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CFU in final product based in CoA of Raw materials


Best Answer kingstudruler1, 02 March 2021 - 11:23 PM

I made an error in the above calculation, here is the corrected calculation:

 

 

If I understand your information correctly, I would approach the calculation as such:

 

Since the Raw material is blended into the final product you are diluting your raw material as follows:

60g/1820g

Then the bacterial count in the finished product would be (assuming the rest of the ingredients are sterile):

(60g)/(1820g) X (100,000 cfu)/(g) = 3297 cfu

 

Then this product would have a bacterial population of

3297cfu/1820g =1.8 cfu/g

If the slice is 80g, then the final count on this slice:

 

1.8 cfu/g X 80g = 144 cfu per slice

 

you were right the first time.   :giggle:

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#1 Lígia

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Posted 02 March 2021 - 06:12 PM

Hello everyone, I was asked to analyse the possibility of the use of a raw material in a product, based on the micro standards of the raw material CoA.This is the scenario:

 

Raw material (X) as defined for TVC 100 000cfu/g

 

Raw material (X)  used final product: 60g 

 

Final product limit for TVC 1000cfug

 

Weight of final product: 1820g divided in slices of 80g (unit of sale)

 

After add the Raw material (X), the product is not submitted to any thermal process.

 

 

1. determine the total amount of cfu "added" to the finished product: 

 

    Raw material (X) as defined for TVC 100 000cfu/g   x   Raw material (X)  used final product: 60g 

 

 

2.deternine de % of Raw material in the Weight of final product:

 

   % = Raw material (X)  used final product: 60g   /  Weight of final product: 1820g

 

 

3.determine cfu in % 

 

   %  x  Raw material (X) as defined for TVC 100 000cfu/g   x   Raw material (X)  used final product: 60g 

 

 

4. determine cfu per slice

 

(  result from point 3  x  80g (unit of sale)  )  / Weight of final product: 1820g

 

 

Is this correct? Can anyone give some feedback? Thank you



#2 juanolea1

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Posted 02 March 2021 - 09:50 PM

If I understand your information correctly, I would approach the calculation as such:

 

Since the Raw material is blended into the final product you are diluting your raw material as follows:

60g/1820g

Then the bacterial count in the finished product would be (assuming the rest of the ingredients are sterile):

(60g)/(1820g) X (100,000 cfu)/(g) = 3297 cfu/g

If the slice is 80g, then the final count on this slice:

 

3297cfu/g X 80g = 263,760 cfu per slice



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#3 juanolea1

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Posted 02 March 2021 - 10:16 PM

I made an error in the above calculation, here is the corrected calculation:

 

 

If I understand your information correctly, I would approach the calculation as such:

 

Since the Raw material is blended into the final product you are diluting your raw material as follows:

60g/1820g

Then the bacterial count in the finished product would be (assuming the rest of the ingredients are sterile):

(60g)/(1820g) X (100,000 cfu)/(g) = 3297 cfu

 

Then this product would have a bacterial population of

3297cfu/1820g =1.8 cfu/g

If the slice is 80g, then the final count on this slice:

 

1.8 cfu/g X 80g = 144 cfu per slice



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#4 kingstudruler1

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Posted 02 March 2021 - 11:07 PM

im not sure i follow - but you might be right with the math.  

 

60 grams of raw material x 100,000 cfu  / 1820 final weight = 3296 cfu added per gram of finished product.   

then you would need to add what the other ingredients add to the cfu.  

 

the COA is exactly 100,000 or is the spec sheet state 100,000 max?   

 

if you are working with accurate numbers it doesn't look like you would be within specification for the final product.   

 

I think you really need to test though.   



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#5 kingstudruler1

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Posted 02 March 2021 - 11:23 PM   Best Answer

I made an error in the above calculation, here is the corrected calculation:

 

 

If I understand your information correctly, I would approach the calculation as such:

 

Since the Raw material is blended into the final product you are diluting your raw material as follows:

60g/1820g

Then the bacterial count in the finished product would be (assuming the rest of the ingredients are sterile):

(60g)/(1820g) X (100,000 cfu)/(g) = 3297 cfu

 

Then this product would have a bacterial population of

3297cfu/1820g =1.8 cfu/g

If the slice is 80g, then the final count on this slice:

 

1.8 cfu/g X 80g = 144 cfu per slice

 

you were right the first time.   :giggle:



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#6 Charles.C

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Posted 03 March 2021 - 04:11 AM

This is actually a statistical question, eg  What is the probability that the average cfu/g of a  random sample of size N taken from the  (mixed) final product X will exceed a limiting value L cfu/g ?

 

However making improbably perfect assumptions everywhere, eg homogeneity, sterility, etc   –

Concentration of "bacterial population" in final product = 6 x 10^6 / 1820 = ca 3300 cfu/g

Which is > 1000 cfu/g so "Not possible".


Kind Regards,

 

Charles.C


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#7 Lígia

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Posted 03 March 2021 - 11:45 AM

Dear all, 

many thanks for your replies!

 

im not sure i follow - but you might be right with the math.  

 

60 grams of raw material x 100,000 cfu  / 1820 final weight = 3296 cfu added per gram of finished product.   

then you would need to add what the other ingredients add to the cfu.  

 

the COA is exactly 100,000 or is the spec sheet state 100,000 max?   

 

if you are working with accurate numbers it doesn't look like you would be within specification for the final product.   

 

I think you really need to test though.   

 

Good morning

ansering your question: the COA is exactly 100,000 or is the spec sheet state 100,000 max?   

It states in the spec sheet 100 000max.



#8 Lígia

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Posted 03 March 2021 - 11:56 AM

This is actually a statistical question, eg  What is the probability that the average cfu/g of a  random sample of size N taken from the  (mixed) final product X will exceed a limiting value L cfu/g ?

 

However making improbably perfect assumptions everywhere, eg homogeneity, sterility, etc   –

Concentration of "bacterial population" in final product = 6 x 10^6 / 1820 = ca 3300 cfu/g

Which is > 1000 cfu/g so "Not possible".

Good morning,

in fact this is a statistical question...

 

My understanding and results were the same:

considering the perfect scenario and assumptions, the "bacterial population" in the final product is higher that the maximum value defined for TVC in the final product, which is <1000cfu/g so " This raw material can not be used in this product considering that TVC> 1000cfu/g.

 

Many thanks for your time and help.

Regards



#9 Bo16

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Posted 08 March 2021 - 01:58 PM

Miro is not equally distributed in many ingredients.  This means that if I have 50 drums of an ingredient from the same batch I may have a 100 cfu/g count from sampling of drum #1 and 300 cfu/g from drum #45.  If you are up at higher counts the differences can be greater.  That is why we do a representative sampling of all incoming ingredients.  Be careful when you try to "math" living organisms.  Also, if the material you are making is able to support growth of microorganisms these numbers can change quickly. 






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