calculations for molecular bio (need a refresher)

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Jeswin

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Jan 11, 2012, 5:11:16 PM1/11/12
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Hey guys, I need to refresh my math skills like for determining
volumes and stuff. It's been a while so where can I go to refresh my
memory? What topics should I look over?

Oh, and if you can help me with this one calculation, it will be very helpful:
Need to prepare a PCR mix with final volume of 30 microliters. I am
told that the stock primer solution concentration is 10 microMolar.
The final concentration needs to be 0.5 microMolar. How much of the
stock solution should I take ? Ohh, as I'm writing this the
(V1)(C1)=(V2)(C2) just came to mind. Is that the way to calculate it?

Thanks everyone

CoryG

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Jan 11, 2012, 5:21:09 PM1/11/12
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0.5uM / 10uM = 0.05
0.05 * 30uL = 1.5uL

or

10uM / 0.5uM = 20
30uL / 20 = 1.5uL

depending how you want to view it

CoryG

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Jan 11, 2012, 5:29:30 PM1/11/12
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In terms of refreshers - I like this one - I think it covers mixing
solutions (not certain of that), but has a good overview of most
organic chem topics (what I learned from it when I started with
organic chem has probably saved my life a few times over):
http://www.amazon.com/Organic-Chem-Lab-Survival-Manual/dp/0471387320/ref=sr_1_2?ie=UTF8&qid=1326320811&sr=8-2

CoryG

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Jan 11, 2012, 6:14:14 PM1/11/12
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Sorry, apparently this is the new version (I just clicked the cover I
recognized after typing the name, but after looking for it on Kindle,
only the new version is available):
http://www.amazon.com/Organic-Chem-Lab-Survival-Manual/dp/0470494379/ref=ntt_at_ep_dpt_1

On Jan 11, 5:29 pm, CoryG <c...@geesaman.com> wrote:
> In terms of refreshers - I like this one - I think it covers mixing
> solutions (not certain of that), but has a good overview of most
> organic chem topics (what I learned from it when I started with
> organic chem has probably saved my life a few times over):http://www.amazon.com/Organic-Chem-Lab-Survival-Manual/dp/0471387320/...

Nathan McCorkle

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Jan 11, 2012, 6:31:32 PM1/11/12
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The way I explain the solution to your problem to students who don't
get the math easily... if you took 30 microliters of the primer
solution, you'd have primers at 10uM, you want the final concentration
to be 0.5uM, which is 20 times less than stock. 30uL/20=1.5uL

Basically what Cory says below

>
> Thanks everyone
>
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Nathan McCorkle
Rochester Institute of Technology
College of Science, Biotechnology/Bioinformatics

Sigma70

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Jan 12, 2012, 11:18:01 AM1/12/12
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Hello Jeswin,
If you search the word "stoichiometry", you will see a bunch of free
training tutorials online.

The following is intended for the DIYBio community that haven't been
formally trained in chem / bio lab calculations, or may be a bit
rusty. If a scientist figures out good methods to answer the two
questions below, they will never need to memorize a formula ever
again, and they'll be able to tackle any conversion calculation.

Q1: How do you build a formula from scratch? How do you know which
side of the equation to place a known value?
Q2: How can you check if your answer is in the correct units of
measurement?

Below are the techniques I use the most:



For Q1: How do you build a formula from scratch? How do you know
which side of the equation to place a known value?

Objective: stock volume = {some relationship between final volume,
final concentration, stock concentration}

Technique: Sanity check. Ask yourself simple questions, based on
your example problem. For this excercise, take a moment and try to
answer each question using common sense before reading the answer
below. Note there is one question for each known quantity in the
original post.

Question: If your stock concentration is higher, will that mean you
need more or less volume of starting material?
.
.
.
.
.
.
.
.
Answer: "less". That means as the concentration rises, you will need
less starting material. This is an inverse relationship, so you know
that the stock concentration should go on the denominator side of the
equation.

Question: If your final volume is higher, will that mean you need
more or less volume of your starting material?
.
.
.
.
.
.
.
.
.
.
.
Answer = "more". That means as the target final volume rises, you
will need more starting material. This is a direct relationship, so
you know that the final volume should go on the numerator side of the
equation.

