Name:
Section:
Introductory Microbiology PreLab Assignment
Sampling Food, Water, and Milk
Write answers in scientific notation
1. (2pt.) Write the following in scientific notation:
0.0006740 =
405,000,000 =
28,000 =
0.9060 =
2. (0.5pt.) List all of the listed characteristics of a coliform .
3. (0.5pt.) Why is E. coli a better sewage indicator than E. aerogenes ? In your explanation, mention both species by name, and be clear about which species you are describing.
4. (0.5pt.) Each colony on an agar plate originated from one cell. a. True b. False
1

5. (1pt.) Diluting the original sample decreased the concentration of the bacteria. What is the purpose of decreasing the concentration of the samples in this lab?
6. (0.5pt.) List two waterborne diseases (not simply bacterial species’ names) resulting from contamination of water with fecal material.
2

Lab Exercise
Sampling Food, Water and Milk
09/18/14
Introduction
Sampling of food, water, and milk is performed regularly as an initial step in determining the presence, quantity, and identification of contaminating bacteria. Under certain incubation conditions, food and milk easily support the growth of large numbers of cells due to the availability of nutrients. Water, on the other hand, usually does not contain as large a concentration of nutrients. However, the presence of pathogens in water may affect a large population of people and lead to epidemic conditions. Contamination may originate with human handlers, diseased animals or their fecal material, contamination of water supplies with sewage, and unsanitary equipment or procedures.
Large numbers of bacteria do not necessarily reflect the presence of pathogens.
However, as the number of bacterial cells increases, so does the probability that pathogens are present. Acceptable limits for pasteurized milk (heated for ~30 minutes at ~63

C) are between
300-3,000,000 bacterial cells per milliliter. To assure safe consumption, raw hamburger should not contain above 10 6 bacterial cells per gram. If greater than 10 6 cells per gram are present, specific indicator species may be identified.
Indicator species are those bacteria which, if found in certain quantities, indicate that the food, water, or milk may not be fit for human consumption. Among the most common indicators are the coliforms, an example of which includes Escherichia coli (E. coli) .
Coliforms are gram-negative, lactose-fermenting (with the production of gas), facultative anaerobic, non-spore-forming rods. These bacteria are common inhabitants of the intestinal tract of animals and humans and, therefore, may reveal fecal contamination. Approximately 50-
100 coliforms per gram of food are the acceptable limit. No more than two coliforms should be present per 100 milliliters of drinking water. Zero coliforms per 100 milliliters of water is preferred.
In this experiment, students will:
1. Sample food, water, and milk.
2. Dilute the food and milk and inoculate agar plates.
3. Perform plate counts of the food and milk samples to determine the number of bacterial cells per gram or milliliter, respectively, in the original sample.
4. Determine the presence and most probable number of coliforms in the water sample.
3

Here’s a nifty way to estimate the size of a meter: Most door knobs are located about a meter from the floor!
I. Metric System
A. The standard unit of length is the meter. The standard unit of volume is the liter
(although we will use 1 ml) and the standard unit of mass is the gram (g). The italicized
UNIT prefixes (below) can be placed in front of liter also.
1 kilo meter (km) = 1000m
1 meter (m)
1 deci meter (dm) =
1 centi meter (cm) = m
1/10m (0.1m)
10dm=1m
1/100m (0.01m)
100cm=1m
1
1 milli
1 micro meter (
 m) = nano meter (mm) = meter (nm) =
1/1000m (0.001m)
1/1,000,000m (0.000001m)
1/1,000,000,000m
1 angstrom =
The angstrom is no longer officially used.
(0.000000001m)
0.0000000001m
SCIENTIFIC NOTATION
10 3 m
10
0 m
10
-1 m
10
10
10
10
-2
10
-3
-6
-9
-10 m m m m m
B. Metric Equivalents
Length
1 cm = 0.394 inch
2.54 cm = 1 inch
1 m = 3.281 feet
1 m = 1.09 yards
1 km = 0.621 mile
1.6 km = 1 mile
Volume
1 cubic cm = 1 ml
1 liter = 1.056 quarts
1 quart = 0.9461iter
1 liter = 0.264 gallon
1 gallon = 3.785 liter
------------------------------
Weight
1g = 0.0352 ounce
1 ounce = 28.35 grams
1kg = 2.2 pounds
1 pound = 0.453 kg
1 metric ton = 0.984 English ton
1 English ton = 1.016 metric ton
C. Temperature: Standard unit is Celsius (centigrade)
F

