Three unrelated channels in the same patch of membrane are studied with tools built…

PHYSICS · Multi-part · Fermi University Physics · Biophysics & Medical Physics · Molecular machines, transport and bioenergetics · Bioelectricity, the senses and the physics of the body

Forces, Machines and Signals, Ch Channels, pumps and the resting potential

Problem

Three unrelated channels in the same patch of membrane are studied with tools built in earlier chapters as well as this one. ((a)) A calcium-selective channel (z=+2z=+2) has ci=100 nMc_i = 100\,\mathrm{nM} and co=2 mMc_o = 2\,\mathrm{mM} at 310 K310\,\mathrm{K} (kBT/e=26.7 mVk_BT/e = 26.7\,\mathrm{mV}). Using the zero-flux condition of the Nernst-Planck equation (built from the drift-diffusion flux J=−D dc/dx−(Dze/kBT)c dψ/dxJ=-D\,\mathrm{d}c/\mathrm{d}x - (Dze/k_BT)c\,\mathrm{d}\psi/\mathrm{d}x and the Einstein relation, both from Chapter 2), find the Nernst potential ECaE_{\rm Ca}. Then show, from the same expression, that the diffusive term −D dc/dx-D\,\mathrm{d}c/\mathrm{d}x and the drift term −(Dze/kBT)c dψ/dx-(Dze/k_BT)c\,\mathrm{d}\psi/\mathrm{d}x are individually non-zero and equal in magnitude but opposite in sign at V=ECaV=E_{\rm Ca}, so long as the potential varies across the membrane. ((b)) A patch clamp resolves a two-state voltage-gated channel by the dwell-time method of Chapter 12. At V1=−40 mVV_1=-40\,\mathrm{mV} the mean open and closed times are τo1=8 ms\tau_{o1}=8\,\mathrm{ms} and τc1=24 ms\tau_{c1}=24\,\mathrm{ms}; at V2=+10 mVV_2=+10\,\mathrm{mV} they are τo2=24 ms\tau_{o2}=24\,\mathrm{ms} and τc2=8 ms\tau_{c2}=8\,\mathrm{ms}. Using this chapter's voltage-gating curve, built from Chapter 6's two-state tilt, in the form Po(V)=1/(1+e−zappe(V−V1/2)/kBT)P_o(V) = 1/(1+e^{-z_{\rm app}e(V-V_{1/2})/k_BT}), extract the apparent gating valence zappz_{\rm app} at T=298 KT=298\,\mathrm{K} (kBT/e=25.7 mVk_BT/e = 25.7\,\mathrm{mV}) from the two open probabilities. ((c)) A mechanosensitive channel in the same patch has an area change on opening ΔA=15 nm2\Delta A = 15\,\mathrm{nm^2} and a midpoint tension σ1/2=8 mN m−1\sigma_{1/2}=8\,\mathrm{mN\,m^{-1}} at 298 K298\,\mathrm{K} (kBT=4.11 pN nmk_BT = 4.11\,\mathrm{pN\,nm}). Using this chapter's tension-gating balance, find the gating free energy ΔG0\Delta G_0 in units of kBTk_BT, and predict the new midpoint tension if a mutation halves ΔA\Delta A while leaving ΔG0\Delta G_0 unchanged.

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