Question

If it costs 9.7 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 -9.7 kcal/mol. Assume that \Delta Δ G°' is -7.3 kcal/mol, the concentration of inorganic phosphate is 3.3 mM, the concentration of ADP is 0.9 mM, the pH is 7.3, and that the temperature is 37 °C. Report your answer in terms of mM concentration to the nearest hundredth.

Answer 1

Hydrolysis of ATP occurs as :

ATP + H₂O = ADP + Pi

The free energy change is given by the equation :

ΔG = ΔG₀ + RT ln ([ADP] [Pi]/ [ATP]) .................(i)

Given : ΔG₀ = -7.3 kcal/mol = 30.5 KJ/mol, ΔG = -9.7 kcal/mol= -40.6 KJ/mol , [ADP]= 0.9mM, [Pi]= 3.3mM, R= 8.314 J/mol.K , T = 37^oC= 310.15 K , [ATP] = unknown

putting the values in equation :

-40.6 = -30.5 + 8.314*310.15 ln {( 0.9*3.3)/ [ATP]}

-40.6 +30.5 = 2578.6 ln {( 0.9*3.3)/ [ATP]}

ln {( 0.9*3.3)/ [ATP]} = -10.1 / 2578.6

ln {2.97)/ [ATP]} = -0.00392

ln 2.97 - ln [ATP] = -0.00392

1.088 -  ln [ATP] = -0.00392

ln [ATP] = 1.088 + 0.00392 = 1.092

2.303 log [ATP] = 1.092

log [ATP] = 1.092/ 2.303= 0.4742

[ATP] = 2.97 mM

the minimum concentration of ATP required to provide just enough energy to run the SR Ca ATPase = 2.97 mM