Question

10. Metabolism. Which of the following reactions is favorable under standard-state conditions (1 M reactants and products; pH = 7)? Calculate the value of AGⓇ for each phosphate transfer reaction. A) ATP + pyruvate ---> phosphoenolpyruvate + ADP B) 3-phosphoglycerate + ATP ---> 1,3-bisphosphoglycerate + ADP C) 1,3-bisphosphoglycerate + creatine ---> creatine phosphate + 3-phosphoglycerate D) glucose 6-phosphate --> glucose 1-phosphate not favorable noer standard

please calculate and show work

Answer 1

A reaction which has a negative ∆G value is very favourable while a reaction with a positive ∆G value is not favourable.

A reaction which has a ∆G value equal to zero is said to be at equilibrium.

Question A:

The reaction given is

ATP + pyruvate → phosphoenolpyruvate + ADP

We know that  ∆G° = ΣG°product - ΣG°reactants [ Equation 1]

The values of ∆G° can be calculated from the table given below:

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

Therefore,

G°product = ATP to ADP = -30.5 kJ mol^-1

G°reactants = Phosphoenolpyruvate = -61.9 kJ mol^-1

Therefore, applying in Equation 1, we get,

∆G° = ΣG°product - ΣG°reactants

∆G° = -30.5 kJ mol^-1 - (-61.9 kJ mol^-1)

∆G° = 31.4  kJ mol^-1

A positive ∆G value means that the reaction is not favourable.

Question B:

3-phosphoglycerate + ATP → 1,3-bisphosphoglycerate + ADP

We know that  ∆G° = ΣG°product - ΣG°reactants [ Equation 1]

Therefore,

G°product = ATP to ADP = -30.5 kJ mol^-1

G°reactants = 1,3-Bisphosphoglycerate = -49.4 kJ mol^-1

Therefore, applying in Equation 1, we get,

∆G° = ΣG°product - ΣG°reactants

∆G° = -30.5 kJ mol^-1 - (-49.4 kJ mol^-1)

∆G° = 18.9  kJ mol^-1

A negative ∆G value means that the reaction is favourable.

Question C:

1,3-bisphosphoglycerate + creatine → creatine phosphate + 3-phosphoglycerate

We know that  ∆G° = ΣG°product - ΣG°reactants [ Equation 1]

Therefore,

G°product = 1,3-Bisphosphoglycerate = -49.4 kJ mol^-1

G°reactants = Creatine phosphate = -43.1 kJ mol^-1

Therefore, applying in Equation 1, we get,

∆G° = ΣG°product - ΣG°reactants

∆G° = -49.4 kJ mol^-1 - (-43.1 kJ mol^-1)

∆G° = -6.3 kJ mol^-1

A negative ∆G value means that the reaction is favourable.

Question D:

Glucose-6-phosphate → Glucose-1-phosphate

We know that  ∆G° = ΣG°product - ΣG°reactants [ Equation 1]

Therefore,

G°product = Glucose-1-phosphate = -20.9 kJ mol^-1

G°reactants = Glucose-6-phosphate = -13.8 kJ mol^-1

Therefore, applying in Equation 1, we get,

∆G° = ΣG°product - ΣG°reactants

∆G° = -20.9 kJ mol^-1 - (-13.8 kJ mol^-1)

∆G°= -7.1  kJ mol^-1

A negative ∆G value means that the reaction is favourable. ​​​​​​​