1
Q
What are the pros and cons of biological NMR
A
pros:
-provides both structure and dynamics
-no need to crystallise sample
-probe molecular interactions
-near physiological conditions (can study protein in solution/even in cell)
cons:
-size limit (35kDa)
-requires isotopic labelling
2
Q
What information can you get from NMR spectra
A
- Chemical shift (δ) – chemical environment
- Integral (area under peak) – number of equivalent spins
- Scalar coupling (J) – local structure
- Relaxation times (T1 and T2) – dynamics
- Dipolar coupling (D) – long range structure
- Nuclear Overhauser effect (NOE) – structure/binding
3
Q
Describe the process of performing NMR
A
- Start with having free protons (all in random spins)
- Apply magnetic field (upward spin, z-axis)
- Proton spins will align w/ external MF - population of protons with upward spins exceed downward spins, so eq/net magnetization points upwards – in z axis
BUT detector is located in X/Y plane - Radiofrequency pulse applied along y axis (90° to the spins) causes the z magnetization to be rotated onto x axis (transverse magnetization)
- Measure how the magnetization precesses & decays along the XY plane to restore its z magnetization
* Measured as Free induction decay in time domain - FT to frequency domain - observe chem. shift peaks
* FID of each proton = depend on the environment they’re in = diff. chem. shift
4
Q
What are the problems with larger molecules
A
- FID of 2 closely spaced peaks is only distinguishable if they take a long time to phase out.
- If the signal decays after a short period of time, there will only be 1 broad average peak of the 2 protons observed
5
Q
Explain what occurs in 2D NMR
A
- A second RF (90°) pulse is generated after a certain amount of time delayed from the first RF pulse.
- The delayed period involves the evolution of magnetization along an indirect dimension.
- After the second RF pulse, an additional mixing period is required for the mix/ transfer of signals between coupled spins.
- The FID after the second pulse is recorded in t2 and then detected to obtain 2D spectra.
For each point to be obtained on the 2D spectra, the whole experiment must be repeated multiple times by varying the evolution time (between pulse 1 & 2) - time consuming
6
Q
What generates a set of FIDs in the time domain
A
different evolution times:
- FT rows for frequencies on x axis
- FT columns for frequencies on y axis
- Result in 2D NMR peaks
7
Q
What are the types of 2D NMR
A
Homonuclear NMR (observe chemical shifts of the same species eg. H-H)
Heteronuclear NMR (observe chemical shifts of 2 different species eg. N-H)
8
Q
Explain heteronuclear NMR
A
- first excitation is followed by an INEPT element to enable the transfer of the excitation from the H to N
- After the evolution time, the polarization is transferred back to H via another INEPT element.
- Like homonuclear 2D NMR, detection of FID is made during t2
9
Q
What is the criteria for a good nucleus for recording heteronuclear NMR
A
- Spin ½, high abundance, high magnetogyric ratio = high sensitivity
- For C & N, abundance is low – must increase sensitivity: growing bacteria in medium containing the labelled isotope
o For 13C, use 13C labelled glucose
o For 15N, use 15N labelled ammonium chloride - The labelled isotopes will be included in the POI produced in the bacteria
- Supplements of vitamins and trace elements of minerals required for the bacteria to grow in the unnatural media
10
Q
Describe H-N HSQC heteronuclear spectra
A
- represents scalar coupling across N-H bond, so only amine groups will appear as a peak, corresponding to 1 aa
- side chains containing NH groups (ex. glutamine, asparagine) will also appear as a peak
11
Q
How is 3D NMR performed
A
- 3 RF pulses applied and record FID in t3
- Can be heteronuclear NMR – measure frequencies of 3 different nuclei eg H, N, C Or H, H, N
12
Q
What are the 4 popular 3D NMR experiments:
A
HNCO
HNCACO
HNCACB
HNCOCACB
13
Q
Describe HNCO
A
- Magnetization is passed from 1H to 15N and then selectively to the carbonyl 13CO
- Magnetization is then passed back via 15N to 1H for detection.
