6.021/Notes/2006-10-27: Difference between revisions
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* <math>G_{K}(V_m,t) = \overline{G_K} n^4(V_m,t)</math> where | * <math>G_{K}(V_m,t) = \overline{G_K} n^4(V_m,t)</math> where | ||
** <math>n(V_m,t) + \tau_n(V_m)\frac{dn(V_m,t)}{dt} = n_\infty(V_m)</math> | ** <math>n(V_m,t) + \tau_n(V_m)\frac{dn(V_m,t)}{dt} = n_\infty(V_m)</math> | ||
* <math>G_{Na}(V_m,t) = \overline{ | * <math>G_{Na}(V_m,t) = \overline{G_{Na}} m^3(V_m,t)h(V_m,t)</math> where | ||
** <math>m(V_m,t) + \tau_m(V_m)\frac{dm(V_m,t)}{dt} = m_\infty(V_m)</math> | ** <math>m(V_m,t) + \tau_m(V_m)\frac{dm(V_m,t)}{dt} = m_\infty(V_m)</math> | ||
** <math>h(V_m,t) + \tau_h(V_m)\frac{dh(V_m,t)}{dt} = h_\infty(V_m)</math> | ** <math>h(V_m,t) + \tau_h(V_m)\frac{dh(V_m,t)}{dt} = h_\infty(V_m)</math> |
Latest revision as of 09:43, 14 November 2006
Hodgkin-Huxley
- assumed conductances on depend on membrane potential and not concentrations
- used this to determine contribution of Na and K currents by fixing membrane potential and changing concentrations which affect Nernst potentials only
- persistent current primarily due to K
- transient current due to Na
- [math]\displaystyle{ J_{Na}(V_m,t) = G_{Na}(V_m,t) \cdot (V_m(t) - V_{Na}) }[/math]
- [math]\displaystyle{ G_{Na}(V_m,t) = \frac{J_{Na}(V_m,t)}{V_m(t) - V_{Na}} }[/math]
- [math]\displaystyle{ G_{K}(V_m,t) = \frac{J_{K}(V_m,t)}{V_m(t) - V_{K}} }[/math]
- [math]\displaystyle{ V_m(t) - V_{Na} }[/math] is constant for [math]\displaystyle{ t \gt 0 }[/math] (step in potential). Same for K
- Thus conductances are simply scaled versions of the current
- Fit the current responses using following parameters
- [math]\displaystyle{ G_{K}(V_m,t) = \overline{G_K} n^4(V_m,t) }[/math] where
- [math]\displaystyle{ n(V_m,t) + \tau_n(V_m)\frac{dn(V_m,t)}{dt} = n_\infty(V_m) }[/math]
- [math]\displaystyle{ G_{Na}(V_m,t) = \overline{G_{Na}} m^3(V_m,t)h(V_m,t) }[/math] where
- [math]\displaystyle{ m(V_m,t) + \tau_m(V_m)\frac{dm(V_m,t)}{dt} = m_\infty(V_m) }[/math]
- [math]\displaystyle{ h(V_m,t) + \tau_h(V_m)\frac{dh(V_m,t)}{dt} = h_\infty(V_m) }[/math]
- [math]\displaystyle{ n_\infty }[/math] and [math]\displaystyle{ m_\infty }[/math] are activating functions
- are about 0 at negative [math]\displaystyle{ V_m }[/math] and has asymptote 1
- [math]\displaystyle{ h_\infty }[/math] is the reverse. =1 for low [math]\displaystyle{ V_m }[/math] and 0 for high [math]\displaystyle{ V_m }[/math]
- [math]\displaystyle{ \tau_m }[/math] (time constant for activating Na) is much smaller than other time constants