Protein Induced Membrane Deformation Numerical study using Surface of Revolution approach Neeraj Agrawal February 28, 2008

Motivation As a tool to study membrane deformations under extreme curvature. As a bridge between Monge and local TDGL formalism. As a validation tool for local TDGL. Outline Theoretical Formalism Numerical Approach Results

Part I Theoretical Derivation of Membrane Equations Theory Solution Results

Surface Evolution R = c o s ( Ã ) z ¡ i n H = Ã + E = R [ H ¡ ] + ¾ d For axisymmetric membrane deformation. Exact minimization of Helfrich energy possible for any (axisymmetric) membrane deformation. Membrane parameterized by arc length, s and angle φ. R = c o s ( Ã ) z ¡ i n H = Ã + s i n ( ) R s=s1 s=0 R For topologically invariant transformations, Inputs from Talid Total curve length, L not known a priori E = R M · 2 [ H ¡ o ] + ¾ d A E = R M ( · 2 [ Ã + s i n ) ¡ H o ] ¾ d u E = 2 ¼ R s 1 ( · [ Ã + i n ) ¡ H o ] ¾ d Theory Solution Results

± E = º ± F = Minimum energy configuration for a membrane is given by subject to Z s 1 o R ¡ c ( Ã ) d = To solve the constrained optimization, we introduce Lagrange multipliers º We minimize, F = 2 ¼ Z s 1 µ · R [ Ã + i n ( ) ¡ H o ] ¾ º c ¶ d using ± F = L = · R 2 [ Ã + s i n ( ) ¡ H o ] ¾ º c Define Theory Solution Results

F = 2 ¼ R L d W e i n t r p F a s f u c o l h v b ; R Ã º d y . P e r So, we get F = 2 ¼ R s 1 L d W e i n t r p F a s f u c o l h v b 1 ; R Ã º d y . P e r f o m i n g a l z d v t , ¢ p = ± + s w ¢ F = Z s 1 µ @ L p i ¡ d ¶ ± + · ¸ A t e q u i l b r m , ¢ F = w h c a d s o @ L p i ¡ d s = a n · ¢ ¸ 1 + µ ¶ Theory Solution Results

D e ¯ n H = ¡ L + p T h e b o u n d a r y q t i s m p l ¯ [ ¡ H ¢ ] + i @ T h e b o u n d a r y q t i s m p l ¯ [ ¡ H ¢ ] 1 + @ L = Membrane minimum energy configuration is given by: p i 2 [ R ; Ã º ] 3 equations @ L p i ¡ d s = [ H ¢ ] 1 · ¸ 1 equation 3 equations Theory Solution Results

Ordinary Differential Equations ¯ n H = Á ( s ) Ã = c o s ( ) i n R 2 ¡ + º · ´ Á p i = Ã : º = · [ Ã ¡ Á ( s ) ] 2 i n R + ¾ p i = R : p i = º : R = c o s ( Ã ) F o r t h e s O D E , 4 b u n d a y c i . l q m 1 Theory Solution Results

H is not conserved along s Boundary Conditions S i n c e , s ¯ x d w h = ¢ . H o t [ ¡ ] 1 r u 6 l a A ( ) · R 2 " Ã µ Á ¶ # ¾ + º H is not conserved along s · h R ³ Ã + s i n ¡ Á ´ ¢ 1 = p i = Ã : [ º ¢ R ] s 1 = p i = R : W h e n p i = º , @ L Theory Solution Results

Case I A t s = , p e c i f y à a n d R . S o ¢ r l z w h v ¯ ; - ( ) µ , p e c i f y à a n d R . S o ¢ r l z w h v ¯ ; 1 - ( ) µ + ¡ Á ¶ º I u m 6 b H g : Theory Solution Results

Case II A t s = , l e ' p c i f y à a n d R w . W h o H ( ) r u º ¾ ; , l e ' p c i f y à a n d R 1 w . W h o H ( ) r u º ¾ ; s=s1 s=0 R R0 Theory Solution Results

Recap à = ¡ + Á º = + ¾ R = c o s ( à ) A t s = : à , R . º ( ) ¾ System of Ordinary Differential Equations à = c o s ( ) i n R 2 ¡ + º · ´ Á º = · [ à ¡ Á ( s ) ] 2 i n R + ¾ R = c o s ( à ) s=s1 s=0 R R0 Boundary Conditions A t s = : à , R . 1 º ( ) ¾ For numerical purpose, we specify R(s=0) = 1 nm. Theory Solution Results

Part II Numerical Solution of Membrane Equations Theory Solution Results

Challenges The ODEs are highly non-linear  uniqueness is not guaranteed. It’s a BVP and not IVP. Curvature Φ(s) is not uniform over the domain [0,s1]. Total system size s1 is not known a priori. s=s1 s=0 R R0 We need a good initial guess to the solution of system of ODEs that is consistent with the boundary conditions. Theory Solution Results

