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NMR Spectroscopy In Structural Analysis Exercises werkcollege 2011 http://nmr.chem.uu.nl/Education/structuralanalysis_notes.php Sabine Gradmann Hans Wienk Rainer Wechselberger NMR Spectroscopy Research Group
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Page 1: NMR Spectroscopy In Structural Analysis · NMR Spectroscopy In Structural Analysis ... Chapter 2. Basic NMR Theory ... Chapter 4. Spin relaxation 14. Assume M z magnetization. Apply

NMR Spectroscopy

In

Structural Analysis

Exercises – werkcollege 2011

http://nmr.chem.uu.nl/Education/structuralanalysis_notes.php

Sabine Gradmann

Hans Wienk

Rainer Wechselberger

NMR Spectroscopy Research Group

Page 2: NMR Spectroscopy In Structural Analysis · NMR Spectroscopy In Structural Analysis ... Chapter 2. Basic NMR Theory ... Chapter 4. Spin relaxation 14. Assume M z magnetization. Apply
Page 3: NMR Spectroscopy In Structural Analysis · NMR Spectroscopy In Structural Analysis ... Chapter 2. Basic NMR Theory ... Chapter 4. Spin relaxation 14. Assume M z magnetization. Apply

EXERCISES – QUESTIONS 98

Notes.

Answers to Questions are included furtheron in this book.

Answer the Question yourself, before turning to the Answers!

For practice, a test exam with answers is found further in the book (p.129–p.138).

Ask questions! The assistants are there to help you.

Chapter 1. Introduction

1. NMR deals with nuclear magnetism, arising from the magnetic moment of the nucleus

of some types of atoms. The everyday form of (macroscopic) magnetism as manifested

by permanent magnets and magnetizable materials, has a different source. Find out

which.

Chapter 2. Basic NMR Theory

2. a) Draw the energy diagrams and indicate the and spin states for the following

nuclei: 3H γ

3H = 2.8535 · 10

8 (T·s)

−1 (I = 1/2)

113Cd γ

113Cd = 5.9340 · 10

7 (T·s)

−1 (I = 1/2)

b) At what frequency do these nuclear spins resonate given a 1.5 Tesla (T) magnetic

field?

c) What are the corresponding angular frequencies (―hoeksnelheid‖) of the precession

motion?

3. A proton at 11.7 T resonates at 500 MHz. To what value must the magnetic field be

raised in order to obtain resonance at 600 MHz?

4. Prove that the energy of a randomly oriented magnetic moment in the static B0 field is

only dependent on the z-component of the magnetic moment (see Reader Eqns. 2.6 and

2.7).

5. Consider the compound chloroform (CHCl3) where all carbon-atoms are of the 13

C-

isotope. The difference in the Larmor frequency of the 13

C spin and 1H spin is 675

MHz. What is field strength of the B0 field? What is the proton Larmor frequency? Use

Table 2.1 of the Reader.

6. The earth magnetic field has a strength of roughly 45 μT.

a) What is the proton resonance frequency in the earth magnetic field?

b) What is the wavelength of the electromagnetic radiation that can be absorbed at this

field?

7. The equation of motion (Reader Eqn. 2.11) is crucial to describe NMR phenomena.

Derive Reader Eqn. 2.12 from Eqn. 2.11 and verify that Eqn. 2.13 is a correct solution.

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EXERCISES – QUESTIONS 99

Chapter 3. An ensemble of Nuclear Spins

8. a) Consider an ensemble of 1H nuclear spins. What is the ratio of the populations of the

upper and lower states at 25 °C and a magnetic field of 2.35 T?

b) What is the population difference between the two spin states per 1·106 spins?

c) What is the population difference for 13

C, under the same conditions, per 1·106 spins?

9. A sample of chloroform is placed in a NMR spectrometer with a field strength

corresponding to a proton resonance frequency of 500 MHz.

a) Sketch the net-magnetization vector of the proton spins in the laboratory frame under

equilibrium conditions.

We want to give a 90° x-pulse to create transverse proton magnetization.

b) What should be the frequency of the B1-pulse?

c) Assume that strength of the B1-field is 0.60 mT. What should be the duration (tp) of

the 90° pulse (i.e. /2)?

d) Sketch the effect of the 90°-pulse in the rotating frame using a vector diagram. What

is the rotation frequency of the rotating frame?

10. Sketch in a vector diagram (rotating frame, on resonance) the effect on the

magnetization of a certain proton for the following cases. For questions 10a−10e we

start with equilibrium (+Mz) magnetization.

a) Apply a 90° x-pulse.

b) Apply a 90° y-pulse .

c) Apply a 180° y-pulse

d) Apply a 90° −x-pulse

e) Apply a 270° x-pulse. What is the difference compared with d)? What if the pulse is

slightly miscalibrated (e.g. if it is slightly too short)?

f ) Start with +Mx magnetization. Apply a 90° x-pulse.

g) Start with +Mx. Apply a 90° y-pulse.

h) Start with +My. Apply a 90° x-pulse.

i) Start with +My. Apply a 90° y-pulse.

11. You want to excite equilibrium magnetization. Unfortunately your excitation pulse is

miscalibrated and is only an 80° x-pulse instead of a 90° x-pulse. Which percentage of

the equilibrium magnetization will be in the transverse plane?

12. Use the equation-of-motion (Reader Eqn. 2.11) to calculate the effect of a y-pulse. The

magnetic moment for a single spin can be replaced by the net magnetisation vector M

combining all spins of the ensemble.

13. Consider exercise 9 again. Suppose you use a B1-pulse with a frequency of 400 MHz.

To examine the effect of this pulse, we need to use a rotating frame with a rotation

frequency of 400 MHz.

a) Sketch in a vector diagram the net-magnetization vector in equilibrium in this

rotating frame.

b) Sketch in a vector diagram the magnetization vector of an individual spin in this

rotating frame. What is the precession frequency of the spin in this rotating field?

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EXERCISES – QUESTIONS 100

c) What is then the effective B0 field in this rotating field? Use Hz as the unit for

magnetic field strength.

d) What is the total effective field Beff

in this rotating field? Assume a B1 strength of 25

kHz.

e) Predict the result of a 10 s B1-pulse.

Chapter 4. Spin relaxation

14. Assume Mz magnetization. Apply a 90° x-pulse.

a) Draw the path that the net-magnetization vector follows back to equilibrium, once in

the laboratory and once in the rotating frame.

b) Same as a) but now only consider T2-relaxation.

c) Repeat this for an 180° x-pulse.

15. Verify that Reader Eqns. 4.2 and 4.3 are valid solutions of Eqn. 4.1.

Chapter 5. Fourier Transform NMR

16. What is the line width at half height (use Reader Eqn. 5.6) of a signal with a T2 of:

a) 1.0 s?

b) 10 ms?

c) 10 s?

d) Associate the T2‘s of a), b) and c) with either a solid protein, chloroform, or a protein

in solution. Explain using the graph on Reader p.22.

e) Sketch the 1D spectra of compounds containing only two protons, with T2 similar to

those of a dissolved protein, chloroform and a solid protein. The difference in the

Larmor frequencies of the two protons is 100 Hz.

17. Sketch roughly the Fourier transforms of the following FIDs (on the left the My part of

the FID, on the right the Mx detection).

a) My Mx

b)

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EXERCISES – QUESTIONS 101

c)

d)

18. The T1 of a certain sample decreases with increasing temperature. At 20 °C it is

necessary to leave 10 seconds between successive scans, while at 40 °C only 5 seconds

are required. Suppose that 3 hours of spectrometer time are available and that the

instrument is set up for operation at 20 °C. It takes 1 hour to warm the sample to 40 °C

and to stabilize the temperature. Assume that the NMR signals are identical at the two

temperatures. What is the best strategy for recording the NMR spectrum: running for 3

hours at 20 °C or warming the sample and running for 2 hours at 40 °C? Use Eqn. 5.7.

Chapter 6. Spectrometer hardware

19. a) What is meant by ‗resolution‘?

b) What are the possibilities to raise the signal-to-noise-ratio (sino) for a 1H

experiment?

c) Compare the ―sino‖ in an NMR experiment using 13

C nuclei at natural abundance

with the ―sino‖ of a 1H in the same molecule.

20. You want to record a spectrum for a compound which has several signals in the range of

–12 kHz to +5 kHz. To what value must the dwell-time be set?

21. If the magnetic field varied by ± 10 nT between the edges of the sample, what level of

error would this introduce in the resonance frequency of 1H nuclei at a magnetic field

Page 7: NMR Spectroscopy In Structural Analysis · NMR Spectroscopy In Structural Analysis ... Chapter 2. Basic NMR Theory ... Chapter 4. Spin relaxation 14. Assume M z magnetization. Apply

EXERCISES – QUESTIONS 102

strength of 11.7 T (500 MHz)? Compare this error to the typical line width of a 1H

signal of 1 Hz.

