Question

To answer this question, please reference the Problem Solving Video: Free Energy, ATP, and Creatine in Resting Muscles. The standard free energy ( Δ G ∘ ′ ) of the creatine kinase reaction is − 12.6 kJ ⋅ mol − 1 . The Δ G value of an in vitro creatine kinase reaction is − 0.1 kJ ⋅ mol − 1 . At the start of the reaction, the concentration of ATP is 6 mM, the concentration of creatine is 12 mM, and the concentration of creatine phosphate is 25 mM.

Using the values given, calculate the starting concentration of ADP in micromolar.

ADP=. µM

Answer 1

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

$\Delta$G   = -0.1KJ/mole

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

[Creatine]                   = 12mM    = 12*10^-3M

[ATP]   =6mM = 6*10^-3M

$\Delta$G     = $\Delta$ G0 + RTlnQc

-12600   = -100 + 8.314*298lnQc

-12600+100 = 8.314*298lnQc

-12500   = 2477.57lnQc

lnQc = -12500/2477.57

lnQc = -5.045

Qc   = 6.44*10^-3

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

6.44*10^-3   = 25*10^-3[ADP]/12*10^-3*6*10^-3

[ADP]   = 6.44*10^-3*12*10^-3*6*10^-3/(25*10^-3)

[ADP]   = 1.85*10^-5 M

[ADP]   = 18.5*10^-6M

[ADP]   = 18.5$\mu$M >>>>>answer