# Dilution Problem! - Making up molar solutions.... (Aug/18/2008 )

Hi all,

I have been having this problem for the past so many days... and i am soo dumb that I cant really understand how to go about doing it...

My stock solution is 500mM. My required concentration in 200nM. And thereafetr I need to do a serial dilution of 100nM, 50nM, 20nM, 10nM, 100pM and 10pM.

The molecular weight of my protein is 1024.

Please please please help me with this, I know these r basics and I have had trouble understanding this before also.. Please explain it step by step.. Thank you.

Look here. The link in it maybe is helpful for you!

http://www.protocol-online.org/forums/inde...st&p=134196

hi,

i will try to resolve this exercice

whatever the MW of your protein, the stock is at 500 mM

1000x dilution makes a 500 nM solution (prefer 2 successive dilutions, one 100x and the other 10x in order to limit pipeting error), then dilute 2.5x to obtain a 200 nM solution

dilute 2x the last solution for 100 nM solution

dilute 2x the 100 nM solution to make a 50 nM solution

dilute 10x the 200 nM solution to obtain a 20 nM solution

and then dilute 2x the 20 nM to 10 nM sol

finally, dilute 100x the 10 nM to 100 pM

and 10x the 100 pM to 10 pM

it could not be simpler!

Hi, thanks for the reply.. Although i am not able to understand the first part of the dilution... I have 10ul aliquots of the 500uM solution, sorry its not mM.... sorry again.... So to how much of this solution do I hav to dilut it 1000x??? Ccan you pleas give me in ul as wel.. sorry idont quite understand the logic behind it... SORRY.. and THANKS A LOT.....

i will try to resolve this exercice

whatever the MW of your protein, the stock is at 500 mM

1000x dilution makes a 500 nM solution (prefer 2 successive dilutions, one 100x and the other 10x in order to limit pipeting error), then dilute 2.5x to obtain a 200 nM solution

dilute 2x the last solution for 100 nM solution

dilute 2x the 100 nM solution to make a 50 nM solution

dilute 10x the 200 nM solution to obtain a 20 nM solution

and then dilute 2x the 20 nM to 10 nM sol

finally, dilute 100x the 10 nM to 100 pM

and 10x the 100 pM to 10 pM

it could not be simpler!

Hi!

the amount you dilute depends on how much of the dilutions you need.

lets see if I can help:

500µM diluted to a 1000 is 500nM, you could pippete 1+999 H20 or to avoid pippeting errors dilute 1:100 first and then 1:10 as folllows:

Mix 1 2 (stock) + 198 H20= 5 µM

Mix 2 20 out of Mix 1 + 180 H2= =0.5µM =500nM

Mix 3 80 out of mix 2 + 120 H20= 200 nM

Mix 4 100 out of Mix 3 + 100 = 100 nM

Mix 5 100 out of mix 4 + 100= 50nM

Mix 6 80 out of mix 5 + 120 = 20nM

Mix 7 100 out of mix 6 + 100= 10nM

Mix 8 2 out of mix 7 + 198 H20= 100 pM=0.1 nM

Mix 9 20 out of mix 8 + 180= 10 pM

hope this helps

If you dont need so much rest dilutions you can half everything I wrote down, the ratios are the same ull just have 100 µl dilutions instead of 200 µl ones.

µ is 10-6 and n is 10-9, thats why you start of with a 1:1000 dilution, nano is a 1000 times smaller

That was perrrrrrrrrfect!!! Thanks a lot...

I have this other doubt... Now that I know to carry out the dilutions, I wud like to know how to go baout doing this...

I am doing a cell culture xperiment... I add 2000 cells per well in a 96well plate... To this i add 50ul media +10ul of my peptide (whose dilution I just asked) + 5ul of an antibody...

I would like to know how to calculate the dilution according to the bath volume.. Because, I might add 10ul of my peptide to a well volume of 55ul and sometimes just 50ul... Wud this interfere with my dilutions that I have just made up??

