To determine the viable cell count of a bacterial culture, you have plated $50 \mu \text{L}$ 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 ______ $\text{CFU mL}^{-1}$. (answer in integer)
This problem requires calculating the viable cell concentration in a bacterial culture using data from a plate count assay.
The final count needs to be in CFU per mL ($\text{mL}^{-1}$), so convert the plated volume from microliters ($\mu \text{L}$) to milliliters ($\text{mL}$).
$ 50 \, \mu \text{L} = 50 \times 10^{-3} \, \text{mL} = 0.05 \, \text{mL} $
Use the formula for viable cell count:
$ \text{Viable Cell Count (CFU/mL)} = \frac{\text{Number of Colonies Counted}}{\text{Volume Plated (mL)}} \times \text{Dilution Factor} $
Substitute the given values into the formula:
$ \text{Viable Cell Count} = \frac{20 \, \text{CFU}}{0.05 \, \text{mL}} \times 100 $
First, calculate the concentration in the plated volume:
$ \frac{20 \, \text{CFU}}{0.05 \, \text{mL}} = 400 \, \text{CFU/mL} $
Now, multiply by the dilution factor to get the count in the original culture:
$ 400 \, \text{CFU/mL} \times 100 = 40000 \, \text{CFU/mL} $
The viable cell count of the bacterial culture is 40,000 $\text{CFU mL}^{-1}$.
| Bioinformatic tool/Database | Utility |
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