**Lesson 10:**

**The Most Probable Number Index**

ObjectiveIn this lesson we will learn the following:

- Understanding dilution plating and colony-forming units.
- What is the MPN method?
- How to determine the amount of bacteria from the MPN method.

Reading AssignmentRead the online lecture.

Lecture

Dilution PlatingIn microbiology we are concerned with determining the

concentration of colony-forming units (CFUs)in our sample, or in other words, the number of CFUs per mL or per gram of the sample. For example, if we were to plate out1mLof a lake water sample and then, after incubation of the plates, find that100 colonieshave arisen, we would then conclude that there were100 CFUs per mLof the lake water.More realistically, as with most lakes, the concentration of CFUs in the water could have been considerably greater. Counting the colonies on a plate inoculated with 1mL of water may be impossible.

We would like to have"countable" plates, containing between 30 and 300 colonies.If fewer than 30, we run into greater statistical inaccuracy. If greater than 300, the colonies would be tedious to count and also would tend to run together.We may try to plate out smaller and smaller amounts, as in the illustration below. With this new lake water sample, a cuontable plate (Plate C: 100 colonies) is achieved with the inoculation of 0.01 mL of sample.

Figuring out the number of CFUs per mL of the sample would go like this:

Whatever the number of CFUs in the inoculum of Plate C (one-hundreth (0.01) of a mL of the sample), there would be one hundred times as many CFUs in one mL of the sample. So, if 100 CFUs are determined to be in the 0.01 mL inoculum, then 100 × 100 CFUs would be present in 1mL; the final answer is 10,000 CFUs/mL of the sample.

There are two drawbacks to this procedure. It is difficult to dispense amounts smaller than 0.1 mL and also the smaller the amount tested, the less representative it is of the sample.

So we now get into "dilution theory" to accomplish the equivalent of plating out succeedingly smaller amounts of sample. Making serial decimal dilutions (for example 1/10 dilutions, each made by adding one part of inoculum to 9 parts of diluent) and inoculating 1mL into each of the plates, we can construct a plating procedure that is equivalent to the image above.

Each test tube below holds 9 mL of lake water (9 parts of diluent).

Illustrating how the concentration of CFUs decreases according to how these cell suspensions are diluted: we determined above that the

sample contains 10,000 CFUs per mL. Taking out 1mL and inoculating it into a 9mL dilution blank (the second tube) would put the10,000 CFUs into a total of 10mL, which is equivalent to 1000 CFUs per mL of the 1/10 dilution of the sample. The density of CFUs continues to decrease ten-fold with each subsequent dilution.Counting 100 colonies in Plate C, note how we can work back to a concentration of 10,000 CFUs per mL of the sample. So, by inoculating 1mL of a 10

^{-2}dilution into the plate which is subsequently counted, we are theoretically doing the equivalent of plating 10^{-2}mL (0.01mL) of the lake water sample.IF 100 colonies arise from plating 1mL of a 1/100 dilution of the lake water, THEN there were 10,000 CFUs (from 100 × 100) per 1mL of the undiluted water sample.

Compare the solution just obtained with the previous solution where Plate C was inoculated with 0.01mL of sample:

IF 100 colonies arise from plating 0.01mL of the water sample, THEN there were 10,000 CFUs per 1mL of the water sample.

Inoculating Other Than 1mLWhen we inoculate plates already containing medium, which is usually the case, we find that it would take too long for a 1mL inoculum to soak into the medium. So, we generally plate 0.1mL 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 1mL inoculum.

In the following illustration, we have built onto the last illustrated example given above by adding inoculations of 0.1mL from each of the dilutions into respective plates. (The numbers of colonies in parenthesis are either "too many" or "too few" for counting. Remember we need between 30 and 300 colonies to count)

Instead of the letters we labeled our plates with earlier, we have labeled each plate with the

dilutionit represents, as if 1mL had been inoculated from that dilution. For example, a plate inoculated with 1mL of a 10^{-2}dilution would have the same label (10^{-2}) as a plate inoculated with 0.1mL of a 10^{-1}dilution, as they are equivalent plates. This value (10^{-2}) has been traditionally called the"plated dilution".You can see from the diagram above that the "plated dilution" is equivalent to the actual amount (in mL) of undiluted sample that is being plated out. For example, a plate labeled 10

^{-2}represents 10^{-2}mL of sample being inoculated onto the plate.

