• Create Account

# 10 millimolar (mM) to 250 micromolar

Bacteria CFU

53 replies to this topic

### #16 pito

pito

Veteran

• Active Members
• 1,033 posts
52
Excellent

Posted 29 August 2012 - 01:20 PM

ascacioc, on 29 August 2012 - 10:49 AM, 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.
I'll add the calculation later.
If you don't know it, then ask it! Better to ask and look foolish to some then not ask and stay stupid.

### #17 ascacioc

ascacioc

Veteran

• Global Moderators
• 271 posts
44
Excellent

Posted 29 August 2012 - 01:32 PM

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

### #18 ascacioc

ascacioc

Veteran

• Global Moderators
• 271 posts
44
Excellent

Posted 29 August 2012 - 02:17 PM

I just found the best explanation here on the forum:
http://www.protocol-...s-calculations/

### #19 ascacioc

ascacioc

Veteran

• Global Moderators
• 271 posts
44
Excellent

Posted 29 August 2012 - 02:18 PM

http://www.protocol-...s-calculations/ explained by the Admin

### #20 pito

pito

Veteran

• Active Members
• 1,033 posts
52
Excellent

Posted 30 August 2012 - 01:42 AM

ascacioc, on 29 August 2012 - 01:32 PM, 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:

Quote

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:

Quote

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.

Quote

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

Quote

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:

C1= concentration of your stock
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.
If you don't know it, then ask it! Better to ask and look foolish to some then not ask and stay stupid.

### #21 pito

pito

Veteran

• Active Members
• 1,033 posts
52
Excellent

Posted 30 August 2012 - 01:44 AM

ascacioc, on 29 August 2012 - 02:17 PM, said:

I just found the best explanation here on the forum:
http://www.protocol-...s-calculations/
This is +- what I mean.

I know that you are right in the end, but its not the best start.
The opening poster needs to understand it, not blindly fill in some formula he doesnt understand.

I'll try to add the calculations needed for this question later today or in the future days.
If you don't know it, then ask it! Better to ask and look foolish to some then not ask and stay stupid.

### #22 ascacioc

ascacioc

Veteran

• Global Moderators
• 271 posts
44
Excellent

Posted 30 August 2012 - 01:51 AM

Maybe you do another thread that we can pin on top of the general techniques forum for having it as a reference forever and ever and not have the same explanation repeated over and over (I am tired of typing) and then we can just refer to that thread, don't you think?

Andreea

### #23 ascacioc

ascacioc

Veteran

• Global Moderators
• 271 posts
44
Excellent

Posted 30 August 2012 - 02:01 AM

I merged the two threads because we are talking about the same kind of calculations (I hope this is not opposed by nobody involved)

PS: Indeed I made a mistake with the 4 uL; I missed a zero. Pito is right, 10x for pointing it out.

### #24 pito

pito

Veteran

• Active Members
• 1,033 posts
52
Excellent

Posted 30 August 2012 - 06:08 AM

ascacioc, on 30 August 2012 - 01:51 AM, said:

Maybe you do another thread that we can pin on top of the general techniques forum for having it as a reference forever and ever and not have the same explanation repeated over and over (I am tired of typing) and then we can just refer to that thread, don't you think?

Andreea

You are the moderator, I am not.
Do what you think is a good idea.
Perhaps its a good idea to make 1 general post with some simple examples in it and pin this somewhere so they can always check this post.

I can only tell you that I have seen hundreds of these kind of posts. People suffer a lot with elementary dilution problems. And almost everytime its the same problem: they have heard of a formula or they even know CV = CV , but have no idea how to use it or what it means.
If you don't know it, then ask it! Better to ask and look foolish to some then not ask and stay stupid.

### #25 prabhubct

prabhubct

Veteran

• Active Members
• 230 posts
14
Good

Posted 30 August 2012 - 07:44 AM

ascacioc, on 28 August 2012 - 12:04 PM, said:

For which final volume? Let us assume you want to have final volume 2 mL for your experiment. You pipette 2 mL = 2000 uL of saline/media whatever.. to which you add X uL of your bacteria at OD600 = 0.4;  X being the solution of the below:

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

2500x = 2000 + x
x = 2000/2499 uL (even though I wouldn't add this little amount < 1 uL) I would do some dilution series before, something like 1 to 100 in order to add 100 times the X volume.

