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

This page is the drill floor for the parent topic. We will hunt down every kind of amino-acid problem — simple ones, acidic and basic side chains, degenerate glycine, the charge-at-a-given-pH twist, an electrophoresis word problem, and a limiting-case exam trap — and solve each one from the ground up.

Before any arithmetic, recall the one machine driving everything — and this time we draw it so the page stands on its own:

Recall The three-form ladder (from the parent)

As pH climbs, an amino acid climbs a ladder of charge: $cation (+ 1) p K_{a 1} zwitterion (0) p K_{a 2} anion (- 1)$ Each rung is a proton leaving. A $p K_{a}$ is the pH at which that rung is half-climbed — half the molecules have lost that proton, half still hold it. pI ::: the pH where the molecule is, on average, at the zwitterion rung → net charge $0$ .

Here is that ladder as a picture — every worked example below is just this diagram with different pKa's plugged in:

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

Figure 1 — text description (for non-visual readers): three rounded boxes sit in a row on a warm cream background. The left box (burnt orange) is the cation, structure $H_{3} N^{+} -CHR-COOH$ , net charge $+ 1$ . The middle box (plum) is the zwitterion, structure $H_{3} N^{+} -CHR-COO^{-}$ , net charge $0$ . The right box (deep teal) is the anion, structure $H_{2} N-CHR-COO^{-}$ , net charge $- 1$ . A black arrow labelled "lose H+, pKa1" runs from the cation to the zwitterion; a second black arrow labelled "lose H+, pKa2" runs from the zwitterion to the anion. An orange arrow along the bottom reads "pH increases →" pointing left-to-right. A plum bracket spans the two arrows above the middle box with the caption "pI = average of the two pKa's hugging the zwitterion."

Look at the plum zwitterion box in the middle: it is bracketed by the two pKa's whose average gives pI. That bracketing is the whole game.

The scenario matrix

Every amino-acid pI/charge problem falls into one of these cells. The worked examples below are labelled by the cell they cover, so together they fill the whole grid.

Cell	Case class	What makes it different	Example
A	Simple neutral AA (2 pKa)	Just average the two	Ex 1 (Alanine)
B	Degenerate: glycine, R = H	No chiral centre — a limiting case	Ex 2
C	Acidic side chain (3 pKa)	Average the two lowest	Ex 3 (Glutamic acid)
D	Basic side chain (3 pKa)	Average the two highest	Ex 4 (Arginine)
E	Charge sign at a given pH (pH > pI)	Predict + / 0 / −	Ex 5
F	Charge sign at a given pH (pH < pI)	The "low pH sounds acidic" trap	Ex 6
G	Real-world word problem + pH = pI edge case	Electrophoresis separation; exact pH = pI	Ex 7
H	Limiting / degenerate pKa values	pH exactly at a pKa; pH → extremes	Ex 8
I	Second acidic side chain: cysteine thiol	3-pKa with a –SH group	Ex 9
J	Exam twist: histidine at blood pH	Side chain pKa near physiological pH	Ex 10

Cell A — Simple neutral amino acid

Worked example

Ex 1 — Alanine. $p K_{a 1} (-COOH) = 2.34$ , $p K_{a 2} (-NH_{3}^{+}) = 9.69$ . Find pI. Forecast: Alanine has an ordinary side chain ( $R = CH_{3}$ , non-ionising). Guess: will pI land near 2, near 6, or near 10?

Step 1 — draw the two-rung ladder. Only two ionisable groups exist (α-COOH and α-NH₃⁺; the methyl side chain is a spectator), so the ladder has exactly three species: $+ 1 H_{3} N^{+} -CH(CH_{3}) -COOH 2.34 0 H_{3} N^{+} -CH(CH_{3}) -COO^{-} 9.69 - 1 H_{2} N-CH(CH_{3}) -COO^{-}$ Why this step? The pI rule only ever uses the two pKa's that flank the neutral (zwitterion) form. Drawing the full structures shows those flanking rungs explicitly.

