Structural Biology - Protein Crystallography

Question 1

What are the applications of studying protein structures?

Answer 1

Membrane proteins
Enzymes
Hormones
Nuclear receptors
Ion channels
Nucleic acids
Drug targets
Explore structure and function
Molecular basis of mutations and disease
Why a mutation is damaging the function of the protein

Question 2

Explain the molecular basis of disease

Answer 2

Find mutations in the protein structure
Example a mutation that causes the pore to become more narrow
Therefore antibiotic couldn’t bind

Question 3

Structure based drug design

Answer 3

Influenza drugs: Design of inhibitors to mimic binding of the substrate
COVID-19: Crystallise main protease of covid that facilitates viral transcription, Identify inhibitors of the protease - Idea for treatment before vaccines were created

Question 4

What are X-Rays?

Answer 4

X-rays are in the range 10^-8 to 10^-12 nm wavelength
Small objects (molecules) need short wavelengths

Question 5

Why can’t we use a microscope to see molecules?

Answer 5

Microscopes are used to see small objects
Only shows images of things with the same size as the wavelength of light (400-700nm)
Protein is 10nm big and atom is 0.1nm big so microscope can’t see a protein
X-rays have a similar wavelength to the size of an atom (0.1nm)
Can’t make X ray microscope due to it’s refractive index

Question 6

How does X-ray crystallography work (basic)?

Answer 6

Grow crystals
Put crystal Infront of X-ray to create an X-ray diffraction pattern
Calculate electron cloud that surrounds an atom (electron density map)
Build model/protein structure

Question 7

What is a crystal, molecule and atom?

Answer 7

Crystal:Ordered array of molecules in three dimensions (3D), many copies of molecules aligned in a specific order
Molecule: set of bonded atoms. The shape of a molecule is defined by the shape of the electron clouds associated with the constituent nuclei
Atom: nucleus (+ve) and electrons (-ve). Electron cloud surrounds the nucleus and is a lot larger than the nucleus of the atom. Volume and shape of atom is defined by the electron cloud. Electron cloud affects the shape of the crystal
A crystal is an ordered 3D array of distinctively-shaped distributions of electrons

Question 8

X-ray as an electromagnetic wave

Answer 8

An X-ray is a travelling electromagnetic wave
There are oscillating electric (E) and magnetic (B) fields
E and B fields oscillate at right angles to each other and the direction of travel
E field: electrostatic force felt by a charged particle due to the presence/motions of other particles
Can be represented by a vector (a quantity with magnitude and direction)
Tells us which way a +ve charge will move

Question 9

X-ray scattering by a single electron

Answer 9

X-ray (wave) is oscillating in an electric field (E)
Electron responds by oscillating
Oscillating electron is re-emitting X-rays over a wide angle (ripple effect)
Scattering of electric field from X-ray in ALL directions

Question 10

X-ray scattering by two electrons

Answer 10

The scattering pattern (observed on the detector) is the result of adding the scattered waves
Waves of both electrons interact with each other
Resultant pattern depends on their relative position
Diffraction: When the X-ray has the same wavelength as the size of the molecule
The diffraction pattern depends on the structure
X-ray scattering by two electrons is like the two-slit experiment
Diffraction of water waves through 2 slits gives an unvarying pattern of peaks/troughs
Pattern depends on slit structure (width, separation) and the wavelength
The relationship between the diffraction pattern and slit structure can give us information of relative position of two atoms (how spaced out they are)

Question 11

Explain circular/wave motion

Answer 11

360 Degrees = 2pi radians
Wave motion is cyclic - cosine and sine waves
Phase angle is the rotated position of line/where you are in the sine/cosine wave and it varies with x
Look at equation in notes
Waves have amplitude, phase and a defined wavelength

Question 12

Adding waves - constructive interference

Answer 12

If two waves of amplitude A are perfectly in phase
(a peak in one wave = a peak in the other)
Then the phase difference is 2npi for n=1,2,3…
Resultant amplitude = 2A
Resultant has the same phase to the components

Question 13

Adding waves - destructive interference

Answer 13

If two waves are perfectly out of phase
The phase difference = (2n-1)pi for n=1,2,3…
This will always give an odd number of p
The resultant amplitude = 0

Question 14

Adding waves - intermediate example

Answer 14

2 waves that are partially out of phase
Phase difference is a fraction of 2pi
Resultant amplitude < 2A
Resultant has a different phase to the components
Amplitude and phase of the resultant wave depends on amplitude and phase of the components

Question 15

How does X-ray scattering by two electrons in different positions affect the amplitude and phase shift?

