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Chapter 13 Problem Set C Key: Serial Dilutions and the C1V1 = C2V2
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1. Suppose you have a stock of a buffer solution that is used in experiments.
The stock is ten times more concentrated than its working concentration
(like orange juice, which is sold in a concentrated form). How would
you prepare 100 mL of the buffer at the working concentration?
10 mL of stock + 90 mL water
2. Suppose you need 105 L of a solution. The solution is made from a
stock solution with a concentration that is ten times the concentration at
which it is normally used. How would you dilute the stock solution?
10 mL of stock + 90 mL water
3. Explain how to prepare a 1/10 dilution of food coloring. Then, use the
1/10 dilution to prepare a final dilution of 1/250 of food coloring. Use
the 1/250 dilution to prepare a final dilution of 1/1000 of food coloring.
Make 10 mL of each dilution. Note that these are SERIAL
DILUTIONS. Each dilution given represents the dilution of the
original food coloring stock.
Dilution 1 (1/10) = 1 mL food coloring stock + 9 mL water
Dilution 2 (1/25 of 1/10 = 1/250) = 0.4 mL of 1/10 dilution + 9.6 mL water
Dilution 3 (1/4 of 1/250 = 1/1000) = 2.5 mL of 1/250 dilution + 7.5 mL
water
4. Explain how a dilution series could be used to prepare a 1/106 dilution of
bacterial cells. Design your own scheme, but make at least three dilution
steps in your process. Make 10 mL of each dilution in your process.
Answer vary but here is an example:
Dilution 1 (1/100) = 0.1 mL of cells + 9.9 mL media
Dilution 2 (1/100 of 1/100 = 1/10,000) = 0.1 mL of 1/100 dilution + 9.9 mL
media
Dilution 3 (1/100 of 1/10,000 = 1/1,000,000) = 0.1 mL of 1/10,000 dilution
+ 9.9 mL of media
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5. Antibodies are frequently very concentrated relative to the concentration
used in an experiment. Design a scheme to dilute an antibody solution
200,000 fold. Assume that you have 1 mL of antibody, but need to retain
at least 0.5 mL of this antibody for future experiments.
Answers vary but all dilutions must multiply to 1/200,000 and first dilution
must use no more than 0.5 mL of antibody
6. To perform a viable cell count of bacteria, a sample of bacterial cells is
first serially diluted. Then, 0.1 mL of diluted cells are spread on a Petri
dish that contains nutrient agar. It is assumed that every living cell in the
0.1 mL plated on the agar divides to form a colony of bacterial cells. A
colony contains so many cells that it is visible to the naked eye. It is also
assumed that each colony originates from a single cell. It is possible to
count the colonies and, therefore, to estimate the number of bacteria in
the 0.1 mL of broth.
Assume you begin with a culture of bacteria that contains 2 X
107 bacteria/mL. Show how you would dilute the culture so
that the final tube has a concentration of 200 bacteria/0.1 mL.
Make 10 mL of each dilution in your scheme.
Answers vary, but each dilution has 10 mL total volume, and all dilutions
must multiply to 1/10,000
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7. You are performing a viable cell count of bacteria. The original broth
has an unknown concentration of bacteria. You dilute the culture as
shown in the diagram below and plate 0.1 mL of the last three dilutions
onto three Petri dishes with nutrient agar. The following day you count
the number of colonies on the plates with the results shown in the
illustration below. What was the concentration of bacteria in the
original tube?
3.2 X 106 cells/mL
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8. You are performing a viable cell count of bacteria. The original broth
has an unknown concentration of bacteria. You dilute the culture as
shown in the diagram below and plate 0.1 mL of the last three dilutions
on three nutrient agar plates. The following day you count the number of
colonies on the plates with the results shown in the illustration below.
What was the concentration of bacteria in the original tube?
6 X 107 cells/mL