Amino acids — zwitterion, isoelectric point pI, classification (essential, non-essential): Exercises

Reminders you will keep reusing:

Level 1 — Recognition

Problem 1.1

At very low pH (strong acid), which form does glycine take: cation, zwitterion, or anion? What is its net charge?

Recall Solution 1.1

WHAT we do: picture the two "hands" of glycine — the –COOH (give-away) and the –NH₂ (grabby). WHY: in strong acid there is a flood of H⁺ everywhere, so every group that can grab a proton will.

The –NH₂ is already holding its proton → –NH₃⁺.
The –COO⁻ now also grabs a proton → –COOH. Both ends protonated: $H_{3} N + - CH_{2} - COOH$ . Answer: the cation, net charge +1.

Problem 1.2

Name the form drawn below and give its net charge: $H_{3} N + - CH_{2} - COO^{-}$

Recall Solution 1.2

One end is + (–NH₃⁺), the other – (–COO⁻). Two opposite formal charges on one molecule that cancel. Answer: the zwitterion, net charge 0.

Problem 1.3

True or false: "The essential amino acids are the chemically strongest ones." Explain in one line.

Recall Solution 1.3

False. "Essential" means only that your body cannot synthesise them, so you must eat them. It is a dietary label, not a ranking of chemical power.

Level 2 — Application

Problem 2.1

Alanine has $p K_{a 1} = 2.34$ (–COOH) and $p K_{a 2} = 9.69$ (–NH₃⁺). Compute its pI.

Recall Solution 2.1

WHAT: alanine has no ionisable side chain, so only two pKa's exist, and they sit on either side of the zwitterion. WHY the average: the zwitterion is created by losing proton 1 and destroyed by losing proton 2. To have exactly equal cation and anion, sit at the midpoint of those two reactions. $pI = \frac{2.34 + 9.69}{2} = \frac{12.03}{2} = 6.02$ Answer: $pI = 6.02$ .

Problem 2.2

Valine: $p K_{a 1} = 2.32$ , $p K_{a 2} = 9.62$ . Find pI and state whether valine is essential or non-essential.

Recall Solution 2.2

$pI = \frac{2.32 + 9.62}{2} = \frac{11.94}{2} = 5.97$ Valine is a neutral amino acid (isopropyl side chain, no ionisable group). From "PVT TIM HaLL", the V stands for Valine → essential (body can't make it). Answer: $pI = 5.97$ ; essential.

Problem 2.3

Serine ( $pI \approx 5.68$ ) is placed in a buffer at pH 3. Which electrode does it migrate toward?

Recall Solution 2.3

Compare pH to pI: $pH 3 < pI 5.68$ . Below the pI → excess H⁺ → net positive → cation. A cation is attracted to the negative electrode. Answer: it moves toward the cathode (–).

Level 3 — Analysis

Problem 3.1

Glutamic acid has three pKa's: $α -COOH = 2.19$ , side-chain $-COOH = 4.25$ , $α -NH_{3}^{+} = 9.67$ . Find pI. Explain which two you average and why.

Recall Solution 3.1

WHAT: map the charge as we raise pH, starting fully protonated. $2 COOH, NH_{3}^{+} + 1 2.19 1 COO^{-} 0 4.25 2 COO^{-} - 1 9.67 - 2$ WHY these two: the neutral (net-zero) species sits between the $2.19$ and $4.25$ ionisations — those are the two pKa's hugging it. The $9.67$ group is still protonated (–NH₃⁺) at that pH, a spectator. Rule check: Glu is acidic → average the two lowest. $pI = \frac{2.19 + 4.25}{2} = \frac{6.44}{2} = 3.22$ Answer: $pI = 3.22$ — the acidic side chain pulls pI down.

Problem 3.2

Arginine has $α -COOH = 2.17$ , $α -NH_{3}^{+} = 9.04$ , guanidinium side chain $= 12.48$ . Find pI and explain the choice.

Recall Solution 3.2

Charge ladder (raising pH): $+ 2 2.17 + 1 9.04 0 12.48 - 1$ The net-zero form lies between $9.04$ and $12.48$ . Those are the flanking pKa's. Rule check: Arg is basic → average the two highest. $pI = \frac{9.04 + 12.48}{2} = \frac{21.52}{2} = 10.76$ Answer: $pI = 10.76$ — the basic side chain pushes pI high.

Problem 3.3 (geometric)

At pH 6, you run glutamic acid ( $pI = 3.22$ ), alanine ( $pI = 6.02$ ), and arginine ( $pI = 10.76$ ) on an electrophoresis gel. Predict the direction and relative speed of each. Use the figure.

Figure — Amino acids — zwitterion, isoelectric point pI, classification (essential, non-essential)

Recall Solution 3.3

Compare each pI to the running pH of 6:

Glu: $6 > 3.22$ → net negative → anode (+). The gap $(6 - 3.22) = 2.78$ is large → strongly negative → fast.
Ala: $6 \approx 6.02$ → essentially at its pI → net ≈ 0 → barely moves.
Arg: $6 < 10.76$ → net positive → cathode (–). Gap $(10.76 - 6) = 4.76$ is very large → strongly positive → fast the other way. Answer: Glu → anode, Arg → cathode (opposite ends), Ala stays near the origin. The bigger the pH–pI gap, the further from neutral, the faster the migration — exactly what the figure's arrows show.

