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)

Question

Which of the following bonds is responsible for maintaining the secondary structure of proteins?

Answer 1

Think of this problem as working backward from what you saw on the plate to what was actually present in the original tube of culture, one dilution step at a time.

You spread $50\ \mu\text{L} = 0.05$ mL of the diluted sample and got 20 colonies. Each colony traces back to one live bacterial cell that landed on the plate, so the diluted sample itself must have held

\[ \frac{20\ \text{colonies}}{0.05\ \text{mL}} = 400\ \text{viable cells per mL} \]

But this 400 cells/mL figure describes the diluted sample, not the original culture. The sample was diluted 100-fold before you plated it, meaning every mL of original culture was spread out to occupy 100 mL of diluted sample. So the original culture is 100 times more concentrated in live cells than what you measured:

\[ 400\ \text{CFU mL}^{-1} \times 100 = 40000\ \text{CFU mL}^{-1} \]

Let's summarize:

Divide colonies by the plated volume (in mL) to get the count in the plated, diluted sample.
Multiply by the dilution factor to undo the dilution and recover the original culture's count.

So the viable cell count of the original bacterial culture is $40000$ CFU mL$^{-1}$.

Answer 2

Think of this problem as working backward from what you saw on the plate to what was actually present in the original tube of culture, one dilution step at a time.

You spread $50\ \mu\text{L} = 0.05$ mL of the diluted sample and got 20 colonies. Each colony traces back to one live bacterial cell that landed on the plate, so the diluted sample itself must have held

\[ \frac{20\ \text{colonies}}{0.05\ \text{mL}} = 400\ \text{viable cells per mL} \]

But this 400 cells/mL figure describes the diluted sample, not the original culture. The sample was diluted 100-fold before you plated it, meaning every mL of original culture was spread out to occupy 100 mL of diluted sample. So the original culture is 100 times more concentrated in live cells than what you measured:

\[ 400\ \text{CFU mL}^{-1} \times 100 = 40000\ \text{CFU mL}^{-1} \]

Let's summarize:

Divide colonies by the plated volume (in mL) to get the count in the plated, diluted sample.
Multiply by the dilution factor to undo the dilution and recover the original culture's count.

So the viable cell count of the original bacterial culture is $40000$ CFU mL$^{-1}$.

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)

Show Hint

Correct Answer: 40000

Solution and Explanation

Questions Asked in GATE BT exam

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)

Show Hint

Correct Answer: 40000

Solution and Explanation

Questions Asked in GATE BT exam

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)