Chapter 7. NMR parameters

22. Consider a compound containing two 1H spins that give signals at 2.0 and 7.0 ppm,

respectively. You perform a 1D experiment at a 500 MHz spectrometer. Calculate the

precession-frequencies in the rotating frame for the two signals, assuming a carrier-

frequency (the zero-frequency of the rotating frame) which is at 4.7 ppm.

23. The chemical shift of two equivalent methylene protons is 3.157 ppm. Their NMR

signal is split in two lines due to the scalar coupling (J-coupling) to the attached 13

C

nuclear spin. The value of the scalar coupling constant is 132.15 Hz. The magnetic field

strength is 0.5 T.

a) Calculate the chemical shift difference between the two methylene signals.

b) Calculate the frequency difference between each methylene signal and the reference

signal of TMS at 0 ppm.

c) What is the chemical shift difference when the field is ten times stronger?

Chapter 8. Nuclear Overhauser effect

24. The following NOEs are observed involving the non-exchangeable protons of a tyrosine

(see appendix p.165 for the atom nomenclature):

atom pair -intensity r (Å)

H2 – H3 0.125 2.45

H5 – H6 0.125 2.45

H2 – Hβ1 0.05 ?

H2 – Hα 0.02 ?

H2 – H? 0.01 ?

H2 – H? 0.005 ?

The distances between the - and -protons are known. Calculate the remaining

distances and suggest assignments for the two unassigned protons.

25. Estimate the molecular weight where the NOE vanishes on a 500 MHz spectrometer

(see Reader p.47).

Chapter 9. Relaxation measurements

26. Derive Reader Eqn. 9.1 from Eqn. 4.2 and prove that at = ln(2) · T1 no signal will be

observed in a inversion-recovery experiment (assume t = ).

27. Assume Mz magnetization. What happens in the following sequence, where the delay is

a certain fixed waiting time: 90x — delay — 180x — delay? Give a vectorial

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EXERCISES – QUESTIONS 103

representation of the evolution of the magnetization for two spins of unequal frequency

(relaxation can be neglected). Use the rotating frame to draw the vector diagrams.

Assume that the frequency with which this frame rotates (the zero-frequency or the

carrier-frequency) is the average of the frequencies of the two spins.

Repeat this for the sequence: 90x — delay — 180y — delay. What is the difference?

28. You measure the T2 of the 15

N-spins in two 15

N-labeled proteins under the exact same

conditions. The average T2 of protein A is 120 ms, while the average T2 of protein B is

60 ms. Give an explanation for this difference.

Chapter 10 Two-dimensional NMR

29. The structure of compound A undergoes a complete change by irradiating with light and

is transformed to compound B: A → B. This reaction can be measured in the SCOTCH

2D experiment (Reader Chapter 10.1). The light pulse is short enough that we can

assume that in the time of the pulse no precession of the spins occurs.

a) A proton A with a resonance ωA has a frequency ωB after the light pulse. Sketch the

2D-spectrum.

b) What is the difference if the reaction is not complete? Sketch the 2D-spectrum again.

30. Figure 1 shows the NMR spectrum of the amino acid tryptophan (W) acquired in D2O

(i.e. the side chain and backbone NH protons are not visible).

Figure 1: 1H-NMR spectrum of structure of tryptophan.

a) Sketch the COSY spectrum without fine structure of the cross-peaks.

b) Sketch the TOCSY spectrum.

c) In the NOESY the β-protons have cross-peaks to just two of the ring protons. To

which of them? Sketch the NOESY spectrum.

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EXERCISES – QUESTIONS 104

31. Figure 2 shows the NMR spectrum of the nucleotide adenosine-monofosfaat (AMP, see

Reader Fig. 12.7 and 12.8) acquired in H2O.

Figure 2: 1H-NMR spectrum of AMP (http://www.bmrb.wisc.edu/metabolomics/).

a) Sketch the COSY spectrum without fine structure of the cross-peaks.

b) Sketch the TOCSY spectrum.

c) Sketch the NOESY spectrum.

32. The three methyl-protons in ethanal are equivalent and are 2.55, 2.78 en 2.88 Å from

the aldehyde proton. The NOE signal intensity V = constant · 1/r6.

a) Calculate the NOE intensity for the distances between each methyl proton and the

aldehyde proton assuming a constant of 1·10–50

.

b) How many cross-peaks will be visible in the 2D NOESY spectrum?

c) Calculate the intensity of this / one of these cross-peaks.

d) Calculate the distance corresponding to this / these cross-peaks.

Chapter 11 The assignment problem

33. Sketch and explain the 1H-NMR spectrum of the following compounds (take only

coupling over 1–3 bonds into account):

a) 1,2-dichloorbenzene

b) 1,3-dichloorbenzene

c) 1,4-dichloorbenzene

34. How many signals and what multiplicity do you expect in the 1H-spectrum for the

following compounds (take only coupling over 1–3 bonds into account):

a) methanol

b) 1-bromo-2,2-dimethyl-propane

c) 2-chloro-1-methoxy-propane

35. Assign the proton spectra belonging to propanal (Figure 3).

Peak data (ppm-value, intensity):

9.798 200 9.793 371 9.789 203

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EXERCISES – QUESTIONS 105

2.499 87

2.495 87

2.475 265

2.470 266

2.451 271

2.446 275

2.426 92

2.422 96

1.137 503

1.113 1000

1.089 484

Figure 3: 1H NMR spectrum of propanal. (SDBSWeb : http://www.aist.go.jp/RIODB/SDBS/ (National Institute

of Advanced Industrial Science and Technology, 07-09-06)).

36. Two proton spectra are shown together with the formulas of the corresponding

compounds (Figure 4). Determine the structure.

Figure 4: 1H-NMR spectra of some organic compounds.

37. Figure 5 shows the 1H-NMR spectrum of the amino acid proline (P) acquired in D2O

(i.e. HN and HO protons are not visible).

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EXERCISES – QUESTIONS 106

Figure 5: 1H-NMR spectrum and structure of proline.

a) Draw the COSY spectrum without fine structure of the cross-peaks.

b) Sketch the TOCSY spectrum.

c) Sketch the NOESY spectrum.

Chapter 12 Biomolecular NMR

38. The following peptide segment is part of a protein sequence: . . . YVGLTSA . . . Figure

6A shows the corresponding TOCSY spectrum of this part. Assign the segment. What is

the origin of the peaks around 7 ppm?

39. Assume this segment has an β-sheet conformation. Add the characteristic peaks you

would expect in the NOESY spectrum in Figure 6B. Use Figure 12.4 in the Reader.

40. Now the segment is located in an α-helix. Add the expected peaks in Figure 6C (use

Reader Figure 12.4).

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EXERCISES – QUESTIONS 107

Figure 6A: 1H,

1H TOCSY spectrum of -YVGLTSA-.

Page 13: NMR Spectroscopy In Structural Analysis · NMR Spectroscopy In Structural Analysis ... Chapter 2. Basic NMR Theory ... Chapter 4. Spin relaxation 14. Assume M z magnetization. Apply

EXERCISES – QUESTIONS 108

Figure 6B: 1H,

1H TOCSY spectrum of -YVGLTSA-.

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EXERCISES – QUESTIONS 109

Figure 6C: 1H,

1H TOCSY spectrum of -YVGLTSA-.

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EXERCISES – QUESTIONS 110

41. Consider again the peptide fragment YVGLTSA of exercises 38–40. Construct the

NOE-pattern for a type-I turn starting at glycine in Figure 7 (use Reader Figure 12.4).

42. We perform an exchange experiment, replacing H2O with D2O. Indicate for Figures

6B–C (Questions 39 and 40) the peaks that are still visible in this solvent (directly after

the exchange of solvent).

For Figure 6B, assume that the HN of the Tyrosine is in a H-bond with an opposite β-

strand, and the β-strand is the outer strand of a β -sheet.

For Figure 6C, first assume that the segment is located in the middle of a long α-helix.

What would be different if the helix starts with the tyrosine (Y) of the segment?

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EXERCISES – QUESTIONS 111

Figure 7: 1H,

1H TOCSY spectrum of -YVGLTSA-

Page 17: NMR Spectroscopy In Structural Analysis · NMR Spectroscopy In Structural Analysis ... Chapter 2. Basic NMR Theory ... Chapter 4. Spin relaxation 14. Assume M z magnetization. Apply

EXERCISES – ANSWERS 112

Answers

Chapter 1. Introduction

1. The type of magnetism of permanent magnets and magnetizable materials is called

ferromagnetism and is caused by a very strong interaction between the magnetic

moments of the electron spin. The magnetic moment of the electron spin is roughly 650

times as large as that of a proton.

Chapter 2. Basic NMR Theory

2. a) The energy diagram consists of two energy levels (both are spin I = 1/2 nuclei), note

that for nuclei with a positive gyro-magnetic ratio the -state (m = 1/2) has the lowest

energy and the -state the highest energy. The energy separation between the two levels

is larger for the 3H spin since it has a higher gyromagnetic ratio and zBE .

b) The resonance frequency can be calculated using Reader Eqn. 2.10a

0

2B0 .

