This question pertains to enzyme kinetics, specifically the Michaelis-Menten model, which is a cornerstone for understanding enzyme function. The model describes how the initial rate of an enzyme-catalyzed reaction (\(v\)) is affected by the concentration of its substrate (\([S]\)). The equation includes two key constants: \(V_{max}\) (the maximum reaction rate) and \(K_m\) (the Michaelis constant).
Understanding the Question
The question asks for the physical or operational definition of the Michaelis constant, \(K_m\).
Key Formula and Approach
The approach is to analyze the Michaelis-Menten equation to understand the relationship between \(v\), \([S]\), and \(K_m\).
The equation is: \[ v = \frac{V_{max}[S]}{K_m + [S]} \]
Detailed Solution
Setting the Condition: To define \(K_m\), let's consider the specific condition where the reaction velocity \(v\) is exactly half of the maximum velocity, i.e., \(v = \frac{V_{max}}{2}\).
Substituting into the Equation: We substitute this condition into the Michaelis-Menten equation:
\[ \frac{V_{max}}{2} = \frac{V_{max}[S]}{K_m + [S]} \]
Solving for \([S]\): We can cancel \(V_{max}\) from both sides:
\[ \frac{1}{2} = \frac{[S]}{K_m + [S]} \]
Cross-multiplying gives:
\[ K_m + [S] = 2[S] \]
Subtracting \([S]\) from both sides yields:
\[ K_m = [S] \]
Conclusion: This derivation shows that \(K_m\) is numerically equal to the substrate concentration at which the reaction rate is half of its maximum. It is also an inverse measure of the enzyme's affinity for its substrate (a lower \(K_m\) implies higher affinity).