A quick example problem:Suppose you inoculate a plate with 0.1mL of a 10^{-1}dilution of a sample of water. After incubation, you find that 80 colonies have arisen on the plate. How many CFUs were there per mL of the water?

Solution:As plating0.1mL of a 10is the equivalent of plating^{-1}dilution1mL of a 10, which is in turn equivalent to plating^{-2}dilution10of water, then you could say that there would have been 80 CFUs per 10^{-2}mL of the original, undiluted sample^{-2}mL of the sample, and proportionately, there would have been10(8000 or 8.0 × 10^{2}times as many CFUs^{3})per mLof the undiluted water sample.Looking at the problem this way:

IF 80 colonies arise from plating 0.01mL of the water, THEN there were 8.0 × 10

^{3}CFUs per 1mL of the water.From the explanation and examples given above, one can figure out the concentration of CFUs in any dilution and plating problem by knowing just

three thingsabout our setup and results:

- The
Colony Count- The
Amount Inoculatedinto the plate that was counted- The
Dilution of the Samplefrom which the inoculation was made

Often it is handy to utilize formulas to work out dilution problems:

Dilutions made × Amount inoculated = "plated dilution"

Dilution factor × #colonies = #CFUs/mL of the original undiluted sample(The dilution factor is simply the

inverseof the plated dilution)

Before we work some problems let's quickly go over

Scientific Notation.The aim of scientific notation is to express any number (for example, 46,700) in this form:

So 46,700 = 467

×100 = 4.67×10,000 = 4.67×10^{4}backwards and forwards.Following the above example, these numbers are converted to scientific notation as follows:

4,000,000 = 4.0

×1,000,000 = 4.0×10^{6}(6 zeros)253,000 = 2.53

×100,000 = 2.56×10^{5}(5 zeros)1000 = 1.0

×1000 = 1.0×10^{3}(3 zeros)1 = 1.0

×1 = 1.0×10^{0}(0 zeros)

For numbers less than 1 (for example 0.0035) a negative exponent will be used:

To change 0.0035 to a number with one digit before the decimal point, one moves the decimal point to the right three places until the decimal point is directly after the first digit (3). To think in simple terms, the power of 10 will equal the amount of decimal places you move the decimal. If you move the decimal point to the right, the power of 10 will be negative. Notice above we move the decimal place 3 spaces to the right, there the scientific notation of 0.0035 equals

3.5 × 10^{-3}.Likewise, these decimals are converted to scientific notation:

0.000043 = 4.3

×10^{-5}(moved 5 places to the right: makes it negative)0.016 = 1.6

×10^{-2}(moved 2 places to the right: makes it negative)0.01 = 1.0

×10^{-2}(moved 2 places to the right: makes it negative)0.00723 = 7.23

×10^{-3}(moved 3 places to the right: makes it negative)

Now that we have covered scientific notation briefly, let's work some problems and use that information to determine the CFUS.

Example ProblemOne mL of a baterial culture was pipetted into a 9 mL dilution blank. One-tenth (0.1) mL of this dilution was pipetted into a 9.9 mL dilution blank. From this dilution, one-tenth (0.1) mL was plated with 25 mL of culture medium. 220 colonies arose after incubation. How many colony-forming units (CFUs) were present per mL of the original culture?

Dilutions made × Amount inoculated = "plated dilution"

0.1 × 0.01 × 0.1 = 0.0001(the 2nd dilution(0.01) is because 2 dilutions were made: 0.1

×0.1 = 0.01)For the "plated dilution" you want the answer in scientific notation so you would move the decimal place until the decimal is after the first significant digit, which is 1. So 0.0001 becomes 1.0

×10^{-4}because you moved the decimal place 4 places to the right. If you move the decimal place to the left the power of 10 would be positive. So the"plated dilutions"is 1.0×10^{-4}or just10.^{-4}

Dilution factor × #colonies = #CFUs/mL of the original undiluted sampleRemember, the dilution factor is simply the inverse of the plated dilutions: which would be

10.^{4}

10^{4}× 220 = ?Here you should put 220 into scientific notation so it will match 10