- CO i-1 is detected
> in each NH strip, there is one CO visible which belongs to the preceding residue
14
Q
Describe HNCACO
A
- Magnetization is transferred from 1H to 15N and then to the 13Cα and to the 13CO
- For detection the magnetization is transferred back from 13CO to 13Cα, 15N and finally 1H
- Because the amide NH is coupled to both Cα of its own residue and of the preceding residue, transfers to both 13CO nuclei occurs
- For each NH group, 2 CO groups are observed in the spectrum. because the coupling between Ni and Cαi is stronger than that between Ni and Cαi-1, the Hi-Ni-COi peak generally ends up being more intense than the Hi-Ni-COi-1 peak.
15
Q
Describe HNCACB
A
- Magnetisation is transferred from 1Hα and 1Hβ to 13Cα and 13Cβ, respectively, and then from 13Cβ to 13Cα. From here it is transferred first to NH and then to HN for detection.
- magnetisation is transferred to 15Ni from both 13Cαi and 13Cαi-1.
- Thus for each NH group there are two Cα and Cβ peaks visible (for same and preceding residues)
16
Q
Describe HNCOCACB
A
- Magnetisation is transferred from 1Hα and 1Hβ to 13Cα and 13Cβ, respectively, and then from 13Cβ to 13Cα. From here it is transferred first to 13CO, then to 15NH and then to 1HN for detection.
- The chemical shift is evolved simultaneously on 13Cαi-1 and 13Cβi-1
17
Q
How are 3D NMR usually used
A
usually used in combination with each other
eg
* An overlay of the HNCO and HN(CA)CO spectra makes it very easy to distinguish between COi and COi-1 for each NH group.
if the peak occurs both in HNCO and in HNCACO, it belongs to preceding residue CO.
If peak occurs only in HNCACO, belongs to same residue CO.
18
Q
How to calculate NMR structure
A
- Utilizes Simulated Annealing (SA)
1. starting structure is heated in a simulation (i.e. the atoms of the starting structure get a high thermal mobility) and then cooled
2. During cooling steps, the starting structure can evolve towards the energetically favorable final structure under the influence of the force field - The force fields involve contributions from input restraints to guide folding of proteins - ideal bond length, ideal bond angles, including information derived from NMR – the NOE-derived distance bounds, and coupling-constant-derived dihedral angle restraints.
Etotal=Ebond + Eangle + Evdw + Edihedral + Enoe
Since there could be multiple ways to satisfy the input restraints, the calculation is repeated many times (>100) to generate an ensemble of structures. If enough/ correct restraint is given in the input, the structures should be similar but not identical.
19
Q
What can you use NMR for
A
- Determine weak protein-protein interactions
o β2m forms dimers
o obtain NMR spectrum of β2m dimers at different concentrations – each conc. = diff chem shifts
o from overlaying diff. spectra can fit a binding curve to determine the weak dimerization constant - Structural approaches to IDPs
o Techniques such as X-ray crystallography are limited in the study of IDPs
o IDPs are virtually impossible to crystalize & even by crystallizing an IDP, the resulting structure would represent only one of the infinite conformations that the protein can adopt
*Disordered proteins can be immediately recognized in a 1H-15N HSQC spectrum for two characteristics:
o limited dispersion (Backbone 1HN restricted in ~1ppm)
o sharp peaks (long T2)
- Thus, even if the spectra are poorly dispersed, the resonances are normally well resolved and present limited overlap
20
Q
Describe high and low energy states of proteins
A
- Low energy states of proteins = highly populated & can be studied easily
- High energy states of proteins (carry biological function) = less populated & can only be studied with some NMR techniques
- Protein dynamics involved in achieving the high energy states = also contribute to the function
21
Q
Explain protein dynamics
A
- Each E state of a protein is composed of a number of local minima
- Diff E state, and diff local minima in 1 E state, are all divided by E barriers – but of different sizes
E barrier between diff. E states = higher barrier, thus result in slower dynamics (µs-ms) and are associated with large conformational changes
E barrier between diff local minima in 1 E state = low barrier, thus result in faster dynamics (ns-ps) and are associated with smaller conformational changes (could just be vibrations in atoms/ bonds, etc.)
