Question

You have a stock of an E. coli cell culture of an unknown concentration. You want to set up a serial dilution experiment so that you can count the viable colonies and determine the concentration of the stock in cells/ml.

In tube A, you add 1 ml of the stock to 999 ml of media. In tube B, you add 1 ml of the material from tube A to 24 ml of media. In tube C, you add 50 microliters of the material from tube B to 9.95 ml of media.

From each of tubes A, B, and C, you take 100 microliters of the material and plate onto corresponding Petri plates A, B, and C. After one-day incubation, you count the colonies: Plate A had way too many to count, since there was nearly a lawn, Plate B had around 3,800 colonies, and Plate C had 42 colonies.

You decide to make another dilution in tube D. You add 1 ml of the material from tube C to 99 ml of media in tube D. You add 100 microliters of the material in tube D to Petri plate D, and after one-day incubation, you saw no colonies at all. Why? Come up with an alternative way to determine the cell concentration in tube D.

Answer 1

the bacterial culture is highly diluted in tube D. As it produced reasonably countable number of colonies in plate C, it must have been diluted very little in D.

So my suggestion would be, make dilution of 1:5 or 1:10, that is add 1 ml of C into 5 or 10 ml of D. This will certainly yield easily countable number of colonies.