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How to determine final amount of 20 ng protein per WB well

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#1 lab2009

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Posted 31 October 2009 - 07:15 AM

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

#2 bob1

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Posted 01 November 2009 - 03:49 PM

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)!

Edited by bob1, 01 November 2009 - 03:50 PM.

#3 lab2009

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Posted 01 November 2009 - 04:26 PM

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

#4 Prep!

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Posted 01 November 2009 - 07:51 PM

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

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

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!!
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#5 tea-test

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Posted 02 November 2009 - 03:25 AM

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.
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#6 almost a doctor

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Posted 02 November 2009 - 03:26 AM

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

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

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.

#7 lab2009

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Posted 02 November 2009 - 08:20 AM

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

lab2009

#8 bob1

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Posted 02 November 2009 - 04:35 PM

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

#9 Feelcontraire

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Posted 18 November 2009 - 06:27 PM

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

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

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.