# 10 millimolar (mM) to 250 micromolar - (Aug/28/2012 )

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Hello,
I am a very new person to cell culture and would need some help with the calculations...
1. I am testing few inhibitors on mouse cellline. I have a chemical compound whose MW is 243. I need to dissolve it in DMSO and get the concentration to 10mM. Then I need to dilute it to 200uM and serial dilute this to 20uM-10um-5um-2.5um and so on...
How do I do this and when serial diluting it, do I have to use the growth media or do I use DMSO? I do not want the concentration of DMSO to go more than 1%. Please help

2. I have BMP4 whose concentration is 10ug/ml. I need 10ul of this, whose concentration is 250pM. How do I calculate?

-cell_bee-

First off - there are many many threads on here about calculations - have a look through some of them, they will tell you how to calculate the answers you want.

You may need to dilute in DMSO, but probably you can dilute in the medium of choice. Note that DMSO is toxic to cells at concentrations above 0.1-0.5% (cell line dependent), so you will have to work out how much you need to add to the cells to keep the DMSO below this concentration range. You should also have what is known as a vehicle only control -this is pure DMSO diluted in your medium to the same concentration as for your treatments, just to make sure that any effect you see isn't just from the DMSO.

-bob1-

Thanks for the reply. I am kinda lost, could you please link me to the thread as there are tons of post and its hard for me to find a particular thread realted to my question.

Thanks again!

-cell_bee-

useful formulae:
concentration (in M)= mass (in grams)/MW (in grams/mol)/volume (in L); since you know your concentration, volume and MW --> mass of substance in grams = volume (in L) * MW (in g/mol) * concentration (in M). This is your stock solution (10 mM), or the most concentrated from which you can take a volume V1 to add to V2 of media to get 200 uM final conc
c1*V1=c2*(V2+V1); where c1 is 10 mM = 10000 uM; c2 is 200 uM and V2 is the final volume on top of which you add your drug (~12 mL for a small flask of cells)
solve for V1 = c2 * V2 /(c1-c2). since you plug in V2 in mL, you will get V1 also in mL. (for V2 = 12 mL you get ~0.245 mL = 245 uL). For 20 uM you need to add 10 times less --> 24.5 uL or you dilute your stock solution 10 times (1 part stock of 10 mM and 9 parts media) (down to 1 mM) and still add 245 uL.

With the two formulae I gave you, you can also answer the second question.

Link to similar discussion with drug soluble in DMSO:
http://www.protocol-online.org/forums/topic/26608-can-i-dilute-the-drug-with-dmem/page__hl__micronagu

Andreea

-ascacioc-

I would really like to know where your formula: 2.5*10^9 CFU/mL * X uL of this = (uL final solution+X uL that you add on top) * 10^6 CFU/mL) comes from!
In symbols: C1 x V1 = (V2 + V1) x C2 or C1 x V1 = V2 x C2 + V1 x C2

Also: plz recheck it, it does not add up.

Many mistakes have been made here.

He wants a final (fixed) end volume, this is not the case here.

I noticed you are a huge fan of c1*V1=c2*(V2+V1), but plz dont use the formula when its not appropriate.

-pito-

ascacioc on Wed Aug 29 18:49:09 2012 said:

useful formulae:
concentration (in M)= mass (in grams)/MW (in grams/mol)/volume (in L); since you know your concentration, volume and MW --> mass of substance in grams = volume (in L) * MW (in g/mol) * concentration (in M). This is your stock solution (10 mM), or the most concentrated from which you can take a volume V1 to add to V2 of media to get 200 uM final conc
c1*V1=c2*(V2+V1); where c1 is 10 mM = 10000 uM; c2 is 200 uM and V2 is the final volume on top of which you add your drug (~12 mL for a small flask of cells)
solve for V1 = c2 * V2 /(c1-c2). since you plug in V2 in mL, you will get V1 also in mL. (for V2 = 12 mL you get ~0.245 mL = 245 uL). For 20 uM you need to add 10 times less --> 24.5 uL or you dilute your stock solution 10 times (1 part stock of 10 mM and 9 parts media) (down to 1 mM) and still add 245 uL.

With the two formulae I gave you, you can also answer the second question.

Link to similar discussion with drug soluble in DMSO:
http://www.protocol-...__hl__micronagu

Andreea

I dont agree with it.
There is no point in giving this formula if he cant understand it or calculate it himself.

-pito-

I am just doing the microbiology approximation, not the biochemistry approximation: I mean 40 uL in 10 mL or in 9.96 mL does not make a difference. Anyhow, how do you pipette 9.96 mL nicely? take 9 mL with the 10 mL pipette and 960 uL with the 1 mL pipette? Do you think that pipetting 9 mL has a smaller error than 40 uL?

