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Chapter 11: Problem 9

In the chapter, we considered an experiment in which a tubule was pulled from a giant unilamellar vesicle and then dynamin was added into the solution. The dependence of nucleation on the radius was inferred by appealing to the force measured in the trap as shown in Figure \(11.36 .\) Here we rederive this force. (a) Write a free energy for the tubule in terms of the bending energy and the tension. (b) Find the force due to the bending energy and the surface tension by evaluating \(f=-a G_{\text {tubule }} / \partial L\) (c) Now consider the effect of the optical trap with stiffness \(k_{\text {trap }}\) on the free energy of the tubule/bead system. Assume that the equilibrium position of the bead corresponds to a tubule of length \(L_{0}\) and write down the expression for the total free energy that includes the free energy of the tubule and the energy of the bead in the trap. Find the equilibrium value of the tubule length \(L^{*}\), and show that it is a result of balancing the tubule force and the force applied by the optical trap.

Short Answer

Expert verified
The free energy of the tubule is \( G_{\text{tubule}} = \kappa \pi R L + 2 \gamma \pi R L \). The force due to bending energy and surface tension is \( f = - a (\kappa \pi R + 2 \gamma \pi R) \). The total free energy of the system \( G_{\text{total}} = \kappa \pi R L + 2 \gamma \pi R L + \frac{1}{2}k_{\text{trap}}(L - L_0)^2 \) depends on the equilibrium tubule length \( L^* \), which is found when the tubule force is balanced with the force applied by the optical trap.

Step by step solution

01

Deriving the Free Energy Expression

Let's denote the length of the tubule by \( L \), surface area by \( A \), and the radius by \( R \). The expression for the free energy of the tubule \( G_{\text{tubule}} \) will be given by the sum of the bending energy \( E_B = \kappa A/2 \) and the tension \( T = \gamma A \). Therefore, \( G_{\text{tubule}} = E_B + T = \frac{\kappa A}{2} + \gamma A \) where \( A = 2 \pi R L \). Substituting \( A \) into the equation for \( G_{\text{tubule}} \) and simplifying, we get \( G_{\text{tubule}} = \kappa \pi R L + 2 \gamma \pi R L \).
02

Evaluating the Force due to Bending Energy and Surface Tension

The force \( f \) is defined as \( f = - a \frac{\partial G_{\text{tubule}}}{\partial L} \). Differentiating \( G_{\text{tubule}} \) with respect to \( L \), we get \(\frac{\partial G_{\text{tubule}}}{\partial L} = \kappa \pi R + 2 \gamma \pi R\). Substituting this result into the equation for \( f \), we have \( f = - a (\kappa \pi R + 2 \gamma \pi R) \).
03

Considering the Effect of the Optical Trap

The energy of the bead in the trap can be expressed with \( E_{\text{trap}} = \frac{1}{2}k_{\text{trap}}(L - L_0)^2 \), where \( L_0 \) is the equilibrium tubule length. Therefore, the total free energy \( G_{\text{total}} = G_{\text{tubule}} + E_{\text{trap}} = \kappa \pi R L + 2 \gamma \pi R L + \frac{1}{2}k_{\text{trap}}(L - L_0)^2 \). To find the equilibrium value of tubule length \( L^* \), we differentiate \( G_{\text{total}} \) with respect to \( L \) and equate it to zero, implying that the tubule force and the force applied by the optical trap are at equilibrium.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Free Energy
In the context of biophysics, free energy serves as an important concept when investigating cellular structures. Free energy is essentially the energy in a system that can be used to perform work. It combines the internal energy of a system and its entropy, giving insight on stability and likelihood of transformations.

For a tubule, the calculation of free energy includes contributions from both bending energy and surface tension. Bending energy is the energy required to deform a structure. This is particularly important in structures like membranes and vesicles because such deformations can affect biological processes.

Mathematically, the free energy of the tubule can be defined as:
  • Bending Energy: \( E_B = \frac{\kappa A}{2} \) where \( \kappa \) is bending stiffness, and \( A \) is the area.
  • Tension: \( T = \gamma A \) with \( \gamma \) as surface tension.
Consequently, the total free energy combining these contributions is\[ G_{\text{tubule}} = \kappa \pi R L + 2 \gamma \pi R L \].

In practice, understanding free energy helps in explaining how physical structures adapt to changes and maintain functionality in a biological environment.
Bending Energy
Bending energy is fundamental when analyzing how structures like vesicles change their shape. It represents the cost, in energy terms, to change the curvature of a membrane.

In our case, the tubule's bending energy contributes directly to its total free energy. When a membrane bends, the molecular arrangement changes, requiring or releasing additional energy to accommodate for shape changes.

This energy, represented as\[ E_B = \frac{\kappa A}{2} \]is dictated by the bending stiffness of the membrane, \( \kappa \), which measures how resistant a structure is to bending. Moreover, the surface area \( A \), vital in this calculation, reflects how the structure's size influences the energy needed for bending.

