1  Counting CFUs

The soil samples obtained from farms A and B were analysed by you and your colleagues in the BM110 Microbiology Laboratory. Each of your groups was assigned a single soil sample, from which you generated serial dilutions according to the experiment as illustrated in Figure 1.1

Figure 1.1: Schematic of the BM110 microbiology laboratory experiment. Each soil sample was added to 10mL of sterile distilled water, vortex-mixed, and incubated. Serial dilutions (100µL in 900µL of sterile distilled water) were performed to obtain \(10^{-1}\) and \(10^{-2}\) dilutions. 100µL of each dilution was plated on nutrient agar, and nutrient agar plus Ampicillin, and incubated. The number of distinct colonies on each plate were counted.

The number counts of individual colonies on each plate were recorded in table 1 of the lab handbook (Figure 1.2).

Figure 1.2: Example filled-in table 1 from the BM110 microbiology laboratory handbook (TNTC, too numerous to count; TFTC, too few to count). This gives us enough data to calculate CFU/mL for both the nutrient agar (NA) and nutrient agar plus Ampicillin (NA/Amp) plates.

1.1 Calculating CFU/mL

The counts in Figure 1.2 allow us to calculate the number of colony-forming units (CFU) there were for each millilitre of the initial soil suspension. We will work through an example calculation below.

TipCalculating CFU for the nutrient agar plate

For the nutrient agar (NA) plate, we saw in Figure 1.2 that there were too many colonies to count at the \(10^{-1}\) dilution, but that 192 colonies were counted at the \(10^{-2}\) dilution. This means that this plate had 192 colony-forming units (CFU).

We only plated 100µL of the \(10^{-2}\) dilution, so we know that there were 192 CFU in 100µL of the \(10^{-2}\) dilution. But we want to know how many CFU there were in 1mL.

There are 1000µL in 1mL, so 100µL is one tenth of 1mL. Therefore, we would expect there to be \(1000/100 \times 192 = 10 \times 192 = 1920\) CFU in 1mL of \(10^{-2}\) dilution.

The \(10^{-2}\) dilution has been diluted 100-fold (i.e. 100 times) compared to the original supernatant in the sample. So, we would expect 1mL of the original sample to have \(100 \times 1920 = 192000 = 192,000 = 1.92 \times 10^5 = 1.92e5\) CFU.

We therefore calculate the original supernatant to have \(1.92 \times 10^5\) CFU/mL.

Callout-challengeCHALLENGE: Calculate CFU for the nutrient agar/Ampicillin plate

From Figure 1.2 we can see that the \(10^{-1}\) dilution gives a count of 37 CFU on the NA/Amp plate.

What is the CFU/mL of the sample supernatant when grown on the NA/Amp plate?

TipHint

Use the example above for the NA plate as a guide.

WarningIf you’re stuck

We plated 100µL of the \(10^{-1}\) dilution, and this resulted in 37 CFU. So how many CFU would there have been if 1mL had been plated out? Call this number X CFU/mL.

The \(10^{-1}\) dilution contains X CFU/mL, but this is a 10-fold dilution of the original sample supernatant. If the sample you plated was at the original concentration it would have had 10 times as many CFU/mL. This value is the answer you’re looking for.

ImportantOnly if you really can’t answer the question…

Plating 100µL of the \(10^{-1}\) dilution resulted in 37 CFU. That volume is 0.1mL so, if 1mL (ten times as much) had been plated out we would have expected to see ten times as many CFU: \(10 \times 37 = 370\) CFU in one mL.

The \(10^{-1}\) dilution has 370 CFU/mL, but is a 10-fold dilution of the original sample supernatant, so we’d expect to see ten times the count of CFU in each mL of the original sample, i.e. \(10 \times 370 = 3700 = 3,700 = 3.7 \times 10^3 = 3.7e3\) CFU/mL.

1.2 Practice Question

If you would like more practice calculating CFU/mL from this kind of data, you can use the question below.

The Reveal answers button to show the answer if you’re stuck, and the Try another question button will deliver a new question with randomly-chosen values.

Note

You can use scientific notation (e.g. 6.25e5), and you need to provide answers correct to three significant figures.