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Biophys J, February 2002, p. 693-712, Vol. 82, No. 2

Membrane Fusion: Stalk Model Revisited

Vladislav S. Markin* and Joseph P. Albanesidagger

 *Department of Anesthesiology and  dagger Department of Pharmacology, University of Texas Southwestern, Dallas, Texas 75390 USA


    ABSTRACT
TOP
ABSTRACT
INTRODUCTION
THEORETICAL
DISCUSSION
REFERENCES

Membrane fusion is believed to proceed via intermediate structures called stalks. Mathematical analysis of the stalk provided the elastic energy involved in this structure and predicted the possible evolution of the overall process, but the energies predicted by the original model were suspiciously high. This was due to an erroneous assumption, i.e., that the stalk has a figure of revolution of a circular arc. Here we abandon this assumption and calculate the correct shape of the stalk. We find that it can be made completely stress free and, hence, its energy, instead of being positive and high can become negative, thus facilitating the fusion process. Based on our new calculations, the energies of hemifusion, of complete fusion, and of the pore in a bilayer were analyzed. Implications for membrane fusion and lipid phase transitions are discussed.


    INTRODUCTION
TOP
ABSTRACT
INTRODUCTION
THEORETICAL
DISCUSSION
REFERENCES

Membrane fusion plays a fundamental role in cell physiology. It is even believed to be a key event in the origin of life (Norris and Raine, 1998). For this reason it has attracted the intense interest of numerous researchers who have attempted to develop a model of this process (see Reviews and References therein, Chernomordik et al., 1995b; Jahn and Sudhof, 1999; Zimmerberg and Chernomordik, 1999; Burger, 2000; Stegmann, 2000; Melikyan et al., 2000). Both biological and artificial membrane fusion involves the merger of two phospholipid bilayers in an aqueous environment. In the early 1980s a qualitative picture of this process emerged suggesting that fusion proceeds via local contact between two lipid bilayers. Hui et al. (1981) termed this contact "a point defect" and proposed that it represented an intermediate stage of fusion. It was clear that membranes could not expose their hydrophobic interiors to water (reflected by Gingel's famous statement: "membranes hate edges." And he went on: "All our ideas of membrane transformations are based on this fact") (Gingell and Ginsberg, 1978). Therefore, to make a connection between two membranes, their monolayers must be strongly bent into an hourglass shape. This structure was called a stalk, and the whole mechanism was called the stalk model. Its mathematical analysis was performed in 1983 (Kozlov and Markin, 1983) with the first English publication appearing in 1984 (Markin et al., 1984).

Mathematical implementation of the stalk model was based on calculation of the elastic energy of the curved monolayers and elucidation of the chain of events leading either to complete fusion or to abortion of the process. The model proved to be very attractive and was adopted in numerous studies, both experimental and theoretical (Leikin et al., 1987; Kozlov et al., 1989; Nanavati et al., 1992; Siegel, 1993; Chizmadzhev et al., 1995, 1999, 2000; Siegel, 1999; Kuzmin et al., 2001). Its success was based on its ability to explain a number of experimental observations (Monck and Fernandez, 1994; Chernomordik et al., 1995a,b, 1997; Chernomordik, 1996; Melikyan et al., 1997; Basanez et al., 1998; Lee and Lentz, 1998; Zimmerberg and Chernomordik, 1999; Goni and Alonso, 2000; Razinkov and Cohen, 2000). Later on the model was further developed to include additional features such as hydrophobic voids (Siegel, 1993), fusion pore dilatation in stages (Chizmadzhev et al., 1995), relative sliding of monolayers (Chizmadzhev et al., 1999), role of membrane tension (Nanavati et al., 1992; Chizmadzhev et al., 2000), et cetera.

However, all these papers had to deal with one significant difficulty: because of the high curvature of the stalk, the calculation of its elastic energy inevitably resulted in very high values, up to hundreds kT. D. Siegel wrote in 1999, "It is troubling that energies predicted for stalk and transmonolayer contact (TMC) intermediates are so high." This energy is too high and raises doubts on the feasibility of the whole model. From the very beginning (1983, 1984) significant efforts were spent in finding a way for the stalk to decrease its energy. The obvious factor to consider was the spontaneous curvature of lipid monolayers. This helped to some extent but at the expense of a necessitating assumption of very high spontaneous curvature at the limit of reasonable values. This struggle with the high bending energy of the stalk continues to this day, sometimes eliciting very ingenious ideas and suggestive terminology (Kuzmin et al., 2001).

This stalk paradox is rather disturbing: the model seems to be intuitively reasonable and agrees with experimental observations, but it suffers from inherent difficulty. We believe that the resolution of this paradox should be found within the stalk model proper rather than by enlisting additional and sometimes artificial considerations. To this end we performed an analysis of the stalk model ab initio. We found that with the correct mathematical treatment of the model all of the difficulties researchers had been struggling with disappeared. Indeed, these difficulties were simply the result of a single unjustified assumption in the stalk model.

The source of these apparent difficulties is the shape of the stalk. Since the original papers (Kozlov and Markin, 1983) the shape of the stalk was not calculated but rather postulated to be the figure of revolution of a circular arc. It was this postulate that brought about a very high bending energy of the stalk. So, following Zimmerberg (2000), we asked the same question: "Are the curves in all the right places?" As we now report, with the right shape of this intermediate structure the stalk can be made completely stress free and its bending energy reduced to zero or even to negative values.

Thus, the numerical results of the original paper (Kozlov and Markin, 1983) are not valid for the properly shaped stalk. Unfortunately, all subsequent theoretical papers devoted to the stalk model followed the lead without questioning the postulate of circular shape. Therefore their conclusions regarding the high bending energy are also wrong. One of us (V.S.M.) is personally responsible for this unjustified assumption and we feel obligated to correct this inaccuracy in the stalk analysis.


    THEORETICAL
TOP
ABSTRACT
INTRODUCTION
THEORETICAL
DISCUSSION
REFERENCES

Original stalk model

In the original model of 1983 (Kozlov and Markin, 1983) the stalk was visualized as an hourglass-shaped local connection between two membranes. It could be comprised of one or two monolayers of opposing membranes, producing either a monolayer or bilayer stalk. Later the monolayer stalk was accepted as a key intermediate in the overall fusion process known as a hemifusion stage, and the bilayer stalk was baptized a fusion pore. The stalk could form after direct closure upon one another of two bulging defects in opposing membranes (layers) having initial curvature, cinit, and a spontaneous curvature, c0.

It was assumed that the stalk and surrounding membranes form an axisymmetrical body of revolution (Fig. 1) with the neutral surface (dotted line) drawn somewhere in the middle of the layer transformed into the stalk. Here a is the shortest distance from the neutral surface to the axis of revolution, r is the marginal radius of the stalk, b is the distance from the axis of revolution to the point where the branches of the stalk become horizontal. Parameters a and b could be called the neck and the width of the stalk. In the original version of the model b = a + r. The coordinates of the contour are x and z, and the angle between the contour and horizontal line is psi .



