The Hodgkin-Huxley potassium activation gate obeys the two-state kinetics

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Biophysics & Medical Physics · Bioelectricity, the senses and the physics of the body

Forces, Machines and Signals, Ch The nerve impulse

Problem

The Hodgkin-Huxley potassium activation gate nn obeys the two-state kinetics n˙=αn(V)(1−n)−βn(V)n\dot n = \alpha_n(V)(1-n) - \beta_n(V)n, with rate constants (in ms−1\mathrm{ms^{-1}}, VV in mV, modern sign convention with rest near −65 mV-65\,\mathrm{mV}): αn(V)=0.01(V+55)1−exp⁡[−(V+55)/10],βn(V)=0.125exp⁡ ⁣[−V+6580].\alpha_n(V) = \frac{0.01(V+55)}{1-\exp[-(V+55)/10]}, \qquad \beta_n(V) = 0.125\exp\!\left[-\frac{V+65}{80}\right]. ((a)) Compute the steady-state occupancy n∞(V)=αn/(αn+βn)n_\infty(V) = \alpha_n/(\alpha_n+\beta_n) and the relaxation time τn(V)=1/(αn+βn)\tau_n(V) = 1/(\alpha_n+\beta_n) at the resting potential V=−65 mVV=-65\,\mathrm{mV} and at V=−20 mVV=-20\,\mathrm{mV}, a potential reached during the rising phase of a spike. State in one sentence what n∞n_\infty and τn\tau_n mean for the two-state gate. ((b)) In the full Hodgkin-Huxley squid model at 6.3 ∘C6.3\,{}^{\circ}\mathrm{C}, the resting state loses stability at a Hopf bifurcation near an injected current of I=9.8 μA cm−2I=9.8\,\mathrm{\mu A\,cm^{-2}}, while a stable large-amplitude firing limit cycle already exists for currents from about I=6.2 μA cm−2I=6.2\,\mathrm{\mu A\,cm^{-2}} (a fold of limit cycles at that current). Someone argues: "the axon can only start firing repetitively once the injected current has passed 9.8 μA cm−29.8\,\mathrm{\mu A\,cm^{-2}}, since that is where rest becomes unstable." Using the geometry of a subcritical Hopf bifurcation with a fold of limit cycles, explain why this argument is not generally correct, and describe what would be seen if the current were raised quasi-statically from zero to 12 μA cm−212\,\mathrm{\mu A\,cm^{-2}} and then lowered back to zero.

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