The portion of a nerve cell that conducts signals is called an axon. Many of the electrical properties of axons are governed by ion channels, which are protein molecules that span the axon's cell membrane. When open, each ion channel has a pore that is filled with fluid of low resistivity and connects the interior of the cell electrically to the medium outside the cell. In contrast, the lipid-rich cell membrane in which ion channels reside has very high resistivity. Assume that a typical open ion channel spanning an axon's membrane has a resistance of 1× 10^11 Ω. We can model this ion channel, with its pore, as a 12-nm-long cylinder of radius 0.3 nm.
(a) What is the resistivity of the fluid in the pore?
(b) If a typical current density (current per unit cross-sectional area) in the cell membrane is 5 mA/cm^2 when the voltage across the membrane (the action potential) is 50 mV, what is the number density of open ion channels in the membrane?
(c) Cell membranes across a wide variety of organisms have a capacitance per unit area of 1 μF/cm^2. For the electrical signal in a nerve to propagate down the axon, the charge on the membrane "capacitor" must change. We still assume that a typical open ion channel spanning an axon's membrane has a resistance of 1×10^11Ω and the number density of open ion channels in the membrane is 100/μm^2. What time constant is required when the ion channels are open?
(a) To find the resistivity of the fluid in the pore, we can use the formula for resistance:
Resistance = Resistivity × Length / Area
We are given the resistance of the ion channel (1× 10^11 Ω) and the length of the channel (12 nm). The area of the channel can be calculated using the formula for the area of a cylinder:
Area = π × radius²
The radius of the channel is given as 0.3 nm. Plugging these values into the equation, we can solve for the resistivity:
Resistivity = (Resistance × Area) / Length
Resistivity = (1× 10^11 Ω × π × (0.3 nm)²) / (12 nm)
Solving this equation will give you the value of the resistivity of the fluid in the pore.
(b) To find the number density of open ion channels in the membrane, we can use the formula for current density:
Current Density = Current / Area
We are given the current density (5 mA/cm^2) and the voltage across the membrane (50 mV). The current can be calculated using Ohm's law:
Current = Voltage / Resistance
We are also given the resistance of the ion channel (1× 10^11 Ω). Plugging these values into the equation, we can solve for the current.
Once we have the current, we can rearrange the formula for current density to find the number density of open ion channels:
Number Density = Current / (Current Density × Area)
The area in this case is given as 1/μm². Plugging in the values will give you the number density of open ion channels in the membrane.
(c) To find the time constant required when the ion channels are open, we can use the formula for the time constant:
Time Constant = Capacitance × Resistance
We are given the capacitance per unit area (1 μF/cm²) and the resistance of the ion channel (1×10^11 Ω).
The area in this case is given as 100/μm². Plugging in the values will give you the time constant required when the ion channels are open.