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FIGURE 1   Steps in membrane fusion. Solid lines represent hydrophilic surfaces, thin lines represent hydrophobic, dotted lines represent neutral surfaces. (A) Parameters of the stalk. (B) Hemifusion, initial stage. (C) Hemifusion, transmonolayers contact. (D) Complete fusion---fusion pore.

According to Helfrich (1973), the density of bending energy accumulated in the stalk is given as
w=<FR><NU>&kgr;</NU><DE>2</DE></FR> (c<SUB><UP>m</UP></SUB>+c<SUB><UP>p</UP></SUB>−c<SUB>0</SUB>)<SUP>2</SUP> (1)
in which kappa  is bending rigidity and cm and cp are principal curvatures along the meridian and parallel to the body of revolution representing the stalk. The energy associated with Gaussian curvature was neglected as it became customary in all subsequent papers. The energy of the stalk was defined as its elastic energy minus the initial elastic energy of two layers without stalks:
W<SUB><UP>s</UP></SUB>=&pgr;&kgr;<FENCE><LIM><OP>∫</OP><LL><UP>stalk</UP></LL></LIM>dA(c<SUB><UP>m</UP></SUB>+c<SUB><UP>p</UP></SUB>−c<SUB>0</SUB>)<SUP>2</SUP>−<LIM><OP>∫</OP><LL><UP>stalk</UP></LL></LIM>dA(c<SUB><UP>init</UP></SUB>−c<SUB>0</SUB>)<SUP>2</SUP></FENCE> (2)
The integrals are taken over the total surface of the stalk. The first integral represents the bending energy of the stalk membrane and the second integral is equal to the bending energy of the initial membrane. The stalk energy Ws was found to be
W<SUB><UP>s</UP></SUB>(a, r)=2&pgr;&kgr;<FENCE><FENCE><FENCE><FR><NU>2</NU><DE>r</DE></FR>+c<SUB>0</SUB></FENCE><SUP>2</SUP>−(2c<SUB><UP>init</UP></SUB>−c<SUB>0</SUB>)<SUP>2</SUP></FENCE></FENCE> (3)

<FENCE>×<FENCE><FR><NU>&pgr;</NU><DE>2</DE></FR> r(r+a)−r<SUP>2</SUP></FENCE>−&pgr;<FENCE><FR><NU>2</NU><DE>r</DE></FR>+c<SUB>0</SUB></FENCE>(r+a)+<FR><NU>2(r+a)<SUP>2</SUP></NU><DE>r<RAD><RCD>a(2r+a)</RCD></RAD></DE></FR> <UP>arctan</UP><RAD><RCD><FR><NU>2r+a</NU><DE>a</DE></FR></RCD></RAD></FENCE>.
If the fusing membranes initially were planar, cinit = 0, then
W<SUB><UP>s</UP></SUB>(a, r)=2&pgr;&kgr;<FENCE><UP>−</UP>4−rc<SUB>0</SUB>(4−&pgr;)+&pgr;ac<SUB>0</SUB>+<FR><NU>2(r+a)<SUP>2</SUP></NU><DE>r<RAD><RCD>a(2r+a)</RCD></RAD></DE></FR> <UP>arctan</UP><RAD><RCD><FR><NU>2r</NU><DE>a</DE></FR>+1</RCD></RAD></FENCE> (4)
In the absence on spontaneous curvature (c0 = 0), the bending energy depends only on the ratio r/a; the function has the minimum equal to Emin = 3.791 kappa , which occurs at r/a = 1.671. Final results depend on the value of the bending rigidity. Chizmadzhev et al. (1995) assumed that bilayer rigidity equals ~10-19 J, giving the minimum monolayer stalk energy of Emin = 45.5 kT. If, as more often assumed, kappa congruent  10 kT, the minimum energy of the stalk is 37 kT.

If spontaneous curvature of the bent layer is not zero, then its energy depends on the variables r and a separately. The initial monolayer stalk has a approx  1 nm. For c0 = -0.01 nm-1 the minimal energy of 3.68, kappa  approx  44.2 kT occurs at r = 1.66 nm. For c0 = -0.1 nm-1 the minimal energy of 2.68 kappa  approx  32 kT occurs at r = 1.54 nm.

Stress-free stalk

Now we shall no longer make the assumption that the stalk is circular in shape but instead its shape will be calculated. Therefore in Fig. 1 A parameter r should be disregarded and the stalk is considered to be a figure of revolution of a certain arbitrary curve. In this curve a is its shortest distance from the axis of revolution, b is the point where the stalk smoothly connects with the rest of the planar membrane. For simplicity, the initial membranes are considered planar. 2H is the distance between neutral surfaces of fusing layers. The contour is supposed to be smooth, and no sharp points are allowed.

The principal curvatures of the stalk (Deuling and Helfrich, 1977; Volkov et al., 1998; Markin et al., 1999) are given by equations
c<SUB><UP>p</UP></SUB>(x)=<FR><NU><UP>sin </UP>&psgr;(x)</NU><DE>x</DE></FR>; c<SUB><UP>m</UP></SUB>(x)=<UP>cos </UP>&psgr;(x) <FR><NU>d&psgr;</NU><DE>dx</DE></FR> (5)
To find the stress-free stalk we impose the condition that the total curvature of the stalk is constant and equal to cstalk:
c<SUB><UP>p</UP></SUB>(x)+c<SUB><UP>m</UP></SUB>(x)=c<SUB><UP>stalk</UP></SUB> (6)
If cstalk = c0, the stalk is stress free.

