The Huxley equation: How to Pull with a Burning Rope and Other Stories about Molecular Motors

Transcription

1 The Huxley equation: How to Pull with a Burning Rope and Other Stories about Molecular Motors J. P. Keener Mathematics Department University of Utah Korea PDE School 217

2 Molecular Motors There are four basic types of molecular motors Cytoskeletal motors such as myosin, kinesin, and dynein Polymerization/depolymerization motors (used for chromosomal segregation) Rotary flagellar motors Nucleic acid motors such as RNA, DNA polymerase

3 Outline of this Talk Main Goal: To show how the Huxley model is used to gain an understanding of several different types of molecular motors Issues to discuss: Load-velocity curves; Huxley model of actin-myosin crossbridges Binding and unbinding of flexible proteins Bidirectional transport Kinetochore motion for Chromosomal segregation

4 Warmup Exercise: Skeletal Muscle

5 Myosin Crossbridges

6 The Power Stroke Kinesin Animation

7 The Huxley Model Suppose a crossbridge can be bound (B) or unbound (U) U kon(x) B, k off (x) Thin filament Thick filament x Unstressed position Let n(x,t) be the density of bound crossbridges with extension x, n t + x [v(t)n] ( = N ) ndx k on(x) k off (x)n v > is contraction Binding site Crossbridge v(t) = filament velocity binding rate: k on, (N n) = free crossbridges unbinding rate k off The force generated is F = kx n(x,t)dx. The Huxley Model is a nonlocal, advection-reaction equation with structure k off k on h koff x

8 Load-Velocity Curve To find the load-velocity curve Pick a velocity v, V = 1..8 n Calculate steady state distribution n(x; v) ñ1.5 ñ1. ñ x/h 1..8 V =.1 V max n ñ1.5 ñ1. ñ x/h 1..8 V =.25 V max n Calculate the force F, ñ1.5 ñ1. ñ x/h 1..8 V = V max n Plot v vs. F. Remark: Although not biophysically accurate, this is a typical load-velocity curve for molecular motors. ñ1.5 ñ1. ñ x/h Speed of shortening Tension

9 Load Dependent Unbinding F Bonds break in a load dependent fashion U k on k off B, k off = βexp( F F ) (Bell s Law). k k G G δx Fx

10 Consequences of Bell s Law Suppose the load is spread uniformly among all bonds S n α n β n S n+1, α n = (N n)k on, β n = nβexp( F nf ) leads to the deterministic equation dθ dt = (1 θ) κθexp(f θ ) with θ = n, κ = β N k on, f = F NF exhibits critical behavior. θ steady state solutions κ =.5 κ =.1.2 Remark: This is an interesting stochastic exit time problem f

11 Position Dependent Binding Unbound flexible proteins undergo an Ornstein-Uhlenbeck process (diffusion with linear restoring force) γdx = kxdt +dξ(t) where ξ(t) represents a white noise stochastic process, and p(x, t) satisfies the Fokker-Planck equation: p t = x (kxp)+ k BT 2 p γ x 2 x with spring constant k, have Gaussian equilibrium distribution, so that the binding rate is also Gaussian k on = κexp( kγx2 2k B T ).

12 Sliding in a Flow Vf V x Let n(x,t) be the density of bound binders with extension x, n t = V n x +α(x) ( N T ) n(x, t)dx β(x)n, α(x) = κexp( kγx2 2k B T ), The force generated is β(x) = k off exp( k x F ), F = k xn(x,t)dx.

13 x And the Answer is n(x,v) Contours Force f V Velocity v with the motion determined by the force balance equation η(v f V) F =. 2 biphasic Force vs. Velocity curve, Velocity vs. V f has hysteretic behavior (explaining the static-kinetic friction transition). Velocity v Velocity v f

14 Bidirectional Transport Goal: To understand the observed switching of transport direction.

15 Motor Density cargo v w - (x) dynein diffusion kinesin w + (x) microtubule x x Let n ± (x,t) = density of + or motors stretched distance x at time t.

