X-ray Crystallography

Question 1

What are x-rays

Answer 1

a traveling electromagnetic wave, which has oscillating electric (E) & magnetic fields (B) at right angles to each other
(wavelength=0.2nm)

Question 2

X-ray crystallography process

Answer 2

1. grow crystal
2. put in front of x-ray beam
3. calculate e density map
4. build protein molecule/structure exactly as how it looks in the crystal

Question 3

What gives diffraction pattern?

Answer 3

interaction of x-rays with e in crystals

Question 4

Crystals

Answer 4

-shape of molecule defined by shape of e clouds association w/constituent nuclei

-crystal = ordered 3D array of distinctively shaped distributions of e

Question 5

Electric field (E)

Answer 5

-describes electrostatic force felt by charged particle due to presence/motions of other charged particles

-can be represented as a vector

-tells you which way a +ve charge will move

Question 6

Diffraction

Answer 6

-objects scattering x-rays at similar wavelengths as that of light
-get diffraction when wavelength = incident ray

Question 7

Waves

Answer 7

-Phase angle θ = indicates position in the wave cycle
-Travelling waves phase varies with position and time
-One cycle corresponds to phase angle 2π (360º) ie. one full rotation of the line
-Y = maximum displacement from the x-axis
-A = amplitude
2πx/λ = wavelength
Y = A.cos(2πx/λ + Φ)

() = phase angle varies with x

Question 8

Adding waves

Answer 8

Constructive interference
-Two waves of amplitude A perfectly in phase = results in 2A in phase with components
-Phase shift = 2nπ

Destructive interference
-Two waves of amplitude A perfectly out of phase = amplitude 0 (they cancel out)
-Phase shift = (2n-1)π

Intermediate example
-Two waves (A) partially out of phase = resultant has different phase to components
-Phase shift = a fraction of 2π

Question 9

X-ray scattering 1e

Answer 9

1. Incident x-ray (oscillating E-field)
2. E responds by oscillating
3. Oscillating electron emits x-rays over a wider angle

Question 10

X-ray scattering 2e

Answer 10

• 2ē start oscillating and emitting x-rays in all directions
• Waves from 2ē in close proximity interfere/interact with one another so we have to add the waves
  -The detector measures the added waves
  -Scattering pattern is the result of adding the two scattered waves
  -Diffraction pattern depends on structure -contains information about the structure of the 2ē, the spacing and relative position to one another

-X-rays come in, interact w/2ē and x-rays scatter w/angle of 2θ
-Incoming x-rays are in phase; top ray travels further before hitting ē = introduces path difference between scattered rays = results in phase shift
-Phase shift is due to structure of ē
-Structure affects A and phase of resultant wave in direction 2θ

-K0 = incident wave vector; k = scattered wave vector
magnitude = l/λ for both; assume that interaction of ē is not altering the wavelength
r = relative position of ē (structure)
path difference between waves scattered by ē at origin and one at position r = AB-OC
phase shift= -2π ((path difference)/λ) * - sign is just a convention
=2πr ( (cos⁡(α)-cos⁡(β) ) /λ)

Question 11

Scattering vectors (S)

Answer 11

-Magnitude and direction of S contain information about λ and scattering angle θ, but it does not point in direction of scattering
-At fixed λ, S varies only with θ
2k because k = k0
length of S |S| = 2k.sin = (2/λ).sinθ

Question 12

Calculating phase shift

Answer 12

-Between ray scattered form O and ray scattered from B: ϕ=2πrS
-Phase shift depends on r and S
-Contains information about the structure (position of 2ē), wavelength, and scattering angle

Question 13

Wave equation

Answer 13

General wave equation:
Y = A.cos(2πx/λ + Φ)

Wave equation using complex numbers:
Y = Ae^{i(2πx/λ + Φ)}

If we are only concerned with phase shifts:
Y = Aei(Φ)

