Understanding of Bacterial Endotoxin Test Calculation

Hello Friends,

This particular blog is about calculation in BET testing. In microbiology, BET is a type of test which requires lot of calculation. It become difficult to understand the calculation as every person has their own style of calculation. I have also done many efforts during the beginning of my carrier to understand the calculation in BET. I know what kind of difficulty a fresher/microbiologist face during the beginning of their carrier.

But such time has gone as I am here to help you with this blog. In this blog we will understand how to prepare the particular potency of endotoxin with labelled claim lysate sensitivity.

Let's start to understand the calculation of BET.

In BET you have requirement of three types of reagents.

1. Lysate

2. Controlled Standard Endotoxin

3. Lal Reagent Water

When you receive the material, you will get a CoA of Lysate and seperate CoA for LRW. In the CoA of lysate you will get the lot no. of lysate, its labelled claim sensitivity and direction of storage, direction of reconstitution. In the same CoA you will get the detail of endotoxin like its lot no., reconstitution volume of CSE, its potency etc. First thing, you should know the labelled claim lysate sensitivity and potency of controlled standard endotoxin. Both are mentioned on the same CoA.

For example if we have a vial of CSE and in CoA it is mentioned that one vial contains 100 EU (endotoxin unit). Reconstitution volume also mentioned in CoA and consider this 5.0 mL here. So, after reconstituting the vial of endotoxin you will get 20 EU/mL potency. 

Your labelled claim lysate sensitivity is 0.125 EU/mL. Labelled claim lysate sensitivity is also denoted by lembda (λ) and if it is 50 test vial then you have to reconstitute the vial with 5.2 mL LRW and if it is 10 test vial then you have to reconstitute the vial with 1.2 mL LRW. Please refer vendor CoA for that.

Preparation of 4 λ endotoxin concentration:

Here 4λ means 4 multiplied by λ (which is lysate sensitivity)

4 X 0.125 =0.5 EU/mL. So you required 0.5 EU/mL concentration of endotoxin which is equivalent to 4λ.

Your initial vial concentration is 20 EU/mL and you have to prepare 0.5 EU/mL. Here is a simple technique. What you have divided by what you required?

You have 20 EU/mL and you required 0.5 EU/mL. So, divide 20 EU/mL with 0.5 EU/mL and you will get the dilution factor.

20 EU/mL = 40
0.5 EU/mL

Here 40 is the dilution factor which means if you dilute 20 EU/mL endotoxin 40 times you will get the concentration of 0.5 EU/mL.

Now, how to dilute the 20 EU/mL concentration to 40 times?

You have to take one part of controlled standard endotoxoin and rest part lal reagent water.

In case of 40 take 1.0 mL Controlled standard endotoxin and 39 mL LRW. Further to reduce the consumption as controlled standard endotoxin is only 5.0 mL available we can do like this, take 0.1 mL controlled standard endotoxin and 3.9 mL LRW and you will get the concentration of 0.5 EU/mL which is also called 4λ.

Preparation of 20 λ endotoxin concentration:

Here 20λ means 20 multiplied by λ (which is lysate sensitivity)

20X 0.125 =2.5 EU/mL. So you required 2.5 EU/mL concentration of endotoxin which is equivalent to 20λ.

Your initial vial concentration is 20 EU/mL and you have to prepare 2.5 EU/mL. Here is a simple technique. What you have divided by what you required?

You have 20 EU/mL and you required 2.5 EU/mL. So, divide 20 EU/mL with 2.5 EU/mL and you will get the dilution factor.

20 EU/mL = 8
2.5 EU/mL

Here 8 is the dilution factor which means if you dilute 20 EU/mL endotoxin 8 times you will get the concentration of 2.5 EU/mL.

Now, how to dilute the 20 EU/mL concentration to 8 times.

You have to take one part of controlled standard endotoxoin and rest part lal reagent water.

In case of 8 take 1.0 mL Controlled standard endotoxin and 7.0 mL LRW. Further to reduce the consumption as controlled standard endotoxin is only 5.0 mL available we can do like this, take 0.1 mL controlled standard endotoxin and 0.7 mL LRW and you will get the concentration of 2.5 EU/mL which is also called 20λ.

Hope this calculation is clear to everyone!

Thanks!!!

Frequently asked questions for microbiology interview

Q 1 : What is the microbial limit of raw water, purified water and water for injection?
Ans: Raw water limit is 500 cfu/mL, Purified water limit is 100 cfu/mL and water for injection limit is 10 cfu/100 mL.

