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Counting and recording TPC Count


Best Answer alpamay, 02 June 2015 - 07:43 AM

total plate count : 25-250

A.  10-2      344, 450

10-3    256, 188

188x1000=188000=19x104 passed

 

B.  10-2     152, 136

10-3   18, 21 

152+136+18+21=327/2.2=149x100=14,90015x103 passed

 

C.  10-2     TNTC, TNTC

 10-3   18, 21

18+21=39/2=19.5x1000=19,500=20x103 passed

 

D.  10-2     TNTC, TNTC

 10-3   30, 55 

30+55=85/2=42.5x1000=42,500=43x103 passed

 

E.  10-2      344, 450 

10-3    256, 259

344+450=794/2=397x100=39,700=40x103 passed

 

 F. 10-2      0, 0

 10-3    0, 1 

answer=<10 passed

 

REFERENCE: US Foods & Drug Administration. January 2001. Bacteriological Analytical Online.

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#1 fuse_23

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Posted 27 May 2015 - 07:34 AM

Hi All,

 

Can help me how to calculate the following readings of TPC:

 

A.  10-2      344, 450

      10-3    256, 188

 

B.  10-2     152, 136

      10-3   18, 21

 

C.  10-2     TNTC, TNTC

      10-3   18, 21

 

D.  10-2     TNTC, TNTC

      10-3   30, 55

 

E.  10-2      344, 450

      10-3    256, 259

 

E. F 10-2      0, 0

      10-3    0, 1

 

Thanks a lot!!!



#2 fuse_23

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Posted 27 May 2015 - 07:49 AM

Additional item to this question:

 

G.  10-2     13, 12

      10-3    2, 3



#3 Charles.C

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Posted 27 May 2015 - 09:00 AM

Hi Fuse,

 

Offhand, from memory of the typical guidelines, you probably only have one directly, semi- valid/usable data set, viz."B"

 

Reason -

 

sets A,C,D,E are internally inconsistent. Should repeat.

sets F.G have insufficient bacteria to give a reliable estimation. Should repeat using lower dilutions.

(If not possible due practical reasons,  approximations for F,G can be made)

 

Was the original solution prepared such that 1ml of solution contains 0.1gram of sample ? If not, how prepared ?

Were inocula of 1ml used with pour plates ? If not, what size of inoculum(a) ?

 

I am asuming TNTC means something like  >> 500 count/plate. Please advise.


Kind Regards,

 

Charles.C


#4 Charles.C

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Posted 27 May 2015 - 09:29 AM

Assuming the answers to my above queries are "Yes", the result (to 2 "significant figures") for B is 17,000 cfu/gram using both dilution pairs.

If only the (strictly) valid pair of "B" were used, the answer would be 14,000cfu/gram.

 

Any other micro. experts out there are only too welcome to comment / correct if necessary. :smile:


Kind Regards,

 

Charles.C


#5 fuse_23

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Posted 28 May 2015 - 01:22 AM

Hi Charles,

 

 

Hi Fuse,

 

Offhand, from memory of the typical guidelines, you probably only have one directly, semi- valid/usable data set, viz."B"

 

Reason -

 

sets A,C,D,E are internally inconsistent. Should repeat.

sets F.G have insufficient bacteria to give a reliable estimation. Should repeat using lower dilutions.

(If not possible due practical reasons,  approximations for F,G can be made)

 

Was the original solution prepared such that 1ml of solution contains 0.1gram of sample ? If not, how prepared ?

Were inocula of 1ml used with pour plates ? If not, what size of inoculum(a) ?----we follow the procedure in BAM using conventional method

 

I am asuming TNTC means something like  >> 500 count/plate. Please advise.

 

I am still studying the your reasons for my questions.   If you say so that we should repeat it, I feel that most of our products or should I say we are doing something wrong with our testing.

 

Regards,

fuse



#6 Charles.C

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Posted 28 May 2015 - 06:06 AM

Hi Fuse,

 

I assumed the data is for a group of 7 arbitrary samples. Perhaps you know of a relationship between them to assist evaluation.

 

Can make 4 practical comments -

(a) APC methods are typically recognized as not very accurate. The accuracy required may relate to the samples.

