I have a protein crystal structure at 1.83 angstroms with 2 molecules in the asymmetric unit. But the B factor of the second molecule is almost twice as large as the first. After running ARP/wARP , the first protein molecule is built in but only 50% of the second molecule is built into the electron density. It appears that I somehow have weak or missing reflections for the second protein molecule. How can this be?
Technically, there's no reason that this could not be the case - in particular if your protein has a lot of disorder-prone regions. I once had a case of one chain completely missing any density - sitting in a solvent channel created by another, well-resolved chain. Having some chains in your asymmetric unit showing much higher B-factors than another chain is a very common phenomenon.
But I'd also check if your space group is definitely correct.
Dear Engi,
Just a month ago we collected the data set to 1.9 A resolution. And I solved structure using SAD (Se_Met). I was able to built one chain all residues. And then used that model to solve the structure. There are two chains in the asymmetric unit. To my surprise I could not drop my R factors. I decided to check maps. The first chai was good, but the second had very fuzzy density. Some places with limited density conversing the model.
When we collected data I knew that this structure will have problems. Because, it was from multiple crystals. Which means that crystal was not perfect.
The protein is a flavin binding protein. Most likely there was limited amount of the substrate, which caused not perfect crystal.
Only option we have – to re-grow the crystal with added flavin.
1. Check if in your case there is a substrate which has to be complexed with your protein. Even if you crystalized an apo- form, It is known that in some cases complexes are better ordered.
2. As suggested above by Michal. Check if your space group was correct. It could be that you processed in the higher symmetry (pseudo-symmetry).
How to check?
a. Let say it is true orthorhombic symmetry.
There are 8 possibilities. So, the easiest to run molecular replacement with all possible groups.
b. Your true symmetry is P2 or P21, with the beta = 90.0.
You have to process data in P1, and try three possible orientations to scale in P21. The lowest in your crystal is so strong, that makes your crystal symmetry a pseudo-symmetry with the higher symmetry (orthorhombic).
c. In hexagonal and trigonal space groups you will have to go to the lowest P3 and check if that improves your maps. Do not worry to have too many chains. The goal is to get best densities, and low R factors.
d. In tetragonal space groups the same idea – lowest symmetry P4.
Good Luck.
And keep in touch, let me know if you solved this mystery.
George.
Dear Michal and George,
Thank you so much for your help. The part that is unresolved in the second protein chain is not particularly disorder-proned. It is resolved in other structures with no problem. Regarding the wrong space group, I actually did process the final structure in Orthorhombic. When I processed in P1 and P2(1) it gave a better total R factor but I did not get a pleasing MR solution. I had overlapping densities in the protein molecules, and again their B factors had large differences between them (when I selected 3 molecules per asymmetric unit 2 molecules had similar B factors whereas the 3rd had twice the value, and when I selected 4 molecules per asymmetric unit 2 molecules had similar B factors, the 3rd had slightly higher, and the 4th had twice the value of the 1st and 2nd) which is why I stayed with the Orthorhombic space group.
This structure is complexed with a ligand (soaked) and actually the apo form also has the exact same problem in the structure! I tried the same workflow on the apo form and again I see overlapping densities in the MR solution and additionally running ARP/wARP gives only around 20% of the entire structure when processed in P2(1).
"The lowest in your crystal is so strong, that makes your crystal symmetry a pseudo-symmetry with the higher symmetry (orthorhombic)."
Can you please elaborate the lowest what? The lowest reflection?
Hi Engi,
Auto correction changed my word.
The phrase was:
"The NCS (non-crystallographic symmetry) in your crystal is so strong, that makes your crystal symmetry a pseudo-symmetry with the higher symmetry (orthorhombic)."
If P1 did not solve your problem, then check images.
If you have double spots, or split spots, sometimes not round spots, could be an indication of problems with data. And that means problem with quality of your crystals.
You have to go back to crystallization.
1. How fast your crystal grow?
If they get too fast, reduce your precipitation agent, or concentration of your protein.
2. If your apo- and complex form both have the same problem, that means that something makes about 25% of your protein in the drops be different. Check how stable is your protein. Maybe crystallization is too slow (long time) and your protein degrades.
In this case opposite approach - increase concentration of precipitant, or protein concentration.
3. Try to crystalize at cold room temperature. Maybe it will stabilize your protein.
4. Check crystal contacts. What makes one chain different from another. Maybe you can save crystal packing by mutating some surface residues.
5. Check N-terminal and C-terminal parts. Maybe removing or adding few residues will make a change.
So many things you can try. The question is if you have time.
In general, you have the structure of your protein and complex (one chain).
If you can drop your R factors to low values, and keep the geometry for the structure, it is not a big problem to have one chain less ordered. You can mention this in your paper, and analyze only ordered chain.
Best regards.
George.
Unfortunately I have run out of time for further experimental work. But thanks a lot for the insight and suggestions, they have been very helpful !! :)
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