a. Dilution problems similar to Exercise 11 – microbial population counts:

Working hypothesis: If we introduce 1 ml of a culture of Escherichia coli containing between 30 – 300 cells/ml then the number of colonies that grow on the plate will give a more precise population count; representing the actual number of cells in the 1 ml (of culture dilution) used.

What assumptions do we make to do this cell count as described in the lab manual?
E. coli don’t die when they are mixed into agar at ~50°C.
Each cell, if culture is mixed well into agar, will grow into one colony.

Discussion:
What do we need to start? We could make a series of dilutions to extinction. (Diluting and plated every dilution until the population was so small nothing grew from the dilution.)That would take much time and material. We could estimate the “ball park” size of our population and use this number to reduce the number of dilutions and pour plates necessary to get a reasonably accurate population count.

Lets assume the value from the spectrophotometer is O.D.₆₆₀ = 0.4 (Why O.D.?)

The lab manual tells us that under these conditions we can make a very crude (but necessary as a starting point) estimate from the following:

An O.D. = 1.0 represents of population size of 3.0 x10⁸ cells/ml

How many cells/ml are do we estimate to be in the sample with O.D.₆₆₀ = 0.4? 1.2 x 10⁸ cells/ml

We also know from our lab manual that for the plates that we are making, the number of colonies must be between 30-300 to be statistically significant.

Based on our spectrophotometer reading, what number between 30 and 300 seems like a logical number for which to aim? 120. So if we put 1ml of a sample that has 60 cells/ml then we expect that 60 colonies will grow in that plate. (What would we expect if 1ml of a 10 fold less concentrated sample were used? How about a 10 fold more concentrated sample?)

By what factor must we reduce the population of 1.2 x 10⁸ cells/ml to get to 120 cells/ml?

So 10⁶ = the dilution factor. The dilution factor is the factor by which we will decrease the concentration.

Another way to say it: We need to make a one million fold dilution. To find the dilution factor – to find out how much we need to reduce the concentration – we divide the starting concentrate by the final concentration (the concentration we wish to reach).

If we reduce the concentration from 1.2 x 10⁸cells/ml to 1.20 x 10²cells/ml we are reducing the concentration by one million or 10⁶ fold. We could also say we are making a 1:1,000,000 or a 1:10⁶ dilution. Since the final concentration is one onemillionth of the initial concentration, we might also say that it is a 1/1,000,000 or 1/10⁶ (one over one million dilution) another waywe can express this is to call this a 10^-6 dilution (since 1/10⁶ = 10^-6) . 10^-6 describes what was done to make the dilution. If we multiply the starting concentration by the dilution description we will end up with the final concentration – but the dilution factor is 10⁶. The dilution factor is the number we multiply the final concentration (or the results of the counts) by to calculate the actual concentration in cells/ml.

b. Other Dilution problems:

This same discussion applies to any dilution problem whether it is counting cells/ml or making a working solution of 12mM from a 1M stock solution – or anything in between. The same principles apply.

To approach a dilution problem (as with any word problem) first determine what information you have been given. There are three big pieces to each dilution puzzle:

1. The Initial concentration (where you start)
2. The final concentration (where you finish)
3. The manipulation that must be made (how to get from start to finish)

Any dilution problem whether a written problem or a practical problem which comes up when working in a lab, will provide parts of the above info and you will calculate the rest. These problems are logical and an exercise in common sense. The math required to do the problems is simple.