This approach was used previously for the analysis of myelin shapes (Deuling and Helfrich, 1977) and beading of nerve fibers under lateral tension (Markin et al., 1999). The contour of the stalk neutral surface can be presented (Volkov et al., 1998; Markin et al., 1999) as
<FR><NU>dz</NU><DE>dx</DE></FR>=<UP>tan </UP>&psgr;(x)=<FR><NU>xc<SUB><UP>p</UP></SUB></NU><DE><RAD><RCD>1−x<SUP>2</SUP>c<SUP><UP>2</UP></SUP><SUB><UP>p</UP></SUB></RCD></RAD></DE></FR> (7)
For the contour in Fig. 2, one can find from Eq. 5 that
xc<SUB><UP>p</UP></SUB>=<FR><NU>1</NU><DE>2</DE></FR> c<SUB><UP>stalk</UP></SUB>x+<FR><NU>a</NU><DE>x</DE></FR><FENCE>1−<FR><NU>1</NU><DE>2</DE></FR> c<SUB><UP>stalk</UP></SUB>a</FENCE> (8)
Then the final equation for the contour takes the form
<FR><NU>dz</NU><DE>dx</DE></FR>=<FENCE><FENCE><FR><NU>1</NU><DE>2</DE></FR> c<SUB><UP>stalk</UP></SUB>x+<FR><NU>a</NU><DE>x</DE></FR> <FENCE>1−<FR><NU>1</NU><DE>2</DE></FR> c<SUB><UP>stalk</UP></SUB>a</FENCE></FENCE><SUP>−2</SUP>−1</FENCE><SUP>−1/2</SUP> (9)
Parameter b of the stalk can be found from the condition dz/dx = 0; then
b=<RAD><RCD>a<SUP>2</SUP>−<FR><NU>2a</NU><DE>c<SUB><UP>stalk</UP></SUB></DE></FR></RCD></RAD>, <UP>or </UP> <FR><NU>b</NU><DE>a</DE></FR> =<RAD><RCD>1−<FR><NU>2</NU><DE>ac<SUB><UP>stalk</UP></SUB></DE></FR></RCD></RAD> (10)
The last equation shows that the ratio b/a depends on a single dimensionless parameter acstalk. The same is true for the shape of the stalk z = z(x) given by the following equation:
<FR><NU>z</NU><DE>a</DE></FR>=<LIM><OP>∫</OP><LL>1</LL><UL><UP>x/a</UP></UL></LIM><FENCE><FENCE><FR><NU>1</NU><DE>2</DE></FR> ac<SUB><UP>stalk</UP></SUB> t+<FR><NU>1</NU><DE>t</DE></FR> <FENCE>1−<FR><NU>1</NU><DE>2</DE></FR> ac<SUB><UP>stalk</UP></SUB></FENCE></FENCE><SUP>−2</SUP>−1</FENCE><SUP>−1/2</SUP>dt (11)
As one can see from here, the shape of the stalk is indeed determined by a single parameter acstalk because the upper limit of this integral b/a is also a function of the same parameter according to Eq. 10. In Fig. 2 A we presented the contour of the stalk neutral surface with parameters acstalk = -0.1. One can see that the contour of the stalk is not a circular arc and that its branches at the end become horizontal. The important point is that if cstalk = c0, then this stalk is completely stress free because its total curvature at every point is equal to the spontaneous curvature c0. Fig. 2 B gives a three-dimensional view of that stalk (of its neutral surface).



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FIGURE 2   Stress-free stalk. (A) Two-dimensional contours for three different spontaneous curvatures shown in the graph. (B) Three-dimensional rendition of the stalk with c0 = -0.1 nm-1.

The height of the stalk L = 2H is found from the integral (Eq. 11) with x = b. In the absence of an analytical solution for the stress free stalk (Eq. 11) it is useful to have a good approximation for it. From the analysis of series expansion of (Eq. 11) at points x = a and x = b one can find a piece-wise approximating function for L:
<FR><NU>L</NU><DE>a</DE></FR>=<FR><NU>2</NU><DE>0.68435−ac<SUB><UP>stalk</UP></SUB></DE></FR>, <UP>if</UP> ac<SUB><UP>stalk</UP></SUB><<UP>−</UP>1 (12a)
and
<FR><NU>L</NU><DE>a</DE></FR>=0.6384 <UP>log</UP>[1+5.4245(<UP>−</UP>ac<SUB><UP>stalk</UP></SUB>)<SUP>−1.5666</SUP>], (12b)

<UP>if </UP>ac<SUB><UP>stalk</UP></SUB>><UP>−</UP>1
Notice that the dimensionless normalized distance L/a depends on a single parameter, acstalk, only. This function is presented in Fig. 3 A together with a numerical solution (Eq. 11). Two lines are indistinguishable at this resolution demonstrating that function (Eq. 12) closely approximates the distance between fusing membranes. Practically, one cannot expect that acstalk would reach a high negative value. Therefore we need only the values between -1 and -0.001, given by Eq. 12b. Hence the distance between membranes (neutral surfaces) is given by
L=2H=0.6384 a <UP>log</UP>⌊1+5.4245(<UP>−</UP>ac<SUB><UP>stalk</UP></SUB>)<SUP>−1.5666</SUP>⌋ (13)
Notice, that this relationship does not depend on membrane stiffness kappa .



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FIGURE 3   Parameters of the stress-free stalk. (A) Normalized height of the stalk L/a as the function of -acstalk. (B) Height of the stalk L as a function of the neck radius a for different values of spontaneous curvature cstalk shown at the curves in nm-1.

Equation 12 and the resulting plot in Fig. 3 A have universal character applicable at any (negative) spontaneous curvature. However, for practical purposes a dimensional Eq. 13, illustrated in Fig. 3 B for a few selected values of c0, is more convenient.

Stalk energy

Now let us determine the energy of the stalk. Because the curvature of the stalk is constant and equal to cstalk, its bending energy is 1/2kappa (cstalk - c0)2 A, in which A is the area of the stalk. If cstalk = c0, bending energy is zero. We have to subtract from here the initial energy of the membrane in the planar state if spontaneous curvature is not zero. In the planar state fusing layers have elastic energy equal to 1/2kappa c02 A. Therefore the energy of the stalk is
W<SUB><UP>s</UP></SUB>=½&kgr;A[(c<SUB><UP>stalk</UP></SUB>−c<SUB>0</SUB>)<SUP>2</SUP>−c<SUB>0</SUB><SUP>2</SUP>] (14)
Notice that the energy of the stress free stalk is -1/2kappa c02A and it is negative. To complete this calculation we have to find stalk area A. If the infinitesimal length of the contour in Fig. 1 A is dl = dx/cos psi , the differential of the area can be found as
dA=2&pgr;x dl=<FR><NU>2&pgr;x dx</NU><DE><UP>cos </UP>&psgr;</DE></FR>=2&pgr;x dx<RAD><RCD>1+<FENCE><FR><NU>dz</NU><DE>dx</DE></FR></FENCE><SUP>2</SUP></RCD></RAD>=<FR><NU>2&pgr; x dx</NU><DE><RAD><RCD>1−<FENCE><FR><NU>1</NU><DE>2</DE></FR> c<SUB><UP>stalk</UP></SUB>x+<FR><NU>a</NU><DE>x</DE></FR> <FENCE>1−<FR><NU>1</NU><DE>2</DE></FR> c<SUB><UP>stalk</UP></SUB>a</FENCE></FENCE><SUP>2</SUP></RCD></RAD></DE></FR> (15)
And hence
A=4&pgr;<LIM><OP>∫</OP><LL><UP>a</UP></LL><UL><UP>b</UP></UL></LIM><FR><NU>x dx</NU><DE><RAD><RCD>1−<FENCE><FR><NU>1</NU><DE>2</DE></FR> c<SUB><UP>stalk</UP></SUB>x+<FR><NU>a</NU><DE>x</DE></FR> <FENCE>1−<FR><NU>1</NU><DE>2</DE></FR> c<SUB><UP>stalk</UP></SUB>a</FENCE></FENCE><SUP>2</SUP></RCD></RAD></DE></FR> (16)
Now the total energy of the stalk can be presented as
W<SUB><UP>s</UP></SUB>=2&pgr;&kgr;<LIM><OP>∫</OP><LL><UP>1</UP></LL><UL><UP>b/a</UP></UL></LIM><FR><NU>⌊(ac<SUB><UP>stalk</UP></SUB>−ac<SUB>0</SUB>)<SUP>2</SUP>−ac<SUB>0</SUB><SUP>2</SUP>⌋t dt</NU><DE><RAD><RCD>1−<FENCE><FR><NU>1</NU><DE>2</DE></FR> ac<SUB><UP>stalk</UP></SUB>t+<FR><NU>1</NU><DE>t</DE></FR> <FENCE>1−<FR><NU>1</NU><DE>2</DE></FR> ac<SUB><UP>stalk</UP></SUB></FENCE></FENCE><SUP>2</SUP></RCD></RAD></DE></FR> (17)
The dependence of the stalk energy on its radius a is presented in Fig. 4 for different values of spontaneous curvature c0. As one can conclude from here, the stalk has negative energy, it tends to expand and to push away fusing membranes. For comparison with previous results we present in Table 1 the energy of the initial monolayer stalk (astalk = 1 nm) both for the old model and the new one in units of kT. One can see a huge difference between the two models.



