Identify the ionisable groups in the peptide

In a peptide, the groups that can contribute to charge are:

• the N-terminal amino group: can be protonated to $-\mathrm{NH_3^+}$
• the C-terminal carboxyl group: can be deprotonated to $-\mathrm{COO^-}$
• any ionisable side chains present in the amino acid residues

From the given peptide structure, the ionisable side chains are:

• one basic side chain (like Lys/Arg type) giving $+1$ in acidic/neutral medium and becoming neutral only at sufficiently high pH depending on group
• one acidic side chain (like Asp/Glu type) giving $0$ in acidic medium and $-1$ in neutral/basic medium

So effectively, the peptide has these charge-bearing groups:

• N-terminus: $+1$ at low pH
• C-terminus: $-1$ at moderate/high pH
• one acidic side chain: $-1$ at moderate/high pH
• one basic side chain: $+1$ at low/neutral pH

Charge at $\mathrm{pH}=2$

At very low pH:

• N-terminus is protonated: $+1$
• basic side chain is protonated: $+1$
• C-terminus is mostly $\mathrm{COOH}$: $0$
• acidic side chain is mostly $\mathrm{COOH}$: $0$

Hence, $Z_1 = +2$ so, $|Z_1|=2$

Charge at $\mathrm{pH}=6$

Around neutral pH:

• N-terminus remains protonated: $+1$
• basic side chain remains protonated: $+1$
• C-terminus is deprotonated: $-1$
• acidic side chain is deprotonated: $-1$

Therefore, $Z_2 = (+1)+(+1)+(-1)+(-1)=0$ so, $|Z_2|=0$

Charge at $\mathrm{pH}=11$

At high pH:

• N-terminus becomes largely deprotonated: $0$
• C-terminus remains $-1$
• acidic side chain remains $-1$
• basic side chain loses proton and becomes neutral: $0$

Thus, $Z_3 = -2$ so, $|Z_3|=2$

Required sum

$|Z_1|+|Z_2|+|Z_3| = 2+0+2 = 4$

Comparison with stored answer

The derived answer is $4$, whereas the stored correct answer is $5$.

Hence, I do not agree with the stored answer.