The outer leaflet of a Gram-negative bacterial membrane carries a signed surface…

PHYSICS · Multi-part · Fermi University Physics · Biophysics & Medical Physics · Macromolecules, membranes and cell architecture · Macromolecules, membranes and cell architecture

Forces, Machines and Signals, Ch Charges in salt water and the assembly of amphiphiles

Problem

The outer leaflet of a Gram-negative bacterial membrane carries a signed surface charge density σ=−0.050 C m−2\sigma = -0.050\,\mathrm{C\,m^{-2}}, negative because it arises from anionic lipid head groups, and sits in a monovalent salt solution of concentration 100 mM100\,\mathrm{mM} at 298 K298\,\mathrm{K} (water's relative permittivity εr=78.4\varepsilon_r=78.4). ((a)) Starting from the one-dimensional Poisson-Boltzmann equation ϕ′′=κ2sinh⁡ϕ\phi'' = \kappa^2\sinh\phi for the dimensionless potential ϕ(x)=eψ(x)/kBT\phi(x)=e\psi(x)/k_BT measured from the charged plane at x=0x=0 out into the salt solution, multiply through by ϕ′\phi' and integrate once, using ϕ→0\phi\to0 and ϕ′→0\phi'\to0 as x→∞x\to\infty, to obtain the first integral 12(ϕ′)2=κ2(cosh⁡ϕ−1)\tfrac12(\phi')^2 = \kappa^2(\cosh\phi-1). ((b)) Combine this first integral with Gauss's law applied at the surface, −ε0εrψ′(0)=σ-\varepsilon_0\varepsilon_r\psi'(0)=\sigma, to derive the Grahame equation σ=8ε0εrc∞kBT sinh⁡ ⁣(eψ02kBT),\sigma = \sqrt{8\varepsilon_0\varepsilon_rc_\infty k_BT}\,\sinh\!\left(\frac{e\psi_0}{2k_BT}\right), where ψ0=ψ(0)\psi_0=\psi(0) is the surface potential. ((c)) Use the Grahame equation to find the surface potential of the bacterial membrane above, stating its sign.

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