Lesson 10_ the Most Probable Number Index

8/12/2019 Lesson 10_ the Most Probable Number Index

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7/11/2014 Lesson 10: The Most Probable Number Index

http://water.me.vccs.edu/courses/ENV108/Lesson10_print.htm 2

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).


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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 the 10,000 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 10-2dilution into the plate which is subsequently

counted, we are theoretically doing the equivalent of plating 10-2mL (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.


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Inoculating Other Than 1mL

When 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 10-2dilution would have the same label (10-2) as a plate inoculated with

0.1mL of a 10-1dilution, 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 -2represents 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 -1dilution of a

sample of water. After incubation, you find that 80 colonies have arisen on the plate. How many


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CFUs were there per mL of the water?

Solution:As plating 0.1mL of a 10-1dilutionis the equivalent of plating 1mL of a 10-2dilution,

which is in turn equivalent to plating 10-2mL of the original, undiluted sampleof water, then

you could say that there would have been 80 CFUs per 10 -2mL of the sample, and

proportionately, there would have been 102times as many CFUs (8000 or 8.0 103) per mL

of 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 103CFUs 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:

1. The Colony Count

2. The Amount Inoculated into the plate that was counted3. 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 inverse of 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:


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So 46,700 = 467 100 = 4.67 10,000 = 4.67 104backwards and forwards.

Following the above example, these numbers are converted to scientific notation as follows:

4,000,000 = 4.0 1,000,000 = 4.0 106(6 zeros)

253,000 = 2.53 100,000 = 2.56 105(5 zeros)

1000 = 1.0 1000 = 1.0 103(3 zeros)

1 = 1.0 1 = 1.0 100(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)


0.01 = 1.0 10

-2

(moved 2 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.


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Example Problem

One 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?


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 10-4or just 10-4.


Remember, the dilution factor is simply the inverse of the plated dilutions: which would be 104.

104 220 = ?

Here you should put 220 into scientific notation so it will match 104so 220 becomes 2.2 102

104 2.2 102 = 2.2 106

To get the 106you simply add the powers of 10 to each other (4 + 2 = 6)

So the number of CFUs per mL is 2.2 106

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

diluted:

Example Problem 2

Five mL of water were pipetted into 45 mL of diluent. One mL of this dilution was pipetted into 9


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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 9parts 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.


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 10-3or just 10-3.


103 180 = ?

Here you should put 180 into scientific notation so it will match 103, so 180 becomes 1.8 102

103 1.8 102 = 1.8 105

To get the 105you simply add the powers of 10 to each other (3 + 2 = 5)

So the number of CFUs per mL is 1.8 105

Most Probable Number

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.


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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 -3dilution shows growth,

but the broth from the 10-4does not, it is then possible to say that there were greater than 1 103

organisms per mL of the original sample but less than 1 104per 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.


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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, 10-3and 10-4dilutions.

To view the 3-tube MPNtable 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 103or 9.3 102.

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
http://water.me.vccs.edu/courses/ENV108/mpntable.htm


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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 10-3dilution is equivalent to inoculating 1 mL of a 10-4dilution; 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 Problem

From 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 Water 1st Dilution

(=10-1)

2nd Dilution

(=10-3)

3rd Dilution

(=10-5)

Amount inoculated into each of three tubesof

Glucose Fermentation Broth1mL 0.1mL 1mL 0.1mL 1mL 0.1mL

Set of tubes (designations used below) A B C D E F

No. of tubes showing growth 3 3 3 3 2 0

No. 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.


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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 with 0.1mL of the 10-3dilution. (As a side note, the

so-called "plated dilution" would then be 10-4and the "dilution factor" would be 104. Recall

that the "plated dilution represents the amount of undiluted sample being inoculated, so you

can imagine 10-4mL of water sample being inoculated into each tube of set D.)

Using basic math, if0.43was in 0.1mLof the 10-3dilution, this is equivalent to 4.3being in

1mLof the same 10-3dilution.

Therefore the most probable number of glucose-fermenting organisms per mL of the original,

undiluted sample was 4.3 103.

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 104. The result is the same: 0.43 104= 4.3 103.

Review

In 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.

Assignments

Complete 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 Projectpaper is due this week. Once you have completed the project either mail,

fax, or email it to your instructor.
http://water.me.vccs.edu/courses/ENV108/projects.htmhttp://water.me.vccs.edu/courses/ENV108/assignment10.htmhttp://water.me.vccs.edu/courses/ENV108/mpntable.htm


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Lab

There is no lab associated with this lesson.

Quiz

Answer 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.
http://water.me.vccs.edu/courses/ENV108/quiz10.htm

Lesson 10_ the Most Probable Number Index

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