Given two equations with the ΔGo′ for each reaction, be able to add or subtract an equation and determine if the reaction is spontaneous.

The concept here is that the free energy for chemical reactions are additive if they have a common intermediate.   This means that we can combine the free energies of two or more steps (reactions) in a pathway and get the overall free energy. The steps in the pathway will occur spontaneously if the overall ΔG is negative.

The add or subtract refers to whether we can simply combine the two equations or whether we have to reverse one or more equations before combining them.

We will use some of the reactions in the pathway for glycogen synthesis as examples (Figure 19.5).

Example 1: Assume standard biological conditions and that all the enzymes are present to catalyze the reactions.  Given the following data:
ATP + Glucose = Glucose-6-P + ADP   ΔGo′=  -4 kcal/mole
Glucose 6-P = Glucose-1-P                 ΔGo′=  +1.7  kcal/mole
Which of the following will occur?
A. Glucose-1-P will be converted to Glucose-6-P
B. Glucose-1-P will be converted to Glucose
C. ADP will be converted to ATP
D. Glucose-6-P will be converted to Glucose-1-P
E. None of the above
Solution: First, We look for the common intermediate, Glucose 6-P. We see that it is on opposite sides of the double arrows (the = sign) so we do not have to subtract an equation and we can combine the two equations. At the same time, we will combine the values for the a ΔGo′s..
ATP + Glucose = Glucose-6-P + ADP   ΔGo′=  -4 kcal/mole
Glucose 6-P = Glucose-1-P                 ΔGo′=  +1.7  kcal/mole

ATP + Glucose + Glucose 6-P = Glucose-6-P + ADP + Glucose-1-P
OR
ATP + Glucose = ADP + Glucose-1-P, ΔGo′= -2.3  kcal/mole

Since the ΔGo′ is negative, we assume that the reaction is spontaneous in the forward direction. The correct answer is D.

Example 2: Assume standard biological conditions and that all the enzymes are present to catalyze the reactions.  Given the following data:
Glucose 6-P = Glucose-1-P,       ΔGo′ = +1.7 kcal/mole
UTP + Glucose-1-P = UDP-Glucose + PPi,   ΔGo′ = +1.6 kcal/mole
Which of the following will occur?
A.  Glucose-1-P will be converted to UDP-Glucose
B.  Glucose-6-P will be converted to UDP-Glucose
C.  UTP will be converted to UDP-glucose
D.  UDP-Glucose will be converted to Glucose-6-P
E.  None of the above
Solution: First, We look for the common intermediate, Glucose-1-P. We see that it is on opposite sides of the duble arrows (the = sign) so no sdubtraction is necessary and we can combine the two equations. At the same time, we will combine the values for the a ΔGo′s..
Glucose 6-P = Glucose-1-P,       ΔGo′ = +1.7 kcal/mole
UTP + Glucose-1-P = UDP-Glucose + PPi,   ΔGo′ = +1.6 kcal/mole

Glucose 6-P + UTP + Glucose-1-P = Glucose-1-P + UDP-Glucose + PPi
OR
Glucose 6-P + UTP = + UDP-Glucose + PPi, ΔGo′= +3.3  kcal/mole

We now see that the equation as written has a ΔGo′ of +3.3 kcal/mole so it is not spontaneous. It will be spontaneous in the reverse direction. So, the correct answer is D.

Example 3: Assume standard biological conditions and that all the enzymes are present to catalyze the reactions.  Given the following data:
Glucose 6-P = Glucose-1-P,                      ΔGo′ =  +1.7 kcal/mole
UTP + Glucose-1-P = UDP-Glucose + PPi    ΔGo′ =  +1.6 kcal/mole
PPi = 2 Pi                                              ΔGo′ =  -4.6 kcal/mole
Which of the following will occur?
A.  2 Pi will be converted to PPi
B.  Glucose-6-P will be converted to UDP-Glucose
C.  UDP-Glucose will be converted to UTP
D.  UDP-Glucose will be converted to Glucose-6-P
E.  None of the above
Solution: First, We look for the common intermediate for the first two equations, Glucose-1-P. We see that it is on opposite sides of the duble arrows (the = sign) so no subtraction is necessary and we can combine the two equations. Next, we look for the common intermediate for the second and third equations, PPi. We see that it is on opposite sides of the duble arrows so no subtraction is necessary and we can combine all three equations. At the same time, we will combine the values for the a ΔGo′s..
Glucose 6-P = Glucose-1-P,                       ΔGo′ =  +1.7 kcal/mole
UTP + Glucose-1-P = UDP-Glucose + PPi,    ΔGo′ =  +1.6 kcal/mole
PPi = 2 Pi,                                              ΔGo′ =  -4.6 kcal/mole

Glucose 6-P + UTP + Glucose-1-P + PPi = Glucose-1-P + UDP-Glucose + PPi + 2 Pi
OR
Glucose 6-P + UTP = UDP-Glucose + 2 Pi, ΔGo′ =  -1.3 kcal/mole

We now see that the equation as written has a ΔGo′ of -1.3 kcal/mole so it is spontaneous. So, the correct answer is B.

Example 4: The metabolites and reactions may also be presented in a form where letters take the place of metabolites. Assume standard biological conditions and that all the enzymes are present to catalyze the reactions.  Given the following data:
A + B = C + D, ΔGo′ =  +1.7 kcal/mole
D + E = G + F,  ΔGo′ =  +1.6 kcal/mole
I + J = F + H ,   ΔGo′ =  +4.6 kcal/mole
Which of the following will occur?
A.  C + D will be converted to A + B
B.  G + F will be converted to D + E
C.  I + J will be converted to F + H
D.  F+ H will be converted to I + J
E.  None of the above
Solution: First, We look for the common intermediate for the first two equations, D. We see that it is on opposite sides of the double arrows (the = sign) so no subtraction is necessary and we can combine the two equations. Next, we look for the common intermediate for the second and third equations, F. We see that F is on same side of the duble arrows so subtraction is necessary. We will reverse the the third reaction and change the value of ΔGo′ from positive to negative. Then, we can combine all three equations. At the same time, we will combine the values for the a ΔGo′s..
A + B = C + D, ΔGo′ =  +1.7 kcal/mole
D + E = G + F,  ΔGo′ =  +1.6 kcal/mole
F + H = I + J ,   ΔGo′ =  -4.6 kcal/mole

A + B + D + E + F + H = C + D + G + F + I + J
OR
A + B + E + H = C + G + I + J, ΔGo′ =  -1.3 kcal/mole

We now see that the equation as written has a ΔGo′ of -1.3 kcal/mole so it is spontaneous. So, the correct answer is D.