Structural Biology. 8/27/10. Why determine structures?. Visualize primary sequence in context of folded protein (buried vs. solvent exposed). Highlight residues important for intermolecular interactions (co-crystals, packing, or computational (docking)).
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Visualize primary sequence in context of folded protein
(buried vs. solvent exposed)
Highlight residues important for intermolecular interactions
(co-crystals, packing, or computational (docking))
Visualize surface features to aid in identifying or designing binding partners
(e.g. clefts, promontories, hydrophobic, or specifics of fold)
Allow for the design of properly folded mutant proteins
Allow use of structural databases to gain insight into function/evolution
Provide a template for modeling studies to understand
the function of related molecules
● Electron microscopy ≈ 5Å ?
● NMR equivalent resolution ≈ 2Å
● X-ray crystallography≈ 1Å
● Hybrid techniques EM + NMR/Crystallography
Center for Biological Sequence analysis DTU
X-ray crystallography utilizes information gleaned from bouncing X-rays off
an ordered array of molecules. NMR utilizes information about magnetic
environment of nuclei with non-zero spin
NMR provides several snapshots of the object of interest all ~equally valid
X-ray crystallography provides one snapshot of the object of interest.
NMR cannot be practically used for large molecules (at least not yet).
X-ray can be used for even very large molecules and complexes.
Most importantly, structures that have been determined using both
techniques are very similar!
(very basic drill)
Analyze Data (make assignments)
Apply distance constraints
and calculate structures
NMR uses the behavior of nuclei with magnetic moments in an applied
Biochemistry 5th edition,Berg, Tymoczko&Stryer
For a given type of nucleus (1H), introduce RF radiation and excite
transitions of nuclei from low to high energy state. Monitor emitted
RF radiation as nuclei descend to low energy state (decay).
FID (Free induction decay)
De-convolute (separate) all the RF emissions from the FID
to get a spectrum
Individual nuclei transition at slightly different frequencies (resonances)
depending on their chemical environments (electron clouds, other nuclei).
The difference in resonance frequencies of nuclei from those of the same
nuclei in a standard compound, are called chemical shifts. Therefore,
Each protein has a unique spectrum for a given nuclei (1H, 13C, 15N,etc)
Example reference compound trimethylsilane (TMS)
protons that are close in space even though they’re not bonded.
A correlation spectroscopy (COSY) experiment results in peaks between
protons that are connected through covalent bonds. In this way, individual
amino-acids have a characteristic signature (i.e. Ala vs. Ser).
Intro to Protein Structure, Branden & Tooze
By using COSY and NOESY experiments, one can identify various AAs and
their neighboring AAs (sequential assignment). Once assignments are made,
NOE info gives distance constraints. Distance constraints between atoms,
once the atoms have been identified, reveal the structure!
and energetic constraints (in addition to the acquired distance constraints).
Because of the limited number of distance constraints and the nature
of solution-structure determination, one ends up with a set of structures
that satisfy the distance criteria. So called “lowest penalty structures”.
Kim et al.,Nature404, 151 - 158 (09 March 2000)
X-ray crystallography between
Collect diffraction data
Protein phase diagram between
(constant temperature, pressure, pH)
How do get a protein crystal? between
This is the hard part!
Start with very pure protein
Get a supersaturated solution
Wait (sometimes a long time!)
Most common technique is vapor diffusion.
Drop(2L protein (20 mg/mL), 2L Reservoir solution)
Reservoir(0.5 mL of 20% PEG 8,000, 200mM MgCl2, 100mM Tris pH 8)
Cover with clear tape and place at RT or 4ºC
Reservoir will slowly pull water out of drop and drop will concentrate.
Hopefully you’ll get crystals. Many commercially available screens.
If you want to make a well-behaved, soluble expression
construct spanning a region of a protein with unknown
structure, you would:
A) Do a data base search to identify other proteins with similar
C) Use several different 2º structure prediction algorithms with
your sequence of interest and any homologs
D) Compare all of these 2º structure predictions (and decide!)
E) Make several different constructs with different starts and stops
F) All of the above
Then you must work out expression, purification details!
X-radiation. Each atom reflects X-rays in all directions. There is
structural information in the “scattered X-rays, but it’s too weak
when the atoms are from just one protein molecule.
A crystal aligns a very large number of molecules in the same orientation.
This provides the potential for a much stronger signal than when using
just one molecule.
reinforce in certain
directions and cancel
in most others
Home X-ray setup between
Crystal is composed of many families of “planes” of atoms.
Each family of planes are parallel and each is separated from
the next by a specific distance “d”. Reflection of X-rays from
these planes is reinforced when the geometric situation pictured
above is achieved.
Bragg’s law - 2dsinθ = n
n usually = 1, is wavelength
and is known
Two dimensional crystal between
“a” and “b” are the lengths of the sides
of the unit cell (each unit cell in black).
