Metodi di indagine di molecole biologiche 2008 Piccioli-Turano Spettroscopia NMR di molecole Biologiche. Aspetti fondamentali della tecnica (Recupero nozioni.

Presentazione sul tema: "Metodi di indagine di molecole biologiche 2008 Piccioli-Turano Spettroscopia NMR di molecole Biologiche. Aspetti fondamentali della tecnica (Recupero nozioni."— Transcript della presentazione:

1 Metodi di indagine di molecole biologiche 2008 Piccioli-Turano Spettroscopia NMR di molecole Biologiche. Aspetti fondamentali della tecnica (Recupero nozioni di base). Esperimenti multidimensionali. NMR eteronucleare. Risonanza Tripla. Assegnamento. Indice del Chemical Shift. Parametri NMR per la determinazione strutturale di molecole biologiche (nozioni di base). Dinamica Interna. Accessibilità al solvente. Idratazione. Interazioni. Sistemi ad alto peso molecolare.

2 Metodi di indagine di molecole biologiche 2008 Piccioli-Turano Laboratorio Lab 1. Lo spettrometro NMR: Set up di un esperimento. Rudimenti. Hardware e configurazioni Lab2. Acquisizione di esperimenti multidimensionali omonucleari ed eteronucleari (tutoriale). Lab3. SOD. Acquisizione spettro HSQC, NOESY ed HSQC-NOESY Lab 4. Processing ed analisi spettri acquisiti (data station) Lab 5. Tripla risonanza. Acquisizione spettro 3D, HNCA, HNCO. Lab 6 Analisi degli spettri. Lab 7. R1 ed R2. Acquisizione di spettri Lab 8 Fitting ed interpretazione Lab 9 Metabolomica Lab 10. TROSY e CRINPET

3 Basics Concepts Nuclear spins Magnetic field B 0 Energy Sensitivity The NMR transition Larmor Frequency Magnetic field B 1 The rotating frame A pulse!

4 Nuclei NMR-attivi Tutti i nuclidi con un numero quantico di spin nucleare I diverso da zero sono NMR attivi, Ovvero possono essere studiati via NMR I=1/2 1 1 H, 13 6 C, 31 15 P, 19 9 F, 15 7 N I=1 2 1 H, 14 7 N I=3/2 23 11 Na 35 17 Cl I=5/2 17 8 O 27 13 Al

5 Magnetic moment I= spin quantum numberm= Magnetic quantum number (-I…+I)  I 2  = (h/2  ) 2 [I(I+1)]Spin magnetic moment  =  I I z =(h/2  )m Spin angular momentum  z =  I z =  (h/2  )m z (-  (h/2  )I….  (h/2  )I) In the presence of a magnetic field B 0

6 Nuclear Magnetic Resonance Many atomic nuclei possess a small magnetic dipole (nuclear magnetic moment) , in the same direction as the angular momentum vector I. In an external magnetic field B 0, the nuclear magnetic moment  is limited by quantum mechanics to a distinct set of orientations or spin states related to the spin quantum number I. This interaction of nuclear magnetic moment with B 0 is called the Zeeman interaction. For spin ½ nuclei, the two possible orientations are  I  m   m   z =  I z =  (h/2  )m z

7 The energy of the NMR Transition Sensitivity B0B0 E m=+1/2 m=-1/2  E=h 0  NUCLEUS B 0 MAGNETIC FIELD Larmor Frequency The two Zeeman level are degenerate at B 0 =0  E=  (h/2  )B 0 E= - B 0

8 Different Isotopes Absorb at Different Frequencies low frequencyhigh frequency 15 N 2H2H 13 C 19 F 1H1H 50 MHz 77 MHz 125 MHz 200 MHz 470 MHz 500 MHz 31 P

9 Nuclear Magnetic Resonance The two orientations for spin ½ nuclei such as 1 H or 13 C have different energy level and therefore slightly different populations at equilibrium. The slightly excess of spins in the +1/2 state is called the Boltzman distribution and is proportional to the square of the magnetic moment m, or (  B 0 2 ) Transitions between states occur with the absorption or emission of a quantum of energy from an electromagnetgic field at precisely the energy difference between the states h (the Larmor frequency ).

10 The sensitivity of NMR transition: Boltzmann distribution N up /N low =e -  E/kT  E=  hB 0 /2  B 0 =9.4 T; T=300 K, 1 H  = 2.675x10 8 rad T -1 s -1  E=2.65x10 -25 J kT= 4.14x10 -21 J. (N low -N up )/(N low +N up ) =  E/2kT  E/kT=10 -4 1:31000  P/P

11 d  /dt=  (t)^B 0 (t)

12  0 =2    0 =    0  0  Mz=-1/2 h  /2  E=-  B 0 = -  hm z B 0 /2   E=  hB 0 /2  STILL, NO NMR EXPERIMENT Mz=1/2 h  /2 

13 B0B0 M0M0 I=1/2 E =  B 0 E =   mB 0  :m = +½    +½   E   ½   B 0  :m =  ½    ½   E   ½   B 0 M 0 =  = M z M x = M y = 0  E =   B 0 =     h 

14 M 0 =  = M z

15 B 1 (t)=  B 1  cos  1 t B 1 is a radiofrequency transmitter A pulse!

16 dM(t)/dt= M(t)^B(t) B(t)= B 0 + B 1 (t) dM(t)/dt= M(t)^B 0 + M(t)^B 1 (t) = M(t)^B 0 + M(t)^|B 1 | cos  1 t (t) Double precession A pulse! Precession around B 0 (z axis) Precession around B 1 (axis defoned in the xyplane and rotating at speed  1)

17 Laboratory Frame Nuclear frequency 1= precession frequency of magnetic field B 1 A pulse!

18 The Rotating frame X’,y’,z’ =laboratory frame X,y,z,=rotating frame (rotating at the frequency 1) In the rotating frame,there is no frequency precession for  and the radifrequency B1 is seen as a static magnetic field The static magnetic field B 0 is not observed in the rotating frame

19 Laboratory Frame jam Fly A (Laboratory Frame) Fly B The movement of Fly B as seen by Fly A Rotating Frame Fly B jam The movement of Fly B as seen by Fly A Fly A (Rotating frame)

20 Precession in the laboratory frame  0 dM/dt=M^  BdM/dt=M^  (B-    ) L.F. R.F. at freq.   If  =  0 dM/dt=0 If    0 dM/dt=M^  B 1 =    If  =  0 +B 1 dM/dt=M^  (B 0 +B 1 -  0  ) Rotation! dM/dt=M^  (B +B 1 -  0  ) dM/dt=M^  (B 1 + (  0  )) dM/dt=M^  (B 1 + (  )) B1B1  0

21 Precession in the laboratory frame  0 M xy from any nuclear spin not exactly on resonance, will also precess in the x’y’ plane at the difference frequency  0.