The concentration of Na^+ inside a squid axon is 0.050 M (50 mM), in comparison with [Na^+] = 0.440 M (440 mM) in the fluid surrounding the cell. a. Calculate the equilibrium potential Eion for the Na^+ ion at 310 K. b. If the membrane potential Emembrane = –0.050 V, is the transport of Na^+ from

Question

The concentration of [latex]Na^{+}[/latex] inside a squid axon is 0.050 M (50 mM), in comparison with [latex][Na^{+}] = 0.440 M (440 mM)[/latex] in the fluid surrounding the cell.

a. Calculate the equilibrium potential [latex]E_{ion}[/latex] for the [latex]Na^{+}[/latex] ion at 310 K.
b. If the membrane potential [latex]E_{membrane} = –0.050 V[/latex], is the transport of [latex]Na^{+}[/latex] from inside to outside the cell membrane spontaneous?

Accepted Answer

Collect, Organize, and Analyze We are given the concentrations of [latex]Na^{+}[/latex] on opposite sides of a cell membrane, and we are asked to calculate [latex]E_{ion}[/latex] for [latex]Na^{+}[/latex] and to determine whether the transport across the membrane is spontaneous. Equation 21.3

[latex]E_{ion}=-\frac{RT}{nF} \ln \left(\frac{[M^{x+}_{outside}]}{[M^{x+}_{inside}]} ight) [/latex] (21.3)

allows us to calculate the equilibrium potential that arises from an unequal concentration of an ion on either side of a permeable membrane. If [latex][Na^{+}_{outside}] > [Na^{+}_{inside}][/latex], then the [latex]\ln[/latex] term in the equation will be greater than 1 and [latex]E_{ion}[/latex] will be negative. Using Equations 21.4 and 21.5,

[latex]E_{transport} = E_{membrane}- E_{ion}[/latex] (21.4)

[latex]\Delta G_{transport} = -nFE_{transport}[/latex] (21.5)

we can calculate the work done during transport; if [latex]\Delta G[/latex] is positive, the process is nonspontaneous; if negative, it is spontaneous.

Solve
a. Substitution into Equation 21.3 gives:

[latex]E_{Na^{+}}=-\frac{RT}{nF} \ln \frac{[Na^{+}_{outside}]}{[Na^{+}_{inside}]} =-\frac{[8.314 \sout{ ext{J}}/ (\sout{ ext{mol}} \cdot \sout{ ext{K}}) ] (310 \sout{ ext{K}})}{(1/\sout{ ext{mol}}) (96,500 \sout{ ext{J}}/V)} \ln \left(\frac{440 \sout{ ext{mM}}}{50 \sout{ ext{mM}}} ight) [/latex]

[latex]= -0.058 V = -58 mV[/latex]

b. Using Equation 21.4, we can calculate [latex]E_{transport}[/latex], which is the driving force available to push the ions across the membrane:

[latex]E_{transport} = E_{membrane}- E_{ion} = -0.050 V -(-0.058 V) = 0.008 V = 8 mV[/latex]

Substituting 8 mV into Equation 21.5:

[latex]\Delta G_{transport} = -nFE_{transport} = -(1/mol)(96,500 J/\sout{ ext{V}})(0.008 \sout{ ext{V}}) = -772 J/mol[/latex]

[latex]=-8 imes 10^{2} J/mol[/latex]

Because [latex]\Delta G < 0 [/latex], the transport is spontaneous.

Think About It The equilibrium potential for [latex]Na^{+}[/latex] under these conditions is negative, consistent with our prediction. The transport of [latex]Na^{+}[/latex] from the inside to the outside of the cell membrane is spontaneous and goes against the concentration gradient under these conditions. The membrane potential (–0.050 V) would not have to vary by much to change the transport from spontaneous to nonspontaneous, and research is ongoing to find ways to modify cell potentials as a means to inhibit both tumor growth and spread.

Question 21.1: The concentration of Na^+ inside a squid axon is 0.050 M (50...

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