# How do we dilute a bacterial culture 50-fold, 100-fold, and 200-fold?

Jul 15, 2015

You could do the dilutions in one step, or you might have to do serial dilutions.

#### Explanation:

There is a limit to how much you can dilute a sample in one step.

You probably wouldn't want to dilute by more than a factor of 100 in a single step, because that would give you a large volume of dilute solution.

Rather, you would use the serial dilution technique.

Remember the formula for dilution factor ($\text{DF}$):

$\text{DF} = {V}_{f} / {V}_{i}$, where

${V}_{i} = \text{aliquot volume}$ and

${V}_{f} = \text{final volume" = "aliquot volume + diluent volume}$

(a) $\text{DF} = 50$

Here we can do a single dilution.

We can add 0.2 mL of sample to 9.8 mL of diluent (normal saline? cells will burst in distilled water because of osmotic pressure).

${V}_{f} = \text{0.2 mL + 9.8 mL" = "10.0 mL}$

"DF" = V_f/V_i = (10.0 cancel("mL"))/(0.2 cancel("mL")) = 50

(b) $\text{DF} = 100$

Here, too, we can do a single dilution.

This time we add 0.1 mL of sample to 9.9 mL of saline.

${V}_{f} = \text{0.1 mL + 9.9 mL" = "10.0 mL}$

"DF" = V_f/V_i = (10.0 cancel("mL"))/(0.1 cancel("mL")) = 100

(c) $\text{DF} = 200$

Here you might want to do a serial dilution.

The formula for serial dilutions is

"DF" = "DF"_1 × "DF"_2 × "DF"_3 ×…

How about a 1:100 dilution followed by a 1:2 dilution?

(1) Add 0.1 mL of sample to 9.9 mL of diluent. ${\text{DF}}_{1} = 100$. Call this "Solution 1".

(2) Add 5 mL of Solution 1 to 5 mL of saline. Call this "Solution 2".

${V}_{i} = \text{5 mL}$

${V}_{f} = \text{5 mL + 5 mL" = "10 mL}$

"DF"_2 = V_f/V_i = (10 cancel("mL"))/(5 cancel("mL")) = 2

"DF" = "DF"_1 × "DF"_2 = 100 × 2 = 200