Question

The activity of an enzyme having Michaelis constant (Km) = x \(\mu\)M will be 80% of Vmax at a substrate concentration of

• A. 2 x \(\mu\)M
• B. 4 x \(\mu\)M
• C. 8 x \(\mu\)M
• D. 10 x \(\mu\)M

Hint:

Use v = 0.8 Vmax in the Michaelis-Menten equation and solve for [S] in terms of Km.

Answer 1

The Correct Option is B

Step 1: Recall the Michaelis-Menten equation.
The rate of an enzyme reaction, v, relates to substrate concentration [S] by
\[ v = \frac{V_{max}[S]}{K_m+[S]} \]
where \(K_m\) is the substrate concentration at which the reaction runs at half its maximum speed.

Step 2: Set up the condition given.
We want the substrate concentration at which the velocity equals 80% of \(V_{max}\), so
\[ 0.8\,V_{max} = \frac{V_{max}[S]}{K_m+[S]} \]

Step 3: Cancel \(V_{max}\) and solve for [S].
\[ 0.8 = \frac{[S]}{K_m+[S]} \]
\[ 0.8\,(K_m+[S]) = [S] \]
\[ 0.8\,K_m + 0.8\,[S] = [S] \]
\[ 0.8\,K_m = 0.2\,[S] \]
\[ [S] = 4\,K_m \]

Step 4: Substitute the given Km.
Since \(K_m = x\ \mu M\), the substrate concentration needed is
\[ [S] = 4x\ \mu M \]

Step 5: Final Answer.
An enzyme reaches 80% of its maximum activity when the substrate concentration is 4 times its Km, so the answer is 4 x \(\mu\)M, option (2).
\[ \boxed{4x\ \mu M} \]