(b) The object of the APC Procedure is to generate Plates with counts typically in a range of  25-250. If otherwise, the accuracy drops significantly. In the absence of any other information, additional use of a 10(-4) dilution might have been informative. 

(c) The different dilutions for each sample are required to give counts which are "plausible" with respect to each other. (for an implausible example, see "A").

(d) Regarding the reasons for the difficulties, many possibilities exist possibly related to sample, procedure, equipment. Difficult to speculate for an anonymous data set.

 

 

You can make some "guesses" for (A,C-F)  by applying the estimation guidelines in BAM. Whether these quesses will be useful may depend on what the micro. specifications for the samples are.

 

Yr microbiologist probably knows all the above already. :smile:


Kind Regards,

 

Charles.C


#7 fuse_23

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Posted 30 May 2015 - 02:53 AM

Assuming the answers to my above queries are "Yes", the result (to 2 "significant figures") for B is 17,000 cfu/gram using both dilution pairs.  ---how come become 17,000 cfu/g?  what is the computation on this?

If only the (strictly) valid pair of "B" were used, the answer would be 14,000cfu/gram.  ----- this is is based from BAM, correct me if I am wrong

 

Any other micro. experts out there are only too welcome to comment / correct if necessary. :smile:

 

Hi Fuse,

 

I assumed the data is for a group of 7 arbitrary samples. Perhaps you know of a relationship between them to assist evaluation.

 

Can make 4 practical comments -

(a) APC methods are typically recognized as not very accurate. The accuracy required may relate to the samples.

(b) The object of the APC Procedure is to generate Plates with counts typically in a range of  25-250. If otherwise, the accuracy drops significantly. In the absence of any other information, additional use of a 10(-4) dilution might have been informative. 

(c) The different dilutions for each sample are required to give counts which are "plausible" with respect to each other. (for an implausible example, see "A").

(d) Regarding the reasons for the difficulties, many possibilities exist possibly related to sample, procedure, equipment. Difficult to speculate for an anonymous data set.

 

 

You can make some "guesses" for (A,C-F)  by applying the estimation guidelines in BAM. Whether these quesses will be useful may depend on what the micro. specifications for the samples are.

 

Yr microbiologist probably knows all the above already. :smile:  ---- we were taught to do micro testing and tabulation of record, our knowledge is only based from what was taught to us.   but im still confused with the computations and reading of TPC That is why I am seeking for help, comments for micro here in this forum

 

Thanks for the reply Charles.C



#8 Charles.C

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Posted 30 May 2015 - 01:33 PM

Hi Fuse,

 

It may help to get more responses if you can expain where/why/how you are confused.


Kind Regards,

 

Charles.C


#9 fuse_23

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Posted 02 June 2015 - 05:45 AM

Hi Charles,

 

Please see my answers to the following, this is how I count and declare into COA.  I am confused because the method of counting/recording of data is different form BAm.  Any micro experts/analyst that could help me solve this.  Thanks A lot.

 

 

Hi All,

 

Can help me how to calculate the following readings of TPC:

 

A.  10-2      344, 450   ===  (344+450)/2 x 100 = 39700

      10-3    256, 188   ===  (256+188)/2 x 1000 = 222000

 

Standard:  1.0 x 10 ^5

Result:  FAILED

          2.2 x 10 ^5

 

B.  10-2     152, 136   ===  (152+136)/2 x 100 = 14,400

      10-3   18, 21  ===  (18+21)/2 x 1000 = 18,900

 

Standard:  1.0 x 105

Result:  PASSED

          1.9 x 104

 

 

C.  10-2     TNTC, TNTC  ===  TNTC

      10-3   18, 21  ===  (18+21)/2 x 1000 = 18,900

 

Standard:  1.0 x 105

Result:  PASSED

          1.9 x 104

 

D.  10-2     TNTC, TNTC  ===  TNTC

      10-3   30, 55  ===  (30+55)/2 x 1000 = 42,500

 

Standard:  1.0 x 105

Result:  PASSED

          4.3 x 104

 

 

E.  10-2      344, 450  ===  (344+450)/2 x 100 = 39,700

      10-3    256, 259  ===  (256 + 259)/2 x 1000 = 257,500

 

Standard:  1.0 x 105

Result:  FAILED

          2.6 x 105

 

 

E. F 10-2      0, 0==== <10

      10-3    0, 1  ==== (0+1)/2 x 1000   = 1000

 

Standard:  1.0 x 105

Result:  FAILED

          2.6 x 105

 

 

Thanks a lot!!!



