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How des a count on a plates get converted to CFUs per gram or ml of sample? Let's illustrate the procedure with an example. Imagine that we perform the following experiment:

Five ml of milk are added to 45 ml of sterile diluent. From this suspension, two serial, 1/100 dilutions are made, and 0.1 ml is plated onto Plate Count Agar from the last dilution. After incubation, 137 colonies are counted on the plate.

This problem may be illustrated as follows:

Note that it is often a good idea to draw out dilution problems until you are comfortable doing them. It will help you to develop a clear picture of what is being done.

The first step in solving this problem is to work out the total dilution of the sample. First 5 ml is added to 45 ml; This is a 1/10 dilution.

Remember, there are many ways to make 1/10 and 1/100 dilutions. A 0.1 ml to 0.9 ml dilution is the same as a 1 ml to 9 ml dilution and a 13 ml to 117 ml dilution. Next, 1 ml of the first dilution is added to 99 ml to make the second dilution, that is a 1/100 dilution. This is repeated with third dilution giving another 1/100 dilution. Then 0.1 ml of the third dilution is plated out on a plate of PCA. The total dilution of the sample is cumulative and can be represented mathematically as....

Notice that the amount put on the plate is also a dilution. Normally CFUs are reported per ml or per gram. In some cases less than 1 ml is put on the plate and this must be taken into account. One way to solve this, is to factor it into the total dilution. In this problem 0.1 ml was added to the plate, or 1/10th of a ml. So multiply the total dilution by 1/10 for the amount added to the plate. This leaves the total dilution as one-one millionth. The next step is to work out the dilution factor. The dilution factor is the reciprocal of the total dilution. In this case it would be......

Finally, multiply the total dilution by the average number of colonies in the plate(s) and report your answer in CFUs/ml or CFUs/gram depending upon where the sample came from; in this case ml because we used milk as a sample.

With enough practice, dilution problems can be worked out quite easily and rapidly. The method described above is just a suggested approach, if you find another way to do these problems which is more intuitive for you, use it. When doing dilution problems, remember the following:

• Note that using this method, the answer in CFUs per one milliliter or per one gram is derived. Answers may need to be adjusted if the number of CFUs per sample (other than a milliliter or gram) is requested. Assume 1 gram = 1 ml. (1 ml of water does indeed weigh 1 gram. That is actually how the ml is defined.)

• Use only those plates with colony counts between 30 and 300. With duplicate or triplicate plating from the same dilution, take the average of the plate counts and then proceed.

• Note that all individual dilutions and the amount plated are multiplied together.

• The initial dilution is often different from the subsequent dilutions. This is generally due to the nature of the sample available for analysis.

• Decimal (1:10) dilutions can be made by adding 1 ml to 9 ml. Proportional amounts can be utilized such as by adding 0.1 ml to 0.9 ml or 11 ml to 99 ml.

• Centimal (1:100) dilutions can be made by adding 1 ml to 99 ml. This can also be done by adding 0.1 ml to 9.9 ml.

• Note that plating 0.1 ml of a 10^-4 dilution results in the same dilution factor (10⁵) as plating 1 ml of a 10^-5 dilution.

Here is another sample problem. Using any method you choose, solve the problem.

One ml of a bacterial culture is pipetted into a 9 ml dilution blank. One-tenth ml of this dilution is pipetted into a 9.9 ml dilution blank. From this dilution one-tenth ml is plated using 25 ml of Plate Count Agar. 219 colonies arise after incubation. How many colony-forming units were present per ml of the original culture?

The correct answer: 2.19 X 10⁶ CFUs / ml. (Note, the 25 ml of Plate Count Agar plated is irrelevant. Why?)

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