Step 1: Understanding the Concept:
The charge on a peptide depends on the pH of the medium and the \(pK_a\) values of its ionizable groups (N-terminal, C-terminal, and side chains). At a pH much higher than the \(pK_a\), the group is deprotonated.
Step 2: Detailed Explanation:
Identify ionizable groups in Gly-Glu-Asp-Tyr at pH 12.5:
1. N-terminal amine (Gly): \(pK_a \approx 9.6\). At pH 12.5, it is deprotonated (\(-NH_2\)). Charge = 0.
2. C-terminal carboxyl (Tyr): \(pK_a \approx 2.2\). At pH 12.5, it is deprotonated (\(-COO^-\)). Charge = -1.
3. Glu side chain carboxyl: \(pK_a \approx 4.2\). At pH 12.5, it is deprotonated (\(-COO^-\)). Charge = -1.
4. Asp side chain carboxyl: \(pK_a \approx 3.9\). At pH 12.5, it is deprotonated (\(-COO^-\)). Charge = -1.
5. Tyr side chain phenol: \(pK_a \approx 10.1\). At pH 12.5, the phenolic group is deprotonated (\(-O^-\)). Charge = -1.
Total negative charge = \(1 (\text{C-term}) + 1 (\text{Glu}) + 1 (\text{Asp}) + 1 (\text{Tyr}) = 4\).
Step 3: Final Answer:
The total number of negative charges is 4.