Question 14

Structures with 2e

Answer 14

Resultant wave:

Y = Aeⁱ⁽⁰⁾ + Y = A.e^i(2πrS)

Wave scattered from origin (r=0) + wave scattered from r

Question 15

Structures with +2e

Answer 15

-Each ē has a different position and vector (r)
-Add all the waves scattered from all electrons
* assume all ē have same amplitude
J = increment of system
N = denoting amount of electrons present

f(S) = Σ ^N _j=1 e ^i2πrj.S

-9 scattered waves of same amplitude go in same direction; their phase depends on the position of the electron it scattered from
-Amplitude and phase of resultant wave depends on that of components

Question 16

Structure factor f(S)

Answer 16

-f(S) describes diffraction pattern = how the diffracted waves in each direction are related to the structure (r)
-f(S) = a complex number
-f(S) is a wave; has amplitude, phase & direction

f(S) = |f(S)| e^iΦ(S)

Question 17

e density function

Answer 17

-E density gives 3D shape of the protein
-E density function = density you can measure at a specific position, r, within the molecule
ρ(r)dxdydz
-By defining ē density at specific position, can map overall ē density on protein molecule
-Protein molecule emits ē in all directions
-Protein consists of different atoms which have different ē clouds
-Proteins have higher ē density at the core
-Amplitude depends on ē density/probability of x-ray interacting with ē/ē cloud at particular positions

Each scattered wave has the form: [ρ(r)dxdydz] e ^i2πrS

Question 18

Total scattering by a molecule in one direction

Answer 18

-add up all contributions
-Have different f(S) in different directions

Question 19

Diffraction pattern of molecules

Answer 19

-Contains information about ē density and phase-shift due to different positions of atoms within molecule

Question 20

Structure factor of a molecule f(S)

Answer 20

-f(S) = sum of phases at different positions of r
-Describes total scattering by the molecule as a function of direction
-Amplitude has to be taken into account = different parts of molecule have different A

-To calculate for every position of the molecule:
-Need to consider all possible dimensions molecule can scatter x-rays
-d(r) = dxdydz

Fourier transform equation:
f(S) = ∫ ^∞ _∞ ρ(r)e ^{i(2πrS) dr}

Question 21

Fourier transform

Answer 21

-Fourier transform = the total scattering in the direction associated with S, from an object described by electron density function is the sum of the waves scattered from every point in the object
-Adding all waves coming from the molecule gives the structure factors

Question 22

Inverse fourier transform

Answer 22

-If you can measure all possible f(S) values (ie. The diffraction pattern in detector), and we know the phase shift, we can calculate ρ(r) ie. structure
-Can apply this theory as long as you can measure every single possible diffraction spot on detector

p(r) = ∫ ^∞ _∞ ρ(r)e ^{-i(2πrS) dS}

-Fourier transform gives information about the total scattered waves coming from a protein molecule that contains electron density information = inverse measures all these waves to work out what the structure looks like

Question 23

Diffraction from crystalline samples

Answer 23

-Diffraction pattern from protein crystal contains same information as scattering from a single molecule = just an enhancement in the signal because you have billion copies of protein in the crystal
-Measure how dark the spots are

Question 24

Why can you not use single molecule for x-ray diffraction?

Answer 24

-Too small; cannot rotate it in 3D for every position
-Diffraction would be too weak
-X-rays cause ionisation radiation damage so molecule will blow apart when electrons get excited and move to different spin

Question 25

Advantages of crystals

Answer 25

-Contain billions of molecules
-Amplify the scattering making it detectable
-FACT: some properties of crystals restrict scattering in certain directions (not a problem)

Question 26

Reflection from semi-transparent crystal planes

Answer 26

-Only ~2% of x-rays are scattered from a protein crystal; most pass through the sample
-Incident angle θ = angle of reflection/scatter θ
-Due to constructive interference, if you have 20,000 waves in phase, you can increase diffraction signal significantly