22
Q
How do proteins use dynamics to perform their functions
A
- Protein dynamics play a role in protein-protein interactions
- also allow for conformational flexibility, giving rise to 2 major properties:
* Biomolecular promiscuity: majority of proteins have more than 1 interacting partner
* IDPs ability to adopt an infinite number of conf. allow it to function as a hub in a network of PPI - Protein dynamics is also related to enzymatic activity
* Enzymes = not functional in lower temperature due to restricted motion
23
Q
Explain NMR relaxation
A
Relaxation is the process by which the spins return to equilibrium
- Longitudinal relaxation (T1) – restoration of Mz at a rate of R1
- Transverse relaxation (T2) – fading of Mxy at a rate of R2
Small proteins:
short T1, long T2
Large proteins:
long T1, short T2
Relaxation is caused by time-dependent fluctuations in the local MF – brought about by protein motion/ protein dynamics
* Short T2 relaxation (fast decay of signal) = broad peaks – which is why NMR is better for small proteins
24
Q
Describe the process of NMR relaxation for dynamics determination
A
- sample preparation (eg. prepare isotope source, etc)
- assignment of protein spectra (assign each peak to each residue)
- carry out dynamics measurements
- interpret results
25
What is the most popular type of relaxation
* 15N relaxation is the most popular, it infers on N-H
26
Explain R1 and R2 measurements
In addition to the 2D heteronuclear NMR method of having INEPT elements after the pulses to enable the transfer of the excitation between nucleus, a new block of pulses is added before the second INEPT element -time for relaxation delay.
Different values of τ give different signal intensities – shorter delay = more intense, but longer delay = less intense, so can form multiple HSQC, can observe decay in signal intensities
The pulse sequences are made in a way that the decay of signal intensities as a function of τ are proportional to either R1 or R2
We fit these decays with single exponential function and obtain R1 or R2 for every peak (hence every residue) in the HSQC.
27
Explain heteronuclear NOE measurements (cross-correlated relaxation)
* The backbone 1H-15N heteronuclear NOE provides information about the motion of individual N-H bond vectors
* Perform 2 experiments: 1 normal HSQC & 1 where H is saturated to measure the effects on N
* Ratio of signal intensities between the 2 experiments = hetNOE
o Rigid NH vector = NOEHX ~ 0.9
o Dynamic NH vector = NOEHX << 1
28
Describe the data interpretation + example
The info. obtained can be used to calculate order parameters –
S2: magnitude of angular fluctuation for chemical bond vector (modelled as theta angle describing how much the bond vector moves in a cone)
S2 = 1 (rigid motion), S2 = 0 (isotropic motion)
Ex. PKA
* has 2 domains – small & large lobe
* interchangeably in open & closed conformation, depending on binding of substrate
* N relaxation experiment performed – can obtain R1, R2, hetNOE value at diff states of proteins (ex. when unbound, when nucleotide-bound, when substrate-bound)
* Shows that protein is very dynamic in all the states
* Can also observe dynamics of the substrate
29
Describe sub-ms dynamics by CPMG (relaxation dispersion)
* Slower (μs – ms) motions are normally associated with larger scale structural rearrangements
* These dynamics are due to chemical exchange between states A and B (native state & excited state)
30
Explain chemical exchange
* process in which a nucleus (of a protein) exchanges between two or more environments (due to being in diff protein conf. states) and results in a change in chemical shift, scalar coupling, and/or NMR relaxation rate
The 2 states are in equilibrium:
kAB: forward rate
kBA: backward rate
kex = kAB + kBA (exchange rate between states A & B)
BUT state B (excited state) has a higher energy than state A, so the equilibrium lies towards state A
* kAB is smaller than kBA
* more population in state A (pA > pB)
* fractional population of B (pB) can be as low as 0.5%
* Many important biological systems rely on chemical exchange between diff. states (ex. opening & closing of ion channels)
* Stabilization of specific states by drugs is an important avenue for drug design
31
Describe chemical exchange & NMR
* Each state will appear as a peak
* The position of the peaks in the spectra depends on how fast exchange is taking place
* If exchange rate is similar to the frequency diff. between the 2 peaks =can result peak broadening
* If exchange rate is too fast (faster than frequency diff. between the 2 peaks) = can result average peak instead of 2 distinct peaks
* Most common way to change the exchange rate is with temperature
32
Explain chemical exchange with skewed populations
* Intensity of each peak corresponds to the population of each state
* Small population will result in the peak disappearing at a fast exchange rate
