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# equimolar amounts

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### #1 claritylight

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Posted 24 December 2009 - 05:20 AM

Hi, can someone show me how to do this:

If I have 5 tubes with the following concentrations, how do I get them so that I have a new tube that is 10uM final concentration, 100ul final volume? Is that even possible? Thanks.

Tube
1. 100uM
2. 100uM
3. 20uM
4. 10uM
5. 15uM

### #2 phage434

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Posted 24 December 2009 - 05:26 AM

I assume you mean 10 mM final concentration of each tube's material. It's not possible, since you have only a 10 mM concentration of tube 4. If you add any amount of liquid to tube 4, the molarity of its component will go down, below your desired final concentration.

### #3 claritylight

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Posted 24 December 2009 - 06:13 AM

I assume you mean 10 mM final concentration of each tube's material. It's not possible, since you have only a 10 mM concentration of tube 4. If you add any amount of liquid to tube 4, the molarity of its component will go down, below your desired final concentration.

Tube 4 has 10uM concentration.

True, if I add any amount of liquid to tube 4, the final concentration of tube 4 will go down. So how would I do it if I wanted the final concentration to just be greater than 1uM. Is it possible then?

### #4 Gerard

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Posted 24 December 2009 - 06:54 AM

What is the volume of a tube?

take 10 ul of tube 1 and add 90 ul liquid or take 100 ul of tube 4
Ockham's razor
Pluralitas non est ponenda sine necessitate
-- "You must assume no plural without necessity".

### #5 claritylight

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Posted 24 December 2009 - 07:06 AM

What is the volume of a tube?

take 10 ul of tube 1 and add 90 ul liquid or take 100 ul of tube 4

The volumes of the 5 sample tubes have 20ul each of liquid.

Not sure what you mean.

I want a new tube that where all the samples are pooled and represented in equimolar amounts in this new tube, so the final concentration of the new tube is at least greater than 1uM (and where each sample is at least greater than 1uM), and new tube has 100ul max final volume (it can be 25ul final volume at the lowest).

Is that possible?

### #6 claritylight

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Posted 24 December 2009 - 07:36 AM

Never mind. I figured it out..below if anyone is interested.

Tube
1. 100uM
2. 100uM
3. 20uM
4. 10uM
5. 15uM

Tube 6 - empty new tube

Solution: Take 1.25ul from tube 1 and 1.25ul from tube 2 and place into tube 6. Take 6.25ul from tube 3 and place into tube 6, take 12.5ul from tube 4 and place into tube 6, take 8.33ul from tube 5 and place into tube 6. This gives final volume as 29.58ul. Add 0.42 ul water for final volume of 30ul (nice round number).

Now, I'll have equimolar amounts of each sample (~4.1uM in the end) mixed into tube 6, at final volume of 30ul. That's what I was asking.

My next question is the final concentration of tube 6 going to be 4.1uM or will it be greater than that?

Thanks.

### #7 Gerard

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Posted 24 December 2009 - 08:20 AM

tube 1 100umol/l * 1.25 ul = 125 pmol
Every tube wil deliver 125 pmol that times 5 gives a total 625 pmol/30 ul ~ 20,8 pmol/ul ~ 20.8 nmol/ml ~20,8 um/l
Ockham's razor
Pluralitas non est ponenda sine necessitate
-- "You must assume no plural without necessity".

### #8 claritylight

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Posted 24 December 2009 - 09:03 AM

tube 1 100umol/l * 1.25 ul = 125 pmol
Every tube wil deliver 125 pmol that times 5 gives a total 625 pmol/30 ul ~ 20,8 pmol/ul ~ 20.8 nmol/ml ~20,8 um/l

thanks!