Chemistry

Problem

A biochemist needs to drive the phosphorylation of glucose to glucose-6-phosphate under physiological conditions. The direct phosphorylation reaction has a standard Gibbs free-energy change of

\Delta G^{\circ}=+13\,\text{kJ mol}^{-1}

. To power this step, the biochemist can choose one of the three reactions in Table 1. Table 1 Candidate Driving Reactions (all values at

298\,\text{K}

pH\,=7.0

)

Reaction (all aqueous)	$\Delta G^{\circ}\,(\text{kJ mol}^{-1})$
(I) ATP $^{4-}$ → ADP $^{3-}$ + P $_\text{i}^{2-}$	−30
(II) PEP → pyruvate $^{-}$ + P $_\text{i}^{2-}$	−62
(III) phosphocreatine $^{2-}$ → creatine + P $_\text{i}^{2-}$	−43

Because the laboratory protocol prohibits direct phosphate transfer between the driving reaction and the glucose substrate, an enzyme must split the driving reaction into two half-reactions that share a phosphorylated intermediate. Only candidate reactions whose mechanism allows formation of such an intermediate are acceptable. (a) Identify which of the three candidate reactions in Table 1 can, in principle, be used to power the phosphorylation of glucose without violating the “no direct phosphate transfer” restriction. Briefly describe the mechanistic requirement that eliminates the other reaction(s). (b) For the permitted driving reaction(s) from part (a), write the overall coupled reaction showing cancellation of the common intermediate. Calculate

\Delta G^{\circ}_{\text{overall}}

for each possibility. (c) Based on your calculations, determine which coupling choice is thermodynamically sufficient yet wastes the least free energy. Justify your selection in one paragraph, referring specifically to the numerical

\Delta G^{\circ}

values and the practical consequence of large negative free-energy changes in metabolic regulation. (a) Identification and mechanistic explanation (b) Construction of overall reaction and calculation of

\Delta G^{\circ}_{\text{overall}}

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