Stupid calculation question-Plz help clarify - (Nov/25/2005 )
Hi I have 2 questions,
1) During Western Blotting sometimes we use 1:5000 dilution and sometimes we use 1:10,000 dilutions of the antibody (Ab), Is it bcos the Abs are having different affinities or is it bcos their stock concentrations are different ??
2) I have 2 nanoMolar solution of a particular protein and I have a 1 mg/ml stock solution but I have only 80 micro liters of the 1mg/ml solution (conc. is 2nanoMoles). How do I make a 80 microliters of 20 microMolar solution of that ??? Is that much stock solution sufficient??
Pls help !!!
may you clarify the question?
stock = 1mg/ml
Kd of protein = ?
with the mw of protein you can calculate the conc in moles per liter of your stock solution.
I recommend you to have a look at this topic :
Thanks a lot Fred33. I checked out the link and found some useful info in that but just be precise with my question the molecular weight is 50,000 Da and from that I calculated the concentration of 1 mg/ml solution of my protein to be 20 nMols. My PI had prepared a 1 ml. solution of the above concentration and used up much of the solution and now theres only about 80 microlitres of the solution present. I want 80 microliters of 20 micromolar solution. Is this possible ???
you have 80 ul of a 20 nanomolar solution and you want to turn it into 80 ul of a 20 micromolar solution?
I do not think this is possible
Ha!!! I thought so that it wouldnt be posible. But now suppose I have a compound that has a molecular weight of 50 KDa and I am dissolving 100 micrograms of this in 100 microlitres of solution. What will the molarity be now ??
hence 100µg in 100µl is the same as 1mg/ml...
thanks to homebrew, i can recommend you this link that may help too.
100µg per 100µl = 1µg per 1µl = 1mg per 1ml = 1g/l, ok?
50kDa = 50000 Da = 50000 U, meaning your protein has a molar mass of 50000g, so you'd have to dissolve 50000g (50kg) in one liter to get a 1M solution. but, as calculated above, you only dissolve the equivalent of 1g in 1 liter, so that just 1/50000th of a mol = 0.00002 mol/l = 0.02 mmol/l = 20 µmol/l = 20µM.
so, in the end, you get 100µl of a 20µM solution (20µM = 20µmol/l = 20nmol/ml = 20pmol/µl) (as you may have had all along ) .
to turn the 20µM solution into a 20nM solution, you'd have to dillute it 1:1000.
Thanks guys, Thanks a lot. I thought that I had a much concentrated solution in my hands and had done a mistake while calculating. Thanks Fred for the link and thanks Jadefalcon for the explanation.