primer concentration (calculation problem) - (Aug/09/2006 )
hye everyone... i need some help in calculation. i tried to solve it but it seems that my brain is not working..its kinda my last resort..
i have a primer which is:
conc: 100 uM working solution of primer
no bases: 23
it only state this. does it means 344 ug/ul?
due to high concentration of primer my professor said that i need to decrease the primer concentration to:
and its been 2days i don't have any idea how to solve this. i did some research on how to calculate but it doesn't work. all i can get after i did some calculation it will end up with pml which is too small of volume. please help.. thank you.
you have a 100 µM solution, right?
you have around 500 µL ?
as well as I understand, there is 344 µg of primer in your tube. i.e. 49.44 nmol
the molecular weight is 6958, so 344 µg is 344/6958=0.049 µmol or 49 nmol.
If your solution is 100 µM, then the volume is 494.4 µL.
now, how to convert 100 µM into µg/µL
you have 100 µmole in 1L
you know that 1 µmol is 6958 µg (molecular weight)
it means you have 695800 µg in 1L
or 0.696 µg in 1µL
the concentration of your solution is 0.696 µg/µL
you need to dilute it to get a 0.5µg/µL solution.
the dilution factor is 0.696/0.5 = 1.392
take 490µL and add (490*1.392)-490 = 192 µL
you have also a toolkit here...
I have a some what similar question. If you have 49.6 kDa protein and the concentration is 1 mg/ml how to calculate the uM weight of the protein. I do not know how to post a new post so sorry I have the do this way. Thanks
In general, first question is about dilutions and the second one about molar concentration, but in a way they overlap. Two lectures I have listed contain all information needed to solve these questions.
i've used the protein section of biomath putting 49.6 in kD and 1 mg (ml) and just clicked calculate.
The third box filled with 20.16pmoles
So in 1ml you have 20.16pm
that are 20.16nmolar.