Dale > If you could measure the continuous function you can directly calculate the > electron density - there is no phase problem. > The paper I referred to earlier gives the details.
I don't think it is quite that easy. The paper you referred to (Miao, et al, Annu. Rev. Phys. Chem. 2008, 59:387-410) gives a good summary but people have struggled to apply this technique for low contrast biological objects. Early tests were on high contrast objects (low entropy images I guess). These probably met the conditions for which classical direct methods would work. Since then, some reconstructions of cells (probably with a high entropy images) have been carried out. However, the images obtained have not yet shown the features obtainable by electron or x-ray tomography with a lens. High non-crystallographic symmetry (e.g. as in spherical viruses) in a crystal gives extra information similar to that which would be obtained by sampling the continuous molecular transform of a single subunit. However it is more complex and perhaps richer because the individual subunits are in different orientations and positions - not a simple sampling of the transform a single subunit. I guess this leads to a question. Does anyone know of a case of a crystal with no non crystallographic symmetry but a very high solvent content for which the structure (at say 2.5A resolution whatever that means) could be solved without resort to additional phasing methods? Colin -----Original Message----- From: CCP4 bulletin board [mailto:CCP4BB@JISCMAIL.AC.UK] On Behalf Of Dale Tronrud Sent: 14 January 2012 08:11 To: ccp4bb Subject: Re: [ccp4bb] Molecular Transform Superimposed on a Dataset On 1/13/2012 11:32 PM, arka chakraborty wrote: > Hi all, > > I would like to ask some questions regarding this thread.. > 1) What is exactly meant by "Fourier transformed electron density"?- > according to my knowldege performing a fourier transform on the > electron density gives you the structure factor back. So, how does it related > to what Prof. James H called "non-lattice-convoluted pattern"? It will be > really nice if somebody can explain the thing in a " decoded" language?! > And also any articles focusing on the concepts discussed in the > entire thread will be very helpful > The real question is "The Fourier transform of what?". All of our usual software takes the electron density of the unit cell and Fourier transforms it assuming that the unit cell is repeated endlessly throughout the universe. This results in non-zero values only at integral values of h, k, and l. If, instead, you assume there is only one unit cell present in the x-ray beam you will get a continuous function for the intensity of scattering. You can use the convolution theorem to show that the value of these two functions are equal at the integral values (except for a scale factor). This thread has been about the relationship between the features one would observe in this continuous function (if it could be measured) and the electron density. If you could measure the continuous function you can directly calculate the electron density - there is no phase problem. The paper I referred to earlier gives the details. Dale Tronrud > Regards, > > ARKO > > On Sat, Jan 14, 2012 at 12:42 AM, Dale Tronrud <det...@uoxray.uoregon.edu > <mailto:det...@uoxray.uoregon.edu>> wrote: > > I think you have to be a little more clear as to what you mean > by an "electron density map". If you mean our usual maps that we > calculate all the time the Patterson map is just the usual Patterson > map. It also repeats to infinity, with the infinitely long Patterson > vectors (infinitely high frequency components) being required to > create the Bragg peaks. If you mean an electron density map of a > single object with finite bounds your Patterson map will also have > finite bounds, just with twice the radius. > > The Patterson boundary is not a sharp drop-off because there aren't > as many long vectors as short ones, but the distribution depends on > the exact shape of your object. Once you have a Patterson map that > has an isolated edge (no cross-vectors) back calculating the original > object is pretty easy. (Miao, et al, Annu. Rev. Phys. Chem. 2008, > 59:387-410) > > Dale Tronrud > > On 01/13/12 10:54, Jacob Keller wrote: > > I am trying to think, then, what would the Patterson map of a > > Fourier-transformed electron density map look like? Would you get the > > shape/outline of the object, then a sharp drop-off, presumably? Is > > this used to orient molecules in single-particle FEL diffraction > > experiments? > > > > JPK > > > > On Fri, Jan 13, 2012 at 12:33 PM, Dale Tronrud > > <det...@uoxray.uoregon.edu <mailto:det...@uoxray.uoregon.edu>> wrote: > >> > >> > >> On 01/13/12 09:53, Jacob Keller wrote: > >>> No, I meant the non-lattice-convoluted pattern--the pattern arising > >>> from the Fourier-transformed electron density map--which would > >>> necessarily become more complicated with larger molecular size, as > >>> there is more information to encode. I think this will manifest in > >>> what James H called a smaller "grain size." > >> > >> I've been thinking about these matters recently and had a nifty > >> insight about exactly this matter. (While this idea is new to me > >> I doubt it is new for others.) > >> > >> The lower limit to the size of the features in one of these > >> "scattergrams" is indicated by the scattergram's highest frequency > >> Fourier component. Its Fourier transform is the Patterson map. > >> While we usually think of the Patterson map as describing interatomic > >> vectors, it is also the frequency space for the diffraction pattern. > >> For a noncrystalline object the highest frequency component > corresponds > >> to the longest Patterson vector or, in other words, the diameter of > >> the object! The bigger the object, the higher the highest frequency > >> of the scattergram, and the smaller its features. > >> > >> Dale Tronrud > >> > >>> > >>> JPK > >>> > >>> On Fri, Jan 13, 2012 at 11:41 AM, Yuri Pompeu <yuri.pom...@ufl.edu > <mailto:yuri.pom...@ufl.edu>> wrote: > >>>> to echo Tim's question: > >>>> If by pattern you mean the position of the spots on the film, I > dont think they would change based on the complexity of > the macromolecule being studied. As far I know it, the position of the > spots are dictated by the reciprocal lattice points > >>>> (therefore the real crystal lattice) (no?) > >>>> The intensity will, obviously, vary dramatically... > >>>> ps. Very interesting (cool) images James!!! > >>> > >>> > >>> > > > > > > > > > > > -- > > /ARKA CHAKRABORTY/ > /CAS in Crystallography and Biophysics/ /University of Madras/ > /Chennai,India/ >