Answer 15

Incident X-rays are in phase
X-ray interacts with the closest electron first and X-ray is scattered at angle of 2 theta
The second electron is further away, so X-ray travels longer
Results in a phase shift - two waves are out of phase
Amplitude is < 2A
Phase shift and amplitude is due to the structure
Scattering object is very small compared to distance to the detector - so rays from two objects can be assumed to be parallel

Question 16

How is scattering from two electrons quantified in equations?

Answer 16

Use wave vectors (direction of wave vector = direction of travel)
K0 = incident wave vector (magnitude 1/l)
K = scattered wave vector (magnitude 1/l) at angle of 2theta
First electron (O) to be hit by the wave vector K0
To hit second electron (B), wave has to travel extra distance AB
R is a positional vector (gives information about position and length)
Use scattering vectors to work out the phase difference
Scattering vector (S) = K - K0
Length of S (|S|) = (2/lambda)sintheta
Triangle is isosceles so angles are equal
At fixed lambda (typical in experiments), |S| (amplitude of S) varies only with scattering angle
Phase difference between the ray scattered from O and the ray scattered from B (at position r): F = 2prS
So phase difference depends on:
r - the relative positions of the 2 electrons
S - the wavelength l and scattering angle theta

Question 17

How are complex numbers used to transform the wave equation?

Answer 17

Wave equation becomes Y=Ae^(i*F)
I is an imaginary number
It is a negative number that stays negative when squared
Wavelength is constant in experiment, so can look only at the phase shift

Question 18

What is the structure factor, f(S)?

Answer 18

Mathematical equation of the diffraction pattern
Describes how the diffracted waves in each direction are related to the structure (the positions of the electrons, r)
It is a wave: has an amplitude and phase
It is a complex number since it is a sum of complex numbers
Longer equation (in notes):
S = scattering vector (depends on angles and wavelengths of incident and diffracted X-rays)
N = number of atoms in unit cell
j = adding new wave in order of 1 (increasing wave by one each time)
R = positional vector
I = imaginary unit
Can be shortened to: f(S) =|f(S)|e^(iF(S))
Shows amplitude (two parallel lines) and phase (e^i)

Question 19

Explain how multiple electrons interact to give amplitude and phase

Answer 19

Multiple scattered waves of the same amplitude going in direction 2 theta
Phase of each depends on the position of the electron it scattered from
Gives a new resultant wave: a new amplitude and new phase
Amplitude and phase depends on the amplitude and phase of the components

Question 20

What is electron density and the electron density function?

Answer 20

In an atom/molecule the positions of the electrons are not well defined
There is an electron cloud around the nucleus
These are the electrons that will interact with the X-rays
Distribution of electrons is described by a 3D electron density function r (r ) = r (x,y,z)
At position r in the molecule there is an amount of electron density: r (r ) = dxdydz
Where dxdydz is an infinitesimally small cube (volume)
Electron density function tells us the probability of finding electrons in that position (r position)
For a single atom, the highest electron density is in the middle
f ( S ) = mathematical equation of describing scattering on detector
r ( r ) = electron density in the molecule

Question 21

Scattering by a molecule in direction, S

Answer 21

Every part of the molecule scatters in every direction
Consider all the scattering in just ONE direction, S (at an angle of 2theta)
Amplitude is proportional to the electron density function as it depends on where the x-rays are coming from
Amplitude assumed to be one before but now defined by electron density function
Electron density is higher in the core of the molecule and lower at the edges
Structure factor for multi-electron system: f(S) =|f(S)|e^I(2pir(S))
Structure factor for molecule: f(S) = integral (r(r)dxdydz) e^I(2pir(S))
Now integrating in XYZ (3-dimensions) because there is so many points (assume infinity)
Add up contributions to find total scattering in one direction
If have different angle of scattering, value of f(S) is different because amplitude is different (because phases are different or less electrons are interacting)
Some dots on result are darker, some lighter