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FIGURE 4   Bending energy of the stress-free stalk with different c0 shown at the curves in nm-1 as a function of its radius a (below abscissa). Above abscissa there are two curves with zero or small positive spontaneous curvature; these stalks are not stress free, however, their bending energy is very small.


                              
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TABLE 1   Stalk energy

Hemifusion

Let us consider the initial hemifusion structure presented in Fig. 1 B. At some point P the trans monolayer peels off of the cis monolayer. Let this point have coordinate xp and angle psi  = psi p. Designate the corresponding points at the neutral surfaces of cis and trans monolayers x1 and x2. The surface between the two monolayers will be considered a reference surface and its curvature will be designated cb. The whole structure can be divided into three parts: the "wings" of the hemi-fused bilayers beyond the point x = x1, the "neck" of the stalk in the range x < x1, and two dimples formed by trans monolayers in the range x < x2. The monolayers of these three parts are smoothly connected to each other. The total energy of hemifusion Wh consists of the bending energy of the neck Wn, of the wings Ww, of the dimples Wd, and hydrophobic energy of two voids Wv. We shall calculate each of these components.

First, let us find coordinate x1. If the stalk has curvature cs and the neck radius a, then from Eqs. 5 and 8 one can find that
x<SUB>1</SUB>=<FR><NU><UP>sin</UP> &psgr;<SUB><UP>p</UP></SUB>−<RAD><RCD>(<UP>sin</UP> &psgr;<SUB><UP>p</UP></SUB>)<SUP>2</SUP>−2ac<SUB><UP>s</UP></SUB>+a<SUP>2</SUP>c<SUP><UP>2</UP></SUP><SUB><UP>s</UP></SUB></RCD></RAD></NU><DE>c<SUB><UP>s</UP></SUB></DE></FR> (19)
and then
x<SUB><UP>p</UP></SUB>=x<SUB>1</SUB>−(h−&lgr;)<UP>sin &psgr;<SUB>p</SUB></UP>,
and
x<SUB>2</SUB>=x<SUB>1</SUB>−2(h−&lgr;)<UP>sin </UP>&psgr;<SUB><UP>p</UP></SUB>. (20)
The energy of the stalk neck Wneck can be readily found from Eq. 17 if the integral is taken from 1 to x1/a.

Calculation of the energy of the wings Ww is more complicated. We select the interface between two monolayers as a reference surface, and coordinates of monolayer neutral surfaces will be related to this reference surface. As in the previous section, we assume that the reference surface has constant total curvature. This assumption will be discussed later. The principal curvatures cm and cp of the reference surface are defined the same way as for a monolayer stalk by Eq. 5. That means that cm < 0 and cp > 0. Designate lambda  the distance from the hydrophilic surface of a monolayer to its neutral surface. Then the principal curvatures of the cis monolayer are given by the equations
c<SUP><UP>cis</UP></SUP><SUB><UP>p</UP></SUB>=<FR><NU>c<SUB><UP>p</UP></SUB></NU><DE>c<SUB><UP>p</UP></SUB>(h−&lgr;)+1</DE></FR> <UP>and </UP>c<SUP><UP>cis</UP></SUP><SUB><UP>m</UP></SUB>=<FR><NU>c<SUB><UP>m</UP></SUB></NU><DE>c<SUB><UP>m</UP></SUB>(h−&lgr;)+1</DE></FR> (21)
The curvatures of the trans monolayer have the opposite sign and can be presented as
c<SUP><UP>trans</UP></SUP><SUB><UP>p</UP></SUB>=<FR><NU>c<SUB><UP>p</UP></SUB></NU><DE>c<SUB><UP>p</UP></SUB>(h−&lgr;)−1</DE></FR>
and
c<SUP><UP>trans</UP></SUP><SUB><UP>m</UP></SUB>=<FR><NU>c<SUB><UP>m</UP></SUB></NU><DE>c<SUB><UP>m</UP></SUB>(h−&lgr;)−1</DE></FR> (22)
Notice that these definitions are consistent for both monolayers. We "look" at a monolayer from the aqueous phase: if the monolayer in a given principal plane is convex, then this curvature is positive and vice versa. The geometry of the wings of the hemi-fused bilayer is determined by a set of three parameters: xp, psi p, and cb. One can find that the principal curvatures of the reference surface of the hemi-fused bilayer are given by
c<SUB><UP>p</UP></SUB>=<FR><NU>1</NU><DE>2</DE></FR> c<SUP><UP>b</UP></SUP>+<FR><NU>x<SUB><UP>p</UP></SUB></NU><DE>x<SUP>2</SUP></DE></FR> <FENCE><UP>sin &psgr;<SUB>p</SUB></UP>−<FR><NU>1</NU><DE>2</DE></FR> c<SUP><UP>b</UP></SUP>x<SUB><UP>p</UP></SUB></FENCE>
and
c<SUB><UP>m</UP></SUB>=<FR><NU>1</NU><DE>2</DE></FR> c<SUP><UP>b</UP></SUP>−<FR><NU>x<SUB><UP>p</UP></SUB></NU><DE>x<SUP>2</SUP></DE></FR> <FENCE><UP>sin</UP> &psgr;<SUB><UP>p</UP></SUB>−<FR><NU>1</NU><DE>2</DE></FR> c<SUP><UP>b</UP></SUP>x<SUB><UP>p</UP></SUB></FENCE> (23)
Instead of Eq. 10 the length of the wings is given by
b<SUB><UP>w</UP></SUB>=<RAD><RCD>x<SUP><UP>2</UP></SUP><SUB><UP>p</UP></SUB>−<FR><NU>2x<SUB><UP>p</UP></SUB><UP>sin</UP>&psgr;<SUB>p</SUB></NU><DE>c<SUP><UP>b</UP></SUP></DE></FR></RCD></RAD> (24)
and the area differential is
dA=<FR><NU>2&pgr; x dx</NU><DE><RAD><RCD>1−<FENCE><FR><NU>1</NU><DE>2</DE></FR> c<SUP><UP>b</UP></SUP>x+<FR><NU>x<SUB><UP>p</UP></SUB></NU><DE>x</DE></FR> <FENCE><UP>sin &psgr;<SUB>p</SUB></UP>−<FR><NU>1</NU><DE>2</DE></FR> c<SUP><UP>b</UP></SUP>x<SUB><UP>p</UP></SUB></FENCE></FENCE><SUP>2</SUP></RCD></RAD></DE></FR> (25)
The bending energy density of the bilayer wings, referred to its reference surface, can be presented as
w<SUB><UP>w</UP></SUB>=<FR><NU>&kgr;</NU><DE>2</DE></FR> <FENCE><FENCE><FENCE><FR><NU>2(h−&lgr;)c<SUB><UP>m</UP></SUB>c<SUB><UP>p</UP></SUB>+c<SUB><UP>m</UP></SUB>+c<SUB><UP>p</UP></SUB></NU><DE>(c<SUB><UP>m</UP></SUB>(h−&lgr;)+1)(c<SUB><UP>p</UP></SUB>(h−&lgr;)+1)</DE></FR>−c<SUP><UP>cis</UP></SUP><SUB><UP>0</UP></SUB></FENCE><SUP>2</SUP></FENCE></FENCE> (26)

