1. Jane Clarke May 2004 How do proteins withstand force? Examining the effect of force on a protein folding landscape by combining atomic force microscopy, protein engineering and simulation

2. Robert Best Susan Fowler Annette Steward Kathryn Scott José Toca Herrera Cambridge University Dept of Chemistry MRC Centre for Protein Engineering Cambridge University Dept of Chemistry MRC Centre for Protein Engineering Wellcome Trust & MRC Emanuele Paci (Zurich) Martin Karplus (Strasbourg/ Harvard) Phil Williams (U. Nottingham)

3. Gene The Protein Folding Problem Misfolded protein sequence A sequence B Folded protein Unfolded protein Proteins fold co-operatively into a unique 3-dimensional structure that is the most stable conformation function Why? How? Why not? How?

4. But proteins don’t just fold one time and that’s it. Mechanical unfolding of proteins may be important in translocation and degradation and in mechanically active proteins For some proteins resisting unfolding may be important

5. Protein folding pathways - and landscapes How does force modify the unfolding landscape? Karplus, Dobson D I N ‡2‡2 ‡1‡1 We can explore the unfolding landscape by folding and unfolding experiments

6. N TS D ∆G TS-N xuxu xfxf ∆∆G TS-N ∆∆G D-N When you add FORCE (F): Relative to the native state, N, the barrier to unfolding (∆G TS- N ) is lowered by: Fx u and the free energy of unfolding (∆G D-N ) is lowered by: F(x u + x f ) The protein is less stable and unfolds more rapidly - the unfolding rate (k u ) is a measure of the height of the barrier between N and TS

7. What does AFM offer? Can investigate the way the energy landscape is perturbed by force Known reaction co-ordinate (N-C length) making it easier to do direct comparison with simulation Single molecule experiments offer the possibility to observe rare events

8. The AFM Experiment Asylum Research MFP

9. 1 2 3 4 ∆L F 1. Non-specific adhesion 2. Unfolding of one domain 3. Unfolded protein stretching 4. Protein detaches

10. Unfolding proteins by AFM is a kinetic measurement: mean unfolding force depends on pulling speed. Unfolding rate constant (extrapolated to 0 force) (k u 0 ) and unfolding distance (x u ) can be estimated by Monte Carlo simulation or analytical techniques. Analysis of AFM data Force (N) Gaub, Fernandez, Evans Slope gives x u Intercept gives k u 0

11. Interpreting the traces: Which traces to choose?

12. The basic reminders about kinetics and thermodynamics The basic reminders about kinetics and thermodynamics Force measurements of protein unfolding are kinetic measurements not thermodynamic measurements So… Beware of the word “stability” - what does it mean in the context of force measurements? “In a protein made up of a number of domains the least stable domains will unfold first and the most stable domains will unfold last”

13. Titin I27 is significantly more stable than I28 (7.6 vs. 3.2 kcal/mol) but I28 unfolds at significantly higher forces

14. But… It is not possible to determine the stability of a protein using AFM folding and unfolding data The basic reminders about kinetics and thermodynamics (2) The basic reminders about kinetics and thermodynamics (2)

15. Carrion Vasquez It is possible to measure refolding rates using AFM BUT - the unfolding and refolding pathway are not necessarily (are unlikely to be?) the reverse of each other.

16. Titin - an elastic protein 1 µm

17. Titin - effect of force Very low force Protein domains straighten out “working” forces Unstructured region unfolds Very high force One or two domains unfold to prevent the protein breaking

18. First experiments - using whole proteins with heterogeneous domains Gaub, Bustamante, Symmonds, Fernandez

19. How do titin domains resist force? Can we characterise the titin I27 forced unfolding pathway in detail?

20. To make multiple repeats of one titin domain A tag to allow easy purification Using molecular biology A tag to allow attachment to AFM

21. In simulations the first step is to form an intermediate by detachment of the A-strand Fernandez, Schulten

22. Humps?

23. V4A When we pull a protein with a destabilising mutation in the A-strand (V4A) it does not affect the unfolding forces at all Fowler et al. JMB 2002 322, 841

24. This intermediate is stable and has essentially the same structure as the native state 1H1H 15 N

25. Increasing force 0 pN>100 pN≈100 pN N N N I I I ‡F‡F ‡F‡F ‡F‡F 3 Å k u ≈10 -4 ∆G ≈ 3 kcal mol -1 k u is the unfolding rate of I to ‡ and x u is the distance between I & ‡ I is populated above 100pN

26. Titin forced unfolding pathway Native state N Intermediate I Transition State ? Unfolded D N I ‡ Free energy profile under force ~3 Å kuku

27. Using protein engineering to analyse forced unfolding pathways: A mechanical  - value analysis Using protein engineering to analyse forced unfolding pathways: A mechanical  - value analysis Best et al. PNAS 2002 99, 12143 V4 V13 V86 I23 F73 L41 L60 L58 N C A A´ G F B E D C G

28. Theory Protein engineering analysis -  = 1 Theory Protein engineering analysis -  = 1 N I ‡ The unfolding force reflects the difference in free energy between I and ‡ U If the mutation removes a side chain that is fully folded in the transition state it will not affect the unfolding force at all.

