Free MCAT-style practice, 1A
MCAT Amino Acid Titration and Isoelectric Point Practice Questions
Eight free MCAT-style practice questions on amino acid pKa values, titration curves, Henderson-Hasselbalch ratios and isoelectric point, with full explanations.
Quick concept primer
An amino acid is a small set of acid-base groups sitting on one carbon, and almost every titration question reduces to the same three moves. First, list the pKa values in increasing order. Second, compare the pH to each pKa: a group is more than 50 percent protonated when pH is below its pKa and more than 50 percent deprotonated when pH is above it. Third, add up charges, remembering that a protonated carboxylic acid is 0, a deprotonated one is -1, a protonated amine is +1 and a deprotonated amine is 0.
The isoelectric point (pI) is the pH at which the average net charge is zero. For an amino acid without an ionizable side chain it is the mean of the two backbone pKa values. With an acidic or basic side chain, average the two pKa values that bracket the neutral species: the two lowest for an acidic amino acid, the two highest for a basic one.
Traps: a buffer works best within about one pH unit of a pKa, not at a pI; and a pH exactly at a pKa means a 1:1 ratio of the two forms, not a neutral molecule.
Free tools: pKa and charge trainer · Amino acid quiz
8 practice questions
Question 1
Alanine has pKa values of about 2.3 (carboxyl) and 9.7 (amino). In a solution of alanine at pH 9.7, which description is most accurate?
- All molecules are zwitterions with net charge 0
- Nearly all molecules are cationic, with net charge +1
- About half the molecules are cations and half are zwitterions, so the average net charge is about +0.5
- About half the molecules are zwitterions and half are anions, so the average net charge is about -0.5
Show answer and explanation
Answer: D. About half the molecules are zwitterions and half are anions, so the average net charge is about -0.5
At pH 9.7 the carboxyl group (pKa 2.3) is long since deprotonated. The amino group has pH equal to its pKa, so it is half protonated by Henderson-Hasselbalch. Molecules with a protonated amino group are zwitterions (net 0); those with a deprotonated amino group are anions (net -1). A 50:50 mix gives an average charge of -0.5. Choice A describes the pI (about 6.0), not this pH, and choices B and C use the wrong pKa.
Question 2
Aspartate has pKa values of approximately 2.0 (alpha carboxyl), 3.9 (side chain carboxyl) and 9.8 (alpha amino). What is its approximate isoelectric point?
- 6.0
- 5.9
- 2.0
- 2.95
Show answer and explanation
Answer: D. 2.95
Aspartate is an acidic amino acid, so the neutral species sits between the two carboxyl pKa values, not at the midpoint of all three. Average the two lowest values: (2.0 + 3.9) / 2 = 2.95. Between pH 2.0 and 3.9 the alpha carboxyl is deprotonated (-1), the side chain is still protonated (0) and the amine is protonated (+1), so the net charge is 0. Choice B averages the lowest and highest values, and choice A is close to the pI of glycine, not aspartate.
Question 3
Lysine has pKa values of about 2.2 (carboxyl), 9.0 (alpha amino) and 10.5 (side chain amino). Which value is closest to its pI?
- 5.6
- 6.35
- 10.5
- 9.75
Show answer and explanation
Answer: D. 9.75
Lysine is basic, so average the two highest pKa values, the ones that bracket the neutral species: (9.0 + 10.5) / 2 = 9.75. Just below pH 9.0 the species carries +2 from two protonated amines and -1 from the carboxylate, for +1; just above 10.5 it carries 0 amines and -1, for -1; between 9.0 and 10.5 the alpha amine has lost its proton and only the side chain amine is +1, giving net 0. Choice A averages the carboxyl and alpha amino values, ignoring the side chain.
Question 4
A student titrates 0.10 M of an unknown amino acid hydrochloride with NaOH and records the center of each buffering plateau.
| Buffering plateau | pH at center |
|---|---|
| First | 1.8 |
| Second | 6.0 |
| Third | 9.2 |
The molecule is net positive before the first plateau. What is its most likely pI, and what is the likely identity?
- 7.6, histidine
- 5.9, glycine
- 10.6, arginine
- 3.0, aspartate
Show answer and explanation
Answer: A. 7.6, histidine
Three plateaus mean three ionizable groups. A side chain pKa near 6.0 fits histidine (imidazole), with the alpha carboxyl at 1.8 and the alpha amino at 9.2. Histidine is classified as basic, so the pI is the mean of the two highest pKa values: (6.0 + 9.2) / 2 = 7.6. Aspartate (D) would show plateaus near 2.0, 3.9 and 9.8, not 6.0, glycine has only two plateaus, and arginine would show a third pKa near 12.5.
Question 5
The glutamate side chain carboxyl has a pKa of about 4.2. At pH 5.2, what is the approximate ratio of deprotonated to protonated side chains?
- 1:10
- 1:1
- 10:1
- 100:1
Show answer and explanation
Answer: C. 10:1
Henderson-Hasselbalch gives pH = pKa + log([A-]/[HA]). Here 5.2 = 4.2 + log(ratio), so log(ratio) = 1 and the ratio is 10:1. Being one unit above the pKa means about 91 percent deprotonated. Choice A reverses the ratio (one unit below the pKa) and choice D would require a two-unit difference.
Question 6
Glycine has pKa values of about 2.3 and 9.6. Which statement about glycine in a solution at its pI (about 6.0) is NOT correct?
- Most molecules are zwitterions
- The molecule is least mobile in an electric field
- The solution resists pH change strongly because the pH is midway between the pKa values
- The average net charge is zero
Show answer and explanation
Answer: C. The solution resists pH change strongly because the pH is midway between the pKa values
A buffer works best near a pKa, where both forms are present in comparable amounts. At pH 6.0 glycine is almost entirely one species (the zwitterion), roughly 3.7 units from each pKa, so it buffers poorly there. Statements D, A and B all describe the pI correctly: zero average charge, mostly zwitterion and little net movement toward either electrode.
Question 7
A chemist starts with 0.10 mol of glycine in its fully protonated form (net charge +1) and adds NaOH. How much NaOH has been added when the pH equals the second pKa (about 9.6)?
- 0.05 mol
- 0.20 mol
- 0.15 mol
- 0.10 mol
Show answer and explanation
Answer: C. 0.15 mol
The first equivalent of base (0.10 mol) converts every cation into the zwitterion, reaching the pI. A further half equivalent (0.05 mol) deprotonates half of the amino groups, which is exactly the point where pH = pKa2. The total is 0.10 + 0.05 = 0.15 mol, 1.5 equivalents. Choice B would be full conversion to the anion.
Question 8
In an electrophoresis buffer at pH 8.0, which of the following amino acids moves toward the cathode (the negative electrode)?
- Cysteine (pI about 5.1)
- Aspartate (pI about 3.0)
- Glycine (pI about 6.0)
- Lysine (pI about 9.75)
Show answer and explanation
Answer: D. Lysine (pI about 9.75)
Direction depends on the sign of the net charge, which depends on pH relative to pI. Lysine's pI is above 8.0, so at pH 8.0 it is net positive and moves toward the cathode. The other three have pI values below 8.0, so each is net negative at this pH and moves toward the anode. Remember that the pH being above the pI means the molecule has lost more protons and is more negative.