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A biologist prepares a culture. After 1 day of growth, the biologist counts 1000 cells. After 2 days of growth, he counts 3000 . Assuming a Malthusian model, what is the reproduction rate and how many cells were present initially?

Short Answer

Expert verified
Reproduction rate \( r = \ln(3) \), initial cells \( N_0 \approx 333.33 \).

Step by step solution

01

Understanding the Malthusian Model

The Malthusian growth model is governed by the equation \( N(t) = N_0 e^{rt} \), where \( N(t) \) is the number of cells at time \( t \), \( N_0 \) is the initial number of cells, and \( r \) is the reproduction rate. Our objective is to find \( N_0 \) and \( r \).
02

Set Up Equations

We have \( N(1) = 1000 \) and \( N(2) = 3000 \). Using the Malthusian model equation: 1. \( 1000 = N_0 e^r \) (from \( t = 1 \))2. \( 3000 = N_0 e^{2r} \) (from \( t = 2 \))
03

Solve for \( e^r \)

Divide the second equation by the first:\[ \frac{3000}{1000} = \frac{N_0 e^{2r}}{N_0 e^r} \] This simplifies to:\[ 3 = e^r \]So, \( e^r = 3 \).
04

Find the Rate \( r \)

Since \( e^r = 3 \), take the natural logarithm on both sides: \[ r = \ln(3) \]
05

Solve for \( N_0 \)

From Step 3, we know \( e^r = 3 \), so substitute \( e^r \) back into the first equation: \[ 1000 = N_0 imes 3 \] Solving for \( N_0 \) gives:\[ N_0 = \frac{1000}{3} \approx 333.33 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reproduction Rate
In the context of the Malthusian growth model, the reproduction rate, often denoted as \( r \), plays a crucial role. It represents the rate at which a population's size increases due to reproduction over time. In a biological setup, like a culture of cells, the reproduction rate tells us how quickly the population can multiply.

To determine \( r \) in our example, we use the Malthusian equation: \[ N(t) = N_0 e^{rt} \]Here, we have observations at two distinct times, which are essential for finding \( r \). By comparing the equations derived from these observations:
  • Day 1: \( N(1) = 1000 \)
  • Day 2: \( N(2) = 3000 \)
These values facilitate finding \( e^r \) by dividing the second equation by the first, simplifying the process to discover that \( e^r = 3 \). The reproduction rate is then determined by taking the natural logarithm: \[ r = \ln(3) \]Understanding the reproduction rate enables us to predict future population sizes when the initial population size and time are known.
Exponential Growth
Exponential growth is a pattern where the growth rate becomes increasingly rapid in proportion to the growing total number or size. In the Malthusian growth model, it is characterized by the constant reproduction rate \( r \).

This model assumes that the population grows continuously and rapidly. The equation that captures this behavior is again: \[ N(t) = N_0 e^{rt} \]Here:
  • \( N(t) \) is the population size at time \( t \).
  • \( N_0 \) is the initial population size.
  • \( e^{rt} \) indicates that any small change in time leads to large changes in population size when \( r \) is positive.
Exponential growth is fundamental in scenarios where resources are abundant, and limiting factors are minimal, allowing the population to grow unchecked initially. In our example, once \( e^r \) was identified as \( 3 \), this indicated a strong exponential growth where the population tripled from Day 1 to Day 2.
Initial Population Size
The initial population size \( N_0 \) is the starting number of individuals in the population prior to any growth occurring. In population dynamics, knowing \( N_0 \) provides a foundation for projecting future population sizes.

For our exercise, \( N_0 \) represents the number of cells in the culture at time \( t = 0 \). Using the Malthusian growth equation and our calculated value of \( e^r \):
  • From the equation \( 1000 = N_0 \times 3 \)
  • We solve for \( N_0 \) to get \( N_0 = \frac{1000}{3} \approx 333.33 \)
This result means there were approximately 333 cells at the very beginning (before the growth observations started on Day 1). Measuring and understanding \( N_0 \) is essential as it affects the scale of the population's future growth projections. The Malthusian model uses this initial size as a baseline to calculate increases driven by the reproduction rate.

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Most popular questions from this chapter

Consider a lake that is stocked with walleye pike and that the population of pike is governed by the logistic equation $$ P^{\prime}=0.1 P(1-P / 10), $$ where time is measured in days and \(P\) in thousands of fish. Suppose that fishing is started in this lake and that 100 fish are removed each day. (a) Modify the logistic model to account for the fishing. (b) Find and classify the equilibrium points for your model. (c) Use qualitative analysis to completely discuss the fate of the fish population with this model. In particular, if the initial fish population is 1000 , what happens to the fish as time passes? What will happen to an initial population having 2000 fish?

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