Chapter 25, Problem 25.19P

Interpretation Introduction

Interpretation:

$∆{\mathrm{G}}^{\mathrm{\text{'}}}$ for ATP hydrolysis in a cell in which the concentrations of ATP, ADP, and phosphate are all $1\mathrm{ }\mathrm{m}\mathrm{M}$, is to be calculated.

Concept Introduction:

Hydrolysis of ATP produces ADP and phosphate. The standard free energy chain for this hydrolysis reaction is known. If the concentrations of the reactants and products are known, the $∆{\mathrm{G}}^{\mathrm{\text{'}}}$ can be calculated.

Answer to Problem 25.19P

$∆{\mathrm{G}}^{\mathrm{\text{'}}}= -48.29\mathrm{ }{\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}$

Explanation of Solution

The hydrolysis reaction of ATP is as follows:

$\mathrm{A}\mathrm{T}\mathrm{P}+{\mathrm{H}}_2\mathrm{O}\mathrm{ }\rightleftharpoons \mathrm{A}\mathrm{D}\mathrm{P}+\mathrm{P}\mathrm{h}\mathrm{o}\mathrm{s}\mathrm{p}\mathrm{h}\mathrm{a}\mathrm{t}\mathrm{e}$

The $∆{\mathrm{G}}^{0^{\mathrm{\text{'}}}}$ for this reaction is $-30.5\mathrm{ }\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$

To calculate the $∆{\mathrm{G}}^{\mathrm{\text{'}}}$, the equation is

$∆{\mathrm{G}}^{\mathrm{\text{'}}}= ∆{\mathrm{G}}^{0^{\mathrm{\text{'}}}}+2.3RT\log \frac{\left[ADP\right]\left[phosphate\right]}{\left[ATP\right]}$

The concentrations of ATP, ADP, and phosphate are all 1 mM, which is equal to $0.001\mathrm{ }\mathrm{M}$.

The value of $2.3\mathrm{R}\mathrm{T}$ at ${37}^0\mathrm{C}$ is $5.93\mathrm{ }\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}{\mathrm{l}}^{-1}$

Substituting all these values in the equation,

$∆{\mathrm{G}}^{\mathrm{\text{'}}}= ∆{\mathrm{G}}^{0^{\mathrm{\text{'}}}}+2.3RT\log \frac{\left[ADP\right]\left[phosphate\right]}{\left[ATP\right]}$

$∆{\mathrm{G}}^{\mathrm{\text{'}}}= -30.5\mathrm{ }{\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}+5.93 kJmo{l}^{-1}\log \frac{\left[0.001\right]\left[0.001\right]}{\left[0.001\right]}$

$∆{\mathrm{G}}^{\mathrm{\text{'}}}= -30.5\mathrm{ }{\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}+5.93 kJmo{l}^{-1}\left(\log 0.001\right)$

$∆{\mathrm{G}}^{\mathrm{\text{'}}}= -30.5\mathrm{ }{\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}+5.93 kJmo{l}^{-1}(-3)$

$∆{\mathrm{G}}^{\mathrm{\text{'}}}= -30.5\mathrm{ }{\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}-17.79 kJmo{l}^{-1}$

$∆{\mathrm{G}}^{\mathrm{\text{'}}}= -48.29\mathrm{ }{\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}$

Conclusion

The $∆{\mathrm{G}}^{\mathrm{\text{'}}}$ for the hydrolysis of ATP is $-48.29\mathrm{ }{\mathrm{k}\mathrm{J}\mathrm{m}\mathrm{o}\mathrm{l}}^{-1}$.