key points Flashcards
1
Q
What problems can be explored in structural biology? (dntk)
A
- Biomolecular self-assembly
- Biomolecular dynamics
- Structure-activity relationships
- Biomolecular interactions
- Biomolecular structure
- Ligand binding/design
- Physical chemistry of biomolecules (e.g. thermodynamic properties)
2
Q
What is a method we can use to help characterise biomolecular interactions?
A
- Isotopic labelling can be used to enrich the system with 13C/15N.
- As these are low abundance, they will stand out in analysis.
3
Q
- What is the size limitation problem in solution NMR? How is solid state affected by this?(IMP)
A
- Rotational correlation time (τc) scales inversely with MW
- As MW ↑, T2 signal-to-noise ↓ and peaks broaden
- This is due to T2 being so short that signals relax quicker than data can be acquired.
- Solid state NMR does not suffer from this, however peaks are generally broader anyway
4
Q
- What is the variable intensity problem? (IMP)
A
- Intermediate exchange regime leads to variable peak intensities
- meaning some peaks will be weak, others intense
- 𝛿A = resonant frequency associated with state A
- Larmor frequency difference (Δω) >> rate of exchange between A/B (kex) à see both states
- Δω << (kex) forms single weighted average state
- Δω ≈ (kex) results in broadening –> variable exchange
6
Q
What is the problem associated with proteins unfolding? What conditions can cause this to occur?
A
- Aggregation on proteins can lead to broad peaks
- Poor chemical shift dispersion of 1H peaks leading to severe overlap, and will appear all in the same environment
- High pressure can cause proteins to unfold and chemical shift envelope to become more compact
7
Q
What is insensitivity in NMR? How can it be overcome?
A
- 1 in 1E+06 signals are being measured of all excited nuclei, meaning experiments can take many days as this is very few signals (insensitive)
- This can be overcome by running at higher temperatures, but biomolecule may not be able to retain its stable fold
8
Q
Why is the knowing stability of a protein fold important?
A
Conditions in which this takes place are important to know as will massively change structure of sample
9
Q
What is the problem with isotopic labelling?
A
- Enriching our organic sample with low abundance 13C/15N isotopes is generally done by growing bacteria and enriching them.
- These feedstocks can be very costly.
10
Q
Advantages of NMR
- NMR provides atomic … for molecules in solution with MW ≤ 1E+05 Da and can also mirror … conditions. Doing this in solution is the best way to validate properties in a biological … ….
- NMR can be used to access … complexes in …-binding studies where … are monitored
- Accessible … nuclei cover all biologically-relevant suspects: … 1H, 15N, …, 31P
A
Advantages of NMR
- NMR provides atomic resolution for molecules in solution with MW ≤ 1E+05 Da and can also mirror physiological conditions. Doing this in solution is the best way to validate properties in a biological cellular environment.
- NMR can be used to access larger complexes in ligand-binding studies where ligands are monitored
- Accessible isotopic nuclei cover all biologically-relevant suspects: 13C, 1H, 15N, 17O, 31P
11
Q
Why do we have a mixing period in 2/3D NMR.
A
- In multi-dimensional NMR we want to know how nuclei interact with one another
- Can be through scalar (bond) or dipolar (space) coupling
- Mixing allows coherence between signals to transfer to find degree of each coupling.
12
Q
How are solvent peaks eliminated from FID acquisition?
A
- Continuous wave decoupling destroys the coherence of large 1H peaks due to water
- A high power 90O acquisition pulse if done quickly before water has a chance to relax back in to spectrum.
13
Q
- A good 1D spectrum has good suppression of large solvent peaks. It can also tell us about the chemical shift dispersion in the spectrum. What is this and what else can a spectrum tell us?
A
- Chemical shift dispersion describes the general spread of the chemical shifts for a given nucleus in the NMR spectrum. If this is very clustered, it would not be wise to proceed with further 2D/3D NMR, due to high overlap.
- A good spectrum will also tell us the concentration of our sample of interest as well as its purity.
14
Q
What is the evolution delay, t1, in 2D NOESY?
A
- Evolution delay creates a second dimension via Fourier transform
- Incrimented in time to capture different chemical shift information in each plane.
- Collect # of planes with a different value for given incremented t1 delay.
15
Q
What is homonuclear NMR and what are some advantages and examples of types of molecules it can be used for
A
- 1H detection in both dimensions
- Circumvents need for expensive and time-consuming isotopic labelling.
- Works with even only a few signals/1Hs
- Can be used for
- Peptides (short proteins)
- Carbohydrates
- DNA/RNA fragments
16
Q
- … … (TOCSY) 2D spectrum yields correlation map for …coupled 1Hs.
- Coupling observed … bonds of intervening 1H couples
- Interrupted only by small/zero …-…coupling or …atoms e.g carbonyl
A
- Total correlation (TOCSY) 2D spectrum yields correlation map for scalar coupled 1Hs.
- Coupling observed ~6 bonds of intervening 1H couples
- Interrupted only by small/zero 1H-1H coupling or hetero atoms e.g carbonyl
17
Q
What are the limitations of TOCSY?
A
- While TOCSY good for isolating spin systems due to coupling within an amino acid side chain.
- Presence of carbonyl shuts of scalar coupling, halting the TOCSY chain
- Can assign by residue type, but if > 1 of the same AA, cannot discriminate easily in sequential assignment of chain.
18
Q
What is spin locking?
A
- A method of maintaining coherence and preventing dephasing during transfer of magnetisation through scalar coupling.