O is the origin.
The sets of planes (green, blue, pink)
are called Miller planes. The green set
intersects the cell edge “a” at a=1/2
and cell edge “b” at b=1. Therefore,
the green set of planes are the (2,1)
of Miller planes. What you do is invert
the 1/2 and it becomes 2. If the planes
intersected “a” at 1/3, and “b” at 1/4,
they would be the (3,4) family of Miller
planes. Etc. You just look at the unit
cell in the upper left corner – The planes
are drawn in all the cells to show they
intersect all the cells in the same way
Note: if you slowly rotated this crystal in the X-ray beam, you would satisfy
the requirements of Bragg’s law. Each set of planes would diffract in different
(X-rays are being shined directly into the side of a unit cell)
h k l
Every reflection arises from a different set of Miller planes.
Every reflection has an index h,k,l – no two are the same.
This diffraction shows that orientation
this crystal has systematic
absenses. But given the
regularity of the diffraction
pattern, we can easily measure
the spacings along “a” and “b”.
So since we know the crystal to “film” distance, the wavelength, and where
the spots are on the film, we can use geometry and calculate the size of
“a” and “b”.
2dsinθ = 1.5418Å (CuK)
d = 1.5418Å/2sinθ
a = 82.2Å
So 1.5/80 = tan 2θ
tan 2θ = 0.01875
2θ = tan-1(0.01875)
θ = 0.537º
sophisticated, you feed any random orientation picture to a program
and it scans the image, finds the spots and uses them to determine
a,b,c, and any angles between them and the lattice type, the symmetry
and the orientation of the crystal! In other words, the program knows
the Miller indices for all the spots.
So you simply start turning the crystal and collecting images. For
example, you turn the crystal 1º and take a 1º oscillation picture.
Do this for 180º, and you have a full data set.
(then subtract from spot)
(Add counts in
Now you must scale the spots from one image to the next (sometimes
your shooting through a thicker part of the crystal etc.)
When all spots have been integrated and scaled, you have a data set.
Now each spot (h,k,l) should really be considered to be a wave. The
intensity of the spot is the amplitude and the number of oscillations
across the unit cell is revealed by its Miller indices. The (1,0,0)
reflection would have one wavelength (of a sinusoidal wave) in the
unit cell along the a direction, the (2,0,0) would be two wavecrests, etc.
These waves can be added together – sometimes reinforcing, some-
times cancelling out. When they’ve all been added together, they
describe the shape of the “thing” that scattered them originally.
X-ray diffraction data gotten so
Bragg’s Law: n = 2d sin
Each data point has index
h k l I
2 0 3 1483.6
3 -1 -3 19999.9
3 -1 -2 6729.6
3 -1 -1 30067.1
3 -1 1 8227.0
3 -1 2 29901.5
3 -1 3 24487.5
3 -1 4 502.1
Now all we need is the “phase”
for each data point (reflection)
Fourier Series gotten so
f(x) = F0cos2(0x + 0)
F1cos2(1x + 1)
F2cos2(2x + 2)
F3cos2(3x + 3)
F4cos2(4x + 4)
Fncos2(nx + n)
f(x) = Fhcos2(hx + h)
Gale Rhodes Crystallography Made Crystal Clear(2nd edition)
waves. In the previous 1D example, the phases were either 0º or
180º. Remember, we have ~40,000 of these “waves”. We know how
tall they are and we know their wavelengths, but we don’t know the
phases. The so-called “Phase Problem”
width of cell
width of cell
Or this ?
One way to address this is to introduce a “heavy” atom into a crystal
and collect another data set (say HG dataset).
Now sinusoidal waves can also be represented as vectors.
The length of the vector is the amplitude of
the wave, the direction is the phase.
Now we have two data sets. One set is HG
the other is native (NAT). We can use a technique
called the Patterson function to locate the
coordinates of the Hg atom. The Patterson
function doesn’t require phases.
Once we locate (in x,y,z) the Hg atom, we actually know its contribution
to each diffraction spot – its little vector!
diffraction from a crystal. A more accurate way of thinking about
what makes a given data point (h,k,l) relatively intense or weak
is given by this formula:
Fhkl is a vector. It is the sum of all the little vectors from all the atoms
in the cell. But we have located the Hg atom so we know its x, y, z.
So we know the direction and the phase for the contribution to the
reflection made by the Hg atom! We will call this Fhg.
So what we have are a bunch of |F abouthkl|s – we have magnitudes
but not directions. So we will represent them as circles with radii that
are proportional to their magnitude.
Native reflection hkl
HG reflection hkl
And for the hkl reflection, we know vector Fhg (Note an Fhg for each hkl)
We also know that |FNAT| + Fhg = |FHG|
Or |FHG| - Fhg = |FNat|
|FHG| is offset by -Fhg
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