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Circuit representation of cell

  1. Jan 3, 2012 #1
    Hello!

    I am confused about this circuit representation. I am not familiar with how it works. I was under the impression charge cannot flow across a capacitor (a wire is required). So how does charge move in the following circuit? Not across the battery - since the charge is supposed to be separated, right? Again, as for a capacitor, a wire is required to provide a path for the electrons (or whatever) to move.
    There is nowhere for it to flow through. Are the points labeled in and out connected somehow?

    http://en.wikipedia.org/wiki/File:Cell_membrane_reduced_circuit.svg

    Thanks in advance.
     
  2. jcsd
  3. Jan 3, 2012 #2

    atyy

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    Ionic charge flows through gnet and Vnet which together represent the "leak channels" and their combined reversal potential (for each type of channel the reversal potential is the Nernst potential). So yes, ionic charge does flow across the battery.

    Although no ionic charge physically flows through the capacitor which represents the lipid bilayer, the induced change in charge on opposite plates of a capacitor allows "effective" current flow for changing currents. This effective charge flow is often called the "capacitive current".

    The in and out points are not connected.

    This equivalent circuit causes there to be charge separation across the capacitor (lipid bilayer) when no current is flowing, due to the presence of the resistor and battery (leak channel which is permeable to potassium).
     
    Last edited: Jan 3, 2012
  4. Jan 4, 2012 #3
    Thanks for the response atyy.
    If it flows across the battery, I do not see where the charge is 'going'. I was under the impression that a battery has a positive and negative terminal, and the electrons (for example) want to move from the negative to the positive terminal, reducing their potential energy. Connecting a wire allows them to do this. If it simply moved within the battery, it would just flow from the negative to the positive terminal (for electrons), and no current would flow through the rest of the circuit. In the above case, what is moving the charge around?

    I'm confused!
     
  5. Jan 4, 2012 #4
    Ah, I think what's happening is that the charge on the capacitor is merely being manipulated by altering the resistance/conductance in the circuit. There is no 'complete' circuit in the representation of the membrane as a circuit. Although ions flow across a membrane in the cell, there is no movement from positive charge to negative charge in the circuit (which would dissipate the charge on the battery), it merely alters the charge on the capacitor. So, in theory, would this be 'everlasting'? In the sense that the battery is not being run down?
    Thanks.
     
  6. Jan 4, 2012 #5

    atyy

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    The battery is due to (1) a potassium concentration difference across the cell membrane and (2) permeability of the cell membrane to potassium.

    A potassium current across the membrane will run the concentration gradient down.

    However, the steady state potential difference is usually reached before any significant change in concentration gradient, so the concentration gradient run down can usually be ignored for currents lasting only a few seconds. However, if the extracellular or intracellular volume is small, even a few seconds of current may change the concentration gradient significantly: http://www.ncbi.nlm.nih.gov/pubmed/7364046, http://www.ncbi.nlm.nih.gov/pubmed/10733967.

    If the membrane is continuously perturbed from the steady state potential difference for several hours and then continuously restored, the run down is potentially significant. However, there are active pumps not explicitly included in the circuit representation that restore the concentration gradient and prevent this run down over the course of several hours. These active pumps require ATP, and if ATP synthesis is prevented, the run down over several hours becomes apparent.
     
    Last edited: Jan 4, 2012
  7. Jan 4, 2012 #6
    In an RC circuit, the capacitor gradually accumulates charge on its plates when connected to a voltage source i.e. a battery. The battery does some work in putting the charge on the plates which is stored as the capacitor's potential energy (1/2 CV2).
    Similarly, the ion channels or pumps act as a voltage source which do work in order to create a potential difference across the membrane (the capacitor), the energy for which they get from ATP.

    Inside a battery there is no literal flow of electrons. The free energy of the chemical reactions that occur inside an electrical cell serve to separate the charges and create a potential difference across its ends.
     
  8. Jan 11, 2012 #7
    Thanks for clearing that up mish. My understanding from this is: the battery separates charge, and connecting a wire provides a potential gradient through the wire for the electrons IN THE WIRE to move: the battery merely establishes the force, the emf, for this process. I am having some difficulty in relating the functioning of a battery to the circuit representation of the cell.

    Thanks for the response atyy. I have some understanding (I think) of the source for the current in a cell, but the circuit representation is confusing. If it was an actual circuit, then the electrons moving through the wire have no complete circuit, is this correct? You said it flows across the battery, in which case the cell battery representation is not analogous to the more 'familiar' circuit and battery and how it functions? Either that or I've misunderstood how a battery functions. I do not see how current can flow through a battery: I imagine it to be like a modified capacitor: current can't flow between the two parallel plates, and the same is true with a battery. By connecting a wire to the plates of a capacitor, current can flow and the potential difference depleted, the same is true for a battery. Except the battery generates the potential difference, whereas a capacitor has to be charged.

    The other thing is currents. The textbooks I have read seem to suggest that the channels and carriers are responsible for the current. From my understanding, the Na-K+ ATPase is also important. Since I mentioned currents... does anyone know how current is "injected" in a voltage or current clamp experiment? All the sources I have read merely state that current is injected. What does that even mean?! I saw a diagram that suggested nothing is injected, just that the electrodes provide a charge, essentially altering the voltage across the membrane, as opposed to actually injecting anything.

    Again, thanks for the responses. Any further help appreciated.
     
  9. Jan 11, 2012 #8

    atyy

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    A battery consists of two electrodes dipped in a solution, and the potential is generated electrochemically, so we get "real" current flow from one electrode into the solution and then into the other electrode. The potential is generated differently in cells by means of a semi-permeable membrane and a concentration difference in ions across the membrane. Nonetheless, in both there is "real" current flow across a battery. This is different from a capacitor, which only permits "effective" current flow for changing currents, not constant currents.

