Question

For the problem in Sample Calculation 10-1, calculate ΔG at 37°C when the membrane potential is (a) -50 mV (cytosol negative) and (b) +150 mV. In which case is Ca²⁺ movement in the indicated direction thermodynamically favorable?

Short answer

(a) The ΔGis $-33.3{\mathrm{kJmol}}^{-1}$. Thus, the movement of Ca²⁺ is thermodynamically favorable.

(b) The ΔG is $+5.2{\mathrm{kJmol}}^{-1}$.

Step-by-step solution

Chemical potential difference

The amount of free energy required to move a chemical over a membrane is determined by the concentrations on both sides. The membrane potential determines the free energy change in the case of ions.

Chemical potential difference is caused by a difference in the concentrations of chemicals on opposite sides of the membrane.The electrochemical potential is defined as the differential in charge trans-membrane transport of ions across a membrane.

In solution (a), Calculation of ΔG at -50mV

Given, Z=2 and T=210K

The intracellular concentration of$\left[{\mathrm{Ca}}^{2+}\right]$ is${10}^{-7}$

The extracellular concentration of$\left[{\mathrm{Ca}}^{2+}\right]$ is${10}^{-3}$

The chemical potential difference is,

$∆\mathrm{G}=\mathrm{RTln}\frac{{\left[{\mathrm{Ca}}^{2+}\right]}_{\mathrm{in}}}{{\left[{\mathrm{Ca}}^{2+}\right]}_{\mathrm{out}}}+\mathrm{ZF}∆\mathrm{\phi }$

On substituting,

$$\begin{gathered} =\left(310\mathrm{K}\right)\ln \frac{\left[{10}^{-7}\right]}{\left[{10}^{-3}\right]}+2\left(96,485{\mathrm{JV}}^{-1}{\mathrm{mol}}^{-1}\right)\left(-0.050\mathrm{V}\right) \\ =-23700{\mathrm{Jmol}}^{-1}-9600{\mathrm{Jmol}}^{-1} \\ =-33.3{\mathrm{kJmol}}^{-1} \end{gathered}$$

The process is thermodynamically favorable as ΔG value is negative.

In solution (b), Calculation of ΔG at +150mV

The intracellular concentration of$\left[{\mathrm{Ca}}^{2+}\right]$ is ${10}^{-7}$

The extracellular concentration of $\left[{\mathrm{Ca}}^{2+}\right]$is ${10}^{-3}$

The chemical potential difference is,

$∆\mathrm{G}=\mathrm{RTln}\frac{{\left[{\mathrm{Ca}}^{2+}\right]}_{\mathrm{in}}}{{\left[{\mathrm{Ca}}^{2+}\right]}_{\mathrm{out}}}+\mathrm{ZF}∆\mathrm{\phi }$

On substituting,

$$\begin{gathered} =\left(310\mathrm{K}\right)\ln \frac{\left[{10}^{-7}\right]}{\left[{10}^{-3}\right]}+2\left(96,485{\mathrm{JV}}^{-1}{\mathrm{mol}}^{-1}\right)\left(+0.150\mathrm{V}\right) \\ =-23700{\mathrm{Jmol}}^{-1}+28900{\mathrm{Jmol}}^{-1} \\ =+5.2{\mathrm{kJmol}}^{-1} \end{gathered}$$

The process is thermodynamically unfavorable as ΔG value is positive.

Hence, the chemical potential difference when the membrane potential is$-50\mathrm{mV}=-33.3{\mathrm{kJmol}}^{-1}$ and the chemical potential difference when the membrane potential difference is $+150\mathrm{mV}=+5.2{\mathrm{kJmol}}^{-1}$.