Thus, the resonance frequency of 3H equals 2.8535·10

8 (T·s)

−1 × 1.5 T / 2 =

0.68122·108 s

−1 = 0.68122·10

2 MHz = 68.1 MHz. Likewise, the resonance frequency of

113Cd is 5.9340·10

7 × 1.5 / 2 = 1.4166·10

7 s

−1 = 14.2 MHz.

Clinical MRI scanners typically operate at 1.5 T. The protons in your brain will then

resonate at 63.8657 MHz.

c) The precession has the same frequency as the resonance frequency. The

corresponding angular frequencies are obtained by multiplying by 2 giving 4.28·108

s−1

(3H) and 8.90·10

7 s

−1 (

113Cd) (NOTE: the unit has to be s

−1 and not Hz).

3. Reader Eqn. 2.10a shows that the resonance frequency depends linearly (―recht-

evenredig‖) on the strength of the magnetic field. Thus, the field should be raised to

11.7 / 500 × 600 = 14.04 T.

4. Reader Eqn. 2.6 denotes the dot-product (―in-produkt‖) of the magnetic moment and the

static magnetic field (two vectors), resulting in the scalar energy:

E B . The

orientation of the magnetic field is normally taken to be parallel with the z-axis, thus

B 0,0,B0 . The magnetic moment will have non-zero x,y,z-projections. Thus:

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EXERCISES – ANSWERS 113

E B x y z 0

0

B0

x 0 y 0 zB0 zB0

5. Use Reader Eqn. 2.10a: the resonance frequency (= Larmor frequency) of the 13

C spins

is

13C 13C2

B0. For the 1H spins

1H 1H

2B0. The difference in the Larmor frequency

is thus:

CH

CH

CH

CHCH B

BBB

131

131

0

0

1310

13

01

131 2222

.

Thus, the field strength is 2× 675·106 Hz / (2.6752·10

8 - 6.7266·10

7 (T·s)

−1 ) =

4.2411500·109 Hz / 2.00254·10

8 (T·s)

−1 = 21.18 T. The Larmor frequency of the protons

is

1H 1H

2B0 = 2.6752·10

8 (T·s)

−1 × 21.18 T / 2MHz.

6. The resonance frequency can be calculated using Reader Eqn. 2.10a

0

2Bz . Thus,

the resonance frequency of 1H equals 2.6752·10

8 (T·s)

−1 × 45·10

-6 T / 2 = 1916.0 Hz =

1.92 kHz. The corresponding wavelength can be calculated using

c (c =

2,99792458·108 m/s), resulting in a wavelength of 156470 m = 1.56·10

2 km.

7. Reader Eqn. 2.11 denotes the cross-product (―uit-produkt‖) of the magnetic moment

and the magnetic field. The cross-product equals the determinant of the matrix shown

on the right-hand side of Reader Eqn. 2.11. The determinant is calculated by expansion

by minors (―ontwikkeling naar de eerste rij‖):

xyyx

xyyxzxzzxyyzzyx

yx

yx

z

zx

zx

y

zy

zy

x

zyx

zyx

zyx

BeBe

BBeBBeBBe

BBe

BBe

BBeBBB

eee

dt

d

00

Note that ex is the unit vector (1,0,0). Similarly, ey=(0,1,0) and ez=(0,0,1). That means

that the ex term shows the change of d/dt in the x-direction, etc.

In the last line, it was used that

B 0,0,B0 . Thus,

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EXERCISES – ANSWERS 114

xyyx

xyyx

BeBe

BeBedt

d

00

00

or

dx

dt B0y

dy

dt B0x

dz

dt 0

Note that the orientation of the resulting vector is given by the right-hand rule (see

Figure on reader p.11).

To verify that Reader Eqn. 2.13 is a correct solution, we back-substitute the solution in

the differential equation and differentiate:

dx

dt

d

dtx (0)cos(Bzt)y(0)sin(Bzt)

x (0)Bz sin(Bzt)y(0)Bz cos(Bzt)

Bz x (0)sin(Bzt)y(0)cos(Bzt) Bzy

dy

dt

d

dtx (0)sin(Bzt)y(0)cos(Bzt)

x (0)Bz cos(Bzt)y(0)Bz sin(Bzt)

Bz x (0)cos(Bzt)y(0)sin(Bzt) Bzx

dz

dt

d

dtz (0) 0

Here you have to use the chain-rule (―ketting-regel‖): d/dt (f(g(t)) = d/dg f × d/dt g.

Note that you have to assume that the answer is correct to establish the correctness of

the answer!

Chapter 3. An ensemble of Nuclear Spins

8. a) Using the Boltzmann equation (Reader Eqn. 3.1), and Reader Eqn. 2.9 and 2.10 to

relate the magnetic field strength to an energy difference and k = 1.38066·1023

J/K, one

finds:

n

n

eE

kT e Bz

kT ehBz

2kT

e2.6752108 T 1 s1 6.626071034 Js2.35T

2 1.380661023 JK 1 298K

e2.6752108 6.626071034 2.35

2 1.380661023 298

e41.65621026

2585.11023 e41.65621026

2585.11023 e41.6562103

2585.1

e0.016113103

0.99998389

b)

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EXERCISES – ANSWERS 115

500004.099998389.1

1199998389.0

)1(

nnn

n

n

n

n

This is short for:

0.99998389 · na = 1 – na na + 0.99998389 · na = 1 na · (1 + 0.99998389) = 1

na · (1.99998389) = 1 na = 1 / 1.99998389

Thus, for an ensemble of 1·106 spins there are 1·10

6 × 0.500004 = 500004 spins in the

-state and 1000000–500004 = 499996 spins in the -state. The population difference

for a real sample, expressed as a fraction of 106 spins, is thus 8.06 spins.

c) For 13

C spins, this is 2.02 spins.

9. a) The net-magnetization vector is aligned with static B0-field under equilibrium

conditions:

b) The frequency of the B1-pulse should match the energy-difference between the two

energy-levels, which means that it should be identical to the Larmor-frequency (see

Reader Eqns. 2.9 and 2.10). Thus, the frequency should be 500 MHz.

c) The magnetization vector precesses around the B1-field with 112

B

(see Reader

Eqn. 3.2). The rotation frequency is: 2.6752·108 (T·s)

−1 × 0.60·10

-3 T / 2 = 25546 Hz.

A 90°-pulse corresponds to a quarter of a rotation. Thus, the length should be ¼ ×

1/25546 = 9.79 s.

d) To describe the effect of the 90°-pulse, the rotating frame must have the same

frequency as the B1-pulse, 500 MHz.

10. a) + My

b) –Mx

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EXERCISES – ANSWERS 116

c) –Mz

d) –My

e) –My

The magnetization vector now travels in the opposite direction to end up at the same

point. If the pulse is slightly too short, a 90° –x-pulse will rotate the magnetization

vector to some point above the xy-plane, whereas a 270° x-pulse will rotate the vector to

some point under the xy-plane. Note that because of the three times longer pulse, this

‗mismatch‘ will be thrice as large, i.e. the effects of miscalibration are more pronounced

for longer pulses.

f) +Mx

g) +Mz

h) –Mz

i) +My

Note that all rotations are clock-wise!

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EXERCISES – ANSWERS 117

11. The projection of the magnetization vector onto the y‘-axis is given by sin(), where

is the flip-angle. Thus, 98.5 % is in the transverse plane.

12. Again we use the equation of motion, Reader Eqn. 2.11. In the rotating frame B =

(0,B1,0):

xzzx

xzzx

yx

z

zx

y

zy

x

yx

yx

z

zx

zx

y

zy

zy

x

zyx

zyx

zyx

MBeMBe

MBeMBe

MM

Be

MMe

MM

Be

MM

BBe

MM

BBe

MM

BBe

MMM

BBB

eee

11

11

11 0000

Thus,

xzzx

xzzx

MBeMBe

MBeMBedt

dM

11

11

, or:

xz

y

z

x

MBdt

dM

dt

dM

MBdt

dM

1

1

0

The solution of the differential equation is analogous to Reader Eqn. 2.13 (just swap the

x,y,z indices):

0)0()(

)cos()0()sin()0()(

)sin()0()cos()0()(

11

11

yy

xzx

xzz

MtM

tBMtBMtM

tBMtBMtM

Which simplifies to:

0)(

)sin()(

)cos()(

10

10

tM

tBMtM

tBMtM

y

x

z

when starting from equilibrium

magnetization. Thus we see a simple rotation around the y-axis going from +z to –x.

