7 replies to this topic

[Mad Researcher]

Hi,

 

I have two antibodies and in the paper it says use 5 µg (Antibody A) and 10 µg (Antibody B). When I checked the data sheet, it says that it is at 100 µg (Antibody A) and for antibody B it is a 200 µl solution. How can I calculate how much antibody to use to make a 10 ml solution?

Edited by Mad Researcher, 10 January 2018 - 06:41 PM.

[bob1]

The information you have provided here is missing a few things - either you are using the antibodies for something like an IP, in which case you usually use a set amount (ug) of antibody, or you are using them for westerns, in which case you probably want a concentration (ug/ml).

 

For both cases you need to get the concentration of the antibody stock - it can be calculated by c = n/v, if you have a volume value for A and an amount for B.

 

From there you would either use v=n/c to get the volume required for your 5 and 10 ug amounts, or you would use C1V1=C2V2 for a concentration conversion.

[Mad Researcher]

I thought so. Could you please look through the webpages and could you please let me know if I am missing something.

https://www.sysy.com...acts-196002.php

http://www.emdmillip...dy,MM_NF-07-432

 

Thank you

The information you have provided here is missing a few things - either you are using the antibodies for something like an IP, in which case you usually use a set amount (ug) of antibody, or you are using them for westerns, in which case you probably want a concentration (ug/ml).

 

For both cases you need to get the concentration of the antibody stock - it can be calculated by c = n/v, if you have a volume value for A and an amount for B.

 

From there you would either use v=n/c to get the volume required for your 5 and 10 ug amounts, or you would use C1V1=C2V2 for a concentration conversion.

[bob1]

The millipore one says (under the biological information section) to refer to the certificate of analysis for the lot-specific concentration. You can find the CoA's under the supporting documentation link and there should be a concentration in mg/ml somewhere in the left hand column of the 1st page.

 

The SySy antibody says to resuspend in 200 ul, then use at a dilution ratio - I note that this particular one is a crude antiserum, so it doesn't have a concentration. However, there is a purified one: https://www.sysy.com...acts-196003.php that does have an amount and a recommended resuspension to make 1 mg/ml.

[Mad Researcher]

Thank you Bob. I found for the Millipore, it says the concentration is at 1 mg/ml. So, when a paper says they used it 5 µg, do they mean they used it 5 µg/ml? In that case, I need to add 5 µl of the antibody + 995 µl of buffer to make a 1ml antibody.

 

For the Sys antibody, I wonder how they determined to use 1:1000 for WB experiments? 

 

 

The millipore one says (under the biological information section) to refer to the certificate of analysis for the lot-specific concentration. You can find the CoA's under the supporting documentation link and there should be a concentration in mg/ml somewhere in the left hand column of the 1st page.

 

The SySy antibody says to resuspend in 200 ul, then use at a dilution ratio - I note that this particular one is a crude antiserum, so it doesn't have a concentration. However, there is a purified one: https://www.sysy.com...acts-196003.php that does have an amount and a recommended resuspension to make 1 mg/ml.

[bob1]

For the Millipore, it's hard to say - for an IP, 5 ug could be the amount that they used to pull down the protein. Westerns are normally done using a concentration or a dilution. However, your calculations are correct, if you want 5 ug/ml, then 5 ul of a 1 ug/ul solution into 995 ul will give the correct concentration.

 

The dilution to use is normally determined by titration - take a sample that you know expresses the protein of interest (this could be a cell line, expressed protein from bacteria/yeast, purified protein etc), and a negative control, run in pairs on a gel, blot, slice membrane into strips so that each contains + and - samples, incubate each strip with one of a series of dilutions (commonly 1:100, 1:200 or 1:250, 1:500, 1:1000, 1:2000) , look for the one that gives the best signal, with the least background.

[Mad Researcher]

For the Millipore, it's hard to say - for an IP, 5 ug could be the amount that they used to pull down the protein. Westerns are normally done using a concentration or a dilution. However, your calculations are correct, if you want 5 ug/ml, then 5 ul of a 1 ug/ul solution into 995 ul will give the correct concentration.

 

The dilution to use is normally determined by titration - take a sample that you know expresses the protein of interest (this could be a cell line, expressed protein from bacteria/yeast, purified protein etc), and a negative control, run in pairs on a gel, blot, slice membrane into strips so that each contains + and - samples, incubate each strip with one of a series of dilutions (commonly 1:100, 1:200 or 1:250, 1:500, 1:1000, 1:2000) , look for the one that gives the best signal, with the least background.

Thanks Bob. In a paper, where the Sys antibody was used, they mention that they used it at 5 ug. How did they find that out?

[bob1]

It's a guess, but I would think that they mean 5 ug/ml, or 1:200 dilution of a 1 ug/ul stock.