The following questions relate to ill eral Chpters 12-15 (nainC.1 26. The Nernst
ID: 541806 • Letter: T
Question
The following questions relate to ill eral Chpters 12-15 (nainC.1 26. The Nernst Equation describes the relationship between the concentration difference of permeating ion across a membrane and the membrane potential at equilibrium. The general form of the equation is Ex-(RT/ zF) × In ((XIow/ [xln), where E.-membrane potential for ion x, R = gas constant, T-absolute temperature, z-valence of ion, F= Faraday's constant (charge per mol ions), [Xout ion concentration outside, and [X]n-ion concentration inside. At room temperature (18 C), and converting from natural logarithms (In) to common logarithms (log), the Nernst equation reduces to E.-(0.058 /2) × log ([Xlod / [X]n). what would the equilibrium potentials be for each of the following ions at the given concentrations? HINT: Use 58 in the Goldman equation for results in mV or 0.058 for results in V (A) [ K'Jout = 3 mM, [K'],-150 mM (B) [Na lou-100 mM, [Nai 10 mM 27. For a typical cell that is 100 times more permeable to K' than to any other ion, use the Goldman equation to determine the potential change that would be produced by a doubling of the extracellular K concentration. (Use appropriate concentrations from the previous question). HINTS: Use 58 in the Goldman equation for results in mV or 0.058 for results in V. Because this problem is about change (one value relative to another), you do not need to know the permeabilities of individual ions; you only need to know that PK is 100 times higher than PNa and PCl Be careful how to handle [CT] because of its negative charge.Explanation / Answer
26. Using the given equatyion and hint, we get the following:
A. Ex = 0.058 * log(3/150) = -0.0985V
B. Ex = 0.058 * log(100/10) = 0.058V
C. Ex = -0.058 * log(110/4) = -0.083V [the ionic charge determines the sign of the membrane potential contribution]
Note: all the given ions were mono valent, therefore z = 1 for all cases.
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