Amino Acid and Peptide Charge: Count the Groups that can Change

Free MCAT content · Biochemistry · Chemistry

A residue name does not determine a whole molecule’s charge. Identify the available groups, compare pH with each pKa, and add their contributions.

About 13 minutes, including the worked example and checks.

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Concept: pH changes protonation, and protonation changes charge

The AAMC outline includes amino acids as dipolar ions, peptide bonds, isoelectric point, and electrophoresis. A useful connection across these topics is the ability to explain a molecule’s charge from its structure and solution conditions.¹

For a protonatable group, pH below its pK_a favors the protonated form; pH above its pK_a favors the deprotonated form. At equality, the two forms have equal populations. The Henderson–Hasselbalch relationship makes that comparison quantitative:

[deprotonated form]/[protonated form] = 10^{(pH − pK_a)}

One pH unit above pK_a gives a 10:1 ratio, not complete deprotonation. Two units above gives 100:1. An integer-charge approximation becomes more useful as pH moves away from the relevant pK_a.²

Group	Protonated charge	Deprotonated charge
Carboxyl group, including Asp or Glu side chains	0	−1
Free amino terminus; Lys, Arg, or His side chain	+1	0
Cys thiol or Tyr phenol	0	−1

Follow the charge change when a group loses one proton

Each deprotonation lowers the charge by one. It does not always create a negative group: an ammonium group becomes neutral when it loses a proton. Cysteine and tyrosine also have ionizable side chains, although they are often neutral around pH 7. Histidine’s side chain is not automatically +1 simply because histidine is classified as a basic amino acid.³

Pattern: distinguish a free amino acid from a peptide

A free amino acid has an α-amino group, an α-carboxyl group, and any ionizable side chain. In a conventional, unmodified linear peptide, count one free N-terminus and one free C-terminus for the whole chain, plus relevant side chains. Internal backbone amino and carboxyl groups have become amides; do not count each residue as if it were a separate free amino acid. Peptides are written from the N-terminus to the C-terminus.⁴

The same residue can supply two different groups at a chain end. A C-terminal glutamate has a terminal carboxyl and a side-chain carboxyl; both matter. By contrast, an internal glutamate retains its side-chain carboxyl but not a free backbone carboxyl.

When a problem specifies a terminal modification, redraw the group before assigning charge. In the changed case below, a terminal carboxyl is replaced by an amide specified to be neutral. That group no longer contributes a carboxyl charge; the unmodified Glu side chain still does.

Rule: locate the zero-charge interval before calculating pI

The isoelectric point, pI, is the pH at which average net charge is zero. It does not mean every group is uncharged. A zwitterion carries both positive and negative formal charges.⁵

List the ionizable groups and their pK_a values in increasing order.
Start with the predominant charge at sufficiently low pH, when these groups are protonated.
Raise pH through each pK_a. Each proton lost lowers the predominant net charge by one.
Find the two dissociations that surround the neutral species. For a simple amino acid or a well-separated charge ladder, their pK_a average estimates pI.

For a free amino acid without an ionizable side chain, those are its two terminal-group pK_a values. With an ionizable side chain, choose the pair by charge accounting rather than averaging every value supplied. For a peptide with several overlapping ionizations, use fractional charges to locate the zero instead of assuming one two-value average is exact.²⁵

Application: charge and pI of Ala–Gly–Glu

Illustrative example, created for this lesson. Consider the unmodified tripeptide Ala–Gly–Glu with free termini. Use the supplied, simplified pK_a values below; they are assumptions for the calculation, not measurements of this particular peptide. Treat the groups as independently titrating.

Group	Assumed pK_a	Predominant charge at pH 7.0
Free N-terminus on Ala	9.0	+1
Free C-terminus on Glu	2.5	−1
Glu side-chain carboxyl	4.2	−1

Count each available group once

Charge at pH 7.0: the pH is below 9.0, so the N-terminus is mostly protonated. It is above both carboxyl pK_a values, so both carboxyls are mostly deprotonated. The predominant net charge is +1 − 1 − 1 = −1. Ala and Gly add no ionizable side chains in this range.

