I am working with a very unstable transcriptional factor. I use DNAse in the lysis buffer and I purify it as inclusion bodies, by using urea and 1M NaCl. After purification, I buffer exchange the protein to 50mM Hepes pH8, 1M NaCl. In this buffer, the protein is stable around 0.5mg/ml. The 260/280 ratio is 0.6.
Nowadays I am working with a N-terminal truncation of this protein. I use the same protocol, but the 260/280 ratio obtained is 1.5. As I purify it by using urea, I assume that it is not possible that my protein has DNA bound. But I cannot explain this 260/280 ratio?
Any ideas?
Thanks
The absorbance of a protein at 280 nm is due to tryptophan and tyrosine residues. If the sequence of the N-terminal truncation has few or no Trp and Tyr residues, the 260/280 ratio will be higher than expected.
Thanks, Adam. In the deleted fragment there is a Trp. Maybe this influenced the 260/280 ratio as you suggested.
Dear Enea,
I agree in principle with Adam's answer. However, both Trp and Tyr residues display an A260/A280 ratio of about 0.5. Thus, unless there are no Trp nor Tyr left nor any UV-absorbing cofactor in your truncated protein, the A260/A280 ratio should remain < 1. Therefore, with an A260/A280 ratio of 1.5, it seems likely that the preparation contains some contaminating material, most likely nucleic acid. Don't forget that nucleic acids absorb ~20-fold stronger than most proteins at 260 nm, hence a small percentage is sufficient to cause a substantial alteration of the A260/A280 ratio. As Sherlock Homes used to say, when you have eliminated the impossible (a pure polypeptide with an A260/A280 ratio >1), whatever remains (contamination), however improbable, must be the truth.
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