16 How does density n ± (x,t) change? (Answer: Use the Huxley model.) n ± t + [{ w ± (x) v(t) } n ±] x ( ) = N ± n ± dx k on(x) ± k ± off (x)n± ( ) Motor stepping rate is load dependent, w(x) = v 1 kx F stall v(t) = cargo velocity binding: unstretched, k on δ(x), (N n) = free motors unbinding: Bell s Law, force unbinding, k off e α x

17 Force Balance cargo velocity: force balance m v +γv = 2k B Tξ(t)+ F + (t)+f (t) forces from motors F ± (t) = kxn ± (x,t;v(t))dx Remark: F ± respond in a time-dependent way to variations of v(t).

18 Ornstein-Uhlenbeck Observation Single run (before unbinding), single motor: t p = x [{w(x) v(t)}p]+d xx p Mean evolution: µ := x(t) µ = w(µ) v(t) Mean force: F = kxp(x,t)dx = kµ Lesson learned: motors track (and generate force) with delayed response to fluctuations in the cargo v(t).

19 Cargo Velocity Evolution OU process suggests delayed force approximation of the form m v +γv = F 1(µ 1)+F 2(µ 2)+ 2k B Tγξ(t), µ 1 = w 1(µ 1) v µ 2 = w 2(µ 2) v where F is the steady state force from the PDE with v = w(µ), i.e., [{ wj (x) w j (µ j ) } ( ) ] n j = N j n j dx kon j x (x) kj off (x)n j, j= 1,2

20 Final Form of Stochastic Differential Equations Take viscous limit (v F/γ) µ 1 = w 1(µ 1) 1 γ [F1(µ1)+F2(µ2)]+ Dξ(t) µ 2 = w 2(µ 2) 1 γ [F1(µ1)+F2(µ2)]+ Dξ(t) Remark: The two motors experience the same noise μj [nm] μ1 μ t [s] Metastability! frequency

21 Symmetry of motor populations 1D invariant manifold (ζ v) described by the 1-D Fokker-Plank equation: where tp = ζ {A(ζ)p}+4D ζζ p A(ζ) = aζ 2 γ is a double-well potential. Bifurcation diagram: [ F 1( ζ +η 2 )+F ζ ] 2( η ). 2 a b v [nm/s] U(ζ) [nm 2 /s] γ [pn s/nm] x1-3 γ [pn s/nm] x1-2 ζ [nm] Bidirectional motion! (including pausing)

22 Switching Time Mean first passage time to switch τ: Aτ +4Dτ = 1, τ(ζ H ) =, (ζ S ) =. τ [s] γ [pn s/nm] x1-3 Conclusion: We have an explanation of the bidirectional motion of these motors as driven by the stochasticity of the cargo.

23 Eukaryotic Chromosomal Segregation Before cells divide their chromosomes are duplicated and then moved to the center (midline) of the dividing cell. Once they are all centered, they are separated into two equal sets, enabling cell division. Question: How are chromosomes moved?

24 Chromosome/microtubule interactions Chromosomes are connected to microtubules(mt) at the Kinetochores where kinetochore structural complexes (such as Ndc8) bind the MT lattice.

25 Chromosome/microtubule interactions Microtubules change their length by polymerizing and depolymerizing at their tip. Dynamic Instability describes the process by which microtubules switch between polymerizing and depolymerizing states. Chromosome velocities are related to the MT polymerization/depolymerization rates. Question: How can depolymerization/polymerization affect chromosomal movement?

26 Ndc8 Complex Question: How can depolymerizing microtubules pull the chromosomes? Remark: It is known to not be due to walking molecular motors (i.e., dynein). Proposed Answer: Biased diffusion of Ndc8 proteins along the MT. Before now there has been no mechanistically satisfactory model of how biased diffusion actually works.