<FENCE><FENCE>−(c<SUP><UP>cis</UP></SUP><SUB><UP>0</UP></SUB>)<SUP>2</SUP></FENCE>[1+(c<SUB><UP>m</UP></SUB>+c<SUB><UP>p</UP></SUB>)(h−&lgr;)+c<SUB><UP>m</UP></SUB>c<SUB><UP>p</UP></SUB>(h−&lgr;)<SUP>2</SUP>]+<FENCE><FENCE><FR><NU>2(h−&lgr;)c<SUB><UP>m</UP></SUB>c<SUB><UP>p</UP></SUB>−c<SUB><UP>m</UP></SUB>−c<SUB><UP>p</UP></SUB></NU><DE>(c<SUB><UP>m</UP></SUB>(h−&lgr;)−1)(c<SUB><UP>p</UP></SUB>(h−&lgr;)−1)</DE></FR>−c<SUP><UP>trans</UP></SUP><SUB><UP>0</UP></SUB></FENCE><SUP>2</SUP>−(c<SUP><UP>trans</UP></SUP><SUB><UP>0</UP></SUB>)<SUP>2</SUP></FENCE>[1−(c<SUB><UP>m</UP></SUB>+c<SUB><UP>p</UP></SUB>)(h−&lgr;)+c<SUB><UP>m</UP></SUB>c<SUB><UP>p</UP></SUB>(h−&lgr;)<SUP>2</SUP>]</FENCE>
in which c<UP><SUB>0</SUB><SUP>cis</SUP></UP> is the spontaneous curvature of the cis monolayer and c<UP><SUB>0</SUB><SUP>trans</SUP></UP> is the same for the trans monolayer.

The total free energy of the bilayer wings is given by the integral over the area of the reference surface from radius xp to bw:
W<SUB><UP>w</UP></SUB>=<LIM><OP>∫</OP></LIM>w<SUB><UP>w</UP></SUB>dA (27)
To estimate the dimple energy we assume that it has a spherical shape. Its radius is rd = x2/sin psi p, and hence the bending energy of two dimples is
W<SUB><UP>d</UP></SUB>=8&pgr;&kgr;(1−<UP>cos &psgr;<SUB>p</SUB></UP>)(1+r<SUB><UP>c</UP></SUB>c<SUP><UP>trans</UP></SUP><SUB><UP>0</UP></SUB>) (28)
Total bending energy of the hemi-fused bilayers is
W<SUB><UP>hb</UP></SUB>=W<SUB><UP>n</UP></SUB>+W<SUB><UP>w</UP></SUB>+W<SUB><UP>d</UP></SUB> (29)
For the sake of comparison with previous numerical results we assume that lambda  = h/2 = 1 nm, kappa  = 10 kT, and psi p = pi /6. Suppose that monolayers have no spontaneous curvature. Then the stress free stalk neck does not contribute any energy to Eq. 26. Bilayer wings contribute Ww = 2.8 kT and dimples have Wd = 33.7 kT. Therefore in total bending energy of Whb (36.5 kT) the main contribution comes from the dimples and constitutes approximately 92% of the total.

The presence of spontaneous curvature in monolayers helps to decrease the bending energy of hemifusion. For example, if c0 of both monolayers is -0.1 nm-1 and psi p = 30°, then Wn = -1.1 kT, Ww = -2.7 kT, Wd = 28.4 kT, and the total is only Whb = 24.6 kT.

The result strongly depends on where the monolayer peel-off occurs, i.e. what is the value of the peel-off angle. If it happens a little farther from the axis of the stalk then the bending energy drastically decreases. For example, if psi p = 15° and there is no spontaneous curvature then Ww = 0.1 kT, Wc = 8.4 kT, and the total is only Whb = 8.5 kT.

Finally, if the peel-off point occurs at psi p = 15° and spontaneous curvature is c0 = -0.1 nm-1, then Wn = -2.5 kT, Ww = -1.6 kT, Wc = 1.4 kT, and the total becomes negative Whb = -2.7 kT.

Therefore, the bending energy of the hemifused bilayers can be very low and even negative if the monolayers have rather pronounced spontaneous curvature. Fig. 5 A presents the bending energy of hemifusion as a function of peel-off angle for two types of bilayer: with no spontaneous curvature of monolayers and with c0 = -0.1 nm-1. As one can see, in the absence of spontaneous curvature the bending energy would be zero if psi p = 0°. This obviously means that the trans monolayer would prefer to remain planar. In the presence of spontaneous curvature c0 = -0.1 nm-1 the bending energy would reach a minimum at psi p = 5° and it would be negative at this configuration and equal to -8 kT.



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FIGURE 5   Free energy of the hemifusion as a function of peel-off angle psi p for two types of bilayer: with no spontaneous curvature and with c0 = -0.1 nm-1. (A) Bending energy only. (B) Total energy of hemifusion, gamma eff = 1.9 mN/m = 0.48 kBT/nm2.

The actual value of the peel off angle is determined by the minimum of the total free energy of the system including the bending energy and the energy of interstices at the stalk axis. It will be considered in the next section.