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EXERCISES – ANSWERS 118

13. a) The net-magnetization is aligned with the B0-field, same as in Exercise 9a.

b) The magnetic moment of an individual spin will make an angle with z-axis and

precesses around the B0-field with 500 MHz in the laboratory frame. When

transforming into the rotating frame, the new precession frequency is = –RF

(Reader Eqn. 5.2), where is the lab-frame Larmor frequency and RF the rotation

frequency of the rotating frame. The spin thus precesses with 500 – 400 = 100 MHz.

c) As the spin still precesses in this rotating field, it must still feel a B0-field. This field

is now much smaller as the precession frequency is also smaller. When we express the

strength of the B0-field in MHz, the strength is 500 MHz in the lab-frame and 100 MHz

in the rotating frame.

d) The total effective field is the vector-sum of the reduced B0 and B1. So the strength is:

Beff B0,reduced

2 B1

2 100106 2

25103 2

100 MHz. The angle of this effective

field with the z-axis is:

arctanB1

B0,reduced

arctan

25 103

100106

arctan 0.25 10

3 0.

e) Since the frequency of the B1-pulse (400 MHz) is much lower than the resonance

frequency of the spins (500 MHz), the spins are not affected by the pulse. So the

orientation of the net-magnetisation vector will not change.

Chapter 4. Spin relaxation

14. a) In the laboratory frame: the magnetization vector starts some where in the xy-plane

and precesses around the z-axis. Due to T2-relaxation, the component in the xy-plane

will shrink. Simultaneously, the vector starts moving back to the positive z-axis due to

T1-relaxation. In the rotating frame: now there is no precession around z-axis since the

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EXERCISES – ANSWERS 119

spin is on resonance. So after the pulse, the vector is along the +y-axis and the vector

rotates around the x-axis back towards the z-axis. Check for yourself that the Figure is

only correct if T1 = T2. What does it look like if T2 < T1?

b) In the laboratory frame: the magnetization vector starts some where in the xy-plane

and precesses around the z-axis. Simultaneously, the length of the vector becomes

smaller due to T2-relaxtion. In the rotating frame, the magnetization starts at the +y-axis

and simply shrinks to eventually vanish.

c) After a 180x pulse the magnetization is along –z. As there is no x- or y-component,

there is no T2-relaxation, only T1 relaxation. The magnetization vector starts at –Mz, will

shrink, at some point be zero, and then start growing again to become +Mz. The

description in the lab and rotating frame are the same since there will be no precession

since there is no transverse magnetization, only longitudinal magnetization.

15.

11

11

)()0(

1

11)0(0

)0(

)0(

1

11

11

1

T

MtMeMM

T

eT

MeT

M

eMdt

deM

dt

dM

dt

d

eMMMdt

d

dt

dM

eqzT

t

eqz

T

t

eq

T

t

z

T

t

eq

T

t

zeq

T

t

eqzeqz

and

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EXERCISES – ANSWERS 120

22

2

)()0(

1

1)0(

)0(

2

2

2

T

tMeM

T

eT

M

eMdt

d

dt

dM

xT

t

x

T

t

x

T

t

xx

Chapter 5. Fourier Transform NMR

16. The line width (l.w. or ½) is related to the T2 via: 2T

1..

wl . a) 0.318 Hz; b) 31.8

Hz; c) 31.8 kHz; d) chloroform = a, dissolved protein = b, solid protein = c. The T2

becomes shorter as the tumbling time of a molecule becomes longer (= slower

tumbling). Chloroform is a very small molecule so tumbles very rapidly, resulting in

very slow T2 relaxation = large T2 values. A solid protein does not tumble at all,

resulting in very fast T2 relaxation = small T2 values.

e)

17. Realize that in the observed FID the present frequency is jRF (Reader Eqn.

5.2). ≠ so when j≠rfIn a), transverse magnetisation starts at +y, then turns to

+x, so turns faster than the rotating frame. Therefore in the spectrum the frequency will

have a positive value. Also, when in and FID intensity goes to zero, T2 relaxation occurs

and line-widths are not zero (Reader Eqn. 5.6). In the FID shown in b) the visible

wavelength is longer, so the visible frequency is smaller than in a). The direction of

rotation is the same as for a), so frequency is positive. The signal decays with similar

speed as in a), so the T2 and line width is the same. In c) the frequency and sign of the

FID are the same as in b), but the signal decays to zero faster. That means that T2 is

shorter and the signal gets broadened. In d), the magnetisation starts at +y and then turns

to –x. This means that is rotates slower than the rotating frame and gives a negative

frequency.

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EXERCISES – ANSWERS 121

a) b)

-25 0 (Hz) +25 0 (Hz) +25

jrf jrf

c) d)

- 0 (Hz) +25 0 (Hz) +25

jrf jrf

18. At 40°C you can record twice as many scans within a the same time as at 20°C. In 2

hours, you thus record 2 × 2/3 = 4/3 as much scans at 40oC as in 3 hours at 20°C. The

signal-to-noise is thus

4

3 better at 40°C.

Chapter 6. Spectrometer hardware

19. a) In the context of NMR resolution is used to refer to the resolving power of the NMR

spectrum, i.e. the degree to which peaks with small differences in resonance frequency

can be distinguished.

b) According to Reader Eqn. 6.1 one can: raise the number of spins (higher

concentration), raise the field, increase the number of scans, lengthen the relaxation

time (higher temperature, lower viscosity medium) or lower the temperature of the

detection circuit (not of the sample).

c) The ―sino‖ of a 13

C experiment will be much lower due to the lower number of spins

and the lower gyro-magnetic ratio. In most cases there will be slightly more favorable

T2 for the 13

C spins than the 1H spin. Disregarding this last effect, the relative ―sino‖ is:

N13C

N1H

13C1H

52

1.11

99.9856.7266107

2.6752108

52

0.0111 0.2514452 0.01 0.03 0.0003519

So the 1H experiment is roughly 3000 times as sensitive!

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EXERCISES – ANSWERS 122

20. The dwell-time has be set according to the Nyquist-equation (see Reader Eqn. 6.2). The

largest absolute value of frequencies is 12 kHz. This frequency has to be sampled twice

per period, so the dwell-time must be 0.5 × 1/12000 = 0.0416 ms.

21. The relative error in the resonance frequency will be equal to the relative error in the

magnetic field, which is 10 nT / 11.7 T ≈ 1·10-9

. This creates an absolute error of 5·108

Hz × 1·10-9

= 0.4–0.5 Hz, roughly half of the natural line width. So even this small

variation in B0 his is unacceptable and must be improved by shimming.

Chapter 7. NMR parameters

22. The precession-frequency in the rotating frame () is given by: = – RF (Reader

Eqn. 5.2), where is the frequency of the signal of interest and RF the rotation

frequency of the rotating frame. In this case it is easy to calculate the frequencies with

respect to an reference signal at 0 ppm. Thus, we need to translate ppm to Hz: at 500

MHz, 1 ppm corresponds to 1 millionth of 500 MHz is 500 Hz.

Thus, the frequency of signal at 2 ppm is 2 × 500 = 1000 Hz (with respect to a signal

at 0 ppm), the frequency of the signal at 7 ppm is 7 × 500 = 3500 Hz (with respect to a

signal at 0 ppm) and the frequency RF of the rotating frame is 4.7 × 500 = 2350 Hz.

Thus, the frequency difference for the signal at 2 ppm is 1000 – 2350 = -1350 Hz (a

counter clockwise rotation) and for the signal at 7 ppm is 3500 – 2350 = +1150 Hz (a

clockwise rotation).

23. a) The frequency difference between the components of the doublet is determined by

the scalar coupling constant and is not influenced by the magnetic field strength. The

frequency difference is thus 132.15 Hz. At 0.5 T, the resonance frequency of 1H equals

2.6752·108

(T·s)−1

× 0.5 T / 2 = 21.28856519 MHz, 1 ppm is thus 21.28856519 Hz,

and 132.15 Hz = 6.207557852 ppm (dus 2 × 3.103778926).

b) The reference signal of TMS is at 0 ppm. The difference to the center of the

methylene doublet is thus 3.157 ppm. At 0.5 T, the resonance frequency of 1H equals

2.6752·108

(T·s)−1

× 0.5 T / 2 = 21.28856519 MHz. 3.157 ppm thus equals 3.157·10-6

×

21.3 MHz = 3.157 x 21.3 Hz = 67.2080003 Hz. The two lines are thus at 67.2080003

+/- 132.15/2 Hz from 0 ppm, which makes the shortest frequency distance equal to 1.13

Hz.

c) Ten times smaller, i.e. 0.6207557 ppm. Therefore the signals appear much closer to

each other in the spectrum (3.46737785 and 2.84662215). So, the J-coupling constant is

y'

2 ppm = –1350 Hz

fast

tumbling 7 ppm =

+1150 Hz

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EXERCISES – ANSWERS 123

in Hz independent of the field strength, but leads to signals at different positions when

the ppm-scale is considered.

Chapter 8. Nuclear Overhauser effect

24. H2 – Hβ1 r = rref × (Vref/V)1/6

= 2.45 × (0.125/0.05)**(1/6) = 2.85 Å.

H2 – Hα 2.45 × (0.125/0.02)**(1/6) = 3.32 Å.

H2 – H? = 3.73 Å.

H2 – H? = 4.19 Å.

