You culture an inoculum of bacteria using the pour plate method, and start by mi

Question

You culture an inoculum of bacteria using the pour plate method, and start by mixing 1 mL of the sample into 9 mLs of medium in a tube (tube 1). You then repeat this process to carry out serial 1/10 dilutions, each time mixing 1 mL of the diluted bacteria with 9 mLs of medium (tubes 2, 3, 4 and 5), and then pouring a plate. You then incubate the plates for 24 hours, and count the resulting colonies on each plate. If you obtain 50 colonies on plate 2, prepared using the 2nd tube, how many bacteria were present in your original sample?

Explanation / Answer

It is given that serial dilution is 1/10.

For first tube we have 1ml of culture and 9ml of medium or the dilution will be given by

1/1+9 = 1/10

For the second tube the dilution will be 1/10 of the first dilution and is given by

1/1+9 multiplied by 1/1+9 = 1/10 * 1/10 = 1/100 or 10-2

So the dilution factor is 1/100 or 10-2

Number of bacterial cells present in the original sample can be calculated as

(Number of colonies formed)/(ml plated)* (dilution before plating) =Number of cells in the original sample/ ml of undiluted culture

or 50/ (0.1 ml plated) * (10-2) = 50/10-3 = 50 * 103 or 5 * 104 colony forming units per ml (cfu/ml)

So the number of colony forming units or the number of bacterial cells present in the original sample is 5 *104 cfu/ml

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