For an experiment with 7 sample tubes and 5 reagents (4 of which are required at the same concentration), the number of required pipette movements are shown. The figure below shows the number of pipette movements necessary without the use of a master mix. For the same experiment, a master mix is now used. Notice the drop in the number of pipette movements - imagine even larger experiments and the time and workload difference achieved.

An illustration of the total number of required pipette movements during an experiment with and without the use of a master mix. 7 empty sample tubes and 5 sample tubes containing different reagents are shown. The reagents are illustrated in different colors and from left to right the reagents are colored yellow, red, green, and blue. A fifth sample tube also containing a reagent is marked with the color gray. Colored lines, corresponding to each reagent color, goes from the reagent tubes to the empty sample tubes, illustrating how each reagent is added to each sample tube. The yellow, red, green, and blue reagent are all added at the same concentration. The gray reagent is added with different concentrations to each of the empty sample tubes. Every line can be counted as a pipette movement, and the total number of pipette movements required for the experiment without a master mix is 35. Beneath the illustration is another illustration of the same experimental setup, but in this case with the use of a master mix. Another sample tube containing the mater mix, which is represented in blue, is added to the setup and the colored lines from the 4 reagent tubes represented in yellow, red, greed and blue, now lead from the reagent tubes to the master mix tube, resulting in fewer pipette movements. Other lines, also illustrating the pipette movements, now goes from the master mix tube and the gray reagent tube to the 7 empty sample tubes. All pipette movements, from the reagents to the master mix and from the master mix to the empty tubes, leads to a considerably lower number of required pipette movements, which is now only 18

Figure 1: Two experimental setups with 7 sample tubes and 5 reagents (4 of which are required at the same concentration). The number of required pipette movements is shown. This figure shows the number of pipette movements necessary with and without the use of a master mix.

When conducting an experiment with a large number of similar samples, it is common to prepare a master mix to ease the workload in the lab. If all your samples have a similar mixture of reagents with only 1 reagent varying from sample to sample, a master mix can be prepared with all the common reagents (of course, this is only possible if all reagents are required in the same concentrations). Now, when setting up your samples, you can pipette from the master mix and then add your variable reagent, instead of mixing e.g., 4 different reagents in correct amounts into each and every tube. The concept is illustrated in the figure above.

In this enzyme kinetics case, at a specific substrate concentration, a master mix, containing NAD+, ethanol, and buffer, can be prepared. When the enzyme is added, the reaction starts, and the enzyme is therefore not included in the master mix. This is not a case where it is crucial to use a master mix; however, the skill is essential for many applications, and it is very useful in enzyme kinetics assays in general.

Calculating substrate concentrations

In the experiment, you will need to calculate how much substrate to add to a tube to obtain the desired substrate concentration. This can be done using the following formula:

C1 · V1 = C2 · V2

where C = concentration and V = volume. Remember to use the same units on both sides of the equality sign!

Example: We have a stock solution of the substrate at 1 M, and we need a final concentration of 200 mM in a volume of 500 μl; what is the volume of stock solution that we need to add for achieving the desired concentration?

If the stock concentration is C1, C2 will be the desired final concentration and V2 will be the final volume. This makes V1 the unknown element. We can now isolate V1, insert the known values and calculate the volume:

Isolate: V1 = (C2 · V2)/C1

Insert and calculate: V1 = (0.2 M · 0.0005l) / 1 M = 0.0001 L = 100 μl

Theory overview