Q.45. The enzyme glucose isomerase catalyzes the inter-conversion of glucose and fructose as shown.
Glucose ⇄ Fructose
The 𝚫𝐆𝟎′ for this reaction is zero 𝐤𝐜𝐚𝐥/𝐦𝐨𝐥. After adding glucose isomerase to a 0.12 M glucose solution
and allowing the reaction to attain equilibrium, the final concentration of fructose in the reaction mixture
will be ____ mM .
ΔG⁰′ = 0 kcal/mol for the glucose-fructose interconversion indicates equilibrium favors equal concentrations of both sugars. At equilibrium, fructose reaches 60 mM in the reaction mixture. This GATE Life Sciences question tests biochemical equilibrium principles.
Reaction Equilibrium Explained
The reversible reaction Glucose ⇄ Fructose has ΔG⁰′ = 0 kcal/mol, meaning K_eq = [Fructose]/[Glucose] = e^(-ΔG⁰′/RT) = 1 at standard conditions (298 K, neutral pH). Equilibrium establishes a 1:1 molar ratio between glucose and fructose. Starting with 0.12 M glucose (no initial fructose), total sugar concentration remains 0.12 M as only isomerization occurs, without side products or volume changes.
Let x = equilibrium [fructose], then [glucose] = 0.12 – x. At equilibrium, x / (0.12 – x) = 1, so x = 0.06 M = 60 mM.
Calculation Breakdown
Initial: [Glucose] = 0.12 M, [Fructose] = 0 M.
Equilibrium: [Glucose] = 0.06 M, [Fructose] = 0.06 M = 60 mM.
This matches industrial glucose isomerase use, where ~50-55% fructose syrups form near 1:1 ratios under enzymatic conditions.
Options Analysis
Typical multiple-choice options for such questions:
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30 mM: Incorrect; assumes half initial concentration without equilibrium ratio.
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60 mM: Correct; derives from K_eq=1 and mass balance.
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120 mM: Incorrect; exceeds total sugar, impossible.
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0 mM: Incorrect; ignores reversible catalysis to equilibrium.
No options deviate here, as ΔG⁰′=0 enforces 50% conversion.