Question 27

Braggs Law

Answer 27

-Path difference = 2dsinθ
-Get strong diffraction only if path difference between waves from adjacent layers = nλ
-No scattering when path difference = odd number of a half wavelength = destructive interference = 2dsinθ = (2n-1)/λ-2
-The positions that you measure on the diffraction spots tells us about the position of the electrons within the protein; can measure the spacing that tells us from which plane the crystal diffraction is coming from

Question 28

Crystalline diffraction

Answer 28

-Structure function of crystal = n x f(S)
-n = amount of molecules in the crystal
-S is only allowed to be discrete positions within the planes, use hkl for distinct space in the crystal
-Hkl = allowable directions that give constructive interference

Question 29

Molecular vs crystalline scattering

Answer 29

-Molecule = diffraction pattern too weak to detect in practice
-Signal:noise ratio = very low (ie. Hard to detect diffraction from background)
-Crystal = same structural information as molecule just amplified based on particular positions allowed by Braggs Law
-Bright enough to be detectable because…
-Signal:noise ratio is much higher (ie. the diffraction spots are clearly distinguished from any background scatter

Question 30

Braggs Law - experimental setup

Answer 30

-For a given λ and d, only certain angles allow constructive interference between planes
nλ = 2dsinθ
-Rotate crystal about horizontal axis perpendicular to x-ray
-Satisfy Braggs Law with scattering angle of 2θ
-Collect an image every ~1º (may do up to 180º)
-All sets of planes in the crystal give a diffraction spot when rotated to satisfy Braggs Law
-Black rings around detector in a picture = ice forming on crystals
-Intensity is proportional to A2 (amplitude)

Question 31

No use crystal example

Answer 31

-A lot of lattices at different orientations so you don’t know where diffraction pattern is coming from, thus cannot work out electron density factor

Question 32

Growing crystals: recombinant technology

Answer 32

-natural sources still used for big complexes since they cannot be easily re-constituted by recombinant expression
-E.g. F0F1-ATP Synthase purified from bovine heart muscle
-Multi complex proteins can be made

-Produces large quantities of protein easily
-Easy to use in lab
-Easier purification via tag technology
-Easy to obtain protein sequence needed for model building via cloning/DNA sequencing
-E. coli easiest and cheapest system
-Disadvantages: no post-translational modifications (e.g. glycosylation, phosphorylation) which may be important for folding and activity of protein in humans
-sf9 cells (insect) = second most common

Question 33

Growing crystals: E. coli advantages and disadvantages

Answer 33

ADVANTAGES:
-Grows rapidly in liquid cultures; doubling time ~30 mins
-Can upscale easily; 1-6L of culture can yield 1-200mg of purified protein

DISADVANTAGES:
-Difficult to make disulphide bonds; e.g in cytoplasm because of reducing environment
-Protein may over-express but be unfolded
-Control folding by slowing down process, lowering temperature, co-expressing w/chaperones
-Can refold proteins by controlled denaturation/renaturation during purification
-Inclusion bodies = aggregates of unfolded protein
-Proteins may be toxic for bacteria (e.g. proteases)
-Either inactivate protein by mutagenesis or use different host

Question 34

Optimising crystallisability

Answer 34

-Proteins that don’t crystallise:
-Multi-subunit domain with linker that is flexible to bind with other proteins
-Best proteins:
-single, rigid domain
-Can separate two domains and crystallise individually, then analyse how they are likely to come together
-Can use bioinformatics to identify the core of the protein to remove any flexible parts
-Can look for homologues: over 20% sequence identity very likely to have similar fold
-Identify highly conserved functional domains
-Look for protein with less flexible regions i.e. shorter loop to connect two domains = better for crystallisation
-Can use limited proteolysis:
-Add non-specific protease to identify stable fragments (folded enough to resist cleavage)
-Increase solubility
-Mutation of hydrophobic residues predicted to be on surface
-Remove/modify solvent-exposed Cys residue to prevent aggregation due to diS bridges
-Presence of ligand may stabilise the protein
-If protein is known to bind a ligand (e.g. ATP) add it before crystallisation
-Useful pre-screening technique –test stabilising effect of ligand on protein melting and use most effective compounds in crystallisation trials
-Purify, remove protein tags, and crystallise

