Question:

The Michaelis constant (Km) in enzyme kinetics represents

Show Hint

Visualize the Michaelis-Menten curve (a plot of reaction rate vs. substrate concentration). Find the V\(_max\) on the y-axis, go down to V\(_max\)/2, and then find the corresponding substrate concentration on the x-axis. That value is K\(_m\).
Updated On: Sep 24, 2025
  • The maximum reaction velocity
  • The substrate concentration at half of Vmax
  • The enzyme concentration
  • The reaction rate at time t=0
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The Michaelis constant (K\(_m\)) is a fundamental parameter in the Michaelis-Menten model of enzyme kinetics. It describes the relationship between the substrate concentration and the reaction velocity.
Step 2: Key Formula or Approach:
The Michaelis-Menten equation is given by: \[ v_0 = \frac{V_{max}[S]}{K_m + [S]} \] where \(v_0\) is the initial reaction velocity, \(V_{max}\) is the maximum velocity, \([S]\) is the substrate concentration, and K\(_m\) is the Michaelis constant.
Step 3: Detailed Explanation:
To understand what K\(_m\) represents, we can analyze the equation under a specific condition. Let's find the substrate concentration when the velocity is exactly half of the maximum velocity, i.e., when \(v_0 = \frac{1}{2}V_{max}\).
Substituting this into the equation: \[ \frac{1}{2}V_{max} = \frac{V_{max}[S]}{K_m + [S]} \] We can cancel \(V_{max}\) from both sides: \[ \frac{1}{2} = \frac{[S]}{K_m + [S]} \] Now, we can solve for \([S]\): \[ K_m + [S] = 2[S] \] \[ K_m = 2[S] - [S] \] \[ K_m = [S] \] This derivation shows that K\(_m\) is numerically equal to the substrate concentration at which the reaction velocity is half of its maximum. It is also an inverse measure of the enzyme's affinity for the substrate; a lower K\(_m\) indicates a higher affinity.
Step 4: Final Answer:
Based on the definition and derivation, K\(_m\) represents the substrate concentration at half of V\(_max\).
Was this answer helpful?
0
0