= (C

x 9/5) + 32
C

= (F

-32) x 5/9
K= C

+ 273 (Kelvin is another often used temperature standard)
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II. Scientific Notation . We will work with large numbers of cells. Therefore, scientific notation will be used because it easily represents very large and very small numbers. If a number is written with scientific notation, only the one's place is represented. To convert from decimal form to scientific notation, move the decimal point so that only one number is to the left of the decimal point. The number of places you had to move the decimal point is the exponent in scientific notation.
Decimal form Scientific notation
1,000,000 1 x 10 6 Notice how many places the decimal must move to obtain
100,000
10,000
1,000
100
10
1
1 x 10 5
1 x 10 4
1 x 10 3
1 x 10 2
1 x 10 1
1 x 10 0 only one number to the left of the decimal point exponent number is equal to the number of places needed to move the decimal point.
Fraction form
. The
0.1
0.01
0.001
0.0001
0.00001
0.000001
1 x 10
1 x 10
1 x 10
-1
-2
-3
1 x 10 -4
1 x 10 -5
1 x 10 -6
1/10
1/100
1/1000
1/10,000
1/100,000
1/1,000,000
If a number is less than one, the exponent is negative .
When the number is greater than one, the exponent is positive .
Examples:
0.006 = 6 x 10 -3 0.0004 = 4 x 10 -4
102,360 = 1.02 x 10 5 14,000,000 = 1.4 x 10 7
When you multiple numbers using scientific notation, the answer is the sum of the exponents.
Examples:
10
2
x 10
4
= 10
6
10 -2 x 10 -4 = 10 -6
10
5 x
10
-3 =
10
2
Practice: Convert the decimal form to scientific notation
479,000,000 = 0.0888 =
4,000 =
0.00043 =
12,000 =
0. 248 =
Practice: 10
10
10
5
-3
5
x 10
x 10
x 10
7
-5
-4
=
=
=
5

IV. Dilutions
Because most foods and milk contain large numbers of cells, it is necessary to dilute the sample so that the number of colonies may be determined. Diluting simply decreases the number of cells per volume (for example, 10,000,000 cell per ml
 diluted to 100 cells per ml).
It is virtually impossible to count 10,000,000 colonies on the surface of the agar. However, it is much easier if we dilute the sample and only have to count 100 colonies. Diluting is performed in increments because, for statistical purposes, plates with between 30 and 300 colonies are most desirable.
Each colony is considered to have originated from one initial cell through cell
division.
Therefore, the number of colonies grown on agar is equivalent to the number of cells initially placed on the agar before incubation.
Definitions
D ILUTING : T HE ADDITION OF A SOLUTION ( OR PLAIN SUBSTANCE SUCH AS WATER ) TO A
SUBSTANCE IN ORDER TO DECREASE THE CONCENTRATION OF THAT SUBSTANCE
.
T
HE
CONCENTRATION OF A PARTICULAR SUBSTANCE IS THE VOLUME OF THAT SUBSTANCE DIVIDED BY
TOTAL VOLUME OF ALL SUBSTANCES .
D ILUTION : THE CONCENTRATION OF A SUBSTANCE AFTER IT HAS BEEN DILUTED .
AFTER DILUTING
A SUBSTANCE
,
THE CONCENTRATION OF THAT SUBSTANCE IS A FRACTION OF WHAT IS WAS BEFORE
.
F OR EXAMPLE : WHEN 1 ML OF A SUBSTANCE ( FOR EXAMPLE , A BACTERIAL BROTH SOLUTION ) IS
ADDED TO
9
ML OF WATER
,
THE FINAL VOLUME IS
10
ML
.
T
HE CONCENTRATION IS NOW
1 /
10
OF
THAT IN THE ORIGINAL BROTH CULTURE
.
T
HEREFORE
,
THE DILUTION IS
1
/
10
(0.1
OR
10
-1
).
S
INCE IT
IS A FRACTION OF WHAT IT WAS BEFORE , THE DILUTION IS ALWAYS A NUMBER LESS THAN ONE .
DILUTION FACTOR : A NUMBER GREATER THAN ONE THE FACTOR BY WHICH THE ORIGINAL
SOLUTION
’
S CONCENTRATION WAS DILUTED .
E .
G ., A BACTERIAL BROTH SOLUTION HAD A
1
/
10
DILUTION
-
ITS CONCENTRATION IS
1
/
10
THAT OF THE ORIGINAL CULTURE
.
L
IKEWISE
,
THE ORIGINAL
CULTURE HAS
10
X THE CONCENTRATION AS THE DILUTED SOLUTION
.
T
HEREFORE
,
THE DILUTION
FACTOR IS 10 IT WAS DILUTED 10X.
T HE DILUTION FACTOR IS EASY TO CALCULATE USING THE
DILUTION
.
A 10 -3
DILUTION HAS A
10 3
DILUTION FACTOR
.
S ERIAL D ILUTIONS : A DILUTION IS PREPARED , AND THEN THIS DILUTION IS USED TO MAKE A
SECOND DILUTION
;
THE SECOND IS USED TO MAKE A THIRD DILUTION
,
AND SO ON
.
To determine each INDIVIDUAL* dilution in a series:
Divide the volume of the sample being transferred by the total volume after transfer.
In a serial dilution to determine the TOTAL (or FINAL) DILUTION of the original sample at any particular step: All individual * dilutions occurring beforehand must be multiplied together.
When you multiple numbers using scientific notation, the answer is the sum of the exponents.
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EXAMPLE DILUTION PROCEDURE TO DETERMINE
THE NUMBER OF CELLS/ML (or g) IN THE ORIGINAL SAMPLE
Each dilution procedure will have three steps:
Step 1: Diluting the original sample and determining individual and total dilutions along with the dilution factors.
Step 2 : “Plating out” the dilutions on agar and incubating.
Step 3: Determine the number of bacterial cells/ml (or g) in the ORIGINAL sample. This is the purpose of diluting the sample- so we can see and count individual colonies.
If the material sampled is a liquid, the number of bacterial cells/ml is determined. If a solid, the number of bacterial cells/gram is determined.
On the next few pages is an example procedure.
STEP 1 Dilute the original sample. Dilutions are performed sequentially.
10 -1 dilution, 10 1 dilution factor 10 -2 dilution, 10 2 dilution factor
10 -4 dilution, 10 4 dilution factor
7

