The number of 

groups present in a tripeptide Asp - Glu - Lys is ____

Each amino acid residue in a peptide may contribute ionisable (titratable) groups that are NOT involved in peptide-bond formation.

Peptide given: Asp - Glu - Lys

Step 1 Identify the terminal (α) groups that remain free
• N-terminus (α-NH3+) comes from the first residue (Asp). ⇒ 1 group
• C-terminus (α-COO-) belongs to the last residue (Lys). ⇒ 1 group

Step 2 Check the side chains of each residue
• Aspartic acid (Asp) side chain: one extra -COOH group. ⇒ 1 group
• Glutamic acid (Glu) side chain: one extra -COOH group. ⇒ 1 group
• Lysine (Lys) side chain: one extra -NH3+ (ε-amino) group. ⇒ 1 group

Step 3 Add all ionisable groups
Total groups = 1 (N-terminal α-NH3+) + 1 (C-terminal α-COO-) + 1 (Asp side -COOH) + 1 (Glu side -COOH) + 1 (Lys side -NH3+) = 5

Hence, the tripeptide Asp - Glu - Lys possesses 5 ionisable groups.

Answer: 5