The distance between the 2 and 5 proton is 2 × 2.45 = 4.9 Å. The distance between

the 2 and 6 proton is 2 × cos(30) × 2.45 = 4.24 Å. So the first distance, must be to the

2-proton, and the second is to the 6 proton. (Use the molecular structure on Reader

p.122 to prove these relations).

25. The NOE vanishes for · c = 1.118 (p.47). Thus, c = 1.118 / (500 × 2 × × 106) =

3.75·10-10

s. This corresponds to a molecular weight of c × 2.4·1012

= 854 Dalton

(Reader Eqn. 4.6).

Chapter 9. Relaxation measurements

26. For the inversion-recovery experiment Mz(0) = -Meq, thus:

M z M eq M z (0)M eq e t

T1

M eq M eq M eq e t

T1

M eq 2M eqe

t

T1

M eq 1 2e

t

T1

and

M z ln(2)T1 M eq 1 2e

ln 2T1

T1

M eq 1 2e ln2 M eq 1 2e

ln 12 M eq 1 2

12 0.

(Using the rule that –ln(x) = ln (x-1

) = ln(1/x)).

27. +My - precession creates an –x component – 180x pulse rotates the +y component to –y

but the leaves the x-component unaffected: effectively the vector is reflected in the xz-

plane – precession rotates the vector along the same angle as in the first part and in the

same direction, thus leaving the vector aligned the –y-axis. This sequence is called a

spin-echo (note that the last three Figures show projections on the x‘y‘-plane).

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EXERCISES – ANSWERS 124

Using a 180y pulse the y-component is unaffected but the –x component is rotated to +x:

a reflection in the yz-plane. Now the same precession-angle ends up at +y.

28. The T2 relaxation time depends on the rotational correlation time which is a measure for

the tumbling time of the molecule. Large molecules have long tumbling times (tumble

very slowly) and have very short T2‘s (see Reader Figure p.21). Thus protein B must be

larger than protein A.

Chapter 10 Two-dimensional NMR

29. a) a peak at coordinate (t1,t2) (A, B).

b) two peaks: one at (A, A) and one at (A, B). Their relative intensities will reflect

their relative concentrations.

30. Diagonal peaks are present, but not shown here!

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EXERCISES – ANSWERS 125

The -protons only ‗see‘ the H4 and H2. Note that the H2 is more than 5 Å from H4

and H7.

31. a) COSY: cross peaks between H1‘×H2‘; H2‘×H3‘; H3‘×H4‘; H4‘×H5‘; H4‘×H5‘‘;

H5‘×H5‘‘.

b) TOCSY: cross peaks between all sugar protons. No cross peak between H2 and H8

c) NOESY: cross peaks between all sugar protons. Possibly also H1‘×H2; H1‘×H8;

H2‘×H8; H3‘×H8, H4‘×H8; H5‘/H5‖×H8, etc. Signal intensities between sugar and

base depend on the base orientation.

32. a) V = C × r-6

. Substituting the values results in signal volumes of 3.64·107, 2.17·10

7 and

1.75·107, respectively.

b) Only one, since the methyl protons are dynamically averaged to equivalence.

c) It will be the sum of the volumes above, so 7.56·107.

d) The corresponding single distance is 2.25 Å, shorter than the three actual distances!

When calculating protein structures from NOESY cross-peak volumes this effect must

be taken into account to prevent distorted structures.

Chapter 11 The assignment problem

33. The chemical shift of the protons in these compounds will be roughly the same, around

6–7 ppm. The number of signals and the splitting will differ.

a) 1,2-dichloorbenzene has two sets of chemically equivalent protons, H3/H6 and

H4/H5. The H3 is coupled to the H4 and the H6 to the H5, thus they have the same

coupling pattern and they are magnetically equivalent. For the H4/H5 is the same thing.

So one expects two signals, each split in a doublet (1:1).

b) 1,3-dichloorbenzene has three different sets of protons: the H2, the H4/H6 and the

H5. The H5 is coupled to the H4 and the H6 so will be a triplet (integral 1/4:1/2:1/4).

The H4 and H6 are split in a doublet (1:1). The H2 is a singlet (1).

c) 1,4-dichloorbenzeen is fully symmetric, so all 4 protons experience the same

chemical environment. The protons are only coupled to the adjacent proton and both

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EXERCISES – ANSWERS 126

have the same coupling pattern, so they are also magnetically equivalent, so one signal

without splitting is observed (integral 4).

34. a) CH3OH

–CH3 doublet (1:1); –OH: quartet(1:3:3:1)

b) (CH3)3–C–CH2Br

–CH3 singlet; –CH2 singlet

c) CH3−CHCl−CH2−O−CH3

–CH3 doublet; –CH quartet of triplets (1:2:1) in case JCH3-CH >> JCH2-CH or a triplet of

quartets in case JCH3-CH << JCH2-CH. If the two J‘s are equal, then a sextet results with

relative intensities 1:5:10:10:5:1; –CH2 doublet; –OCH3 singlet.

35. The triplet at 9.7 ppm –COH; the quartet of doublets at 2.4 ppm –CH2, the triplet at 1.1

ppm –CH3.

36. a) 2-chloro-propane; c) diethyl-para-benzenedicarboxylate (note: in theory the CO- and

O-groups could also be reversed).

37.

Chapter 12 Biomolecular NMR

38. see Figure below. The peaks around 7 ppm are the aromatic ,-protons of Y.

39. add dN(i,i+1) strong peak and dNN(i,i+1) weak peak. Note that the dNN(i,i+1) contacts

should be symmetrical with respect to the diagonal.

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EXERCISES – ANSWERS 127

40. add dN(i,i+1) weak peak, dNN(i,i+1) strong peak, dNN(i,i+2) weak peak, dN(i,i+3)

medium peak, dN(i,i+4) weak peak. Note that d contacts are not visible in this region

of the spectrum!

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EXERCISES – ANSWERS 128

41.

42. If the β-strand is on the outside of a β-sheet, the amide proton of every second residue is

exposed to the solvent. E.g. if Y‘s HN is in a H-bond, the amide protons of V,L,S will

be unprotected and will exchange rapidly disappear from the spectrum. If the segment is

in the middle of a long helix, all peaks remain visible because the amide protons are

protected by hydrogen bonds (at least initially, they will be exchanged only very

slowly). If the helix starts at Y, the amide protons Y,V,G will be unprotected and will

exchange rapidly and disappear from the spectrum.

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OEFENTENTAMEN – OPGAVEN 129

OEFENTENTAMEN STRUCTUUR-ANALYSE / NMR

1. Imaging voor beginners

Magnetic Resonance Imaging (MRI) is eigenlijk NMR, maar dan gebruikt om afbeeldingen te

maken van objecten. Het principe van MRI onderzoeken we in deze opgave.

Stel je een NMR-buis voor die gevuld is met watten en twee glazen bolletjes (Figuur 1A).

Beide bolletjes zijn gevuld met water en ze bevinden zich allebei in het gevoelige gedeelte

van de detectiespoel. Het water in de bolletjes geeft een signaal in het 1H-spectrum; de watten

niet.

Figuur 1

Nu wordt in de z-richting een extra magneetveld aangelegd waarvan de sterkte variëert met de

z-coördinaat (de hoogte in de NMR-buis). Dit wordt ook wel een veldgradiënt G(z) genoemd.

Voor deze gradiënt geldt: G(z) = G·z waarbij G gelijk is aan de gradiëntsterkte in mT/cm. De

spins in bolletje A ervaren dus een andere sterkte van het magneetveld dan de spins in bolletje

B (voor A is dat B0 + G·zA en voor B is dat B0 + G·zB).

Je voert dit experiment uit. Neem aan dat B0 gelijk is aan 14.1 T en G = 0.0282 mT/cm. In het

NMR spectrum zie je twee signalen (beide zonder opsplitsing en met gelijke intensiteit).

a) Leg uit waarom je twee signalen ziet.

b) Welk bolletje (A of B) geeft een signaal bij hogere frequentie? Leg uit.

Je meet het verschil in chemical shift tussen de twee signalen: = 1.274 ppm.

c) Bereken het hoogteverschil (z) tussen bolletje A en B.

Stel nu dat we een derde bolletje water toevoegen precies tussen A en B in.

d) Schets het NMR spectrum.

Je stopt vervolgens een normale NMR-buis gevuld met water in de spectrometer. De buis is

gevuld tot de de hoogte van bolletje A (Figuur 1B). Je zet de gradiënt weer aan en neemt het

spectrum op.

e) Schets het NMR spectrum en leg uit waarom het er zo uitziet.

Tenslotte stop je een stuk weefsel in een NMR-buis. In het midden tussen de plek van bolletje

A en B bevindt zich een stuk bot, ter hoogte van bolletje B bevindt zich een holte gevuld met

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OEFENTENTAMEN – OPGAVEN 130

water, en de rest is spierweefsel (Figuur 1C). In bot bevindt zich vrijwel geen water, in

spierweefsel een beetje. Neem aan dat je alleen signalen afkomstig van water kunt zien in het

uiteindelijke spectrum.

f) Schets het NMR spectrum en leg uit.