Find the neutral interval: at very low pH, the N-terminus contributes +1 and both carboxyls contribute 0. Deprotonating the C-terminus at pK_a 2.5 changes the predominant total to 0; deprotonating the Glu side chain at 4.2 changes it to −1. The final terminal-amino deprotonation near 9.0 would change it to −2.

Estimate pI: the neutral species lies between the 2.5 and 4.2 dissociations, so pI ≈ (2.5 + 4.2)/2 = 3.35. Averaging 2.5 and 9.0 would ignore the side chain that removes neutrality first.

When the integer approximation is too rough

Suppose a separate example specifies a histidine side-chain pK_a of 6.0 at pH 7.0. Deprotonated:protonated = 10:1, so its average contribution is about +1/11, or +0.09, rather than exactly 0. At pH 6.0 it contributes +0.5 on average. These fractional values describe a population; an individual group is not literally one-tenth protonated.²

For independent groups, a basic group contributes +1/[1 + 10^{(pH − pK_a)}], and an acidic group contributes −1/[1 + 10^{(pK_a − pH)}]. Add all contributions. This follows directly from the population ratio and is useful near pK_a or when a precise average charge is requested.

Apply charge to migration, with the method specified

In a native charge-dependent electrophoresis setup, ignoring electroosmotic flow, a negatively charged molecule moves toward the positive electrode, the anode; a positively charged molecule moves toward the negative cathode. Above pI the molecule is net negative; below pI it is net positive. The example peptide at pH 7 therefore moves toward the anode. This predicts direction, not migration speed.⁵

Use the pK_a values and conditions supplied by a passage. Values for free amino acids are not universal constants for the same groups in peptides or folded proteins; experimental work on peptide model compounds illustrates that distinction.⁶ Before opening the checks, name the groups you counted and explain why each one can carry its assigned charge.

Check your reasoning

Open an answer to see the reasoning. Each check stands on its own.

Apply it

Apply the charge ladder. An illustrative problem gives free glutamic acid these pK_a values: α-carboxyl 2.2, side-chain carboxyl 4.2, and α-ammonium 9.7. What is its approximate pI? Use the usual dominant-species approximation.

A. 3.2

Correct.

At low pH the net charge is +1. The first carboxyl deprotonation makes it 0; the second makes it −1. The neutral species is therefore bracketed by 2.2 and 4.2: pI ≈ (2.2 + 4.2)/2 = 3.2.

B. 5.4

Review this choice.

This approximately averages all three pK_a values. pI is not the average of every ionization value; identify the two dissociations surrounding the neutral species.

C. 6.0

Review this choice.

This approximately averages the α-carboxyl and α-ammonium values. That shortcut omits the ionizable Glu side chain, which changes the net charge before the amino group deprotonates.

D. 7.0

Review this choice.

This approximately averages 4.2 and 9.7. Between those transitions the predominant molecule has two negative carboxyls and one positive ammonium group: its net charge is −1, not 0.

The principle to carry forward

Select the pK_a pair by locating the neutral species. The residue's extra ionizable group can change which pair belongs in the calculation.

Try a different application

Changed case for a later visit. An unmodified Ala–Gly–Glu peptide has free N- and C-termini. At pH 7.0, use pK_a values 9.0 for the N-terminus, 2.5 for the C-terminus, and 4.2 for the Glu side chain. The C-terminal carboxyl is then converted into a neutral amide. Assume the remaining groups' pK_a values are unchanged. How does the predominant net charge change?

A. From −1 to −2

Review this choice.

Amidation removes a negative terminal-carboxyl contribution under these conditions. Removing −1 makes the total less negative, not more negative.

B. From −1 to 0

Correct.

Initially, N-terminus +1, C-terminus −1, and Glu side chain −1 sum to −1. After C-terminal amidation, +1 + 0 − 1 = 0 in the dominant-state approximation. The Glu side-chain carboxyl is still present.

C. From 0 to +1

Review this choice.

This omits one negative group before modification. A C-terminal Glu has both a terminal carboxyl and a side-chain carboxyl, giving an initial total of −1.

D. From +1 to 0

Review this choice.

The initial peptide is predominantly negative at pH 7. Converting a carboxyl to an amide also does not remove the positive N-terminal contribution.

The principle to carry forward

Recount available groups after a structural change. A predominant net charge of 0 at one pH does not establish the exact pI, because the average charge can include fractional contributions.