Animated Solution for Chemistry - Organic Chemistry: The structure of a peptide is given below If the absolute values of the net charge of the peptide at pH = 2, pH = 6, and pH = 11 are $|z_1|$, $|z_2|$ and $|z_3|$, respectively, then what is $|z_1|+|z_2|+|z_3|$?

Imagine you are looking at a molecular chain, a tripeptide, composed of three distinct amino acid building blocks. To solve this problem, our first mission is to identify these blocks and, more importantly, spot the functional groups that can gain or lose a proton.

Starting from the left, we have Tyrosine, which features an N-terminal amino group ($-N{H}_2$) and a phenolic hydroxyl group ($-OH$) on its aromatic ring. Moving to the center, we find Glutamic acid, characterized by its acidic side chain containing a carboxyl group ($-COOH$). Finally, on the right, we have Lysine, which brings a basic side chain amino group ($-N{H}_2$) and the C-terminal carboxyl group ($-COOH$). In total, we have five ionizable groups that will dictate the net charge of the peptide depending on the environment.

The behavior of these groups is governed by a simple but powerful rule involving $pH$ and $pKa$. Think of $pKa$ as the threshold of a functional group. If the surrounding $pH$ is lower than the $pKa$ (an acidic environment rich in $H^{+}$ ions), the group will hold onto its proton, remaining protonated. Conversely, if the $pH$ is higher than the $pKa$ (a basic environment), the group will lose its proton, becoming deprotonated.

Let's take our peptide on a journey through three different pH environments.

First, we drop it into a highly acidic solution at $pH=2$. Here, protons are abundant. The N-terminal $-N{H}_2$ and the Lysine side chain $-N{H}_2$ both eagerly accept a proton to become $-N{H}_3^{+}$, contributing a $+1$ charge each. The $-COOH$ and $-OH$ groups remain neutral in their protonated states. Thus, the net charge $z_1=+1+1=+2$. The absolute value $|z_1|$ is $2$.

Next, we shift to a mildly acidic/neutral environment at $pH=6$. The amino groups (with $pKa\approx 9-10$) are still protonated, giving us a $+2$ charge. However, the carboxyl groups (with $pKa\approx 2-4$) now lose their protons, becoming $-CO{O}^{-}$, which contributes a $-2$ charge. The phenolic $-OH$ (with $pKa\approx 10$) remains neutral. The positive and negative charges perfectly balance out: $z_2=+2-2=0$. The absolute value $|z_2|$ is $0$. This state, where the net charge is zero, tells us that $pH=6$ is very close to the peptide's isoelectric point (pI).

Finally, we plunge the peptide into a highly basic solution at $pH=11$. Now, protons are scarce. The amino groups lose their extra protons and become neutral ($-N{H}_2$). Both carboxyl groups remain deprotonated, contributing a $-2$ charge. Additionally, the phenolic $-OH$ group finally loses its proton to become $-O^{-}$, adding another $-1$ charge. The net charge $z_3=-2-1=-3$. The absolute value $|z_3|$ is $3$.

To wrap it up, we simply sum the absolute values we found:

The final answer is $5$. By carefully tracking the protonation state of each group, even the most complex peptide charge problems become straightforward.