2. T1 en waterondrukking

Met de onderstaande puls-sequentie kun je in een NMR experiment T1-tijden bepalen:

a) Laat met een vectordiagram zien wat de uitkomst van het experiment is als =0 en beide

pulsen langs de +x-as gegeven worden. Je begint met evenwichtsmagnetisatie.

b) Hoe verhoudt de intensiteit van het meet-signaal zich tot de intensiteit van de

evenwichtsmagnetisatie als =0?

De algemene formule voor het herstel van de evenwichtsmagnetisatie door T1-relaxatie is:

1)0()(T

eqzeqz eMMMM

.

c) Hoe lang is ten opzichte van T1 als je met deze sequentie geen signaal-intensiteit ziet?

Deze puls-sequentie kan ook gebruikt worden om het signaal van water te onderdrukken

wanneer je een spectrum van een eiwit in oplossing wilt opnemen. De T1 van een water-

proton is ca. 4s; de T1 van een eiwit-proton ca. 1s.

d) Waarom kan de T1 van een proton-spin in verschillende moleculen zo verschillend zijn?

e) Schets de relatieve intensiteit van het signaal voor een water-proton en voor een eiwit-

proton als functie van de wachttijd .

f) Wat is de optimale waarde van om het water-signaal te onderdrukken? En hoeveel

procent van de maximale signaalintensiteit verlies je in dit geval voor het eiwit-proton?

3. Reacties in 2D

Stel je voor dat je een lichtgevoelige stof onderzoekt. De stof kan met behulp van laserlicht

van verschillende golflengten in verschillende vormen worden gebracht. Een bepaald proton

kan onder invloed van deze lichtpulsen in drie verschillende chemische omgevingen verkeren:

A220nm

B450nm

C

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OEFENTENTAMEN – OPGAVEN 131

Neem aan dat aan het begin van een experiment alleen A aanwezig is. Neem verder aan dat

voor de Larmor-frequentie van het proton in de drie omgevingen A, B en C het volgende

geldt: A >C > B.

Schets 2D spectra voor de volgende puls-sequenties:

a) laserflits220nm – 90x – t1 – laserflits450nm – t2 (onvolledige omzetting bij 220nm; volledig bij

450nm).

b) laserflits220nm – 90x – t1 – laserflits220nm & laserflits450nm – t2 (alle omzettingen onvolledig)

Neem nu aan dat in de situatie van b) de conversie voor de reactie van A naar B 65% is en die

voor de reactie van B naar C 90% is.

c) Wat zijn dan de relatieve intensiteiten van de pieken in het spectrum bij b)?

4. Een stukje eiwit

Een segment van een eiwit, een peptide, bevat 8 residuen:

…T A G S L E V Y…

1 2 3 4 5 6 7 8

De volgende figuren laten COSY en TOCSY spectra van het peptide zien. De spectra tonen in

F2 alleen het gebied tussen 6.4 en 10.4 ppm.

a) Ken zoveel mogelijk signalen toe; geef in het TOCSY spectrum voor alle pieken aan wat

de toekenning is (één-lettercode bij de diagonaal-piek en , etc. bij de kruis-pieken.

Stel dat het hele segment een -helix conformatie heeft.

b) Teken in het TOCSY spectrum de H-HN en HN-HN contacten die je in het NOESY

spectrum verwacht te zien.

Stel nu dat het segment een turn bevat (zie het NOESY spectrum).

c) Welke residuen maken deel uit van de turn? BONUS-punt: Welk type turn is dit?

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OEFENTENTAMEN – OPGAVEN 132

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OEFENTENTAMEN – OPGAVEN 133

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OEFENTENTAMEN – OPGAVEN 134

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OEFENTENTAMEN – ANTWOORDEN 135

ANTWOORDEN OEFENTENTAMEN

1. (12p)

a) Omdat de magneetveldsterkte bij de twee bolletjes verschillend is, heeft het water ook

verschillende Larmor-frequenties. Daarom krijg je twee signalen in het spectrum. (2p)

b) Bolletje A zit bij een postieve z-coordinaat terwijl bolletje B zich bij een z-coordinaat ≈ 0

bevindt. De magneetveldsterkte Beff = B0 + G·z is dus bij A hoger dan bij B en dus is de

Larmorfrequentie van A hoger. (2p)

c) ppm Hz: = B/2 <=> = 2.6752·108 /2 = 6.00·10

8 Hz = 600 MHz (1p). 1 ppm

is dus 600 Hz. Het frequentieverschil tussen de twee pieken is daarom 600*1.274 = 764.8 Hz

(1p). Nu moeten we deze frequentie relateren aan een magneetveldsterkte, dus weer: =

B/2 <=> B = 2*764.83/2.6752·108 = 17.96 T (1p). (Dit kan je ook in één

keer berekenen door bovenstaande formules te combineren tot: B = B0*1.274·10–6

).

Dit is het verschil in magneetveldsterkte tussen bolletje A en B. Dit moeten we relateren aan

de z-coordinaat. Daarvoor moet je de gradient-sterkte G gebruiken: 0.0282 mT/ cm => 1 mT

= 35.46 cm <=> 1 T = 35.46·10–3

cm. Dus 17.96 T komt overeen met 17.96*35.46·10–3

cm

= 0.637 cm (1p). (4p totaal)

d) Er is nu een derde signaal precies tussen de twee andere signalen (2p).

e) Omdat er nu op iedere z-coordinaat water is, krijg je als het ware een superpositie van alle

signalen voor ieder denkbeeldig bolletje voor iedere waarde van z. De optelsom is een brede

lijn (1p). De breedte van het signaal komt overeen met de hoogteverschil tussen de onderkant

en de bovenkant (1p). (2p totaal)

f) Bij het bot zul je geen signaal zien omdat er geen water in zit, bij het spierweefsel is er wel

signaal, maar niet zo intens als bij de holte. Het spectrum is als het ware een afbeelding van

de waterdichtheid van het object langs de z-as en kan daarmee gebruikt worden om de bouw

van het weefsel aan te duiden.

2. (12 p)

a) +Mz –Mz –My. (3p)

b) De intensiteit van het gemeten signaal is even groot als maar tegengesteld in teken aan de

evenwichtsmagnetisatie. (1p)

c) Omdat Mz(0) gelijk is aan –Meq wordt de vergelijking Mz() = Meq*(1–2e–/T1

) (1p). Als er

geen signaal meer te zien is, is Mz() gelijk aan 0. Dus Meq(1–2e–/T1

) = 0 dus 2e–/T1

= 1; dus

e–/T1

= 0.5 dus –/T1 = ln(0.5) dus = –T1*ln(0.5) = T1*ln2 (1p). Oftewel: dan is precies

ln(2) zo lang als T1. (2p totaal)

d) Verschillen in T1 worden veroorzaakt door verschillen in bewegelijkheid van het molecuul

(1p). Dit wordt veroorzaakt door een verschil in massa van het molecuul (1p). (2p totaal)

e) Beide curves beginnen bij –1 als = 0 en lopen exponentieel op tot +1 als >>T1 (ca. 5*T1,

dus 20s voor het water en 5s voor het eiwit (1p)). De nuldoorgang is rond 2.8s (ln(2)*4) voor

water en 0.7s (ln(2)*1) voor het eiwit proton (1p). (2p totaal)

f) De beste water-onderdrukking krijg je bij de nuldoorgang voor het water, dus bij 2.8s (1p).

Dan is de relatieve intensiteit van het eiwit: Mz() = Meq*(1–2e–/T1

) = 1*(1–2e–2.8/1

) = 0.875.

Dus je hebt dan 12.5% minder intensiteit voor het eiwit-signaal (1p). (2p totaal)

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OEFENTENTAMEN – ANTWOORDEN 136

3. (12 p)

a) De 220nm flits creëert een fractie toestand B (onvolledige omzetting), dus tijdens t1 heb je

evolutie van het proton met A en B. De 450nm flits zet vervolgens alle B volledig om in C,

terwijl er niets verder met A gebeurt. Daarom heb je tijdens t2 dus evolutie met C en A.

Schematisch:

begin flits1 t1 flits2 t2 piek (F1, F2)

A A A A A A, A

B B C C B, C

Je krijgt twee pieken in het 2D-spectrum. (2p voor pieken, 2 voor juist tekenen) (4p totaal)

b) Omdat er nu ook een flits van 220nm na t1 is, wordt het niet omgezette A opnieuw

gedeeltelijk omgezet in B. Schematisch:

begin flits1 t1 flits2 t2 piek (F1, F2)

A A A A A A, A

B A, B

B B B

C C B, C

Nu krijg je dus vier pieken in het spectrum (2p voor pieken, 2 voor juist tekenen) (4p totaal)

c) Schematisch:

begin flits1 t1 flits2 t2 piek (F1, F2)

A A A A A A, A (rel. int. 12.3%)

100 35.0 12.3

B A, B (rel. int. 22.8%)

22.8

B B B

65.0 6.5 (rel. int. 6.5%)

C C B,C

58.5 (rel. int. 58.5%)

(stap 1 = 2p en stap 2 = 2p) (4p totaal)

4. (12p)

a) toekenning alle residuen op diagonaal (3p); toekenning alle kruis-pieken (3p) (6p totaal)

b) op juiste plaats en ook intensiteit in verhouding: dNN en dNN(i,i+2) aan beide zijden van de

diagonaal (2p); dN, dN(i,i+3) en dN(i,i+4) (2p) (4p totaal)

c) juiste residuen turn (2p) BONUS: +1 voor herkennen turn

Totaal 4*12= 48p. tel je score op, tel er 6 bij op en deel door 5.4 je cijfer.