Rudimentary Approach: Initial Guess (σ=0) Rudimentary Approach: We expect H= Φ(s) be a solution of the equation since for this conformation, membrane energy is exactly zero. However, it’s true only for a free membrane – a pinned membrane need not satisfy H= Φ(s). Also, H= Φ(s) is an IVP while for pinned membrane, system of ODE is BVP. The Fluke: F o r à t s a i f y b h e u n d c , R 1 z . W Á = p v w g l 6 ¯ ² T k ( ¡ ) N I x Theory Solution Results

Initial Guess (σ=0) Case I: Case II: Á = a e x p ( ¡ s b ) c + u h t R 1 e x p ( ¡ s b ) 2 c + u h t R d à = : 5 p ¼ h a 1 c ³ e r f ( b ) ¡ s + ´ 2 i Case II: Á = a 1 e x p ( ¡ s b ) 2 c + u h t R d 6 à = : 5 p ¼ · a 1 c µ e r f ( b ) ¡ s + ¶ 2 ¸ Theory Solution Results

Initial Guess (σ=0) Case III: Á = a w h e n b < s H ( s ) ¼ + t a n 1 w h e n b < s 2 Ã = ¡ a 1 ( b 2 ) s + 8 > < : i f For numerical purposes, we approximate step function as: H ( s ) ¼ 1 2 + t a n h k where H(s) is a Heaviside step function and we use k=10 Theory Solution Results

Numerical Solution Protocol (σ=0) When ∫ Φds >> 0, start from ∫ Φds ≈ 0 and repeat the complete exercise with small increments in Φ Calculate initial guess for Ψ  Ψig Calculate Rig based on Ψig Calculate νig based on Rig and Ψig “Assume” s1 (≈ R0) R(s1) ≠ R0 S o l v e à = c s ( ) i n R 2 ¡ + º · Á [ ] ¾ Manually Check R(s1) ? along with R(s1) = R0 A t s = : à , R . 1 º ( ) ¾ Done !! Theory Solution Results

Parameters · = 4 k B T ¾ R 5 n m For σ ≠ 0, convergence of solution is difficult Theory Solution Results

Part III Preliminary Results Theory Solution Results

Case I Spatial organization of epsin on circular pinned membrane. Epsin induced spontaneous curvature modeled as a Gaussian function. Different results obtained by varying parameters of Gaussian function. Epsin leads to tubulation of vesicles. The outer diameter of tubules is ~20 nm  C0 ~ 0.05 nm-1. -Ford, Nature, 419, 361, 2002 Theory Solution Results

Results I = 2 ¼ R ³ Ã + ´ d b2 Prefactor = 0.05 1/nm Std. Dev. = 20 nm g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d Theory Solution Results

Results I Membrane Deformation profiles b2 Prefactor = 0.05 1/nm Std. Dev. = 20 nm Membrane profiles not very interesting !! Membrane deformation profile for different values of b2 What if prefactor is increased ? Theory Solution Results

Results II = 2 ¼ R ³ Ã + ´ d b2 Prefactor = 0.09 1/nm Std. Dev. = 20 nm b2 E n e r g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d Theory Solution Results

Results II b2 Membrane Deformation profiles Prefactor = 0.09 1/nm Std. Dev. = 20 nm Results can not be obtained for higher value of prefactor since membrane will pinch-off !! What happens if we reduce std dev ? Theory Solution Results

Results III = 2 ¼ R ³ Ã + ´ d b2 Prefactor = 0.09 1/nm Std. Dev. = 10 nm b2 E n e r g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d Theory Solution Results

Results III b2 Membrane Deformation profiles Prefactor = 0.09 1/nm Std. Dev. = 10 nm Theory Solution Results

Results IV = 2 ¼ R ³ Ã + ´ d Prefactor = 0.05 1/nm Std. Dev. = 30 nm E g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d Prefactor = 0.05 1/nm Std. Dev. = 30 nm Membrane deformation profile for different values of b2 Theory Solution Results

Results V = 2 ¼ R ³ Ã + ´ d Prefactor = 0.02 1/nm Std. Dev. = 20 nm E g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d Prefactor = 0.02 1/nm Std. Dev. = 20 nm Membrane deformation profile for different values of b2 Theory Solution Results

Results VI Effect of System Size n e r g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d Prefactor = 0.05 1/nm Std. Dev. = 20 nm R0 = 300 nm Membrane deformation profile for different values of b2 Theory Solution Results

Results VII Effect of Surface Tension g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d Prefactor = 0.05 1/nm Std. Dev. = 20 nm σ = 0.1 μN/m Theory Solution Results

Monge Gauge for Results I Prefactor = 0.05 1/nm Std. Dev. = 20 nm σ = 0 μN/m E n e r g y · = 1 2 R A ¡ H ¢ d x We need to increase number of discrete points to remove asymmetry in membrane profile at 300 nm. The Energy decreases as we move away from the boundary. Theory Solution Results

Monge Gauge for Results II y · = 1 2 R A ¡ H ¢ d x Prefactor = 0.09 1/nm Std. Dev. = 20 nm σ = 0 μN/m Surface revolution predicts vesicle formation for these parameter values. Theory Solution Results

