protein crystals have different symmetry forms. they are hardly visible by eye
DESCRIPTION
Crystals are made from very large numbers of small units: Unit cells. The unit cell may contain more than one protein. The packing of the unit cells gives rise to a 3D structure with clear symmetry: a protein crystal. - PowerPoint PPT PresentationTRANSCRIPT
Crystals are made from very large numbers of small units: Unit cells. The unit cell may contain more than one protein. The packing of the unit cells gives rise to a 3D structure with clear symmetry: a protein crystal.
Protein crystals have different symmetry forms. They are hardly visible by eye.
Good crystals diffract X-ray radiation (wavelength in the order of 0.5 tot 2.0 Angstrom) very well in a regular pattern. In the diffraction pattern (b) more than 100.000 diffracted beams are collected.
Protein crystals contain large channels and holes filled with solvent. Usually around 50% of the total volume is solvent.
The usual method to obtain protein crystals is via the hanging drop method. About 10 l of a 10 mg/ml protein solution is put on a thin glass plate. In the precipitant solution the concentration of ammonium sulfate (the most frequently used salt) is higher than in the protein solution. Water evaporated from the top is taken up by the precipitant. After some time (weeks) the proteins in the solution start to crystallize (depending on temperature, bufferpH).
A narrow beam of X-rays (generated via arotating anode (university based) or viaa synchroton (national or European facilities))hits the crystal and diffract when electron density is encountered.
Most of the primary beam, when hitting a protein crystal, passes straight through it.Some X-ray interact with electrons on each atom and the x-ray beam is diffracted. Becauseof the symmetry in a crystal the diffracted beams interfere with one another.The relationship between the distance between the planes d and the wavelength is given byBragg’s law: 2d sin = . This relationship enables one to determine the size of the unit cell.
The reflection angle for a diffracted beam can be calculated from the distance r between the diffracted spot and the position where the primary beam hits the film. From the geometry the Tangent of the angle 2 = r/A. A is the distance between crystal and film.
Two diffracted beams are shown. These are defined by amplitude ( this is the strength/intensity of the beam, the higher the amplitude the more signal), the phase (this is relatedto the interference, negative/positive, with other beams) and wavelength (which is determined by the x-ray source). The better and intenser the monochromatic light the higher the quality of the diffraction pattern. The best quality at the moment is obtainedusing the synchotron facility in Grenoble, with a good second best EMBL Hamburg.The wavelength and the amplitude of the beams can be measured but the phase is a difficultproblem which can usually not be solved directly from the measurements.
A general method has been introduced to solve the phase problem: multiple isomorphousreplacement (MIR). This requires the introduction of new x-ray scatterers in the unit cell.These additions are heavy atoms, there need to be only a very few and they should not disturb the 3D structure. With some luck p.e. SH groups exposed in the solvent channelsbind heavy metals. Because heavy atoms contain many electrons they diffract the x-raybeams more strongly than H,C,N,O or S.
The diffraction patterns of DNA (A-DNA left) and B-DNA (right)
The amplitudes and phases are used to calculate an electron-density map. This map has to be fittedwith the amino acid sequence. The interpretation of the map is complicated by limitations. First,the map contains errors, usually due to errors in phase angles. In addition, the quality of the map depends on the resolution of the diffracted data, and this depends on the quality of the crystal.Resolution is measured in Angstroms, the smaller this number, the better. At 2.0 Angstrom resolution the difference between sidechains are visible, but only at very high resolution (1.0Angstrom or better) we can directly determine what amino acid is observed in the diffraction pattern.
One dimensional 1H-NMR spectra.a) Ethanol. The chemical shifts are distinctfor the different hydrogen atoms. Resonancesare split into three or four peaks due to their neighbours (CH2 by the CH3 group and viseversa).
b) 1H-NMR spectrum of 36 amino acid protein.
The overlap in the 1D NMR spectrum is diminished using 2D NMR spectroscopy. The peakson the diagonal from upper right to lower left are identical to the 1D spectrum in the previous slide. The off-diagional peaks represent through-space interaction between hydrogen atoms.
The interaction between atoms can be measured using different NMR methods:Through-bonds: COSY (a)Through-space: NOESY (b)
The information present in the NMR spectra can be used to calculate a 3D structure, which by virtue of the nature of the NMR data always is an ensemble of structures.The NMR data reveal more than X-ray data the flexibility present in protein structures.See the disorder in a few loops in the structure. See also the next slide.
The standard dimensions (Angstroms, degrees) of a planar trans-peptide group. The NH-COgroups form the peptide plane.
The Cis-peptide group.
A polypeptide chain in its fully extended form showing the planarity of the peptide groups
The torsional degrees of freedom ofa peptide unit. The only resonablemovements are around the Ca-N() bond and the Ca-C bond
Newman projections
Steric interactions between CO and the amide Hydrogen (NH) on adjacent residuesprevents the occurrence of the conformation with= -60, = +30
The Ramachandran diagram.
The conformation angle distribution of all residues (except Gly and Pro) in 12 very welldetermined X-ray structures.
A Ramachandran diagram of Gly residues. Note the large degree of freedom.
A right handed helix. The values are - 57 and = -47, 3.6 residues per turn and a pitch (distancebetween two points after complete turn) of 5.4 Angstrom.Hydrogen bonds between CO en NH are indicated by dashed lines. These hydrogen bonds are between residuesN and N+4. This results in a strong hydrogen bond in a near optimal N....O distance of 2.8 Angstrom. The coreof the helix is tightly packed, the sidechain groups alldirect outward and avoid steric interference with the peptide backbone and with each other (next slide).A left-handed helix has a major problem: the side chainscontact the backbone too closely. However, the Van der Waalsoverlap is not severe enough to avoid that 1-2 % of all non-Gly residues adopt this conformation.
The hydrogen bond pattern of several polypeptide helices.
The 3-10 helix has 3.0 peptide units per turn with a pitch of 6.0 Angstrom.The helix has 3.6 units per turn and a pitch of 5.4 AngstromThe helix has 4.4 residues per turn and a pitch of 5.2 AngstromAlle helices can be observed in proteins.
The hydrogen bond associations in pleated sheets.
A two-stranded -antiparellel pleated sheet.Dashed lines indicate hydrogen bonds. Note that the R groups (purple) on each polypeptide chainaltenatively extend to opposite sides of the sheet.
Schematic representationof silk -sheets.