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OEFENTENTAMEN – ANTWOORDEN 137

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OEFENTENTAMEN – ANTWOORDEN 138

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APPENDICES 139

Appendix A: Vector and matrix features and other mathematical functions

Vector properties

It is convention to print vectors in boldface to distinguish them from scalars.

Graphically speaking, an arrow can conveniently and effectively represent a vector: the length

of the arrow is proportional to its magnitude while the tip of the arrow shows its direction

(Figure A.1).

Figure A.1

A vector can also be specified by its components:

(A.1).

The number of the components basically determine the dimension of a vector: 2-dimensional

(2 components) vs. 3-dimensional (3 components). The components are related to the

coordinate system in use. Figure A.2 shows vectors described in Cartesian coordinates.

Figure A.2

One can describe vector algebra in both graphically (using arrows) and with components.

The magnitude of the vector is the same as the length of the arrow shown in Figure A.2:

(A.2).

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APPENDICES 140

A vector whose magnitude is unity (= 1) is called a unit vector (―eenheidsvector‖). The unit

vector of vector v shown in Figure A.2 can be obtained by dividing the vector by its own

magnitude.

(A.3).

The unit vector of the axes in a coordinate system (unit coordinate vectors) can be expressed

as

(A.4),

where i, j & k (or ex, ey, ez) are the unit vectors of the X, Y & Z axes, respectively.

Vector algebra

For the vectors a, b, and c:

(A.5).

If two vectors are equal to each other, all the corresponding components are equal to each

other:

(A.6).

Vectors of the same type can be added to yield the resultant vector. Adding two

vectors is equivalent to adding their components (Figure A.3):

Figure A.3

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APPENDICES 141

(A.7).

As shown in Figure A.3, vectors a and b were connected tip-to-tail, forming two sides of a

triangle. The third side of the triangle is the resultant vector (c) drawn from the tail of the first

vector to the tip of the second vector. The sequence of addition is not important

(commutative):

(A.8).

Addition of more than two vectors can be performed similarly, two at a time:

(A.9).

As shown in Eqn. A.9, vector addition is associative. One can move the vectors freely to

connect tip-to-tail. The sequence of addition is not important.

Vector subtraction is equivalent to subtraction of the components:

(A.10).

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APPENDICES 142

Figure A.4

Figure A.4 shows two equivalent graphical methods to find c. As shown in Eqn. A.10, vector

subtraction is a special case of vector addition.

When a vector is multiplied by a scalar, it magnifies (Figure A.5):

Figure A.5

(A.11)

where d = a scalar multiplier. The vector lengthens when d > 1 while it shortens if d < 1. If d

is negative, the direction of the vector reverses. The vector operation shown in Eqn. A.11 is

commutative:

(A.12).

Scalar multiplication is also distributive:

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APPENDICES 143

(A.13).

Finally, let's revisit Figure A.2. As shown in Figure A.6, a vector can be expressed as

the sum of the component vectors:

Figure A.6

(A.14),

where i, j & k are the unit coordinate vectors of the X, Y & Z axes, respectively.

Vector multiplication

There are two different kinds of vector multiplication: the scalar product

(―inproduct‖) and the vector product (―uitproduct‖). First, the scalar product is defined:

(A.15).

The result of the scalar product is a scalar. It tells us something about the influence of

one vector on another vector. The scalar product is also called the dot product following the

product symbol used. It is commutative:

(A.16)

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APPENDICES 144

and distributive:

(A.17).

From Eqn. A.15:

(A.18)

As shown in Eqn. A.18, the square of the magnitude of a vector is the same to the scalar

product of the vector with itself. Let's apply this to the triangle formed by the three vectors

shown in Figure A.7:

Figure A.7

(A.19).

From the Pythagorean Theorem:

(A.20).

From Eqns. A.19 and A.20:

(A.21),

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APPENDICES 145

where is the angle between the two vectors. So the scalar product can also tell us something

about the angle between the vectors. Eqn. A.21 is an alternative definition of the scalar

product, and from Eqn. A.21:

(A.22).

In words: the components of a vector are its projections to the axes.

Another important vector multiplication is the vector product. It is also called as the

cross product following the symbol used and is defined as:

(A.23).

It is understandable why it is called the vector product since the result of this multiplication is

another vector. From Eqn. A.23:

(A.24)

and

(A.25).

From Eqn. A.23:

(A.26).

In other words, the cross product of a vector with itself is always 0.

From Eqns. A.18, A.21 and A.23:

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APPENDICES 146

(A.27).

One very interesting thing here is the fact that the resulting vector from the cross product is

always perpendicular to the two vectors involved in the cross product:

(A.28).

The right-hand rule: the direction of the resulting vector is the direction a right-handed

screw advances when it is rotated from the direction of the first vector to the direction of the

second vector through the smallest angle between them.

From this right-hand rule the unit vectors are described as follows (see Figure A.6):

(A.29).

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APPENDICES 147

Matrix properties

A matrix is composed of several rows and columns of numbers:

(A.30).

As shown in Eqn. A.30, a m x n (dimension) matrix has m rows and n columns. The elements

of the matrix are denoted by aij where i = the row number, and j = the column number. 3 x 3

and 3 x 1 matrices are the most commonly used types. A matrix with m = n is called a square

matrix. In a diagonal matrix which is also a square matrix, all elements except the diagonal

ones (i = j) are zero.

Matrices of the same dimension can be added or subtracted, element by element:

(A.31).

When a matrix is multiplied by a scalar, all the elements of the matrix are multiplied by the

scalar value:

(A.32).

This operation is commutative as shown in Eqn. A.32].

Matrix multiplication

One of the most useful properties of the matrix is the matrix multiplication. Imagine a

system of linear equations:

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APPENDICES 148

(A.33),

where a's & c's = scalars, and x, y & z = the unknowns. Eqn. A.33 can be rewritten as:

(A.34).

Eqn. A.34 is expressed in matrix multiplication form. The general form of matrix

multiplication is

(A.35)

or

(A.36).

Note that the number of columns in the left vector (n in Eqn. A.35) must be the same to the

number of rows in the right matrix. The dimension of the resulting vector is m (rows of the

left matrix) x p (columns of the right matrix) as shown in Eqn. A.35.

Unlike the scalar multiplication, the matrix multiplication is not commutative:

[aij] [bij] ≠ [bij] [aij] (A.37).

However, it is distributive:

c(a+b) = ca + cb (A.38)

and matrix multiplication is also associative:

a(bc) = (ab)c (A.39).

For any scalar d:

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APPENDICES 149

(A.40).

Matrix determinant

The determinant of a 2 x 2 matrix is defined as:

(A.41).

The determinant of a 3 x 3 matrix can be reduced to a series of the determinants of 2 x 2

matrices:

(A.42).

Other useful mathematical rules

Differentiating:

d/dx ef(x)

= f‘(x) · ef(x)

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APPENDICES 150

Goniometry:

Using a circle the following relations can be explained:

The tangens of an angle is defined as the ratio between opposite and adjacent sides of a

rectangled triangle en is therefore:

Thus:

From Pythagoras‘ theorem it follows:

Derivatives: d/dx (sinx) = cosx

d/dx (cosx) = −sinx

Logarithms: ln(1/x) = ln (x-1

) = –ln(x)

Rotations:

Figure A.8

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APPENDICES 151

Appendix B: 1H and 13C chemical shifts of common functional groups

Page 57: NMR Spectroscopy In Structural Analysis · NMR Spectroscopy In Structural Analysis ... Chapter 2. Basic NMR Theory ... Chapter 4. Spin relaxation 14. Assume M z magnetization. Apply

