Problem 6 - Calculus for Biology and Medicine

Question

Initially you measure that a colony of bacterial cells contains 1000 cells. 2 hours later you measure the colony again, and count 2000 cells. (a) How many cells would you expect the colony to contain 4 hours after the start of the experiment? (b) In fact, you realize that the hemocytometer that you used to count the cells for both measurements is only accurate to \(10 \%\), meaning that if you count 1000 cells, the real number of cells is somewhere between \(1000-100=900\) cells and \(1000+100=\) 1100 cells. What is the largest possible number of cells in the colony 4 hours after the start of the experiment? And what is the smallest possible number of cells at 4 hours?

Step-by-Step Solution

Answer

(a) 4000 cells; (b) between 3600 and 4400 cells.

Step 1: Determine Growth Rate

In 2 hours, the number of cells doubled from 1000 to 2000. To find the growth rate, assume exponential growth, where \[ N(t) = N_0 \times e^{kt} \] where \(N_0 = 1000\), \(t = 2\), and \(N(t) = 2000\). Substitute these values into the equation:\[ 2000 = 1000 \times e^{2k} \]Simplifying gives:\[ 2 = e^{2k} \]Take the natural logarithm of both sides:\[ \ln(2) = 2k \]Solve for \(k\):\[ k = \frac{\ln(2)}{2} \approx 0.3466 \]

Step 2: Predict Bacterial Count After 4 Hours

To find the expected number of cells after 4 hours, use the exponential growth formula again with \( t = 4 \):\[ N(4) = 1000 \times e^{k \times 4} \]Substitute \( k = 0.3466 \):\[ N(4) = 1000 \times e^{0.3466 \times 4} \]\[ N(4) = 1000 \times e^{1.3864} \]\[ N(4) = 1000 \times 4 \] (because \( e^{\ln(4)} = 4 \))\[ N(4) = 4000 \]Thus, we expect the colony to contain 4000 cells after 4 hours.

Step 3: Adjust for Measurement Error

The hemocytometer is accurate to \( 10\% \), so the initial measurement of 1000 cells could be as low as 900 or as high as 1100, and the same applies to the measurement of 2000 cells, which could actually be between 1800 and 2200.Calculate the growth rate for the smallest possible increase (from 900 to 1800) and the largest (from 1100 to 2200).1. For smallest increase: - \(2 = \frac{1800}{900} = e^{2k}\) - \(k = \frac{\ln(2)}{2} = 0.3466\) (Growth rate remains the same)2. For largest increase (same reasoning): - Growth rate remains 0.3466.

Step 4: Calculate Smallest and Largest Predictions

Using the growth rate \( k = 0.3466 \) and initial count \( N_0 = 900 \) or \( N_0 = 1100 \):1. Smallest possible count: - \( N(4) = 900 \times e^{0.3466 \times 4} \) - \( N(4) = 900 \times 4 \) - \( N(4) = 3600 \)2. Largest possible count: - \( N(4) = 1100 \times e^{0.3466 \times 4} \) - \( N(4) = 1100 \times 4 \) - \( N(4) = 4400 \)Thus, at 4 hours, the number of cells could range from 3600 to 4400.

Key Concepts

Exponential Growth

In biology, exponential growth refers to the increase in a population size where the rate of growth is proportional to the current size of the population. This occurs when resources are abundant, and each member of the population is reproducing at a constant rate. In the context of bacterial growth, exponential growth means that bacteria multiply rapidly, doubling in number over constant time intervals.

In mathematical terms, exponential growth can be represented by the formula:

• \[ N(t) = N_0 \times e^{kt} \]

Where:

• \( N(t) \) is the number of bacteria at time \( t \).
• \( N_0 \) is the initial number of bacteria.
• \( k \) is the growth rate constant, and \( e \) is the base of the natural logarithm.

This formula assumes ideal conditions without limitations on nutrients or space, which is often the case for bacteria in laboratory settings before they reach the carrying capacity of their environment. It is important to recognize that this ideal model does not incorporate external factors that may impede growth, such as nutrient depletion or waste accumulation.

Measurement Error

When conducting experiments, especially in biology, it's crucial to account for potential measurement errors. Measurement error refers to the difference between the measured value and the true value. It's an inherent part of any scientific measurement due to limitations in equipment precision or human error. In the given exercise, the measurement tool, a hemocytometer, has an error margin of \(10\%\).

This means the counted number of bacterial cells may vary within \(10\%\) of the measured value. To calculate the true possible values:

• For an initial count of 1000 cells, the true number could be as low as 900 or as high as 1100 (\(1000 \pm 100\)).
• Similarly, for a measurement of 2000 cells, the actual number could range from 1800 to 2200 (\(2000 \pm 200\)).

Considering this error helps provide a more realistic range of values, offering insight into the precision and reliability of experimental results. Understanding how measurement error might affect data is essential in drawing accurate conclusions from experiments.

Growth Rate Calculation

Calculating the growth rate is a fundamental aspect of understanding bacterial growth patterns. In this scenario, it involves determining how quickly a bacterial population doubles over a specific time period. To calculate the growth rate \( k \), we utilize the information that the population doubles from 1000 to 2000 cells in 2 hours.

• Set up the equation based on exponential growth: \[ N(t) = N_0 \times e^{kt} \] where the known values are \( N(t) = 2000 \), \( N_0 = 1000 \), and \( t = 2 \).
• Rearrange to find the growth rate equation: \[ 2 = e^{2k} \]
• Solving for \( k \), first take the natural logarithm of both sides: \[ \ln(2) = 2k \]
• Solve for \( k \): \[ k = \frac{\ln(2)}{2} \approx 0.3466 \].

This growth rate \( k \) indicates the speed at which the bacteria are multiplying. Equipped with this information, one can predict the future size of the bacterial population, provided the conditions remain stable. Understanding the growth rate is crucial for predicting how fast bacterial populations can increase under specific conditions, which has implications in various fields like microbiology, medicine, and environmental science.