Andreea

C1*V1 = C2* V2, Problem original stock conc. 2.5*10^9 CFU/mL  required conc. 10^6 CFU/ml

2.5*10^9 CFU/mL * X uL = 10^6 * 2000uL
so, X uL = (10^6 * 2000uL)/2.5*10^9
X    = 1.25 uL

I would add 1.25 uL of 2.5*10^9 CFU/mL to 1998.75 uL of saline / media to get 10^6 CFU/ml

I agree with Andreea that adding small volume to big may not be good idea why don't you go to back to stored serial dilutions and pick

X uL = (10^6 * 2000uL)/2.5*10^7
= 125 uL from 2.5*10^7 CFU/mL ( from 10^4 serial diluted sample) to 1875 uL of saline / media to get 10^6 CFU/ml

Edited by prabhubct, 30 August 2012 - 08:12 AM.

“Those who mind don't matter, and those who matter don't mind.”
--  Bernard M. Baruch

### #26 pito

pito

Veteran

• Active Members
• 1,033 posts
52
Excellent

Posted 30 August 2012 - 08:26 AM

prabhubct, on 30 August 2012 - 07:44 AM, said:

ascacioc, on 28 August 2012 - 12:04 PM, said:

For which final volume? Let us assume you want to have final volume 2 mL for your experiment. You pipette 2 mL = 2000 uL of saline/media whatever.. to which you add X uL of your bacteria at OD600 = 0.4;  X being the solution of the below:

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

2500x = 2000 + x
x = 2000/2499 uL (even though I wouldn't add this little amount < 1 uL) I would do some dilution series before, something like 1 to 100 in order to add 100 times the X volume.

Andreea

C1*V1 = C2* V2, Problem original stock conc. 2.5*10^9 CFU/mL  required conc. 10^6 CFU/ml

2.5*10^9 CFU/mL * X uL = 10^6 * 2000uL
so, X uL = (10^6 * 2000uL)/2.5*10^9
X    = 1.25 uL

I would add 1.25 uL of 2.5*10^9 CFU/mL to 1998.75 uL of saline / media to get 10^6 CFU/ml

I agree with Andreea that adding small volume to big may not be good idea why don't you go to back to stored serial dilutions and pick

X uL = (10^6 * 2000uL)/2.5*10^7
= 125 uL from 2.5*10^7 CFU/mL ( from 10^4 serial diluted sample) to 1875 uL of saline / media to get 10^6 CFU/ml

Pls check your calculations...

This is an excellent example of using a formula without thinking (= you use it on auto-pilot rather then reasoning, thinking the entire time) or checking it.

Also: in the second part of your post I dont see what you are trying to do.
You use 2.5*10^7cfu/ml ? but the conc is 2.5*10^9 cfu/ml in his stock ... not sure where you got this factor 100 difference. I know you can use the less diluted ones, but thats not the issue here and its confusing. In this you are changing the "stock" solution with the "diluted (100 times)" solution.
+you make the same mathematical mistake as in the first calculation.

Edited by pito, 30 August 2012 - 08:34 AM.

If you don't know it, then ask it! Better to ask and look foolish to some then not ask and stay stupid.

### #27 prabhubct

prabhubct

Veteran

• Active Members
• 230 posts
14
Good

Posted 30 August 2012 - 09:48 AM

ascacioc, on 28 August 2012 - 11:49 AM, said:

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

Andreea

pito, on 30 August 2012 - 01:42 AM, said:

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

Here I suppose that 100 µl of culture taken from 10^6 dilutions to get

(250colonies/100 µl)* 10^6 = 2.5 * 10^9 colonies/ml
i.e 2.5 * 10^9 CFU/ml

From Pito's post ""You use 2.5*10^7cfu/ml ? but the conc is 2.5*10^9 cfu/ml in his stock ... not sure where you got this factor 100 difference. I know you can use the less diluted ones, but thats not the issue here and its confusing. In this you are changing the "stock" solution with the "diluted (100 times)" solution."