Hydrophobic voids

Siegel was the first to pay attention to the fact that at the ends of a monolayer stalk there should be void interstices (Markin and Hudspeth, 1993; Siegel, 1993) because three lipid monolayers meeting here cannot fit together without gaps. The shape of the interstices at the initial stage of hemifusion is presented in Fig. 6. The voids carry two additional contributions to the total energy of the system associated with their volume, V, and area, A:
W<SUB><UP>void</UP></SUB>=W<SUB><UP>void,V</UP></SUB>+W<SUB><UP>void,A</UP></SUB> (30)
Assuming that the void is nothing but vacuum, one can estimate the first term as
W<SUB><UP>void,V</UP></SUB>=P<SUB><UP>atm</UP></SUB>V. (31)
in which Patm is atmospheric pressure. However, 1 atm is equivalent to 0.025 kT/nm3. Therefore even if the void volume amounts to a few dozens of nm3 (and this is all we deal with in the stalk model), Wvoid,V barely reaches 1 kT and hence can be safely neglected.



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FIGURE 6   Shape of interstices at the initial stage of hemifusion.

The hydrophobic energy of the void was initially estimated by Siegel in 1993 (Siegel, 1993) and subsequently corrected in 1999 (Siegel, 1999). The estimate was based on the assumption that the hydrophobic interstice surface is equivalent to the surface of long chain alkanes with vacuum, which is known to be in the range 20 to 27 mN/m. However, using these numbers to find the energy of interstices in the HII phase one would come up with energy that is an order of magnitude higher than what is known for the HII phase (Siegel, 1993). The reason is that much of the "surface area" of the interstice lies at the rim of the interstice, where two lipid/vacuum interfaces would be less than 0.1 nm apart. In this case one cannot use the surface tension of the free surface lest the energy be strongly overestimated. There are two solutions to this dilemma. One might try to take into consideration the existence of the other hydrophobic surface in close proximity and calculate the energy of two surfaces as the function of the distance between them. However, this path involves rather poorly known functions and in the complex geometry of interstices these calculations could give a very approximate result. One can of course try to approximate the interstice geometry with a simple geometrical shape, like a cylinder (Kuzmin et al., 2001), but this approach completely neglects the interstices at the initial stage of hemifusion (Fig. 6).

A different approach was proposed in Siegel (1999) that was based on surface area scaling. Siegel suggested using an "effective surface tension" and found it to be gamma eff = 1.9 mN/m = 0.48 kBT/nm2. This approach seems quite reasonable and convenient in practical implementation. We shall use it below.

Based on the previous equations for the shape of trans and cis monolayers we calculated the area of voids Avoid and found their hydrophobic energy Wvoid = gamma eff Avoid. Total energy of the hemifusion is equal to
W<SUB><UP>h</UP></SUB>=W<SUB><UP>hb</UP></SUB>+W<SUB><UP>void</UP></SUB> (32)
and it is a function of the stalk radius as and peel-off angle psi p. Actual shape of the hemifusion structure is determined by interplay between bending energy and hydrophobic energy. At a given as angle psi p is determined by the minimum of the total energy.

Fig. 5 B presents the total energy of hemifusion as a function of peel-off angle for two cases: c0 = 0 and c0 = -0.1 nm-1. Both curves display a minimum. However, positions of these two minima are quite different. The bilayer with no spontaneous curvature has a minimum of 50 kT at psi p = 26.4°, whereas in the second case there is much lower minimum of 28.8 kT at psi p = 22.2°.

When the stalk enlarges, the energy of hemifusion as well as the peel-off angle change. Fig. 7 presents the free energy of hemifusion as a function of stalk radius as and peel-off angle psi p for different spontaneous curvatures of monolayers. Effective hydrophobic surface tension was assumed equal to gamma eff = 0.48 kBT/nm2. The lines connect the points with equal free energy of the system, and following the tradition of using Greek names for such types of curves (e.g., isochors and isobars), we shall call them isergons. (The term was proposed by A. J. Hudspeth.)



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FIGURE 7   Isergones, lines of equal free energy at the plane with coordinates stalk radius as-peel-off angle psi p. Effective hydrophobic surface tension gamma eff = 0.48 kBT/nm2. (A) c0 = 0. (B) c0 = -0.348 nm-1.

As one can see here the energy of the hemifusion structure quickly increases with its radius unless spontaneous curvature is very high as in Fig. 7 B where we took dioleoylphosphatidylethanolamine (DOPE) spontaneous curvature c0 = -0.348 nm-1. This is somewhat puzzling because this would virtually preclude the initial hemifusion structure from evolving into transmembrane contact as in Fig. 1 C and would be even less favorable for an extended trilaminar structure.

However, there is a possibility that the actual void hydrophobic energy can be drastically decreased. The void energy can be very large for pure bilayers of a single lipid component, but it will significantly decrease for bilayers formed from lipid mixtures or containing small amount of impurities, which may fill the voids. Siegel (1999) noted that in biological membranes, the voids forming around the stalk might have much lower energy because they could be filled up with some impurities always present in the membrane. Even minute fractions of apolar lipid like triglycerides and dolichol could lower the energy of TMCs by up to 50 to 80 kT. This might practically eliminate the hydrophobic energy of interstices, reducing the total energy of intermediate structures to the bending energy only.

But is there enough oil in the membrane to fill the hemifusion interstices? The equilibrium solubility of long chain alkanes or other long chain apolar oils in a bilayer is typically a few volume percent. For our estimate let us take the lower limit of 1% only. Then for the initial hemifusion structure with as = 1 nm, the necessary amount of oil would be found in two fusion bilayers inside the circle with radius of 3.13 nm. This hardly exceeds the bent portion of the bilayers involved in hemifusion. Therefore, enough oil can be squeezed just from the stalk proper to fill up the interstices. This result holds for a moderate enlargement of the hemifusion so that the hydrophobic void energy can be drastically decreased if not eliminated completely. To illustrate this we repeated the previous calculation for effective surface tension reduced to 1/5 of its tentative value and c0 = -0.1 nm-1. Then the initial hemifusion structure has less than 5 kT of energy; when the radius as increases from 1 to 2 nm, the energy grows only to 10 kT. In this case all the evolutions leading to stalk enlargement, to transmembrane contact, and to complete fusion become eminently possible.

Deformation of tilt

Another way to eliminate the voids in the hemifusion structure is to shift lipid molecules to the void space. In principle, lipid molecules can displace in the direction normal to the monolayer, and by doing so they could fill up the voids of hemifusion (Fig. 8). Of course this would happen at a cost. Normal displacement of lipid molecules represents a different kind of membrane deformation---deformation of tilt. During the last decade, the deformation of tilt attracted the attention of many researchers (Mackintosh and Lubensky, 1991; May, 2000; Hamm and Kozlov, 2000 and references therein). Kuzmin et al. (2001) suggested that this deformation can occur during membrane fusion and hence should be taken into consideration. Let us estimate the energy cost that could be expected for this deformation and if it will be compensated by the release of hydrophobic energy after void collapse.



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FIGURE 8   Tilt deformation in HII phase. (A) Cross-section of the void space between three lipid cylinders. (B) Collapse of the void space.