Q 2: What is the TOC limit of purified water and water for injection?
Ans: Not more than 500 ppb.

Q 3: What is the inoculum size used for growth promotion test?
Ans: Not more than 100 cfu

Q 4: What is the inoculum size used for growth inhibition test?
Ans: More than 100 cfu

Q 5: What is biological indicator?
Ans: A biological indicator is well characterized preparation of specific microorganism with a known resistance to a specific sterilization process.

Q 6: How many biological indicators are used for total viable spore count?
Ans: At least four

Q 7: Which microorganism is used for the validation of steam heat sterilizer?
Ans: Geobacillus stearothermophillus

Q 8: What is the acceptance criteria for disinfectant efficacy validation?
Ans: Not less than 3 log reduction in vegetative cells and not less than two log reduction in spore cells.

Q 9: What is endotoxin?
Ans: Endotoxin is a lipopolysaccharide or LPS. LPS consists of the lipid A portion containing fatty acids and disaccharide phosphates, core polysaccharides and the O-antigen. Endotoxin is a type of pyrogen and is a component of the exterior cell wall of Gram-negative bacteria, like E. coli.

Q10: Why we expose the settle plate for four hours?
Ans: Settle plate method is one on the technique of environment monitoring to check the contamination level of the air. Microorganisms are not freely movable as they are attached with particles. We monitor the particles of 0.5-5.0 micron. The settlement of particles depends upon the size of the particle. The heavier the particle the more easily it settles down and the lighter the particle the more difficult for it to settle down. For these range of particles, it generally takes 4 hours for their settlement. The another factor is the dryness of the plate. If we expose the plates for more time, chances of its dryness become higher. However, validation study is required to prove that plates are not dried during or after completion of incubation time. By considering all these factors plates are exposed for 4 hours.

Q 11: Why 0.5-5.0 micron particles measured by non viable particle counter?
Ans: The product contamination is mainly caused by this range of particles because this size of particles are not easily settle down and become threat to the product.






Thanks!!!

Frequently asked questions for microbiology interview

Hello everyone,

My today's post is quite different, interesting and very helpful for all industrial microbiologists. Here, I am posting few frequently asked questions as knowledge booster for interview preparation of microbiology/quality assurance. The level of questions will be such that it will helpful for fresher as well as experienced microbiologist also.

Q 1: What is the acceptance criteria for growth promotion test?

Ans: ±factor of two (for example if we have inoculum size of 50 cfu then it could be 25-100 cfu)

Q 2: How many pathogens are tested as per USP?

Ans: Total seven pathogens (Escherichia coli, Salmonella, Staphylococcus aureus, Pseudomonas aeruginosa, Clostridia, Candida albicans and test for bile tolerant gram negative bacteria).

Q 3: Why Salmonella (pathogen) testing requires 10.0 g sample? 

Ans: Due to non uniform distribution of salmonella in sample more sample is required for identification.

Q 4: Why MacConkey broth is incubated at 42-44°C and MacConkey agar further incubated at 30-35°C?

Ans: Click on the link

Q 5: How many passage of cultures are acceptable?

Ans: Not more than five passage from the original culture.

Q 6: What is the difference between disinfectants, antiseptic and sanitizers?

Ans: Clink on the link

Q 7: Why 70% IPA is more effective then 90%,95% or 100%?

Ans: 70% IPA is more effective than other percentage of IPA because IPA destroy the bacterial cell by denaturing the protein content present in their cell wall. There is balance in penetration of IPA into interior of the bacterial cell and denaturation of proteins. If percentage of IPA will be high then protein quickly denatures and it form a protective covering outside the cell wall and IPA not able to penetrate into interior of the cell wall to completely kill the bacterial cell. However, if concentration remains low then it is not able to denature the protein content present in the cell wall of bacterial. That's why 70% IPA is more effective than 90%, 95% or 100% concentration.

Q 8: Which chemical is used for fogging and what is its concentration?

Ans: Hydrogen peroxide (H2O2) with silver ions. The combination of these two ingredients gives a synergistic broad spectrum of activity on all kinds of viruses, bacteria, fungi, yeast, mold, protozoa and algae. Generally 20% v/v concentration is used for fogging.

Q 9: What is the role of gram's iodine in gram's staining?

Ans: Gram's iodine is used as a mordant to form crystal violet-iodine complex so that the dye can not be removed easily.

Thanks!!!