How to determine final amount of 20 ng protein per WB well - (Oct/31/2009 )

Hello

I am not very good with calculations. I used the nanodrop instrument to determine my sample protein concentration as 2 mg/ml. I would like to add 20 ul into each well such that the final protein amount is 20 ng (per one well). Would any one please show in a step by step manner the calculation to derive this (I am confused with mg, ug, ng and the amount per well).

Since I need to load the protein in duplicate or triplicate, what would the calculations be for three wells: i.e 20 ul per well and there are three wells and the protein amount is still 20 ng per well. Remember, my original stock is 2 mg/ml. If this is too low, just use a suitable original concentration as an example. Please, can any one help me with this, as I am rather confused.

Thanks
lab2009

go here, here or any other of the similar concentration/dilution threads around.

Remember to keep your units the same - if you are working with ng keep all units as ng (e.g. ng/ml)!

Hello bob1

Thank you for the equation. Please refer to the example by biogirl1230.
Do you consider the 50ug in 100ul final volume (i.e 5ul stock + 95ul water) as the working solution? (This is an example)

For a western blot, if I add 20 ul from the working solution into one well, would the amount be 10ug (i.e 1/5). Is this correct?

lab2009

lab2009 on Oct 31 2009, 09:45 PM said:

20ng/20ul is nothing but 1ug/ml
Dilute your stock (2mg/ml to 1 ug/ml) and aliquot 20 ul in each well and u have 20 ng in each well!!!
If you want triplicates from the same preparation, prepare atleast 60 ul of 1ug/ml and load it three times. If you want separate preparations, then prepare 20ul and load!!!
Best luck!!
Hope its clear!!

Are you sure that you just want to apply 20ng on the gel and not 20µg?
2mg/ml is the same like 2µg/µl. So considering that we need to apply 20µg we have to add 10µl of your protein solution to each well.

lab2009 on Nov 2 2009, 12:26 AM said:

Hi there, that old post from biogirl1230 is actually wrong.

The equation for concentrations is definitely right C1V1=C2V2, but in that old post there is a confusion between concentrations and total µg. On that particular example biogirl1230 should have taken 50µl of her original stock + 50µl of water, for her to have 50µg of DNA in 100µl of solution (ie. 0.5µg/µl = 50µg in 100µ).

For your current question, I agree with Pradeep. 20ng in 20µl is no other than 1ng/µl = 1µg/ml, so if your stock is at 2mg/ml you need to dilute it 1:2000 to have a 1µg/ml solution. You can now take 20µl from this solution to have your desired amount of protein.

Use C1V1=C2V2
Now, remember that mg/ml = µg/µl so 2mg/ml = 2µg/µl and 1µg/ml = 1ng/µl (or 0.001µg/µl)

So in this case you need: V1= (0.001µg/µl * V2)/2µg/µl
for example, to prepare 1ml (1000µl) at 1µg/ml, you need: V1= (0.001*1000)2= 0.5µl of stock concentartion + 999.5µl H2O (or buffer)

As this are really low volumes it might be worth considering doing serial dilutions, for example dilute your sample 1:100 first, and then dilute this new sample 1:20.

Hope this helps.

Thank you all for helping me. The explanation is clearer.

lab2009

Sorry, should have checked that the answer was right... just knew that there had been a few threads earlier in the year.

lab2009 on Oct 31 2009, 07:15 AM said:

Just keep in mind 1mg/ml =1ug/ul=1ng/nl
You have 2ng/nl. If you want 20ng that is 10 times that amount of ng and 10 times that amount of nanoliters, ie 10nl.

As you can't probably pippete less than 1ul I recommend you to dilute 1 ul of your sample in 100ul of buffer and load 1ul of that.

I recommend doing the less serial dilutions as possible as low concentrated proteins adhere to surfaces leading to incorrect dilutions.