crystallography

Question 1

what is the storage form of DNA in chromatin

Answer 1

a long piece of DNA wrapped around a histone octamer

Question 2

a beta 1 - 42

Answer 2

peptide that form fibres and is deposited in plaques in the brains of those with Alzheimers disease

• structure was determined via solid state NMR

Question 3

optical microscope vs impossible xray microscope

Answer 3

optical microscope
lens recombine scattered visible light to reconstruct image

x rays shone onto a crystal
refracted beams
but can only measure intensity of waves = amplitude

phase info about relative time of arrival at the detector is lost

there cannot form an image
no lenses capable of focusing x ray

analytical procedures
comp calculations = phase info
then can construct electron sensitive maps
can put electron models = structure

cannot measure phases
can amplitude
phases contain bulk of structural info in image formation in X-ray crystallography
but all we can do is collect amplitudes

amplitude is hight of wave above mean

Question 4

x rays

why do e use ansgstroms in crystallography

Answer 4

high frequency
energetic
good penetrating power
electromagnetic radiation w short wavelength

1A = 10 to the minus 8 cm
0.1nm
close to the distances between bonded atoms C-C = 1.5A

Question 5

x ray diffraction

Answer 5

• must produce crystals of the protein or fibres = difficult - protein crystals made by repetition of a unique block many times into a strict lattice
  unit cell hs to be repeated by translation in all dimensions
  alll unit cells oriented in same way = 3d block
  proteins highly solvated in the crystal
  texture of cheesee crumble under manipulation

a static image is produced which is the average of all the molecules which we presume are all the same in the crystal to determine atomic positions

the phase problem associates with the way that we measure the diffraction data

any size protein
as small as sodium chloride and as big as 12 kDa

drug discover - can produce imaged of binding sites for ligands

Question 6

bimolecular techniques to look at what?

Answer 6

DNA and replicate 
enzymes and catalysis 
virus organisation 
nucleosomes and DNA storage
Ribosomes and protein synthesis 
Receptors and signalling 
Antibodies
ion channels
photo reaction centre
chaperones
transcription factors

Question 7

b dna

Answer 7

hydrated form of DNA

Question 8

calculate how much solvent is likely to be in a crystal
crystal solvent content
V = 1 / ( 1.23M / (V/Z) )
v = solvent content

M  = molecular weight of protein
V = vol of unit cell = x x y x z
Z = no of protein molecules in unit cell

Answer 8

10% 90 %

early stage check check its sensible

Question 9

how do the spots on diffraction patterns arise?

Answer 9

x rays interact w atoms 
scattered by cloud electrons
waves shoot of in all directions 
most goes str8 through 
due to diffraction by electron cloud 
disperses rays over a significant anlgle scattered waves interfere w each other 
constructive and destructive

depending on symmetry of the atom

Question 10

constructive interference

Answer 10

amplitude doube

phase is same

Question 11

and destructive

Answer 11

annihilate each other - no signal

Question 12

braggs law

n lambda = 2d sin theta
where theta = angle of incidence
d = distance

tells us when u get a diffraction spot occurring

stops considered existing as planes in a crystal

Answer 12

diffracted waves from planes of atoms one produce constructive interference when their path difference is an integral number of wavelengths

Question 13

Ewald sphere

Answer 13

constructive interference only occurs as reciprocal lattice points pass through the sphere

after diffraction occurs
diffraction spots exist on an inverse lattice
edge a in real unit cell = length of reciprocal lattice = 1/a

Question 14

Fourier transformations

enables dissection or construction of complex wave forms from simple constituents

the diffraction pattern of an object is its Fourier transform

Answer 14

enables domain inversions

from real space to diffraction space
=1/distance

time to frequency 1/time in NMA

optical diffraction patterns stimulate Fourier transforms

Question 15

lattice sampling of the molecular transform

Answer 15

small spacings between molecules give wide spacings in the diffraction pattern

big spacings put spots close together

in the diffraction pattern

in a salt crystal - very spread not many of the

compete data set captured in shorter time

measuring intensity of acc spot bit easier

Question 16

why do we need phases

Answer 16

use Fourier transform to generate electron density

as long as we have
phase info needed to supplement intensity information

diffraction pattern comes from interference
couple waves which are out of phase with each other which would be interrfering

