Serial Dilutions Made Easy
Author: Jan Hilten and Carol Sanders
Woodrow Wilson Biology Institute
1993
Introduction:
Many areas of science use serial dilutions in the preparations for different experiments. This exercise is presented as an aid for the teacher in helping his/her students improve their skills and more quickly understand the particular application for which serial dilutions are a tool. Serial dilutions are often used in microbiology, biotechnology, and in chemistry classes, to name just a few. Therefore a clear, concise, and nonthreatening approach to the learning of this very important concept is essential.
Serial dilutions are usually made in increments of 1000, 100 or 10. The concentration of the original solution and the desired concentration will determine how great the dilutions need to be and how many dilutions are required. Important also is the total volume of solution needed. If only small quantities of solutions are needed then greater numbers of dilutions are necessary.
The most common examples deal with concentration of cells or organisms, or the concentration of a solute. The approximate concentration should be known at the start of the experiment before the appropriate number and amount of dilutions can be made. In order to arrive at the desired concentration, use serial dilutions, instead of making one big dilution, in order to finally arrive at the desired concentration. This method is not only cost effective but it also allows for small aliquots to be diluted instead of unnecessarily large quantities of materials.
This technique involves the removal of a small amount of an original
solution to another container which is then brought up to the original
volume using the required buffer or water. In the example below, if you
have 1 mL of your original solution, and you remove 10 µL and place
it in a tube containing 990 µL of water or media you have made a
1:100 dilution. If the original solution contained 5 x 10^{8}
organisms or cells/mL, we now have a concentration of 5 x 10^{6}
cells/mL, because we have simply divided our concentration by 100. Now,
if we want to dilute this by a factor of 1:1000, we must remove 1 µL
of the second solution and place it in a tube containing 999 µL
of media. We have now diluted our secondary concentration by 1000, and
would then divide our concentration by 1000 to yield a 5 x 10^{3}
cells/mL.
Practice Exercises
 You are given a test tube containing 10 mL of a solution with 8.4
x 10^{7} cells/mL. You are to produce a solution that contains
less than 100 cells/mL. What dilutions must you perform in order to
arrive at the desired result?
ANSWER: You should perform a series of three 1:100 dilutions
to yield 84 cells/mL.
1 mL of original solution to 99 mL of water = 8.4 x 10^{5}
cells/mL.
1 mL of second solution to 99 mL of water = 8.4 x 10^{3} cells/mL.
1 mL of third solution to 99 mL of water = 8.4 x 10^{1} or
84 cells/mL.
 You have a microtube containing 1 mL of a solution with 4.3 x 10^{4}
cells/mL and you are to produce a solution that contains 43 cells/mL.
What dilutions must you perform?
ANSWER: You could perform the following dilutions: 10 µL
of original solution to 990 µL of water = 4.3 x 10^{2} cells/mL.
100 µL of second solution to 900 µL of water = 4.3 x 10^{1}
or 43 cells/mL.
 You are given a container with 5 mL of a solution containing 5.1 x
10^{3} cells/mL. You are to produce a solution that contains
approximately 100 cells/mL.
ANSWER: You would perform the following dilutions:
0.5 mL of original solution to 4.5 mL of water = 5.1 x 10^{2}
cells/mL 1 mL of second solution to 4 mL of water = 1.02 x 10^{2}
cells/mL or 10^{2} cells/mL
 You are given a container of yeast cells for which Klett units have
been determined on a Klett Summerson Colorimeter. The container contains
a population whose concentration is 2.6 x10^{6} cells/mL. You are to prepare
a suspension which, when you spread 1 mL of the suspension on appropriate
media, will result in about 100 cells.
ANSWER:
10 µl of original solution to 990 µl (or 1.0 mL) of
sterile water = 2.6 x 10^{4} cells/mL
10 µl of second solution to 990 µl (or 1.0 mL) of sterile
water = 2.6 x 10^{2} cells/mL
0.5 mL of third solution to 0.5 mL of sterile water = 1.3 x 10^{2}
or 130 cells/mL 0.77 mL of fourth solution to 0.23 mL of sterile water
= 100 cells/mL
Note: Corrected 9 March 2005
References
Thomas R. Manney and Monta L. Manney. pp. 2829.
Evelyn Morholt and Paul F. Brandwein pp. 458460. Harcourt Brace Jovanovich, Inc.
David A. Micklos and Greg A. Freyer. pp. 244245. Cold Springs Harbor
