by Oliver Yu

**me:**Hey MJ, what are the chances that there's a bug in the sample, but that your tests just happen to not catch it?

**MJ:**I tell you what.... if there's one CFU (colony forming unit) in there, my test is going to pick it up.

So suppose the final sample I hand over has exactly 1 CFU in the entire sample.

If you are using 40mL bottles to collect samples, that's a concentration of 1 CFU/40mL =

**0.025 CFU/mL**.

In a 12,000-liter bioreactor... a.k.a 12,000,000-mL bioreactor, you're looking at 300,000 colony forming units floating around in your production culture before your QC methods are sensitive enough to pick it up.

Knowing this 0.025 CFU/mL is crucial in estimating the contamination time-window.

### Contamination TimeWindow

Anytime you have a bioreactor contamination, one (good) question that gets asked is: "So when did the contamination happen?"This is because the signs of bioreactor contamination show up long after the insult as it takes time for the microbial contaminants to consume detectable amounts of oxygen and nutrients to crash the dO2 and pH signals.

All you need to compute this time-window is a spreadsheet of your contamination timeline:

And the equation for exponential growth:

X = Xwhere:_{0}e^{μ(t - t0)}

**X**is the concentration at time**t****X**is the concentration at time_{0}**t**_{0}**e**is the natural log constant**μ**is the growth rate

If we want to know the time of microbial contamination, we're interested in solving for

**t**.

_{0}X is given to us by QC Micro...in this example, QC Micro counted the last sample and found the concentration to be:

The time of contamination is known to us:X = 2.2 x 10^{5}CFU/mL

t = 4.5 days

And if we want to be uber-conservative, we assume that the initial insult was simply 1 CFU. So if our bioreactor is 12,000-liters, the initial concentration is 1 CFU/12,000,000mL or:

X_{0}= 8.3 x 10^{-8}CFU/mL

**X**, we know

**t**, we know

**e**, but we don't know

**μ**, so at this point we have 1 equation, but 2 unknowns (

**μ**and

**t**).

_{0}One way to estimate

**μ**is to assume that the "last clean sample" was just short of the detection limit: 0.025 CFU/mL (assuming 40 mL sample bottle). Solving for the growth rate:

μ = ln ( X/X_{0}) / ( t - t_{0})

μ = ln ( 2.2 x 10^{5}/ 0.025 ) / ( 4.5 - 3.5 ) = 16 day^{-1}

Since we now know the growth rate (

**μ**), we can flip the equation around and solve for the time of the initial insult (

**t**):

_{0}tUsing a simple plug 'n chug of the exponential growth equation and plate counts from QC Micro, one can estimate the time at which the microbial contamination actually took place._{0}= t - ln ( X/X_{0}) / μ

t_{0}= 4.5 - ln ( 2.2 x 10^{5}/8.3 x 10^{-8}) / 16 = 3.14 days (culture duration)

Question: What are the implicit assumptions of this method?

See also:

## 3 comments:

Question: What are the implicit assumptions of this method?

=============================

There are:

1)μ=const;

2)We have asynchronous culture of microorganisms.

Sergey Klykov,PhD

Question: What are the implicit assumptions of this method?

1st: Are you up-estimating the inicial concentration of contaminants?. sampling is one of the most critical points and it's prone to contaminate the culture, so 1 ufc/12.000.000 ml it's a good assumption at the last non-contaminated sample!

2nd: cross-inhibition might affect μ as sergey point out.

3rd: you consider zero lag phase, and constant growth rate!

4th: inoculum was clean and no contamination was introduced from the very beginning.

Question: What are the implicit assumptions of this method?

1st: Are you up-estimating the inicial concentration of contaminants?. sampling is one of the most critical points and it's prone to contaminate the culture, so 1 ufc/12.000.000 ml it's a good assumption at the last non-contaminated sample!

2nd: cross-inhibition might affect μ as sergey point out.

3rd: you consider zero lag phase, and constant growth rate!

4th: inoculum was clean and no contamination was introduced from the very beginning.

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