are the major forces of evolution?

Answer: There are four major forces:

• Natural selection
• Drift
• Gene flow
• Mutation

Our goal is to develop models of individual forces and explore them to develop expectations of evolutionary behavior.

Let's follow the ODD protocol, even though it was designed for agent-based models.

Purpose: To understand the effects of natural selection in the absence of all other evolutionary forces.

Entities

Definition: A locus is a chromosomal or genomic location of a gene.

Definition: An allele is an alternate form of a gene at a given locus.

Definition: Gene frequency is the relative proportion of an allele.

Definition: The gene pool is the set of all alleles in a population at all loci.

Assumptions and Simplifications

Discuss: What assumptions do we need to make to remove the forces of drift, gene flow, and mutation?

Answer:

• Infinite population (Removes drift)
• Closed system (Removes gene flow)
• Fixed number of alleles (Removes mutation)

Assumptions and Simplifications

Discuss: In terms of the entities (loci and alleles), what simplifications would be helpful at this stage?

Answer: First focus on one loci and two alleles.

Discuss: Before adding selection, do we have a model of evolution to start with?

Answer: Yes! We can start from Hardy-Weinberg Equilibrium.

Assumptions and Simplifications

Model: Hardy-Weinberg Equilibrium states that allele and genotype frequencies in a population will remain constant from generation to generation in the absence of other evolutionary influences.

State variables

Suppose our population is diploid.

Let \( A_{1} \) and \( A_{2} \) denote the two alleles at our focus locus.

Let \( p \) and \( q \) denote the proportion of \( A_{1} \) and \( A_{2} \), respectively (Note: \( p + q = 1 \))

Assumptions and Simplifications

State variables - Time

Discuss: What is our unit for time? Is it discrete or continuous?

Answer: Time is measured in generations, which is discrete.

Under the assumption of random mating, what are the genotype frequencies for \( A_{1}A_{1} \), \( A_{1}A_{2} \), and \( A_{2}A_{2} \)?

This follows since the number of \( A_{1} \) alleles in a genotype is a binomial random variable.

For a variable to be a binomial random variable, ALL of the following conditions must be met:

• There are a fixed number of trials \( N \) (N=2 chromosomes)
• On each trial, there are two possible outcomes - “success” or “failure” (\( A_{1} \) or \( \ A_{2} \))
• The probability of each outcome is the same on each trial (\( p \) and \( q=1-p \))
• Trials are independent of one another (random mating)

The probability of having \( k \) successes in \( N \) trials is

\[ \textrm{Prob[k successes in N trials]} = \left(\begin{array}{c}N \\ k\end{array}\right)p^{k}q^{N-k} \]

Question: How do allele frequencies change over generations?

\[ p_{0} \rightarrow p_{1} \rightarrow p_{2} \rightarrow \cdots \]

\[ p_{t+1} = F(p_{t}), t \geq 0 \]

Note: Subscript corresponds to generation number.

Hardy-Weinberg

\[ p_{0} \rightarrow p_{0} \rightarrow p_{0} \rightarrow \cdots \]

\[ p_{t+1} = p_{t} \]

How to construct \( F \)? To introduce selection, we need to model fitness, which for a diploid population occurs on the genotype and not on the allele. We will model relative fitness - i.e., we only care about the fitness of each genotype relative to the other genotypes.

Selection

Note: Rather than model the change of allele \( p \), we will model the change of allele \( q \), as we are interpreting this model in the case of mutation, where \( A_{2} \) is the new mutant allele. This just changes perspective, not the resulting biological interpretation.

Selection

For model derivation purposes, it helps to imagine a finite population of individuals \( N_{t} \).

Discuss: We have an implicit assumption here that we haven’t mentioned yet. Can you spot it?

Answer: Fitness is independent of time (parameters).

Selection

Discuss: Thinking of the weights \( w_{ij} \) as multiplicative growth factors, what would Count\( _{t+1} \) be?

Selection

Discuss: How many \( A_{2} \) alleles are there in the \( t+1 \) generation? How many total alleles are there in the \( t+1 \) generation?

Selection

Answer: \[ \begin{align} N_{2,t+1} & = 2p_{t}q_{t}N_{t}w_{12} + 2q_{t}^2N_{t}w_{22} \\ N_{t+1} & = 2\bigl(p_{t}^2N_{t}w_{11} + 2p_{t}q_{t}N_{t}w_{12} + q_{t}^2N_{t}w_{22}\bigr) \end{align} \]

Discuss: So what is the frequency of allele \( A_{2} \) in generation \( t+1 \)?

\[ \begin{align} N_{2,t+1} & = 2p_{t}q_{t}N_{t}w_{12} + 2q_{t}^2N_{t}w_{22} \\ N_{t+1} & = 2\bigl(p_{t}^2N_{t}w_{11} + 2p_{t}q_{t}N_{t}w_{12} + q_{t}^2N_{t}w_{22}\bigr) \end{align} \]