Question 35

Growing crystals

Answer 35

-Precipitate protein slowly so that it crystallizes
-Cannot do this by evaporation
-Use precipitating agents (e.g. Ammonium Sulphate, PEG)
-Use vapour diffusion method: type of controlled evaporation that does not allow complete drying of sample
-Other techniques
-Micro dialysis –loses a lot of protein; also removes water = metastable phase
-Microfluidics –have thin capillaries of protein and precipitant and let them diffuse
-LCP plate crystallisation –sandwich protein and precipitant between 2 glass planes

Question 36

Growing crystals: vapour diffusion technique

Answer 36

-Hanging drop
-Advantage: can increase size of drop
-Sitting drop
-Advantage: can transfer on another mixture of chemicals
-Both techniques have very similar equilibrium kinetics
1. Place precipitant solution in reservoir
2. Mix equal volumes of precipitant and protein solution
3. Initial precipitant concentration in well is half that in the reservoir
4. Seal the chamber
5. System slowly reaches equilibrium
6. Because of unequal volumes, water is extracted from the well, increasing protein & precipitant conditions –this favours precipitation to get a crystal

Question 37

Phase diagram of protein crystallisation

Answer 37

Metastable zone for crystallization
-Start by increasing [protein] or [precipitant]
-By removing some water molecules, proteins come closer together and form small crystals (disulphide bonds, hydrogen bonds, etc)
-State where protein is close to aggregation zone but remains in solution
-Protein can go back to being soluble monomers, microcrystals, or aggregation –have no control over that (relies on how good the sample is)

Wrong need right conditions for formation of large 3D crystals
-Wrong conditions lead to microcrystals
-If abundant nucleation sites, you are inducing nucleation of protein next to it, but not getting bigger crystals
-High [precipitant] + high [protein] = protein will aggregate and not form crystals
-Low [precipitant] + high [protein] = protein is still in solution

Question 38

Optimisation of crystal

Answer 38

-In each well: buffer, salt, organic molecule or secondary salt that tries to precipitate or make protein metastable
-If you get crystals, can start varying reagents and concentrations
-Use additive screens –can screen 2000 conditions in single experiment
-If you don’t get crystals
-Check for activity, functionality, folding of protein
-Try homologues
-Try functionally related distant proteins
-If all fails move to next target
-Practical issues:
-Cannot predict which precipitant to use but have experience to see what works/doesn’t
-Do trial and error; lots of set-ups
-Try other techniques: batch crystallisation, dialysis, free interface diffusion

Question 39

Lysozyme

Answer 39

-Small robust protein in saliva
-Grows within 15mins of setting up

Question 40

Properties of crystals

Answer 40

-Soft and wet (not hard)
-Contain large solvent channels (30-70% solvent content)
-Important to keep protein in functional state
-Disadvantage: different physiological conditions of crystal and protein (e.g. different pH) –take care with biochemistry
-If it dries: crystal integrity is lost, protein may denature, hydrogen bonds stabilised by water molecules = break down = protein starts to aggregate
-Crystals still have enzymatic activity
-Can run assay e.g. fluorescent spectroscopy to observe crystal structure is physiologically relevant
-Can add ligand to crystallised protein “soaking” to determine structure of protein-ligand complex

