End of Chapter, Problem 50 Use the standard free energies provided in Table 12.4

Question

End of Chapter, Problem 50 Use the standard free energies provided in Table 12.4 and information provided below to ascertain the impact of reaction conditions on the conversion of glucose-1 phosphate to glucose-6-phosphate. For the isomerization of glucose-1-phosphate to glucose-6-phosphate, the Go. [Express your answer in kJ/mol using 2 significant figures. the tolerance is+7-2% At 37°C and concentrations of 5 mM glucose-6-phosphate and 0.1 mM glucose-1-phosphate, the significant figures.] [Express your answer in k)/mol using 2 = the tolerance is +/-296 Under these differing conditions, the isomerization of G1P to G6P is spontaneous under th modified conditions (37°C) biological standard states (25 C)

Explanation / Answer

Consider the following isomerization reactions of some simple sugars and values for their standard Gibbs free energy Delta G^not:

Reaction A:
Glucose-1-phosphate-->glucose-6-phosph...
Delta G= -7.28 kJ/mol

Reaction B:
Fructose-1-phosphate -->glucose-6-phosphate
Delta G= -1.67 kJ/mol

Delta G^not=??? kJ/mol?

solution:

G-1-P G-6-P
G = –7.28 kJ/mol

G-6-P F-1-P . (Reverse of Reaction B)
G = +1.67 kJ/mol
–––––––––––––––––
G-1-P F-1-P . (add the two reactions)
G = –7.28 + 1.67 kJ/mol = –5.61 kJ/mol.

The isomerisation of glucose-6-phosphate (G6P) to fructose-6-phosphate (F6P) is a key step in the metabolism of glucose for energy. At 298 K, G6P F6P G = 1.67 kJ mol–1 Calculate the equilibrium constant for this process at 298 K. Marks 6 Using G° = -RTlnK, 1.67 × 103 = -(8.314) × (298) × lnK K = 0.510 Answer: K = 0.510

A key step in the metabolism of glucose for energy is the isomerism of glucose-6-phosphate (G6P) to fructose-6-phosphate (F6P); G6P F6P At 298 K, the equilibrium constant for the isomerisation is 0.510. Calculate G at 298 K. Marks 4 Using G° = -RTlnK: G° = -(8.314 J K-1 mol-1 ) × (298 K) × ln(0.510) = +1670 J mol-1 = +1.6 kJ mol-1 Answer: +1.6 kJ mol-1 Calculate G at 298 K when the [F6P] / [G6P] ratio = 10. Using o = + ln G G RT Q , when the reaction quotient Q = [F6P] 10 [G6P] : -1 -1 -1 = (+1670J mol ) +(8.314J K mol ) (298K) ln(1 G 0) = +7400 J mol-1 = +7.4 kJ mol-1

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