Dilution Theory – Page 2:

More Dilution Plating

John L's Bacteriology Pages > 
Selected General Topics  >  Dilution Theory:
Dilution Theory per se:
• Page 1 – Dilution Plating
• Page 2 – More Dilution Plating
• Page 3 – The MPN Method
Supplementary Pages:
• Five-Tube MPN Table
• Practice Set 1 (Plating)
• Practice Set 2 (Plating&MPN)

A quick review of highlights from the previous page ("Dilution Theory–Page 1"):

  • Inoculating plates from increasing dilutions (decreasing concentrations) of a sample is equivalent to plating successively smaller amounts of the sample.
  • We treat the units grams and milliliters as equivalents. This is done for convenience. One ml of water does indeed weigh one gram and vice versa. In real life however, the same may not apply to other things – especially solid samples.
  • We need to have "countable" plates – having preferably between 30 and 300 colonies.

On this page:

  • Inoculating other than 1.0 ml into/onto our plates.
  • Using handy dilution formulas which introduce the terms "plated dilution" and "dilution factor." We note the fact that the actual amount of original, undiluted sample plated is the "plated dilution" and vice-versa.
  • Checking your answer!

When we inoculate plates already containing medium (the usual case in our lab courses), we find that it would take too long for a one ml inoculum to soak into the medium. So, we generally plate 0.1 ml from each dilution made. For each plate, you can readily see that we are then inoculating one-tenth the number of CFUs there would have been in a one ml inoculum.

In the following diagram, we have built on the last illustrated example given on Page 1 by adding inoculations of 0.1 ml from each of the dilutions into respective plates. (The numbers of colonies in parentheses are either "too many" or "too few" for counting – remembering our "30-300 rule" above.)

Instead of the letters we labeled our plates with previously, we have labeled each plate with the dilution it represents – as if one ml had been inoculated from that dilution. For example, a plate inoculated with one ml of a 10–2 dilution would have the same label (10–2) as a plate inoculated with 0.1 ml of a 10–1 dilution, as they are equivalent plates. This value (10–2) has been traditionally called the "plated dilution." (A more fitting term we have come up with – and may officially substitute some day – is "virtual dilution"!)

Remembering our discussion on Page 1, you can see that the value of the "plated (virtual) dilution" is equivalent to the actual amount (in ml or g) of undiluted sample that is being plated out. For example, a plate labeled "10–2" represents 10–2 ml or gram of sample being inoculated onto the plate.

A quick example problem:  Suppose you inoculate a plate with 0.1 ml of a 10–1 dilution of a sample of milk. After incubation, you find that 80 colonies have arisen on the plate. How may CFUs were there per ml of the milk?

Solution:  As plating 0.1 ml of a 10–1 dilution is the equivalent of plating 1 ml of a 10–2 dilution which is in turn equivalent to plating 10–2 ml of the original, undiluted sample of milk, then you could say that there would have been 80 CFUs per 10–2 ml of the sample, and – proportionately – there would have been 102 times as many CFUs (i.e., 8000 or 8.0 X 103) per ml of the undiluted milk sample.

Looking at the problem this way:
IF 80 colonies arise from plating 0.01 ml of the milk,
THEN there were 8.0 X 103 CFUs per one ml of the milk.


From the foregoing explanation and examples, one can figure out the concentration of CFUs (i.e., the number of CFUs per ml or gram of the sample) in any dilution and plating problem – by knowing just three things about our setup and results:

  • The COLONY COUNT
  • The AMOUNT INOCULATED into the plate that was counted
  • The DILUTION OF THE SAMPLE from which the inoculation was made

Often it is handy to utilize formulas to work out dilution problems. In Bacteriology 102, we show continuously that the following set of formulas always work (if they are used properly). We could use one "universal" formula, but we have traditionally used these: the first (already used above) to find what portion of our sample is being analyzed (expressed as our so-called plated or virtual dilution) and the second to inflate our colony count proportionately, resulting in the number of CFUs that were in one gram or ml of the original, undiluted sample. (Don't just take our word for it. Spend a little time here and see how ultimately we can always come up with the number of CFUs per one ml or one gram of the undiluted sample.)

dilutions made  X  amount inoculated  =  "plated dilution"

dilution factor
(simply the inverse of
the plated dilution)
 X  # colonies  =  # CFUs/ml(or gram)
of the original
undiluted sample