electron density 1/v = volume of unit cell

h+K+L = reciprocal lattice co ordinates each diffraction that we measure will have a h k l value hitch will tell u which plan in the crystal gave rise to that

f h k l intensity of black spots by takin square root

each spot
take square root
exponential component of equation

sampling function
tells u what layer in the unit cell this summation is relevant to

imainary component
the phases for each measure h k l

fhkl
allows to move from diffraction space to real space

Question 17

x ray diffraction procedure

Answer 17

1. purify protein
2. grew crystals
3. measure interference pattern of diffracted waves
   = get highest resolution
4. determine the lost phases
   methods:
5. isomorphous replacement - soak heavy metal ion into crystal of protein and use that to regain phase info - used for myoglobin lysosyme
6. anomalous dispersion to solve phase problem
7. molecular replacement - use one structure to solve a new structure
8. calculate electron density map from Fourier sum of measured wave amplitudes and phases from above -fit model to electron density chicken wire model
9. refine model - alter model to give best fit of electron density
10. validation

Question 18

high purity needed

Answer 18

exploit feutre o protein t purify

solubitly -
adding ammonium sulphate to protein sol to fractionally precipitate out

charge - ion exchange choromotogaph

size - gel filtration ch
size exclusion chromatography to remove aggregates before crystallisation trial

hydrophobicity - hydrophobic interaction chromatography green fluorescent protein

attach affinity tags during exp

sds or native electrophoresis to examine purity

assay for specific activity to confirm identity

Question 19

crystallisation procedure

Answer 19

prepare a supersaturated sol (more dissolved solute than thermodynamically stable) sol of protein by adding precipitants adjusting pH and controlling temp which will effect solubility

then enable nucleation which can be homogenoeoou - aggregation process to produce protein parcels or a particular size - crystal growth = spontaneous or hetero something dropped into crystallisation while setting it up or components that form templates for crystal growth

so that excess supersaturation can be translated into a crystal

Question 20

catch 2 situation

Answer 20

the eel of spuersutration requires for homogenous nucleation higher than that required for sustained growth (metastable growth)
problem is showers of little crystals as metastable zone isn’t the optimum for growing homogenous

solution
1. slow approach to interface
2. take bad crystals and crush them up centrifuge add small quanitited of seed solution
add seeds into metastable zone without reaching metastable zone = large crystals

Question 21

protein crystallisation via hanging drop

Answer 21

ammonium sulphate = precipitating agent

Question 22

native to hanging drop

Answer 22

sitting drop

Question 23

how to find the right conditions

Answer 23

dunno what temp to use

screen w an array of experimental conditions
known to be effective

commercially available precipitating solution

use robotic nanoliter dispensers to fill plastic trays to speed up process
96 well plae
oblong part of container
solvent resolver

protein in well
tray stored in
crystal hotels with photgraphic review

lare image of drops as their stored

mosquito crystallisation robotic peptte - 96 50 nanoliter drops in a minute

Question 24

membrane proteins crystallisation s

Answer 24

in order to crystalise need to be taken out of the membrane using detergents

have an apolar end and a polar ed
lipids
but
can interact with membrane proteins to make

have to be careful with conc
at high conc hydrophobic tails collapse together

Question 25

the unit cell

Answer 25

basic repeating uni of the crystal

can only be moved by translation not rotation

7 crystal systems

w centring = 14 bravais lattices = allowed symmetry operators have 32
2 3 6fold symmetry

cat have 5 fold symmetry

230 space groups that all crystal must fit into
only 65 allowed for chiral molecules like crystals due to the L amino acids

