| Dilution of the food | 10–2 | 10–3 | 10–4 | 10–5 |
| Result | + | + | – | – |
At least one organism able to produce a population of cells in the medium (ultimately visible as growth) was present in the one ml inoculum from the 10–3 dilution. Therefore, there were at least 1000 (or 1 X 103) per gram of the original, undiluted food product. Likewise there were none ("less than 1") in the one ml inoculum from the 10–4 dilution, so we can say there would have been less than 10,000 (or 1 X 104) per gram of the original, undiluted food product. Therefore, the indicated number per gram of the food product would be equal to or greater than 1 X 103 and less than 1 X 104.
| Dilution of the food | 1/10 | 1/100 | 1/1000 | 1/10,000 | 1/100,000 |
| Result | + | + | – | + | – |
Switching the results for the 1/1000 and 1/10,000 dilutions (invoking "Phelp's Index") and using the same reasoning as in the above problem, the indicated number per gram of the food product would be equal to or greater than 1 X 103 and less than 1 X 104.
| Dilution of the food | 10–1 | 10–2 | 10–3 | 10–4 | |
| Amount of dilution inoculated into each of 3 tubes of broth |
1.0 ml | 1.0 ml | 1.0 ml | 1.0 ml | 0.1 ml |
| Number of tubes showing growth | 3 | 3 | 3 | 2 | 0 |
Using a 3-tube MPN table, determine the most probable number (MPN) of microorganisms per gram of the food.
(Note that in the table above, the tubes in the last set were each inoculated with 0.1 ml of the 10–4 dilution – which is of course equivalent to being inoculated with 1.0 ml of a 10–5 dilution. The rule that we must have decimally decreasing amounts being inoculated into the tubes is maintained.)
Choose 3-2-0 for your results. Using the 3-tube MPN table, 3-2-0 corresponds to 0.93. So, according to the table, there would be 0.93 microorganism (able to grow in the medium) per inoculum of the middle set of tubes – which received 1.0 ml of a 10–4 dilution. Therefore, in the original, undiluted food, we would expect 104 times as many organisms per gram, so the MPN per gram of the food would be 0.93 X 104 or 9.3 X 103.
| Dilution of the food | 10–1 | 10–2 | 10–3 | 10–4 |
| Amount of dilution inoculated into each of 3 tubes of broth |
0.1 ml | 0.1 ml | 0.1 ml | 0.1 ml |
| Number of tubes showing growth | 3 | 3 | 1 | 0 |
Using a 3-tube MPN table, determine the MPN of microorganisms per gram of the food.
(Note that we added the middle row of the above table for this key. Don't forget the effect of inoculum size on the results, and it was not figured into the top row which just indicates the dilution we made of the food itself, according to the text of the problem.)
Choose 3-1-0 for your results. Using the 3-tube MPN table, 3-1-0 corresponds to 0.43. So, according to the table, there would be 0.43 microorganism (able to grow in the medium) per inoculum of the middle set of tubes – which received 0.1 ml of a 10–3 dilution. This is equivalent to 1.0 ml of a 104 dilution of the food. Therefore, the MPN per gram of the food would be 0.43 X 104 or 4.3 X 103.
| Amount of hamburger added to each of three flasks of broth |
50 g | 5.0 g | 0.5 g |
| No. of flasks from which Salmonella was isolated |
3 | 3 | 0 |
Calculate the Salmonella count (i.e., the most probable number of Salmonella) per gram of the hamburger.
Checking the table, 3-3-0 indicates an average of 2.4 organisms (in this case, Salmonella) were inoculated into each of the middle set of flasks – i.e., those inoculated with 5 grams of hamburger. If 2.4 were in 5 grams of the hamburger, this is equivalent to 0.48 being in 1 gram.
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Page last modified on 7/20/00 at 6:00 PM, CDT. John Lindquist, Department of Bacteriology, University of Wisconsin – Madison |