$word_{count}$: 200

Three facts solve all electrophoresis problems: (1) At pI, amino acid has zero net charge → no migration. (2) At pH < pI, net positive charge → migrates to cathode (negative electrode). (3) At pH > pI, net negative charge → migrates to anode (positive electrode). Calculating pI: Neutral amino acids (Gly, Ala, Val, Leu): pI = $\frac{pKa1 + pKa2}{2}$ ≈ $\frac{2.3 + 9.7}{2}$ ≈ 6.0. Acidic amino acids (Asp, Glu): pI = $\frac{pKa1 + pKaR}{2}$ ≈ $\frac{2.0 + 3.9}{2}$ ≈ 3.0 (use the TWO acidic pKa values). Basic amino acids (Lys, Arg): pI = $\frac{pKa2 + pKaR}{2}$ ≈ $\frac{9.0 + 10.5}{2}$ ≈ 9.7 (use the TWO basic pKa values). At pH 7: acidic amino acids (pI ~3) are above pI → anion → anode. Basic amino acids (pI ~10) are below pI → cation → cathode. Neutral amino acids (pI ~6) are near neutral → slight negative charge → slow migration to anode. This framework handles any JEE electrophoresis question in under 30 seconds.