Anyhow: feel free to have more detailed explanations with more correct calculations I don't mind being corrected and learning smth new everyday

-ascacioc-

I just found the best explanation here on the forum:
http://www.protocol-online.org/forums/topic/7050-useful-dilution-techniques-calculations/

-ascacioc-

-ascacioc-

ascacioc on Wed Aug 29 21:32:12 2012 said:

I am just doing the microbiology approximation, not the biochemistry approximation: I mean 40 uL in 10 mL or in 9.96 mL does not make a difference. Anyhow, how do you pipette 9.96 mL nicely? take 9 mL with the 10 mL pipette and 960 uL with the 1 mL pipette? Do you think that pipetting 9 mL has a smaller error than 40 uL?

Anyhow: feel free to have more detailed explanations with more correct calculations I don't mind being corrected and learning smth new everyday

I know that you are helping and that you can indeed use your formula however it is not the correct way to learn it.
Your formula is ok when used by an experienced person and someone who knows what he/she is doing and when to (not) use it.
However you dont learn the basics and the fundamental math behind if you use this formula from the start without knowing/understading it.

About the pipetting thing: yes you are right, I understand and agree with what you say, however this is not the correct way if thinking for new people in the lab or students in general.
Also: its dangerous to explain it the way you did it because people tend to make it a general assumption and when they are faced with quantities that need to be 100% correct then they will also use this kind of reasoning and make major mistakes.

I am exaggerating a lot and making it a bit harsh, but this is just to make a certain point and to make you think about the mistakes!

But first to start:
Being a bit harsh here:

4 µl of a 2.5 × 10*9 cfu/ml stock , how much is this?

2.5 x 10^9 cfu/ml , for 4µl this is: 2.5 × 10*9 cfu/1000µl X 4µl = 10^7 , not 10^6 as you requested.

I hope you spot your mistake...
BTW: it allready started at the begin of your opening post, where you dont specify what the volume was you used to plate out to count.
Ascacioc allready stated this, see below (the quote starting with "your calculation...")

ALso:

So

2.5 *10^9 *X = (10000+X)*10^6 CFU/ml

2500 X = 10000+X

x = 10000/2499 uL = 4 ul

Am I right

Its completely "wrong"! If you do this on an exam you risk getting a -10 ! Not even a 0!

Why?

you use 10000 ? 10000 what??? I am assuming 10000µl ? But you also use 10^6 CFU/ml , in ml this time and you use 2.5x 10^9 ???? 2.5x 10^9 what? 2.5x 10^9 monkeys in a plastic jar? Or 2.5x 10^9 pigs in a pigstable?

You are just lucky here that you can "erase" the ml "from 2.5 *10^9" (actually it does not state its in ml) with the 10^6 CFU/ml , if this wasnt the case, it would all be wrong.

(I know that here it works out fine , because on both sides you have cfu/ml you dan "erase" from eachother, but for your own sake: use the correct terminology/units, it will prevent you from making mistakes in the future in other calculations)

This sentence is important, but overlooked during the entire discussion:

Your calculation is correct only if you plated 100 uL of your dilution

+ in the end its not correct at all because the both of you assume "something" ... Guess what?

+ the use of the formula is a bit wrong and makes no point in the end (if you want a fixed endvolume) + creates what I call "ignorant people entering numbers in formulas without knowing what they mean".

Also: practise some more examples? On what? Filling in this formula ? Thats not practising at all, thats just doing what I blame in the previous sentence (ingnorant..)

Anyway

Its "semi-correct", but not the ideal way (in fact its not correct at all, its just a quick tool to quickly calculate something that is usable in specific situations)
+ the final volume you want is not what you get.
If you want a final volume of 10 then its 10ml and not 10ml + Xml

Aslo: In my opinion the formula is way to complex for someone who is struggling with dilutions.
This does not really help.
He needs to learn/understand the principles of serial dilutions, not learn some "magic" formula.

Hi
I am trying to calculate colony forming units calculation. I cultured bacteria overnight (20ml) and did O.D reading at 0.4 and did serial dilution (10*1 to 10*9) and I got 250 colonies at 10*6 dilution. So I got 2.5 × 10*9 cfu/ml (original, if I am right). But I need 10*6 or 10*7 CFU/ml (for my experiment). How can I reduce it down to it. Please give example calculation.

