Understanding the Question

Phosphoglucoisomerase catalyzes the reversible isomerization:

Glu‑6‑P ⇌ Fru‑6‑P

At the start, only Glu‑6‑P is present with some initial concentration C₀.

At equilibrium, only 0.05% of the original Glu‑6‑P remains unconverted.

This means:

• Fraction of Glu‑6‑P left = 0.05% = 0.05/100 = 5×10⁻⁴
• Fraction converted to product Fru‑6‑P = 1−5×10⁻⁴ = 0.9995

Step-by-Step Solution

Let the initial concentration of Glu‑6‑P = C₀ and initial Fru‑6‑P = 0.

At equilibrium:

• [Glu‑6‑P]_eq = 0.0005 C₀ (because only 0.05% remains).
• The amount converted to Fru‑6‑P is C₀ − 0.0005 C₀ = 0.9995 C₀, so
• [Fru‑6‑P]_eq = 0.9995 C₀.

For the reaction

Glu‑6‑P ⇌ Fru‑6‑P,

the equilibrium constant expression is

K = [Fru‑6‑P]_eq / [Glu‑6‑P]_eq.

Substitute the equilibrium concentrations (the C₀ cancels):

K = (0.9995 C₀) / (0.0005 C₀) = 0.9995 / 0.0005 ≈ 1999 ≈ 2.0×10³.

So, the equilibrium constant K is about 2×10³.

This large value means the reaction strongly favors Fru‑6‑P at equilibrium, with almost all Glu‑6‑P converted.

SEO-Friendly Explanation

In biochemical thermodynamics, calculating the phosphoglucoisomerase equilibrium constant for Glu‑6‑P to Fru‑6‑P helps quantify how strongly a reaction favors product formation. In this example, only 0.05% of glucose‑6‑phosphate is left at equilibrium, so almost all substrate is converted into fructose‑6‑phosphate.

Because the equilibrium constant is defined as the ratio of product concentration to reactant concentration, K = [Fru‑6‑P]/[Glu‑6‑P], the very small remaining fraction of Glu‑6‑P leads to a very high K value of about 2×10³. This high equilibrium constant clearly indicates that, under standard conditions, the phosphoglucoisomerase reaction is strongly product‑favored toward fructose‑6‑phosphate.