^{4}so 220 becomes 2.2×10^{2}

10^{4}× 2.2 × 10^{2}= 2.2 × 10^{6}To get the 10

^{6}you simply add the powers of 10 to each other (4 + 2 = 6)

So the number of CFUs per mL is2.2 × 10^{6}

Now, let's work a problem in which we start out with something other than 1 mL of sample being diluted:

Example Problem 2Five mL of water were pipetted into 45 mL of diluent. One mL of this dilution was pipetted into 9 mL diluent. From this dilution, 0.1 mL was plated. After incubation, 180 colonies were counted. Determine the number of colony-forming units (CFUs) per mL of the original water sample.

A 1/10 dilution is achieved when 5 mL of sample are added to 45 mL of diluent. Remember than 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. Also remember that the formulas will always give the answer as number of CFUs per 1mL no matter what amount we start with.

Dilutions made × Amount inoculated = "plated dilution"

0.1 × 0.1 × 0.1 = 0.001

Move the decimal place 3 places to the right and put in scientific notation: So the plated dilution is 1.0

×10^{-3}or just10.^{-3}

Dilution factor × #colonies = #CFUs/mL of the original undiluted sample

10^{3}× 180 = ?Here you should put 180 into scientific notation so it will match 10

^{3}, so 180 becomes 1.8×10^{2}

10^{3}× 1.8 × 10^{2}= 1.8× 10^{5}To get the 10

^{5}you simply add the powers of 10 to each other (3 + 2 = 5)

So the number of CFUs per mL is1.8× 10^{5}

Most Probable NumberTo envision the theoretical basis for the Most Probable Number (MPN) Method, think of a ten-fold dilution series being made of a water sample with one mL of each dilution being inoculated into a separate tube of an all-purpose broth medium.

After incubation, the broth tubes are observed for the presence or absence of growth. Theoretically, if

at least one organismhad been present in any of the inocula, visible growth should be seen for that particular tube. If the broth inoculated from the 10^{-3}dilution shows growth, but the broth from the 10^{-4}does not, it is then possible to say that there were greater than 1 × 10^{3}organisms per mL of the original sample but less than 1 × 10^{4}per mL.Bacteria are rarely, if ever, distributed evenly in a sample. For example, if a 10 mL sample contains a total of 300 organisms, not every one mL divisor will contain 30 organisms; some will contain more or fewer, but the average of all ten divisors (parts) in the entire 10 mL sample will be 30. This also applies to any of the dilution tubes which inocula are taken.

To increase the statistical accuracy of this type of test, more than one broth tube can be inoculated from each dilution. Standard MPN procedures use a minimum of 3 dilutions and 3, 5 or 10 tubes per dilution. The statistical variability of bacterial distribution is better estimated by using as many tubes as possible or practical. After incubation, the pattern of positive and negative tubes is noted, and a standardized MPN Table is consulted to determine the most probable number of organisms (causing the positive results) per unit volume of the original sample.

In the following example, a set of 3 tubes of an all purpose broth medium is inoculated from each of the ten-fold dilutions, with each tube being inoculated with one mL.

After incubation, the number of tubes showing growth is recorded. As the succeeding dilutions were made, the organisms were diluted to such an extent that none were in the inocula of seven of the tubes (marked negative). In order to estimate the number of organisms per mL of the sample which would cause this kind of growth response, we locate the three sets of tubes which show dilution of the organisms "to extinction" - for example, those tubes which were inoculated from the 10

^{-2}, 10^{-3}and 10^{-4}dilutions.To view the 3-tube MPN table which allows us to determine the average of organisms being inoculated into each of the tubes. Look until you see the combination we have found above (3, 2, 0) and you will see that it suggest an average of 0.93 organisms from the middle set of test tubes. Therefore the most probable number of organisms per one mL of the original, undiluted sample would be 0.93 × 10

^{3}or 9.3 × 10^{2}.This method, with the associated table, only works if there is a succession of 1/10 dilutions being tested (in an appropriate medium) for growth, and the tubes chosen to compare with the table are in consecutive order of increasing dilution.