Bending energy impacts cellular processes like the formation of vesicles and other curved structures. As bending involves transitioning from one structural state to another, the energetic contributions must be understood to appreciate how cells manage shape alterations with minimal energy expenditure.
Optical Trap
An optical trap, often called "optical tweezers," leverages laser light to manipulate microscopic particles with precision. This tool is particularly useful in experiments involving biological macromolecules and cellular components.

Optical traps work by exerting forces using focused laser beams to hold and move cellular or sub-cellular structures. In the experiment described, an optical trap assists in understanding how the tubule's length, \( L \), balances against mechanical forces.

The free energy concerning the trap includes two primary components:
  • The structure's energy: \( G_{\text{tubule}} \)
  • The potential energy of the optical trap: \( E_{\text{trap}} = \frac{1}{2}k_{\text{trap}}(L - L_0)^2 \)
Here, \( L_0 \) denotes the equilibrium length, where the balance of forces is achieved, and \( k_{\text{trap}} \) represents the trap's stiffness, indicator of how strongly the trap resists deviations from equilibrium.

By integrating the bending energy with the trap's force, researchers can determine why and how structures stabilize at particular lengths, providing insights into the mechanical stability of cellular structures under various conditions.

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Most popular questions from this chapter

The height function used in the chapter can be used to compute the change in area of a membrane when it suffers some deformation. Consider the height function \(h(x, y)\) defined over an \(L \times L\) square in the \(x-y\) plane. The function \(h(x, y)\) tells us the deformed state of the \(L \times L\) square. To compute the change in area, we divide the square into small \(\Delta x \times \Delta y\) squares and consider how each of them is deformed on the surface \(h(x, y)\) (see Figure 11.17 A). The function \(h(x, y)\) maps the small square into a tilted parallelogram. (a) Show that the two sides of the parallelogram are given by the vectors \\[ \Delta x\left(\mathbf{e}_{x}+\frac{\partial h}{\partial x} \mathbf{e}_{z}\right) \text { and } \Delta y\left(\mathbf{e}_{y}+\frac{\partial h}{\partial y} \mathbf{e}_{z}\right) \\] where \(\mathbf{e}_{x}, \mathbf{e}_{y},\) and \(\mathbf{e}_{z}\) are unit vectors. (b) The magnitude of the vector product of two vectors is equal to the area subsumed by them, that is, a \(x \mathbf{b}=a b \sin \theta\) where \(\theta\) is the angle between vectors a and b. Using this fact, show that the area of the parallelogram is \\[ d^{\prime}=\sqrt{1+\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}} \Delta x \Delta y \\] (c) Now assume that the partial derivatives appearing in the above equation are small and expand the square-root function using \(\sqrt{1+x^{2}} \approx 1+\frac{1}{2} x^{2},\) to demonstrate the useful formula for the area change, \\[ \Delta a=a^{\prime}-a=\frac{1}{2}\left[\left(\frac{\partial h}{\partial x}\right)^{2}+\left(\frac{\partial h}{\partial y}\right)^{2}\right] \Delta x \Delta y \\] By integrating this expression over the entire domain, we can find the total area change associated with the surface. Divide both sides by \(a=\Delta x \times \Delta y\) to find a formula for the relative change of area \(\Delta a / a\).

Consider a two-dimensional generalization of Equation 11.36 (a) Show that the free energy associated with membrane deformation may be written as \\[ G_{\text {hydrophobic }}=\frac{K_{\mathrm{b}}}{2} \int\left(\nabla^{2} u\right)^{2} \mathrm{d}^{2} r+\frac{K_{\mathrm{t}}}{2} \int\left(\frac{u}{a}\right)^{2} \mathrm{d}^{2} r \\] To this add the change in free energy due to the change in tension, namely, \\[ G_{\text {tension }}=G_{0}-r \pi R^{2} \\] where \(G_{0}\) is a constant that sets the reference value for the energy of the loading device. Explain why the free-energy expression for the composite system of channel and loading device can now be written as \\[ G_{\mathrm{MscL}}=G_{0}-\tau \pi R^{2}+\frac{K_{\mathrm{b}}}{2} \int\left(\nabla^{2} u\right)^{2} \mathrm{d}^{2} r+\frac{K_{\mathrm{t}}}{2} \int\left(\frac{u}{a}\right)^{2} \mathrm{d}^{2} r \\] To compute this free energy explicitly, we need to know the profile of the field \(u(\mathbf{r})\) around the membrane. (b) Find the Euler-Lagrange equation for this free-energy functional. (c) Solve the partial differential equation for the deformation profile due to the membrane protein and imitate the various steps leading up to Equation 11.58 , but now done for the full two-dimensional deformation profile. (d) Compare this solution with that of the one-dimensional approximation given in the chapter.