As a specific example let us consider the interstices in the HII phase (Fig. 8 A) formed by DOPE (Siegel, 1993; Fuller and Rand, 2001). Assuming that the lipidic phase has the shape of circular cylinders Siegel called the hydrophobic voids between them trilaterally symmetric voids. In cross section (Fig. 8 A) these voids have the shape of curvilinear triangles. Siegel estimated the energy required to produce a unit length of such void configuration equal to 41.6 pN or 5 kT/nm.

Is it possible that the void space between three HII cylinders would collapse, i.e., that the three monolayers in Fig. 8 A deform in such a way that they would fill up this void space? To answer this question one has to compare the energy of the void WTV including the hydrophobic energy of the tails and bending energy of initial monolayers with deformation energy of the collapsed void Wcollapse. The system will acquire configurations with lower energy.

Deformation of tilt can be described by parameter of tilt that in a simple case can be visualized as an angle theta  between the axis of lipid molecule and the normal to the monolayer. Departure of this angle from 0° exposes a portion of hydrophobic surfaces of lipid molecules and hence increases the energy of the monolayer. To take this into consideration one has to add to the bending energy (Eq. 1) an additional term with tilt energy 1/2kappa theta theta 2. In line with Eq. 1, one can also include into this expression a certain spontaneous value of tilt theta 0, changing it to 1/2kappa theta (theta  - theta 0)2. However, here we limit ourselves to the simplest form of the tilt energy.

Collapse of the void in Fig. 8 means that the upper monolayer CB shifts to position AB and becomes planar. The other monolayers lining the void are transformed in a similar way. The orientation of lipid molecules---tilt angle theta (x)---in the new, planar configuration of the monolayer between A and B changes from pi /6 to 0:
&thgr;<SUB><UP>A</UP></SUB>=<FR><NU>&pgr;</NU><DE>6</DE></FR><UP> and </UP>&thgr;<SUB><UP>B</UP></SUB>=0. (33)
Let us find the deformation energy of planar monolayers in the collapsed configuration. At first glance it might seem counterintuitive but the planar monolayer AB has not only tilting but also bending energy. This statement deserves some additional explanation although it was already presented in Hamm and Kozlov (2000) and Kuzmin et al. (2001).

What is the nature of the bending deformation? It is actually the change of the shape of lipid molecules. Bending results in different changes of the cross section (not simply the size) of heads and tails of lipid molecules. The curvature of the initial monolayer CB is negative so that lipid heads are compressed and tails are expanded. Axes of lipid molecules have different orientation in space, although they remain normal to the monolayer surface: there is no tilt. Now let as perform a gedanken experiment: transform monolayer CB into monolayer AB in such a way that all orientations of lipid molecules are preserved. In this transformation the lipid molecules slide along each other (along their axis) without changing their molecular shapes. Therefore, the bending energy determined by the shape of the molecules is preserved. Now from comparison of molecular orientation of the monolayers CB and AB it is not difficult to figure out that in the new, planar configuration AB the role of curvature c is played by the derivative dtheta /dx. Transition from CB to AC involves a certain work due to exposure of lateral surfaces of lipid molecules to aqueous surrounding, which is proportional to theta 2. The density of deformation energy in the collapsed monolayers can be presented as
w<SUB><UP>def</UP></SUB>(x)=<FR><NU>1</NU><DE>2</DE></FR> &kgr;<FENCE><FR><NU>d&thgr;</NU><DE>dx</DE></FR>−c<SUB>0</SUB></FENCE><SUP>2</SUP>+<FR><NU>1</NU><DE>2</DE></FR> &kgr;<SUB>&thgr;</SUB>&thgr;<SUP>2</SUP> (34)
Here, as before, kappa  is the bending modulus, whereas kappa theta is the tilt modulus.

The total deformation energy of the collapsed void per unit length is
W<SUB><UP>collapse</UP></SUB>=6<LIM><OP>∫</OP><LL><UP>A</UP></LL><UL><UP>B</UP></UL></LIM>w<SUB><UP>def</UP></SUB>(x)dx, (35)
where coefficient 6 accounts for six equivalent parts of the total perimeter of the void cross section.

Of course orientation of lipid molecules in the planar monolayer AB will not remain the same as in the original monolayer CB. It will relax to the new positions theta (x) to minimize the energy of the total monolayer. According to calculus of variations, the function theta (x) that minimizes the integral (Eq. 35) is determined by the Euler equation
&kgr; <FR><NU>d<SUP>2</SUP>&thgr;</NU><DE>dx<SUP>2</SUP></DE></FR>−&kgr;<SUB>&thgr;</SUB>&thgr;=0 (36)
As one can see from the dimension analysis, this equation defines a characteristic constant of length,
&tgr;=<RAD><RCD><FR><NU>&kgr;</NU><DE>&kgr;<SUB>&thgr;</SUB></DE></FR></RCD></RAD>, (37)
that gives the range of the monolayer where tilting and bending of lipid molecules occur. Beyond this range the planar monolayer is free of both tilting and bending. Notice that spontaneous curvature c0 that was present in the deformation energy Eq. (34) disappeared from the Euler Eq. 36. Boundary conditions for this differential equation are given by Eq. 33.

Solution of the Euler Eq. 36 is
&thgr;(x)=<FR><NU>&thgr;<SUB><UP>A</UP></SUB><UP> sinh</UP>[(l−x)/&tgr;]</NU><DE><UP>sinh</UP>[l/&tgr;]</DE></FR> (38)
in which l is the distance between points A and B and deformation energy is
W<SUB><UP>collapse</UP></SUB>=6<FENCE><FR><NU>1</NU><DE>2</DE></FR> &thgr;<SUP><UP>2</UP></SUP><SUB><UP>A</UP></SUB><RAD><RCD>&kgr;&kgr;<SUB>&thgr;</SUB></RCD></RAD> <UP>coth </UP><FR><NU>l</NU><DE>&tgr;</DE></FR>+&kgr;c<SUB>0</SUB>&thgr;<SUB><UP>A</UP></SUB>+<FR><NU>1</NU><DE>2</DE></FR> &kgr;c<SUP>2</SUP><SUB>0</SUB>l</FENCE> (39)
To estimate the collapse energy Wcollapse one needs to know the value of the tilt modulus kappa theta . Unfortunately there are no direct experimental measurements of this material constant. In the absence of direct measurements Hamm and Kozlov (2000) estimated it at 40 mN/m, whereas Kuzmin et al. (2001) came up with 33 mN/m. Although these estimates were based on different models, the results are rather close and can be summarized as 9 kT/nm2. That gives the characteristic length tau  = 1.05 nm.

The other parameters can be taken from Fuller and Rand (2001): Rh = 4 nm, RN = 2.9 nm, RW = 2.18 nm, lambda  = 0.73 nm, and c0 = -1/RN. Then the energy of collapsed void is 4.1 kT/nm. This result is not very different from the open void energy of 5 kT/nm found by Siegel (1999); therefore, these two configurations are energetically virtually equivalent.

