3 ms·
Naive question: these images are made at low but finite temperatures, so each atom would fluctuate within its position such that there should be an uncertainty.
by edna314 6y ago
Naive question: these images are made at low but finite temperatures, so each atom would fluctuate within its position such that there should be an uncertainty. Is that uncertainty lower than 1 angstrom? Otherwise, how can one talk about atomic resolution if fluctuations are larger?
- natechols 6y agoThe uncertainty also comes from averaging many images together. But yes, it's less than 1 Angstrom; the deposited structures (for example https://www.rcsb.org/structure/7A6A https://www.rcsb.org/structure/7A6A) record atomic uncertainties, "B-factors" or "temperature factors" in crystallographer-speak. In the example I pulled out the B-factors are in the low-double digits for the main-chain atoms, and if I'm remembering the math correctly, this will correspond to atomic displacements with a radius around 0.5 Angstrom. (Of course that's just an estimate by the software that performed the model optimization, but my understanding is that this part is relatively straightforward.)
- edna314 6y agoHmm, but then the b-factor (and therefore also resolution) is dependent on the flexibility of molecule you want to look at? Does that mean that these guys just found a system for which the main chain was rigid enough such that the thermal fluctuations are low enough?
- natechols 6y agoOh absolutely, they picked a model system that is both very rigid and has extremely high internal symmetry, and I assume they did rotational averaging to use fewer images. But these ideal systems are very useful for determining the limitations of the underlying technology like electron detectors; 15 years ago the best you could do with a molecule like this was around 4 Angstrom resolution.
- edna314 6y agoOk, that’s fair. But, then the number makes more sense in relative terms than in absolute. Still, absolutely impressive.