Reading equilibrium pot 2

1:

Here are the answers for Figure 7:

• For bulk electroneutrality to apply, the extracellular concentration of chloride ions, which are negatively charged, has to be equal to the concentration of the positively charged sodium and potassium ions, and thus $\left[\text{Cl}^-\right]_\text{out} = \left[\text{Na}^+\right]_\text{out} + \left[\text{K}^+\right]_\text{out} = 120\ \text{mM} + 5\ \text{mM} = 125\ \text{mM}.$
• The total number of particles on the outside is thus 125 + 120 + 5 = 250 milliosmolar.
• The equation describing the total number of particles on the inside of the cell (the left hand side) is then $250\ \text{mOsm} = \left[\text{K}^+\right]_\text{in} + \left[\text{Cl}^-\right]_\text{in} + [P].$
• The requirement for bulk electroneutrality implies that the intracellular concentration of potassium ions has to be equal to the concentration of chloride ions, or that $\left[\text{K}^+\right]_\text{in} = \left[\text{Cl}^-\right]_\text{in}.$
• The fact that the Nernst potentials for chloride and potassium must be equal to one another at equilibrium leads to Equation 7: $\left[\text{Cl}^-\right]_\text{in}\ \left[\text{K}^+\right]_\text{in} = \left[\text{K}^+\right]_\text{out}\ \left[\text{Cl}^-\right]_\text{out}.$
• Since we have worked out the concentrations of the potassium and chloride ions on the outside, we can substitute in those values to obtain $\left[\text{Cl}^-\right]_\text{in}\ \left[\text{K}^+\right]_\text{in} = 5\ \text{mM} \times 125\ \text{mM} = 625\ {\text{mM}^2}.$
• But since we have just established, based on bulk electroneutrality, that the concentrations of chloride and potassium ions inside the cell are identical, we can further simplify this to $\left[\text{Cl}^-\right]_\text{in}^2 = 625\ {\text{mM}^2}.$
• This immediately implies that $\left[\text{Cl}^-\right]_\text{in} = 25\ \text{mM}.$
• From bulk electroneutrality, this also implies that $\left[\text{K}^+\right]_\text{in} = 25\ \text{mM}.$
• Osmotic balance then implies that $

\begin{align} 250\ \text{mOsm} &= \left[\text{K}^+\right]_\text{in} + \left[\text{Cl}^-\right]_\text{in} + [P] \\

                  &= 25\ \text{mOsm} + 25\ \text{mOsm} + [P],

\end{align} $ so that $[P] = 200\ \text{mM}$.

• This solves the problem: the membrane is in osmotic balance, has bulk electroneutrality on both sides, and is at an equilibrium.
• What is the value of the equilibrium potential across the membrane? We could calculate either the chloride ion or the potassium ion equilibrium; both should give the same value if we have done our calculations correctly. To use Equation 6, we will calculate the potassium ion equilibrium: $E_{\text{K}^+} = 58\ \text{mV} \times \log_{10} \frac{5\ \text{mM}}{25\ \text{mM}} = -40.5\ \text{mV}.$
• To check our work, set up the equivalent Nernst equation for chloride ions; you will immediately see whether we've gotten the right answer or not.

2: Here are the answers for Figure 8, when sodium is not permeable:

• Solving for the concentration of the chloride ions outside of the cell (the right-hand side) can be done easily by applying the rule of bulk electroneutrality.
• The total concentration of positive charges on the outside is 120 mM + 5 mM = 125 mM, so the concentration of chloride ions must also be $\left[\text{Cl}^-\right]_\text{out} = 125\ \text{mM}$.
• The total number of particles on the outside is thus 120 mOsM + 5 mOsM + 125 mOsM = 250 mOsM.
• In turn, this implies a constraint on the number of particle inside the cell, which must equal $250\ \text{mOsm} = \left[\text{Na}^+\right]_\text{in} + \left[\text{K}^+\right]_\text{in} + 5\ \text{mOsm} + 110\ \text{mOsm},$ which in turn implies that $250\ \text{mOsm} - 5\ \text{mOsm} - 110\ \text{mOsm} = 135\ \text{mOsm} = \left[\text{Na}^+\right]_\text{in} + \left[\text{K}^+\right]_\text{in}.$
• We also have Equation 7, $\left[\text{Cl}^-\right]_\text{in}\ \left[\text{K}^+\right]_\text{in} = \left[\text{K}^+\right]_\text{out}\ \left[\text{Cl}^-\right]_\text{out}.$
• Since we know the concentrations of the potassium and chloride ions outside the cell, and the concentration of chloride ions inside the cell, this becomes $5\ \text{mM} \times \left[\text{K}^+\right]_\text{in} = 5\ \text{mM} \times 125\ \text{mM},$ so that $\left[\text{K}^+\right]_\text{in} = 125\ \text{mM}.$
• Electroneutrality provides another equation: $\left[\text{Na}^+\right]_\text{in} + \left[\text{K}^+\right]_\text{in} = 5\ \text{mM} + 1.2 (110\ \text{mM}) = 137\ \text{mM},$ which implies that $\left[\text{Na}^+\right]_\text{in} = 137\ \text{mM} - 125\ \text{mM} = 12\ \text{mM}.$
• Now we have a problem: the number of particles on the inside of the cell is 12 mOsm + 125 mOsm + 5 mOsm + 110 mOsm = 252 mOsm; but we already established that the number on the outside of the cell is 250 mOsm!
• So, this implies that water will move from where it is more concentrated (i.e., on the outside) to where it is is less concentrated (i.e., to the inside, where the concentration of particles is higher).
• Assuming initial equal volumes of 1 liter, the change in volume will be (using Equation 3) $(250 \times 1 - 252 \times 1)/(252 + 250) = -2/502$, or a change in volume on either side of about 0.4 percent (an increase in volume on the inside, and a corresponding decrease on the outside). We will therefore neglect this, although if we wanted a precise answer, we would need to include this change.
• An estimate of the membrane potential is provided by solving the Nernst equation for the potassium ion, or $E_{\text{K}^+} = 58\ \text{mV} \times \log_{10} \frac{5\ \text{mM}}{125\ \text{mM}} = -81\ \text{mV}.$