Question 41

Braggs Law and resolution

Answer 41

-Max theoretical diffraction angle: θ = 90º (2θ=180º)
-Sinθ = nλ/2d, for n=1 and θ = 90º and typical λ=1.0Å in an x-ray experiment
-Min theoretical distance (dmin) = 0.5Å = resolution limit of data
-Highest resolution at highest diffraction angles (edge of detector)
-C atom ~ 1.4Å
-Very long distances are very rare, thus 0.5Å is rare because crystals are not perfect
-Low resolution gives information of structure and overall shape but loses information about sidechains, ligand, water molecules present, etc

Question 42

Crystals are not perfect

Answer 42

-Crystal growth defects
-Inherent flexibility (side chains, domain structure, radiation damage)
-Results in unit cells not all perfectly aligned
-Crystals made of mosaic blocks oriented slightly different from one another
-Within one block all unit cells align
-Lower resolution diffraction

Question 43

Real protein crystals

Answer 43

2θmin ~ 30-40º
n =1 and λ = 1.0Å
dmin = 1/2sinθmax = 1.9Å

Question 44

Producing x-rays

Answer 44

-Rotating anode generator
-Accelerates electrons into a copper anode
-Low intensity (long exposure times)
-Fixed wavelengths (depends on anode; for Cu = 1.54Å)
-Synchrotron (particle accelerator)
-Electrons accelerated around a circle and emit radiation
-High intensity (v. short exposure times 0.01-10s)
-Tuneable; can select wavelength (0.7-2.5Å)

Question 45

Cryo-cooled crystals

Answer 45

-To reduce free radicals from x-ray radiation travelling in water
-Need to add cryo-protectant solvent (anti-freeze e.g. glycerol)
-Prevents formation of ice crystals within solvent channels
-Crystal placed in nylon loop
-Frozen in liquid N for storage/transport
-At beamline, crystal maintained in stream of N gas at ~100K
-Cooling prevents drying out and protects against radiation damage

Question 46

Radiation damage

Answer 46

-Starts affecting high resolution (long planes)
-Free radicals break bonds/lattice
-Free radicals propagate in water
-Photoelectric effect –releases electrons

Question 47

Phase problem

Answer 47

-In diffraction experiment, can only measure intensity of scattered waves:
I(hkl) = F(hkl)2  I = A2
-Cannot measure phase so don’t have phase shift information; needed to solve structure
-3 methods: MIR, MAD, MR
-MIR = put heavy atom (e.g Mercury, gold, platinum, etc) of known structure to guess phase of atom to improve maps
-MAD = altering wavelength of experiment
-Molecular Replacement (MR) = most common

Question 48

Calculation of maps

Answer 48

-E density map:
-Amplitude F(hkl) from diffraction pattern (h, k, l = determined spot positions)
-Phase by solving phase problem MR
-Use inverse Fourier transform to give ρ(r)
-Errors
-Rmerge = average error in intensity measurements
~ 2-4% for good data
~10-12% is tolerable
-Impacts interpretation of data

Question 49

Importance of high-resolution data

Answer 49

-At low resolutions (6-7Å) = helices as rods and sheets as slabs of density
-Cannot build atomic model
-Main chain carbonyl-O “bumps” = barely visible  high likelihood of misfitting model
-Automatic building feasible at ~2Å

Question 50

Model building

Answer 50

1. Trace the polypeptide chain (by connecting Ca atoms)
2. Fit amino acid sequence to density

Question 51

Model refinement

Answer 51

-Gives better fit to observed data = better maps
-Helps ensure atomic model conforms to ideal stereochemistry; corrects distortions introduced from manual model building
-Can calculate improved phases

Question 52

Observation/parameter ratio

Answer 52

-Limitation
-Affected by resolution
-Increase ratio by introducing stereochemistry (prior knowledge about protein/bonds)
-Know bond length, allowable bond angles, dihedral angles, chiral centres, aromatic planar groups, Van der waals radii, etc
-Research protein data banks
-During refinement minimize target function that expressed:
-Discrepancy between structure factors Fobs¬ and Fcalc
-Discrepancy between model geometry and ideal geometry