Equation 39 can be used for estimation of the energy of the trilaminar structure that could appear after further expansion of the stalk. Using the fact that the open void has approximately the same energy as the collapsed one, we shall use this equation for our estimation. Notice that the last term in Eq. 39 represents the energy of the monolayer in the initial, planar state. If we refer the energy of the trilaminar structure to the planar state, then this amount should be subtracted from the final energy. Besides, we assume that the length l is large in the characteristic scale of the system, i.e., it considerably exceeds 1 nm. Then the energy of the unit of length of the trilaminar contact is given by
f<SUB><UP>tri</UP></SUB>=6<FENCE>½ &thgr;<SUP><UP>2</UP></SUP><SUB><UP>A</UP></SUB><RAD><RCD>&kgr;&kgr;<SUB>&thgr;</SUB></RCD></RAD>+&kgr;c<SUB>0</SUB>&thgr;<SUB><UP>A</UP></SUB></FENCE> (40)
The physical meaning of this quantity is a linear tension, which is why the designation f is used here.

To estimate the linear tension of the trilaminar contact, consider bilayers without spontaneous curvature. Then ftri = 7.8 kT/nm. The presence of negative spontaneous curvature noticeably decreases this tension; for example, if c0 = -0.1 nm-1 then ftri = 4.7 kT/nm. This linear tension can even become zero, if c0= -(theta A/2 )<RAD><RCD> &kgr;<SUB>&thgr;</SUB>&kgr;</RCD></RAD> = -theta A/2tau . Numerically it gives c0 = -0.25 nm-1, which is not very far from the spontaneous curvature of DOPE that readily forms the HII phase.

In the absence of spontaneous curvature there is only one term, 1/2 theta <UP><SUB>A</SUB><SUP>2</SUP></UP><RAD><RCD>&kgr;&kgr;<SUB>&thgr;</SUB></RCD></RAD> , left in the braces of Eq. 40. This term represents both bending and tilting energy, which give equal contributions to the total trilaminar energy. This can be demonstrated in the following way. Angle theta  exponentially decreases with coordinate: theta (x)=theta A exp (-x/tau ) . Therefore, the tilting energy changes as wtilt (x)=(kappa theta theta <UP><SUB>A</SUB><SUP>2</SUP></UP>/2) exp (-2x/tau ) . After integration it becomes 1/4 kappa theta tau theta <UP><SUB>A</SUB><SUP>2</SUP></UP> = 1/4 <RAD><RCD><IT>&kgr;&kgr;<SUB>&thgr;</SUB></IT></RCD></RAD> . This is exactly one-half of the total energy of thetrilaminar structure. The other one-half is contributed by the bending energy.

Complete fusion-fusion pore

The next step after formation of the hemifusion structure is rupture of the trans monolayers and completion of the fusion process (Fig. 1 D). At this moment a fusion pore is formed, and the fusion process is completed. Here we analyze the energetic cost of the fusion pore and of its consequent enlargement. In the previous sections we have considered the free energy associated with bilayer wings included in the hemifusion structure. It was found that its energy contribution was insignificant. However, this does not mean that completely fused bilayers have negligible energy. The reason is that the bending energy is concentrated at the neck of the fusion pore.

In the literature there are different approaches to the calculation of this energy. In the simplest one (Markin et al., 1984; Chizmadzhev et al., 1995) the bilayer is visualized as a single layer with bending rigidity equal to the sum of the bending rigidity of the monolayers. Originally this model was called a bilayer stalk and it resulted in the same equations that were derived from a monolayer stalk. In the old model the minimal energy of such a bilayer stalk with bending rigidity 10-19 J was found to be 90 kT (Chizmadzhev et al., 1995). However, this is a gross underestimation because two monolayers are bent quite independently, and their elastic energy should be calculated separately. If this were done, then this value would increase to 150 kT.

So the old model predicts very high elastic energy for a fusion pore due to the assumption of a circular shape for its neutral surface. In the present approach the shape of the neutral surface is calculated with constant total curvature. As we have seen above, for a single monolayer this results in the stress-free stalk. However, this result cannot be extended to a bilayer because its monolayers cannot be made stress free simultaneously. However, their energy can be made much lower than in the old model.

We calculate the bending energy of the bilayer using the equations derived in the Hemifusion section. We select the interface between two monolayers as a reference surface (Fig. 1 D). Its principal curvatures cm and cp are defined the same way as for a monolayer stalk by Eq. 5. The principal curvatures of the cis and trans monolayer are defined at their respective neutral surfaces and are presented by Eqs. 21 and 22. As we mentioned before these definitions are consistent for both monolayers: if the monolayer in a given principal plane is convex, then the given curvature is positive and vice versa. However, two monolayers are bent in the opposite directions. For an example, let us consider parallel curvatures at the equatorial plane. According to our convention, the bilayer parallel curvature cp here is positive, and the radius of the bilayer parallel curvature obviously exceeds the monolayer thickness: Rp = 1/cp > h. From Eqs. 21 and 22 one finds that c<UP><SUB>p</SUB><SUP>cis</SUP></UP> > 0 and c<UP><SUB>p</SUB><SUP>trans</SUP></UP> < 0, just confirming that two monolayers are bent in the opposite directions and determining the sign of their curvature.

Now let us see how the curvature of both monolayers varies along the bilayer. We shall look for the solution where the reference surface of the bilayer has a constant total curvature cm + cp cb and the radius of its "waist" is ab. Then the principle curvatures of the bilayer are
c<SUB><UP>p</UP></SUB>=<FR><NU>1</NU><DE>2</DE></FR> c<SUP><UP>b</UP></SUP>+<FR><NU>a<SUB><UP>b</UP></SUB></NU><DE>x<SUP>2</SUP></DE></FR> <FENCE>1−<FR><NU>1</NU><DE>2</DE></FR> a<SUB><UP>b</UP></SUB>c<SUP><UP>b</UP></SUP></FENCE>
and
c<SUB><UP>m</UP></SUB>=<FR><NU>1</NU><DE>2</DE></FR> c<SUP><UP>b</UP></SUP>−<FR><NU>a<SUB><UP>b</UP></SUB></NU><DE>x<SUP>2</SUP></DE></FR> <FENCE>1−<FR><NU>1</NU><DE>2</DE></FR> a<SUB><UP>b</UP></SUB>c<SUP><UP>b</UP></SUP></FENCE> (41)
These equations are similar to the stalk equations, but here it is explicitly stressed that the radius of the bilayer waist is ab. As one can see, the sum of these two curvatures (total curvature, ctot = cp + cm) equals cb, but for separate monolayers this sum is not a constant but rather varies along the bilayer. Fig. 9 illustrates how the total curvature of trans and cis monolayers varies with x for different values cb equal correspondingly to -0.1, -0.01, and -0.001 nm-1. A horizontal line in each panel presents the total curvature of the bilayer reference surface.