Diffusion in the cell

Transcription

1 Diffusion in the cell Single particle (random walk) Microscopic view Macroscopic view

2 Measuring diffusion Diffusion occurs via Brownian motion (passive) Ex.: D = 100 μm 2 /s for typical protein in water t ~ L 2 /D time to cross cell (10 μm): ~1 s time to travel down a 1-m neuron: ~300 years! PBoC

3 Measuring diffusion 1D: <r 2 > = 2Dt FRAP experiment 2D: <r 2 > = 4Dt FRAP experiment 3D: <r 2 > = 6Dt PBoC

4 Fick s First Law j = D c dc/dx < 0 j > 0 x particles flow from high to low concentration! PBoC 13.2

5 Diffusion Equation N box t = c t x y z c t = j x conservation of mass 1D diffusion equation (Fick s second law) c t = D 2 c x 2 PBoC 13.2

6 Diffusion (single-particle perspective) for a single particle, what are its options? move left at some rate k move right (same rate! random motion!) stay put (probability has to add up to 1) also known as a random walk with drunken pauses MICROstates, all equally probable (no energy dependence on position) PBoC 13.2

7 Diffusion (single-particle perspective) What s the probability of finding a particle at x at time t + Δt? i.e., p(x,t + Δt)? p(x,t+δt) = (1-2kΔt) p(x,t) + kδt p(x-a,t) + kδt p(x+a,t) stay put from left from right sum over microscopic trajectories p(x,t+δt) p(x,t) + Δt p/ t p(x±a,t) p(x,t) ± a p/ x + (a 2 /2) 2 p/ x 2 more Taylor expansions! p/ t = (a 2 k) 2 p/ x 2 Which is the diffusion equation again, with a 2 k = D. PBoC 13.2

8 Stochastic approach: Brownian motion what if you want to calculate ACTUAL particle trajectories? Langevin Equation (1908) slow, deterministic forces mẍ = γẋ + σξ(t) rapid, stochastic forces ξ(t) =0 ξ(t 1 )ξ(t 0 ) = δ(t 1 t 0 ) Gaussian white noise

9 Going from stochastic to deterministic mẍ = γẋ + σξ(t) γẋ >> mẍ γẋ = σξ(t) (Limit of strong friction) Every stochastic process has a corresponding Fokker- Planck equation on its probability t p(x, t x 0,t 0 )= σ2 2γ 2 2 x 2 p(x, t x 0,t 0 ) conditional probability (Markov process) initial condition: boundary condition: Solution: (Green s function) p(x, t t 0 x 0,t 0 )=δ(x x 0 ) p(x,t x 0,t 0 )=0 p(x, t x 0,t 0 )= 1 (x x 0 ) 2 e 4D(t t 0 ) 4πD(t t0 ) (D = σ2 2γ 2 )

10 Diffusion Equation c(x, t) = N e x2 /4Dt 4πDt PBoC

11 Diffusion Equation Fluorescence recovery after photobleaching (FRAP) PBoC

12 PBoC Diffusion in the presence of a force J 2 = D dc dx J = D dc dx + F γ c steady state, J(x) = 0 (no flux) J 1 = vc = F γ c No acceleration! Force (scaled by the drag γ) leads to drift v D dc dx = F γ c c(x) =c(0)e (U(x) U(0))/γD D = kt/γ Einstein relation

13 Fluctuation-dissipation theorem Langevin equation γẋ = F (x)+σξ(t) t p(x, t x 0,t 0 )=(D 2 x 2 x F (x) γ )p(x, t x 0,t 0 ) corresponding Fokker-Planck equation (Smoluchowski Equation) D = kt/γ D = σ2 2γ 2 Einstein relation (previous slide) from Langevin equation Therefore: 2 =2kT Dissipative force (friction) derives from random fluctuations!

14 Ex: membrane-protein insertion Model growth of nascent protein as a freely-jointed chain r 2 = N r L 2 ; N r = t/τ Range of TM helix in membrane due to ribosome tether grows with t as a function of synthesis rate, subject to potential: U (r, t ) = U 1 (r ) + U 2 (r, t ) = U 1 (r ) + αr 2 ; α = 3kT t 2 U1(r) comes from thermodynamics U2(r,t) comes from ribosome tether Novel diffusion-elongation model of thermodynamically AND kinetically driven membrane insertion τ L 2 diffusion simulations : t p(r,t) = De βu(r,t) e βu(r,t) p(r,t) irreversible COMMITMENT J. Gumbart et al. (2013) JACS. 135:

15 Ex: membrane-protein insertion force? Recent cryo-em based structure shows part of the nascent protein unfolded between ribosome and channel, just as predicted Structure of the SecY channel during initiation of protein translocation. E Park, JF Ménétret, JC Gumbart, SJ Ludtke, W Li, A Whynot, TA Rapoport and CW Akey. Nature, 506: , force?

16 (Perfectly) Absorbing sphere Derive PBoC dn dt =4πDac 0 assuming c(a) =0 c( ) =c 0 (perfect absorber) c t =0

17 (Finitely) absorbing sphere # of receptors absorption rate dn dt = Mk onc(a) Finite rate of absorption c(a) = c 0 1+Mk on /4πDa k on 0? k on? dn dt = Mk on c 0 1+Mk on /4πDa 4πDac 0 rate limited PBoC

18 Optimal number of receptors Assume 90% of maximum (diffusionlimited) is sufficient for the cell dn dt =0.9(4πDac 0) dn dt =0.9(4πDac 0)= Mk on c 0 1+Mk on /4πDa 0.1Mk on =0.9(4πDa) If each receptor is 10 nm 2 and the cell surface is 1200 μm 2... M =9 4πDa k on receptors only of the surface needs to be covered! (more receptors doesn t help) PBoC

19 Real chemoreceptors cryo tomogram of receptor array MDFF fit of atomic structure to averaged map Briegel, Ames, Gumbart et al. The mobility of two kinase domains in the Escherichia coli chemoreceptor array varies with signalling state (2013) Mol Microbio. 89:

20 mean first passage time PBoC For a random walker, when does it reach a given position? p t = D 2 p x 2 diffusion eq. j = D p x = 1 τ Fick s 1st law U(x) = 0 x = 0 x = δ δ p(δ) =0 0 dx p(x) =1 p(x) = 2 2 x + 2 Assume particle escapes at exit MFPT τ = δ2 2D 1D diffusion x = 0 j = D p x F γ p p(x) =Ae Fx/kT j 0 F x = δ U(x) = -Fx from PMF MFPT τ = ktγ F 2 (efδ/kt 1) γδ F