Question

If it costs 9.9 kcal/mol to “run” the sarcoplasmic reticulum (SR) Ca ATPase (i.e., pumping one mol of Ca2+ ions from the cytoplasm to the sarcoplasmic reticulum of the muscle cell per 1 ATP hydrolyzed), then what is the minimum concentration of ATP required to provide just enough energy to run the SR Ca ATPase? In other words, what does the concentration of ATP need to be so that the free energy of ATP hydrolysis is -9.9 kcal/mol. Assume that \DeltaΔG°' is -7.3 kcal/mol, the concentration of inorganic phosphate is 3.5 mM, the concentration of ADP is 0.7 mM, the pH is 7.4, and that the temperature is 37 °C. Report your answer in terms of mM concentration to the nearest hundredth.

Answer 1

Ans> We know the ATP hydrolysis equation: ATP + H₂O -------> ADP + P_i Also G = G⁰ + (RT)ln(K_c) where  G represents the gibbs free energy and K_c represents the equillibrium constant of the ATP hydrolysis process. Now R = 1.986cal/k/mol ; T= 310K ; G⁰ = -7.3kcal/mol ; [P_i] = 3.5mM ; [ADP] = 0.7mM

Now to have just sufficient amount of energy to run the CaATPase th free energy from the ATP hydrolysis process should be -9.9kcal/mol = G

Therefore substituting these values in the above equation we get; -9900 = -7300 + (1.986 x 310)ln{[ADP][P_i]/[ATP]} ln[(3.5 x 0.7)/[ATP]} = -4.223 => 2.45/[ATP] = (-4.223) = 0.01465 Therefore [ATP] = 2.45/0.01465 = 167.235mM = 1.67 x 10² mM = 167 mM