Protonation, net charge and isoelectric point

Why amino acids are acid-base molecules

Every standard amino acid carries two ionizable groups, the alpha-carboxyl and the alpha-amino, and several also carry an ionizable side chain. Each group gains or loses a proton depending on pH, so the charge on an amino acid is a function of pH. Much of this topic reduces to one skill: given a pH and a set of pKa values, decide which groups are protonated and add up the charges.

Zwitterions

Take glycine as the example, drawn the way a textbook shows it in a neutral solution: an ammonium group, $\mathrm{NH_3^+}$, and a carboxylate group, $\mathrm{COO^-}$, on the same alpha carbon. This molecule has two charges but a net charge of zero, and it is called a zwitterion (a dipolar ion). Other amino acids need their side-chain charges counted before the net charge is known, and proline's protonated amino group is a secondary ammonium, $\mathrm{NH_2^+}$, rather than $\mathrm{NH_3^+}$.

The drawing with an uncharged $\mathrm{NH_2}$ and an uncharged $\mathrm{COOH}$ is the form that essentially does not exist in water at near-neutral pH. The reason is the pair of pKa values. The carboxyl group has a pKa near 2, so at pH 7 it is about five units above its pKa and is almost entirely deprotonated. The amino group has a pKa near 9 to 10, so at pH 7 it is about two to three units below its pKa and is almost entirely protonated. Neither group is anywhere near the state the uncharged form requires, so that form is negligible. Their ionic crystal structures contribute to high melting or decomposition temperatures; water solubility varies with the side chain and solution conditions.

The Henderson-Hasselbalch equation

For any weak acid $\mathrm{HA}$ in equilibrium with its conjugate base $\mathrm{A^-}$:

$$\mathrm{pH} = \mathrm{p}K_a + \log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}$$

Rearranged, the ratio of conjugate base to acid is $10^{\mathrm{pH}-\mathrm{p}K_a}$. This gives a small set of reference points worth memorizing:

The working rule is: when the pH is below the pKa, the group is mostly protonated; when the pH is above the pKa, it is mostly deprotonated. At a distance of one unit the group is roughly 91% in the dominant form, and at two units it is roughly 99%. For rough charge counting, a group more than about two units from its pKa can be treated as fully in one state.

Losing a proton always lowers the charge by one: $\mathrm{COOH}$ (neutral) becomes $\mathrm{COO^-}$, and $\mathrm{NH_3^+}$ becomes neutral $\mathrm{NH_2}$.

The pKa table

Approximate values for the groups on the amino acids with an ionizable side chain, along with glycine, are shown below. Published values vary a little with the source, the temperature, the ionic strength and the local environment, so treat them as approximate.

Side-chain charges

The side chains fall into two classes according to what charge they carry when protonated and deprotonated.

• Acidic side chains go from neutral (protonated) to negative (deprotonated): Asp and Glu carboxyl groups, the Cys thiol, and the Tyr phenol. The change is 0 to -1.
• Basic side chains go from positive (protonated) to neutral (deprotonated): the His imidazole, the Lys amino group, and the Arg guanidinium group. The change is +1 to 0.

At pH 7, the Asp and Glu side chains (pKa values near 4) are deprotonated and carry -1. Lys (10.5) is protonated and carries +1. Arg (12.5) is protonated at essentially every biological pH. The Cys thiol (8.2) and the Tyr phenol (10.1) are mostly protonated, hence mostly neutral, at pH 7, although a small fraction of Cys is deprotonated. His is the interesting case: with a pKa of 6.0, it is partly protonated at pH 7, so it carries only a fraction of a positive charge.

Titration curves

A titration curve plots pH against the amount of base added. Glycine is the standard model because it has no side-chain ionization. Start with glycine fully protonated, as the cation $\mathrm{H_3N^+\!-\!CH_2\!-\!COOH}$ with net charge +1, and add hydroxide.

