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(D) The peptide is $Tyr-Asp-Lys$. The ionizable groups are the $N$-terminal $-NH_2$ $(pK_a \approx 9)$,the $C$-terminal $-COOH$ $(pK_a \approx 2)$,the

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$5$

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(D) The peptide is $Tyr-Asp-Lys$. The ionizable groups are the $N$-terminal $-NH_2$ $(pK_a \approx 9)$,the $C$-terminal $-COOH$ $(pK_a \approx 2)$,the side chain $-COOH$ of $Asp$ $(pK_a \approx 3.9)$,the side chain $-OH$ of $Tyr$ $(pK_a \approx 10)$,and the side chain $-NH_2$ of $Lys$ $(pK_a \approx 10.5)$.At $pH = 2$: The $N$-terminal $-NH_2$ is protonated ($-NH_3^+$,$+1$),the $C$-terminal $-COOH$ is neutral $(0)$,the $Asp$ side chain $-COOH$ is neutral $(0)$,the $Tyr$ side chain $-OH$ is neutral $(0)$,and the $Lys$ side chain $-NH_2$ is protonated ($-NH_3^+$,$+1$). Net charge $z_1 = +2$,so $|z_1| = 2$.At $pH = 6$: The $N$-terminal $-NH_2$ is protonated ($-NH_3^+$,$+1$),the $C$-terminal $-COOH$ is deprotonated ($-COO^-$,$-1$),the $Asp$ side chain $-COOH$ is deprotonated ($-COO^-$,$-1$),the $Tyr$ side chain $-OH$ is neutral $(0)$,and the $Lys$ side chain $-NH_2$ is protonated ($-NH_3^+$,$+1$). Net charge $z_2 = (+1) + (-1) + (-1) + (+1) = 0$,so $|z_2| = 0$.At $pH = 11$: The $N$-terminal $-NH_2$ is neutral $(0)$,the $C$-terminal $-COOH$ is deprotonated ($-COO^-$,$-1$),the $Asp$ side chain $-COOH$ is deprotonated ($-COO^-$,$-1$),the $Tyr$ side chain $-OH$ is deprotonated ($-O^-$,$-1$),and the $Lys$ side chain $-NH_2$ is neutral $(0)$. Net charge $z_3 = (-1) + (-1) + (-1) = -3$,so $|z_3| = 3$.Therefore,$|z_1| + |z_2| + |z_3| = 2 + 0 + 3 = 5$.

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 $|z_1| + |z_2| + |z_3|$?

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