#10 alpamay

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Posted 02 June 2015 - 07:43 AM   Best Answer

total plate count : 25-250

A.  10-2      344, 450

10-3    256, 188

188x1000=188000=19x104 passed

 

B.  10-2     152, 136

10-3   18, 21 

152+136+18+21=327/2.2=149x100=14,90015x103 passed

 

C.  10-2     TNTC, TNTC

 10-3   18, 21

18+21=39/2=19.5x1000=19,500=20x103 passed

 

D.  10-2     TNTC, TNTC

 10-3   30, 55 

30+55=85/2=42.5x1000=42,500=43x103 passed

 

E.  10-2      344, 450 

10-3    256, 259

344+450=794/2=397x100=39,700=40x103 passed

 

 F. 10-2      0, 0

 10-3    0, 1 

answer=<10 passed

 

REFERENCE: US Foods & Drug Administration. January 2001. Bacteriological Analytical Online.



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#11 alpamay

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Posted 02 June 2015 - 07:50 AM

its better if you will do 10-1 to 10-4 dilution...



#12 Charles.C

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Posted 02 June 2015 - 07:59 AM

Hi Fuse,

 

If you are going to compare the results of yr data to a  product specification, it is necessary that yr measurement / interpretation Procedure matches that for which the quoted Standard was based on.  If you use a different Procedure, you need to validate that the methodologies are equivalent.

 

Do you know which Standard was used for the Specification ?

 

PS - Discussions in the literature over the appropriate interpretation of plate counts are innumerable so don't be too surprised. :smile:


Kind Regards,

 

Charles.C


#13 Charles.C

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Posted 02 June 2015 - 10:38 PM

Hi Fuse,

 

As I mentioned before IMO several of the data sets are internally not consistent when comparing the 2 dilutions.

 

I have attached a typical set of measurements as published by FAO for you to compare with yr own data . Their calculation rules are similar to BAM but more expanded.

 

IMO the inconsistency is reflected in the ratio of the calculated colony counts / gram between the 2 dilutions.  For several samples it is considerably different to the expected level.  Cause unknown.

 

Attached File  fao counts.png   626.88KB   0 downloads


Kind Regards,

 

Charles.C


#14 fuse_23

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Posted 12 June 2015 - 03:09 AM

Thank you very much for the inputs. :smile:



#15 Charles.C

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Posted 12 June 2015 - 12:56 PM

Hi Fuse,

 

Thanks yr post.

 

Here is a little more info. regarding my previous comments concerning “consistency”. Quoting a Micro. Lab. Manual -

 

Plates in which microbial growth is proportionately greater in the highest dilutions

 

This situation may occur as a result of accidental contamination of the sample during plating, incorrect identification of the sample dilution ratio on the plates or be caused by the presence of inhibitory substances in the sample. Consider the result as a “laboratory accident” and repeat the test. If the suspicion of the presence of inhibitory substances inn the sample is high, repeat the test using an adequate procedure to eliminate or reduce the influence of these components on the result.

 

 

 

IMO, the data sets A, E may exhibit a feature such as the above although absence of a higher dilution makes any conclusion difficult, ie which dilution is correct ?. :smile:

 

In addition, but maybe dependent on yr numerical usage of "TNTC", the data sets C, D seem to exhibit the reverse effect of the above.

 

I assess this “consistency” characteristic via the ratio of the counts for the respective dilutions. I use a maximum value of 2.0 as a  limit for the ratio.

My opinions for samples (A-G) are –

 

Sample      Ratio        Data*               Result

 A                5.6              LA                    ?

 B                1.3           Satisfactory      Pass

 C              >5.0             LA?                    ?

 D              >2.4           Usable(?)         Pass(?)

 E                6.5              LA                      ?

 F                  -             Satisfactory         Pass

 G                  -             Satisfactory        Pass

 

(LA = Laboratory Accident)

* Assessed in the Context of the Specification (G was assumed to be 10^5cfu/g)


Kind Regards,

 

Charles.C





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