1 / 49

Jane Clarke May 2004

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. Cambridge University Dept of Chemistry MRC Centre for Protein Engineering. Robert Best Susan Fowler

hayes-travis
Télécharger la présentation

Jane Clarke May 2004

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

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

  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 proteinsFor some proteins resisting unfolding may be important

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

  6. TS ∆∆GTS-N ∆GTS-N D xf ∆∆GD-N xu N When you add FORCE (F): Relative to the native state, N, the barrier to unfolding (∆GTS-N) is lowered by: Fxu and the free energy of unfolding (∆GD-N) is lowered by: F(xu + xf) The protein is less stable and unfolds more rapidly - the unfolding rate (ku) 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. 2 3 4 1 ∆L F 1. Non-specific adhesion 2. Unfolding of one domain 3. Unfolded protein stretching 4. Protein detaches

  10. Analysis of AFM data • Unfolding proteins by AFM is a kinetic measurement: mean unfolding force depends on pulling speed. • Unfolding rate constant (extrapolated to 0 force) (ku0) and unfolding distance (xu) can be estimated by Monte Carlo simulation or analytical techniques. Slope gives xu Force (N) Intercept gives ku0 Gaub, Fernandez, Evans

  11. Interpreting the traces:Which traces to choose?

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

  15. It is possible to measure refolding rates using AFM Carrion Vasquez 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 Protein domains straighten out Very low force Unstructured region unfolds “working” forces One or two domains unfold to prevent the protein breaking Very high force

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

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

  22. Humps?

  23. 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 V4A

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

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

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

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

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

  29. Protein engineering analysis - F = 0 • The unfolding force reflects the difference in free energy between I and ‡ ‡ • 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 N NB only works if the barrier we are examining is the same in WT & mutant (xu must remain the same) I

  30. The A’ strand is partly detached in ‡

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

  32. BUT xu changes - can’t do a F-value analysis but this mutation clearly lowers the force, ie F must be <1

  33. 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?

  34. MD simulations - the protein unfolds via an intermediate

  35. intermediate transition state V4 V13 Proportion of native contacts Analysing structures from the simulations - experimental F-values are reproduced

  36. N I ‡ D Mechanical unfolding pathway: Native State N G- strand C 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 Step 1: A-strand pulled off to form I A’ A N C Transition State

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

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

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

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

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

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

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

  44. In the mutant V86A ku is rate constant for unfolding from N to ‡F and xu 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 In wildtype ku is rate constant for unfolding from I to ‡F and xu is the distance between I & ‡F ‡F ‡F ‡F I N I I N N 3 Å 0 pN ≈100 pN >100 pN Increasing force

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

  46. How do titin domains resist force?

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

More Related