10. Numerical exercises
10.1. Calculation of membrane potential
10.1.1. Membrane conductance values of a resting neuron are (in relative units) gK+: gNa+: gCl- = 1: 0.005: 0.1.
([Na+]e=140 mM, [Na+]i=14 mM, [K+]e=4 mM, [K+]i=140 mM, [Cl-]e=110 mM, [Cl-]i=5 mM.)
What is the resting membrane potential?
The reversal potentials for the three ions are (Nernst equation):
VK=-93 mV, VNa=+60 mV, VCl=-81 mV. Hence
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10.1.2. At the peak of the action potential, due to opening of Na+ channels, relative conductances are altered as follows:
gK+: gNa+: gCl-= 1: 20: 0.1.
What is the membrane potential at the peak of the action potential?
Using the new conductance values as weight factors
10.1.3. Depolarization was caused by Na+ influx. How much did intracellular [Na+] increase, if the cell is spherical with a diameter of 20 μm, and the specific capacitance of the membrane is 0.01 pF/μm2 ?
From the data the cell volume is
the cell surface is
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hence the total capacitance is
The imported charge is
How many moles of Na+ ions does this charge correspond to?
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What is the change in concentration due to this amount of ions?
Thus, intracellular [Na+] increases from 14 mM to 14.0045 mM.
The activity of ion channels can alter the membrane potential without significantly affecting ionic concentrations!
10.2. Na+/H+ exchanger:
electroneutral secondary active transport
10.2.1. Under what conditions is the transporter at equilibrium?
The catalyzed reaction: 1 H+i + 1 Na+e → 1 H+e + 1 Na+i If ΔG<0, the process goes
in the forward direction, if ΔG>0, the process goes
in the reverse direction, if ΔG=0, there is no net transport.
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The free energy change for the above reaction:
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(electroneutral ⇒ does not depend on membrane potential).
In the cell [H+]e=10-7.4 M, [H+]i=10-7.1 M, [Na+]e=140 mM, [Na+]i=14 mM. Thus,
Because zH=zNa=1,
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10.2.2. Why is this transporter necessary?
From the Nernst equation the reversal potential for H+ is Vrev= –18 mV. The resting membrane potential is far more negative (approximately –90 mV), therefore, the
thermodynamic driving force for H+ is inward directed, and would eventually cause acidification of the cell.
Note: On the other hand, the Na+/H+ exchanger would reach equilibrium only at [H+]i=10-8.4 M (i.e., upon alkalinization of the cytosol!). Therefore, the activity of the Na+/H+ exchanger is
10.3. Na+/Ca2+ exchanger:
electrogenic secondary active transport
10.3.1. Under what conditions is the transporter at equilibrium?
The catalyzed reaction: 1 Ca2+i + 3 Na+e → 1 Ca2+e + 3 Na+i If ΔG<0, the process goes
in the forward direction, if ΔG>0, the process goes
in the reverse direction, if ΔG=0, there is no net transport.
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The free energy change for the above reaction:
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(electrogenic ⇒ membrane potential sensitive).
Since zCa=2 and zNa=1,
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In the cell [Ca2+]e=10-3 M, [Ca2+]i=10-7 M, [Na+]e=140 mM,
[Na+]i=14 mM. Thus, the reversal potential of the transporter is Vrev=-60 mV. The resting membrane potential is more negative (~
–90 mV), i.e., Vm<Vrev, and ΔG<0.
Therefore, in a resting cell the transport cycle proceeds as written: the Na+/Ca2+
exchanger pumps Ca2+ ions out of the
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10.3.2. Why is this transporter necessary?
From the Nernst equation the reversal
potential for Ca2+ is Vrev= +120 mV. At the normal resting membrane potential
(~ –90 mV) Ca2+ experiences a large inward directed thermodynamic driving force. The cell can maintain its extremely small intracellular Ca2+ concentration only with the help of active mechanisms – the Na+/Ca2+ exchanger is one of these.
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10.3.3. Under certain conditions intracellular [Na+] can rise
significantly in a cell, and this can lead to "reversal" of the NCX, i.e. to extrusion of Na+ from the cell in exchange for Ca2+ uptake.
Calculate – assuming that the other parameters ([Ca2+]e, [Ca2+]i, [Na+]e, and Vm) remain unchanged – the intracellular [Na+] at
which the transporter reverses.
Expressing [Na+]i from the above equation
10.4. Na+/K+ pump: electrogenic primary active transport 10.4.1. Under what conditions is the transporter at equilibrium?
The catalyzed reaction:
3 Na+i + 2 K+e + ATPi → 3 Na+e + 2 K+i + ADPi + Pi If ΔG<0, the process goes
in the forward direction, if ΔG>0, the process goes
in the reverse direction,
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The free energy change for the above reaction:
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Since zNa= zK=1,
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The standard free energy change for ATP hydrolysis is
Therefore, at body temperature,
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In the cell [Na+]e=140 mM, [Na+]i=14 mM, [K+]e=4 mM, [K+]i=140 mM, [ATP]i=5 mM, [P]i=5 mM, [ADP]i=10-3 M (check unit!!).
Thus, Vrev=-131 mV.
The resting membrane potential is more positive (~ –90 mV), i.e., Vm>Vrev, and ΔG<0. Therefore, under resting conditions the transport cycle proceeds as written: the Na+/K+ pump pumps Na+ ions out of the cell.
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