Convert uM/L to ng

Hi - I am trying to dilute primers for a qPCR reaction. I'm a bit lost. Here is an example of one of my primers:

Came as 97.7 nmol to which I added 977 ul ... I believe this gives me a stock solution of 100 uM/L

I need to figure out how much of this solution I should use so that I am adding 0.5ng of primer to each reaction.

Any help would be much appreciated!

[JohnS]

Originally Posted by Unregistered

Hi - I am trying to dilute primers for a qPCR reaction. I'm a bit lost. Here is an example of one of my primers:

Came as 97.7 nmol to which I added 977 ul ... I believe this gives me a stock solution of 100 uM/L

I need to figure out how much of this solution I should use so that I am adding 0.5ng of primer to each reaction.

Any help would be much appreciated!

M (Molar) is shorthand for mole per liter, so you have 100 µM or 100µmol/L. M/L is redundant.

However, any conversion between grams and moles requires knowing the molecular weight of the desired species, which is also the molar mass in grams per mole. Without it, the conversion is not possible, so you need to find it in a reference or compute it from the chemical formula.

Thanks - I do have the MW in g/mol, for this sample it is 6357.2 g/mol

[JohnS]

Originally Posted by Unregistered

Thanks - I do have the MW in g/mol, for this sample it is 6357.2 g/mol

So your 100 µmol/L solution is
100 µmol/L x 6357.2 g/mol =635.72 mg/L

You want 0.5 ng so

0.5 ng x 1 L/635.72 mg = 0.7865 nL

I have no idea how you would measure that and you may wish to start by diluting your stock solution to result in a more convenient to measure amount.

Okay - does this logic work out:

100umol/L stock diluted 1:100 = 100nmol/mL working solution

100nmol/mL working solution x MW (6357.2g/mol)/1000/1000 = 0.64 ng/ul

ul needed for 0.5ng = 0.5/0.64 = 0.79ul

[JohnS]

Originally Posted by Unregistered

Okay - does this logic work out:

100umol/L stock diluted 1:100 = 100nmol/mL working solution

100nmol/mL working solution x MW (6357.2g/mol)/1000/1000 = 0.64 ng/ul

ul needed for 0.5ng = 0.5/0.64 = 0.79ul

100 µmol/L is 100 nmol/mL without dilution (numerator and denominator both divided by 1000.

If you dilute 100:1, you get 1 µmol/L and you would need 79 nL, probably still too small to measure.

If you dilute stock by 1000, you get 100 nmol/L, and 790 nL or 0.79 µL.

Perhaps Mrs X will wander through and assist. She is a chemist. I can do the conversions but don't have the practical experience as I am not a chemist.

[Mrs X]

Originally Posted by JohnS

100 µmol/L is 100 nmol/mL without dilution (numerator and denominator both divided by 1000.

If you dilute 100:1, you get 1 µmol/L and you would need 79 nL, probably still too small to measure.

If you dilute stock by 1000, you get 100 nmol/L, and 790 nL or 0.79 µL.

Perhaps Mrs X will wander through and assist. She is a chemist. I can do the conversions but don't have the practical experience as I am not a chemist.

Sorry i didn't answer this earlier, brain not working properly today.

Probably easier to do two sequential dilutions, 1:100, then 1:100 again. (10µL of stock made up to 1mL, then again 10µL of this second solution made up to 1mL)

This would mean you need to take a 7.9µL aliquot of the last solution to get 5ng. It is a bit easier to dial up 7.9µL on an eppindorf than 0.79µL.

Note, brain really isn't functioning, check these calculations before using.