https://file.aiquickdraw.com/mathbot/file/1726267798793n9f8ahnq.jpg

Let's solve the bacterial growth problems from the image.

### 1. **E. coli Growth Calculation (First problem)**

Given:
- Initial cell count: \(1 \times 10^6\) (1 million cells)
- Doubling time: 20 minutes
- Growth duration: 15 hours

#### Formula:
The number of cells after time \(t\) (in minutes) is given by the formula:
\[
N(t) = N_0 \times 2^{\frac{t}{T_d}}
\]
Where:
- \(N_0\) is the initial number of cells
- \(t\) is the total time in minutes
- \(T_d\) is the doubling time in minutes

#### Step 1: Convert 15 hours to minutes.
\[
t = 15 \times 60 = 900 \text{ minutes}
\]

#### Step 2: Calculate the number of doublings.
\[
\frac{t}{T_d} = \frac{900}{20} = 45
\]

#### Step 3: Calculate the final number of cells.
\[
N(900) = 1 \times 10^6 \times 2^{45}
\]
I will now compute this.

### 2. **E. coli Growth in TSB (Second problem)**

Given:
- Initial cell count: \(1 \times 10^6\)
- Doubling time: 20 minutes (assumed since it's E. coli)
- Growth duration: 3 hours

#### Step 1: Convert 3 hours to minutes.
\[
t = 3 \times 60 = 180 \text{ minutes}
\]

#### Step 2: Calculate the number of doublings.
\[
\frac{t}{T_d} = \frac{180}{20} = 9
\]

#### Step 3: Calculate the final number of cells.
\[
N(180) = 1 \times 10^6 \times 2^9
\]

### 3. **C. perfringens Growth Calculation (Third problem)**

Given:
- Initial cell count: \(5.0 \times 10^5\)
- Doubling time: 10 minutes
- Growth duration: 5 hours

#### Step 1: Convert 5 hours to minutes.
\[
t = 5 \times 60 = 300 \text{ minutes}
\]

#### Step 2: Calculate the number of doublings.
\[
\frac{t}{T_d} = \frac{300}{10} = 30
\]

#### Step 3: Calculate the final number of cells.
\[
N(300) = 5.0 \times 10^5 \times 2^{30}
\]

Let me compute these results.Here are the solutions for each problem:

### 1. **E. coli Growth after 15 hours**:
- Final number of cells: **\(3.52 \times 10^{19}\)** cells

### 2. **E. coli Growth after 3 hours**:
- Final number of cells: **512 million cells** (or \(5.12 \times 10^8\))

### 3. **C. perfringens Growth after 5 hours**:
- Final number of cells: **536.87 trillion cells** (or \(5.37 \times 10^{14}\))

Would you like more details on any of the steps, or have any questions?

Here are 5 follow-up questions you might find interesting:
1. How does the growth rate change if the temperature affects the doubling time of bacteria?
2. What would be the effect if growth inhibitors were present during the incubation period?
3. Can this exponential growth continue indefinitely in real-life conditions? Why or why not?
4. How does the initial number of cells impact the final count after a specific growth period?
5. How do you calculate the growth when bacteria experience a lag phase before starting exponential growth?

**Tip:** Always convert your time units to match the doubling time before performing calculations to avoid errors.

If a culture of Escherichia coli containing 1 million cells is allowed to grow for an additional 15 hours under ideal growth conditions, how many total cells are in the culture? The doubling time of E. coli is 20 minutes. You inoculate a TSB (tryptic soy broth) with 10^6 E. coli and incubate it at 37 degrees Celsius. Assuming the maximum growth rate for E. coli, how many bacterial cells are in the culture after 3 hours? If a culture of Clostridium perfringens containing 5.0 x 10^5 cells is allowed to grow for an additional 5 hours under ideal growth conditions, how many total cells are in the culture? The doubling time of C. perfringens is 10 minutes.

Learn how to calculate the exponential growth of Escherichia coli and Clostridium perfringens under ideal conditions. Explore how to use doubling time and exponential growth formulas for real-life bacterial cultures. Suitable for advanced biology students.

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