Level 4 — Synthesis

Problem 4.1

Using Henderson–Hasselbalch, show from scratch that for a simple amino acid $pI = \frac{1}{2} (p K_{a 1} + p K_{a 2})$ . Then verify: at the pI, is the cation/anion ratio exactly 1?

Recall Solution 4.1

WHAT: write H–H for each of the two equilibria (see Henderson–Hasselbalch equation). $pH = p K_{a 1} + lo g \frac{[ zwit ]}{[ cation ]}, pH = p K_{a 2} + lo g \frac{[ anion ]}{[ zwit ]}$ WHY add them: adding cancels the awkward $[zwit]$ term. $2 pH = p K_{a 1} + p K_{a 2} + lo g (\frac{[ zwit ]}{[ cation ]} \cdot \frac{[ anion ]}{[ zwit ]}) = p K_{a 1} + p K_{a 2} + lo g \frac{[ anion ]}{[ cation ]}$ At the pI, net charge $= 0$ requires $[cation] = [anion]$ , so $lo g (1) = 0$ : $2 pH = p K_{a 1} + p K_{a 2} \Rightarrow pI = \frac{p K _{a 1} + p K _{a 2}}{2} ✓$ Ratio check: yes — the derivation assumed $[cation] = [anion]$ , i.e. ratio $= 1$ ; that is precisely the definition of net-zero charge. Consistent.

Problem 4.2

Histidine ( $p K$ 's: $1.82, 6.00, 9.17$ ) is placed at pH 7.4 (blood pH). Compute pI, then decide the net charge and migration direction at pH 7.4.

Recall Solution 4.2

Charge ladder: His has a basic imidazole side chain (pK $6.00$ ). $+ 2 1.82 + 1 6.00 0 9.17 - 1$ Net-zero sits between $6.00$ and $9.17$ → basic → average the two highest: $pI = \frac{6.00 + 9.17}{2} = \frac{15.17}{2} = 7.585$ At pH 7.4: $7.4 < 7.585$ → below pI → net positive (slightly) → migrates toward the cathode (–), but weakly since $7.4$ is close to the pI. Answer: $pI = 7.585$ ; at blood pH histidine is faintly positive → cathode.

Level 5 — Mastery

Problem 5.1

A peptide Gly–Asp–Lys is a tiny protein fragment (see Proteins — peptide bond and structure). Only the free N-terminus (–NH₃⁺, $p K \approx 8.0$ ), the free C-terminus (–COOH, $p K \approx 3.1$ ), the Asp side chain (–COOH, $p K \approx 3.9$ ), and the Lys side chain (–NH₃⁺, $p K \approx 10.5$ ) ionise (the internal amino/carboxyl are locked into peptide bonds). Estimate the pI.

Recall Solution 5.1

WHAT: collect all ionisable groups and sort their pKa's: $3.1 (C-term COOH), 3.9 (Asp COOH), 8.0 (N-term NH_{3}^{+}), 10.5 (Lys NH_{3}^{+})$ Build the charge ladder starting fully protonated (2 amino groups +, 2 acids neutral → net $+ 2$ ): $+ 2 3.1 + 1 3.9 0 8.0 - 1 10.5 - 2$ WHY these two: net-zero sits between the $3.9$ and $8.0$ ionisations. Average that flanking pair: $pI = \frac{3.9 + 8.0}{2} = \frac{11.9}{2} = 5.95$ Answer: $pI \approx 5.95$ . Notice the peptide's pI is set by whichever groups bracket net-zero — the extreme pK's ( $3.1$ and $10.5$ ) are spectators there.

Problem 5.2

Two proteins have $pI = 4.5$ and $pI = 8.5$ . You must separate them cleanly by charge at a single pH. (a) Pick a good pH. (b) State each protein's charge and electrode there. (c) Why does this matter for solubility? (link the idea).

Recall Solution 5.2

(a) Choose pH midway, e.g. pH 6.5 (between the two pI's). Any pH strictly between $4.5$ and $8.5$ works; midway maximises the charge difference. (b)

Protein A ( $pI 4.5$ ): $6.5 > 4.5$ → net negative → anode (+).
Protein B ( $pI 8.5$ ): $6.5 < 8.5$ → net positive → cathode (–). They run to opposite electrodes → clean separation. (c) Solubility: a molecule is least soluble at its own pI (net charge 0 → no charge–charge repulsion → molecules clump and precipitate). Running at pH $6.5$ keeps both proteins charged and therefore soluble while they migrate. See Buffers — why pI matters for solubility and Electrophoresis and separation techniques.

Problem 5.3

Explain, using chirality, why 19 of the 20 standard amino acids are optically active but glycine is not (link the concept).

Recall Solution 5.3

A carbon is a chiral centre only if it holds four different groups (see Optical isomerism and chirality). For a general amino acid the α-carbon carries: –NH₂, –COOH, –H, and –R → four different groups → chiral → optically active (naturally the L-form). Glycine has $R = H$ . Now the α-carbon holds: –NH₂, –COOH, and two identical H's. Only three distinct groups → not a chiral centre → glycine is optically inactive. Answer: glycine's side chain is itself an H, destroying the four-different-groups requirement.

Recall One-line self-check before you close the page

Migration rule ::: pH above pI → negative → anode; pH below pI → positive → cathode; pH = pI → no movement. Which pKa's set the pI ::: only the two that hug the net-zero (zwitterion) form. Least soluble pH ::: exactly at the pI (net charge 0, molecules clump).