Thank you

1) You got 250 colonies at the 10^6 dilution (advise: use ^ and not *). Then you conclude you have 2.5 x 10^9 cfu/ml ? Where did this come from?
I understand: if you have 250 colonies at the 10^6 dilution you would have 250/1000 colonies in a 10^9 dilution. (you diluted it 1000 times more compared with the 10^6 dilution, so its not 2.5, its 0.25 colonies) + I am assuming you do mean 10^-6, 10^-9 etc and always added the same amount to dilute into the same dilutant. ? Or do you start working from 10^9 as "startpoint" and work your way down to 10^0 ?
So where does the 2.5 × 10*9 cfu/ml come from? Its one factor off, unless you indeed plated out 100µl, and it does not make sense.
You dont even specify that you had 250 colonies in (0.)1ml , for all we know, you poured 999999000 ml on your petri dish (and not just (0.)1 ml).
You need to specify this! There is a difference between 250 colonies in a 10^6 dilution and 250 colonies in a 10^6 dilution prepared with 1ml.
Remember what I said about mentioning units!
Its important!

2) the formula, altough you can use it, its not entirely correct and not really a good advice

2.5*10^9 CFU/mL * X uL of this = (uL final solution+X uL that you add on top) * 10^6 CFU/mL

No idea what it means

It doesnt make sense at all: how can you have a final solution (final volume) and then still add something to the top?
Makes no sense at all.
I can make 1 liter final volume and then still add 4444liter on top ?
(I know, I am exaggerating here, but dont use formulas like this, its not correct).

The mistake you make here is simple: you "redefine" what V1 and V2 is!
So its "dangerous" to link it with C1V1=C2V2
Definition of C1V1=C2V2 :
C1= concentration of stock 1 (or initial conc)
V1 = volume of stock 1 you take and you fill up (with dilutant) to end up with V2 (or initial volume)
C2= conentration of stock 2 (or end conc)
V2= end/final volume of stock 2 (or end volume)

While in the longer formula it is:
C1 x V1 = (V2 + V1) x C2

C1= concentration of stock 1
V1 = volume of stock 1 you take and add to V2
C2= conentration of stock 2
V2= clear water/broth you take without cells
In fact all you do is calculate a certain V3 (the endvolume)
So V1 and V2 are not the same as before.
So this formula is not correct for what chandch needs.

BTW: do you know what is behind the C1V1=C2V2 logic? It comes from chemistry... moles...see later.

Anyway, both formulas are in principle correct. However in the second formula you define the endvolume as a function of the startingvolume (endvolume V1 and what you add, in this case V2)
However (!) chemically speaking there is a bit of a problem.
Because in formula 2 you speak about adding volumes (fixed ones, you select them) together and you just add them up (10 ml + 0.04ml) and then you say: V1 + V2 = endvolume, sum of both. This is not always true.
Sometimes V1 + V2 is not just the sum of both. You can have some for example 1liter of substance A + 1liter if substance B and end up with 1.8 liter in total and not 2liter.

I know I am going in detail here, but its best to really understand what happens and not just memorise formulas that are +- correct.

So you cant really state that both formulas are correct and linked. This is not entirely true.
Altough, both can be linked with the fundamentals: #moles before and after stays the same ==> #moles = CxV so Moles prior to diluting is: C1xV1 and after: C2xV2 , meaning that both are equal thus:
C1xV1 = C2xV2

3) use your logical reasoning to solve problems like this.

Lets assume you have 250 colonies in a 10^-6 dilution and you used 1ml to plate, so you counted 250 cells on a plate covered with 1ml.
This means: 250 colonies in 1ml from the 10^-6 dilution. This means that, not diluted, you have 250 x 10^6 colonies in 1 ml. If you have problems grassping this, make it easier for yourself with a simple example: 1 colony in a sample you diluted 100 times(10^-2) (in 1ml) means ==> not diluted 100 times as many colonies in a not diluted ml , meaning 100 colonies not diluted ml.

Now the problem where you need 10^6 colonies.

Well, what do we know? We know the following:

250 x 10^6 colonies in each ml of your sample.

but you want 1 x 10^6 colonies per ml.
So what does this mean?

It simple means you need to dilute 250 times.
Now you can use the CV =CV rule, altough, I would advice you not to use it blindly, rather understand it.

250 x 10^6 colonies in 1 ml
and you want 10^6 colonies in 1 ml (10ml final volume, in another, new test tube, makes 10^7 colonies in total).

So the question is: how much of the original sample (250 x 10^6 colonies in 1 ml, in a total volume of, lets say, 5ml in your test tube) do I need to add to a new test tube to have a final concentration of 10^6 colonies in a final volume of 10 ml?
And you know you need to dilute 250 times...

