Question:hard
The activity of an enzyme having Michaelis constant (Km) = x \(\mu\)M will be 80% of Vmax at a substrate concentration of
The activity of an enzyme having Michaelis constant (Km) = x \(\mu\)M will be 80% of Vmax at a substrate concentration of
Show Hint
Use v = 0.8 Vmax in the Michaelis-Menten equation and solve for [S] in terms of Km.
Updated On: Jul 8, 2026
- 2 x \(\mu\)M
- 4 x \(\mu\)M
- 8 x \(\mu\)M
- 10 x \(\mu\)M
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
Enzyme velocity follows $v=\dfrac{V_{max}[S]}{K_m+[S]}$. This question fixes the velocity at 80% of $V_{max}$ and asks what substrate concentration produces that, in terms of $K_m$.
Step 2: Key Formula or Approach:
Set $v=0.8V_{max}$ and solve the Michaelis-Menten equation for $[S]$ in terms of $K_m$.
Step 3: Detailed Explanation:
\[ 0.8V_{max}=\frac{V_{max}[S]}{K_m+[S]} \]
Cancel $V_{max}$ from both sides:
\[ 0.8=\frac{[S]}{K_m+[S]} \]
Cross multiply:
\[ 0.8K_m+0.8[S]=[S] \]
Move terms with $[S]$ to one side:
\[ 0.8K_m=0.2[S] \]
Divide both sides by $0.2$:
\[ [S]=4K_m \]
Given $K_m=x\ \mu M$, this becomes $[S]=4x\ \mu M$. This makes sense because the Michaelis-Menten curve is a rectangular hyperbola that approaches $V_{max}$ slowly, so getting to a high fraction like 80% needs a substrate level well above $K_m$, specifically four times as much.
Step 4: Final Answer:
\[ \boxed{4x\ \mu M} \]
The enzyme reaches 80% activity at a substrate concentration of 4 x \(\mu\)M, option (2).
Enzyme velocity follows $v=\dfrac{V_{max}[S]}{K_m+[S]}$. This question fixes the velocity at 80% of $V_{max}$ and asks what substrate concentration produces that, in terms of $K_m$.
Step 2: Key Formula or Approach:
Set $v=0.8V_{max}$ and solve the Michaelis-Menten equation for $[S]$ in terms of $K_m$.
Step 3: Detailed Explanation:
\[ 0.8V_{max}=\frac{V_{max}[S]}{K_m+[S]} \]
Cancel $V_{max}$ from both sides:
\[ 0.8=\frac{[S]}{K_m+[S]} \]
Cross multiply:
\[ 0.8K_m+0.8[S]=[S] \]
Move terms with $[S]$ to one side:
\[ 0.8K_m=0.2[S] \]
Divide both sides by $0.2$:
\[ [S]=4K_m \]
Given $K_m=x\ \mu M$, this becomes $[S]=4x\ \mu M$. This makes sense because the Michaelis-Menten curve is a rectangular hyperbola that approaches $V_{max}$ slowly, so getting to a high fraction like 80% needs a substrate level well above $K_m$, specifically four times as much.
Step 4: Final Answer:
\[ \boxed{4x\ \mu M} \]
The enzyme reaches 80% activity at a substrate concentration of 4 x \(\mu\)M, option (2).
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