Entirely different question then -- if you want 200 ng/uL of a primer solution in this case, and you are starting with a 100 uM solution of a molecule with MW of 9281 (or, in other words: 9281 g/mole), you can do the following:

Since 100 uM = 100 umoles/L

And 100 umoles/L = 0.0001 moles/L

and, since moles = g/MW, (0.0001 moles/L)*9281 g/mole = 0.9281 g/L

which = 928100000 ng/L, which = 928.1 ng/uL

So, to get 200 ng/uL solution from a 928.1 ng/uL solution,

you need to dilute by a factor of (928.1/200) = 1:4.6405

So, to prepare 100 uL of 200 ng/uL solution from your 100 uM stock:

add 21.549402 uL of 100 uM stock + 78.4506 water = 200 ng/uL primer solution.

Hope this helps.

Add 5 uL of the 100 uM stock to 111.0125 uL of diluent (nuclease-free water or TE pH 8.0), then take 5 uL of that resulting solution and dry it down.  That will give you 200 ng of this primer.

Thank you. but what´s the logic ?

Actually I need 200 ng/ul to perform T4 kinase 3´ primer labeling with gama-ATP. I just don´t get it how to take 200 ng from the 100 uM primer Stock Solution.

Entirely different question then -- if you want 200 ng/uL of a primer solution in this case, and you are starting with a 100 uM solution of a molecule with MW of 9281 (or, in other words: 9281 g/mole), you can do the following:

Since 100 uM = 100 umoles/L

And 100 umoles/L = 0.0001 moles/L

and, since moles = g/MW, (0.0001 moles/L)*9281 g/mole = 0.9281 g/L

which = 928100000 ng/L, which = 928.1 ng/uL

So, to get 200 ng/uL solution from a 928.1 ng/uL solution,

you need to dilute by a factor of (928.1/200) = 1:4.6405

So, to prepare 100 uL of 200 ng/uL solution from your 100 uM stock:

add 21.549402 uL of 100 uM stock + 78.4506 water = 200 ng/uL primer solution.

Hope this helps.

Thank you very much!!! This helps a lot!

Glad it helped Rodrigo - and please forgive the impossible number of decimal places I used to get the point across ;]