Here I suppose changing stock to less diluted one will be good idea. I am not changing "stock" solution with the "diluted (100 times)" solution but to concentrated 100 times. you have 10^1 to 10^9 diluted series use 10^4 instead of 10^6 one to get 10^6 CFU/ml
“Those who mind don't matter, and those who matter don't mind.”
--  Bernard M. Baruch

### #28 pito

pito

Veteran

• Active Members
• 1,033 posts
52
Excellent

Posted 30 August 2012 - 10:46 AM

prabhubct, on 30 August 2012 - 09:48 AM, said:

ascacioc, on 28 August 2012 - 11:49 AM, said:

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

Andreea

pito, on 30 August 2012 - 01:42 AM, said:

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

Here I suppose that 100 µl of culture taken from 10^6 dilutions to get

(250colonies/100 µl)* 10^6 = 2.5 * 10^9 colonies/ml
i.e 2.5 * 10^9 CFU/ml

From Pito's post ""You use 2.5*10^7cfu/ml ? but the conc is 2.5*10^9 cfu/ml in his stock ... not sure where you got this factor 100 difference. I know you can use the less diluted ones, but thats not the issue here and its confusing. In this you are changing the "stock" solution with the "diluted (100 times)" solution."

Here I suppose changing stock to less diluted one will be good idea. I am not changing "stock" solution with the "diluted (100 times)" solution but to concentrated 100 times. you have 10^1 to 10^9 diluted series use 10^4 instead of 10^6 one to get 10^6 CFU/ml

You are mixing things!

Look back at your numbers.

You first state that you would take  1.25 uL from a stock solution that is, as you wrote it, original stock conc. 2.5*10^9 CFU/mL

Then you state 125 uL from 2.5*10^7 CFU/mL (which is correct) however you also state

Quote

Here I suppose changing stock to less diluted one will be good idea. I am not changing "stock" solution with the "diluted (100 times)" solution but to concentrated 100 times. you have 10^1 to 10^9 diluted series use 10^4 instead of 10^6 one to get 10^6 CFU/ml

going from 10^9 to 10^7 is not going from less concentrated to more concentrated....

Read the bleu text: changing stock to less diluted? Stock is always the "less diluted" one!
The stock is what you dilute.

Also: a problem here is that we speak about 10^9 and 10^7 and dilutions etc...
While its clear that you might be speaking about 10^-9 and 10^-7 because you mention the 10^4 diluted series.. thus it should be 10^-4 you are mentioning....

I think you are mixing things up because you are not using the "-" sign in your dilution series!

Also: check your math again
==>
2.5*10^9 CFU/mL * X uL = 10^6 * 2000uL
so, X uL = (10^6 * 2000uL)/2.5*10^9
X = 1.25 uL

This is NOT correct.

2.5*10^9 CFU/mL * X uL = 10^6 * 2000uL
==>
X uL = 10^6 * 2000uL/2.5*10^9 CFU/mL =/= 1.25 !

the numerator is smaller then the denominator, so how can you end up with 1.25? You see this immediatly by just looking at the numbers (thinking).

You should directly see that the numerator has 10^6 and the denominator  10^9 , so 1000 times bigger while the numerator has only 2000 left to multiply and the denomintor 2,5.....
If you don't know it, then ask it! Better to ask and look foolish to some then not ask and stay stupid.

### #29 prabhubct

prabhubct

Veteran

• Active Members
• 230 posts
14
Good

Posted 30 August 2012 - 11:09 AM

Thanks Pito for correcting math error,

2.5*10^9 CFU/mL * X uL = 10^6 * 2000uL
so, X uL = (10^6 * 2000uL)/2.5*10^9
= 0.8 uL
add 0.8 uL of 2.5*10^9 CFU/mL to 1999.2 uL of saline / media to get 10^6 CFU/ml

@ Pito going from 10^9 to 10^7 is not going from less concentrated to more concentrated....

Read the bleu text: changing stock to less diluted? Stock is always the "less diluted" one!
The stock is what you dilute.

I think we have serial dilution stock of 10^1 , 10^2.......10^9.
and 10^7 must be more concentrated original stock than 10^9 because its serially diluted.

@ pito : 0.8 uL from a stock solution that is, as you wrote it, original stock conc. 2.5*10^9 CFU/mL

Then you state 80 uL from 2.5*10^7 CFU/mL
I Mean to take 80 uL from original stock of 10^4 as you are having 2.5*10^7 CFU/mL in 10^4 stock.