The pipette aspiration experiment described in the chapter can be used to measure the bending modulus \(K_{\mathrm{b}}\) as well as the area stretch modulus. Lipid bilayer membranes are constantly jostled about by thermal fluctuations. Even though a flat membrane is the lowest-energy state, fluctuations will cause the membrane to spontaneously bend. The goal of this problem is to use equilibrium statistical mechanics to predict the nature of bending fluctuations and to use this understanding as the basis of experimental measurement of the bending modulus. (Note: This problem is challenging and the reader is asked to consult the hints on the book's website to learn more of our Fourier transform conventions, how to handle the relevant delta functions, the subtleties associated with the limits of integration, etc.). (a) Write the total free energy of the membrane as an integral over the area of the membrane. Your result should have a contribution from membrane bending and a contribution from membrane tension. Write your result using the function \(h(\mathbf{r})\) to characterize the height of the membrane at position \(\mathbf{r}\) (b) The free energy can be rewritten using a decomposition of the membrane profile into Fourier modes. Our Fourier transform convention is $$h(\mathbf{r})=\frac{A}{(2 \pi)^{2}} \int h(\mathbf{q}) \mathrm{e}^{-\mathrm{i} \mathbf{q} \cdot \mathbf{r}} \mathrm{d}^{2} \mathbf{q}$$, where \(A=L^{2}\) is the area of the patch of membrane of interest. Plug this version of \(h(\mathbf{r})\) into the total energy you derived above (that is, bending energy and the energy related to tension) to convert this energy in real space to an energy in \(q\) -space. Note that the height field in \(q\) -space looks like a sum of harmonic oscillators. (c) Use the equipartition theorem in the form \((E(\mathbf{q}))=k_{\mathrm{B}} T / 2,\) where \(E(\mathbf{q})\) is the energy of the \(q\) th mode and the free energy can now be written as $$F[\tilde{h}(\mathbf{q})]=\frac{A}{(2 \pi)^{2}} \int E(\mathbf{q}) \mathrm{d}^{2} \mathbf{q}$$. Use this result to solve for \(\left(|\tilde{h}(\mathbf{q})|^{2}\right\rangle,\) which will be used in the remainder of the problem. (d) We now have all the pieces in place to compute the relation between tension and area and thereby the bending modulus. The difference between the actual area and the projected area is $$A_{\mathrm{act}}-A=\frac{1}{2} \int[\nabla h(\mathbf{r})]^{2} \mathrm{d}^{2} \mathbf{r}$$. This result can be rewritten in Fourier space as \\[ \left(A_{\mathrm{act}}-A\right)=\frac{1}{2} \frac{A^{2}}{(2 \pi)^{2}} \int_{\pi / \sqrt{A}}^{\pi / \sqrt{a_{0}}} q^{2}\left(|\hbar(\mathbf{q})|^{2}\right) 2 \pi q \mathrm{d} q \\]. Work out the resulting integral, which relates the areal strain to the bending modulus, membrane size, temperature, and tension. The limits of integration are set by the overall size of the membrane (characterized by the area \(A\) and the spacing between lipids \(\left(a_{0}\) is the area per \right. lipid), respectively. This result can be directly applied to micropipette experiments to measure the bending modulus. In particular, using characteristic values for the parameters appearing in the problem suggested by Boal (2002) such as \(\mathrm{r} \approx 10^{-4} \mathrm{J} / \mathrm{m}^{2}, K_{\mathrm{b}} \approx 10^{-19} \mathrm{J},\) and \(a_{0} \approx 10^{-20} \mathrm{m}^{2},\) show that $$\frac{\Delta A}{A} \approx \frac{k_{\mathrm{B}} T}{8 \pi K_{\mathrm{b}}} \ln \left(\frac{A \tau}{K_{\mathrm{b}} \pi^{2}}\right)$$, and describe how this result can be used to measure the bending modulus. (For an explicit comparison with data, see Rawicz et al. (2000) . For further details on the analysis, see Helfrich and Servuss \((1984) .\) An excellent account of the entire story covered by this problem can be found in Chapter \(6 \text { of Boal }(2002) .)\)

Consider the function \(h\left(x_{1}, x_{2}\right)=x_{1}^{2}+x_{1} x_{2}-2 x_{2}^{2},\) which we assume describes the shape of a deformed lipid bilayer membrane. As shown in Figure \(11.14, x_{1}\) and \(x_{2}\) are the coordinates of the reference plane below the membrane. (a) Make a plot of the height as a function of \(x_{1}\) and \(x_{2}\) (b) Compute the principal radii of curvature as functions of \(x_{1}\) and \(x_{2}\) (c) Compute the bending free energy for the piece of membrane corresponding to the square \(0 \leq x_{1} \leq 1\) and \(0 \leq x_{2} \leq 1\) in the reference plane.
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