Problem 45 - Calculus for Biology and Medicine

Question

Population genetics is the study of how the frequency of particular traits changes within a population over time. We are studying a gene that comes in two alleles (i.e., variants) \(A\) and \(a\). The \(A\) allele makes individuals reproduce a little faster than the \(a\) allele. So we expect the \(A\) alleles to take over the population with time. Suppose that a proportion \(p\) of all individuals within the population carry the \(A\) allele (with the remaining proportion, \(1-p\), carrying the \(a\) allele). If the \(A\) allele boosts reproduction rate by an amount \(s\) it can be shown under some assumptions that the proportion of \(A\) -allele individuals obeys a differential equation $$ \frac{d p}{d t}=\frac{1}{2} s p(1-p) $$ (a) Use separation of variables and partial fractions to find the solution of \((8.31)\), assuming \(p(0)=p_{0}\). (b) Show that if \(p_{0} \neq 0\), then \(\lim _{t \rightarrow \infty} p(t)=1\). Explain why this behavior makes sense biologically. (c) Suppose \(p_{0}=0.1\) and \(s=0.01\); how long will take until \(p(t)=0.5 ?\)

Step-by-Step Solution

Answer

(a) \( p(t) = \frac{\frac{p_0}{1-p_0} e^{\frac{1}{2} s t}}{1 + \frac{p_0}{1-p_0} e^{\frac{1}{2} s t}} \); (b) \( \lim_{t \to \infty} p(t) = 1 \); (c) \( t \approx 440.5 \).

Step 1: Separate Variables

We want to solve the differential equation \( \frac{d p}{d t} = \frac{1}{2} s p (1-p) \) by separating variables. To do this, we'll rewrite it as \( \frac{d p}{p(1-p)} = \frac{1}{2} s \, dt \). This separates \( p \) and \( dt \) to opposite sides of the equation.

Step 2: Integrate Both Sides

Integrate both sides of the equation: \[ \int \frac{1}{p(1-p)} \, dp = \int \frac{1}{2} s \, dt \]. This requires partial fraction decomposition for the left side:\[ \frac{1}{p(1-p)} = \frac{1}{p} + \frac{1}{1-p}\], so the integral becomes:\[ \int \left( \frac{1}{p} + \frac{1}{1-p} \right) \, dp = \frac{1}{2} s t + C \], where \( C \) is a constant of integration.

Step 3: Solve the Integrals

The integral on the left is \[ \ln |p| - \ln |1-p| = \frac{1}{2} s t + C \].Simplifying gives us\[ \ln \left( \frac{p}{1-p} \right) = \frac{1}{2} s t + C \].

Step 4: Solve for p(t)

To find \( p(t) \), exponentiate both sides to remove the natural logarithm:\[ \frac{p}{1-p} = e^{\frac{1}{2} s t + C} \]. Let \( K = e^C \) (for simplicity), so:\[ \frac{p}{1-p} = K e^{\frac{1}{2} s t} \].Solving for \( p \):\[ p = \frac{K e^{\frac{1}{2} s t}}{1 + K e^{\frac{1}{2} s t}} \].

Step 5: Determine K Using Initial Condition

Use the initial condition \( p(0) = p_0 \) to find \( K \). At \( t = 0 \),\[ p_0 = \frac{K}{1+K} \].Solving for \( K \), we get \( K = \frac{p_0}{1-p_0} \).

Step 6: Show Limit as t Approaches Infinity

To show \( \lim_{t \to \infty} p(t) = 1 \), observe that as \( t \to \infty \), \( e^{\frac{1}{2} s t} \to \infty \), making the term \( 1+K e^{\frac{1}{2} s t} \approx K e^{\frac{1}{2} s t} \). So,\[ p(t) \approx \frac{K e^{\frac{1}{2} s t}}{K e^{\frac{1}{2} s t}} = 1 \].

Step 7: Solve for p(t) = 0.5

We need to solve for \( t \) such that \( p(t) = 0.5 \). With \( K = \frac{0.1}{0.9} \) and \( p(t) = 0.5 \),\[ 0.5 = \frac{\frac{0.1}{0.9} e^{\frac{1}{2} \cdot 0.01 \cdot t}}{1 + \frac{0.1}{0.9} e^{\frac{1}{2} \cdot 0.01 \cdot t}} \].Set up the equation:\[ 1 = \frac{0.1}{0.9} e^{\frac{1}{2} \cdot 0.01 \cdot t} \].Solve for \( t \):\[ e^{\frac{1}{2} \cdot 0.01 \cdot t} = 9 \].Taking the natural log of both sides:\[ \frac{1}{2} \cdot 0.01 \cdot t = \ln 9 \]\[ t = \frac{2 \ln 9}{0.01} \approx 440.5 \].

