# 6.021/Notes/2006-10-16

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Jump to navigationJump to search- [math]\displaystyle{ V_m^o = \sum_n \frac{G_n}{G_m}V_n }[/math]
- 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
- [math]\displaystyle{ J_n^a+J_n^p=0 }[/math] for all [math]\displaystyle{ n }[/math] where [math]\displaystyle{ J_n^a }[/math] is the active current (pump) and [math]\displaystyle{ J_n^p }[/math] is the passive current (electro-diffusion).
- Define
**quasi-equilibrium**: no net flux but requires energy - [math]\displaystyle{ J_m = 0 = \sum_n G_n(V_m^o-V_n) + \sum_n J_n^a }[/math]
- [math]\displaystyle{ V_m^o = \sum_n \frac{G_n}{G_m}V_n - \frac{1}{G_m}\sum_n J_n^a }[/math]
- 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