Question

Total Heterotrophic Plate Count: A serial dilution was conducted using standard methods, then selected dilutions were used to set up pour plates with total plate count agar, and plates incubated for 48 – 72 hours. Each dilution was plated in duplicate, with 1.0 ml of inoculum placed into Petri plates, then pour plates prepared. After incubation, you counted the following:

10*-3 dilution: Too many to count on both plates.
10*-4 dilution: 378 colonies; 427 colonies.
10*-5 dilution: 53 colonies; 47 colonies.
10*-6 dilution: 4 colonies; 5 colonies.

Select the appropriate dilution and use that dilution to calculate the number of colony forming units per ml of sample. Remember to take into account the volume of liquid used to inoculate the agar plates.

Answer 1

There are four dilutions here. The 10*-3 dilution has too many colonies to count for. The 10*-4 dilution has countable colonies but still there are way too many and hence very tedious. The 10*-5 has around 50 colonies and hence can be counted with ease. The 10*-6 has very few colonies around 4-5 and thus errors can be huge in this case.

Thus, the best plate to choose is the one with 10*-5 dilution. The volume of the dilution that was inoculated on the plate was 1.0mL. Hence, 1.0mL of 10*-5 dilution had (47+53)/2 = 100/2 = 50 colonies. Therefore, the number of colonies in 1.0mL of original culture will be 50*10^5 = 5*10^6 = 5 million cfu.