Some things to keep in mind:

- Other amounts than 1 mL can be inoculated into the broth tubes. For example, inoculating 0.1 mL of a 10
^{-3}dilution is equivalent to inoculating 1 mL of a 10^{-4}dilution; the "plated dilution" (10^{-4}) is the same in each case.- It doesn't matter what amount of medium there is in the tubes being inoculated. For example, the same growth response should be evident if we doubled the amount of broth in each tube.
- Also, we don't have to inoculate our tules from dilutions. For example, one can set up a series of tubes where 10 grams are inoculated into each tube in the first set, 1 gram is inoculated into each tube in the second set, and 0.1 gram is inoculated into each tube in the third set. The sequence of decimally-decreasing amounts is maintained.

Example MPN ProblemFrom a water sample obtained from "Lake Microbe", you inoculated 10 mL into 90mL of sterile diluent; this is the "first dilution" indicated in the table below. After thorough mixing, 1mL of this dilution was added to 99mL of sterile diluent, and a third dilution was made the same way. From each of these dilutions, tubes of Glucose Fermentation Broth were inoculated with amounts as shown in the table below. After appropriate incubation, the tubes were checked for growth and acid production, and the data are summarized below.

Dilution of Lake Water1st Dilution

(=10^{-1})2nd Dilution

(=10^{-3})3rd Dilution

(=10^{-5})Amount inoculated into each of three tubesof Glucose Fermentation Broth 1mL 0.1mL 1mL 0.1mL 1mL 0.1mLSet of tubes (designations used below) A B C D E FNo. of tubes showing growth 3 3 3 3 2 0No. of tubes showing acid production 3 3 3 1 0 0

What was the most probable number of

glucose-fermenters per mLof the original sample of water? Here is one way of finding the solution:

- Choose the three consecutive sets of tubes that show "dilution to extinction" of glucose-fermenting organisms - for example, sets C, D and E. We can indicate "3-1-0" to represent the number of positive tubes.
- Checking the MPN table (print it out) 3-1-0 indicates that an average of
0.43 organism(causing the determining reaction) was inoculated into each of the tubes in the middle set (D) - for example, the tubes inoculated with0.1mL of the 10. (As a side note, the so-called "plated dilution" would then be 10^{-3}dilution^{-4}and the "dilution factor" would be 10^{4}. Recall that the "plated dilution represents the amount of undiluted sample being inoculated, so you can imagine 10^{-4}mL of water sample being inoculated into each tube of set D.)- Using basic math, if
0.43was in0.1mLof the10, this is equivalent to^{-3}dilution4.3being in1mLof the same 10^{-3}dilution.- Therefore the most probable number of glucose-fermenting organisms per mL of the original, undiluted sample was
4.3 × 10.^{3}An alternate (easier) method would be to multiply the MPN value from the table (0.43) by the dilution factor of the "D" set of tubes which is 10

^{4}. The result is the same: 0.43×10^{4}=4.3 × 10.^{3}

ReviewIn microbiology we are concerned with determining the

concentration of colony-forming units (CFUs)in our sample, or in other words, the number of CFUs per mL or per gram of the sample.We would like to have"countable" plates, containing between 30 and 300 colonies.Sometimes it makes it easier in determing the CFU by using Scientific Notation which moves the decimal place so that it falls directly after the first significat digit. The power of 10 will be positive if the decimal place moves to the left and negative if it moves to the right.To envision the theoretical basis for the Most Probable Number (MPN) Method, think of a ten-fold dilution series being made of a water sample with one mL of each dilution being inoculated into a separate tube of an all-purpose broth medium. To increase the statistical accuracy of this type of test, more than one broth tube can be inoculated from each dilution. Standard MPN procedures use a minimum of 3 dilutions and 3, 5 or 10 tubes per dilution. The statistical variability of bacterial distribution is better estimated by using as many tubes as possible or practical.

AssignmentsComplete Assignment 10 on Most Probable Number Index. You may do the Assignment online to get credit or print it out and send it to the instructor.

Also, the fifth Project paper is due this week. Once you have completed the project either mail, fax, or email it to your instructor.

LabThere is no lab associated with this lesson.

QuizAnswer the questions in the Lesson 10 quiz . When you have gotten all the answers correct, print the page and either mail or fax it to the instructor. You may also take the quiz online and directly submit it into the database for a grade.