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Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glyco...

Glycolysis is the process by which energy is harvested from glucose by living things. Several of the reactions of glycolysis are thermodynamically unfavorable (nonspontaneous), but proceed when they are coupled with other reactions.

Which of these reactions is (are) unfavorable? Select all that apply. Which of these reactions can be coupled so that overall reaction is favorable? Select all that apply. What is the net change in free energy if one selection from part (b) is coupled so that the overall reaction is favorable?
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Answer #1
Concepts and reason

The problem is based on the concept of spontaneity of the reaction. For a reaction to be spontaneous, change in Gibbs free energy is less than zero. Gibbs free energy is related to enthalpy of the system as follows:

ΔG=ΔHTΔS\Delta G = \Delta H - T\Delta S

Here, ΔH\Delta H is the enthalpy change of the system, T is absolute temperature and ΔS\Delta S is the entropy change of the system.

Fundamentals

For two reactions with different Gibbs free energy, the reaction with negative value of change in Gibbs free energy is spontaneous and favorable. If the value of Gibbs free energy is positive then the reaction is non-favorable.

(a)

Reactions showing glycolysis are as follows:

Pi+glucoseglucose6phosphate+H2OΔG=13.8kJmol1{P_i} + {\rm{glucose}} \to {\rm{glucose}} - 6 - {\rm{phosphate}} + {{\rm{H}}_{\rm{2}}}{\rm{O }}\Delta G = 13.8{\rm{ kJ mo}}{{\rm{l}}^{ - 1}} …… (A)

Pi+fructose6phosphatefructose1,6biphosphate+H2OΔG=16.3kJmol1{P_i} + {\rm{fructose}} - 6 - {\rm{phosphate}} \to {\rm{fructose}} - 1,6 - {\rm{biphosphate}} + {{\rm{H}}_{\rm{2}}}{\rm{O }}\Delta G = 16.3{\rm{ kJ mo}}{{\rm{l}}^{ - 1}} …… (B)

ATP+H2OADP+PiΔG=30.5kJmol1{\rm{ATP}} + {{\rm{H}}_2}{\rm{O}} \to {\rm{ADP}} + {{\rm{P}}_i}{\rm{ }}\Delta {\rm{G}} = - 30.5{\rm{ kJ mo}}{{\rm{l}}^{ - 1}} …… (C)

For negative value of change in Gibbs free energy reactions are favorable. Therefore, those reaction which have positive value of change in Gibbs free energy are unfavorable. From above three reactions, the value of Gibbs free energy is positive for reaction (A) and reaction (B) therefore both the reactions are unfavorable.

(b)

Reactions are couples in such a way that final change in Gibbs free energy is negative. From reaction A, B and C, Since, value of Gibbs free energy is positive for both A and B the coupling of A and B will give positive value only.

Since, the value of Gibbs free energy is negative for only C whose magnitude is greater than Gibbs free energy of both A and B. Thus, the correct coupling is of reaction C with both A and B because for both the coupling, value of change in Gibbs free energy is negative means favorable reaction.

(c)

Since, the correct coupling is of reaction C with both A and B calculate change in Gibbs free energy of either of one set.

For A and C set,

Calculate net change in Gibbs free energy as follows:

ΔGT=ΔGA+ΔGC\Delta {G_{\rm{T}}} = \Delta {G_{\rm{A}}} + \Delta {G_{\rm{C}}}

Substitute 13.8kJmol113.8{\rm{ kJ mo}}{{\rm{l}}^{ - 1}} for ΔGA\Delta {G_{\rm{A}}} and 30.5kJmol1 - 30.5{\rm{ kJ mo}}{{\rm{l}}^{ - 1}} for ΔGC\Delta {G_{\rm{C}}} thus,

ΔGT=(13.8kJmol1)+(30.5kJmol1)=16.7kJmol1\begin{array}{c}\\\Delta {G_{\rm{T}}} = \left( {13.8{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}} \right) + \left( { - 30.5{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}} \right)\\\\ = - 16.7{\rm{ kJ mo}}{{\rm{l}}^{ - 1}}\\\end{array}

Ans: Part a

Reaction A and B are unfavorable.

Part b

Reaction C is coupled with both A and B.

Part c

Net change in Gibbs free energy is 16.7kJmol1 - 16.7{\rm{ kJ mo}}{{\rm{l}}^{ - 1}} .

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