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

2. 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

3. Part I Theoretical Derivation of Membrane Equations Theory Solution
Results

4. 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 ] + d A E = R M ( 2 [ Ã + s i n ) H o ] d u E = 2 R s 1 ( [ Ã + i n ) H o ] d
Theory
Solution
Results

5. ± 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
We minimize, F = 2 Z s 1 R [ Ã + i n ( ) H o ] c d
using F = L = R 2 [ Ã + s i n ( ) H o ] c
Define
Theory
Solution
Results

6. 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 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 d y . P e r f o m i n g a l z d v t , p = + s w F = Z s 1 @ 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 +
Theory
Solution
Results

7. 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 ]
3 equations @ L p i d s = [ H ] 1
1 equation
3 equations
Theory
Solution
Results

8. Ordinary Differential Equationsn H = Á ( s ) Ã = c o s ( ) i n R 2 + Á p i = Ã : = [ Ã Á ( s ) ] 2 i n R + p i = R : p i = : 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

9. 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 Á # +
H is not conserved along s h R Ã + s i n Á 1 = p i = Ã : [ R ] s 1 = p i = R : W h e n p i = , @ L
Theory
Solution
Results

10. 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 - ( ) + Á I u m 6 b H g :
Theory
Solution
Results

11. 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 ;
s=s1
s=0 R R0
Theory
Solution
Results

12. 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 + Á = [ à Á ( s ) ] 2 i n R + R = c o s ( à )
s=s1
s=0 R R0
Boundary Conditions A t s = : Ã , R . 1 ( )
For numerical purpose, we specify R(s=0) = 1 nm.
Theory
Solution
Results

13. Part II Numerical Solution of Membrane Equations Theory Solution
Results

14. 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

15. 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 T k ( ) N I x
Theory
Solution
Results

16. Initial Guess (σ=0) Case I: Case II: Á = a e x p ( ¡ s b ) c + u h t R1 e x p ( s b ) 2 c + u h t R d à = : 5 p h a 1 c 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 Ã = : 5 p a 1 c e r f ( b ) s + 2
Theory
Solution
Results

17. Initial Guess (σ=0) Case III: Á = a w h e n b < s H ( s ) ¼ + t a n1 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 ) 1 2 + t a n h k
where H(s) is a Heaviside step function and we use k=10
Theory
Solution
Results

18. 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 + Á [ ]
Manually
Check R(s1) ?
along with
R(s1) = R0 A t s = : Ã , R . 1 ( )
Done !!
Theory
Solution
Results

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

20. Part III
Preliminary Results
Theory
Solution
Results

21. 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

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

23. Results I Membrane Deformation profiles b2
Prefactor = /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

24. 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 1 R s à + i ( ) d
Theory
Solution
Results

25. Results II b2 Membrane Deformation profiles
Prefactor = /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

26. 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 1 R s à + i ( ) d
Theory
Solution
Results

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

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

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

30. Results VI Effect of System Sizen e r g y = 2 1 R s à + i ( ) d
Prefactor = /nm
Std. Dev. = 20 nm
R0 = 300 nm
Membrane deformation profile for different values of b2
Theory
Solution
Results

31. Results VII Effect of Surface Tensiong y = 2 1 R s à + i ( ) d
Prefactor = /nm
Std. Dev. = 20 nm
σ = 0.1 μN/m
Theory
Solution
Results

32. Monge Gauge for Results I
Prefactor = /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

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

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

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

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

37. 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 1 R s à + i ( ) d
Theory
Solution
Results

38. 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

39. 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 1 R s à + i ( ) d
We limit the simulations to b2 < 150 nm, at b2 = 150nm, the second epsin shell lies at 300 nm.
Theory
Solution
Results

40. 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 1 R s à + i ( ) d
We limit the simulations to b2 < 150 nm, at b2 = 150nm, the second epsin shell lies at 300 nm.
Theory
Solution
Results

41. 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

42. 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 6 = , s y t e m n d i z g E 2 R 1 . S c x a w v l u h f L b H e n c , w a m i z t h r f l g y b o s : B d 1 R . 6 = x p u v -
Theory
Solution
Results

43. 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 ~ 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 nm and few coats in range nm.
Ehrlich, Cell. 118, 591, 2004
Theory
Solution
Results

44. Membrane Compartments
Membrane has active patches of 400 nm (or nm) 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

45. 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 kBT. .
s=s1
s=s0
R(s0) R0
Theory
Solution
Results