* The other peak that has a higher population will still be present (and still experience peak broadening & peak averaging – but not as prominent if both = equal in abundance)
Therefore:
* the exchange regime is determined by the ratio of the exchange rate (kex) to chemical shift difference (Δω)
33
Explain CPMG (relaxation dispersion)
* Provides information at a residue specific level
* Not all the peaks of the spectrum will be suitable for CPMG
o This is because not all the parts of a protein normally are in conformational exchange
o Only the peaks that correspond to residues involved in the protein dynamics
34
Explain the CPMG experiment
* Also involve adding a series of additional pulse before the 2nd INEPT element (like R1/R2 measurement)
* Vary the frequency of additional pulses
* Total delay time = same
* Pulses applied = 180° pulses which will reverse the phase angles and initiate re-phasing of the magnetization
* So, the magnetization must spend more time on x/y plane, resulting in a smaller magnetization detected in the end position because of T2 relaxation (but not by a lot)
35
Explain the CPMG experiment if there is only 1 state (no exchange)
* more pulses within same delay time will not change the R2 relaxation detected
* Because the protein experiences the same chem. shift throughout the diff. pulses
-magnetization will precess and end up to the same location
36
Explain the CPMG experiment if there is an exchange between 2 states
* chem. shifts of the 2 states = diff, & will cause changes in precession speed during the delay time, so the magnetization will not end up in the same location at the end
-vector also smaller because more T2 relaxation AND exchange
* More pulses = less time for exchange between states (suppress chem. exchange)
* Less pulse = more time for chem. exchange
* SO more pulses = R2 detected will be more similar to intrinsic relaxation & result in better peak, less pulses = R2 detected will deviate from the intrinsic relaxation and result in a smaller peak
* plot R2 measured vs frequency of pulses to see if chemical exchange occurs
37
Explain the exchange parameter: chemical shift difference between the states + example
* If known = able to deduce structure of the other state, characterize interactions
* ex. L99A T4 Lysozyme
o All Xray structures cannot explain how the ligand enters the binding pocket
o NMR able to deduce an “Invisible” conformational state that is only in 3% abundance - at 298K, 1ms lifetime
o Then, use Rosetta method: structure calculation from chemical shifts
o Once know structure, can mutate the transient state to be more abundant to be able to observe peak in spectra
o Can also perform double mutant – make excited state more stable than native state
38
Explain the following exchange parameters: population and rate
pB : Population of state B
* relates to equilibrium position Ke, Kd
* can observe temperature dependence of Ke or Kd (Van’t Hoff analysis) to give ΔΗ, ΔS
kAB: forward rate, temperature dependence gives size of barrier
kBA: backward rate, temperature dependence gives size of barrier
39
What can you obtain from CPMG experiments
1. Structure
2. Kinetics
3. Thermodynamics
* All at per-residue level
* For a transition that involves states present at only a few % in solution
* Does not have to be 2-state!
40
Describe the Ribonuclease A example
* 14 kDa enzyme
* transphosphorylation of single-stranded RNA
* no cofactors or metal ions
* rate-limiting step in the catalysis is product release (Kcat =~ Koff)
* Add 2 diff. ligands to the enzyme to form 2 complexes: ES complex & EP complex
* Perform CPMG experiments on both Apo and ligand-bound enzyme
-can characterize AE barrier between open & closed state of each form
-result = same dynamics but reversed population of the states
41
Describe the 2 dihydrofolate reductase example
* reduces dihydrofolic acid (vit B9) to tetrahydrofolic acid using NADPH as cofactor
* has many enzymatic states
* the Met 20 loop changes conf in each state
* using NMR, can probe all the conf. changes to know where we are in the enzymatic cycle
* recorded CPMG relaxation dispersion experiment in the presence of all enzymatic states
1. calculate Δω (from CPMG dispersion fitting)
2. measure chemical shifts in different forms of the enzyme (HSQC spectra)
* compare observed chem shift and actual chem shift of diff enzyme forms – correlate perfectly
* before the complete conf. change, proteins = already start populating the next state
* conclusion: the dynamic energy landscape efficiently funnels the enzyme through its catalytically competent conformations along a preferred kinetic path, where the number and heights of the energetic barriers between consecutive conformations have been minimized
42
What is the key summary of CPMG experiments
* CPMG is sensitive to dynamics ranging in the ms-to-µs timescale
* CPMG is most effectively used when two conformations are in intermediate exchange
* CPMG provides three major data: populations, kex and Δω.