Question 22

Fourier transform to measure scattering

Answer 22

The total scattering in the direction (S) from an object ( r (r ) ) is the sum of the waves scattered from every point in the object
Measuring addition of all the waves coming from all the atoms within the molecule
Look at equation in notes

Question 23

Inverse Fourier transform to measure scattering

Answer 23

Inverse Fourier transform: used to determine the real structure r ( r ) from diffraction pattern f ( S)
We have to measure the scattering in all directions (at all values of S to work out the structure, r ( r )
Rotate the molecule by 2theta every time
The theory works for any structure

Question 24

Why are crystals used instead of single molecules?

Answer 24

Individuals molecules give a weak signal that is hard to distinguish from the background noise of the detector. They scatter very few photons
Crystals contain billions of molecules. They amplify the scattering, making it detectable. Properties of crystals restrict scattering to certain directions
In a crystal: orientation of all molecules is the same –> all waves are added together –> constructive interference –> strong signal on the detector

Question 25

Explain the properties of crystals and what a unit cell is

Answer 25

There are huge numbers of molecules in a crystal
Crystal = 0.2mm, Molecule = 10nm
So there is 20,000 molecules along one side and 8,000,000,000,000 in total
Repetitive nature: each molecule is aligned in the same direction
Each molecule occupies a unit cell
A unit cell is the smallest repeating unit of the crystal with translational symmetry (like a brick)
Unit cells can have more than one molecule
Crystals have many unit cells

Question 26

Reflection of X-rays from a semi-transparent layer of molecules

Answer 26

Crystal planes are semi-transparent
Only 2% of X-rays are scattered from a protein crystal
So each layer of molecules in the crystal scatters far less
Every plane contributes more or less equally to the scattering from the crystal
Crystals are made of several layers
X-rays penetrate every crystal layer and every layer diffracts the X-ray which amplifies the signal
Diffracted x-rays interfere between layers but signal is not strong enough to measure

Question 27

Explain Bragg’s law

Answer 27

Incident angle theta(i) and reflected angle theta(r) are equivalent
Path difference = 2dsin(theta)
Bragg’s Law: Strong diffraction if path difference is a whole number of wavelength
2dsin(theta) = n(lambda)
n = integer/whole number
d = spacing between crystals
If the two layers are in phase, there is strong interference

Question 28

Explain Bragg’s law with destructive interference

Answer 28

Different incident angle
2dsin(theta) = (2n-1)lambda/2, where 2n-1 is odd
When path difference is an odd number of half a wavelength
Leads to destructive interference
No scattering on the detector
Applies to layers of a crystal:
At layer 1: path difference is almost a whole number (will see diffraction)
At layer 50: wavelength is an odd number of half a wavelength which doesn’t obey bragg’s law
Destructive interference and no diffraction
Last layers could give a diffraction

Question 29

How are layers in crystals connected?

Answer 29

Layers connected via H-bonds, salt bridges, hydrophobic interactions
Crystalline lattice connected in different dimensions (3D)

Question 30

Molecular vs Crystalline scattering

Answer 30

Single molecule:
Diffraction pattern is related to the structure but too weak to detect in practice
Diffraction is too weak to distinguish from background of detector
Background noise of detector due to scattering of air
Signal to noise ratio is very low
Crystalline:
Much higher signal to noise ratio
Diffraction spots are clearly distinguished from the background scatter
Diffraction pattern is the same as that for the molecule but only observed in particular directions (made up of spots)

Question 31

Explain how crystalline diffraction works

Answer 31

Scattering only occurs if the path difference between waves from adjacent layers is a whole wavelength
Law is effective because there are huge numbers of molecules in a crystal (20,000 molecules along one side)
Crystalline samples amplify the scattered intensity
Makes it measurable in allowed directions
The positions of observed rays tells us about the orientation and spacing of the planes within the crystals
The variation in intensity (spots/reflections) is the same as the variation you would see in a single molecule
Some areas have no diffraction (no black spots on detector) due to destructive interference or weak electron density

