Question

In simple enzyme kinetics, when V becomes Vmax, Km is equal to __________.

• A. \(\frac{1}{2}\text{Vmax}\)
• B. \(\frac{\text{Vmax}}{3}\)
• C. \(\frac{\text{Vmax}}{4}\)
• D. \(\frac{\text{Vmax}}{6}\)

Hint:

Always remember: \(K_m\) is the substrate concentration required to achieve exactly half of the maximum velocity (\(\frac{1}{2}V_{\text{max}}\)).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Enzyme kinetics describe how the rate of an enzyme-catalyzed reaction varies with the concentration of substrate. This relationship is mathematically represented by the Michaelis-Menten equation.

Step 2: Detailed Explanation:
The Michaelis-Menten equation is expressed as:
\[ V = \frac{V_{\text{max}}[S]}{K_m + [S]} \] where:
- \(V\) is the initial reaction velocity.
- \(V_{\text{max}}\) is the maximum reaction velocity at saturating substrate concentrations.
- \([S]\) is the substrate concentration.
- \(K_m\) is the Michaelis constant.
Let us analyze the definition and significance of \(K_m\):
By definition, the Michaelis constant \(K_m\) is equal to the substrate concentration at which the reaction velocity (\(V\)) reaches exactly half of the maximum velocity (\(\frac{1}{2}V_{\text{max}}\)).
Let us prove this mathematically by substituting \(V = \frac{1}{2}V_{\text{max}}\) into the equation:
\[ \frac{1}{2}V_{\text{max}} = \frac{V_{\text{max}}[S]}{K_m + [S]} \] Divide both sides by \(V_{\text{max}}\):
\[ \frac{1}{2} = \frac{[S]}{K_m + [S]} \] Cross-multiply and solve:
\[ K_m + [S] = 2[S] \implies K_m = [S] \] Therefore, when the reaction rate is half of \(V_{\text{max}}\), the substrate concentration is equal to \(K_m\).
In simple enzyme kinetics exams, the conceptual value associated with the defining threshold of \(K_m\) is \(\frac{1}{2}V_{\text{max}}\).

Step 3: Final Answer:
Therefore, the Michaelis constant \(K_m\) is conceptually defined at a velocity of \(\frac{1}{2}V_{\text{max}}\).