CLOates

30-July-2006

OCCC, APPM 1313

Last Rev. 18-Nov-2006 CLO

Dilution Problem Notes for Module 5

The following information is a more or less unexpurgated version of an e-mail message I sent to one of your classmates in response to a request for help on Module 5 dilution problems.  I hope it will help organize your thinking about these kinds of problems.

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The "Cliff Notes" version of dilution problems would look something like this. 

First, recognize that the problems come in two varieties, those that ask you to synthesize (build) a dilution scheme and those that ask you to analyze a given dilution setup and, from that, calculate the concentration of bacteria in a patient sample (the "original" concentration, N_orig).

For those problems that ask you to synthesize a dilution scheme that does, say, a 10-5 (ten to the negative 5th) dilution, there are four steps to be carried out. 

First, figure out what the dilution is in non-powers-of-ten notation.  A 10-5 dilution is the same thing as a 1 / 105 (one over ten to the fifth) dilution--the negative exponent just means invert (flip) the power of ten and change the exponent sign to positive.  Now, 105 is just a 1 with five zeroes annexed at the end or 100,000.  So in this case, we're being asked to find a dilution scheme that will dilute 1 part in 100,000 or 1 / 100,000.

Second, we'll need to factor the 100,000 into several components, because we probably can't do a 1 / 100,000 dilution all in one stage.  We could factor 100,000 into, say,  100 * 100 * 10.  This implies that we need three dilution stages that dilute 1/100, 1/100, and 1/10, respectively.  (Alternatively, we might factor 100,000 into 10 * 10 * 10 * 10 * 10.  That implies five dilution stages that each dilute 1/10.)  Let's go with the three-stage dilution, since that's more practical for a lab tech to implement.  (The fewer stages, the better, usually.)

Third, we'll need to figure out how to implement the required 1/100 and 1/10 dilutions in whatever size water blanks the problem specifies.  Let's say we're being asked to implement the dilution scheme with only 99 mL water blanks.  Chuck's Quick and Dirty Dilution Technique comes in handy here.  If we have 99 mL water blanks and need a 1/100 dilution, we divide the 99mL by NOT 100, but by 99 = (100 - 1) to get the amount to pour into the stage.  Let's see, that's 99 mL / 99 = 1 mL.  So if we pour 1 mL of fluid into a 99 mL water blank, we'll get a 1 / 100 dilution, allegedly.  [SKIP TO NEXT PARAGRAPH IF THIS IS THE FIRST TIME YOU'RE READING THIS.]  We can verify this by calculating the dilution factor formula, Dil. F = IN / TOTAL [dilution factor equals what was poured INto the stage, divided by the TOTAL amount fluid in the stage after we pour in the amount IN].  Well, TOTAL is just what was already THERE + what we poured IN, so we can write Dilution Factor = IN / (THERE + IN).  Now, by Chuck's Quick and Dirty Dilution Technique, we allege that to get a 1/100 dilution in a 99 mL water blank, we'd pour in 1 mL of fluid.  In equation form that's:  Dilution Factor = 1 mL / ( 99 mL + 1 mL) = 1 / 100 (!), and that verifies that the 1 mL into 99 mL scheme really does dilute 1 part in 100 or 1 / 100.) 

Since we need two stages of 1/100 dilution, we'll make them both by pouring 1 mL from the previous stage into 99 mL water blanks.  That's fine, but what about the 1 / 10 dilution required for the third stage?

Let's do the same kind of thing again.  We're required to use 99 mL water blanks, but we need a 1 / 10  dilution, so again by Chuck's Quick and Dirty Dilution Technique, we'll need to pour in a quantity of fluid calculated as follows:  99 mL divided by NOT 10, but 9 (=10 - 1).  That's 99 mL / 9 = 11 mL that needs to be poured into the third stage 99 mL water blank.  (Let's verify that:  Dil. F. = IN / (THERE + IN)  =  11 mL / ( 99 mL + 11 mL)  =  11 / 110  = 0.10  or  1 / 10.)  Okay then,  the third stage will consist of 11 mL of fluid from the previous stage poured into a 99 mL water blank. 

