A dilution series is a group of solutions that have the same components at different concentrations. This strategy can be used to:

1. Generate standard curves. One example is to make a standard curve of absorbances from known amounts of BSA (0.25 mg; 0.5 mg; 1.0 mg and 2.0 mg) to determine protein content of an unknown protein using Beer's Law (A=ecl).

2. Working with bacteria to get a reasonable number of bacteria to count on a plate.

This can be done in two ways

 1. Independently dilute each sample. Pipette out the appropriate amount by removing a decreasing amount of from the stock solution. Note that you need to open the stock solution 4x in the example below:

Example- Make a 1/2 dilution series of BSA ranging from 0.25 - 2 mg of protein

This works wells for some situations but not always. It would be difficult to dilute bacteria to 1/1000,000 to get a reasonable dilution to count on a plate.

2. Serial dilutions of consecutive dilutions. These dilutions are dependent on one another. This strategy is useful if you only want to use a small amount of your original stock concentration. It is also handy if measuring out dilutions independently is not possible (such as taking out 0.000001 ml of your stock or original solution).

Example- make 10 ml series to dilute a bacterial stock 1/ 100,000. First you prepare 10 ml of dH20 into 3 test tubes. Remove 0.1 ml of water from each 3 tubes - now you have 9.9 ml of water (diluent)

Original Broth -> Take out 0.1 ml add to Test Tube #1- 1/100 dilution (MIX WELL!)

Tube #1 -> Take out 0.1 ml add to Test Tube #2- 1/100 dilution of a 1/100 dilution = 1/10,000 total dilution relative to the original broth . (MIX WELL)

Tube #2 -> Take out 0.1 ml (100 µl) and add to Tube #3- 1/100 dilution of a 1/10,000 dil = 1/100,000 dilution in Tube #3 relative to the orig. solution d(MIX WELL)

Note that you end up w/ 10 ml of solution in Tube 3, and 9.9 ml in Tube 1 & 2

Practice - Set up a series of 3 test tubes containing 10 ml of 1/10 diluted bacteria. What is your final dilution?

Answer:

Dr. Sheperd, a wonderful prof who is now retired from SDSU, showed me this neat trick (that I think that he came up with himself!) to get all the solutions to equal the same exact volume in each test tube. Simply use this equation & simplify using algebra (Ah Hah- Something you learned from high school math is applicable!):

X/ (X + Y) = Z

X= volume to be transferred to successive tubes

Y= volume needed in each tube at the end of the dilution scheme

Z = the dilution factor.

Example: So lets say you need 3 ml of a 1/2 dilution.

X=?

Y= 3 ml

Z = 1/2

X/(X + 3ml) = 1/2 Do some algebra times (X + 3) on both sides

X = (X + 3 ml)/2 Do some algebra tricks to simplify (just multiple both sides x 2 to get rid of that pesky 1/2 fraction)

2X= X + 3ml (Do some more algebra tricks (subtract 1X on each side to get the Xs to be on the same side of the equation)

X = 3 ml

Add 3 ml of diluent (water) to each tube and transfer 3.0 ml successively until you reach the last tube- then just discard the last 3.0 ml transfer. Now each tube will have the same amount of the dilution series. Just to check that over

Original tube- Take out 1.5 ml + 3 ml of water (Mix Well) Tube #1

Hum is that right? 3.0 ml/ (3 ml + 3.0 ml) total = 3.0 ml / 6.0 ml = 1/2 dilution

Ok- then take out 3 ml from Tube #1 + 3 ml of water Tube 2 Mix well

Then take out 3 ml from Tube #2 + 3 ml of water Tube #3 Mix well.

Note that there is a 1/2 dilution that increases at each step.

Practice Problem. What volume do you need to transfer to make 3 ml of a set of 3 serial dilutions of 1/3?