27 Kinetochore Motion How can depolymerizing microtubules pull? kmt n kt binders T l z L y Let n(z,y,t) be the density of bound kinetochore binders with rest position y and binding position z, (a Huxley model) n t = ( z (vn)+kon(z,y) nt L k on = κexp ( k(y ) z)2 H(l z), 2k B T l ) ndz k off (z,y)n, k off = κ off exp ( k z y F ),

28 Protofilament Shape What is the shape of the depolymerizing protofilament? i) l < ii) < l < L iii) l > L f > f > f < u θ f z Total energy for a protofilament is E = l ( α 2 ( θ φ) 2 + k ) lat 2 u2 +zf ds, du ds = sinθ, To find the equilibrium shape, set δe =. dz ds = cosθ.

29 Shape Equations Shape Equations (linearized): d 2 w dz 2 = k latθ, with boundary conditions α d2 θ dρ = w +ρθ, dz2 f(z) = k L (y z)n(z,y)dy n p dz = f, θ = φ, w =, ρ =, at z = l θ =, u =, at z = Remark: With f = these reduce to the classical beam equation d 4 θ dz 4 k lat α θ =.

30 Depolymerization Rate Curled or loaded filaments break more easily: k break (z) = βexp(κ b θ ρ ρ ) because of curvature and load - (Bell s law). Consequently, the depolymerization velocity is v d = l and total force generated by the bonds is F = k L l k break (z)dz, (y z)n(z,y,t)dzdy.

31 Depolymerization Rate For an end-loaded protofilament: 3 1 Displacement u (nm) z (nm) Relative Depolymerization Rate End Load (nondimensional) Typical of catch-bonds

32 Finding the solution Pick a value of v and l; Find n(z,y;v): = z (vn) + kon(z,y) ( ntl l ndz ) k off (z,y)n Calculate f(z): Find protofilament shape, f(z) = np k L (y z)n(z,y)dy Find v d : v d = l k break(z)dz Adjust l until v d = v (a fixed point problem) Calculate the total force: F = k L l (y z)n(z,y,t)dzdy. Make lots of plots.

33 And the Answer is Load (pn) a b c Velocity (µ m s -1 ) a c b Velocity (µ m s -1 ) δ = l/l Remark: There is no stall force. There is disconnect at high loads.

34 How Does that Work? 1 8 protofilament load force density (pn µ m 1 ) a 5 5 b c y/l u i) l < ii) z < l < L iii) l > L f > f > f < θ f load(pn µ m 1 ) z(nm) Microtubule protofilament shape 25 u, Displacement (nm) a z(nm) a b b c c

35 Effects of Parameters? γ = L2 k α (protein stiffness/protofilament bending stiffness): γ=7 6 1 Load (pn) γ=14 Velocity (µ m s -1 ) 1-1 γ= γ= Velocity (µ m s -1 ) δ = l/l k (Ndc8 spring constant): There is an optimal k κ=6 6 1 Load (pn) κ=1 Velocity (µ m s -1 ) 1-1 κ=1 κ=6 κ= κ= Velocity (µ m s -1 ) δ = l/l

36 Effects of Parameters? ρ (Bell s law coefficient for protofilaments): ρ= 6 ρ=1 1 Load (pn) ρ=5 Velocity (µ m s -1 ) 1-1 ρ= ρ=1 1 ρ= Velocity (µ m s -1 ) δ = l/l Remark: Notice the flat load-velocity profile. (This has been observed experimentally.)

37 Conclusion Protein flexibility plays an important role in their binding and unbinding, etc., and may lead to critical behaviors (i.e., thresholds, bifurcations, etc.). Biased diffusion of Ndc8 proteins enables depolymerizing microtubules to pull chromosomes, answering the question of how to pull with a burning rope. cf. Shtylla and Keener, J. Theor. Biol. 263, pp (21) cf. Shtylla and Keener, SIAM J. Appl. Math. 71, pp (211).

38 Thanks! Thanks to Blerta Shtylla, Pomona College Chris Miles, Utah NSF (funding)