Let's solve the problem in **3 steps**: --- ### **Step 1: Identify Ionizable Groups and Their Charges at pH 7** **Peptide:** Gly-Lys-Tyr-His-Val - **N-terminus** (amino group, Glycine): **+1** at pH 7 - **C-terminus** (carboxyl group, Valine): **-1** at pH 7 - **Side chains to consider:** - **Lysine (Lys)**: Side chain pKa ≈ 10.5, **positively charged (+1) at pH 7** - **Tyrosine (Tyr)**: Side chain pKa ≈ 10.1, **neutral at pH 7** - **Histidine (His)**: Side chain pKa ≈ 6.0, **mostly neutral at pH 7** (minor fraction protonated, but typically considered 0 at pH 7) **Explanation:** At pH 7, the N-terminus is protonated (+1), the C-terminus is deprotonated (-1), Lys has a +1 charge, Tyr and His are neutral. --- ### **Step 2: Add Up the Charges** - N-terminus: **+1** - C-terminus: **-1** - Lysine side chain: **+1** - Tyrosine side chain: **0** - Histidine side chain: **0** **Explanation:** Sum the charges of all ionizable groups identified in Step 1. --- ### **Step 3: Calculate the Net Charge** Net charge = (+1) + (-1) + (+1) + (0) + (0) = **+1** **Explanation:** Adding up all the charges gives us a net charge of **+1**. --- ## **Final Answer:** **+1**

The concept used in answering this question is the understanding of the ionization states of amino acids and terminal groups at a specific pH, based on their pKa values. By identifying which groups are protonated or deprotonated at pH 7, we determine their charges. Summing these charges provides the net charge of the peptide. This involves applying principles of acid-base chemistry, particularly the relationship between pKa values and the degree of ionization of functional groups in peptides.

The net charge of the peptide Gly-Lys-Tyr-His-Val at pH 7 is **+1**.