give answer in 2 step with explanation at the end of each step and final answer at the end:5. You are given a flask of a bacterial broth culture that has a concentration of 36x10 CFU/mL. You transfer

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:5. You are given a flask of a bacterial broth culture that has a concentration of 36x10 CFU/mL. You transfer 500 µl of the culnurn into tube A, which contains 4.5 ml of wat You then transfer 100 jul. from tube A imo tube II, which contains 9.9 md. of wa Finally, you transfer Inmd from nube B into nube C, which contains 9.0 ml, of water. You then inoculate a CBA agar plate with 100 µl. from tube C.
What is the dilution factor between tube A and tube 11? Show your work
b. What is the total dilution of the bacteria in tube C relative to the original fisk? Show your work and express your answer either as a ratio or using scientific notation
.How many colanies would you expect to grow on the CBA agar plate that was inoculated with 100 µl. from tube C? Show your work

VIPSolutions · Accepted Answer

# Bacterial Dilution and Colony Calculation

## Step 1: Dilution Calculations

### a. **Dilution Factor Between Tube A and Tube II**

- **From tube A, 100 µL (.1 mL) is transferred into tube II containing 9.9 mL water.**
- **Total final volume in tube II:** $.1 \text{ mL} + 9.9 \text{ mL} = 10 \text{ mL}$
- **Dilution factor:**  
  $$
  \text{Dilution factor} = \frac{\text{Volume transferred}}{\text{Final volume}} = \frac{.1}{10} = \frac{1}{100}
  $$
- **Explanation:** 1 part sample is diluted into 99 parts diluent for a 1:100 dilution.

---

### b. **Total Dilution in Tube C Relative to Original Flask**

Let's break down each dilution step:

1. **Flask → Tube A:**  
   - 500 µL (.5 mL) into 4.5 mL water → final volume = .5 + 4.5 = 5 mL  
   - Dilution: $ \frac{.5}{5} = \frac{1}{10} $

2. **Tube A → Tube II:**  
   - Already calculated: $ \frac{1}{100} $

3. **Tube II → Tube C:**  
   - 1 mL into 9 mL water → final volume = 1 + 9 = 10 mL  
   - Dilution: $ \frac{1}{10} $

- **Total dilution (multiply each step):**
  $$
  \text{Total dilution} = \frac{1}{10} \times \frac{1}{100} \times \frac{1}{10} = \frac{1}{10,000}
  $$
- **Scientific notation:** $ 1 \times 10^{-4} $
- **Explanation:** The sample is diluted by a factor of 1/10,000 relative to the original flask after all three steps.

---

## Step 2: Expected Colony Calculation

### c. **Number of Colonies Expected on Plate**

- **Original concentration:** $ 36 \times 10 $ CFU/mL = 360 CFU/mL
- **Dilution in tube C:** $ 1 \times 10^{-4} $
- **Inoculated volume:** 100 µL = .1 mL

**Calculate the number of CFU in .1 mL from tube C:**
$$
\text{CFU in 1 mL of tube C} = 360 \times 1 \times 10^{-4} = .036 \text{ CFU/mL}
$$
$$
\text{CFU in .1 mL} = .036 \times .1 = .0036 \text{ CFU}
$$

**Explanation:** After all dilutions, the number of bacteria plated is much less than 1, so it is highly unlikely that any colonies will be observed.

---

## **Final Answers**

**a.** The dilution factor between tube A and tube II is **1:100** (or $ 1 \times 10^{-2} $).

**b.** The total dilution in tube C relative to the original flask is **1:10,000** (or $ 1 \times 10^{-4} $).

**c.** The number of colonies expected to grow on the CBA agar plate is **much less than 1** (specifically, .0036), so you would expect **zero colonies** under normal circumstances.

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