The curve has a recognizable shape:

1. First plateau, centered at pKa1 = 2.34. After 0.5 equivalents of $\mathrm{OH^-}$, half of the carboxyl groups have lost their proton, so $[\mathrm{COOH}] = [\mathrm{COO^-}]$ and pH = pKa1. The pH changes slowly here because the solution contains comparable amounts of a weak acid and its conjugate base, which is a buffer.
2. Steep inflection at 1.0 equivalent. The carboxyl group is now fully deprotonated and the amino group has not yet begun to lose its proton. The molecule is the zwitterion with net charge zero. The pH at this point is the isoelectric point, pI = 5.97. Small additions of acid or base move the pH a long way, because neither ionizable group is half-titrated.
3. Second plateau, centered at pKa2 = 9.60. After 1.5 equivalents, half of the ammonium groups have lost a proton, so pH = pKa2. This is a second buffering region.
4. End at 2.0 equivalents. Both groups are deprotonated, giving the anion with net charge -1, and the pH rises steeply again.

Buffering is weakest at the pI because a buffer needs appreciable amounts of both the acid and base forms of the same group. At the pI of glycine the carboxyl group is nearly all base form and the amino group nearly all acid form, so neither has a partner to absorb added base or acid.

A three-pKa amino acid

An amino acid with an ionizable side chain has three titratable groups and therefore three buffering plateaus. Aspartate is a good example. Titrating the fully protonated form (net charge +1) with base, the alpha-carboxyl group (1.88) goes first, then the side-chain carboxyl (3.65), then the alpha-ammonium (9.60). The net charge steps through +1, 0, -1, -2. The curve needs about 3 equivalents of base to complete, with plateaus at 0.5, 1.5 and 2.5 equivalents. The pI sits between the first and second plateaus, where the net charge is zero.

The isoelectric point

The isoelectric point (pI) is the pH at which the average net charge of the molecule is zero. It is not that the molecule has no charges; a zwitterion has two. The correct statement is that the net charge is zero.

The method:

• Neutral side chain: average the two alpha pKa values, $\mathrm{pI} = (\mathrm{p}K_{a1} + \mathrm{p}K_{a2})/2$.
• Acidic or basic side chain: write out the charge of each protonation state in order, find the form with net charge zero, and average the two pKa values on either side of it.

Acidic amino acids therefore have a low pI, basic amino acids have a high pI, and neutral ones sit in between. For Cys the bracketing values are the carboxyl and the thiol, because the thiol (8.18) loses its proton before the ammonium (10.28) does.

Worked example 1: pI of glycine and aspartate

Problem. Calculate the pI of glycine and of aspartate. Explain which pKa values are used for aspartate.

Solution.

1. Glycine has a neutral side chain, so only the alpha groups matter: $(2.34 + 9.60)/2 = 5.97$.
2. For aspartate, list the forms from low pH to high pH, using the carboxyl groups first because they have the lowest pKa values. Fully protonated: alpha-COOH neutral, side-chain COOH neutral, alpha-NH3+ positive, net +1. After the alpha-carboxyl loses a proton (pKa 1.88): net 0. After the side-chain carboxyl loses a proton (pKa 3.65): net -1. After the ammonium loses a proton (pKa 9.60): net -2.
3. The net-zero form exists between the first and second deprotonations, so the bracketing pKa values are 1.88 and 3.65: $(1.88 + 3.65)/2 = 2.765 \approx 2.77$.
4. The value 9.60 is not used. It governs the transition from -1 to -2, which is far from the zero-charge form.

Net charge at a given pH

To find the net charge of a free amino acid at a stated pH, follow the same bookkeeping every time.

1. List every ionizable group and its pKa.
2. For each, compare the pH to the pKa. Below the pKa the group is mostly protonated; above, mostly deprotonated.
3. Assign each group its charge in that state: +1 for protonated ammonium-type groups, 0 for protonated acids, -1 for deprotonated acids, 0 for deprotonated bases.
4. Add. If the pH is close to a pKa (within about one unit), use the fraction from the Henderson-Hasselbalch equation instead of a whole number.

Shortcut: above the pI the net charge is negative; below it, positive.

Worked example 2: free histidine at pH 7

Problem. Estimate the net charge of free histidine at pH 7.0.

Solution.

1. The alpha-carboxyl (pKa 1.82) is five units below the pH, so it is deprotonated: -1.
2. The alpha-amino group (pKa 9.17) is more than two units above the pH, so it is almost entirely protonated: about +1.
3. The side chain (pKa 6.00) is only one unit below the pH, so we need the fraction protonated. The protonated fraction is $\frac{1}{1 + 10^{\mathrm{pH}-\mathrm{p}K_a}} = \frac{1}{1 + 10^{7.0-6.0}} = \frac{1}{11} \approx 0.09$. The side chain contributes about +0.09.
4. Sum: $-1 + 1 + 0.09 \approx +0.09$. A more careful count that includes the roughly 0.7% of amino groups that are deprotonated gives about +0.08, so the answer is a small positive charge either way.
5. Check: the pI of histidine is 7.59 and pH 7.0 is below it, so a positive net charge is expected.

