CLOates

OCCC, APPM 1313

Last Rev.

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.

______________________________

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 / 10^{5} (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, 10^{5} 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.

______________________________

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.