43
How do proteins exist
* Proteins exist in continuum of functional dynamics: from mainly ordered proteins, proteins with folded domains and disordered regions (IDRs) or completely disordered proteins (IDPs)
44
How much of the proteome is disordered
* At 30% of the proteome is disordered and often the disordered regions are those that are functionally important.
45
Discuss tools to characterize IDPs
* Current AI based tools (like Alpha Fold) completely fail to characterize IDRs/IDPs
Powerful techniques such as X-ray crystallography are inherently limited in the study of IDPs.
* IDPs are impossible to crystalize.
* even by crystallizing an IDP under particular conditions, the resulting structure would represent one of the infinite conformations that the protein can adopt in solution.
> NMR is still able to measure observables in the case of IDPs, however, the use of traditional (e.g. NOE based) NMR is ineffective and new methods must be developed.
46
What can you use to obtain residual secondary structure using NMR
can predict chemical shifts associated with secondary structure elements.
* This is because chemical shifts are sensitive to the local backbone structure:
47
Discuss chemical shift anisotropy (CSA) + example
* Same nucleus in diff. secondary structure elements = in diff. local environments = diff chemical shifts
* Therefore, by comparing measured and predicted chemical shifts, can assess the amount of secondary structure in a protein (folded or disordered)
Ex. chemical shifts of the diff. nucleus of alanine in diff. secondary structures
CA of alanine in a-helix = distributed ~55 ppm vs in b-sheet = distributed ~52 ppm
48
What does chemical shifts allow us to estimate and why
* Since chemical shifts are averaged, the secondary structure element does not have to be present 100% of the time. SO chemical shifts allow us to estimate propensities/populations of helices, strands, poly-proline helices, β-turns even if they are not fully formed.
49
Describe the process of CSA
1. measure 2D and 3D NMR spectra (e.g. combinations of HNCA, HNcoCA, CBCAcoNH, HNCACB, HNCO, HNcaCO)
2. Assign the backbone chemical shifts (which aa corresponds to which peak)
3. Use these data in programs/servers such as Talos, d2D to determine residue-specific secondary structures
50
How do you determine the topology of the secondary structure propensities
Paramagnetic Relaxation Enhancement (PRE) (another NMR experiment)
* attach paramagnetic labels (carry lone electrons) on proteins via cysteine modification
* when observe spin label NMR spectra, residues of proteins in the vicinity of the lone electron will experience a decrease in their peak intensity in comparison to reference spectra
o Distance-dependent effect: only residues that are closer than ~ 5 Å will have their peaks disappear
* Can attach spin labels at various places on a protein and see which regions are affected which allows you to determine the topology of secondary structure elements relative to the other known regions
51
What is a powerful tool to investigate residual structure in IDPs
chemical shifts (report on secondary structure) + PREs (reports on topology of secondary structure)
52
Describe the example of using NMR to determine IDP region of DNAJB6 protein
* a key component of the chaperone network involved in the Hsp40-Hsp70 cycle, that protects from many neurodegenerative diseases.