Question 32

Explain crystalline diffraction in terms of equations

Answer 32

Fxtal (S) = Nfmol (S)
N is the amount of molecules in the crystal (signal is amplified by N)
Express structure factors as Fxtal(hkl), where h, k, l obey Bragg’s law
Only discrete values of S are allowed by Bragg’s law
Molecule: electron density function includes information about amplitude and phase, use integration
Crystal: Only consider points h, k, l so don’t integrate, use summation
Order of diffraction is n = 1 in protein crystallography

Question 33

Experiment set up and Bragg’s law in a real experiment

Answer 33

To record all possible reflections (scattering in every direction) the crystal is rotated during the experiment about a horizontal axis perpendicular to the beam
Incident X-ray angle is theta and X-ray is deflected through 2theta (diffracting angle)
X-ray still penetrates through other layers so all sets of planes in the crystal will give a diffraction spot when rotated into the Bragg condition
Collect one image every 1 degrees
For a complete dataset the crystal may have to be rotated by up to 180 degrees

Question 34

Compare Natural Vs recombinant sources to grow crystals

Answer 34

Natural sources are useful for big complexes since these can’t be easily reconstructed by recombinant expression Ex. ATP synthase purified from bovine heart muscle (cows)
Need to use recombinant sources for higher organisms (human proteins)

Question 35

What are the advantages and disadvantages of using E. coli to make recombinant crystals?

Answer 35

Well characterised organism
Grows rapidly (doubling time less than 30 minutes)
1-6L of culture can yield 1-200mg of purified protein
Problems: Difficult to make proteins with disulphide bridges. Reducing environment of the cytoplasm prevents formation of disulphides
Protein may over-express well but be unfolded. Control folding by expressing at lower temperatures. May be able to refold proteins by controlled denaturation/re-naturation
Proteins may be toxic for bacteria (ex. proteases). Try to get around this by tight control of expression. Or inactivation of enzyme by mutagenesis

Question 36

How is recombinant DNA technology used to make a protein more crystallisable?

Answer 36

Isolate gene of interest and clone into expression plasmid
E. coli expression system is the most common
Use fusion tags to improve expression and solubility of recombinant protein

Not all proteins are crystallisable as proteins are very dynamic/mobile
Best case: single, rigid domain
Avoid (or modify): Multi-domain proteins with flexible linkers and proteins with flexible N-and or C-termini
Use prior information (alpha fold/databases) to identify core, structured domains in the protein
Only focus on part that has enzymatic activity to obtain better crystals
Limited proteolysis: Use non specific protease to cleave off mobile parts of the protein
Usually best to remove protein tags added to enable affinity purification

Question 37

How is sequence analysis/bioinformatics used for protein crystallisation?

Answer 37

If cannot crystallise protein, use homologues
If sequence identify is over 20%, the structures will be very similar
Align sequences
Functional domains will be very highly conserved
Linkers have fewer structural constraints and are likely to be more variable in sequence between homologues
Increase solubility
Proteins can aggregate so won’t be able to crystallise
Mutagenesis to replace hydrophobic residues (must be similar in size) to make protein more hydrophilic
Use chemicals - Disulphide bridges can make protein aggregate so remove Cys residues by iodoacetamide
A ligand may stabilize the protein
If protein is known to bind a ligand (ex. ATP, drug), add to the protein before crystallisation
Ex. ATP reduces motion that otherwise interferes with generating the crystal
Use ligand that makes protein more thermostable
Protein has higher melting temperature = more rigid = more likely to crystallise

Question 38

How are crystals grown?

Answer 38

Dissolve protein in a solvent
Precipitate the protein slowly so it crystallises
CANNOT do this by open evaporation which works for salt crystals
Slow alteration of the solvent that the protein is dissolved in Ex. rising of salt concentration to remove water from the surface of the protein
Use precipitating agents (Ammonium sulphate (salt), Polyethene glycol) to reduce solubility of solute (protein) leading to crystallisation
Commonly use vapour diffusion methods
Type of controlled evaporation that does not allow complete drying out of the sample

Question 39

Vapour diffusion technique for controlled precipitation

Answer 39

Sitting drop or hanging drop technique
Dissolve protein of interest in buffer
Create reservoir with a higher concentration of precipitant/salt than the protein solution
Protein solution is mixed with equal volume of reservoir solution in a hanging or sitting drop / well
Chamber is sealed and placed in a controlled environment
Precipitant concentration in well is half of that in the reservoir
Water evaporates from drop and system reaches equilibrium
This increases protein and precipitant concentrations in the drop leading to supersaturation
Nucleation can occur, forming crystals
Sitting drops allows robotics usage to control drop size and make very small crystals

Question 40

What are the different techniques to make crystals?

