Strenuous muscle exertion (as in the 100 -meter dash) rapidly depletes ATP levels. How long will 8 m \(M\) ATP last if 1 gram of muscle consumes \(300 \mu\) mol of ATP per minute? (Assume muscle is \(70 \%\) water.) Muscle contains phosphocreatine as a reserve of phosphorylation potential. Assuming [phosphocreatine] \(=40 \mathrm{m} M,[\text { creatine }]=4 \mathrm{m} M,\) and \(\left.\Delta G^{\circ \prime} \text { (phosphocreatine }+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \text { creatine }+\mathrm{P}_{\mathrm{i}}\right)=-43.3 \mathrm{kJ} / \mathrm{mol}\) how low must [ATP] become before it can be replenished by the reaction: phosphocreatine \(+\mathrm{ADP} \rightleftharpoons \mathrm{ATP}+\) creatine? [Remember \(\Delta G^{\circ \prime}\) (ATP hydrolysis) \(=-30.5 \mathrm{kJ} / \mathrm{mol} .\)]

First, calculate how long 8 mM ATP will last if 1 gram of muscle consumes 300 μmol of ATP per minute. 1 mM of ATP in 1 g of muscle is 1 mmol/kg, since 1 g is 0.001 kg. Then, for 70% of muscle weight that consists of water, 1 mM of ATP is equal to 1 mmol/L or 1 μmol/ml. Therefore, 8 mM of ATP equals 8 μmol in 1 ml of muscle. So, given the consumption rate of 300 μmol/min, time in minutes will be \( \frac{8}{300} \) or approximately 0.027 minutes.

Now, calculate the minimum [ATP] for ATP replenishment to occur via the reaction of phosphocreatine and ADP. Here, apply the equation for calculating Gibbs free energy in a reaction: \( \Delta G = \Delta G^{\circ \prime} + RT \ln(Q) \), where R is the universal gas constant (8.314 J / (mol × K), T is the absolute temperature (assume 298 K for human body temperature), ∆G°’ is the standard Gibbs free energy change for a reaction, and Q is the reaction quotient. For the reaction phosphate creatine → creatine + Pi, we know \( \Delta G^{\circ \prime} = -43.3 kJ/mol \). Similarly, for the ATP → ADP + Pi reaction, we know \( \Delta G^{\circ \prime} = -30.5 kJ/mol \). The Gibbs free energy change for the reaction of Phosphocreatine + ADP -> ATP + Creatine must therefore equal \( \Delta G_{Phosphocreatine->Creatine} + \Delta G_{ATP->ADP} = -43.3 kJ/mol + -30.5 kJ/mol = -73.8 kJ/mol \). Since the reaction ATP + Creatine -> ADP + Phosphocreatine is the reverse of this, its \( \Delta G^{\circ \prime} \) will be 73.8 kJ/mol. But the reaction must be in equilibrium to shift to the left and make ATP, so ∆G must equal 0. Therefore, set \( \Delta G = 0 \) in the equation from before and rearrange to derive the formula for Q: \( Q = e^{-\Delta G^{\circ \prime}/(RT)} \). Use this formula to find Q at equilibrium (Q = [ADP][Creatine]/[ATP][Phosphocreatine]). Then, assume constant [ADP] and [Creatine] and solve the equation for [ATP].

Plug in the values to calculate Q. Here, \( \Delta G^{\circ \prime} = 73.8×10^3 J, R = 8.31 J/K/mol, T = 298 K \). Thus, \( Q = e^{-73800/(8.31*298)} = 4.06×10^-14 \). At equilibrium, Q = [ADP][Creatine]/[ATP][Phosphocreatine]. Assuming constant [ADP] and [Creatine] and that [Creatine] = 4 mM and [Phosphocreatine] = 40 mM, solve for [ATP]: [ATP] = [ADP]*[Creatine]/(Q*[Phosphocreatine]) = [ADP]*4 mM / (4.06×10^-14 * 40 mM). Assume [ADP] = 0.2 mM as reasonable physical condition and insert it to the equation, to get [ATP] ≈ 1.23 mM.