- This is done by locking magnetisation along an axis using an RF (pulse) field, causing magnetisation to flip and reacquire coherence.
19
Q
The Nuclear Overhauser (NOE) effect is a form a … (…-…) relaxation, in which … information is coupled between nuclei through …(<… Å)
A
The Nuclear Overhauser (NOE) effect is a form a dipolar (spin-lattice) relaxation, in which electronic information is coupled between nuclei through space (<5 Å)
20
Q
- Outline NOE’s origin
A
- Perturbation of one nucleus (spin I) from equilibrium can cause changes in relaxation rate (R1) and population distribution of nearby spins (S)
- Cross relaxation generates an increase or decrease in intensity on neighbouring spins (NOE).
22
Q
- (IMP) Describe a 1H-1H NOESY pulse program
A
- Three π/2/90o pulses
- 1: Creates transverse magnetisation precessing during incremented evolution period, t1
- 2: Creates longitudinal magnetization that mixes (population distribution changes via dipolar relaxation of excited spins) during mixing period, tm which varies with size
- 3: Creates transverse magnetization from remaining longitudinal magnetization for detection.
23
Q
How does molecular size affect NOE?
A
- Small molecules tumble rapidly (small τc) à +ve NOE
- Vice versa with large/highly viscous molecules, giving – ve NOEs approaching 100%
- Certain NOE’s may have zero intensity
- Can be avoided by changing T or viscosity
24
Q
How can NOESY data be used in combination to TOCSY spectra
A
- Spectra can be overlayed on to one another, the differences of which give an indication of 1Hs that are close in space but not in bonds
- Namely, the coupling between residues in the fingerprint region, walking from one residue to another via NH-Ha dipolar coupling
25
Q
Give a solution to the size effect problem in NOESY.
A
- Use ROESY (rotating frame Overhauser effect spectroscopy)
- Operates in same plane (XY) and through space to give map of dipolar coupled 1Hs
- Always gives a + ve value and leads to increase in signals regardless of size
29
Q
- Assigning CSI data
- If Δ𝛿 > |0.1| residue assigned +1 or -1
- If Δ𝛿 < |0.1| residue assigned value of 0
- Stretch of 4 or more +ve/-ve values gets assigned as a helix/sheet
- Mixed /0 assigned as random coil
- 75% accuracy if only Hα used
- >90% accuracy if 13C considered
A
37
Q
- In 1H-15N HSQC, every …-bearing … yields a peak (now labelled …)
- Every AA in backbone + additional AA’s with …-bearing …
- Polarisation transfer via … (through …) from 1H spins (high γH) to spins with low γx
- Enhances signal from … … …nuclei.
A
- In 1H-15N HSQC, every H-bearing N yields a peak (now labelled 15N)
- Every AA in backbone + additional AA’s with N-bearing sidechains
- Polarisation transfer via J-coupling (through bonds) from 1H spins (high γH) to spins with low γx
- Enhances signal from low-sensitivity heteronuclei.
44
Q
How would we interpret chemical shift perturbations in a spectrum?
A
- As [ligand] increases, chemical shifts of nuclei at/near binding site affected
- Given as a weighted change Δtotal
- More smeared peak from blue to red show CS with increased concentration, indicating direct involvement with binding, or indirect conformational change due to binding.
45
Q
Discuss the drawbacks of using Chemical shift perturbation, and what types of interactions it is best suited for.
A
- No direct way to know if interaction if due to direct binding or structural rearrangement upon complex formation
- Spread of data gives an indication of which it is (tight –> binding site, diffuse –> conformational change)
- This approach is best suited to weak interactions in low mM range showing clear fast exchange, where we know the old and new peak location and how they are related.
51
Q
Describe the CBCANH 3D correlation method
A
- Cα are positive
- Does not tell us which residue is i and which is i-1
52
Q
Describe the CBCA(CO)NH 3D correlation method
A
- Always paired with CBCANH, to determine difference between residues i/i+1
- Strips of these resolved peaks can be used to sequentially assign backbone (not identity of AA though)
- Allows magnetisation only through C=O carbon
53
Q
Describe the HNCA/HN(CO)CA pair that could be run to reinforce speculations made so far.
A
- HNCA: In each NH strip, 2 Cα peaks visible from residue i and i+1
- HN(CO)CA: In each NH strip, 1 Cα visible from preceding residue (i-1)
54
Q
Describe the HNCO/HN(CA)CO pair that could be run to reinforce speculations made so far.
A
- HNCO: In each NH strip, 1 CO visible from residue i-1
- HN(CA)CO: In each NH strip, 2CO peaks visible from residue i and i-1
55
Q
- What 2D experiments should be done to assign side chains of our protein first?
A
- HCCH-TOCSY: At each 13C chemical shift, see 1H atoms attached to given C and all other Cs in side chain
- HCCH-COSY: At each 13C chemical shift, see 1H atoms attached to given C and neighbouring Cs only.
- Nitrogen methods not used as not present in most sidechains
56
Q
What additional experiments can be done to assign side chains of our protein?
A
- Use CC(CO)NH and H(CCO)NH in combination with our HCCH-TOCSY/COSY
- CC(CO)NH: in each NH strip, see all C nuclei in sidechain of residue i-1
- H(CCO)NH: in each NH strip see, all H nuclei in sidechain of residue i-1
62
Q
What response is generated from plotting STD AF values vs [ligand] in STD NMR
A
Michaelis-Menten hyperbolic dose-response