3: Here are the answers for Figure 8, when sodium is permeable:

• In the equilibrium condition, we computed that the internal sodium ion concentration was $\left[\text{Na}^+\right]_\text{in} = 12\ \text{mM}$, and we were given that its external concentration is $\left[\text{Na}^+\right]_\text{out} = 120\ \text{mM}$.
• Using the Nernst equation, the predicted equilibrium potential for sodium across the membrane is $E_{\text{Na}^+} = 58\ \text{mV} \times \log_{10} \frac{120\ \text{mM}}{12\ \text{mM}} = +58\ \text{mV},$ which is very different from the $-81\ \text{mV}$ equilibrium potential that we calculated for the chloride and potassium ions.

4:

• We found that the values of sodium, potassium and chloride ion concentrations inside the cell were 12 mM, 125 mM, and 5 mM (respectively), and the concentrations of these ions outside the cell were 120 mM, 5 mM, and 125 mM (respectively).
• Then the estimated resting potential across the membrane is $

\begin{align}

       V_m &= 58\ \text{mV} \times \log_{10} \frac{1 \cdot 120\ \text{mM} + 1 \cdot 5\ \text{mM} + 1 \cdot 5\ \text{mM}} {1 \cdot 12\ \text{mM} + 1 \cdot 125\ \text{mM} + 1 \cdot 125\ \text{mM}} \\
           &= 58\ \text{mV} \times \log_{10} \frac{130}{262} \\
           &= -17.7\ \text{mV},

\end{align} $ which is midway between the Nernst or equilibrium potentials of $-81\ \text{mV}$ that we calculated for the chloride and potassium ions, and $+58\ \text{mV}$ that we calculated for the sodium ions.

5:

• The sum of particles on the inside of the squid giant axon is $400\ \text{mOsm} + 50\ \text{mOsm} + 52\ \text{mOsm} + 385\ \text{mOsm} = 887\ \text{mOsm}$.
• The sum of the particles outside of the squid giant axon is $20\ \text{mOsm} + 440\ \text{mOsm} + 560\ \text{mOsm} = 1020\ \text{mOsm}$.
• The difference is 133 mOsm, which is a very large amount!
• If these were the only particles, water would leave the axon because it is more concentrated on the inside (where the number of particles is smaller); thus, the neuron would shrink.
• However, there are other ions that are not in this list, such as magnesium and calcium, that are also important for osmotic balance, as well as impermeable particles on either side that contribute to allowing the axon to remain in osmotic balance.

6:

• On the inside, there are $400\ \text{mM} + 50\ \text{mM}$ of positive charge $= 450\ \text{mM}$, and there are $52\ \text{mM} + 385\ \text{mM}$ of negative charge $= 437\ \text{mM}$; therefore, to maintain electroneurtrality, there must be $13\ \text{mM}$ of negative charges that are unaccounted for on the inside of the cell. These may be other ion species, charged proteins, or something else.
• On the outside, there are $20\ \text{mM} + 440\ \text{mM}$ of positive charge $= 460\ \text{mM}$, and there are $560\ \text{mM}$ of negative charge; this implies there are $100\ \text{mM}$ of positive charges that are unaccounted for outside the cell to maintain electroneutrality. Again, these could be other ion species, like magnesium or calcium.

7:

• For the sodium Nernst potential, using Equation 8, you should obtain $+54.78\ \text{mV}$, which is about $+55\ \text{mV}$.
• For the potassium Nernst potential, using Equation 6, you should obtain $-75.46\ \text{mV}$, which is about $-75\ \text{mV}$.
• For the chloride Nernst potential, using Equation 5, you should obtain $-59.87\ \text{mV}$, which is about $-60\ \text{mV}$, close to the resting potential of the nerve cell.
• Because the membrane is permeable to both ions with positive Nernst potentials and ions with negative Nernst potentials, the concentration gradients will tend to run down if they are not energetically maintained. Thus, the squid giant axon is not in equilibrium; rather, it is in a steady state condition.