Question 53

Temperature factor (B-factor refinement)

Answer 53

-Models static disorder and vibrational motions of atoms in structure (in x, y, z, space)
-B=8π[u^2] , u = atomic displacement
-High B-factors = electron density is smeared out and disappears from contoured maps
-Refinement determines optimum B-factor for each atom
-B-factors should be lowest in centre of structure = increase moving towards surface
-Surface loops = high B-factors (more mobility)
-Difficult to build loops because atoms may appear as weak density
-Atoms should be omitted if there is no density
-High B-factors can indicate problems in the model

Question 54

R factor

Answer 54

-Fractional difference between observed and calculated diffraction
-Want to be as low as possible ~24%
-Overfitting = reduce R but not improving the model
e.g. placing atom in density blob that is just noise

Question 55

Monitoring progress of refinement

Answer 55

-Check R
-Calculate new electron density map after each refinement
-Check for unusually high B-factors
-Check stereochemistry fits
-Check Ramachandran plot for phi-psi angles
-Re-build model where necessary

Question 56

Finished model

Answer 56

-R minimised
-Minimised unaccounted density
-Ramachandran plot has no outliers
-No stereochemical outliers

Question 57

Examples of membrane protein related diseases

Answer 57

• Cystic fibrosis (point mutation in ABC-transporter)
• Sideroblastic anaemia (ABC-transporter)
• Chron’s disease

Question 58

Membrane facts

Answer 58

• 30% of human genome encodes for membrane proteins
• Signalling, transport of substrates, extrusion of harmful substrates
• Ideal drug targets

Question 59

Working with membrane proteins

Answer 59

• Protein is in oil-like environment
• Try to get it out of aqueous solution but still maintain its function
  o Very hydrophobic so must use detergent
  o Careful because removal from membrane may result in loss of function

Question 60

Lipids

Answer 60

• Hydrophilic head (can be glycosylated, phosphorylated, etc)
• Hydrophobic tail

Question 61

Detergents

Answer 61

• Mimicking structure of lipid
• By mixing with membrane protein, can extract protein out of membrane
• When extracted, covered by detergent
  o Barrier prevents crystallisation
• Lipid-like detergents
  o Amphipathic and zwitterionic (can switch to being +ve or -ve)
  o Good for functional work; bad for structural work
  o During crystallisation, change pH → change charge of lipid
• Mimicking structure of lipid
• By mixing with membrane protein, can extract protein out of membrane
• When extracted, covered by detergent
  o Barrier prevents crystallisation
• Lipid-like detergents
  o Amphipathic and zwitterionic (can switch to being +ve or -ve)
  o Good for functional work; bad for structural work
  o During crystallisation, change pH * Mimicking structure of lipid
• By mixing with membrane protein, can extract protein out of membrane
• When extracted, covered by detergent
  o Barrier prevents crystallisation
• Lipid-like detergents
  o Amphipathic and zwitterionic (can switch to being +ve or -ve)
  o Good for functional work; bad for structural work
  o During crystallisation, change pH → change charge of lipid→ protein becomes unstable→loses activity
• Non-ionic detergents
  o Preferred
  o Mimicking hydrophilic head and hydrophobic tail
  o Can alter chain length
  change charge of lipid protein becomes unstable loses activity
• Non-ionic detergents
  o Preferred
  o Mimicking hydrophilic head and hydrophobic tail
  o Can alter chain length
  protein becomes unstable loses activity
• Non-ionic detergents
  o Preferred
  o Mimicking hydrophilic head and hydrophobic tail
  o Can alter chain length