Question: If your final target concentration is higher, will that
mean you need more or less volume of your starting material?
.
.
.
.
.
.
.
.
.
.
Answer: "more". That means as the target concentration rises, you
will need more starting material. This is a direct relationship, so
you know that the final concentration should go on the numerator side
of the equation.

Write the equation based on the answers of those questions:

stock volume needed = (final volume x final concentration) / stock
concentration


Congrats! you just invented a formula from scratch!




Q2: How can you check if your answer is in the correct units of
measurement?

Objective: We need to make sure that the units of measurement match
up. Example real-life problem: If stock concentration is in g/L, and
we need a final concentration in mM, we need to convert the numbers
into the appropriate units.

Technique: In our formula from Q1, I like to cancel the units of
measurement when I'm multiplying and dividing numbers, the same way we
would cancel variables in algebra: X = (4 * Y) / (3 * Y), the "Y"
cancels out: X = 4/3. I've heard this called "dimensional analysis"
or "unit analysis".

Using your example:

30 uL x0.5 uM /10uM =
The "uM" cancels out, as if it was a variable in an algebra problem

30 uL x0.5 /10= 1.5 uL stock

This technique is very useful if you're doing multiple unit
conversions in a single equation (like converting uM to ug, with a
given molecular weight and target volume), or switching between nano,
micro, milli, K. It keeps the equation organized, and prevents common
mistakes, like multiplying a number by 1000 when we should have
dividided the number by 1000 instead.

example problem: 5 uL of 15mM stock = ? nmol

solution:
5 uL x (15 mmol / L) x (1 L / 1,000,000 uL) x (1,000,000 nmol / 1
mmol) =

cancel out the "uL", "L", "mmol", and the "1,000,000" cancels too:

5 x 15 x (1 nmol) = 75 nmol



On Jan 11, 5:11 pm, Jeswin <phillyj...@gmail.com> wrote:

Jeswin

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Jan 12, 2012, 9:22:00 PM1/12/12
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cool, thanks. There's one more thing I don't fully understand. What is
the meaning of dilutions written like "1:1000"? If I have a solution
of ampicillin at 1:1000 and I have to add it to 2 mL of LB-solution,
how do I do that?

jlund256

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Jan 13, 2012, 2:35:05 PM1/13/12
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On Jan 12, 8:22 pm, Jeswin <phillyj...@gmail.com> wrote:
> cool, thanks. There's one more thing I don't fully understand. What is
> the meaning of dilutions written like "1:1000"? If I have a solution
> of ampicillin at 1:1000 and I have to add it to 2 mL of LB-solution,
> how do I do that?

Typically, you will have a concentrated stock of ampicillin (say 100
mg/ml) that is used at a 1:1000 dilution. So 2 ul of the ampicillin
stock gets added to 2 ml (2000 ul) of LB.

Jim Lund

mad_casual

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Jan 13, 2012, 2:54:46 PM1/13/12
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Caveat emptor!
If I'm over complicating things, please shout me down, but care needs
to be exercised in notation:
1:1000 dilution = 1 uL of (presumably) solute in 1000 uL of
(presumably) solvent. 1001 uL total volume.
1/1000, 1000X, or 1000-fold dilution = 1 uL of solute in 999 uL of
solvent. 1000 uL total volume.

It's usually not a problem at 1:1000, usually. But at 50X, 10X, and
lower, you can really misinterpret what is being done (1:1 = 1/2 = 2-
fold dilution). I only say this as I've worked with "differently
specialized" professionals (e.g. medical technologists, engineers, or
bartenders) who didn't make or understand the distinction.

Sigma70

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Jan 18, 2012, 10:33:26 AM1/18/12
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mad_casual is totally correct.

In practice, scientists (myself included) are usually lazy and notate
1:100 when we really meant 1/100. Typically, if the denominator isn't
a nice round number (like 100, as opposed to 101), you can be
confident that the original author wasn't making the distinction.
This can get really confusing when we're talking about serial
dilutions. Just use your best judgement, and check if the implied
concentrations match up with reported results
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