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 !!!
Thanks
Prao123

-prao-

QUOTE

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 ??

It's because affinity is different.

QUOTE

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??

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 :

http://www.protocol-online.org/forums/inde...725&#entry31725

-fred_33-

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 ???

-prao-

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

-aimikins-

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 ??

-prao-

well

QUOTE

1 mg/ml solution of my protein to be 20 nMols.

hence 100µg in 100µl is the same as 1mg/ml...

thanks to homebrew, i can recommend you this link that may help too.
http://www.graphpad.com/quickcalcs/Molarityform.cfm

-fred_33-

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.

mike

-jadefalcon-

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.
Prao

-prao-