    You can think of the Na-K ATPase as just setting up the concentration gradient across the membrane. The concentration gradient and the semi-permeability of the membrane together generate the membrane potential, which is represented as the battery in the circuit. Subsequent current flows change the concentration gradient, but usually so little that the change in the concentration gradient can be ignored over an experiment lasting just a few minutes. Over the course of several hours, current flows change the concentration gradient, if there is no Na-K ATPase. You can think of the Na-K ATPase as replenishing the concentration gradient, or rechargeing the battery.

    You can think of a voltage or current clamp experiment as simply taking two wires (R1,R2), one which is poked into the inside of the cell, while the other is placed in the extracellular fluid. A battery is connected across the two wires in a voltage clamp experiment, while a current source is connected across the two wires in a current clamp experiment. There are more refined versions of such experiments, but this is the basic idea.

    So the circuit diagram for a voltage clamp experiment becomes a complete ciruit of these elements in series:

    external battery for voltage clamp - R1 - cell - R2.

    R2 is connected back to the external battery, and the cell represented as in your OP. In a current clamp experiment, the external battery is replaced with an external current source. Additionally, a voltage clamp experiment has an ammeter in series with R1 to measure current flow, while a current clamp experiment has a voltmeter in series with R1 and R2, ie. across the the membrane resistance, to measure membrane voltage. Since the voltage across the membrane resistance is wanted, but we have R1 and R2 in the way, the errors from the R1 and R2 have to be taken into account in trying to determine the voltage across the membrane resistance.
     
    Last edited: Jan 11, 2012
  10. Jan 11, 2012 #9
    Thanks again atyy.
    With the current clamp experiment, when you inject current, the voltage changes, but then it levels off. Thats why I thought the current clamp used two charged electrodes (other then believing I saw it in one of the many sources I went through), was because the voltage reaches what in the various sources is described as a steady state. If a current was continuously applied, then wouldn't the voltage keep changing? The only way it would reach a steady state is if the change induced in the voltage causes a change in ion movements across thee cell that balanced the injected current. This would mean no net current and therefore no change in voltage. I do not see how you can have a net current and no continuous change in voltage.
    Unless,... an initial increase in concentration (negligible overall but enough to alter voltage) takes place when current is injected, and then the flow of charge out is matched the introduction of charge from the electrode. This would essentially mean no net current. For example, the introduction of negative ions into the cell will hyperpolarise the cell; as the cell hyperpolarises, the anions will move out. At a certain voltage the movement of the anions out of the cell matches the introduction from the electrode. Overtime I'm guessing this would deplete the positive charge in the extracellular fluid, unless the anions are recycled or their contribution to the fluid is negligible.
    Thanks for all the help. I can't put information together unless I can imagine whats happening. The sources just say what is observed, not really what is actually happening.
     
  11. Jan 11, 2012 #10

    atyy

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    Let's suppose a steady voltage is possible, and is reached after we wait for long enough. Since the voltage is steady, the capacitor will just be an open circuit, we so can ignore it. Then all we have is a current source, battery, and resistor in series. Let's ignore the battery for the moment. So it's just a current source I and a resistor R in series in a closed circuit. We can have a steady voltage V across the resistor R as long as V=IR.
     
  12. Jan 15, 2012 #11
    Thanks atyy.
    I just don't see where the charge is going. A current moves charge from one point to another, in which case, unless the net current is zero, charge is being transferred. The difference in charge between two points is then changing. I think I need to read some books on circuits.
     
  13. Jan 15, 2012 #12

    atyy

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    From the point of view of circuit theory, there is no charge accumulation anywhere in the circuit. The charge is just going round and round.

    First, let's say you have a resistor and a current source in a closed circuit, the charge is just going round and round in a loop.

    Next, let's consider a capacitor and resistor in parallel, with both connected in series to a current source in a closed circuit. Let's start with the capacitor uncharged. The current deposits positive charge on side A of the capacitor. This excess positive charge on side A repels positive charge from the opposite side B of the capacitor, so now there is excess negative charge on the side B of the capacitor. There is increased charge separation across the capacitor, but considering both sides of the capacitor, the positive charge on side A and the negative charge on side B sum to zero, so there is no charge accumulation on the capacitor.

    The positive charge repelled from side B will go in a closed loop and deposit even more positive charge on side A, which repels even more positive charge from side B, leaving side B even more negatively charged. Thus the charge separation across the capacitor is increasing, while the total charge on both sides of the capacitor together are zero. Although no charge really moves in the space between the capacitor sides, the deposition of positive charge on side A together with the flow of positive charge repelled away from side B is treated mathematically as a flow of positive charge "through" the capacitor.

    As more positive charge gets deposited on side A, it becomes harder and harder to deposit more positive charge on side A, since positive charges repel each other. Thus the current "through" the capacitor must eventually stop. However, since we have a resistor in parallel to the capacitor, the current from the current source can be diverted through the resistor. At steady state, the capacitor is "charged" (constant non-zero charge separation, zero total charge), and the capacitor acts as an open circuit and no current flows through it - all of the steady current flows in a loop from the current source through the resistor and back to the current source.

    The potential difference across the capacitor is proportional to the amount of charge separation across its sides, while the total charge on the capacitor is always zero. The constant of proportionality between potential difference and charge separation is called the "capacitance" of the capacitor.
     
    Last edited: Jan 15, 2012
  14. Jan 19, 2012 #13
    Thanks atyy, that cleared up MANY questions. I was always under the impression that there was a loss of voltage, not that current could be made to go round continuously.
     
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