Answer 40

Vapour diffusion
Micro dialysis
Microfluidics - tiny capillaries
LCP plate crystallisation - membrane proteins

Question 41

Explain the crystallisation theory

Answer 41

Stable state: protein and precipitant concentrations are low, protein remains soluble, no nucleation, drops are clear
Metastable state:
Heterogenous nucleation (growth) - won’t lead to crystal growth, precipitation, molecules stick together, hydrogen bonds, salt bridges and hydrophobic interactions between molecules
Homogenous spontaneous nucleation - crystal nuclei form from aggregation of protein molecules that have a regular structure needed to support growth
Unstable state: Proteins sticking together (aggregation), no nucleation, decomposition
Want slow nucleation - otherwise crystal might be unstructured
Look at graphs in notes

Question 42

What happens when precipitant and protein concentrations are too high or low?

Answer 42

If precipitant and protein concentrations are very high: formation of micro crystals. Rapid uncontrolled crystallisation. Consumed the protein so doesn’t form any proper crystals. Not useful for diffraction experiment
If protein or precipitate concentration is too low: metastable state not reached, insufficient nucleation, No crystals
Look at graphs in notes

Question 43

How is crystal formation optimised?

Answer 43

Use different combination of reagents
Screen for pH, salt, organic solvents
10 - 20 plates with different combinations of buffer
1000 experiments running in parallel
If no success, consider: Is there a flexible part that prevents the crystal from forming, looking for homologue

Question 44

Practical issues with crystallisation

Answer 44

We cannot predict which precipitant to use, so use trial and error
Commercial crystallisation kits are available
Other techniques: Batch crystallisation, Dialysis, Free interface diffusion

Question 45

How are robots used for crystallisation?

Answer 45

Liquid-handling robots
Can set up 96 plates in 2 minutes
Use 0.1mL of protein per drop

Question 46

What are the properties of crystals and why are solvents so important?

Answer 46

Form a regular array
Crystal is soft and wet NOT dry and hard
Crystals contain large solvent channels (30-70% is solvent = not all space in crystals is occupied by protein)
If allowed to dry out, crystal integrity is lost and proteins denature
Crystals are very fragile, require gentle handling
Presence of solvent allows crystals to keep their native properties: proteins are nearly in solution in a crystal Ex. can observe enzymatic activity
Can add ligand that will bind to the apo enzyme by travelling through channels
To determine the structure of the protein-ligand complex
Protein can still do chemistry: substrate –> product

Question 47

Why is it important to study the structure of membrane proteins?

Answer 47

50% of all drug targets are membrane proteins

Question 48

Examples of membrane protein related diseases

Answer 48

Cystic Fibrosis (ABC-transporter) - Single mutation affects trafficking
Sideroblastic anemia (ABC-transporter)
Chron’s Disease (MFS transporter)
Thyroid dyshormonogenesis (MFS transporter)
Glucose uptake in the intestine (Na,K-ATPase pump)
Multi drug resistance (RND-pump): Efflux of drugs out of gram -ve bacteria, involved in antibiotic resistance. Inhibiting a single transporter does not work as it mutates quickly so have to modify proteins so efflux pumps can’t export antibiotics

Question 49

Why are membrane proteins important?

Answer 49

30% of human genome codes for membrane proteins
Signalling, transport of substrates, extrusion of harmful substances, Ideal trug targets
To date there are almost 200,000 structures in PDB (cryo-EM or crystallography)
Only 6000 belong to membrane protein structures
Have limited info on membrane proteins

Question 50

What is important to consider when working with membrane proteins?

Answer 50

The protein is in an oil like (hydrophobic) environment
Must get it out in an aqueous solution
Removal from membrane may result in loss of function
Using a detergent allows oil (hydrophobic) and water to mix
Be careful when choosing a detergent because inner and outer membrane of gram negative bacteria are different composition

Question 51

What are the properties of lipids?

