Question

The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential difference of about 70 mV exists across the membrane. The thickness of the membrane is 8 nm. A) If the membrane were empty (filled with air), what would be the magnitude of the electric field inside the membrane? (Assume that this field would be constant in magnitude). B) If the dielectric constant of the membrane is K = 3, what would the field be inside the membrane?3) Cells can carry ions across a membrane against the field ("uphill") using a variety of active transport mechanisms. One mechanism does so by using up some of the cell's stored energy (converting ATP to ADP, for those who remember their biochemistry). How much work does it take to carry one sodium ion (charge = +e) across the membrane against the field?

Answer 1

The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about 0.070 V exists across the membrane. The thickness of the cell membrane is 8.0 x 10-9 m.

Answer 2

A) Magnitude of the electric field with air filling the membrane: 8.75 × 10^7 V/m.

B) Magnitude of the electric field with realistic dielectric constant: 2.92 × 10^7 V/m.

C) Work done by the active transport mechanism: 5.0 × 10^-19 J.

Step-by-step explanation:

A) Electric field with air:

Calculate the electric potential difference (V): V = Ed (voltage = electric field * thickness).

Rearrange for E: E = V/d.

Substitute values: E = 70 mV / 8 nm (convert nm to m: 8 nm * 10^-9 m/nm = 8 × 10^-9 m).

Calculate the electric field: E ≈ 8.75 × 10^7 V/m.

B) Electric field with realistic dielectric constant:

Modify the formula for the electric field to account for the dielectric constant: E = V / (Kd).

Substitute values: E ≈ 70 mV / (3 * 8 × 10^-9 m).

Calculate the electric field: E ≈ 2.92 × 10^7 V/m.

C) Work done by active transport:

Calculate the potential energy change for one sodium ion across the membrane: ΔEp = eV (change in potential energy = charge * potential difference).

Substitute values: ΔEp = (1.602 × 10^-19 C) * (70 mV).

Convert millivolts to volts: 70 mV * 10^-3 V/mV = 0.07 V.

Calculate the work done: W = ΔEp ≈ 5.0 × 10^-19 J.

Therefore, with air filling the membrane, the electric field would be much stronger, highlighting the crucial role of the membrane's structure and dielectric properties in maintaining a stable environment for cellular processes. The active transport mechanism needs to overcome this energy barrier to move ions against the field, utilizing the stored energy from ATP hydrolysis.