Be able to explain how the free energy in ATP can be used to synthesize glucose-6-phosphate from glucose and phosphate even though this reaction is endergonic.

The concept here is that chemical reactions are additive if they occur at the same time and at the same active site.  That is, one reaction cannot occur without the other so the free energy in ATP or some other high-energy compound can be used for the biosynthesis of another compound as long as the overall ΔG is negative.

Remember, that you may change the sign of ΔGo′ or ΔG by reversing the direction of an equation.  That is,

Glucose + Pi ---> Glucose-6-P + H2O, ΔGo′ = +3.3 kcal/mol

Can be rewritten as

Glucose-6-P + H2O ---> Glucose + Pi, ΔGo′ = -3.3 kcal/mol

Remember, that you may add the partial reactions together to get the final reaction and, at the same time, combine the ΔGo’s or ΔGs to get the overall ΔGo’ or ΔG as shown below.

Glucose + Pi = Glucose-6-P + H2O,  ΔGo′ = +3.3 kcal/mol

ATP + H2O = Pi + ADP,                  ΔGo′ = -7.0 kcal/mole

Glucose + Pi + ATP + H2O = Glucose-6-P + H2O + Pi + ADP, ΔGo′= -3.7

Removing H2O and Pi from both sides,

Glucose + ATP = Glucose-6-P + ADP, ΔGo′= -3.7

Here are some examples:

Example 1: Assume standard biological conditions and that all the enzymes are present to catalyze the reactions.  Given the following data:
Glucose + Pi = glucose-6-P + H2O,  ΔGo′ = +3.3 kcal/mol
ATP + H2O = Pi + ADP,                ΔGo′ = -7.0 kcal/mole
Calculate the ΔGo′ for the following reaction and determine whether the reaction is spontaneous
Glucose + ATP --> Glucose-6-P + ADP
A.  –10.3 kcal/mol and spontaneous
B.  +10.3 kcal/mol and not spontaneous
C.  –3.7 kcal/mol and not spontaneous
D.  +3.7 kcal/mol and not spontaneous
E.  –3.7 kcal/mol and spontaneous
The answer is worked out above.

Example 2: Assume standard biological conditions and that all the enzymes are present to catalyze the reactions.  Given the following data:
Phosphoenolpyruvate + H20 = Pi + Pyruvate,       ΔGo′ = -15 kcal/mole
ATP + H20 = Pi + ADP,                                     ΔGo′ = -7 kcal/mole
Calculate the delta ΔGo′ for the following reaction and determine whether the reaction is spontaneous
Phosphoenolpyruvate + ADP = ATP + Pyruvate
A.  -22 kcal/mol and spontaneous
B.  +22 kcal/mol and not spontaneous
C.  -8 kcal/mol and not spontaneous
D.  +8 kcal/mol and not spontaneous
E.  -8 kcal/mol and spontaneous
Solution: First, since ADP is a product, we must reverse the second equation above and change the sign of the ΔGo′ from minus to plus.
ATP + H20 = Pi + ADP,  ΔGo′ = -7 kcal/mole
becomes
ADP + Pi = ATP + H20,  ΔGo′ = +7 kcal/mole
Next, we combine the two partial equations
Phosphoenolpyruvate + H20 = Pi + Pyruvate, ΔGo′ = -15 kcal/mole
ADP + Pi = ATP + H20,   ΔGo′ = +7 kcal/mole

Phosphoenolpyruvate + H20 + ADP + Pi = ATP + H20 + Pi + Pyruvate
OR
Phosphoenolpyruvate + ADP = ATP + Pyruvate, ΔGo′ = -8 kcal/mole

We now see that the equation as written has a ΔGo′ of -8 kcal/mole so it is spontaneous. It would be not be spontaneous in the opposite direction. So, the correct answer is E.

Example 3: Assume standard biological conditions and that all the enzymes are present to catalyze the reactions.  Given the following data:
Glycerol-3-P + H20 = Pi + Glycerol,   ΔGo′ = -2 kcal/mole
ATP + H20 = Pi + ADP,                    ΔGo′ = -7 kcal/mole
Calculate the delta ΔGo′ for the following reaction and determine whether the reaction is spontaneous
Glycerol + ATP = ADP + Glycerol-3-P

A.  -9 kcal/mol and spontaneous
B.  +9 kcal/mol and not spontaneous
C.  +5 kcal/mol and spontaneous
D.  -5 kcal/mol and not spontaneous
E.  -5 kcal/mol and spontaneous

Solution: First, since Glycerol-3-P is a product, we must reverse the first equation above and change the sign of the ΔGo′ from minus to plus.
Glycerol-3-P + H20 = Pi + Glycerol,   ΔGo′ = -2 kcal/mole
becomes
Pi + Glycerol = Glycerol-3-P + H20,  ΔGo′ = +2 kcal/mole
Next, we combine the two partial equations
Pi + Glycerol = Glycerol-3-P + H20,  ΔGo′ = +2 kcal/mole
ATP + H20 = Pi + ADP,                    ΔGo′ = -7 kcal/mole

Pi + Glycerol + ATP + H20 = Glycerol-3-P + H20 + Pi + ADP
OR
Glycerol + ATP = Glycerol-3-P + ADP, ΔGo′ = -5 kcal/mole

We now see that the equation as written has a ΔGo′ of -5 kcal/mole and is spontaneous. It would not be spontaneous in the opposite direction. So, the correct answer is E.