6.021/Notes/2006-10-13

• injury potential V (when cell is broken open )is less than 0
  • V depends on extracellular concentration of potassium [math]\displaystyle{ c^o_K }[/math]
  • higher [math]\displaystyle{ c^o_K }[/math] means higher V
  • V does not depend on [math]\displaystyle{ c^o_{Na} }[/math]
• Bernstein model(1902)
  • new concept: rest: [math]\displaystyle{ J_m=0 }[/math]
  • time to reach rest much smaller than steady state
  • [math]\displaystyle{ J_m = 0 = J_K = G_K(V_m-V_K) }[/math]
  • Thus [math]\displaystyle{ V_m=V_K }[/math]
  • membrane is selectively permeable to K and has the potential needed to counteract diffusion
• Baker, Hodgkin, Shaw (1962), squid giant axon data
  • [math]\displaystyle{ c^i_K\uparrow\rightarrow V^o_m \downarrow }[/math], [math]\displaystyle{ c^o_K\uparrow\rightarrow V^o_m \uparrow }[/math], [math]\displaystyle{ c^o_K=c^i_K\rightarrow V^o_m\approx 0 }[/math]
  • measurements supported Bernstein model
• Data doesn't fit exactly with Bernstein model for all cells
• Multiple ionic species
  • [math]\displaystyle{ J_m = J_1 + J_2 \ldots J_n }[/math]
  • Define [math]\displaystyle{ V_m^o }[/math] as the membrane voltage at rest [math]\displaystyle{ J_m = 0 }[/math]
  • [math]\displaystyle{ J_m = \sum_n G_n(V_m^o-V_n) = 0 }[/math]
  • [math]\displaystyle{ \sum_n G_nV_m^o=\sum_n G_nV_n }[/math]
  • [math]\displaystyle{ G_m=\sum_n G_n }[/math]
  • [math]\displaystyle{ V_m^o = \sum_n \frac{G_n}{G_m}V_n }[/math]
    • The membrane potential is the weighted sum of Nernst potentials
  • Assume K, Na, and all other ions
  • Nernst potentials: K = -72mV, Na = +55mV, other (leakage) = -49mV
  • [math]\displaystyle{ V_m^o = -60mV }[/math]
• But change in concentration not only changes [math]\displaystyle{ V_n }[/math], also changes [math]\displaystyle{ G_n }[/math]
• Hodgkin-Huxley model (to be discussed in more detail later)
  • [math]\displaystyle{ \sum_n G_n(V_m^o)\cdot (V_m^o-V_n) = 0 }[/math]
• Rest is not equilibrium
  • rest is that there's no change in charge but they doesn't imply no flux
  • The flow of sodium can compensate for the flow of K at rest