Biomedical Engineering Reference
In-Depth Information
Solving for the membrane potential,
g
ΔΦ
g g
gg g
+
ΔΦ
+
ΔΦ
K
K
Na
Na
Cl
Cl
ΔΦ
=
(3.22)
m
++
K
a
Cl
Since K
+
, Na
+
, and Cl
−
ions are not at their equilibrium potentials, (3.22) gives a
steady state potential and not an equilibrium state potential. There is a continuous
flux of those ions at the resting membrane potential. Often the equilibrium poten-
tial is not computed explicitly but defined as an independent leakage potential
ΔΦ
L
and determined by the resting potential. If the sum of current across the membrane
is not zero, then charge builds up on one side of the membrane while changing the
potential. Then, measuring the current flowing across a cell helps in determining
the capacitative and the resistive properties if the membrane potential is known.
Conductance properties of two ionic species can also be obtained in terms of the
membrane potentials. Consider a cell with a transfer of K
+
ions and Na
+
ions. Then
(
)
(
)
()
()
g
ΔΦ
− ΔΦ
t
+
g
ΔΦ
− ΔΦ
t
=
0
K
m
K
Na
m
Na
Rearranging this gives
(
)
−ΔΦ −ΔΦ
g
g
m
K
Na
=
(3.23)
(
)
ΔΦ
− ΔΦ
K
m
Na
Thus, in steady-state, the conductance ratio of the ions is equal to the inverse
ratio of the driving forces for each ion.
EXAMPLE 3.7
Electrical activity of a neuron in culture was recorded with intracellular electrodes. The ex-
periments were carried out at 37°C. The normal saline bathing the cell contained 145-mM
Na
+
ions and 5.6-mM K
+
ions. In this normal saline, the membrane potential was measured
to be
−
46 mV, but if the Na was replaced with an impermeant ion (i.e., the cell was bathed
in a Na-free solution) the membrane potential increased to
−
60 mV. In the normal saline
the
E
Na
was
+
35 mV, so that the membrane is impermeable to Cl
−
. To simplify the issue you
can ignore Na efflux in the Na-free saline condition. What ratio of
g
Na
to
g
K
would account
for the normal membrane potential?
Solution: From (3.23),
(
)
g
g
−−−
46
( 60)
Na
K
0.17
=
=
)
(
46
( 35)
−−+
