6. Two days ago you sponge sampled a chilled pork carcass as part of your compan
ID: 83127 • Letter: 6
Question
6. 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 on E. coli/Coliform Petrifilm (1 ml of sample 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 54 blue colonies with gas on the plate. From the last 1:10 dilution, there were 9 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, cm^2, or g).
Explanation / Answer
The microbilogical techniques of performing serial dilution and counting the number of colonies on the plate is performed to calculate the actual microbial load in a sample which is otherwise TNTC. Here, it can be seen that two serial dilutions of 1:10 are made from the sample. The first dilution gave 54 colonies whereas the second dilution gave only 9 colonies. It has been stated experimentally that any growht plate containing number of colonies between 30 to 300 represents a significant count and can be used for further calculations and experimentations. Thus, it can be clearly seen that the first dilutions gave significant number of colonies and hence it can be used for further calculations.
Secondarily, the calculation of total microbial load in the sample can be performed as below:
Microbial load = (Number of colonies * volume plated) / Dilution factor
Microbial load = (54 * 1)/ 0.1 = 5.4 colonies/ml
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.