Case II Clathrin induced deformation of circular pinned-membrane Clathrin induced spontaneous curvature modeled as a step function. Theory Solution Results

Results = 2 ¼ R ³ Ã + ´ d b2 Prefactor = 0.02 1/nm E n e r g y · 1 s i ³ Ã + i ( ) ´ d Theory Solution Results

Results b2 Membrane Deformation profiles b2 is the circumference of clathrin coat. b2 Membrane Deformation profiles Prefactor = 0.02 1/nm We can’t increase b2 > 280 nm  membrane pinches-off. Membrane deformation profile for different values of b2 Theory Solution Results

Results II = 2 ¼ R ³ Ã + ´ d Prefactor = 0.04 1/nm Membrane deformation profile for different values of b2 E n e r g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d Theory Solution Results

Case III Spatial organization of epsin on circular pinned membrane. Epsin induced spontaneous curvature modeled as a Gaussian function. Different results obtained by varying parameters of Gaussian function. b2 2b2 Theory Solution Results

Results I = 2 ¼ R ³ Ã + ´ d Prefactor = 0.05 1/nm Std. Dev. = 20 nm Membrane deformation profile for different values of b2 E n e r g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d We limit the simulations to b2 < 150 nm, at b2 = 150nm, the second epsin shell lies at 300 nm. Theory Solution Results

Results II = 2 ¼ R ³ Ã + ´ d Prefactor = 0.02 1/nm Std. Dev. = 20 nm Membrane deformation profile for different values of b2 E n e r g y · = 2 ¼ 1 R s ³ Ã + i ( ) ´ d We limit the simulations to b2 < 150 nm, at b2 = 150nm, the second epsin shell lies at 300 nm. Theory Solution Results

Future Work · R H d o r ( ¡ ) Attempt to get convergence for σ ≠ 0. To study the effects of boundary conditions. Which definition of energy is relevant ? Suggestions ? 1 2 · R s H d o r ( ¡ ) Theory Solution Results

Low σ Limit F o r ¾ 6 = , s y t e m n d i z g E 2 ¼ R . S c ¯ x a w v We restrict ourselves to σ = 0 or σ~0 for following reasons: Since the ODE is non-linear, we do not know if the obtained solution is infact a global minimum and not a local minimum. However, when σ = 0 or σ~0 , the H computed from solution matches very closely with H0 thus telling us that obtained solution is indeed a global minimum. For σ>>0 , the H computed from solution does not match nicely with H0. So, we can not conclude if our solution is a global minimum. F o r ¾ 6 = , s y t e m n d i z g E 2 ¼ R 1 . S c ¯ x a w v l u h f L b ³ ¡ ´ H e n c , w a m i z t h r f l g y b o s : B d 1 R . ¾ 6 = x p u v - Theory Solution Results

Clathrin Coat Clathrin forms the honeycomb lattice on the cytoplasmic surface of coated pits found both on plasma membrane & even at trans-Golgi network. Purified clathrin and AP2 were assembled into cages. The variety of cage arrangements (tennis ball, hexagonal barrel) depends on size of cage, smaller the size, smaller is the variety. The smallest symmetrical cage type is hexagonal barrel. For an hexagonal barrel cage, the width of whole assembly at the widest points is about 70 nm. This width indicate that C0 ~ 0.057 nm-1. However, in this hexagonal barrel cage, the adaptor layer is very tightly packed leaving no room for a membrane vesicle in the central cavity. Smith, et. al. EMBO J. 17, 4949, 1998 In weakly acidic, Ca2+ containing buffers of low ionic strength, clathrin triskelions spontaneously self-assemble forming heterogeneous population of cages. Schmid, Annu. Rev. Biochem. 66, 511, 1997 In vivo experiments suggests clathrin coats of diameter 90-100 nm and few coats in range 100-200 nm. Ehrlich, Cell. 118, 591, 2004 Theory Solution Results

Membrane Compartments Membrane has active patches of 400 nm (or 230-750nm) surrounded by inactive patch of 200 nm width. This indicates that membrane has size of 400 nm, beyond which, it is pinned by cytoskeleton. (Ehrlich, Cell, 118, 2004, 591) AP2 + Clathrin alone do not lead to clathrin polymerization on PIP2 monolayer (Ford, Science, 291, 2001, 1051) . Theory Solution Results

Epsin Energetic Affinity of ENTH domain of epsin-1 with Ins(1,4,5)P3 is 3.6 ± 0.4μM at 100C measured by calorimetric titration. (Ford, Science, 291, 2001, 1051) This indicates binding energy of ~ -4.9*10-20 J/epsin at 100C. We assume epsin are located along a ring at s0. A ring of epsin at s0 has perimeter of 2πR(s0). Assuming each epsin occupies 3 nm, maximum 2πR(s0)/3 epsin can fit in the ring. Now, R(s0) ~ 100 nm (from Fig. 4 in the manuscript). Hence, at max. 200 epsin can fit, which corresponds to energy of -2500 kBT. . s=s1 s=s0 R(s0) R0 Theory Solution Results