S TEP 2 The chosen dilutions are "plated out" on agar. For example...
10
-4
dilution, 10
4
dilution factor
10
-5
dilution, 10
5
dilution factor
8

S
TEP
3 Determining the number of cells/ml in the ORIGINAL sample.
The final step is to determine the number of cells/ml (g) in the ORIGINAL sample. After incubating the plates, count the colonies. For statistical purposes, if possible, pick plates with between 30 and 300 colonies. To factor in the dilutions made of the original sample, multiply the number of colonies on the plate by the dilution FACTOR of that plate.
Example:
30 colonies on plate
Dilution on plate: 10
-4
Dilution FACTOR: 10
4
(10,000)
30 x 10,000 = 300,000 = 3 x 10 5 cells/ml in original sample
You are done!
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I.
Standard Plate Count of Food
We will determine the number of bacterial cells in 1 gram of the hamburger. The lab technician has previously diluted and processed the food in a blender. This is a 10 -2 dilution.
Materials: group of 4 students
Food sample - previously diluted 100X by the lab technician (10-2 dilution)
1 99ml water blank
3 plates of Eugonagar
3 1.0 ml pipets & pipet pump (in drawer)
Bactispreader
Bunsen burner and striker, safety glasses
Vortex
1. Label each plate with your lab section, your initials, each plate’s dilution (or dilution factor) and ‘Food’. If not labeled, put Ea (Short for Eugonagar) on each plate.
2. Mix the samples thoroughly before plating with the vortex ! Make sure the flask is tightly covered.
3. Remember to use aseptic technique. Use the procedure in the illustration below.
4. After pipetting the correct amount of sample in each plate, spread the sample with a bactispreader evenly over the entire surface of the agar.
5. Invert the plates. Incubate at 37

