NMR in Biophysical Chemistry. Preview: Ligand Binding and Allosterics

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1 NMR in Biophysical Chemistry Erik Zuiderweg and David Case Preview: Ligand Binding and Allosterics 1

2 Simple Ligand Binding: Two-site exchange 2

3 3

4 f A =f B =0.5 k ex =10-1 s -1 4

5 f A =f B =0.5 k ex =10 0 s -1 5

6 f A =f B =0.5 k ex =10 1 s -1 6

7 f A =f B =0.5 k ex =10 2 s -1 7

8 f A =f B =0.5 k ex =10 3 s -1 8

9 f A =f B =0.5 k ex =10 4 s -1 9

10 f A =f B =0.5 k ex =10 5 s -1 10

11 f A =f B =0.5 k ex =10 6 s -1 11

12 Fast exchange titration 12

13 k ex =10 6 s -1 f A =1.00 f B =0.00! = f FREE! BOUND + f BOUND! BOUND 13

14 k ex =10 6 s -1 f A =0.67 f B =0.33! = f FREE! BOUND + f BOUND! BOUND 14

15 k ex =10 6 s -1 f A =0.50 f B =0.50! = f FREE! BOUND + f BOUND! BOUND 15

16 k ex =10 6 s -1 f A =0.33 f B =0.67! = f FREE! BOUND + f BOUND! BOUND 16

17 k ex =10 6 s -1 f A =0.00 f B =1.00! = f FREE! BOUND + f BOUND! BOUND 17

18 SIMPLE USE: Chemical Shift Mapping 14N 14 N Amide 15N 15 N 15N Amide H Complex: 15 N- 2 H Flavodoxin 14 N- 1 H Methionine 18 Synthase

19 Quantitating Fast Exchange Binding Constants ( ) ± ([L] TOT + [P] TOT + K D ) 2! 4[L] TOT [P] TOT [LP] = [L] TOT + [P] TOT + K D 2 19

20 Slow exchange Titration 20

21 k ex =0.1 s -1 f A =1.00 f B =0.00 I A = f FREE I FREE I B = f BOUND I BOUND 21

22 k ex =0.1 s -1 f A =0.67 f B =0.33 I A = f FREE I FREE I B = f BOUND I BOUND 22

23 k ex =0.1 s -1 f A =0.50 f B =0.50 I A = f FREE I FREE I B = f BOUND I BOUND 23

24 k ex =0.1 s -1 f A =0.33 f B =0.67 I A = f FREE I FREE I B = f BOUND I BOUND 24

25 k ex =0.1 s -1 f A =0.00 f B =1.00 I A = f FREE I FREE I B = f BOUND I BOUND 25

26 Quantitating Slow Exchange Binding Constants Saturation = I BOUND I BOUND + I FREE 26

27 Range of Quantitation 27

28 Range of Quantitation 28

29 Range of Quantitation 29

30 Intermediate exchange titration 30

31 kex=10 3 f A =1.0 f B =0.0 LW = f A LW A + f B LW B + f A f B! A "! A ( ) 2 k ex 31

32 kex=10 3 f A =0.9 f B =0.1 LW = f A LW A + f B LW B + f A f B! A "! A ( ) 2 k ex 32

33 kex=10 3 f A =0.8 f B =0.2 LW = f A LW A + f B LW B + f A f B! A "! A ( ) 2 k ex 33

34 kex=10 3 f A =0.5 f B =0.5 LW = f A LW A + f B LW B + f A f B! A "! A ( ) 2 k ex 34

35 kex=10 3 f A =0.2 f B =0.8 LW = f A LW A + f B LW B + f A f B! A "! A ( ) 2 k ex 35

36 kex=10 3 f A =0.1 f B =0.9 LW = f A LW A + f B LW B + f A f B! A "! A ( ) 2 k ex 36

37 kex=10 3 f A =0.0 f B =1.0 LW = f A LW A + f B LW B + f A f B! A "! A ( ) 2 k ex 37

38 Quantitating Kinetics Slow Exchange : Upper limit on k ex (typical: k ex << 10 s -1 ) Fast Exchange : Lower limit on k ex (typical: k ex >> 10 4 s -1 ) Intermediate Exchange : quantify k ex (typical: 10 2 s -1 < k ex < 10 4 s -1 ) 38

39 Simulating two-site exchange dm A x (t) dt dm A y (t) dt dm B x (t) dt dm B y (t) dt =!" A M A y (t)! ( R A 2 + k ) AB M A x (t) + k BA M B x (t) = +" A M A x (t)! ( R A 2 + k ) AB M A y (t) + k BA M B y (t) =!" B M B y (t)! ( R B 2 + k ) BA M B x (t) + k AB M A x (t) = +" B M B x (t)! ( R B 2 + k ) BA M B y (t) + k AB M A y (t) 39

40 Simulating two-site exchange Steady-state solution # i! A " R A 2 " k AB k BA & $ % k AB i! B " R B 2 " k BA ' ( # $ % A M + B M + & ' ( = # p M A tot & $ % p B M tot ' ( 40

