Question

Enumeration of Escherichia coli and Total Coliforms by Plate Count: Chromogenic agar was used to enumerate E. coli and total coliforms. There are various formulations of chromogenic agar, which can be used for different target organisms. The chromogenic agar used in this analysis is selective for the growth of coliform bacteria. It is differential in that E. coli appears purple, and other coliforms appear pink. You used the same dilution series as above to prepare spread plates, with 0.1 ml of inoculum placed onto each plate. After incubation, you counted the following number of colonies:

10*-3 dilution: Plate 1: 48 pink, 96 purple. Plate 2: 46 pink, 106 purple.
10*-4 dilution: Plate 1: 3 pink, 9 purple. Plate 2: 6 pink, 11 purple.
10*-5 dilution: Plate 1: 2 pink, 2 purple. Plate 2: 1 pink, 1 purple.
10*-6 dilution: Plate 1: 0 pink, 0 purple. Plate 2: 0 pink, 0 purple.

Select the appropriate dilution and use to calculate the number of colony forming units per ml of sample for both E. coli and all coliforms (in answering this question you should consider whether E. coli is a coliform). Remember to take into account the volume of liquid used to inoculate the agar plates.

Answer 1

The pink is the count of E.coli and purple is the count of other coliforms.

In 10*-3 dilution, each plate has around 50+100 = 150 colonies which are very tedious to count.

In 10*-4 dilution, plate 1 has 3+9 = 12 colonies and plate 2 has 6+11 = 17 colonies. These are easy to count and a fairly large number of colonies.

In 10*-5 dilution, plate 1 has 2+2 = 4 colonies and plate 2 has 1+1 = 2 colonies. As can be seen, one plate has half the number of colonies from the other plate and hence the readings will be variable thereby resulting in high error.

In 10*-6 dilution, both plates have 0 colonies and hence cannot be used to estimate the number of cfu.

Therefore, the appropriate dilution here is 10*-4.

For this dilution, the average number of E.coli colonies = average number of pink colonies = (3+6)/2 = 9/2 = 4.5.

And the average number of other chromogenic bacteria = average number of purple colonies = (9+11)/2 = 20/2 = 10.

0.1mL of the sample was plated here. Hence, the average number of pink colonies in 1.0mL of dilution will be 4.5*10 = 45 and the average number of purple colonies in 1.0mL of dilution will be 10*2 = 20.

Therefore, the colony-forming units (cfu) of E.coli = 45*10^4 = 0.45*10^6 = 0.45 million.

Now, E.coli is also a coliform bacteria. Therefore, the total coliforms in the dilution are 45+20 = 65. Hence, the cfu of total coliforms =65*10^4 = 0.65*10^6 = 0.65 million.