[riboswitch]

I'm having a hard time solving a problem proposed by my Cell Physiology Book. I don't know if I'm in the right forum (since I'm having a biology problem and this forum is a haven for chemists) but I need help because my book apparently doesn't have any hints on how to solve the problem, and I would appreciate help from others. Obviously I only need a hint on how to solve it...

Problem:

A small spherical cell has a total membrane surface area of 10 μm² (1 μm = 10^-4 cm) and a capacitance of 1 × 10^-6 F/cm². Initially the cell has a membrane potential V_m= 0 mV, and there are no open channels in the membrane. Assume that [Cl^-]_i = 10 mM, [Cl^-]_o = 140 mM, RT/F = 26.7 mV and γ_Cl = 10^-11 S.

• What happens to V_m when a single Cl^- channel opens?
• Draw a graph of V_m as a function of time following the opening of the channel. Indicate the values of initial and final V_m. What is the value of the membrane time constant in the presence of the open Cl^- channel?

My solutions:

The first one is easier than the second question. In this case, the difference between the membrane potential and the equilibrium potential of the chloride ion will determine the direction of the ion flow. Initially there are more chloride ions outside the cell than inside, so chloride ions would naturally flow down its concentration gradient. However, no current flow will occur as the chloride ion gates are closed. When the ion channel opens, chloride ions will flow in such a way as to drive the membrane potential toward its own equilibrium potential. In other words, in order to determine the direction of the chloride ion flux, I must calculate first the equilibrium potential of the ion using the Nernst equation:

[tex]E_{Cl^{-} }  \ = \  \frac{RT}{zF}ln \frac{[Cl^{-}]_{o}}{[Cl^{-}]_{i}} [/tex]
[tex]E_{Cl^{-} }  \ = \  (-26.7 \ mV) \ ln \frac{(140)}{(10)} \ = \ -70.5 \ mV[/tex]

When the chloride ion channel opens, the value of V_m (initially 0 mv) shifts towards the more negative value of E_Cl (-70.5 mV), which means that chloride ions will enter the cell rendering the intracellular side of the membrane more negative than the exterior. An inward flux of chloride ions will occur.

I'm having problems with the second question. I need to draw a graph of the membrane potential as a function of time. In order to do that, my guess is that I should use the following equation:

[tex]\Delta V_{m}(t) \ = \  \Delta V_{m, \infty } \ (1 \  - e^{\frac{-t}{ \tau_{m}} })[/tex]

with:

[tex]\tau_{m} \ = \ R_{m} \  \times \ C_{m}[/tex]

τ is supposed to be the membrane time constant, while R and C represent resistance and capacitance, respectively. I have the value of capacitance (1 ×10^-6 F/cm²) and I probably need to convert the value of capacitance based on the cell surface area to determine its so called total cell capacitance.

I need to calculate the value of the resistance using the classical Ohm's Law, but it requires having the value of the current flow and the potential difference across the membrane. I don't have the value of the current flow, but I do have the value of the conductance of the single chloride channel (γ). Resistance is the inverse value of the conductance, so I can say that R_m = 1/γ_Cl, assuming that only one Cl^- channel is open.

[tex] \tau _{m} \ = \  \frac{C_{m}}{ \gamma _{Cl^{-}}} [/tex]

Another value that I have to determine is:

[tex]\Delta V_{m, \infty } \ = \ I_{m} \  \times \ R_{m} \ = \  \frac{I_{m}}{ \gamma _{Cl^{-}}} [/tex]

I'm blocked here. I have no idea how to calculate I_m necessary to solve the equation above. I could say that the current is equal to the product of conductance and the potential difference across the membrane, but I need to calculate the variation of V_m as a function of time, and ΔV_m,∞ is supposed to be a constant value.

Am I in the right track? Did I miss something pretty important? What should be my next move?

[mjc123]

Sorry, post was incomplete, pressed post instead of preview.

It seems to me that in your equations
V = V_∞(1-e^-t/τ)   and    τ = C_m/γ_Cl^-
you have all you need.
Is not V_∞ = -70.5mV. Why are you bothered about I? I is not constant as V is changing.

[riboswitch]

Sorry, post was incomplete, pressed post instead of preview.

It seems to me that in your equations
V = V_∞(1-e^-t/τ)   and    τ = C_m/γ_Cl^-
you have all you need.
Is not V_∞ = -70.5mV. Why are you bothered about I? I is not constant as V is changing.

I thought I have to calculate ΔV_m,∞ using the following equation:

[tex]\Delta V_{m, \infty } \ = \  \frac{I_{m}}{ \gamma _{Cl^{-}}} [/tex]
I re-examined the problem, and I finally figured out that I actually have everything I need. I already have the value of ΔV_m,∞, which is just the difference between the final membrane potential after the opening of the chloride ion channel and the initial membrane potential (V_m = 0 mV)

[tex]\Delta V_{m, \infty } \ = \  V_{m,final} \ - \ V_{m,initial} \ = \ -70.5 \ mV \ - \ 0 \ mV \ = \ -70.5 mV [/tex]
The value of the membrane time constant can be determined using the equation I wrote earlier:

[tex]\tau _{m} \ = \  \frac{C_{m} \  \times \ A_{cell}}{ \gamma _{Cl^{-}}} \ = \  \frac{10^{-13} \ F \ \times \  \Omega }{10^{-11}} \ = \ 10^{-2} \ sec.[/tex]
The membrane time constant is the amount of time that is needed to reach 63% of V_m,∞.

I'll use the equation below and draw a graph of it using whatever software/online tool is available for free:

[tex]y \ = \ -70.5 \ (1 \ - \ e^{-100x} )[/tex]
I hope I did everything right. Thanks for the help.