Question

Only a tiny fraction of the diffusible ions move across a cell membrane in establishing a Nernst potential (see Focus On 20: Membrane Potentials), so there is no detectable concentration change. Consider a typical cell with a volume of \(10^{-8} \mathrm{cm}^{3},\) a surface area \((A)\) of \(10^{-6} \mathrm{cm}^{2},\) and a membrane thickness \((l)\) of \(10^{-6} \mathrm{cm}\) Suppose that \(\left[\mathrm{K}^{+}\right]=155 \mathrm{mM}\) inside the cell and \(\left[\mathrm{K}^{+}\right]=4 \mathrm{mM}\) outside the cell and that the observed Nernst potential across the cell wall is \(0.085 \mathrm{V}\). The membrane acts as a charge-storing device called a capacitor, with a capacitance, \(C,\) given by $$C=\frac{\varepsilon_{0} \varepsilon A}{l}$$ where \(\varepsilon_{0}\) is the dielectric constant of a vacuum and the product \(\varepsilon_{0} \varepsilon\) is the dielectric constant of the membrane, having a typical value of \(3 \times 8.854 \times 10^{-12}\) \(\mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-2}\) for a biological membrane. The SI unit of capacitance is the firad, \(1 \mathrm{F}=1\) coulomb per volt \(=1 \mathrm{CV}^{-1}=1 \times \mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-1}\) (a) Determine the capacitance of the membrane for the typical cell described. (b) What is the net charge required to maintain the observed membrane potential? (c) How many \(\mathrm{K}^{+}\) ions must flow through the cell membrane to produce the membrane potential? (d) How many \(\mathrm{K}^{+}\) ions are in the typical cell? (e) Show that the fraction of the intracellular \(K^{+}\) ions transferred through the cell membrane to produce the membrane potential is so small that it does not change \(\left[\mathrm{K}^{+}\right]\) within the cell.

Short answer

The capacitance of the membrane for the cell is given by the formula \(C=\frac{\varepsilon_{0} \varepsilon A}{l}\) and yields a specific value when computed with the given values. The net charge required to maintain the observed membrane potential can be calculated using the relation \(Q = C \cdot V\). Using the charge of a single ion, we can find the number of \(K^{+}\) ions required to establish this potential. By considering the \(K^{+}\) concentration inside the cell, the total number of \(K^{+}\) ions inside the cell can be computed. Lastly, by comparing the number of ions that crossed the membrane with the total number of ions inside, we can confirm that the transfer was negligible.

Step-by-step solution

Calculate the Membrane's Capacitance

First, let's calculate the capacitance of the membrane using the given equation \(C=\frac{\varepsilon_{0} \varepsilon A}{l}\) where, \(\varepsilon_{0} \varepsilon = 3 \times 8.854 \times 10^{-12} \mathrm{C}^{2} \mathrm{N}^{-1} \mathrm{m}^{-2}\), \(A = 10^{-6} \mathrm{cm}^{2}\) and \(l = 10^{-6} \mathrm{cm}\). Substituting these values will give us the capacitance.

Determine the Net Charge Required

The capacitance obtained in the first step can be used to calculate the net charge required to maintain the observed membrane potential using Ythe equation \(Q = C \cdot V\), where V is the voltage across the membrane which is given as \(0.085 V\).

Calculate the Number of Ions Required

The number of \(K^{+}\) ions required to maintain the membrane potential can be obtained by dividing the charge obtained in step 2 by the charge of a single \(K^{+}\) ion. The charge of a single \(K^{+}\) ion is equal to the charge of a proton (\(1.6 \times 10^{-19}\) coulombs).

Calculate the Number of \(K^{+}\) ions in the Cell

The number of \(K^{+}\) ions in the cell can be determined using their concentration, Avogadro's number, and the volume of the cell. The given concentration of \(K^{+}\) ions inside the cell is \(155 mM\) and the cell volume is \(10^{-8} \mathrm{cm}^{3}\).

Verify the Ion Transfer is Negligible

Finally, we check if the fraction of the intracellular \(K^{+}\) ions transferred through the cell membrane to produce the membrane potential significantly affects the concentration inside the cell by dividing the number of ions that crossed the membrane (step 3) by total ions in the cell (step 4). If the fraction is very small, it can effectively be treated as negligible.

Key concepts

Cell Membrane Capacitance

The cell membrane capacitance is a critical concept in understanding how cells manage electrical signals. Capacitance is a measure of a system's ability to store charge per unit voltage. In biological terms, this relates to how the cell membrane can hold opposing charges on either side, thus storing electrical energy that can be used for various cellular processes.

In the context of a typical cell, the capacitance (\(C\)) of the membrane can be calculated using the formula \(C = \frac{\varepsilon_{0} \varepsilon A}{l}\), where \(\varepsilon_{0}\) is the permittivity of free space, \(\varepsilon\) is the dielectric constant of the cell membrane, \(A\) is the surface area, and \(l\) is the thickness of the membrane. This value essentially reflects how much electric charge the cell membrane can store, and is fundamental to the control of ionic flows across the membrane.

Membrane Potential Calculation

Calculating the membrane potential is vital for understanding the electrical condition of a cell. The membrane potential is the difference in electric charge between the interior and exterior of a cell. This electrical potential allows for the transmission of signals within and between cells and is crucial for functions such as muscle contractions and neuron firing.

The Nernst potential is a particular measure of membrane potential that considers a single ion type. To find the net charge (\(Q\)) required to maintain a membrane potential (\(V\)), we use the capacitance (\(C\)) calculated previously and the formula \(Q = C \cdot V\). This allows us to determine the number of ions that must shift across the membrane to maintain that potential, which is essential in neuronal activity and muscle contractions.

Ionic Flow Across Cell Membrane

Ionic flow across the cell membrane is the process by which ions, such as potassium (\(K^+\)), sodium (\(Na^+\)), calcium (\(Ca^{2+}\)), and chloride (\(Cl^-\)), move through the membrane. This movement is critical for maintaining the membrane potential and for conducting electrical signals.

The flow is regulated by ion channels and is driven by concentration gradients and the electrical potential across the membrane. When ions move from a high to low concentration, this movement can either depolarize or hyperpolarize the cell, which are states of increased and decreased membrane potential, respectively. Understanding how many \(K^+\) ions must flow through the membrane to produce a specific membrane potential gives insight into the cell's electrical activity and its ability to respond to stimuli.

Concentration Changes in Cells

Changes in ion concentrations within a cell are regarded as signals that can initiate various cellular actions. The concentration of ions like \(K^+\) is tightly regulated, as significant changes can affect the cell's function. However, it's intriguing that the establishment of a membrane potential, such as the Nernst potential, typically involves quite a small number of ions relative to the total number of ions inside the cell.

As a result, even though these ions play a crucial role in generating the membrane potential, their movement across the membrane does not lead to a detectable change in the overall concentration in the cell. This underscores an important aspect of cellular dynamics; a small number of ions can dramatically influence the electrical properties of a cell without altering the cell's internal ionic environment.