Key Concepts

Alleles

In population genetics, genes can come in different forms called **alleles**. These are variants of a gene that determine specific traits in organisms. For example, if you are studying a gene that influences the color of flowers, one allele might produce red flowers, while another allele might produce white flowers. This concept is critical in understanding how diversity exists within a population.
In the exercise, we are considering a gene with two alleles, represented as \(A\) and \(a\). Each individual in the population carries two alleles: one from each parent. The combination of these alleles determines certain characteristics of the individuals. If the \(A\) allele allows individuals to reproduce faster than the \(a\) allele, it is likely to become more prevalent over time as it gives a reproductive advantage. Essentially, this is a simplified model of natural selection, where advantageous traits become more common in subsequent generations.
**Key Points:**

• Alleles are different forms of the same gene.
• They are critical in understanding genetic variation and inheritance patterns.
• Natural selection can favor one allele over another, depending on the advantage it offers.

Differential Equation

A differential equation is a mathematical equation that involves the rates at which quantities change. They are used extensively in fields such as physics, engineering, and of course, biology. When applied to population genetics, differential equations help model how gene frequencies change over time: a critical component in understanding evolution.
In this exercise, the differential equation we are dealing with is \(\frac{d p}{d t} = \frac{1}{2} s p(1-p)\). Here, \(p\) is the proportion of the population carrying the \(A\) allele, \(s\) is the selection coefficient that measures how much the \(A\) allele increases reproduction, and \(t\) is time. This equation basically states that the rate of change of the proportion of alleles depends on several factors, including the current proportion of the alleles and their reproductive advantage.
Solving the differential equation involves several mathematical techniques, such as separation of variables and integration, which might be unfamiliar at first but are fundamental to understanding complex biological systems.
**Understanding the Components:**

• \(\frac{d p}{d t}\) represents the rate of change of the allele frequency.
• \(s\) is the selection coefficient, indicating the reproductive advantage.
• \(p(1-p)\) captures the effects of genetic drift and selection.

Reproduction Rate

The reproduction rate is a crucial concept in population genetics. It determines how quickly a population can grow. When an allele increases the reproduction rate of organisms carrying it, natural selection will tend to increase the frequency of that allele in the population over time.
In the given problem, the \(A\) allele boosts an individual's reproduction by an amount \(s\), which is known as the selection coefficient. This small advantage can, over numerous generations, lead to significant shifts in allele frequencies. The core idea is that even a slight increase in reproduction efficiency can have a profound impact on which traits become dominant.
A higher reproduction rate means that individuals with the \(A\) allele will have more offspring who, in turn, carry the same allele. This creates a positive feedback loop, gradually causing the \(A\) allele to overtake the \(a\) allele in the population.**Considerations:**

• Small changes in reproduction rate can lead to big effects over time.
• The selection coefficient \(s\) quantifies how much an allele improves reproduction.
• The long-term outcome is the potential fixation of beneficial alleles.

Logarithmic Functions

Logarithmic functions play a significant role in solving the differential equations found in population genetics. In these equations, natural logarithms are often utilized due to their special property of converting multiplicative relationships into additive ones, which simplifies problem-solving.
For the provided exercise, logarithmic functions appear when integrating the differential equation: specifically,\[\ln \left( \frac{p}{1-p} \right) = \frac{1}{2} s t + C\]. Here, the logarithmic function helps to find an explicit expression for \( p(t)\), the proportion of \( A\) alleles over time. The solution involves exponentiating both sides to eliminate the logarithm and isolate \( p(t)\).
This process transforms complex relationships between time, rate of change, and allele frequency into a more manageable algebraic form, making it easier to calculate specific values such as how long it takes for certain allele frequencies to be achieved.**Benefits of Using Logarithms:**

• Facilitates the unraveling of complex exponential relationships.
• Transforms multiplicative relationships into additive ones for simplicity.
• Crucial for calculating changes over time in population genetics.