protein band is shifted after boiling sample with sample buffer - (Jul/02/2008 )
My target protein is 100KD, monomer receptor. I used regular Laemmli sample buffer before running SDS PAGE. Without boiling sample it showed a strong band at 100KD. But when the sample was boiled at 65 and 95 degree, the 100KD band disapeared, instead of there is a 50KD band showing up. Does that mean the protein is broken down into two 50KD parts? The boiling can cause protein degradation?
-deventx-
QUOTE (deventx @ Jul 2 2008, 09:35 AM)
My target protein is 100KD, monomer receptor. I used regular Laemmli sample buffer before running SDS PAGE. Without boiling sample it showed a strong band at 100KD. But when the sample was boiled at 65 and 95 degree, the 100KD band disapeared, instead of there is a 50KD band showing up. Does that mean the protein is broken down into two 50KD parts? The boiling can cause protein degradation?
Are you immunoprecipitating your receptor? If so, what you’re seeing could be the IgG chains of your IP antibody. Without boiling the heavy and light chains can run together on a gel, you will find them around 75 kDa. When you boil, they run separately and the heavy chain is found around 50kDa while the light chain runs around 25kDa. When blotting your primary antibody won’t interact with these chains, but your secondary will. If available, I would recommend using a primary antibody against your receptor that is directly conjugated to HRP. That way you won’t have to add secondary.
Also, sometimes boiling can cause receptors to oligomerize. That would explain why you don’t see your 100kDa band anymore (it may not be getting into the gel). But that doesn’t really explain the appearance of the 50kDa band.
-jlolsen-
Are you sure that your receptor is a monomer? A lot of receptor have subunits that are assembled together with non-cavalent bonds. If so, the boiling denatures the bonds and you have two subunits of 50 kDa.
-Madrius-
QUOTE (jlolsen @ Jul 2 2008, 10:55 AM)
QUOTE (deventx @ Jul 2 2008, 09:35 AM)
My target protein is 100KD, monomer receptor. I used regular Laemmli sample buffer before running SDS PAGE. Without boiling sample it showed a strong band at 100KD. But when the sample was boiled at 65 and 95 degree, the 100KD band disapeared, instead of there is a 50KD band showing up. Does that mean the protein is broken down into two 50KD parts? The boiling can cause protein degradation?
Are you immunoprecipitating your receptor? If so, what you’re seeing could be the IgG chains of your IP antibody. Without boiling the heavy and light chains can run together on a gel, you will find them around 75 kDa. When you boil, they run separately and the heavy chain is found around 50kDa while the light chain runs around 25kDa. When blotting your primary antibody won’t interact with these chains, but your secondary will. If available, I would recommend using a primary antibody against your receptor that is directly conjugated to HRP. That way you won’t have to add secondary.
Also, sometimes boiling can cause receptors to oligomerize. That would explain why you don’t see your 100kDa band anymore (it may not be getting into the gel). But that doesn’t really explain the appearance of the 50kDa band.
Thank you.
Yes. I am sure it is monomer based on 100 KD molecular weight calculated by amino acid sequence. The sample is membrane fraction protein, without immunoprecipitating. Therefore it is not heavy or light chain.
-deventx-
QUOTE (Madrius @ Jul 2 2008, 12:04 PM)
Are you sure that your receptor is a monomer? A lot of receptor have subunits that are assembled together with non-cavalent bonds. If so, the boiling denatures the bonds and you have two subunits of 50 kDa.
Thank you
Based on 100 KD molecular weight calculated by amino acid sequence, I am sure it is monomer. That is why I do not understand why I got 50KD band after boiling.
-deventx-