      # Enumeration of yeasts and molds

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### #1 Janie

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• Serbia Posted 04 January 2018 - 04:58 PM

Hello everybody. I have a question about yeasts and molds. By our national standard for this types of microorganisms,we add an inoculation of sample on a two DRBC agar plates. Formula for enumeration of mo is N=number of mo counted on plates/volume of inoculation X dilution of inoculation that shows microogranisams (d).
Well,my question is,if I added O,1 ml of 10-6 dilution,is the number for d in formula above,10-6 or 10-7?
I think it should be 10-6,because that is the last dilution I added,but after putting 0,1ml on a plate,theoryticly the dilution became 10-7.
What is your take on it?

### #2 FurFarmandFork

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Posted 04 January 2018 - 05:06 PM

Hi Janie,

Hoping I'm following correctly, but yes. If your final units are CFU/mL, then if you only had 0.1mL of your final dilution, you have essentially another ten-fold reduction in the organisms you would count on the plate. Or to make an obvious example, if I plate 0.1 mL of a sample and count 4 CFU, then I would assume at 1 mL of my sample would contain 40 CFU. Then move on to whatever dilution that sample is already at.

Austin Bouck
Owner/Consultant at Fur, Farm, and Fork.
Consulting for companies needing effective, lean food safety systems and solutions.

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### #3 Charles.C

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Posted 04 January 2018 - 05:43 PM

Hello everybody. I have a question about yeasts and molds. By our national standard for this types of microorganisms,we add an inoculation of sample on a two DRBC agar plates. Formula for enumeration of mo is N=number of mo counted on plates/volume of inoculation X dilution of inoculation that shows microogranisams (d).
Well,my question is,if I added O,1 ml of 10-6 dilution,is the number for d in formula above,10-6 or 10-7?
I think it should be 10-6,because that is the last dilution I added,but after putting 0,1ml on a plate,theoryticly the dilution became 10-7.
What is your take on it?

Hi Janie,

IMEX, for self-belief,  it is easier to (a) Pick a specific example like as in previous post, (b) interpret the data in respect to weights of sample (not liquid) on a plate by using  simple arithmetic.

Then compare the answer to that from yr interpretation of the perpetually confusing albeit simpler looking textbook formulae.

Kind Regards,

Charles.C

### #4 Janie

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• Serbia Posted 04 January 2018 - 07:03 PM

Thank you both for fast replies :)

Ok, you are right, concrete example is the best way to go. I plated 0,1 ml from final dilution 10-6, on two drbc agar pre prepared plates (0,2 ml in total) . I counted 179 cfu in total (100 cfu on one plate and 79 cfu on other ). How do I present my result then? I want it it to be in units: cfu /ml

Best regards :)

### #5 FurFarmandFork

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Posted 04 January 2018 - 08:36 PM

In that situation, I would average the two plates so that you have an average value of 89.5 CFU/0.1mL of 10^6 dilution.

Therefore 895 CFU/mL of 10^6 dilution

Therefore 895,000,000 CFU/mL of original liquid.

Austin Bouck
Owner/Consultant at Fur, Farm, and Fork.
Consulting for companies needing effective, lean food safety systems and solutions.

Subscribe to the blog at furfarmandfork.com for food safety research, insights, and analysis.

### #6 Charles.C

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Posted 05 January 2018 - 07:51 AM

Thank you both for fast replies :)

Ok, you are right, concrete example is the best way to go. I plated 0,1 ml from final dilution 10-6, on two drbc agar pre prepared plates (0,2 ml in total) . I counted 179 cfu in total (100 cfu on one plate and 79 cfu on other ). How do I present my result then? I want it it to be in units: cfu /ml

Best regards :)

Hi Janie,

Should state that my experience/comments are mainly based on routine APC, not Y&M, but i would anticipate interpretation not hugely dissimilar.

(1) What was the original sample size/dilution ? As in typical  standard methodologies,  i would avoid such a small plating volume as 0.1 ml if possible.

(2) i recommend you choose one standard methodology. These usually define a counting method/rules (which can vary) and give an example (BAM, ISO, APHA, USDA are 4 popular options of which IIRC 1st/4th are on-line). Otherwise you may end up with "incompatible" results with those from other people. I have encountered endless arguments over such aspects.

(3) It is usually recommended to have usable paired counts at a minimum of  2 dilutions to enable a validation of the dilution procedure, ie, whether the data is in fact usable. IMEX operator mistakes do happen. (there is a detailed APC example of such a case elsewhere on this forum).

(4) Regardless of details, it looks like you have a prodigiously large amount of Y&M in yr product. What is the maximum limit, 10,000/g ?

Kind Regards,

Charles.C

### #7 Janie

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• Serbia Posted 07 January 2018 - 07:50 PM

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?

### #8 Charles.C

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Posted 08 January 2018 - 09:05 PM

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 H20 > 10ml C (dilution 2), dr=10-2

>> 0.01g A/ml C

Suppose take 1ml C/9ml H20 > 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

>> 10-4g A generated Y cfu

>> 1gram A  will generate Y x 104cfu/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 107 = 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

Kind Regards,

Charles.C

### #9 Janie

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• Serbia Posted 11 January 2018 - 08:01 AM

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 H20 > 10ml C (dilution 2), dr=10-2
>> 0.01g A/ml C

Suppose take 1ml C/9ml H20 > 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
>> 10-4g A generated Y cfu

>> 1gram A will generate Y x 104cfu/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 107 = 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. ### #10 Janie

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• Serbia Posted 11 January 2018 - 08:02 AM

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 H20 > 10ml C (dilution 2), dr=10-2
>> 0.01g A/ml C

Suppose take 1ml C/9ml H20 > 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
>> 10-4g A generated Y cfu

>> 1gram A will generate Y x 104cfu/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 107 = 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. ### #11 Janie

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• Serbia Posted 11 January 2018 - 08:07 AM

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 H20 > 10ml C (dilution 2), dr=10-2
>> 0.01g A/ml C

Suppose take 1ml C/9ml H20 > 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
>> 10-4g A generated Y cfu

>> 1gram A will generate Y x 104cfu/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 107 = 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. Hi Charles,
Thank you so much for putting it like this,step by step. It makes so much more sense and I finally understood it.
Once again,thank you for your patience and fast replies :)

### #12 Janie

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• Serbia Posted 11 January 2018 - 08:07 AM

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 H20 > 10ml C (dilution 2), dr=10-2
>> 0.01g A/ml C

Suppose take 1ml C/9ml H20 > 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
>> 10-4g A generated Y cfu

>> 1gram A will generate Y x 104cfu/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 107 = 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. Hi Charles,
Thank you so much for putting it like this,step by step. It makes so much more sense and I finally understood it.
Once again,thank you for your patience and fast replies :)

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