The total number of negative charge in the tetrapeptide, Gly-Glu-Asp-Tyr, at pH 12.5 will be _________ (Integer answer)

We begin by recalling that, for any ionisable group, the Henderson-Hasselbalch equation $$\text{pH}= \text{p}K_a + \log\!\left(\dfrac{[\text{A}^-]}{[\text{HA}]}\right)$$ tells us the protonation state. When the pH is much greater than the pKa, the term $$\log([\text{A}^-]/[\text{HA}])$$ is large and positive, meaning the conjugate base $$\text{A}^-$$ predominates, so the group is de-protonated and carries a negative charge if it is an acid. Conversely, basic groups become neutral when de-protonated.

The tetrapeptide is $$\text{Gly} - \text{Glu} - \text{Asp} - \text{Tyr}$$. After peptide-bond formation only two termini remain free: the N-terminus of Gly and the C-terminus of Tyr. Along with these, certain side chains can also ionise. We list every potentially ionisable group with its approximate pKa value:

$$ \begin{aligned} \text{N-terminal } \mathrm{NH_3^+} &: \text{p}K_a \approx 9.0 \\ \text{C-terminal } \mathrm{COOH} &: \text{p}K_a \approx 2.0 \\ \text{Glu side-chain } \mathrm{COOH} &: \text{p}K_a \approx 4.3 \\ \text{Asp side-chain } \mathrm{COOH} &: \text{p}K_a \approx 3.9 \\ \text{Tyr side-chain } \mathrm{OH} &: \text{p}K_a \approx 10.1 \end{aligned} $$

Now the solution pH is $$12.5$$, which is considerably higher than every pKa listed above. Therefore:

1. The N-terminal group: at $$\text{pH}=12.5 \gt 9.0$$, it is de-protonated, converting $$\mathrm{NH_3^+}$$ to neutral $$\mathrm{NH_2}$$. Hence it contributes $$0$$ charge.
2. The C-terminal group: at $$12.5 \gt 2.0$$, the $$\mathrm{COOH}$$ is fully de-protonated to $$\mathrm{COO^-}$$, giving $$-1$$ charge.
3. The Glu side chain: at $$12.5 \gt 4.3$$, it exists as $$\mathrm{COO^-}$$, contributing $$-1$$ charge.
4. The Asp side chain: at $$12.5 \gt 3.9$$, it also becomes $$\mathrm{COO^-}$$, contributing $$-1$$ charge.
5. The Tyr phenolic OH: at $$12.5 \gt 10.1$$, the phenolate form $$\mathrm{O^-}$$ predominates, giving another $$-1$$ charge.

Adding all individual charges algebraically:

$$ \text{Total charge}=0 + (-1) + (-1) + (-1) + (-1)= -4 $$

The magnitude of negative charge is the absolute value, so the peptide carries four negative charges.

Hence, the correct answer is Option D.