Consider the following intracellular and extracellular ion concentration for a t

Question

Consider the following intracellular and extracellular ion concentration for a typical cell: A therapeutic effect of an uncharged synthetic inhibitor of an intracellular an enzyme requires the ratio of its intracellular to extracellular concentration be at least 4 (C_in C_out = 4). The inhibitor's entry into cells occurs upon brief depolarization of the membrane to nabla_p = -50 mV accomplished by opening of the Na^+ channels, and is coupled lo Na^+ transport into the cell. How many moles of will need lo enter the cell for the desired ratio of the inhibitor's concentration to be established? Assume constant transmembrane potential of - 50 mV during this event. Using the typical information cell and the Nernst equation a estimate the equilibrium potential for car: b) based on the result of your calculation, if the Ca^2+ cannels are open or closed in this cell, and e) calculate Gibbs free energy for transfer of mole of Ca^2+ into the cell.

Explanation / Answer

Solution:

We know, the change in energy across the cell membrane can be given by:

G = RT ln(c2/c1) + ZFV = 2.303RT log(c2/c1) + ZFV

Where V is the potential in volts across the plasma membrane, Z is the electrical charge of the transported species, and F stands for the Faraday constant (96500 J/V.mol).

G for Na+:

G = 2.303RT log(10/140) + ZFV

      = 2.303x8.315x310xlog(1/14) + 1x96500x(-0.06)

     = -6803.8 – 5790 J/mol = -12593.8 J/mol = -12.6 kJ/mol

G for the inhibitor:

G = 2.303RT log(c2/c1) + ZFV

       = 2.303x8.315x310xlog(4) + 0x96500x(-0.05)

      = 3574 J/mol = 3.6 kJ/mol

Negative G implies the direction of travel i.e from intracellular to extracellular for Na+ ions, because the channel can only allow exergonic transport, not drive endergonic transport. A positive G, for inhibitor implies that travel must be in the other direction i.e from extracellular to intracellular. The energy exchange is to be balanced (i.e. in and out)

In other words, 3.6 kJ/mol energy is needed to drive the inhibitor inside the cell. This energy can be provided from the Na+ moving from intercellular to the extracellular. So the driving force for the active transport of inhibitor is the force provided by the movement of sodium ions.

Therefore moles of Na+ need to enter the cell = 3.6 kJ mol-1/12.6 kJ mol-1 = 0.285 moles

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