Step 2 — average the flanking pKa's. The middle (zwitterion) box is bracketed by 2.34 and 9.69: $pI = \frac{p K _{a 1} + p K _{a 2}}{2} = \frac{2.34 + 9.69}{2} = \frac{12.03}{2} = 6.015$ Why this step? At exactly this pH there is as much cation as anion, so the two $\pm 1$ charges cancel to net zero.

Verify: $6.015$ sits neatly between $2.34$ and $9.69$ — that matches the "midpoint" intuition. Forecast "near 6" ✔.

Cell B — The degenerate case: glycine

Worked example

Ex 2 — Glycine. $p K_{a 1} = 2.34$ , $p K_{a 2} = 9.60$ . Find pI. Then: why is glycine special? Forecast: Glycine's side chain is just an H. Does that change the pI formula at all?

Step 1 — draw the two-rung ladder. Two ionisable groups (α-COOH, α-NH₃⁺), so three species with $R = H$ : $+ 1 H_{3} N^{+} -CH_{2} -COOH 2.34 0 H_{3} N^{+} -CH_{2} -COO^{-} 9.60 - 1 H_{2} N-CH_{2} -COO^{-}$ Average the two flanking the middle box: $pI = \frac{2.34 + 9.60}{2} = 5.97$ Why this step? $R = H$ contributes no proton, so glycine behaves like any simple AA for pI — the ladder is identical in shape to alanine's.

Step 2 — the degenerate twist. Glycine's α-carbon carries: H, H, COOH, NH₂ → two identical H's. Four distinct groups are needed for chirality; glycine only has three distinct ones. Why this step? This is why glycine is the only standard amino acid that is not chiral (see Optical isomerism and chirality) — a genuine degenerate case.

Verify: $5.97$ lies between the two pKa's ✔. And no chiral centre ⇒ optically inactive ⇒ no L/D forms.

Cell C — Acidic side chain (three pKa's)

Worked example

Ex 3 — Glutamic acid. $p K_{a} (α -COOH) = 2.19$ , $p K_{a} (side-COOH) = 4.25$ , $p K_{a} (α -NH_{3}^{+}) = 9.67$ . Find pI. Forecast: Glutamic acid has two acidic groups. Should its pI be pulled low or high?

Step 1 — build the full-structure charge ladder. Start fully protonated (most acidic pH). Net charge $= + 1$ (NH₃⁺ is +1, both COOH neutral). $+ 1 H_{3} N^{+} -CH(R)-COOH 2.19 0 H_{3} N^{+} -CH(R)-COO^{-} 4.25 - 1 H_{3} N^{+} -CH(R^{-}) -COO^{-} 9.67 - 2 H_{2} N-CH(R^{-}) -COO^{-}$ where $R = (CH_{2})_{2} COOH$ turns to $R^{-} = (CH_{2})_{2} COO^{-}$ at pKa 4.25. Why this step? We must locate which two ionisations surround the net-zero form.

Step 2 — find the neutral rung. Net charge $0$ occurs after losing the first proton (at 2.19) but before the second (at 4.25). So the flanking pKa's are the two lowest. $pI = \frac{2.19 + 4.25}{2} = \frac{6.44}{2} = 3.22$ Why this step? Average the pKa's on either side of the neutral species — here both are the acidic ones.

Verify: $3.22$ is well below 6, confirming the forecast: an extra acid group drags pI down ✔.

Cell D — Basic side chain (three pKa's)

Worked example

Ex 4 — Arginine. $p K_{a} (α -COOH) = 2.17$ , $p K_{a} (α -NH_{3}^{+}) = 9.04$ , $p K_{a} (guanidinium side chain) = 12.48$ . Find pI. Forecast: Two basic groups → do we average the lowest two or the highest two?