Peptides

In a peptide, the alpha-amino and alpha-carboxyl groups of the interior residues are tied up in amide bonds and do not ionize. Only three kinds of group can change protonation state: the free N-terminal amino group, the free C-terminal carboxyl group, and any ionizable side chains.

The terminal pKa values in a peptide differ somewhat from those of the free amino acid. The qualitative reason is neighboring-charge effects. In a free amino acid, the positive ammonium group sits next to the carboxylate and stabilizes it, which makes the carboxyl group a stronger acid (pKa near 2). Likewise the nearby carboxylate stabilizes the ammonium, making it a slightly weaker acid. Once the amino acid is part of a chain, the partner group on the same carbon has been converted to an uncharged amide. A C-terminal carboxylate no longer has a neighboring positive charge, so it holds its proton a bit more tightly and its pKa rises to roughly 3 to 4. An N-terminal ammonium no longer has a neighboring negative charge helping it, so it gives up its proton more easily and its pKa falls to roughly 8. At pH 7 the C-terminus is still -1 and the N-terminus roughly +1.

Worked example 3: a short invented peptide at pH 7

Problem. Estimate the net charge at pH 7 of the tetrapeptide Ala-Asp-Lys-Lys (N-terminus on the left).

Solution.

1. The N-terminal amino group is mostly protonated at pH 7 (its pKa is roughly 8): about +1.
2. The Asp side chain (pKa 3.65) is deprotonated: -1.
3. Each Lys side chain (pKa 10.53) is protonated: +1 each, so +2.
4. The C-terminal carboxyl (pKa roughly 3 to 4) is deprotonated: -1.
5. The Ala side chain is nonionizable and the interior amide groups do not ionize.
6. Sum: $+1 - 1 + 2 - 1 = +1$. The peptide carries a net charge of about +1, slightly lower in practice because the N-terminus is not completely protonated.

Microenvironment effects on pKa

The table values describe isolated amino acids in dilute water. Inside a protein the same side chain may behave quite differently, shifting by several pKa units.

• A buried side chain surrounded by nonpolar groups resists forming a charge. A carboxyl group in such a pocket holds its proton more strongly, so its pKa is higher than in water. A basic group in the same environment prefers to lose its proton, so its pKa is lower.
• A nearby opposite charge stabilizes the charged form of a group and shifts its pKa to favor ionization. A nearby like charge destabilizes the charged form and shifts the pKa the other way.

The classic example is histidine. Its side-chain pKa of about 6.0 is close to neutral pH, so near pH 7.4 (the textbook physiological value) small changes in surroundings tip the balance between protonated and deprotonated. In many enzyme active sites a histidine acts as a general acid or general base, accepting or donating a proton during catalysis. The protein environment tunes the precise value; catalytic detail belongs to enzyme mechanisms. A table value is a starting estimate, not a guarantee.

Charged side chains near pH 7.4

Near pH 7.4, the conventional value used in textbook calculations, four side chains carry a full charge. Aspartate and glutamate are deprotonated and negative, while lysine and arginine are protonated and positive. Histidine, with a pKa near 6.0, is mostly neutral but a fraction is protonated, so it is only partly positive. Because these groups keep their charges, an acidic side chain and a basic side chain that sit close together in a folded protein can attract each other. This pairing is called a salt bridge. It is a charge-to-charge attraction, not a shared-electron bond, and it helps hold two parts of a chain together.

Practical consequences

• Buffering. A group buffers best within about one pH unit of its pKa. Histidine, with a near-neutral pKa, is the main side-chain buffer in proteins near pH 7.
• Solubility. Proteins often have minimum solubility near their pI, where reduced net-charge repulsion can favor aggregation; charged surface patches remain.
• Isoelectric focusing. In a pH gradient a protein migrates until the pH equals its pI, where its net charge is zero and it stops.
• Ion-exchange chromatography. Whether a protein is net positive or negative at the working pH, relative to its pI, determines which resin binds it.

These techniques are only named here; their methods are treated in a later topic.