* Structured N & C-term domains are linked by a disordered region
>NMR chem. shift determined an a5 helix in disordered region
>PRE used to determine topology of the a5 helix:
-if the spin label is attached N-terminally to helix 5, PRE effects are observed in helix 2
-if the spin label is attached C-terminally to helix 5, PRE effects are observed in helix 3
Conclusion: found a 10-residue segment within a 70-residue long disordered region that forms a helix which interacts with the J domain. This has huge implications for function as it inhibits binding of the partner protein and keeps the protein in an auto-inhibited form
53
Explain using NMR to determine the disordered regions in membrane proteins
* Membrane proteins = larger molecules
Solution (liquid) NMR is not suitable for determining structural propensities/ topologies of disordered regions in larger proteins because:
In every NMR experiment, Dipolar coupling (interaction of magnetic dipoles) & CSA (change in chem shift due influences from local environment) will cause broad peaks & bad signals
* for small molecules in solution NMR: fast molecular tumbling of the molecule within the liquid allows dipolar coupling & CSA to be averaged out so peaks = sharp & good signal
* for larger molecules, although solution NMR allows tumbling, its large size will still result in slow molecular tumbling
o Slow tumbling allows for enhanced dipolar coupling & CSA, which cause time-dependent field fluctuations
o leads to faster relaxation of transverse magnetization (short T2)
o cause fast signal decay, result in broad peaks & bad signal
54
What is the problem with solid-state NMR and how can you overcome it
* Sample = in powder form
* Spectra = worse for both small & large molecules because no tumbling is allowed - enhanced dipolar coupling & CSA
Overcome it by magic angle spinning:
* Pack powdered sample in rotor that has a fast rotation (4000-5000 Hz) around the Magic Angle Spinning of 54.74o allows averaging out the DC and CSA effects – able to get sharp peaks
55
What information does solid-state NMR record + example
Solid-state NMR can record both info about rigid and flexible domains of membrane proteins:
eg. alpha-synuclein
* completely IDP in solution but can be membrane-bound
able to bind membrane because:
* + charged residues in N-ter + Imperfect KTKEGV motif repeated throughout res. 1-70
o Allow formation of amphipathic helical modules – can elicit membrane binding
56
How can you study the membrane binding properties of aS
* using synaptic-like vesicles as membrane model (SUV)
Goal is to study:
* Structure and dynamics of αS bound to synaptic-like vesicles
* Topological properties
* Molecular bases for the membrane affinity
57
How can you determine aS affinity to different types of lipids
By CD:
* Use diff types of lipids to form SUV
* For DOPC/ DOPE/ DOPS – need high conc .of lipid to achieve stronger alpha helix conformation
* For POPG, does not need as high of lipid conc.
* Shows affinity three times higher for POPG than DOPC/DOPE/DOPS lipids
-Difference = the other 3 lipids have + charged head groups but POPG = - charged --> correspond to binding with + charged N-terminal of aS
Then perform Solid-state NMR on the system - result = observe:
* Rigid helix - N-terminal 25 residues
* Intermediate Behavior (dynamics/structure) - residues 26-97
* Unstructured and highly dynamical - C-terminal residues 98-140
PRE used to determine topology of the rigid helix/ flexible region relative to membrane
* attach spin label to lipid incorporated in membrane & observe changes in NMR spectrum of the rigid N-terminus & the flexible C-terminus
N-ter = a lot of signals lost in spin labelled spectrum - indicates binding with membrane
C-ter = not much signal diff between spin labelled & ref spectrum - doesn’t come close to membrane
58
Describe chemical exchange in weak binding
* process in which a nucleus (of a protein) exchanges between two or more environments (due to formation of a complex by weak binding). This results in a change in chemical shift, scalar coupling, and/or NMR relaxation rate
o measured by CEST/DEST - works well if B has higher molecular weight than A
* diff from chemical exchange when protein is still a monomer, but in diff. conformation
o measured by CPMG - works well if A and B have similar molecular weights
* same principles: state B has a higher energy than state A, so the equilibrium lies towards state A - more population in state A (pA > pB) in slow exchange with state B
59
Describe CEST/DEST
* Based on the application of a weak saturating field (B1) at various frequencies on each peak of the spectrum
60
Explain CEST for proteins that form a minor oligomeric/complex state in solution
* although NMR spectrum records a sharp peak for the major monomeric state, there will also be an unobservable broad peak (associated with high Mw the oligomeric form) underneath.
