As far as I know, dilution factor = volume of sample/total volume and CFU = (no colony x dilution factor)/volume plated in mL. For example, if I added 10gr of sample into 90mL sterilized Ringer's solution. Is it 10-1? After that, i made serial dilutions, and transfer 1mL from 10-1 to second tube (9mL as well). Then this will be10-2? Then, for example, I take 1cc from the first tube (not from the diluted primary sample) and overlay it on a petri with a medium and the final colonies are 25. The right logcfu/mL is (25*1)/0.1?
If your CFU is 25, your aliquot is 1cc, and your dilution factor is 10-2, the calculation of the log CFU/mL is:
Log CFU/mL = Log10(CFU / (dilution factor*aliquot)) = Log10(25/(10-2*1)) = 3.40.
Look at the problem in grams of sample. Adding 10 grams of sample into 90 ml of diluent is as you note a 1:10 dilution. Thus, each ml of the 1:10 contains 0.1 gram of sample.
An additional 1:10 dilution results in each ml containing 0.01 gm of sample or as you note 10-2 gm.
Therefore, if the 0.01/10-2 gm yields 25 colonies, then the original sample must have 25/0.01 or 25 x 100 colony forming units per gram.
I think so, the answer of Antima is correct, because the dilution is 1:10
If your CFU is 25, your aliquot is 1cc, and your dilution factor is 10-2, the calculation of the log CFU/mL is:
Log CFU/mL = Log10(CFU / (dilution factor*aliquot)) = Log10(25/(10-2*1)) = 3.40.
Dear Shima
i saw your question today.I think Your last sentences "The right cfu/mL is (25*1)/0.1) is right that is equal with 25* 10 = 250 cfu/ml so log 250 = 2.4 .
If you measuring log CFU/mL = 3.4
If you measuring log CFU/g = 2.4
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