Reaction Kinetics & Michaelis-Menten
Reaction Kinetics & Michaelis-Menten
Enzyme $E$ binds substrate $S$ forming complex $ES$, which releases product $P$ and free enzyme. At low $[S]$ the rate is approximately first-order; at high $[S]$ the enzyme saturates and the rate plateaus at $V_{\max}$. The Michaelis-Menten equation captures this with two parameters: $V_{\max}$ and $K_m$.
The quasi-steady-state approximation (QSSA) — $d[ES]/dt\approx0$ — is the standard derivation. Its validity requires $[E]_0\ll K_m+[S]_0$. Extensions include competitive and uncompetitive inhibition and the cooperative Hill equation.
Michaelis-Menten Rate Law
$$v=\frac{V_{\max}[S]}{K_m+[S]},\quad K_m=\frac{k_{-1}+k_2}{k_1},\quad V_{\max}=k_2[E]_0.$$ Half-maximum rate occurs at $[S]=K_m$.
Lineweaver-Burk Linearisation
$$\frac{1}{v}=\frac{K_m}{V_{\max}}\cdot\frac{1}{[S]}+\frac{1}{V_{\max}}.$$ Slope gives $K_m/V_{\max}$; y-intercept gives $1/V_{\max}$; x-intercept $=-1/K_m$.
Example 1
$V_{\max}=10\,\mu\text{M/s}$, $K_m=2\,\mu\text{M}$, $[S]=8\,\mu\text{M}$. Find $v$.
Solution: $v=10\times8/(2+8)=8\,\mu\text{M/s}$.
Example 2
A competitive inhibitor increases apparent $K_m$. How does this appear on Lineweaver-Burk?
Solution: Slope increases (steeper); y-intercept ($1/V_{\max}$) is unchanged — lines intersect on the y-axis.
Practice
- What does $K_m$ tell us about enzyme-substrate affinity?
- State the QSSA validity condition for Michaelis-Menten.
- Write the Hill equation and state its relevance to cooperativity.
- How does uncompetitive inhibition affect both $K_m$ and $V_{\max}$?
Show Answer Key
1. $K_m$ is the substrate concentration at which the reaction rate is half-maximal ($v = V_{\max}/2$). A small $K_m$ indicates high affinity — the enzyme reaches half-saturation at low substrate concentration.
2. QSSA requires $[E]_0 \ll [S]_0$ (enzyme much less abundant than substrate) so that $[ES]$ reaches quasi-steady state quickly: $d[ES]/dt \approx 0$. More precisely, $\epsilon = [E]_0/([S]_0 + K_m) \ll 1$.
3. $v = V_{\max}[S]^n / (K_{0.5}^n + [S]^n)$. The Hill coefficient $n$ measures cooperativity: $n=1$ (no cooperativity, reduces to Michaelis-Menten), $n>1$ (positive cooperativity, sigmoidal curve), $n<1$ (negative cooperativity).
4. Uncompetitive inhibitor binds only to the ES complex, not free enzyme. Both $K_m$ and $V_{\max}$ decrease by the same factor $(1+[I]/K_i)$. The ratio $V_{\max}/K_m$ remains unchanged, unlike competitive inhibition.