29. Protein engineering analysis -  = 0 N I ‡ If the mutation removes a side chain that is fully unfolded in the transition state it will reduce the unfolding force by a significant amount - that we can predict U The unfolding force reflects the difference in free energy between I and ‡ NB only works if the barrier we are examining is the same in WT & mutant (x u must remain the same) NB only works if the barrier we are examining is the same in WT & mutant (x u must remain the same)

30. The A’ strand is partly detached in ‡

31. Most  -values are ≈ 1 Most of the protein is intact in the transition state Most  -values are ≈ 1 Most of the protein is intact in the transition state

33. BUT x u changes - can’t do a  -value analysis but this mutation clearly lowers the force, ie  must be <1 BUT x u changes - can’t do a  -value analysis but this mutation clearly lowers the force, ie  must be <1

34. Results: The only part of the protein completely detached in ‡ is A strand A’ and G are partly disrupted. How? Can molecular dynamics simulations help? Results: The only part of the protein completely detached in ‡ is A strand A’ and G are partly disrupted. How? Can molecular dynamics simulations help?

35. MD simulations - the protein unfolds via an intermediate

37. Analysing structures from the simulations - experimental  -values are reproduced Proportion of native contacts intermediatetransition state V13 V4

38. Native State Transition State Step 1: A-strand pulled off to form I Step 2: G-strand pulled off, breaking main chain & sidechain contacts with A’ & sidechain contacts with A- B loop. A’ loses contacts with G & E-F loop C N N C G- strand A’ A Mechanical unfolding pathway: NI‡DNI‡D

39. Transition State ‡ Native state N Intermediate I Unfolded D A’ G G A N C G G G N N C C Titin forced unfolding pathway N I ‡ Free energy profile under force

40. Increasing force 0 pN>100 pN≈100 pN N N N I I I ‡F‡F 3 Å ‡F‡F ‡F‡F k u ≈10 -4 ∆G ≈ 3 kcal mol -1 Force induced unfolding pathway

41. Increasing force 0 pN>100 pN≈100 pN N N N I I I ‡F‡F N ‡D‡D ‡F‡F ‡F‡F k u ≈10 -4 Denaturant (0 pN) Does force change the energy landscape? Do these transition states have the same structure? 3 Å

42. Force The A’ G region remains partly folded in TS The A strand unfolds very early The core plays no role in withstanding force and remains fully folded in TS The core is partly unfolded in TS The A’ G region is fully unfolded in TS Chemical denaturant The A strand remains partly folded in TS

43. Increasing force 0 pN>100 pN≈100 pN N N N I I I ‡F‡F N ‡D‡D ‡F‡F ‡F‡F k u ≈10 -4 Denaturant (0 pN) Force changes the energy landscape Transition states have different structures 3 Å ? ? when? k u ≈10 -4

44. Cannot measure at rates below ~100 nm/s Experimental limitations

45. These mutants have a significantly longer x u (~6Å) than wt (~3Å) All these mutants destabilise the protein significantly BUT These very destabilising mutants have apparently a lower k u than wild type Simplest model: these mutants are unfolding directly from N So x u = x N ->‡ & k u = k N-‡ Simplest model: these mutants are unfolding directly from N So x u = x N ->‡ & k u = k N-‡ Williams et al: Nature 2003

46. Increasing force 0 pN>100 pN≈100 pN N N N I I I ‡F‡F ‡F‡F ‡F‡F 3 Å In wildtype k u is rate constant for unfolding from I to ‡ F and x u is the distance between I & ‡ F In the mutant V86A k u is rate constant for unfolding from N to ‡ F and x u is the distance between N & ‡ F This will happen if the mutation allows the protein to unfold before I is populated (by destabilising I &/or lowering the unfolding barrier ‡ F ) 6 Å

47. Increasing force 0 pN>100 pN≈100 pN N N N I I I ‡F‡F N ‡D‡D ‡F‡F ‡F‡F k u ≈10 -4 Denaturant (0 pN) Force changes the energy landscape Transition states have different structures 6 Å k u ≈ 10 -7 3 Å When? k u ≈10 -4 At v. low forces the “physiological” barrier may be the important one

48. How do titin domains resist force?

49. Why are some proteins more mechanically stable than others? Titin I27 (muscle) Spectrin (cytoskeleton) T4 Lysozyme Barnase Tenascin fnIII (intracellular matrix) (enzymes)