What is the calculation formula to count Spore strip population for Biological indicator strips?
As per supplier’s COA we perform heat shocked treatment for BI and calculate average colonies per plate. But how to find out Spore population per Strips?
I have read provided SOP as well as USP chPter <55>for Biological indicator. But in this, testing Method has been defined but not mentioned Calculation formula.
In USP 42 chapter <55>, Pg no. : 6386 has mentioned one sentence, as like this :" calculate the average no. Of spores per test sample from the results, using the appropriate dilution factor. "
According to this, our sop have already defined one formula:
Spore population = avg. Colonies (cfu) x Reciprocal of Dilution /No. Of Strips
But in that, if
For 10^-4Dilution obtained avg. colonies 373 cfu/ml, then calculated spore population will be 1243333.33 spores/ strip.
For 10^-5 dilution obtained avg. Colonies 106 cfu/ ml, then calculated Spore population will be 3533333.33 spores /strip.
For 10^-6 dilution obtained avg. Colonies 48 cfu /ml, then calculated spore population will be 16000000 spores /strip.
Then how is possible , Higher dilution gives calculated maximum spores count per one strip in comparison to 10^-4 Dilution.
Because if 10^-4 dilution gives more no. Of colonies, then spores count must be more in tjis dilution.
But we obtained false reault.
If anyone have idea regrading to this, Kindly correct our calculation formula to count spore population.
madam formula is correct but calculation is wrong.
for 10^-4 dilution= 373x10^4=3.73x10^6=6.57 spore
for 10^-5 dilution=106x10^5=1.06x10^7=7.02 spore
Kindly explain me Formula.
For BI strip verification , we have taken 3 strips for single test. In which, this 3 strips together get cold and heat shock treatment. So, in formula we consider 03 no. Of strips.
Because i get different answer in which 10^-4 dilution obtained 1.24 x 10^6 spores, while 10^-5dilution obtained 3.53x 10^6 spores. Which are more than the dilution of 10^-4 .
For 10-4 dilution =obtained colonies in the plate are = 373 cfu /plate,
Reciprocal dilution = 10^4
No. Of strips = 03.
Here, calculated answer is 1.24x 10^6 spores.
For 10^-5 dilution! Obtained colonies in the plate are= 106 cfu /plate,
Reciprocal dilution =10^5.
No. Of strips= 03,
Here, calculated answer is 3.73x10^6 spores.
Kindly solve my query.
Calculation is right but there may be problem in procedure , i have got 85 cfu/ml in 10^5 dilution and 5 cfu/ml in 10^6 dilution.
what is the number of spores per strip which is used for confirmation of spore count?
if your process is correct then check all results are under acceptable criteria? if it is out of limit then there may be problem in strip.
Sorry for late response.
2.32 x 10^6 spores per strip
Procedure is same as per define in indian pharmaguideline.
I also got 373 cfu in 10^4 dilution, 106 cfu in 10^5 dilution and 48 cfu in 10^6 dilution.
And if calculations and formula is also correct then why number of spores are high in 10^-6 dilution insteda of 10^-4 dilution? Which I can’t understand…where is my method / formula / calculation is wrong ?
Yes sir, procedure is correct and i got result in acceptable limit of 50-300 % colonies.
And i also check this BI strip in autoclave and then after incubate it at 55°c temperature for 7 days.there is no any turbidity / growth found in medium, too. Then either is possible to get problem in strip?
There are 16x10^6 spores for 10 ^6 dilution while upper acceptable limit is 6.96x10^6 , Number of cfu/ml should be less than 6.96 at 10^6 dilution , while it is 48 in this test. this is too much. so this strip is fail in test.
Actually we use strip in autoclave top middle bottom then sterilised put in broth for strips separately then we use another one broth taken then directly transfer ur strips to broth then incubate in 55-60 c for seven days after seven we get result negative is positive to shows other three shows positive I.e no turbidity three broths(Top,middle,bottom)
Hi,
Is there any guidance quoting how many strips should be used when performing each validation test?
Thank you!
Hi ,
May I know how you guys got that spore count as 6.57
3.73X 10Power 6 = 6.57
Is there any calculation to get that 6.57 spore value ?
log (3.37 x 10 ^6) = log 3.37+ log 10^6 = 0.57+ 6 = 6.57
Thanks a lot !!!
I have another doubt .
While calculating the population of the ampules .
According to the USP we need to break the ampules.
In such cases, how we can break the ampule without any contaminants getting in to the test-tube.
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