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 $$\mid 𝑧_1 \mid, \mid 𝑧_2 \mid$$, and $$\mid 𝑧_3 \mid$$, respectively, then what is $$\mid 𝑧_1 \mid + \mid 𝑧_2 \mid + \mid 𝑧_3 \mid$$?

The peptide shown in the question has six ionisable centres:

– one N-terminal $$NH_3^+$$ group (basic)
– one Lys side-chain $$\varepsilon\text{-}NH_3^+$$ group (basic)
– one Arg side-chain guanidinium $$NH\!-\!C(NH_2)_2^+$$ group (basic)
– one Asp side-chain $$COOH$$ group (acidic)
– one Glu side-chain $$COOH$$ group (acidic)
– one C-terminal $$COOH$$ group (acidic)

Let us evaluate the net charge at the three specified pH values using the usual p$$K_a$$ values:
basic groups : $$pK_a(N\text{-term})\approx 9.0,\; pK_a(Lys)\approx 10.5,\; pK_a(Arg)\approx 12.5$$
acidic groups : $$pK_a(C\text{-term})\approx 3.5,\; pK_a(Asp)\approx 4.0,\; pK_a(Glu)\approx 4.2$$

At such an acidic pH every acidic group is still protonated (neutral) while every basic group is protonated (positively charged).

Net charge $$z_1 = (+1)\times3 = +3$$ ⇒ $$|z_1| = 3$$

This pH is higher than all the acidic p$$K_a$$ values, so the three $$COOH$$ groups are de-protonated (each −1). It is still below the basic p$$K_a$$ values, so all three basic groups remain protonated (each +1).

$$z_2 = (+1)\times3 + (-1)\times3 = 0$$ ⇒ $$|z_2| = 0$$

Now the Lys side chain ($$pK_a\;10.5$$) and the N-terminal group ($$pK_a\;9.0$$) are de-protonated (neutral), but the Arg side chain ($$pK_a\;12.5$$) is still protonated (+1). All three acidic groups stay de-protonated (each −1).

$$z_3 = (+1)\times1 + (-1)\times3 = -2$$ ⇒ $$|z_3| = 2$$

Finally,

$$|z_1| + |z_2| + |z_3| = 3 + 0 + 2 = 5$$

Therefore the required sum is 5.