Question

To determine the viable cell count of a bacterial culture, you have plated 50 \(\mu\)L of a 100-fold diluted sample of the culture on a nutrient agar plate and obtained 20 colonies after overnight incubation. The viable cell count of the culture is CFU mL^-1. (answer in integer)

Hint:

CFU/mL = (colonies divided by volume plated in mL) times the dilution factor; do not forget to multiply back by the dilution factor.

Answer 1

Correct Answer: 40000

Step 1: Recall the viable count formula.
Each colony on the plate grew from one viable cell present in the volume of diluted sample spread on it. The viable count of the ORIGINAL culture comes from
\[ \text{CFU mL}^{-1} = \frac{\text{number of colonies}}{\text{volume plated (mL)}} \times \text{dilution factor} \]

Step 2: List the known values.
Number of colonies counted = 20.
Volume of diluted sample plated = \(50\ \mu\text{L} = 0.05\ \text{mL}\).
Dilution factor = 100, since the sample was diluted 100-fold before plating.

Step 3: Find the count in the diluted sample.
\[ \text{CFU mL}^{-1}\ \text{(diluted sample)} = \frac{20}{0.05} = 400\ \text{CFU mL}^{-1} \]

Step 4: Scale back up to the original, undiluted culture.
Since the plated sample was 100 times more dilute than the original culture, the original culture has 100 times more viable cells per mL:
\[ \text{CFU mL}^{-1}\ \text{(original culture)} = 400 \times 100 = 40000\ \text{CFU mL}^{-1} \]

Final Answer:
The viable cell count of the culture is \[ \boxed{40000\ \text{CFU mL}^{-1}} \]