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JEE Advanced levelSingle correct optionDifficulty 3 of 3bm075

The pI values are computed with pI = mean of the two pKa values flanking the neutral species. Data: aspartic acid pKa = 2.0 (alpha-COOH), 3.9 (side-chain COOH), 10.0 (); lysine pKa = 2.2, 9.0, 10.5 (alpha-COOH, alpha-, side-chain ); alanine pKa = 2.3, 9.7. A mixture of the three is subjected to electrophoresis in a buffer of pH 6.00. Which statement is correct?

  1. ALysine moves to the cathode, aspartic acid moves to the anode, alanine stays at the origin
  2. BAspartic acid moves to the cathode, lysine moves to the anode, alanine stays at the origin
  3. CLysine and alanine move to the cathode, aspartic acid stays at the origin
  4. DAll three move to the anode

Answer: A

pI(Asp) = , pI(Lys) = , pI(Ala) = . At pH 6.00, Asp (pH above pI) is negative and goes to the anode; Lys (pH below pI) is positive and goes to the cathode; Ala (pH = pI) has zero net charge and does not move.
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