Step 1 — build the full-structure charge ladder. Let $R^{+}$ be the protonated guanidinium side chain, $R$ its neutral form. Fully protonated: COOH (0) + NH₃⁺ (+1) + $R^{+}$ (+1) $= + 2$ . $+ 2 H_{3} N^{+} -CH(R^{+}) -COOH 2.17 + 1 H_{3} N^{+} -CH(R^{+}) -COO^{-} 9.04 0 H_{2} N-CH(R^{+}) -COO^{-} 12.48 - 1 H_{2} N-CH(R)-COO^{-}$ Why this step? Drawing each species (matching the depth of Ex 3) shows exactly where net charge passes through $0$ .

Step 2 — locate net-zero. Charge hits $0$ after losing protons at 2.17 and 9.04, before the 12.48 loss. The flanking pKa's are 9.04 and 12.48 — the two highest. $pI = \frac{9.04 + 12.48}{2} = \frac{21.52}{2} = 10.76$ Why this step? Neutral form is bracketed by these two ionisations, so we average them.

Verify: $10.76$ is well above 6 — an extra base group pushes pI up ✔. Mirror image of Cell C.

Cell E — Charge sign when pH > pI

Worked example

Ex 5 — Alanine at pH 9. Using $pI = 6.02$ (from Ex 1), what is the net charge sign? Forecast: pH 9 is above pI. Positive or negative?

Step 1 — compare pH to pI. $pH 9 > pI 6.02$ . Why this step? pI is the pH of zero charge. Move above it → the solution is more basic than the balance point.

Step 2 — reason with protons. Higher pH means fewer H⁺ around, so groups lose protons. The –NH₃⁺ (pKa 9.69) is partly deprotonated, and –COO⁻ is fully deprotonated → the molecule leans anionic. $pH > pI \Rightarrow net charge negative (-)$ Why this step? Above pI the average charge always tips negative.

Verify (Henderson–Hasselbalch, see Henderson–Hasselbalch equation): for the amino group, $pH = p K_{a} + lo g \frac{[ NH _{2} ]}{[ NH _{3}^{+} ]}$ gives $9 = 9.69 + lo g \frac{[ NH _{2} ]}{[ NH _{3}^{+} ]}$ , so $lo g \frac{[ NH _{2} ]}{[ NH _{3}^{+} ]} = - 0.69$ , ratio $\approx 0.204$ . Carboxyl is essentially all COO⁻. Net leans − ✔. Migrates to anode (+).

Cell F — Charge sign when pH < pI (the trap)

Worked example

Ex 6 — Lysine at pH 7. $pI(Lys) = 9.75$ . Net charge sign? Forecast: pH 7 "sounds acidic-ish/neutral." Trap: many students say "neutral → charge 0." What's true?

Step 1 — compare pH to pI. $pH 7 < pI 9.75$ . Why this step? Only the comparison to pI, not to 7, decides charge.

Step 2 — reason with protons. Below pI there is an excess of H⁺ relative to the balance point, so basic groups stay protonated (–NH₃⁺). Extra positive charge survives. $pH < pI \Rightarrow net charge positive (+)$ Why this step? Below pI the average charge always tips positive — regardless of whether pH is "7."

Verify: At pH 7, lysine's two amino groups (pKa 9.0, 10.5) are almost fully protonated; only one COO⁻ is negative. Rough net: $(+ 1) + (+ 1) + (- 1) \approx + 1$ . Positive ✔. Migrates to cathode (−). Trap avoided.

Cell G — Real-world word problem: electrophoresis + the exact pH = pI edge case

Worked example

Ex 7 — A lab mixture of glycine (pI 5.97), aspartic acid (pI 2.95) and lysine (pI 9.75) is run in a buffer at pH 5.97 (exactly glycine's pI). Predict where each migrates. (See Electrophoresis and separation techniques.) Forecast: At a pH set exactly to glycine's pI, three amino acids, one pH. Will glycine's net charge be truly zero, and does "zero net charge" mean only one species is present?

