氨基酸等电点 Isoelectric point of an amino acid — 公式推导、符号与局限

\[\mathrm{pI} = \frac{\mathrm{p}K_{a1} + \mathrm{p}K_{a2}}{2}\]

酸性 / 碱性氨基酸取夹着中性形式的两个 pKaacidic/basic residues: average the two pKa values flanking the neutral form

\[\mathrm{pI} = \tfrac12(\mathrm{p}K_{a,\alpha\text{-COOH}} + \mathrm{p}K_{a,\text{侧链}})\ (\text{酸性氨基酸})\]\[\mathrm{pI} = \tfrac12(\mathrm{p}K_{a,\alpha\text{-COOH}} + \mathrm{p}K_{a,\text{side chain}})\ (\text{acidic amino acid})\]

含义Meaning

分子净电荷为零时的 pH。电泳中等电点处的分子不移动，溶解度通常最低。

The pH at which the net charge is zero; molecules do not migrate in electrophoresis and are usually least soluble.

符号说明Symbols

符号Symbol	含义 Meaning	单位 Unit
\(\mathrm{p}K_{a1}, \mathrm{p}K_{a2}\)	两侧（α-COOH 与 α-NH₃⁺）的 pKapKa values of α-COOH and α-NH₃⁺	1

推导Derivation

简单氨基酸存在 H₂A⁺ ⇌ HA（两性离子）⇌ A⁻。净电荷为零要求 [H₂A⁺] = [A⁻]。
For a simple amino acid H₂A⁺ ⇌ HA± ⇌ A⁻; zero net charge requires [H₂A⁺] = [A⁻].
\([\mathrm{H_2A^+}] = [\mathrm{H^+}][\mathrm{HA}]/K_{a1}\)，\([\mathrm{A^-}] = K_{a2}[\mathrm{HA}]/[\mathrm{H^+}]\)。
Write both in terms of [HA].
令二者相等：\([\mathrm{H^+}]^2 = K_{a1}K_{a2}\)，取负对数即得。
Equate: \([\mathrm{H^+}]^2 = K_{a1}K_{a2}\); take −lg.

适用条件与局限Assumptions & limitations

多个可电离基团时只取电中性形式两侧的两个 pKa；蛋白质的 pI 需数值求解所有基团的电荷和。
With several ionisable groups use the two pKa values bracketing the neutral form; proteins need a numerical sum over all groups.
pKa 受环境（蛋白内部、离子强度）影响。
pKa values shift with environment.

例题Worked example

甘氨酸 pKa₁ = 2.34、pKa₂ = 9.60：pI = 5.97。谷氨酸 pI = (2.19 + 4.25)/2 = 3.22。

Glycine: pI = (2.34 + 9.60)/2 = 5.97. Glutamate: (2.19 + 4.25)/2 = 3.22.