Answer 51

Hydrophilic head, hydrophobic tail
Glycosylation/phosphorylation affects charge of head group and how protein behaves during purification
Lipids are structurally and chemically different in different parts of the body

Question 52

How are detergents used to extract proteins from the membrane?

Answer 52

Detergents have similar structure to the lipid (hydrophilic head, hydrophobic tail)
Aggregate so hydrophobic interior is protected from aqueous environment
Lipid-like/charged detergents:
Charge altered by changing pH of buffer
Bad for crystallisation because can affect structure of the membrane lipid or can lead to aggregation
Good for functional analysis
Non-ionic detergents: Uncharged
Contain glucoside sugar moiety
Use for crystallisation

Question 53

Why is protein engineering important and how is it used for membrane protein crystallisation?

Answer 53

GPCRs and transporters are very unstable proteins
Don’t crystallise or diffract well
Engineering is necessary
Ex. T4 lysozyme, BRIL cytochrome (soluble proteins are attached to the membrane protein)
Ex. Stabilisation by random mutagenesis (increase melting temperature/stability)
Can use scaffolds like antibodies or nanobodies

Question 54

What are the 3 ways of creating membrane proteins?

Answer 54

Type II 3D crystal: Mix protein with precipitant
Hydrophobic part of membrane protein is masked by detergent
So crystal contact (charged residues) is at the ends of the membrane protein
2D crystal: Remove detergent and place pure protein back in lipid environment
Change ratio of lipid:protein
Generate 2D crystals that are 1-2 layers thick
Hydrophobic layers close enough for VDW interactions
Can’t use for X-ray diffraction, only for electron diffraction (crystals are very small)
Type I 3D crystal: Add lipidic cubic phase
Still have hydrophobic contacts
Crystals grow in 3 dimensions
Can use X-ray crystallography

Question 55

How are scaffolds (antibodies) used for membrane protein crystallisation?

Answer 55

To increase hydrophilic surface of the protein, to make crystal contacts, to stabilise protein
Antibodies can be raised against a specific antigen in the hydrophilic region of the protein
Antibodies have heavy and light chains
Constant domain: same amino acid sequence in all antibodies
Variable domain: consists of different amino acids, region where antigen/epitope binds, gives specificity
Fab antibody: when flexible hinge region is removed, contains V and C domain so still a little flexible
Fv antibody: only contains V domain, used for the crystallisation of GPCR, cytochrome oxidase and potassium channels, increases surface where crystal contacts occur

Question 56

How are scaffolds (nanobodies) used for membrane protein crystallisation?

Answer 56

Nanobodies: antibodies found in camels
Constant and Variable domain
Remove constant domain due to flexibility
Variable domain binds antigen/epitope
Crystal lattice is highly ordered
Not changing properties of protein (nanobody can still bind)
Stabilizes protein, promotes crystal contacts, epitope tagging (membrane proteins engineered to include tag recognised by epitope)

Question 57

What is Lipidic Cubic Phase (LCP) crystallisation

Answer 57

Using monoolein (artificial lipid)
Hydrophobic tail adapts different conformations depending on lipid:water ratio and temperature
At 30% water and 80 degrees monoolein forms cubic ordered structure with symmetry
Hydrophobic tails point outwards and pack against each other
Hydrophilic heads point towards hydrated channel
Add purified membrane protein to cubic structure
Protein inserts into the membrane due to hydrophilic interactions
It is very sticky like toothpaste
Must use sandwich method to grow crystal:
Sticky cubic phase protein + buffer between 2 glass plates
Diamond cutter to cut glass, tweezers to lift plate and loop to pick up crystals, then freeze in liquid nitrogen
Using diamond cutter can release glass fragments so be careful

Question 58

How do you interpret the diffraction pattern on the detector of a soluble protein?

Answer 58

High resolution spots on the outside
White line is the beam stock (lead put in front of the crystal) that will catch 98% of crystals that have not been scattered
Dark region on outside comes from the water within the crystal
Fuzzy dark patches: detergent diffraction, not ordered enough to give diffraction
The darker the spot on the detector, the higher the intensity (Intensity=Amplitude^2)

Question 59

How is diffraction data collected?

