2. Two days ago, you sponge sampled a chilled beef carcass as part of your compa

Question

2. Two days ago, you sponge sampled a chilled beef carcass as part of your company’s generic E. coli regulatory testing program. You sampled 100 cm^2 each from the brisket, flank, and rump for a 300 cm^2 total composite sample. After sample collection, you returned to the laboratory, added 65 ml of total diluent to the sponge bag, and stomached to homogenize the sample. Two successive 1:10 dilutions were made, and each dilution was plated on E. coli/Coliform Petrifilm (1 ml per plate). After 48 hours of incubation at 35°C, you have now removed the plates from the incubator and have counted them. Generic E. coli on these Petrifilm plate appear as blue colonies that are associated with gas. The sample plated directly from the bag was TNTC (Too Numerous To Count). From the first 1:10 dilution, there were 99 blue colonies with gas on the plate. From the last 1:10 dilution, there were 3 blue colonies with gas. Which is the appropriate dilution from which to calculate the number of E. coli from the sample? Calculate how many colony forming units of generic E. coli were cultured from the sample and give the appropriate measurement unit (per ml, cm2, or g).

Explanation / Answer

The appropriate dilution to calculate the number of E.coli from the sample is from last 1:10 dilution, because from first dilution you may get mixture of other coliforms on the plate whille upon diluting further you will get exactly E.coli colonies on the plate. The first dilution may contain fungal contamination also, but above stated that no fungal contamination, but you will get more colonies due to other bacteria present along with recquired bacteria, so choose last dilution for calculation.

The result is as follows:

dilution factor is 102

CFU = no.of colonies X dilution factor / volume of culture plate

= 3 X 102 / 1 = 3X102 per gram.

CFU will be in ml for liquid samples whereas for solid samples it will be in grams.

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