Answer: \[ \begin{align} q_{t+1} & = \frac{N_{2,t+1}}{N_{t+1}} \\ & = \frac{2p_{t}q_{t}N_{t}w_{12} + 2q_{t}^2N_{t}w_{22}}{2\bigl(p_{t}^2N_{t}w_{11} + 2p_{t}q_{t}N_{t}w_{12} + q_{t}^2N_{t}w_{22}\bigr)} \\ & = \frac{p_{t}q_{t}w_{12} + q_{t}^2w_{22}}{p_{t}^2w_{11} + 2p_{t}q_{t}w_{12} + q_{t}^2w_{22}} \end{align} \]

Replace \( p_{t} = 1-q_{t} \) to arrive at

\[ q_{t+1} = \frac{(1-q_{t})q_{t}w_{12} + q_{t}^2w_{22}}{(1-q_{t})^2w_{11} + 2(1-q_{t})q_{t}w_{12} + q_{t}^2w_{22}}. \]

Since we are concerned with relative fitness, we have

We consider 5 common types of selection:

• Codominance (genic selection):

• Complete dominance (\( A_{2} \) is dominant):

• Complete recessiveness (\( A_{2} \) is recessive):

• Overdominance (\( s > 0 \), \( s > t \), heterozygote highest fitness):

• Underdominance (\( s < 0 \), \( s < t \), heterozygote lowest fitness):

Codominance (genic selection):

\[ \begin{align} q_{t+1} & = \frac{(1-q_{t})q_{t}(1+s) + q_{t}^2(1+2s)}{(1-q_{t})^2 + 2(1-q_{t})q_{t}(1+s) + q_{t}^2(1+2s)} \\ & = F(q_{t},s) \end{align} \]

Simulate: \( s = 0.01 \), \( q_{0} = 0.04 \)

Simulate: \( s = 0.01 \), \( q_{0} = 0.04 \)

N <- 3000
s <- 0.01
q <- rep(0.04,N)
for (i in 1:(N-1)) {
  qi <- q[i] # For readability
  q[i+1] <- ((1-qi)*qi*(1+s) + qi*qi*(1+2*s))/((1-qi)*(1-qi) + 2*(1-qi)*qi*(1+s) + qi*qi*(1+2*s))
}
plot(1:N,q)

Codominance (left); Dominant (center); Recessive (right)

Definition [Wikipedia]: Balancing selection refers to a number of selective processes by which multiple alleles (different versions of a gene) are actively maintained in the gene pool of a population at frequencies longer than expected from genetic drift alone.

Overdominance (\( s > 0 \), \( s > t \), heterozygote highest fitness):

Question: How can we find fixed points algebraically and determine their stability?

Answer: We are going to use symbolic math! Symbolic math is a computer algebra system that will rigorously solve algebraic equations (and perform calculus) using a computer.

We will be using SymPy, a Python library for doing symbolic mathematics. In reality, we can use the package in R by using the rSymPy package.

library(rSymPy)

q <- Var("q")
w11 <- Var("w11")
w12 <- Var("w12")
w22 <- Var("w22")
s <- Var("s")
t <- Var("t")

\[ \begin{align} q_{t+1} & = \frac{(1-q_{t})q_{t}(1+s) + q_{t}^2(1+2s)}{(1-q_{t})^2 + 2(1-q_{t})q_{t}(1+s) + q_{t}^2(1+2s)} \\ & = F(q_{t},s) \end{align} \]

sympy("F = ((1-q)*q*w12 + w22*q**2)/(w11*(1-q)**2 + 2*(1-q)*q*w12 + w22*q**2)")

Substitute weights and simplify:

(CD <- sympy("CD = simplify(F.subs([(w11, 1), (w12, 1+s), (w22, 1+2*s)]))"))

[1] "(q + q*s + s*q**2)/(1 + 2*q*s)"

[1] "(q + q*s + s*q**2)/(1 + 2*q*s)"

\[ F(q_{t},s) = \frac{q_{t}(1 + s) + sq_{t}^2}{1+2sq_{t}} \]

Where are the fixed points? \( F(q_{t},s) = q_{t} \Rightarrow F(q_{t},s) - q_{t} = 0 \)

sympy("solve(CD - q, q)")  # Solve CD - q = 0 for q

[1] "[1, 0]"

Let's determine stability by finding the derivative \( \frac{dF}{dq_{t}}(q_{t},s) \)

sympy("dF = simplify(diff(CD, q))")

[1] "(1 + s + 2*q*s + 2*q**2*s**2)/(1 + 4*q*s + 4*q**2*s**2)"

\[ \frac{dF}{dq_{t}}(q_{t},s) = \frac{1 + s + 2sq + 2s^2q^2}{1 + 4sq + 4s^2q^2} \]

Now evaluate at fixed points:

sympy("simplify(dF.subs(q, 0))")  # Substitute q = 0 and simplify

[1] "1 + s"

sympy("simplify(dF.subs(q, 1))")  # Substitute q = 1 and simplify

[1] "(1 + s)/(1 + 2*s)"

Now evaluate at fixed points:

\[ \begin{align} \frac{dF}{dq_{t}}(0,s) & = 1+s \\ \frac{dF}{dq_{t}}(1,s) & = \frac{1+s}{1+2s} \end{align} \]

This, if \( s > 0 \) (advantageous), then 0 is unstable and 1 is stable, whereas if \( -1 < s < 0 \) (deleterious), then 0 is stable and 1 is unstable.

Exercise: Perform the same analysis yourself with the remaining 4 types of selection.