I have dilutions from 200nM to 10pM... so for exampl, if I add 10ul of 200nM into a 55ul or 50ul well volume, wud it make a differece????

Please help me and thanks to all who have explained their best.. Thanks guys, Bioforum rocks...

the amount you dilute depends on how much of the dilutions you need.

lets see if I can help:

500µM diluted to a 1000 is 500nM, you could pippete 1+999 H20 or to avoid pippeting errors dilute 1:100 first and then 1:10 as folllows:

Mix 1 2 (stock) + 198 H20= 5 µM

Mix 2 20 out of Mix 1 + 180 H2= =0.5µM =500nM

Mix 3 80 out of mix 2 + 120 H20= 200 nM

Mix 4 100 out of Mix 3 + 100 = 100 nM

Mix 5 100 out of mix 4 + 100= 50nM

Mix 6 80 out of mix 5 + 120 = 20nM

Mix 7 100 out of mix 6 + 100= 10nM

Mix 8 2 out of mix 7 + 198 H20= 100 pM=0.1 nM

Mix 9 20 out of mix 8 + 180= 10 pM

hope this helps

If you dont need so much rest dilutions you can half everything I wrote down, the ratios are the same ull just have 100 µl dilutions instead of 200 µl ones.

µ is 10-6 and n is 10-9, thats why you start of with a 1:1000 dilution, nano is a 1000 times smaller

Uhm so youre saying the your END concentrations have to be 200nM, 100nM, etc?

Or......do you have to add 10 microliters of those different dilutions to your reaction?

In any case yes, you are diluting them a bit more.

If you have a solution that is 200nM thats 200 nmol per liter that means 200 nm pero 1000 microliters. If you are taking 10 microliters out of that you are taking 2nmol and putting that in a new volume 50+5+10 and yeah diluting it.

If it means that the end concentrations have to be a certain way you couldnt really have a fixed volume that you have to add. It would depend. Say you had 200nm solution and the end should be 100nm that would be a 1:2 dilution, half of your total volume would have to be that 200nm solution, no way.

So I would say that you make those dilutions and add 10 microliters of each to your reaction, even though you are diluting them a bit more it doesnt matter, you are diluting everyone of them in the same volume (65 microliters) and the ratios between dilutions stay that same.

Hope your experiment works!

Hi,

Thanks a lot for the reply... where were u all these days????????????????????????? h ehe hehe...

One final question and thats it!!!

Like you said, yeah I am diluting all of it in the same volume so the ratio will be same...

Can u tell me how to calculate for the end concentration??

tat is, my stock is 500uM, and i need to add 200nM, 100nM, 50nM etc etc all these concentrations are the end concentration... my well volume is 50 or 55 and i need to add 10ul of this concentrated solution... I hope i made some sense...

Please help me out.. thanks a lot.... U r a star!!

Or......do you have to add 10 microliters of those different dilutions to your reaction?

In any case yes, you are diluting them a bit more.

If you have a solution that is 200nM thats 200 nmol per liter that means 200 nm pero 1000 microliters. If you are taking 10 microliters out of that you are taking 2nmol and putting that in a new volume 50+5+10 and yeah diluting it.

If it means that the end concentrations have to be a certain way you couldnt really have a fixed volume that you have to add. It would depend. Say you had 200nm solution and the end should be 100nm that would be a 1:2 dilution, half of your total volume would have to be that 200nm solution, no way.

So I would say that you make those dilutions and add 10 microliters of each to your reaction, even though you are diluting them a bit more it doesnt matter, you are diluting everyone of them in the same volume (65 microliters) and the ratios between dilutions stay that same.

Hope your experiment works!

Sometimes calculations can be tiring, at least for me.

So I usually check my results with an online calculator.

http://www.graphpad.com/quickcalcs/Molarityform.cfm

try it.