The hexapeptide P has an isoelectric point (pI) of 6.9. Hexapeptide Q is a variant of P that contains valine instead of glutamate at position 3. The two peptides are analyzed by polyacrylamide gel electrophoresis at pH 8.0. Which one of the following statements is CORRECT?
P will migrate faster than Q towards the anode
P will migrate faster than Q towards the cathode
Both P and Q will migrate together
Q will migrate faster than P towards the anode
ANSWER
The correct answer is (A): P will migrate faster than Q towards the anode.
Official answer: ASource: Official answer key, GATE 2019 BT (IIT Madras)
Rule: at running pH > pI, a peptide is net negatively charged; at pH < pI it is net positively charged. Here P has pI = 6.9 and the gel runs at pH 8.0 > 6.9, so P carries a net negative charge during the run.
Q differs from P only at position 3, where the acidic residue glutamate (negatively charged side chain at pH 8.0) is replaced by valine (a neutral, nonpolar side chain). This removes one negative charge from the molecule.
Losing an acidic (negative-contributing) residue makes the peptide less acidic overall, so its isoelectric point rises: pI(Q) > pI(P) = 6.9.
Losing just one acidic residue out of six is not enough to push the pI past 8.0, so Q is still net negatively charged at pH 8.0 — but its net negative charge is smaller in magnitude than P's, because 8.0 is closer to Q's (higher) pI than it is to P's pI.
Rule: negatively charged species migrate towards the anode in electrophoresis (positively charged species migrate towards the cathode); a larger net charge magnitude gives faster migration when size/shape are essentially unchanged, as is the case for these near-identical hexapeptides.
Therefore both P and Q migrate towards the anode, but P, carrying the larger net negative charge, migrates faster than Q.
Checking the options: (A) matches this conclusion exactly — correct. (B) wrongly directs migration to the cathode, which applies to net positive charge — incorrect. (C) would require equal net charge, which is false here — incorrect. (D) reverses which peptide is faster; Q, with the smaller charge, is the slower one — incorrect.
KEY IDEA
A protein/peptide run in a gel above its pI carries a net negative charge and migrates towards the anode; the size of that net negative charge (and hence migration speed) depends on how far the running pH is above the pI. Removing an acidic (negatively charged) residue raises the pI, which — at a fixed running pH above both pI values — reduces the net negative charge and therefore the migration speed towards the anode.
WATCH OUT FOR
Confusing which peptide is "more negative": students often assume that because Q has one fewer negative side chain it must be the more mobile one, or they send the negatively charged species towards the cathode instead of the anode (cathode attracts cations, anode attracts anions).
From the GATE 2019 BT paper. Every question of this paper · Download the full paper · All 9 Assays & electrophoresis questions, with notes
