Enzyme kinetics formulas

Constant-volume batch. $k$ does not depend on concentration; it varies with temperature. Its units are $(\text{mol/L}{)}^{1-n}{\text{time}}^{-1}$.

Constant volume. $C_{A0}$ is the initial concentration of reactant A.

The rate does not depend on concentration. Valid only while $C_A>0$: A is used up completely at $t=C_{A0}/k$ and stays at zero after that.

Equivalently $C_A=C_{A0}{e}^{-kt}$, or $-\ln (1-X_A)=kt$ in terms of conversion. The half-life does not depend on $C_{A0}$, and A never quite reaches zero.

Applies when $-r_A=k{C}_A^2$, for example $2A\to P$ with $k$ defined on A, or $A+B\to P$ with $-r_A=k{C}_A{C}_B$ and $C_{A0}=C_{B0}$ (so $C_B=C_A$ throughout). The half-life is inversely proportional to $C_{A0}$.

Any $n\ne 1$. Gives the zero-order result for $n=0$ and the second-order result for $n=2$. For $n<1$ it holds only until A is used up, at $t=\frac{C_{A0}^{1-n}}{(1-n)k}$.

Any $n\ne 1$. The proportionality is the basis of the half-life method for finding the order.

$T$ in kelvin, $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$, $E_a$ in J/mol. A plot of $\ln k$ against $1/T$ is a straight line of slope $-E_a/R$.

Eliminates $A$. When $T_2>T_1$ and $E_a>0$, $k_2>k_1$.

$K_m$ has concentration units, the same as [S] (mM, mol/L or g/L). $v$ and $V_{max}$ are concentration per time (e.g. µmol L${}^{-1}$ min${}^{-1}$).

From $d[ES]/dt\approx 0$. Because $k_2>0$, $K_m$ is always larger than the dissociation constant $k_{-1}/k_1$.

$k_{cat}$ (s${}^{-1}$) is the turnover number: substrate molecules converted per active site per unit time when the enzyme is saturated. $[E{]}_0$ is the molar concentration of active sites.

Units L mol${}^{-1}$ s${}^{-1}$, a second-order rate constant. It compares substrates for the same enzyme. For the one-intermediate mechanism, $k_{cat}/K_m=k_1{k}_2/(k_{-1}+k_2)$, which is less than $k_1$, so it cannot exceed the rate constant for enzyme and substrate meeting by diffusion, of the order of $10^8$ to $10^9$ L mol${}^{-1}$ s${}^{-1}$.

At low substrate the reaction is first order in S with rate constant $V_{max}/K_m$. At high substrate it is zero order.

At $[S]=K_m$, $v=V_{max}/2$. This is the working definition of $K_m$ under either assumption.

Plot $1/v$ against $1/[S]$: slope $K_m/V_{max}$, $y$-intercept $1/V_{max}$, $x$-intercept $-1/K_m$.

Plot $v$ against $v/[S]$: slope $-K_m$, $y$-intercept $V_{max}$, $x$-intercept $V_{max}/K_m$.

Plot $[S]/v$ against [S]: slope $1/V_{max}$, $y$-intercept $K_m/V_{max}$, $x$-intercept $-K_m$.

Competitive: ${\alpha }^{\prime }=1$. Uncompetitive: $\alpha =1$. Pure non-competitive: $K_i=K_i^{\prime }$, so $\alpha ={\alpha }^{\prime }$.

The inhibited reaction still follows the Michaelis–Menten form, with these apparent constants at a fixed [I].

I binds only free E, competing with S for the active site. $K_{m,app}=K_m(1+[I]/K_i)$ increases and $V_{max}$ is unchanged, so enough substrate overcomes the inhibition.

I binds only ES. $V_{max,app}$ and $K_{m,app}$ both fall by the factor $1+[I]/K_i^{\prime }$, so their ratio $V_{max}/K_m$ is unchanged. The inhibition grows as [S] rises.

I binds E and ES with equal affinity, at a site other than the active site. $V_{max,app}=V_{max}/(1+[I]/K_i)$ and $K_m$ is unchanged. The fractional inhibition is the same at every [S].

A second substrate molecule binds ES to form an inactive $E{S}_2$, with $K_{SI}=[ES][S]/[E{S}_2]$. At low [S] this reduces to Michaelis–Menten. At very high [S], $v\approx V_{max}{K}_{SI}/[S]$, so the rate falls. On a Lineweaver–Burk plot the points curve upward near the $1/v$ axis.

From $dv/d[S]=0$. $v_{peak}$ is the highest rate the enzyme can reach and is always below $V_{max}$.

Natural log. $V_{max}$ in concentration per time; $K_m$ and [S] in the same concentration unit.

Same assumptions. Gives the batch time for a target conversion $X$.