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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 con-centration 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 108 organisms or cells/mL, we now have a concentration of 5 x 106 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 103 cells/mL.




Practice Exercises

  1. You are given a test tube containing 10 mL of a solution with 8.4 x 107 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 105 cells/mL.
    1 mL of second solution to 99 mL of water = 8.4 x 103 cells/mL.
    1 mL of third solution to 99 mL of water = 8.4 x 101 or 84 cells/mL.

  2. You have a microtube containing 1 mL of a solution with 4.3 x 104 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 102 cells/mL.
    100 µL of second solution to 900 µL of water = 4.3 x 101 or 43 cells/mL.

  3. You are given a container with 5 mL of a solution containing 5.1 x 103 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 102 cells/mL
    1 mL of second solution to 4 mL of water = 1.02 x 102 cells/mL or 102 cells/mL

  4. 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 x106 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 104 cells/mL
    10 µl of second solution to 990 µl (or 1.0 mL) of sterile water = 2.6 x 102 cells/mL
    0.5 mL of third solution to 0.5 mL of sterile water = 1.3 x 102 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. 28-29.

Evelyn Morholt and Paul F. Brandwein pp. 458-460. Harcourt Brace Jovanovich, Inc.

David A. Micklos and Greg A. Freyer. pp. 244-245. Cold Springs Harbor


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