Question

Many biochemical reactions that occur in cells requirerelatively high concentrations of potassium ion (K+ ).The concentration of $K^{+}$in muscle cells is about 0.15 M. The concentration of $K^{+}$in blood plasma is about 0.0050 M. The high internal concentration in cells is maintained by pumping $K^{+}$from the plasma. How much work must be done to transport 1.0 mole of $K^{+}$from the blood to the inside of a muscle cell at 37°C (normal body temperature)? When 1.0 mole of $K^{+}$is transferred from blood to the cells, do any other ions have to be transported? Why or why not? Much of the ATP (see Exercise 84) formed from metabolic processes is used to provide energy for transport of cellular components. How much ATP must be hydrolyzed to provide the energy for the transport of 1.0 mole of $K^{+}$?

Short answer

The ATP that must be hydrolyzed to provide the energy for the transport of 8.8 KJ of 1.0 mole of $K^{+}$is 0.29 mol of ATP.

Step-by-step solution

Introduction to the Concept

The relation between change inGibbs free energy and equilibrium constant is as
follows:
$\Delta G=\Delta G^o+RTlnK$
Here,
$\Delta G=$Gibbs free energy changes
$\Delta G°=$ Gibbs free energy changes under standard conditions.
R = Universal gas constant
T = temperature
K = equilibrium constant

Determination of  ΔG

1. It has a concentration of $K^{+}$in muscle cells

The concentration of $K^{+}$in blood plasma

$K^{+}\left(blood\right)=K^{+}\left(muscle\right)$

$\Delta G^o=0$

$\Delta G=RT\times \ln \times \frac{{\left[K^{+}\right]}_{muscle}}{{\left[K^{+}\right]}_{blood}}$

$=8.314J/molK\times 310K\times \ln \left(\frac{0.15}{0.0050}\right)$

$=8.8\times {10}^{3}J/mol$

$\Delta G=w_{\max }$

Thus work that must be done to transport 1.0 mole of $K^{+}$should be at least 8.8 kJ.

Other ions should be delivered alongside it to preserve electroneutrality. Anions should be carried into cells, or cations in cells should be delivered into the bloodstream. In practice, $N{a}^{+}$in the cell is transferred into the bloodstream.

ATP should be hydrolyzed$=\frac{1\text{mol of ATP}}{30.5\text{kJ}}\times 8.8\text{kJ}$

$=0.29molofATP$

Therefore, to transport 1.0 mole of K, 8.8 KJ of work is required. The amount of ATP that should be hydrolyzed is 0.29 mol of ATP .