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Bridging Time and Length Scales in Materials Science and Bio-Physics. Workshop I: Multiscale Modelling in Soft Matter and Bio-Physics . September 26-30, 2005. The Enigma of Biological Fusion A comparison of two routes. With Kirill Katsov (MRL, UC Santa Barbara)
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Bridging Time and Length Scales in Materials Science and Bio-Physics Workshop I: Multiscale Modelling in Soft Matter and Bio-Physics September 26-30, 2005
The Enigma of Biological FusionA comparison of two routes With Kirill Katsov (MRL, UC Santa Barbara) Marcus Mueller (Institute fur Theoretische Physik, Gottingen)
Why is Fusion Important? Cell Trafficking Excocytosis/Endocytosis Viral Entry
Why is Fusion Difficult to Understand? Stability: long-lived holes must be difficult to form Fusion: long-lived holes must be easy to form
The Physicist’s View Kozlov and Markin 1983
Why does rate of hole formation go up? Presumably, due to reduced line tension
Why does rate of hole formation go up? Presumably, due to reduced line tension
Analytic Approach to FusionSelf-Consistent Field Theory • Investigate many possible configurations • Calculate free energy barriers of each • Change architecture easily • Analogous to Hartree Theory • Highly Non-Linear Set of Equations
Two Consequences 1. Main Barrier in Old Mechanism is Expansion
SCF Calculation of New Mechanism Line tension of extended stalk favors small R and a
SCF Calculation (cont) Reduced line tension of hole favors large a Membrane tension favors large R
IMI Just before F1(R,a) = aFIMI(R) +FS
IMI and its free eneregy g/g0=0.0 g/g0=0.4
IMI Just before F1(R,a) = aFIMI(R) +FS Just after F2(R,a) = aFHI(R) +(1-a)FH(R-d)+Fd F1(R,a) = F2(R,a) defines a ridge a(R)
Free energy barriers in new and old mechanism new old barriers decrease with decreasing f and increasing g
Resolving the enigma of fusion Membranes are stable because line tension of holes is large
Resolving the enigma of fusion Membranes are stable because line tension of holes is large But if hole forms next to stalk, line tension is reduced
Dependence of free energy on line tension Energy of hole 2plHR-gpR2 Energy of critical hole plH2/g Boltzmann factor PH= (AH /s2)exp(- plH2/gkT)
Boltzmann factor PH=(AH/s2) exp(- plH2/gkT) EXPONENTIAL DEPENDENCE ON SQUARE OF LINE TENSION: ENSURES STABILITY OF NORMAL MEMBRANES
Boltzmann factor PH=(AH/s2) exp(- plH2/gkT) EXPONENTIAL DEPENDENCE ON SQUARE OF LINE TENSION: ENSURES STABILITY OF NORMAL MEMBRANES Example: In simulation p lH2/gkT = 8.76, AH/s2=39 PH~ 6x10-3
Boltzmann factor PH=(AH/s2) exp(- plH2/gkT) EXPONENTIAL DEPENDENCE ON SQUARE OF LINE TENSION: ENSURES STABILITY OF NORMAL MEMBRANES ENABLES FUSION TO OCCUR BY REDUCING THAT LINE TENSION
Reducing the line tension from lH to ldr = alsh+(1-a) lH PH-->Psh = (Nsas/s2) exp(-pl2dr/gkT) so Psh/PH = (Nsas/AH) exp(pl2H/gkT)(1-l2dr/l2bare) = (Nsas/AH) (AH/s2 PH)x x= (1-l2dr/l2bare) Stability implies PH<<1 Therefore rate of hole formation near stalk Psh/PH>>1
EXAMPLE: IN SIMULATION ldr=lH/2, Nsas/AH~0.3 Pdressed/Pbare~ 14 P~ exp(-pl2/gkT) PH~ 6x10-3
In Biological Membranes, Effect is Greater lH~2.6x10-6 erg/cm g ~ 20 erg/cm2 PH~1.7 x 10-11(AH/s2) very stable
In Biological Membranes, Effect is Greater lH~2.6x10-6 erg/cm g ~ 20 erg/cm2 PH~1.7 x 10-11(AH/s2) very stable ldr/ lH= 0.5, Nsas/AH~0.3 Psh/PH=0.3(1/ 1.7 x 10-11)7/16 ~1x104 four orders of magnitude
Conclusion: The Enigma’s Solution Because fusion is thermally excited and excitation energy proportional to l2
Conclusion: The Enigma’s Solution Because fusion is thermally excited and excitation energy proportional to l2 Membranes can both be stable and undergo fusion
Furthermore Any process which affects the line tension slightly affects the rate of fusion greatly i.e. exquisite control