Question

To answer this question, please reference the Problem Solving Video: Free Energy, ATP, and Creatine in Resting Muscles.

Suppose a sprinter's muscle tissue contains creatine phosphate at a concentration of 120 mM after dietary supplementation. The sprinter's muscle tissue also contains 4 mM ATP, 0.013 mM ADP, and 13 mM creatine.

Use the table of the standard free energies of hydrolysis of phosphorylated compounds and the given concentrations to calculate the free energy change, ΔG, of the creatine kinase reaction in this sprinter at 25 °C

Standard Free Energy of Hydrolysis

Compound	kJ⋅mol−1
phosphoenolpyruvate	−61.9
1,3‑bisphosphoglycerate	−49.4
creatine phosphate	−43.1
ATP (to ADP)	−30.5
glucose 1‑phosphate	−20.9
pyrophosphate	−19.3
glucose 6‑phosphate	−13.8
glycerol 3‑phosphate	−9.2

ΔG= __________ kJ⋅mol−1

Answer 1

Creatine kinase reaction:

Creatine + ATP Creatine phosphate + ADP

This reaction can be broken into two half reactions:

Creatine Creatine phosphate (deltaG = + 43.1 kJ/mole, as the reverse reaction, i.e. hydrolysis of creatine phosphate has -43.1 value)

ATP ADP (deltaG = - 30.5 kJ/mole)

Thus adding these two half reactions we get the original creatine kinase reaction. The free energy changes are also additive in this case. So the net free energy change of the reaction will be

DeltaG = (+43.1 - 30.5) kJ/mole = +12.6 kJ/mole

The free energy change is positive for this reaction.

(The given concentrations are not needed to calculate free energy change here. Free energy change remains constant in every concentration values.)