SOME EXAMPLES:

  1. The following is a sample problem from Bacteriology 102 worked out with the formulas:  One ml of a bacterial culture was pipetted into a 9 ml dilution blank. One-tenth ml of this dilution was pipetted into a 9.9 ml dilution blank. From this dilution, one-tenth ml was plated with 25 ml of culture medium. 220 colonies arose after incubation. How many colony-forming units were present per ml of the original culture? (Does the amount of medium in the plate matter in the calculations?)

    dilutions
    made
     X  amount
    inoculated
     =  "plated dilution"
    1/10 X 1/100  X  1/10  =  1/10,000 or 10–4

    dilution factor  X  # colonies  =  # CFUs/ml
    104  X  220  =  2.2 X 106

    You can also look at the problem this way: If 220 colonies arose from plating (the equivalent of) 10–4 ml of the culture, then (proportionally) there would have been 220 X 104 or 2.2 X 106 CFUs per one ml of the culture. This is the reasoning behind the second of the two dilution formulas.


    When checking your answer, which you can always do as follows for such problems:  Start with the sample which you have determined to contain 2.2 X 106 CFUs per ml. Then, see if you wind up with the stated number of colonies on the plate, after making the specified dilutions in which the number of CFUs per ml is sequentially reduced.

    • The first, 1/10 dilution contains 2.2 X 105 CFUs per ml.
    • The second, 1/100 dilution contains 2.2 X 103 CFUs per ml.
    • As 0.1 ml of the second dilution was inoculated into the plate, we would expect the number of CFUs in the inoculum to then be 2.2 X 102 which is 220, the number of colonies we counted on the plate.

  2. Here is a problem where we start out with something other than 1 ml or 1 gram of sample being diluted:  Five ml of milk were pipetted into 45 ml of diluent. One ml of this dilution was pipetted into 9 ml of diluent. From this dilution, 0.1 ml was plated. After incubation, 180 colonies were counted. Determine the number of colony-forming units per ml of the original milk sample.

    A 1/10 dilution is achieved when 5 ml of sample are added to 45 ml of diluent. Remember that a 1/10 dilution can be made in a variety of ways – as long as there is one part of sample added to 9 parts of diluent. Even if we had a dilution we could not so reduce – e.g., something like 3 grams of hamburger added to 80 ml of diluent which would result in a 3/83 dilution – we can still simply "plug it into" the formula and we could wind up with the answer. And remember that the formulas will always give the answer as no. of CFUs per one ml (or one gram) no matter what amount we start with. Wouldn't this problem have the same answer if we had put 1 ml of milk into 9 ml of diluent?

    dilutions
    made
     X  amount
    inoculated
     =  "plated dilution"
    1/10 X 1/10  X  1/10  =  1/1000 or 10–3

    dilution factor  X  # colonies  =  # CFUs/ml
    103  X  180  =  1.8 X 105

    As for the above problem, you can look at this one as follows: If 180 colonies arose from plating (the equivalent of) 10–3 ml of the milk, then (proportionally) there would have been 180 X 103 or 1.8 X 105 CFUs per one ml of the original, undiluted milk sample.


  3. For a problem presented on the previous page (as no. III) in which no dilutions were made, we can still work it out with the formulas:  Five ml of an undiluted spring water sample were added to a petri dish to which 15 ml of melted Plate Count Agar were then added. Fifty colonies were counted after incubation. How many CFUs were present per ml of the original, undiluted spring water sample?

    dilutions
    made
     X  amount
    inoculated
     =  "plated dilution"
    1  X  5  =  5

    dilution factor  X  # colonies  =  # CFUs/ml
    1/5  X  50  =  10
    Note that when there are no dilutions, we indicate "1" – not zero! – for the dilutions made. As always, the "dilution factor" is the inverse of the so-called "plated dilution" (according to how we defined our terms), and the "plated dilution" always represents the amount of sample being plated.


Here's a treat for the general public: Click here for some more practice problems from the Bacteriology 102 Lab Manual with the solutions given on a separate page.


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• Index to the Dilution Theory pages.
• Site Outline of related pages.

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Page content was last modified on 2/12/12 at 7:00 PM, CST.
John Lindquist, Department of Bacteriology,
University of Wisconsin – Madison