Not for something this large. If you are thinking about MW cut off filters, they tend to come in pore sizes for specific MW cut-offs, 10 kD, 40kD, 100kD, 150kD are common cut-off sizes. Your 8kb fragment will be more than 5,000 kD and that does not even include the proteins bound. Gel filtration might work, but I am not sure this is worth it if you are doing a chromatin IP.

Thanks Gregory for the answer. Is there a way I can calculate the size in Dalton of this digested crosslinked fragment (I am not sure what proteins are bound to it though)? I am not doing ChIP with this experiment. My goal is to look at the proteins bound to this fragment using MS.

In general, calculate 660 per base for double stranded DNA that would give you a MW of about 5,200,000 g/mole just for the fragment. More importantly, calculate how many copies of this fragment you have, given 2 copies per cell. Even if you could purify this fragment, will you have enough bound proteins to see by MS? For every 1 million cells, you would get about 1.7x10-13 moles of a 50 kD protein if one copy is bound to each fragment. That is 0.17 pmoles, or about 9 nanograms. Ask your MS facility how much pure protein they need in order to get some peptide sequences.

Sorry, my calculations are off. If one copy of a protein bound to your fragment you would get 2 million proteins per 1 million cells. That works out to 3.3x10-18 moles or about 1.7x10-13 grams for a 50kD protein, or about 0.17 picograms. You need about 10,000 fold more to even consider getting some MS data.

Yes that is very true. My current goal is to get the crosslinked fragment isolated from the rest of the chromosomal DNAs. Right now I am trying an approach with biotinylation. If it works, I will then scale up and combine multiple sets to get MS data, even a weak signal will do great. It is gonna be labor intensive though