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

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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 am getting confused with the formula used and where did i get v2- 12 ml. I dont think this is making any sense to me.

-cell_bee-

cell_bee on Thu Aug 30 20:49:23 2012 said:

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 am getting confused with the formula used and where did i get v2- 12 ml. I dont think this is making any sense to me.

Just forgot that formula ... there is no point in it if you dont understand the math/logic behind it.

Anyway: I want you to write down 2 things => first you write down everything you know (write down what you know with the correct units and what it means + the meaning of those things, what are moles ? What is MW ? What do you do to go from not diluted to 10^-3) and secondly: write down what you need (need to calculate).

-pito-

I took 12 mL as an example of final volume (totally random) As Pito says: state everything you know about your system: final volume would be something important to mention

-ascacioc-

plate count
dilution factor
overall dilution factor
sample concentration
250
106
0.1
107
2.50 x 109

Reference-- http://www.mansfield.ohio-state.edu/~sabedon/biol4039.htm

C1V1 = C2V2
2.50 x 109 * X = 106 * 10ml

X = 10^7/ 2.5* 10^9
= 0.4 * 10^-2
= 0.004ml
= 4 ul

Dilution factor is written as 10^6 with positive and concentration factor as 10^-6

-Inbox-

prabhubct on Fri Aug 31 06:46:21 2012 said:

plate count
dilution factor
overall dilution factor
sample concentration
250
106
0.1
107
2.50 x 109

Reference-- http://www.mansfield...on/biol4039.htm

C1V1 = C2V2
2.50 x 109 * X = 106 * 10ml

X = 10^7/ 2.5* 10^9
= 0.4 * 10^-2
= 0.004ml
= 4 ul

Dilution factor is written as 10^6 with positive and concentration factor as 10^-6

know what you mean, but you are not reading what I wrote down.

You state:

"and 10^7 must be more concentrated original stock than 10^9 because its serially diluted"
and at the same time:
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.

If the dilution factor is written with a positive power then both your sentences are about dilution factors? And thus you are making a mistake in one of them.

==> the higher the dilution factor, the lower the concentration!

Or?

-pito-

It was mistake.
I think
This means
10^9 dilution stock is of 2.5*10^6 CFU/mL.
10^8 dilution stock is of 2.5*10^7 CFU/mL.
10^7 diltion stock is of 2.5*10^8 CFU/mL.
10^6 dilution stock is of 2.5*10^9 CFU/mL.
10^5 dilution stock is of 2.5*10^10 CFU/mL.
10^4 dilution stock is of 2.5*10^11 CFU/mL.
10^3 dilution stock is of 2.5*10^12 CFU/mL.

10^8 dilution stock is of 2.5*10^7 CFU/mL.
Add 40ul of 10^8 dilution stock is of 2.5*10^7 CFU/mL. to 9960ul to make upto 10ml.

For cross checking check below,
2.5*10^7 CFU/ml ------ 1000 ul
so Y ------ 40 ul

Y = 10^9/ 10^3
= 10^6CFU/ml

-Inbox-

prabhubct on Fri Aug 31 10:25:38 2012 said:

It was mistake.
I think
This means
10^9 dilution stock is of 2.5*10^6 CFU/mL.
10^8 dilution stock is of 2.5*10^7 CFU/mL.
10^7 diltion stock is of 2.5*10^8 CFU/mL.
10^6 dilution stock is of 2.5*10^9 CFU/mL.
10^5 dilution stock is of 2.5*10^10 CFU/mL.
10^4 dilution stock is of 2.5*10^11 CFU/mL.
10^3 dilution stock is of 2.5*10^12 CFU/mL.

10^8 dilution stock is of 2.5*10^7 CFU/mL.
Add 40ul of 10^8 dilution stock is of 2.5*10^7 CFU/mL. to 9960ul to make upto 10ml.

For cross checking check below,
2.5*10^7 CFU/ml ------ 1000 ul
so Y ------ 40 ul

Y = 10^9/ 10^3
= 10^6CFU/ml

Ok, I understand this.
But you see now how difficult it can be if you dont define what you mean from the start and if you dont use the powers correctly.

-pito-

pito on Fri Aug 31 12:42:48 2012 said:

prabhubct on Fri Aug 31 10:25:38 2012 said:

It was mistake.
I think
This means
10^9 dilution stock is of 2.5*10^6 CFU/mL.
10^8 dilution stock is of 2.5*10^7 CFU/mL.
10^7 diltion stock is of 2.5*10^8 CFU/mL.
10^6 dilution stock is of 2.5*10^9 CFU/mL.
10^5 dilution stock is of 2.5*10^10 CFU/mL.
10^4 dilution stock is of 2.5*10^11 CFU/mL.
10^3 dilution stock is of 2.5*10^12 CFU/mL.

10^8 dilution stock is of 2.5*10^7 CFU/mL.
Add 40ul of 10^8 dilution stock is of 2.5*10^7 CFU/mL. to 9960ul to make upto 10ml.

For cross checking check below,
2.5*10^7 CFU/ml ------ 1000 ul
so Y ------ 40 ul

Y = 10^9/ 10^3
= 10^6CFU/ml

Ok, I understand this.
But you see now how difficult it can be if you dont define what you mean from the start and if you dont use the powers correctly.

Yeah Pito, Its quiet simple dilution problem. just I missed power (-Ve, +ve confusion) and some wrong math in hurry. your critical remarks do helped a lot.

-Inbox-

Hi
I tried to understand your message (from point 3), but not sure. I will do it here to make sure I am doing in right way...

original concentration (O.D 0.4)
(Did serial dilution 10^1 to 10^9)
250 colonies at 10^6 dilution (100 ul)
In 1 ml 2500 colonies at 10^ 6 dilution. So it's 2500 X 10^6 cfu/ml
I need 1 X 10^6 cfu/ml. So I need to dilute 2500 times (I need 10 ml as total volume)
1/2500 dilution means 1 + 2499
1/250 + 2499/250 = 0.004 + 9.996 ml

Do I need to take 4ul from original stock
Am I right? Please tell me

-chandch-

chandch on Fri Aug 31 17:00:46 2012 said:

Hi
I tried to understand your message (from point 3), but not sure. I will do it here to make sure I am doing in right way...

original concentration (O.D 0.4)
(Did serial dilution 10^1 to 10^9)
250 colonies at 10^6 dilution (100 ul)
In 1 ml 2500 colonies at 10^ 6 dilution. So it's 2500 X 10^6 cfu/ml
I need 1 X 10^6 cfu/ml. So I need to dilute 2500 times (I need 10 ml as total volume)
1/2500 dilution means 1 + 2499
1/250 + 2499/250 = 0.004 + 9.996 ml

Do I need to take 4ul from original stock
Am I right? Please tell me

Yes.
You are right.

However 1 more question: would you really do it like this?
Would you add 4µl into 9,996 ml?

-pito-
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