Answer 59

Crystals are very fragile to handle so use a nylon loop matched to shape of crystal to pick it up
Cryoprotection to prevent radiation damage (occurs due to high water content)
Use robots to collect data from a crystallisation plate without needing to harvest or freeze the crystal
Get 5-6A diffraction (if lucky)
Usually diffraction is 10-15A
Optimising crystals gets 8A
Keep screening a lot of crystals and conditions
Usually screen over 300 crystals in 24H
1 crystal out of 100 gives <8A
Use robots to mount many crystals

Question 60

Anisotropic vs Isotropic diffraction

Answer 60

Isotropic diffraction (edges): for soluble protein
Diffraction intensity is uniform due to cubic symmetry
Anisotropic diffraction (centre): for membrane protein
Intensity is not uniform, non cubic symmetry because detergent micelles create gaps in between regions of crystal structure
Gives weak signal
Missing information
Overcome by adding antibody, fusion or nanobody to create crystal contacts

Question 61

X-ray free electron laser (XFEL)

Answer 61

Use multiple microcrystals
Intense X-ray beam so crystal explodes due to radiation damage
Noisy diffraction pattern
No rotation

Question 62

Explain resolution in terms of Bragg’s law

Answer 62

Bragg’s law: path difference has to be a whole wavelength
Path difference = 2dsintheta
Strong diffraction if 2dsintheta = n*lambda (where n is an integer)
Max theoretical diffraction angle: 90 degrees (2theta =180 degrees)
For N = 1 (order of diffraction) and wavelength of 1 Angstrom
Theoretical spacing between layers (dmin) = 0.5Angstrom
dmin corresponds to the resolution limit of the data-how finely we are probing the structure in the crystal
The smaller the spacing, the higher the resolution (crystal is diffracting well)
Highest resolution is obtained at the highest diffraction angles
Crystal is rotated in front of the X-ray to satisfy angles of 2theta so obtain different d-spacings
Long distance interactions between layers (diagonal)= small d-spacing
Lowest d-spacing determines the resolution
Highest resolution for protein is 0.8A, small molecules can achieve 0.5A

Question 63

Why does the actual resolution of a protein crystal differ from the theoretical value?

Answer 63

Protein crystals have much lower resolution than theoretical value
Typically, 2thetamax = 30-40 degrees
40 degrees gives 1.9A resolution limit (for n = 1 and l = 1A) because sintheta = nlambda/2d
The highest resolution data are obtained at the highest angles
High resolution diffraction spots are on the outside of the detector

Question 64

How do misalignments in crystal impact diffraction on the detector?

Answer 64

Theory assumes crystals are perfect, but in reality they are not because: Crystal growth defects (missing molecules) + inherent flexibility (side chains, domain structure, radiation damage)
So, unit cells are not perfectly aligned
Crystals are composed of mosaic blocks (multiple unit cells lined up) which are oriented slightly different from each other
Leads to elongated spot on detector as one block can overlap with neighbouring block

Question 65

Explain where high and low resolution information is found on the detector

Answer 65

High angle=smaller spacing between layers (d-spacing)=high resolution=detailed structural information=close to edge of detector
Low angle= low resolution=can detect overall structure but water molecules or ligands will be unclear

Question 66

How are X-rays made?

Answer 66

In the lab: rotating anode generator
Produces X-rays by accelerating electrons into a copper target (anode)
Low intensity (long exposure times)
Fixed wavelength (depends on anode)
1.5A for copper anode: ideal for C-C bonds (also 1.5A in length)
OR Synchrotron: particle accelerator
Electrons are accelerated around a circle and emit radiation
High intensity (short exposure times)
Tuneable: can select wavelength (between 0.7 and 2.5A)

Question 67

How is X-ray crystallography data in practice? Rotation and cry-cooling

Answer 67

We need to measure as many diffraction spots as possible
Rotate crystal through 180 degrees to collect data (1degree per image)
At synchrotron, collection of complete dataset (180 - 360 images) takes 2 minutes
Usually, data are collected from cryo-cooled crystals
X-rays induce radiation damage
Add a cryo-protectant solvent (anti-freeze ex. glycerol)
Prevents formation of ice crystals within solvent channels of protein crystal
Pick up crystal in a thin nylon loop
Freeze in liquid nitrogen for storage/transport
Slows down the free radicals and protects against radiation damage
In modern labs this is done by a robot

