```El potencial de
membrana, las pilas,
el equilibrio Donnan,
la ecuacion de Nernst
y la bomba de Na y K.
Que NO ES una celula:
no es una pila
no es una pila de concentracion
no es algo en equilibrio Donnan
Una pila de
concentracion
Fe2+ 100mM
Fe3+ 10mM
Fe2+ 4mM
Fe3+ 150mM
E = 0.059 log[(4.10)/(100.150)]= -149 mV
Una celula
E  -65 mV
[K+] 143 mM
[Na+] 33.4 mM
[Cl-] 11.5 mM
[K+] 11mM
[Na+] 133mM
[Cl-] 144mM
[P-] 165 mM
m1K+ = mK+0 + RT ln [K+]1 + FE
si m1K+ = m2K+ 
m2K+ = mK+0 + RT ln [K+]2
E = RT/F ln [K+]2 / [K+]1 = 66 mV
Una celula
[K+] 143 mM
[Na+] 33.4 mM
[Cl-] 11.5 mM
[K+] 11mM
[Na+] 133mM
[Cl-] 144mM
[P-] 165 mM
PK+ = 1
PNa+ = 0.01
PCl- = 0.1
flujos proporcionales a p y Dm
Una celula
mK+0 + RT ln [K+]1 + FE = mK+0 + RT ln [K+]2
mNa+0 + RT ln [Na+]1 + FE = mNa+0 + RT ln [Na+]2
mCl-0 + RT ln [Cl-]1 - FE = mCl-0 + RT ln [Cl-]2
[K+]1 + [K+]2 154 mM
[Na+]1 + [Na+]2 166.4 mM
[Cl-]1 + [Cl-]2 = 155.5 mM
[Cl-]2 = [K+]2 + [Na+]2
NO !
Una celula
[K+] 143 mM
[Na+] 33.4 mM
[Cl-] 11.5 mM
[K+] 11mM
[Na+] 133mM
[Cl-] 144mM
[P-] 165 mM
La celula es un capacitor electrico
Una celula
[K+] 103.8 mM
[Na+] 111.7 mM
[Cl-] 50.5 mM
[K+] 50.2mM
[Na+] 54.3 mM
[Cl-] 104.5 mM
[P-] 165 mM
E = -18.6mV
este es el potencial Donnan
Entonces, por que no se
mide el potencial que
calculamos ?
La bomba de Na+ / K+
[K+] 143 mM
[Na+] 33.4 mM
[Cl-] 11.5 mM
[P-] 165 mM
2 K+
[K+] 11mM
[Na+] 133mM
[Cl-] 144mM
3 Na+
Como podemos calcularlo ?
Valores iniciales [Na],[K],[Cl],E
dt
FNa(12) = PNa. DmNa + 3fbomba
FK(12) = PK. DmK - 2fbomba
FCl(12) = PCl. DmCl
Modifican valores [Na],[K],[Cl]
E = qexc / C
(qexc = [Na+]+[K+]-[Cl-]-[P-])
160
140
120
K + in
N a + in
100
[ ] / mM
C l- in
K+ out
80
N a+ out
C l- o u t
60
40
a ca se a p a g a la b o m b a N a /K
20
0
0
200
400
600
800
1000
1400
1600
1200
1400
1600
1800
2000
t / ms
0
0
200
400
600
800
1000
-1 0
-2 0
E / mV
1200
-3 0
-4 0
-5 0
-6 0
-7 0
t / ms
1800
2000
Cuando la bomba deja de
ser electrogenica ...
t/s
0
0
500
1000
1500
2000
e q uilib rio D o nna n
-2 0
E = -1 8 .5 5 m V
[K ]i = 1 0 3 .7 m M
[N a ]i = 1 1 1 .8 m M
-4 0
E / mV
[C l]i = 5 0 .6 m M
-6 0
E = -9 5 .9 m V
E = -7 9 .3 9 m V
[K ]o = 5 0 .3 m M
[K ]i = 1 5 0 .6 m M
[K ]i = 1 4 7 .7 m M
[N a ]o = 5 4 .2 m M
[N a ]i = 1 7 .9 m M
[C l]i = 3 .4 2 m M
[N a ]i = 2 3 .9 m M
[C l]i = 6 .6 7 m M
[C l]o = 1 0 4 .4 m M
-8 0
[K ]o = 6 .3 3 m M
[N a ]o = 1 4 2 .1 m M
-1 0 0
[C l]o = 1 4 8 .3 m M
[K ]o = 3 .4 2 m M
[N a ]o = 1 4 8 .1 m M
[C l]o = 1 5 1 .4 m M
-1 2 0
2500
B o m b a vs . c a n a le s
60
60
E s te s p ik e s e p ro d u c e
a l lle va r la p e rm e a b ilid a d
a l s o d io d e 0 .0 1 a 1 0 0
40
40
20
20
0
E / mV
0
0
200
400
600
800
1000
1200
1 2 3134.1
5
00
-2 0
-2 0
-4 0
-4 0
-6 0
-6 0
E s ta s u b id a s e p ro d u c e a l a p a g a r la
bom ba de N a y K
-8 0
-8 0
tie m p o / m s
1 2 3 3 .1 7
1 2 3 3 .1 9
34.6
34.4
34.2
34
Na+
33.8
33.6
142.7
142.6
142.5
142.4
142.3
142.2
K+
142.1
142
1233.14
1233.15
1233.16
1233.17
1233.18
1233.19
1233.2
60
40
20
0
-20
-40
-60
-80
E/mV
Conclusiones:
1 - La Celula es un capacitor que
acumula carga a los lados de la
membrana.
2 - La carga del capacitor se
establece por acción de la bomba
de K+/Na+
solamente un concepto para
calcular facilmente el potencial.
```