At substrate concentration equal to Km, what percentage of enzyme active sites are occupied by substrate molecules?

Explanation

## Definition of Km **Key Point:** Km (Michaelis constant) is defined as the substrate concentration at which the reaction velocity is exactly half of Vmax (V₀ = Vmax/2). ## Enzyme Saturation at [S] = Km The Michaelis-Menten equation is: $$V_0 = \frac{V_{max} \cdot [S]}{Km + [S]}$$ When [S] = Km: $$V_0 = \frac{V_{max} \cdot Km}{Km + Km} = \frac{V_{max} \cdot Km}{2Km} = \frac{V_{max}}{2}$$ **High-Yield:** Since V₀ is directly proportional to the fraction of enzyme active sites occupied by substrate, when V₀ = Vmax/2, exactly **50% of active sites are occupied**. ## Clinical Significance **Clinical Pearl:** Km is a useful measure of enzyme affinity for substrate—lower Km indicates higher affinity (enzyme reaches half-maximal velocity at lower substrate concentration). This principle is exploited in competitive enzyme inhibition assays. ## Mnemonic **HALF-SAT:** At Km, the enzyme is **HALF-SATurated** — 50% of sites bound, 50% velocity of maximum.