41 Simulating two-site exchange Steady-state solution ( ) = Im M tot p A! "!! B S! #% $ &% ( ) + p B (! "! A ) + i( k AB + k BA ) (! "! A ) + ik AB ( ) + ik BA ( )(! "! B ) + k AB k BA '% ( )% 41

42 More Possibilities in 2D 42

43 Ligand Binding: Three-site exchange 43

44 Curved traces in sequential 3-site exchange 2L + P! " L + LP! "L 2 P P PL PL2 44

45 Sometimes, a two-site exchange must be three-site exchange 31 P NMR K D =10-9 M Δω=200 r/s IHP Free IHP Bound 45

46 KD =10-9 M IHP kon_max =108 M-1 s-1 IHP koff_max =0.1 s-1 Hemoglobin Hemoglobin!" =100 rs-1 46

47 Sometimes, a two-site exchange must be three-site exchange 31 P NMR K D =10-9 M Δω=200 r/s IHP Free IHP Bound 47

48 Solution: one ligand, two consecutive fast processes KDa =10-4 M k =108 M-1 s-1 KD =10-9 M KDb =10-5 k =10 9 s-1 k =104 s-1 k =104 s-1 entry site 48

49 Simulating three-site exchange d dt! M A + (t) $ # M B + (t) & = " # M C + (t)% &! i' A ( R A 2 ( k AB ( k AC k BA k CA $! M A + (t) $ # k AB i' B ( R B 2 ( k BA ( k BC k & # CB M B + (t) & " # k AC k BC i' C ( R C 2 ( k CA ( k CB % & " # M C + (t)% & 49

50 Solution 50

51 Ligand Binding: four-site exchange 51

52 Ligand-binding-induced conformational change = 4-site exchange k =105 M-1 s-1 k =103 s-1 k =10-2 s-1 k =10-1 s-1 k =10-1 s-1 k =10-3 s-1 k =108 M-1 s-1 k =103 s-1 52

53 k =105 M-1 s-1 k =103 s-1 k =10-2 s-1 k =10-1 s-1 k =10-1 s-1 k =10-3 s-1 k =108 M-1 s-1 k =103 s-1 53

54 k =105 M-1 s-1 k =103 s-1 k =10-2 s-1 k =10-1 s-1 k =10-1 s-1 k =10-3 s-1 k =108 M-1 s-1 k =103 s-1 54

55 k =105 M-1 s-1 k =103 s-1 k =10-2 s-1 k =10-1 s-1 k =10-1 s-1 k =10-3 s-1 k =108 M-1 s-1 k =103 s-1 55

56 Allosterics 56

57 Allosterics is wide-spread Site b knows whether site a is occupied Intra-molecular signal transduction 57

58 e.g. signal transduction 58

59 NMR allows delineation of the allosteric mechanism of a cytidylyltransferase Stevens, S.Y., Sanker, S., Kent, C. and Zuiderweg, E.R.P., Nature Structural Biology 8, (2001)

60 Crystal structure of CTP:glycerol-3- phosphate cytidylyltransferase (GCTase)+ 2CTP Courtesy of Martha Ludwig Chris Weber 60

61 An Allosteric Enzyme CTP CTP 300 µm (300 µm) CTP CTP CTP G3P 1 µm CTP (1500 µm) G3P CTP CTP G3P G3P (1 µm) G3P CTP G3P (100 µm) (100 µm) G3P G3P G3P G3P 61

62 CTP CTP 300 µm (300 µm) CTP CTP CTP G3P 1 µm CTP (1500 µm) G3P CTP CTP G3P G3P (1 µm) G3P CTP G3P (100 µm) (100 µm) G3P G3P G3P G3P 62

63 CTP CHEMICAL SHIFT MAPPING First CTP K D < 5 µm Second CTP K D = 500 µm 63

64 CTP CTP CTP 64

65 DEFINING THE KNF-ALLOSTERIC MODEL or +S or or +S 65

66 Where is the negative cooperativity coming from? First CTP ΔG = -7.3 Kcal/M Second CTP ΔG = -4.6 Kcal/M 66

67 Binding sites identical, Expect identical local binding free energies Thus: 2.7 Kcal/M of binding free energy lost on interface First CTP ΔG = -7.3 Kcal/M Second CTP ΔG = -4.6 Kcal/M 67

68 15 N R 2 relaxation with and without exchange broadening suppression GCT GCT(CTP) GCT(CTP) 2 68

69 GCT GCT(CTP) GCT(CTP) 2 Dynamic Dynamic Rigid ΔS = 0 ΔS = neg Therefore, the allosteric free energy of negative cooperativity has an entropic component. 69

70 Other example: Allosteric energy is (partially) dynamic in origin

71 Mäler, L., Blankenship, J., Rance, M. & Chazin, W. J. Nature Struct. Biol. 7, (2000).

72 1 L 2 72

73 Calb Calb(Ca 2+ ) Calb(Ca 2+ ) 2 Dynamic Rigid Rigid ΔS = neg ΔS = 0 The allosteric free energy of positive cooperativity has an entropic component. 73

74 Cooper A, Dryden DT: Allostery without conformational change. A plausible model. Eur Biophys J 1984, 11:

75 Thank you for your attention I hope this was useful 75