Question 62

Membrane protein crystallisation

Answer 62

1. Get protein out of membrane using detergent (3 methods)
   a. Protein-detergent complex
   - Purified protein → crystallise it (e.g. ammonium sulphate, vapour diffusion, etc)
   - Crystal contacts = charged area of molecule; rest is masked by detergent → unstructured; no regular pattern
   b. Extract membrane lipids and mix with protein ie. exchange detergent w/lipid
   - Use different solvents → mix → 2D crystal
   - Crystal packing happens along hydrophobic portion
   - Advantage: get layer of crystals
   - Disadvantage: cannot use x-ray crystallography need to use ē microscope
   - Not commonly used: hard to control formation of 2D crystals; can only cover ±30º so get a noisy model
   c. Lipid Cubic Phase
   - Synthetic lipid
   - Mix protein w/lipid
   - Allows to grow 3D crystals because of structure of lipid

Question 63

Protein engineering

Answer 63

• GPCR’s and transporters are extremely unstable proteins
  o Do not crystallise or diffract well
  o Engineer to introduce soluble proteins within membrane protein: T4 lysozyme, BRIL cytochrome
  o Can stabilise protein by random mutagenesis
  o Use scaffolds (e.g. antibodies, nanobodies)

Question 64

Scaffolds

Answer 64

• Use other protein to bind to the protein of interest
  o Commonly use antibodies (Fab or Fv)
  o Can use nanobodies: antibodies found in camelids families; simpler structure
• Increase hydrophilic surface of protein and make crystal contacts → increases crystallisable area
• Successful for crystallisation of GCPR (Fab), cytochrome oxidase (Fv) , potassium channels (Fab)

Question 65

Fusions

Answer 65

• Successful for GPCR structural biology
  o In GPCR: intracellular loop 3 is v. unstable → replace with soluble protein

Question 66

Lipid cubic phase (LPC) crystallisation

Answer 66

• Synthetic lipid (monoolein)
• Occurs at 25% water and 80ºC
• Use glass plate format: mixture + crystallisation components
• Problem: taking crystals out of glass plate format → use diamond cutter

Question 67

Collecting diffraction data -membrane proteins

Answer 67

• Pick up crystals with nylon loop
  
  • Crystals usually 5-100μ
  • Cryoprotection to prevent radiation damage
  • Hard to freeze because detergent likes organic molecule used for cryocooling
  • Alternative: take crystallisation plate and mount in front of x-ray beam → rotate the plate → limited by ±30º otherwise start hitting plastic which will absorb x-ray
  • Lucky to get 5-6Å; usually get 10-15Å
  • Optimise crystals to get 8Å; keep screening crystals in varying conditions
  • Screen +300 crystals in 24hrs
  • 1/100 gives high resolution
  • No control of how these molecules pack
  • Severe anisotropic diffraction

Question 68

E density maps - membrane proteins

Answer 68

• Lots of errors because missing a lot of data

Question 69

Model building and refinement -membrane proteins

Answer 69

• Cannot see side chains at 4Å
  
  • Refine and go back to model
  • α-helices: charged residues will point towards interior of protein (water-soluble portion); hydrophobic residues sit outside of helix and point towards the membrane as it makes contact w/lipid

Question 70

X-ray free e laser (XFEL)

Answer 70

• Intense x-ray source → one exposure of crystal makes it explode because of radiation damage
• Can use microcrystals
• Use different magnets ~3km
• Can measure diffraction pattern right before crystal disintegrates
• Used for GPCRs, transporters, etc

Question 71

Future of protein crystallography

Answer 71

• Cryo-EM
  o Can crystallise protein directly on grid with EM
  o Disadvantage: slow technique; data processing takes +24hrs; limited by size of protein (>100kDa)
• Alphafold
  o AI for predicting structures
• Crystallography
  o Rapid screening of samples for drug development (collect data in <2mins)
  o Time resolved experiments at room temperature: can collect data before radiation damage as enzyme is turning over catalysis
  o Can do photosensitive experiments if protein is excitable by particular wavelength
  o X-ray data collection while performing other measurements on crystals e.g validate oxidation state of metal bound to crystal
  o Processing data is fast
  o Easy access