Question 26

data collection

Answer 26

determination of space group is the first step

can predict orientation of crystal in machine

hkl reciprocal lattice co ordinates
box around whee spot is predicted
estimate backgeound locally in that region

experimental mesurent of intensity

Question 27

crystal formed by unit cells stacked up in 3d like bricks

only allowed to have translation between 1 brick n next
bricks can have diff shapes
diff organisation of internal content

many diff types of bricks

7 rbravais latticies
general shape of the these bricks

work comes form classic studies of minerals from 20th century

lattices have internal symmetry that defines space group

Answer 27

diffraction process causes certain type of diffracted rays to be annihilated some stronger than other
symmetry in crystal revealed in that diffraction pattern

crystallography magnifies atoms so that we can see atomic structure

has parralelels with microscopy

X-rays only get half way through image formation
measure all scattered waves but cannot refocus them like in a light microscope focuses by lenses onto a plane where u see the waves recombined in an appropriate manner

can only measure the energy of the interference pattern

measure intensity of black spots

regaining lost phases from diffraction

Question 28

10 to the power 20

molecule v weak
amplify it = duplicates of the same brick throughout lattice same

Answer 28

Gail Rhodes crystallography made crystal clear

Question 29

electron density eq

Answer 29

summation of all the waves scattered by unique scattered waves

Question 30

f

Answer 30

structure factor amplitude

comes from the square root of the intensity of the black spots measured in the diffraction pattern

Question 31

sampling function

Answer 31

tells u which plane in the crystal has generated that particular F

because we’ve collected data from thousands of planes each black spot is a plane
Fourier transform of a spot is a set of interference finger
2d array of dots on a card = diffraction pattern

Question 32

what is the product of the diffraction exp

Answer 32

electron density map

Question 33

the X-ray structure that you determine is actually an average of 10 to the 20 proteins scattering simultaneously

Answer 33

if u assume the diffraction pattern of an object is a Fourier transform off the object the inverse transform is the object itself

Question 34

isomorphous replacement

Answer 34

grow protein crystal
measure a day set for the native protein
then soak heavy metal complexes into the crystal and then remeasure the data to calculate the difference in the intensities for each hkl which are followed soaking in heavy metal

calculate the foriour transform with intensity differences squared and phases set to 0 - in a tube of fouriour map known as a paterson map
calculation gives rise to a map of interatomic vectors the size of which are dependent on the number of electrons in the atoms involved

Patterson showed the this calculation gives rise to a map of interatomic vectors

the size of the interatomic vectors is dependent on the number of electron in atoms involved

when using heavy metal complexes we get dominant peaks in this delta f squared synthesis

we use Patterson maps to find the positions of heavy atoms by inspection of the nap
by hand
>4 sites = tricky
modern software packages employed to sort through combinations of heavy atom species

can see vectors between atoms distributed on paterson map. huge peak at origin corresponding to
vector between each atom and itself

you can determine x y z positions of heavy atoms in real unit cell

for small molecules powerful for getting structure solutions

but hopeless on big proteins would be hopeless because there are n squared minus n peaks where n is the number of atoms involved
if u have 2 atoms you can see you’ve got 3 peaks you’ve got to find in the map
1000 atoms gets v complex
not suitable for big proteins

platinum tetra chloride widely used reacts w methionine residues well

heavy metal it is soaked in = mercury

Question 35

once u know the poistion of the heavy atom in the crystal - can use those positions to calculate heavy atom structure factor

argand diagram

Answer 35

structure factors are vectors w amplitude and phase plotted w real and imaginary axes

amplitude = real
phase - imaginary

1. vector for a particular h k l for the native protein called f p
2. another vector for protein heavy atoms derivative complex fpH
3. fH

measured fp and fph and know the size of fh
and we know phase angle for fh

info for determining the phase of fp by some trigonometry

Question 36

phase ambiguity solved

Answer 36

draw axis
draw circle w set radius from origin equal to the value of fp

then minus fh going from the origin = centre of a circle whoms radius is equal to fph

where it intersects with fp circle in 2 places
therefore phase is ambiguous

in order to get an unambiguous estimation of what we have to so is do another heavy atom to go through to procedure again
if we now plot -fh
the second heavy derivative and make the circle origin from this oo ordinate
this circle should intersects at one of the intersections causing the prior ambiguity enabling you to set the phase for fp at this particular angle known as alpha p
harker construction

Question 37

anomalous scattering to solve the wave problem

Answer 37

near the edge of an x ray absorption edge the normal rules od diffraction break down

pairs of religions roared by centre of symmetry in the diffraction pattern has the same intensity
hkl and bar h bar k bar l