APPENDICES 152

Appendix C: Random coil 1H chemical shifts for the common amino acids

3.28 2.96 4.69 8.31 C

3.88 4.50 8.38 S

2.64 2.13

CH 2 CH 3

2.15 2.01 4.52 8.42 M

2.03 3.68 3.65

CH 2 CH 2

2.28 2.02 4.44 P

7.15 6.86

7.30 7.39 7.34

1.23

1.48 0.95 0.89

1.64 0.94

0.97

H2,6 H3,5

3.13 2.92 4.60 8.18 Y

H2,6 H3,5 H4

3.22 2.99 4.66 8.23 F

CH 3 4.22 4.35 8.24 T

1.19 CH 2 CH 3 CH 3

1.90 4.23 8.19 I

0.90 H CH 3

1.65 4.38 8.42 L

0.94 CH 3 2.13 4.18 8.44 V

1.39 4.35 8.25 A

3.97 8.39 G

H H NH

3.28 2.96 4.69 8.31 C

3.88 4.50 8.38 S

2.64 2.13

CH 2 CH 3

2.15 2.01 4.52 8.42 M

2.03 3.68 3.65

CH 2 CH 2

2.28 2.02 4.44 P

7.15 6.86

7.30 7.39 7.34

1.23

1.48 0.95 0.89

1.64 0.94

0.97

H2,6 H3,5

3.13 2.92 4.60 8.18 Y

H2,6 H3,5 H4

3.22 2.99 4.66 8.23 F

CH 3 4.22 4.35 8.24 T

1.19 CH 2 CH 3 CH 3

1.90 4.23 8.19 I

0.90 H CH 3

1.65 4.38 8.42 L

0.94 CH 3 2.13 4.18 8.44 V

1.39 4.35 8.25 A

3.97 8.39 G

H H NH

8.12 7.14

H2 H4

3.26 3.20 4.63 8.41 H

1.70 3.32 7.17

1.45 1.70 3.02 7.52

2.38 6.87

7.59 6.91

2.31 2.28

7.24 7.65 7.17 7.24 7.50 10.2

6.62

CH 2 CH 2 NH

1.89 1.79 4.38 8.27 R

CH 2 CH 2 CH 2 NH 3 +

1.85 1.76 4.36 8.41 K

7.59 CH 2 NH 2

2.13 2.01 4.37 8.41 Q

NH 2 2.83 2.75 4.75 8.75 N

CH 2 2.09 1.97 4.29 8.37 E

2.84 2.75 4.76 8.41 D

H2 H4 H5 H6 H7 NH

3.32 2.99 4.70 8.09 W

H H NH

8.12 7.14

H2 H4

3.26 3.20 4.63 8.41 H

1.70 3.32 7.17

1.45 1.70 3.02 7.52

2.38 6.87

7.59 6.91

2.31 2.28

7.24 7.65 7.17 7.24 7.50 10.2

6.62

CH 2 CH 2 NH

1.89 1.79 4.38 8.27 R

CH 2 CH 2 CH 2 NH 3 +

1.85 1.76 4.36 8.41 K

7.59 CH 2 NH 2

2.13 2.01 4.37 8.41 Q

NH 2 2.83 2.75 4.75 8.75 N

CH 2 2.09 1.97 4.29 8.37 E

2.84 2.75 4.76 8.41 D

H2 H4 H5 H6 H7 NH

3.32 2.99 4.70 8.09 W

H H NH

For X in GGXA, pH 7, 35ºC (Bundi and Wüthrich 1979) .

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APPENDICES 153

Random Coil Chemical shifts (in ppm) for the 20 common amino acids in acidic 8 M urea (from Wright, Dyson

et. al., Journal of Biomolecular NMR, 18: 43–48, 2000).

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APPENDICES 154

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APPENDICES 155

Appendix D: Nuclear Overhauser Effect

The population change of the -state, d/dt(n), after a disturbance from equilibrium

can be expressed using the W0, W1a, W1b and W2 rates. The rate equation is:

)()(

)())((

11

2211

eq

AAB

eq

AAA

eqeq

BB

nnWnnW

nnWnnWWWndt

d

(D.1).

Thus, the -state looses magnetization (first term), but also gains some magnetization from

the -, the - and the -states.

Similar expression can be found for the time dependence of the other three states.

Now we look at the net population differences of spin A and B, nA and nB, resp. These are

)()(

)()(

nnnnn

nnnnn

B

A

(D.2a,b).

From this we can calculated the time derivative

ndt

dn

dt

dn

dt

dn

dt

dn

dt

d

ndt

dn

dt

dn

dt

dn

dt

dn

dt

d

B

A

(D.3a,b).

If we now introduce in Eqns. D.3 the results from Eqn. D.1 (and the expressions for the other

spin states) we can derive the dependency of the population from the transition rates:

))(())(2(

))(())(2(

02210

02210

eq

AA

eq

BBBB

eq

BB

eq

AAAA

nnWWnnWWWndt

d

nnWWnnWWWndt

d

(D.4a,b).

It can be derived that the first terms of Eqns. D.4a and D.4b describe the T1 relaxation of spins

A and B, respectively:

)2(1

)2(1

210

1

210

1

WWWT

WWWT

BB

B

AA

A

(D.5a,b).

The second term of Eqn. D.4 results in transfer of magnetization from A to B. We define the

cross-relaxation between A and B as

= W2 – W0 (D.6).

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APPENDICES 156

With this definition, from Eqn. D.4 we arrive at the Solomon-Bloembergen equations:

)()(

)()(

eq

AA

eq

BBBB

eq

BB

eq

AAAA

nnnnndt

d

nnnnndt

d

(D.7a,b).

Now let us return to the steady-state NOE experiment of Chapter 8. Spin B was

selectively saturated (nB = 0). After some time the two-spin system will reach a steady state

with

0Andt

d (D.8).

Eqn. D.7a then becomes

)()(0 eq

BB

eq

AAAA nnnnndt

d (D.9),

so that the NOE (Eqn. 8.1) becomes

AA

eq

A

eq

A

eq

A

eq

AA BB

n

n

n

nn

)( (D.10).

Here we exploited the fact that the macroscopic magnetization MA is proportional to the

population nA, and that the populations are themselves proportional to the gyro-magnetic ratio

(Eqns. 2.8 and 3.1).

For identical nuclei A = B, and A = , and Eqn. D.10 reduces to the simple form

(D.11).

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APPENDICES 157

Appendix E: 2D NOESY experiment

In mathematical terms the 2D NOE experiment can be described as follows. During the

evolution period the transversal magnetization of nucleus B can be written as

)sin()cos( 111 titMeMM BB

eqeqB B

B

B

ti

(E.1).

This results at time point c (Figure 10.3) in a z-component

)cos( 1, tMM Beq

zB B (E.2).

According to the Solomon equation (Eqn. D.7) we have for MA a dependency from MB

)()( eqBBAB

eqAAaA MMMMM

dt

d (E.3).

For short mixing times 11

Am T we can neglect spin-lattice relaxation and Eqn. E.3

becomes approximately

)( eqBBABA MMM

dt

d (E.4).

Using Eqn. E.2: )cos(1 1tMMdt

dB

eqBABA (E.5).

For short mixing times m we can approximate this by

)cos(1 1tMM

BeqBAB

m

A

(E.6),

and thus a fraction of the A-magnetization MA is modulated with B

)cos(1 1tMM BeqBmABA (E.7).

During the detection period this evolves with 2ti Ae

and after Fourier transformation

we will have a cross-peak at (B,A) in the 2D spectrum (F1, F2), correlating the protons A

and B in the 2D NOESY spectrum. The intensity of this cross-peak is

mABABI ~),( (E.8).

Since for biomolecules we have 6

~AB

cAB

r

(E.9)

we get for the cross-peak intensity from Eqn. E.8:

6

~),(AB

cmAB

rI

(E.10).

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APPENDICES 158

Appendix F: Amino acid COSY, TOCSY and NOESY cross-peaks.

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APPENDICES 159

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APPENDICES 160

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APPENDICES 161

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APPENDICES 162

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APPENDICES 163

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APPENDICES 164

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APPENDICES 165

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APPENDICES 166

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APPENDICES 167

Page 73: NMR Spectroscopy In Structural Analysis · NMR Spectroscopy In Structural Analysis ... Chapter 2. Basic NMR Theory ... Chapter 4. Spin relaxation 14. Assume M z magnetization. Apply

APPENDICES 168

Appendix G: Typical chemical shift values found in nucleic acid

Page 74: NMR Spectroscopy In Structural Analysis · NMR Spectroscopy In Structural Analysis ... Chapter 2. Basic NMR Theory ... Chapter 4. Spin relaxation 14. Assume M z magnetization. Apply

APPENDICES 169

Appendix H: Typical short proton–proton distances for B-DNA

All distances are given in Å. Sequential distances (to its 3' neighbor) below the diagonal,

intraresidual distances above the diagonal.

H6/8 3.8 2.1 3.6 4.1 4.9 3.4 4.4 4.8

3.5 1' 3.0 2.3 3.9 3.6 4.5 3.8

3.8 4.1 2' 1.8 2.4 3.9 3.8 4.0 2.1

2.3 2'' 2.7 4.0 4.9 3.6

4.9 3' 2.7 3.7 2.9 4.1

4.2 4' 2.7 2.3 4.9

1.8 4.3 3.2 4.5 3.6 5' 1.8 3.4

3.3 4.0 4.7 4.1 5'' 4.4

4.8 3.5 3.8 2.3 H6/8

fast

tumbling 7 ppm =

+1150 Hz