o Therefore, although the saturating B1 field is not in resonance with the peak’s frequency, a lot of peak intensity will be lost – because the field is still in the range of the unobservable high Mw state broad peak
o Will still result in 0 intensity when in exact resonance with the sharp monomeric state peak
o Results in broad CEST profile
* This happens because saturation of the large, exited state is transferred to the small NMR observable main state through chemical exchange
observe a peak in N-ter of asyn - binds to membrane (has exchange with high Mw state)
61
Explain CEST for monomeric proteins with no chemical exchange
when the saturating B1 field applied is in resonance with the peak’s frequency w/ 0 offset, the peak’s intensity will be diminished
o If the applied pulse is not in exact resonance with the peak’s frequency, the intensity will mostly be maintained
o Results in a narrow CEST profile
62
Draw the CEST profile with & without exchange
w/exchange is broad
w/o exchange is narrow
63
How can you get more information out of the CEST profile
Can probe the exchange per residue basis – & stack the profiles on top of each other to generate 3D plots
* Can determine the regions that are bound and the kinetics of binding
64
Describe the CEST on aS bound to liposomes
* During binding, N-ter becomes helix and bind to membrane while C-ter remains disordered
* NMR data of aS bound to liposomes without saturation (reference spectrum) vs. with saturation at diff offsets – observe loss in peak intensity across diff offsets
* Residues in the N-ter, when stacked = all broad profiles
* Residues in the C-ter, when stacked = all narrow profiles
* If perform CEST saturation on protein alone = no difference because no CEST effect (no residues involved in binding)
65
Explain the double anchor mechanism of aS
* Chaperone SNARE complex assembly for docking of pre-synaptic vesicles to pre-synaptic membrane before release
* For fast transport of vesicles across synapse
66
How do you make aS mutant with enhanced activity
Primary requisites:
* Same aminoacidic composition
* Same overall charge
just swapping of charges at 2 places (E46K/K80E)
Intended requisites:
* Same membrane affinity
* More exposed NAC region – region involved in binding to 2nd vesicle SO if more exposed = increase affinity to bind to 2nd vesicle
* result in enhanced binding of the 2 vesicles - bigger and more assemblies of liposomes with mutant
67
Discuss how the aS can disrupt the membrane
* formation of intermediate oligomeric species (in the form of lewy bodies) = toxic - can bind, disrupt membrane:
o metabolize membrane and result in Ca2+ influx
o lyse mitochondrial membrane
o lead to cell death
o toxicity related to Parkinson’s & neurodegenerative diseases
* eventually oligomeric aggregates can also form amyloid fibrils
68
What does ssNMR do for aS
Elucidating detailed structural properties:
* ssNMR = can look at both rigid region (with DARR) and disordered regions (with INEPT)
* found that asyn forms 2 diff. oligomers:
o Type a (non-toxic) structure:
-N-ter = rigid
-C-ter = flexible
o Type b (toxic) structure:
-N-ter and C-ter are dynamic
-NAC region = rigid
From this experiment = able to link structural properties to toxicity
69
How do different NMR techniques help to study membrane-protein interactions
* ssNMR in the cross polarisation regime (rigid regions of membrane-bound proteins)
* ssNMR in the INEPT regime (floppy regions of membrane-bound proteins)
* CEST probing the interaction from the unbound state perspective
70
Explain CEST MRI
* a relatively new MRI technique involving exogenous or endogenous compounds containing exchangeable protons
* Exchangeable protons on a solute (minor state) are in exchange with bulk water (main state)
* Proton in bulk water signal has a frequency of 4.75 ppm and another characteristic frequency when it is on the solute
* Soluble solute protons are selectively saturated (irradiate pulse in resonance with frequency of proton on solute), but because they are in exchange with bulk water, the bulk water signal will also decrease
* The transfer of this saturation is detected indirectly through the water signal with enhanced sensitivity
o Measure normal bulk water signal (S0) and bulk water signal upon saturation of solute proton (Ssat)
* Saturation of bulk water, will result in loss of water signal due to direct saturation– assigned to 0 ppm in Z spectra
* Magnetization Transfer Ratio corrected with asymmetry analysis will generate another CEST peak in the Z spectra - corresponding to the exchangeable proton at a certain position SO can identify where the exchangeable proton is located
71
What are the medical applications of CEST MRI
Ex. exchange of amide protons with bulk water
* exchange of proton is sensitive to pH – locate tissue acidosis
* Require low magnetic fields – applicable to clinical fields
* can locate tumor cells & discriminate tumor grades
72
Explain gluCEST MRI
Exchange of bulk water and –NH2 protons of glutamate (endogenous imaging agent)
* image glutamate location in body
* Sensitive to [Glu] and pH
* Glutamate = a major neurotransmitter in the brain
* Applications – perform gluCEST on stroke patients’ brain – observe altered conc. of glutamate in diff brain lobes (absence in one and excess in another)
* Can also map glutamate conc. in Alzheimer’s and Huntington’s disease patients’ brains
PS
flashcards
Decks in class (3)
# Cards