Step 1 — treat the exact pH = pI case for glycine. At $pH = pI = 5.97$ the average net charge is exactly $0$ , so glycine does not migrate. But this is a balance, not a single species: cation and anion coexist in equal amounts, and they simply cancel. $[cation] = [anion] \Rightarrow average charge = (+ 1) (x) + (- 1) (x) + 0 = 0$ Why this step? This is the critical edge case. "Net zero" does not mean every molecule is a zwitterion — it means the $+ 1$ and $- 1$ populations are exactly balanced (with the zwitterion dominant). The molecule sits still because the pulls left and right are equal.

Step 2 — compute charge signs for the other two at pH 5.97:

Aspartate: pH $5.97 >$ pI $2.95$ → net − → moves to anode (+).
Lysine: pH $5.97 <$ pI $9.75$ → net + → moves to cathode (−). Why this step? Migration direction is set purely by the sign of net charge, which the pH–pI comparison delivers.

Step 3 — read the plate. Glycine stays exactly on the origin (its pH = pI), aspartate runs to the anode, lysine to the cathode → clean three-way separation with one species pinned at zero. Why this step? Buffers hold pH fixed (see Buffers — why pI matters for solubility); loading a mixture at one amino acid's pI parks it while the others separate — and, being least charged, glycine is also least soluble here.

The figure below is the plate you would actually see.

Figure 2 — text description (for non-visual readers): a horizontal electrophoresis plate on cream background. A dashed vertical line at the centre marks the origin (loading well) at pH 5.97. The left edge is labelled "CATHODE (−)" in plum; the right edge "ANODE (+)" in teal. Three coloured dots represent the amino acids: a plum dot labelled "Lysine, pI 9.75, net +" has moved left toward the cathode (a plum arrow points from origin to it); an orange dot labelled "Glycine, pI 5.97, net 0" sits on the origin with no arrow; a teal dot labelled "Aspartate, pI 2.95, net −" has moved right, farthest, toward the anode (a teal arrow points from origin to it). The horizontal axis reads "migration distance (− toward cathode | + toward anode)."

Verify: at pH 5.97, glycine's average charge $= 0$ (stationary), $5.97 > 2.95$ (Asp −), $5.97 < 9.75$ (Lys +) ✔.

Cell H — Limiting / degenerate pKa cases

Worked example

Ex 8 — Two limiting checks on glycine ( $p K_{a 1} = 2.34$ , $p K_{a 2} = 9.60$ ). (a) What fraction is deprotonated exactly at $pH = p K_{a 1} = 2.34$ ? (b) As pH → 0 and pH → 14, what is the charge?

Forecast: At pH exactly equal to a pKa, is the group mostly protonated, half, or mostly not?

Step 1 (a). Henderson–Hasselbalch: $pH = p K_{a} + lo g \frac{[ base ]}{[ acid ]}$ . Set $pH = p K_{a}$ : $0 = lo g \frac{[ base ]}{[ acid ]} \Rightarrow \frac{[ base ]}{[ acid ]} = 1$ So the carboxyl is exactly half –COO⁻ and half –COOH → fraction deprotonated $= 0.5$ . Why this step? $lo g 1 = 0$ ; pKa is by definition the half-way pH. This is the limiting boundary between cation and zwitterion.

Step 2 (b). As $pH \to 0$ (drowning in H⁺): everything protonated → COOH + NH₃⁺ → net +1 (the pure cation). As $pH \to 14$ (drowning in OH⁻): everything deprotonated → COO⁻ + NH₂ → net −1 (the pure anion). Why this step? These are the extreme limits of the charge ladder — the endpoints beyond which charge cannot go for a simple AA.

Verify: ratio at pH = pKa is $1$ (fraction 0.5) ✔; limits are $+ 1$ and $- 1$ ✔ — matching the ladder's endpoints.

Cell I — Second acidic side chain: cysteine thiol

Worked example

Ex 9 — Cysteine. $p K_{a} (α -COOH) = 1.96$ , $p K_{a} (α -NH_{3}^{+}) = 8.18$ , $p K_{a} (side -SH thiol) = 10.28$ . Find pI. (A standard test case — the thiol is a weak acid, not a base.) Forecast: The –SH is acidic like a carboxyl, but with a high pKa. Does it behave like an acidic AA (average the low pair) or does its lateness change things?

