53 replies to this topic

[cell_bee]

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-...-added-to-dmem/
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.

[pito]

cell_bee, on 30 August 2012 - 12:49 PM, said:

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-...-added-to-dmem/
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.

I forgot all about your question.

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

I'll try to answer your question in the following 2days.

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).

[ascacioc]

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

[prabhubct]

plate count
dilution factor
volume added to plate
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

[pito]

prabhubct, on 30 August 2012 - 10:46 PM, said:

plate count
dilution factor
volume added to plate
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.

This does not add up.


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?

[prabhubct]

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

Edited by prabhubct, 31 August 2012 - 02:27 AM.

[pito]

prabhubct, on 31 August 2012 - 02:25 AM, 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.

[prabhubct]

pito, on 31 August 2012 - 04:42 AM, said:

prabhubct, on 31 August 2012 - 02:25 AM, 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.

[chandch]

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

Edited by chandch, 31 August 2012 - 09:13 AM.

[pito]

chandch, on 31 August 2012 - 09:00 AM, 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?

[chandch]

pito, on 31 August 2012 - 12:55 PM, said:

chandch, on 31 August 2012 - 09:00 AM, 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?

Hi Thank you.  I really don't how to calculate concentrations. I have taken as this example.  Again I have one question, if the above answer is right how can we add 4µl into 9,996 ml. Because for most of experiments I do O.D reading 0.4 and from that I have to dilute it down to 1 X 10 ^6 or 10^7. And I have to take 1ml for my experiments
What is the best way? I know it's difficult if we want add 4µl into 9.996 ml (if we take this as example) Please help.

One person taught me that if I get 1.5 X  10^8 CFU/ml (original concentration), I should take 1 ml original stock and add 9ml water so it makes 10^7cfu/ml. Is it right? again taking 1 ml from this and add 9ml water, so it makes 10^6 cfu/ml. Is it right?

Thank you

[pito]

chandch, on 31 August 2012 - 01:11 PM, said:

pito, on 31 August 2012 - 12:55 PM, said:

chandch, on 31 August 2012 - 09:00 AM, 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?

Hi Thank you.  I really don't how to calculate concentrations. I have taken as this example.  Again I have one question, if the above answer is right how can we add 4µl into 9,996 ml. Because for most of experiments I do O.D reading 0.4 and from that I have to dilute it down to 1 X 10 ^6 or 10^7. And I have to take 1ml for my experiments
What is the best way? I know it's difficult if we want add 4µl into 9.996 ml (if we take this as example) Please help.

One person taught me that if I get 1.5 X  10^8 CFU/ml (original concentration), I should take 1 ml original stock and add 9ml water so it makes 10^7cfu/ml. Is it right? again taking 1 ml from this and add 9ml water, so it makes 10^6 cfu/ml. Is it right?

Thank you

If you take 1 ml of 1.5 X 10^8 CFU/ml and you add 9ml dilutant, you end up with 1.5 X 10^7 CFU/ml , you dilulted 10 times.

and you can indeed dilute it one more time and end up with 1.5 X 10^6 CFU/ml


And did you read my question about the 4µl and 9,996 ml?

[chandch]

pito, on 31 August 2012 - 01:22 PM, said:

chandch, on 31 August 2012 - 01:11 PM, said:

pito, on 31 August 2012 - 12:55 PM, said:

chandch, on 31 August 2012 - 09:00 AM, 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?

Hi Thank you.  I really don't how to calculate concentrations. I have taken as this example.  Again I have one question, if the above answer is right how can we add 4µl into 9,996 ml. Because for most of experiments I do O.D reading 0.4 and from that I have to dilute it down to 1 X 10 ^6 or 10^7. And I have to take 1ml for my experiments
What is the best way? I know it's difficult if we want add 4µl into 9.996 ml (if we take this as example) Please help.

One person taught me that if I get 1.5 X  10^8 CFU/ml (original concentration), I should take 1 ml original stock and add 9ml water so it makes 10^7cfu/ml. Is it right? again taking 1 ml from this and add 9ml water, so it makes 10^6 cfu/ml. Is it right?

Thank you

If you take 1 ml of 1.5 X 10^8 CFU/ml and you add 9ml dilutant, you end up with 1.5 X 10^7 CFU/ml , you dilulted 10 times.

and you can indeed dilute it one more time and end up with 1.5 X 10^6 CFU/ml


And did you read my question about the 4µl and 9,996 ml?

Thank you. Yes. I think it's difficult to add 4µl to 9,996 ml saline. Is there any way to do this?

[cell_bee]

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?

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[pito]

chandch, on 31 August 2012 - 01:28 PM, said:

pito, on 31 August 2012 - 01:22 PM, said:

chandch, on 31 August 2012 - 01:11 PM, said:

pito, on 31 August 2012 - 12:55 PM, said:

chandch, on 31 August 2012 - 09:00 AM, 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?

Hi Thank you.  I really don't how to calculate concentrations. I have taken as this example.  Again I have one question, if the above answer is right how can we add 4µl into 9,996 ml. Because for most of experiments I do O.D reading 0.4 and from that I have to dilute it down to 1 X 10 ^6 or 10^7. And I have to take 1ml for my experiments
What is the best way? I know it's difficult if we want add 4µl into 9.996 ml (if we take this as example) Please help.

One person taught me that if I get 1.5 X  10^8 CFU/ml (original concentration), I should take 1 ml original stock and add 9ml water so it makes 10^7cfu/ml. Is it right? again taking 1 ml from this and add 9ml water, so it makes 10^6 cfu/ml. Is it right?

Thank you

If you take 1 ml of 1.5 X 10^8 CFU/ml and you add 9ml dilutant, you end up with 1.5 X 10^7 CFU/ml , you dilulted 10 times.

and you can indeed dilute it one more time and end up with 1.5 X 10^6 CFU/ml


And did you read my question about the 4µl and 9,996 ml?

Thank you. Yes. I think it's difficult to add 4µl to 9,996 ml saline. Is there any way to do this?

Sorry for the late reply, but you need to make in between solutions...

for example: if you need a 1/100 dilution, its best to make a 1/10 dilution and dilute this again 10 times... to avoid the need to add a very small amount to a very large amount.