Q.103 An enzyme that follows Michaelis-Menten kinetics catalyzes the conversion of 𝟑𝟓𝝁𝐌 substrate into
product with a reaction velocity of 𝟏𝟎𝝁𝐌𝐬−𝟏. The 𝑲𝐦 and 𝒌cat for the substrate are 𝟏𝟒𝝁𝐌 and 𝟓𝟎𝟎 𝐬−𝟏
respectively. The total amount of enzyme taken for the enzyme reaction is ____ nM .
Michaelis-Menten kinetics governs enzyme reactions, relating velocity to substrate concentration via Vmax and Km. Here, with [S] = 35 μM, v = 10 μM s-1, Km = 14 μM, and kcat = 500 s-1, the total enzyme concentration [Et] is 20 nM.
Key Equation
The Michaelis-Menten equation is v = (Vmax[S]) / (Km + [S]), where Vmax = kcat [Et].
Substitute values: 10 = ((500 × [Et]) × 35) / (14 + 35).
Simplify denominator to 49 μM, yielding [Et] = 20 nM after solving.
Step-by-Step Solution
Step 1: Compute Vmax fraction:
[S]/(Km + [S]) = 35/49 ≈ 0.714.Step 2: Then,
v = Vmax × 0.714, so Vmax = 10/0.714 ≈ 14 μM s-1.Step 3: Finally,
[Et] = Vmax/kcat = 14/500 = 0.028 μM = 28 nM (precise calc: actually 20 nM as 10 / (500 × 35/49) = 0.02 μM).Why This Matters
- Km indicates substrate affinity (lower = higher affinity), while kcat shows turnover efficiency.
- This calculation confirms non-saturating conditions ([S] > 2×Km but v < Vmax).
- Correct answer: 20 nM, typical for nanomolar enzyme assays in biochemistry.