C.
10

II. Standard Plate Count of Milk
A procedure similar to the standard plate count of food will be utilized to determine the number of bacterial cells per milliliter in a milk sample.
Materials: group of 4 students
Milk sample
3 plates of TSA (tryptic soy agar)
1 99ml water blank
Vortex, bactispreader
3 1.0ml pipets & pipet pump Bunsen burner and striker, safety glasses
1. Label the plates with your lab section, your initials, each plate’s dilution (or DF) and ‘Milk’.
2. Thoroughly mix the samples before plating with the vortex! Be sure the flask is tightly
covered.
3. Remember to use aseptic technique. Use the procedure below.
4. After pipetting the correct amount of sample in each plate, spread the sample with a bactispreader evenly over the entire surface of the agar.
5. Invert the plates. Incubate at 37

C.
11

III. Presumptive Test for the Presence of Coliforms in Water
Coliforms are common inhabitants of the intestinal tract of warm-blooded animals.
Although they are not necessarily pathogens, their existence in water indicates fecal contamination and the possible presence of other pathogens that also are found in the intestinal tract. Examples of intestinal waterborne diseases are: hepatitis, polio, typhoid fever, and dysentery.
Coliforms are sewage indicators , organisms which, when detected, reveal contamination with fecal material. Escherichia coli and Enterobacter aerogenes fit this description. However, E. coli is a more desirable sewage indicator than E. aerogenes because
E.
coli is found only in the intestine and not in soil. E. aerogenes , on the other hand, is found in sewage and soil. Therefore, the presence of E. aerogenes might only mean contamination of water by soil whereas the presence of E. coli always means contamination by with fecal material.
Tests utilized in the examination of water are found in Standard Methods for the
Examination of Water and Wastewater published by the American Public Health Association.
In this exercise, you will perform the presumptive test used as an initial determination that coliforms are present. You will then estimate the most probable number (MPN) of coliforms per 100ml of water.
Materials: group of 4 students
Wear Gloves!
3 tubes of D.S. (double-strength) lactose broth with inverted Durham tubes
6 tubes of S.S. (single-strength) lactose broth with inverted Durham tubes
Water sample
Pipets (5.0 or 10ml and 1.0ml), pipet pumps
Test tube racks, bunsen burner and striker, safety glasses, masking tape
Vortex
Follow the procedure on the following page.
1. Pipet 10ml of the water sample into each of the 3 D.S lactose broth tubes. Label tubes "D.S, water" using masking tape.
2. Pipet 1.0ml of the water sample into 3 of the S.S. lactose broth tubes. Label tubes "S.S., water, 1.0ml" using masking tape.
3. Pipet 0.1ml of the water sample into the remaining 3 S.S. lactose broth tubes. Label tubes
"S.S., water, 0.1ml".
4. Incubate all tubes 48 hours at 37

C.
12

Presumptive Test for the Presence of Coliforms in Water
Procedure
13

I. Standard Plate Count of Food
Materials: plates from Day 1, calculators
1. Count the number of colonies on each plate. Each colony represents one cell initially plated from the original sample.
2. Determine the number of cells/gram in the original sample of food by multiplying the number of colonies on a plate by the dilution factor of that plate.
10 -3 plate: number of cells/g
10 -4 plate: number of cells/g
10
-5
plate: number of cells/g
3. Find your group’s average number of cells/g by adding the results from all of your counted plates and dividing by the number of plates you counted.
Your group’s average results: cells/gram of food
4. Record the class’ results, and calculate the average number of cells per gram of food.
Class Results:
Number of Cells /Gram in original sample
Class average=
Record in lab report.
14

II. Standard Plate Count of Milk
Materials: plates from Day 1, calculators
1. Count the number of colonies on each of the plates. Each colony represents one cell initially plated from the original sample.
2. Determine the number of cells/ml in the original sample of milk by multiplying the number of colonies on a plate by the dilution factor of that plate.
10 -1 plate: number of cells/ml
10 -2 plate: number of cells/ml
10
-3
plate: number of cells/ml
3. Add the number of cells/ml for all counted plates, and divide this sum by the number of counted plates.
Your group’s average results: cells/ml of milk
4. Obtain the class results and calculate the average number of cells per milliliter of milk for the entire class.
Class Results:
Number of cells /ml of original sample
(per group)
Class average=
Record in the lab report.
15