So how do you solve this?
Final volume has to be 10ml , but it has to be diluted 250 times (in 1ml, not in the entire endvolume 10ml)... what does this mean? simple: you remember this => 1 ml + 9ml means 1/10 dilution.
So 1/250 dilution means?
1 ml + 249 ml (250ml in total) or because you want 10ml as a total volume ==> 1/25 ml + 249/25 ml (divided by 25 becaue 250ml/25 equals 10 ml)==> 0,04 ml+ 9,96 ml

I know this seems complicated and a bit stupid since if you simple use C1V1 = C2V2 , you can do it a lot faster ==>

250 x 10^6 X V1(?) = 10^6 x 10 ml
Meaning X = 0.04 ml or 40µl

And this is not the 4µl you guys found because you missed a factor 10 at the start and you didnt get 4µl, but 4,0016..... (its a different number)
(you guys also got 4 µl for 10ml, this is due to the fact that you miss a 10 at the start where you reason that 250 in 10^6 equals 2.5 in 10^9 , which is 10 off when assuming you used 1ml, but you didnt since you used 0.1ml)

A little bit different approach:

Again, reasoning, not using formulas you dont understand.
250 x 10^6 colonies per 1 ml in 5ml , means : 250 x 10^6 colonies X 5 = 1250x10^6 colonies in total (or: 125x 10^7 colonies)
And I want: 10^6 colonies per 1ml in 10 ml, meaning: 10^7 colonies in total.
So how much ml of the stock (125x 10^7 colonies) do I need to take to get 10^7 colonies in total?
1/125 the part (10^7 is 1/125the of 125x 10^7 colonies) , meaning 1/125 x 5ml = 0.04 ml

I know that 40µl does not make a big difference in the 10ml (or even in 1ml) in the end (the pipetting errors etc are bigger, but its about the math and the correct thinking! Too much people use "stupid" formulas without knowing what it means. Eg: final volume + another volume.. this is just idiotic and not correct at all.
+ this time it does not make a difference since its a small volume added in a larger one 10 000 µl or 10 040 µl, nobody will know the difference.. but this is not always the case...

The correct use of the C1V1 = C2V2 formule is:

V1= volume of your stock you need to take to end up with the desired or final (C2) contration (and desired/final volume), this is the unknown factor

C2= desired contration you want
V2= Final volume you want.

Eg: I have a stock of 5 bacteria per ml (C1= 5cells/ml) and I want a final volume of 4 ml (V2) with 1bacterium in total (C2)

gives= 5 bacteria/ml X V1 = 1 bacterium/4ml X 4 ml ==> V1 = 0,2 ml of V1 added to a new test tube, which I then fill up to an end volume of 4ml.

Or: different question, you want a final concentration of 1 bacterium per 1ml in 4 ml total volume.
gives: 5 bacteria/ml X V1 = 1 bacterium/ml X 4 ml ==> V1 = 0,8 ml of V1 , filled up with 3.2 ml of water or broth or whatever it us you use.

Lets say we use ascacioc formula on this last one:

5 CFU/mL * X mL of this = (mL final solution+X mL that you add on top) * 1 CFU/mL

would give: 5 CFU/mL * X mL of this = (4mL final solution+X mL that you add on top) * 1 CFU/mL
gives: 5X = 4+X ==> X = 1

So you would add 1 ml of a 5 cfu/ml counting stock into 4ml ==> final volume would be 5 ml with 5 cells , meaning 5 cells/5ml , meaning 1cell/ml .
So it does seem to work.... however its not entirely correct and its not how this formula should be used in this case.
He wants a fixed volume, you dont archieve this here!
The concentration of cells is right, but thats not the issue here (it is, but we also need a fixed volume)
And its confusing wih the general formula C1V2=C2V2

And a common mistake here would be (if you memorise the formula and not understand it): the found X is 1 ,so they take 1 ml of the stock and fill this up to 4ml, meaning => 1ml stock + 3 ml water, meaning 5cells in total of 4ml, meaning 1 cell to much since you wanted 4 cells in total (1 cell per ml).

So my advice: completely forget this formula from ascacioc at this time.
Start by trying to understand what diluting is and then link this with C1V2=C2V2
(and figure out from where C1V2=C2V2 comes, hint moles) When you do this you will most likely never forget what it means and you will be able to play with it and make dilutions without problems.

It seems a bit harsh, but if you need a fixed endvolume for some type of reaction then often you do really need this end volume and realising you just made more then your tube can hold, could be a problem....

its very dangerous to start working with C1V1 = C2V2 and C1 x V1 = (V2 + V1) x C2 if you dont fully understand it.
Too many people here know the C1V1 = C2V and still arent able to use it correctly because they dont know what C1 V2 etc means!

So my advise: forget the formula , start reasoning and think about it. No point in using a formula you learned by hearth but didnt understand.

-pito-
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