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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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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))

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


Structural Biology techniques

● Electron microscopy ≈ 5Å ?

● NMR equivalent resolution ≈ 2Å

● X-ray crystallography≈ 1Å

● Hybrid techniques EM + NMR/Crystallography


Resolution particulars







4.0 Å


3.0 Å


1.8 Å


1.0 Å

Center for Biological Sequence analysis DTU


X-ray crystallography & Nuclear Magnetic Resonance (NMR)

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)

Purify protein

Collect data

Analyze Data (make assignments)

Apply distance constraints

and calculate structures


NMR - How it works

NMR uses the behavior of nuclei with magnetic moments in an applied

magnetic field.

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)


NMR (continued)

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)


A nuclear Overhauser effect (NOE) experiment give peaks between

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!


Refinement is used in conjunction with known geometric

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

(basic drill)

Grow crystals

Collect diffraction data

Solve structure


Protein phase diagram

(constant temperature, pressure, pH)









How do get a protein crystal?

This is the hard part!

 Start with very pure protein

 Get a supersaturated solution

 Wait (sometimes a long time!)

 Keep trying….



Most common technique is vapor diffusion.


Drop(2L protein (20 mg/mL), 2L 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.


But remember, before you try crystallizing…..

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

protein sequences

  • Use several different sequence alignment algorithms to align
  • any homologous sequences

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!


When X-rays shine on atoms, the atoms become new sources of

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.

Scattered X-rays

reinforce in certain

directions and cancel

in most others




Another way of thinking about it…

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

“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



This is a real diffraction pattern of a crystal in a special orientation

(X-rays are being shined directly into the side of a unit cell)



h k l

Green (2,1,0)

Blue (1,1,0)

Pink (1,-1,0)

Orange (-4,4,0)

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

this crystal has systematic

absenses. But given the

regularity of the diffraction

pattern, we can easily measure

the spacings along “a” and “b”.

1.5 mm

Direct beam

Where (1,0,0)

would be

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”.





1.5 mm


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)

= 1.074º

θ = 0.537º


We can do the same for b and c. Actually, programs have gotten so

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.

Integrate background

(then subtract from spot)

Integrate spot

(Add counts in



Do this for all (e.g.) ~40,000 spots in you data set.

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

Bragg’s Law: n  = 2d sin

Each data point has index

and intensity

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

3 dimensions

Now all we need is the “phase”

for each data point (reflection)


Fourier Series

(1D example)

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)


The only trouble is, we must know the offset (phase) for each of the

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”

1 wavelength



width of cell

width of cell

This ?

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.

= 45º

0, 360

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!


Now Miller indices are a very convenient way of thinking about

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 |Fhkl|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



Structure solved at CAMD

IQGAP1 “GAP-related domain”