# 6.021/Notes/2006-10-16

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
• ${\displaystyle V_{m}^{o}=\sum _{n}{\frac {G_{n}}{G_{m}}}V_{n}}$
• Rest is not equal to equilibrium
• eventually all ions would equilibrate inside and outside and there would be no membrane potential
• For this to not happen, must have pumps (primary active transport)
• Pumps allow us to go from rest to equilibrium
• treat as current source
• ${\displaystyle J_{n}^{a}+J_{n}^{p}=0}$ for all ${\displaystyle n}$ where ${\displaystyle J_{n}^{a}}$ is the active current (pump) and ${\displaystyle J_{n}^{p}}$ is the passive current (electro-diffusion).
• Define quasi-equilibrium: no net flux but requires energy
• ${\displaystyle J_{m}=0=\sum _{n}G_{n}(V_{m}^{o}-V_{n})+\sum _{n}J_{n}^{a}}$
• ${\displaystyle V_{m}^{o}=\sum _{n}{\frac {G_{n}}{G_{m}}}V_{n}-{\frac {1}{G_{m}}}\sum _{n}J_{n}^{a}}$
• The first term is the "indirect effect" whereas the second term is the "direct effect" of the pump
• Both terms depend on the pump as without the pump, both would be 0
• All pumps have indirect effect but only some pumps have direct effect
• A non-electrogenic pump has no net charge change (e.g. trade one sodium ion for one potassium ion)
• An electrogeneic pump such as 3 sodium for 2 potassium would have net active current
• Active transport gets energy from glucose metabolism (namely ATP)
• Experiment shows that sodium and potassium transport are linked
• Sodium is needed for potassium transport and vice-versa
• Sodium is pumped out, potassium pumped in
• The direct effect on membrane potential is negative