Question

5 of 17 > To answer this question, please reference the Problem Solving Video: Free Energy, ATP, and Creatine in Resting Muscles. The standard free energy (AG) of the creatine kinase reaction is -12.6 kJ. mol-'. The AG value of an in vitro creatine kinase reaction is -0.1 kJ. mol". At the start of the reaction, the concentration of ATP is 5 mM, the concentration of creatine is 17 mm, and the concentration of creatine phosphate is 25 mM. Using the values given, calculate the starting concentration of ADP in nanomolar. (ADP) = nM

Answer 1

creatine + ATP ------> ADP + creatine phosphate    $\Delta$ G⁰   = -12.6KJ/mole

$\Delta$G    = -0.1KJ/mole

$\Delta$G⁰   = -12.6KJ/mole   = -12600J/mole

$\Delta$G    = -0.1KJ/mole    = -100J/mole

$\Delta$G   = $\Delta$ G⁰ + RTlnQc

-12600   = -100 + 8.314*298lnQc

8.314*298lnQc = -12600+100

lnQc   = -12500/(8.314*298)

lnQc   = -5.045

Qc   = 9.02*10^-6

[ATP] = 5mM = 5*10^-3M

[creatine] = 17mM = 17*10^-3M

[creatine phosphate] = 25mM = 25*10^-3M

Qc = [creatine phosphate][ADP]/[ATP][creatine]

9.02*10^-6   = 25*10^-3*[ADP]/5*10^-3*17*10^-3

[ADP]    = 9.02*10^-6*5*10^-3*17*10^-3/(25*10^-3)

[ADP]   = 30.668*10^-9M

             = 30.668nM >>>>answer