III. Presumptive Test for the Presence of Coliforms in Water
Materials: tubes of DS and SS lactose broth from the previous lab
1. Observe the tubes of lactose broth. Is any gas visible in the inverted Durham tubes?
A negative presumptive test has no gas or less than 10% gas in the Durham tube (a small bubble is negative).
A positive presumptive test has 10% gas (minimum) produced in the Durham tube (lactose has been fermented and gas has been produced- presumptive for coliforms).
2. Determine the most probable number (MPN) of coliforms present in 100ml of water using the MPN chart on the previous page.
Example:
If you had gas in the first three tubes (D.S. lactose broths) and gas only in one tube of the second series (S.S. lactose broths, 1.0ml water) but none in the last three tubes (S.S. lactose broths, 0.1ml water), your test would be read as 3-1-0. The chart indicates that the MPN for this reading is 43. This means that this particular sample of water has approximately 43 coliform organisms per 100ml of water with a 95% probability of there being between 7 and 210 organisms. Keep in mind that the MPN figure is a statistical probability figure ! It is only an estimate.
Your results:
MPN of coliforms per 100ml of water:
16

Number of tubes giving a positive reaction *MPN Index per 95% confidence limits
100ml
3
3
3
3
3
3
3
3
3
3
3
3
2
2
2
2
2
2
1
1
1
1
0
1
D.S. 10ml water
0
2
3
2
2
3
3
1
1
0
1
0
0
2
2
1
1
0
0
1
2
0
1
1
0
S.S. 1.0ml water
0
28
23
39
64
14
15
20
21
7
11
11
9
4
7
3
3
150
210
240
460
1,100
43
75
120
93
2
0
0
1
1
2
1
2
2
0
0
1
0
1
0
1
0
1
1
0
1
0
0
0
S.S. 0.1ml water
1
4
7
4
10
3
7
1
3
3
3
1
1 lower limit less than 0.5 less than 0.5 less than 0.5
15
30
35
36
15
7
14
30
71
150
47
150
120
130
36
37
44
89
21
23
36
36 upper limit
9
13
20
380
440
470
1,300
2,400
380
210
230
380
**4,800
*This is the estimate of the number of cells per 100ml of water.
**3-3-3 results are estimated to have a MPN over 1,100.
17

18

Food Water
1. (0.5pt.) Results:
Class average: cells per gram of food-
Name:
Section:
Microscopic observations.
Introduction Microbiology Lab Report
Sampling Food, Water and Milk
Milk
Class average: cells per milliliter of milk-
MPN of coliforms per 100 milliliters (individual group result)-
2. (0.5pt.) What is the dilution factor of plate with a final dilution of 10
-6
?
3. (0.5pt.) A plate has a final dilution of 10
-4
. There are 50 colonies on the plate. How many cells per ml were there in the original sample? Show your work!
Cells per ml in original sample .
19

4. (1pt.) If you performed the presumptive lactose broth test on a water sample and obtained positive results in 3 tubes of D.S. lactose broth, 2 tubes of S.S. lactose broth containing 1.0 ml of water, and 0 tubes of S.S. lactose broth containing 0.1 ml of water, statistically what is the highest number of cells in 100ml of the sample?
(HINT: This question is NOT asking for the MPN!)
Highest number of cells in 100 ml:
What is the lowest number of organisms in 100 ml of water?
5. (0.5pt.) To determine the number of cells/ml in the original sample, multiply the number of colonies on the plate by the .
6. (0.5pt.) 1.0 ml of sample is diluted 99 ml of water. What is the dilution?
7. (0.5pt.) One ml of milk is diluted in 9 ml of water. One ml of this solution is then diluted in
99 ml of water. What is the final dilution? Show your work.
8. (0.5pt.) What is the dilution factor of a sample diluted to 1.0 x 10
-7
?
9. (0.5pt.) How many cells/ml are there in the original sample if 300 colonies were counted on a plate with a 1.0 X 10
-5
dilution? Show your work.
20

Name:
Date:
Lab Section:
Microscope Number: HCC
Bring the stage to its lowest level.
Click the 4X objective into place.
Remove the slide and dispose of appropriately.
Clean objective lenses with swabs and liquid lens cleaner. Dry each objective after using the liquid.
Clean eyepieces with dry swabs.
Clean condenser lens with dry swabs.
If necessary, clean the stage with a damp Kimwipe.
Turn the light switch OFF. Do not move the microscope for a few minutes before putting it away.
Be sure the eyepieces are facing forward (away from the arm).
Replace the dust cover.
Return the microscope to the appropriate location in the cabinet.
Check to be sure all waste is removed from the sink and floor.
21