This means 10^6 diltion stock is of 2.5*10^9 CFU/mL.
10^5 diltion stock is of 2.5*10^8 CFU/mL.
10^4 diltion stock is of 2.5*10^7 CFU/mL.
10^4 diltion stock is of 2.5*10^6 CFU/mL.
10^3 diltion stock is of 2.5*10^5 CFU/mL.

Edited by prabhubct, 30 August 2012 - 11:19 AM.

“Those who mind don't matter, and those who matter don't mind.”
--  Bernard M. Baruch

### #30 pito

pito

Veteran

• Active Members
• 1,033 posts
52
Excellent

Posted 30 August 2012 - 12:43 PM

prabhubct, on 30 August 2012 - 11:09 AM, said:

Thanks Pito for correcting math error,

2.5*10^9 CFU/mL * X uL = 10^6 * 2000uL
so, X uL = (10^6 * 2000uL)/2.5*10^9
= 0.8 uL
add 0.8 uL of 2.5*10^9 CFU/mL to 1999.2 uL of saline / media to get 10^6 CFU/ml

@ Pito going from 10^9 to 10^7 is not going from less concentrated to more concentrated....

Read the bleu text: changing stock to less diluted? Stock is always the "less diluted" one!
The stock is what you dilute.

I think we have serial dilution stock of 10^1 , 10^2.......10^9.
and 10^7 must be more concentrated original stock than 10^9 because its serially diluted.

@ pito : 0.8 uL from a stock solution that is, as you wrote it, original stock conc. 2.5*10^9 CFU/mL

Then you state 80 uL from 2.5*10^7 CFU/mL
I Mean to take 80 uL from original stock of 10^4 as you are having 2.5*10^7 CFU/mL in 10^4 stock.

This means 10^6 diltion stock is of 2.5*10^9 CFU/mL.
10^5 diltion stock is of 2.5*10^8 CFU/mL.
10^4 diltion stock is of 2.5*10^7 CFU/mL.
10^4 diltion stock is of 2.5*10^6 CFU/mL.
10^3 diltion stock is of 2.5*10^5 CFU/mL.

I see what you mean...

But this is exactly my point from the start...

You are working "the other way around" (and still mixing things up, your explenation is contradictory, see below)

Normally people work like this:

Stock -> 10^-1 -> 10^-2 -> 10^-3 -> and so on...

The opening poster was indeed doing this, however rather then writing down 10^-9 he wrote down 10^9 , which caused you to make this mistake.

Remember what he said in his opening post:

Quote

I got 250 colonies at 10*6 dilution. So I got 2.5 × 10*9 cfu/ml (original, if I am right
He goes from 250 colonies at 10^6 to 2.5 x 10^9 colonies per ml in the original one (plating out 0,1ml).
this is only possible if the 10^6 actually means 10^-6 !

See what I mean?

Your explenation is "correct" (the reasoning is) but its not right in the end....(+mixing things up)

Altough, I do find it weird you write this:

Quote

I think we have serial dilution stock of 10^1 , 10^2.......10^9.
and 10^7 must be more concentrated original stock than 10^9 because its serially diluted.

10^7 is more concetrated then 10^9? 10^7 is a lower number then 10^9.
Unless you mean that 10^7 means you diluted it 10^7 times and 10^9 means diluting 10^9 times... But I have never seen people work like this...
dilutions are always given in negative powers.

Also: if you link what you stated (the quouted part) with what you write afterwars:

Quote

This means 10^6 diltion stock is of 2.5*10^9 CFU/mL.
10^5 diltion stock is of 2.5*10^8 CFU/mL.
10^4 diltion stock is of 2.5*10^7 CFU/mL.
10^4 diltion stock is of 2.5*10^6 CFU/mL.
10^3 diltion stock is of 2.5*10^5 CFU/mL.

Then something went wrong again.

you are mixing things up over and over.
===> in the bold reasoning you state that a 10^5 dilution is less diluted then a 10^4 dilution, meaning 10^5 is more concentrated (more CFUs) then 10^4, but at the same time (see quoted part) you state that "10^7 must be more concentrated original stock than 10^9 because its serially diluted"

So what is it now?
Either "and 10^7 must be more concentrated original stock than 10^9 because its serially diluted." is correct or the last part (the bold part) of your post is the correct one.
But they cant be both right.

Things are becoming more and more complicated because the same mistakes are made over and over.
If you don't know it, then ask it! Better to ask and look foolish to some then not ask and stay stupid.