Step 1 — build the full-structure charge ladder. Let $R = CH_{2} SH$ (neutral thiol) turning to $R^{-} = CH_{2} S^{-}$ when deprotonated. Fully protonated at very low pH: COOH (0) + NH₃⁺ (+1) + SH (0) $= + 1$ . $+ 1 H_{3} N^{+} -CH(R)-COOH 1.96 0 H_{3} N^{+} -CH(R)-COO^{-} 8.18 - 1 H_{2} N-CH(R)-COO^{-} 10.28 - 2 H_{2} N-CH(R^{-}) -COO^{-}$ Why this step? The –SH ionisation (pKa 10.28) sits above the amino ionisation (pKa 8.18), so it is the last proton to leave. The neutral form is therefore reached by losing only the carboxyl proton — the thiol has not yet ionised at that pH.

Step 2 — locate net-zero and average. Net charge $0$ occurs between the 1.96 loss and the 8.18 loss. The flanking pKa's are the two lowest (1.96 and 8.18): $pI = \frac{1.96 + 8.18}{2} = \frac{10.14}{2} = 5.07$ Why this step? Even though cysteine has three pKa's, the zwitterion is bracketed by carboxyl and amino — the thiol only ionises after the molecule is already negative, so it is a spectator at the pI.

Verify: $5.07$ lies between 1.96 and 8.18 ✔, and is slightly below 6 because the extra acidic thiol nudges pI down — consistent with an acidic side chain ✔.

Cell J — Exam twist: histidine at blood pH

Worked example

Ex 10 — Histidine. $p K_{a} (α -COOH) = 1.82$ , $p K_{a} (imidazole side chain) = 6.00$ , $p K_{a} (α -NH_{3}^{+}) = 9.17$ . (a) Find pI. (b) At blood pH 7.4, roughly what charge is on the imidazole side chain? Forecast: Histidine's side-chain pKa (6.00) is unusually close to physiological pH. Does that make histidine a good biological buffer?

Step 1 (a) — build the full-structure charge ladder. Let $R^{+}$ be protonated imidazolium, $R$ neutral imidazole. Fully protonated: COOH (0) + $R^{+}$ (+1) + NH₃⁺ (+1) $= + 2$ . $+ 2 H_{3} N^{+} -CH(R^{+}) -COOH 1.82 + 1 H_{3} N^{+} -CH(R^{+}) -COO^{-} 6.00 0 H_{3} N^{+} -CH(R)-COO^{-} 9.17 - 1 H_{2} N-CH(R)-COO^{-}$ Net-zero form is bracketed by the middle two pKa's (6.00 and 9.17). $pI = \frac{6.00 + 9.17}{2} = \frac{15.17}{2} = 7.585$ Why this step? The neutral species appears after losing protons at 1.82 and 6.00, before 9.17 — so those two flank it.

Step 2 (b) — side chain at pH 7.4. Henderson–Hasselbalch on the imidazole (pKa 6.00): $7.4 = 6.00 + lo g \frac{[ neutral ]}{[ protonated ^{+} ]} \Rightarrow lo g (\dots) = 1.4 \Rightarrow ratio \approx 25$ So only about $1/26 \approx 3.8%$ stays protonated → side chain is mostly neutral at blood pH, but a meaningful minority is charged. Why this step? A pKa sitting near physiological pH means the group flips between charged/neutral with small pH changes — exactly what makes histidine an excellent buffer near pH 7.

Verify: $pI = 7.585$ lies between 6.00 and 9.17 ✔. Protonated fraction $= \frac{1}{1 + 1 0 ^{7.4 - 6.0}} = \frac{1}{1 + 25.1} \approx 0.0383$ ✔ — small but non-zero, so histidine buffers blood.