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
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.
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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 106 bacterial cells per gram. If greater than 106 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.
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 prefixes (below) can be placed in front of liter also.
UNIT / SCIENTIFIC NOTATION /1 kilometer (km) = / 1000m / 103m
1 meter (m) / m / 100m
1 decimeter (dm) = / 1/10m (0.1m) / 10-1m
10dm=1m
1 centimeter (cm) = / 1/100m (0.01m) / 10-2m
100cm=1m
1 millimeter (mm) = / 1/1000m (0.001m) / 10-3m
1 micrometer (mm) = / 1/1,000,000m (0.000001m) / 10-6m
1 nanometer (nm) = / 1/1,000,000,000m
(0.000000001m) / 10-9m
1 angstrom =
The angstrom is no longer
officially used. / 0.0000000001m / 10-10m
B. Metric Equivalents
Length / Volume / Weight /1 cm = 0.394 inch / 1 cubic cm = 1 ml / 1g = 0.0352 ounce
2.54 cm = 1 inch / 1 liter = 1.056 quarts / 1 ounce = 28.35 grams
1 m = 3.281 feet / 1 quart = 0.9461iter / 1kg = 2.2 pounds
1 m = 1.09 yards / 1 liter = 0.264 gallon / 1 pound = 0.453 kg
1 km = 0.621 mile / 1 gallon = 3.785 liter / 1 metric ton = 0.984 English ton
1.6 km = 1 mile / ------/ 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)
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 106 Notice how many places the decimal must move to obtain
100,000 1 x 105 only one number to the left of the decimal point. The
10,000 1 x 104 exponent number is equal to the number of places
1,000 1 x 103 needed to move the decimal point.
100 1 x 102
10 1 x 101
1 1 x 100 Fraction form
0.1 1 x 10-1 1/10 If a number is less than one,
0.01 1 x 10-2 1/100 the exponent is negative.
0.001 1 x 10-3 1/1000
0.0001 1 x 10-4 1/10,000 When the number is greater
0.00001 1 x 10-5 1/100,000 than one, the exponent is positive.
0.000001 1 x 10-6 1/1,000,000
Examples:
0.006 = 6 x 10-3 0.0004 = 4 x 10-4
102,360 = 1.02 x 105 14,000,000 = 1.4 x 107
When you multiple numbers using scientific notation, the answer is the sum of the exponents.
Examples:
102 x 104 = 106
10-2 x 10-4 = 10-6
105 x 10-3 = 102
Practice: Convert the decimal form to scientific notation
479,000,000 = 0.0888 =
4,000 = 12,000 =
0.00043 = 0. 248 =
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Practice: 105 x 107 =
10-3 x 10-5 =
105 x 10-4 =
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
Diluting: The addition of a solution (or plain substance such as water) to a substance in order to decrease the concentration of that substance. The concentration of a particular substance is the volume of that substance divided by total volume of all substances.
Dilution: 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.
For 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. The concentration is now 1/10 of that in the original broth culture. Therefore, the dilution is 1/10 (0.1 or 10-1). Since 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. Likewise, the original culture has 10x the concentration as the diluted solution. Therefore, the dilution factor is 10 - it was diluted 10X. The dilution factor is easy to calculate using the dilution. A 10-3 dilution has a 103 dilution factor.
Serial Dilutions: 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.
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.
Step 2 The chosen dilutions are "plated out" on agar. For example...
Step 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: 104 (10,000)
30 x 10,000 = 300,000 = 3 x 105 cells/ml in original sample
You are done!
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DAY 1 Lab Procedures
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.
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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 1 99ml water blank
3 plates of TSA (tryptic soy agar) 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.
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.
Presumptive Test for the Presence of Coliforms in Water
Procedure
Day 2