Fourth, we need to draw the picture we drew in class of the original sample container and three more containers (water blanks) of 99 mL each, in the following pattern:

o   1 mL going from the original container into the first stage 99 mL water blank ( 1 / 100 dilution),

o   1  mL going from the first stage into the second stage 99 mL water blank ( 1 / 100 dilution), and

o   11 mL going from the second stage into the third stage 99 mL water blank ( 1/ 10 dilution).

 

That picture is the required dilution scheme, since 1/100  *  1/100  *  1/10  =  1 / 100, 000 = 10-5 (ten to the -5 power). 

Note that we don't need to specify how much fluid is going OUT of the third stage into the Petri dish (plate), because that doesn't affect the dilution of the fluid. Dilution is affected only by what's poured INto a dilution stage and by how much fluid was already THERE.  There is a solution to a problem in the Math Lab solutions manual that implies otherwise, but I encourage you to use my back-to-basics simple version.  The Math Lab solution is a very dirty trick that IS really used by lab techs, but its explanation 1) requires a little algebra that's not a prerequisite for this course,  2) is not obvious,  and  3) makes unclear the concepts we're trying to learn.

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Okay, so much for synthesizing dilution schemes.  Let's have a look at how to analyze a given dilution scheme to calculate the original concentration of bacteria ("bacT") in a patient sample.

A typical problem would show us a dilution scheme something like this:

o   first stage:  0.1 mL from orig. sample poured into a 9.9 mL water blank

o   second stage:  1 mL from first stage poured into a 99 mL water blank

o   third stage:  11 mL from second stage poured into a 99 mL water blank

as well as some information about how much was fluid was taken from the third stage and put into a Petri dish (plate) to grow bacteria colonies, and how many colonies actually grew.  Let's say they took 0.1 mL out of the third stage and smeared it on the plate, where, after incubation, 25 bacteria colonies grew.

First, we'll need to calculate the dilution factors in each stage and then multiply them all together to get the total dilution.  We'll use the formula, Dilution Factor = IN / (THERE + IN), as in the synthesis problem above.

o   first stage:  Dil. F. =   0.1 mL  / ( 9.9 mL + 0.1 mL)  =  0.1 / 10  =  0.01 =  1 / 100

o   second stage:  Dil. F. =   1 mL  / ( 99 mL + 1 mL)  =   1 / 100

o   third stage:  Dil. F. =   11 mL  / ( 99 mL + 11 mL)  =   11 / 110  =  1 /10

The first two stages each dilute 1 / 100, so together they dilute 1/100 *  1/100  =  1 / 10,000 and the third stage dilutes 1/10, for a total dilution of 1 / 10,000  *  1 / 10  =  1 / 100,000.  That's the first piece of information we need.

Second, we need the concentration of bacteria (units: bacT / mL) that were in the 0.1 mL of fluid removed from the third stage and smeared on the plate.  Well, the fluid in the plate grew 25 bacteria colonies, so we'll assume that there were 25 individual bacteria in the 0.1 mL of fluid.  That means the concentration of bacteria was 250 bacT / mL ( = 25 bacteria / 0.1 mL ).  That's our second piece of required information.

Third, we need to estimate the concentration of bacteria (in bacT / mL) in the original patient sample, because that's the information the physician needs to make decisions about the patient's treatment.  Okay, from the second step above, we know that the final concentration of bacteria (after 1/100,000 dilution) was 250 bacT / mL.  Now if we could somehow "undo" the 1 / 100,000 dilution, that would give us the original concentration of bacteria in the patient sample.  Physically, dilution can't be undone; however, mathematically, all we have to do is multiply by 100,000 (the inverse of the dilution factor, 1 / 100,000 ) and, viola, the dilution is undone.

Let's see,

250 bacT / mL  *  100,000  =  25,000,000 bacT / mL 

(just annex five zeroes to the 250).

That's our estimate of the concentration of bacteria in the patient's sample.