Well, all the smaller dilutions were uncountable so we had to work only with one. In our lab we use ISO standars, and the one considering emuneration and calculation of the colonies states that if we have one dilution, we must use the formula: N = number of enumereted colonies / added volume x the smallest countable dilution.

But,I have an issue with all spread plate techniques. This is how we do it in my lab: for y&m analysis, we always add 0,1 ml of two dilutions on agar plates (in duplicate), And, for example, if I plated 0,1ml of 10-3 dilution, I write on that plate 10-4. And then we find colonies, and use the formula in which we enter that dilution was 10-4. For me,that does not make sense. The dilution was 10-3, and it does not matter if I plated 0,1 ml, or 1 ml, or whatever, it stays the same. The result vaires for one degree, when we compare it. I asked my bosses about it, and the answer..well, I did not get any.

So I wanted to check how the people in other countries do their calculations. And how far off base are we. If we are? Maybe I am wrong?

Hi Janie,

As I understand yr question, I agree with you. I daresay the procedure you are using has been notationally "adjusted" for the convenience of calculation.

I give one (my method) simple example below (Note - >> means "implies that"), (dr = dilution ratio)

Taking a mean value (= Ycfu) of yr "duplicates" >89.5 cfu

I assume you started with 25gram sample A /225ml water for initial solutionB (dilution 1), eg dr = 10^{(-1)}

>> 0.1gram A __contained __in 1ml B, ie there is 0.1g A/ml B

Suppose take 1ml B/9ml H_{2}0 > 10ml C (dilution 2), dr=10^{-2}

>> 0.01g A/ml C

Suppose take 1ml C/9ml H_{2}0 > 10ml D (dilution 3), dr = 10^{-3}

>> 0.001g A/ml D

Suppose take **0.1ml D** for plating which generates __average __Y cfu

then 0.1 ml D contains 10^{-4} gram A

>> 10^{-4}g A generated Y cfu

>> 1gram A will generate Y x 10^{4}cfu/g .........**[or equivalently = Y /(0.1 x dr) **= Plate count for A in cfu/g

How this relates to yr data depends on meaning of your dilution ratio 10^{-3},10^{-4} etc

So if plating 0.1 ml and yr dr = 10^{-6} matches mine, my answer would be 89.5 /[0.1 x 10^{-6}] = 89.5 x 10^{7} = 895M = same as Post 5

This is why IMO it's easier to work with weights when understanding the process. Afterwards can just use a formula.

**Edited by Charles.C, 08 January 2018 - 09:37 PM.**

correction