In circumstances of duplicate plates, you need to average the plate counts befor

Question

In circumstances of duplicate plates, you need to average the plate counts before proceeding with the calculation. ONLY average plates together WITHIN THE SAME DILUTION. Choose the dilution with plates that are within the countable range. Average the two plates from that same dilution, and then proceed with the calculation.

9. Two days ago you sponge sampled a chilled pork carcass as part of your company’s generic E. coli regulatory testing program. You sampled a total of 300 cm^2 from the carcass. After sample collection, you returned to the laboratory, added 50 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 (in duplicate; 1 ml of sample was plated on each plate) on E. coli/Coliform Petrifilm. 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 54 and 72 blue colonies with gas on the plates. From the last 1:10 dilution, there were 3 and 12 blue colonies with gas on the duplicate plates. 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, cm^2, or g).

Explanation / Answer

In order to calculate the microbial load of a TNTC sample, dilution of the sample to appropriate range and subsequent plating of the same on a medium plate for appearance of countable colonies is performed. According to the information, two set of 1:10 dilutions were performed. The higer dilution gave 54 and 72 (or 63 in average) colonies whereas the lower dilutions gave 3 and 12 (or 7 in average) colonies in duplication. Conventionally, it has been stated that any plate containing number of colonies between 30 to 300 is a significant choice to calculate with bare eyes and thus, the first diultion which gave 54 and 72 or average of 63 colonies would be appropriate to count the microbial load.

Secondarily, in order to calculate the total microbial load of the sample, one can directly use the following mathematical formula:

Microbial load = (Average number of colonies appearing * dilution factor) / Volume plated

Microbial load = (63 * 0.1) / 1 = 6.3 colonies/ml

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