单位点配体结合 Single-site ligand binding — 公式推导、符号与局限

\[\theta = \frac{[\mathrm{L}]}{K_d + [\mathrm{L}]},\qquad K_d = \frac{[\mathrm{P}][\mathrm{L}]}{[\mathrm{PL}]}\]

配体不过量时的严格解（二次方程）exact (quadratic) solution when ligand is not in excess

\[[\mathrm{PL}] = \frac{([\mathrm{P}]_0 + [\mathrm{L}]_0 + K_d) - \sqrt{([\mathrm{P}]_0 + [\mathrm{L}]_0 + K_d)^2 - 4[\mathrm{P}]_0[\mathrm{L}]_0}}{2}\]

含义Meaning

蛋白质单一位点结合配体的饱和曲线，形式与米氏方程相同；\(K_d\) 是半饱和时的游离配体浓度。

Saturation of a single binding site; same form as Michaelis–Menten. \(K_d\) is the free-ligand concentration at half saturation.

符号说明Symbols

符号Symbol	含义 Meaning	单位 Unit
\(K_d\)	解离常数dissociation constant	mol·L⁻¹
\([\mathrm{L}]\)	游离配体浓度free ligand concentration	mol·L⁻¹

推导Derivation

\(\theta = [\mathrm{PL}]/([\mathrm{P}] + [\mathrm{PL}])\)，用 \([\mathrm{P}] = K_d[\mathrm{PL}]/[\mathrm{L}]\) 消去 [P]。
\(\theta = [\mathrm{PL}]/([\mathrm{P}] + [\mathrm{PL}])\); eliminate [P] with \(K_d\).
若配体总量不远大于蛋白，[L] ≠ [L]₀，需由守恒列二次方程求解（“配体耗竭”）。
If ligand is not in large excess, solve the quadratic from mass balance (ligand depletion).
结合自由能 \(\Delta G^\circ = RT\ln K_d\)（以 1 mol/L 为标准态）。
Binding free energy \(\Delta G^\circ = RT\ln K_d\) (1 M standard state).

适用条件与局限Assumptions & limitations

假设单一、独立、等价位点；多位点或协同需用亚代尔方程或希尔方程。
Single independent site; multiple or cooperative sites need Adair or Hill equations.

例题Worked example

\(K_d = 1\ \mathrm{nmol/L}\) 对应 \(\Delta G^\circ = 8.314\times298\times\ln 10^{-9} = -51\ \mathrm{kJ/mol}\)。

\(K_d = 1\) nM corresponds to \(\Delta G^\circ = -51\) kJ/mol.