Question 68

Explain radiation damage by X-rays

Answer 68

X-rays damage biological samples
Photoelectric effect: release of electrons that generates of free radicals
Damages molecules within crystals
Loss of high resolution spots
Low distances are affected the most so lose high resolution information
Reduce by cryo-cooling

Question 69

Explain the phase problem

Answer 69

F(S) = structure factor of all diffracted waves (is represented by a complex number). It has an amplitude and phase, it is a wave
In a diffraction experiment, we can only measure the intensity of the scattered waves, F(hkl)
The darker the spot, the higher the intensity (Intensity = Amplitude^2)
We cannot measure the phase
We don’t know which part of the molecule gave rise to the phase shift/diffraction pattern
Need to solve the phase problem: By molecular replacement/MIR/MAD, homologous structure or introduce a heavy metal that binds to specific residues

Question 70

How is an electron density map calculated and what are the origins of errors?

Answer 70

Calculation of maps requires |F(hkl)| and phi(hkl)
Get |F(hkl)| (amplitude) from diffraction pattern
Get phi (hkl) by solving the phase problem
H, k, l are determined from spot positions
But there will be errors
Intensity errors: Can’t measure spots on detector very accurately as intensity is weak at high resolution
Rmerge = average error in intensity measurements, usually 2-4% for good data, 10-12% is tolerable
Phase errors

Question 71

Explain the importance of high resolution data

Answer 71

At 6-7 Angstroms: can see helices as rods and sheets as slabs of density but can’t build an atomic model into maps
At 1.5 Angstroms: atomic resolution. Carbon atom is around this size so can build atomic model
Can check if model is correct Ex. in a membrane protein hydrophobic groups must be on the outside

Question 72

How does model building and refinement work to determine the structure?

Answer 72

Manual model building does not build a completely accurate model but it is required to get to the starting point for refinement
First fit protein sequence to the electron density
Refinement adjusts atomic model to give a better fit
Computational process
Ensures atomic model has ideal stereochemistry
Calculates improved phases which will give better maps
Structure determination requires several rounds of map calculation, model building refinement

Question 73

Explain the observations to parameters ratio in crystallographic refinement

Answer 73

Parameters: Have 4 parameters per atom (X, Y, Z, B)
B factor: vibrational measurement
= 18,000 reflections in total
Observations = 23,000 amplitude measurements
Observation to parameter ratio is 1.25
So don’t have sufficient parameters to refine the structure
Parameters are NOT independent: protein structure is made of atomic arrangements, we know bond lengths, bond angles, dihedral angles, chiral centres, planar groups, VDW radii
Knowledge about ideal stereochemistry increases number of observations = higher ratio = higher resolution

Question 74

What do we want to minimise with refinement?

Answer 74

During refinement we minimize a target function that expresses the difference between Fobs and Fcalc
F(obs) = structure factors measured on detector
F(calc) = structure factors calculated
This is the difference between model geometry and ideal geometry
We want to minimize the function

Question 75

Explain what positional refinement is

Answer 75

In each cycle the program determines the direction each atom should move to improve the fit of the data
To satisfy electron density
Each atom moves a small step (max 0.5A)

Question 76

Explain what temperature factor (aka B-factor) refinement is

Answer 76

B factors model the static disorder and vibrational motions of atoms in the structure
It affects the diffraction pattern
4th parameter in our models
If side chain is ordered/not moving = low B factor = good diffraction pattern on detector
If side chain is labile/rotating = high B factor = no diffraction pattern on detector
Centre of structure: low B factor
Outside + surface loops: high B factor
Nothing to do with thermal stability

Question 77

Why is adding bound water molecules important?

Answer 77

Waters are added at high resolution (<2.7A)
Waters in binding site are important for enzyme chemistry or protein-ligand interactions

Question 78

What does the R factor tell us?

Answer 78

During refinement, calculate the R factor to monitor the progress
Tells us how good the fit is
Try to minimize the R factor
Includes Fobs and Fcalc

Question 79

How is the progress of refinement monitored?

Answer 79

Stop refinement when R factor stops dropping
Check for unusually high B factors
Look at Ramachandran plot for bad phi-psi angles (don’t want outliers)
Look at notes for data collection of refinement