The total number of ionizable groups present in a tripeptide Asp–Glu–Lys is .

Question

The total number of ionizable groups present in a tripeptide Asp–Glu–Lys is  .

Canvas Classes · Accepted Answer

Ionizable groups in Asp–Glu–Lys:

A tripeptide has:
1. One free N-terminus (–NH₂/–NH₃⁺)
2. One free C-terminus (–COOH/–COO⁻)
3. Side chain ionizable groups

Counting for each residue:

| Residue | Side chain ionizable group |
|---|---|
| Asp (Aspartic acid) | –CH₂COOH (pKa ~3.9) → 1 ionizable group |
| Glu (Glutamic acid) | –CH₂CH₂COOH (pKa ~4.1) → 1 ionizable group |
| Lys (Lysine) | –(CH₂)₄NH₂ (pKa ~10.5) → 1 ionizable group |

Total ionizable groups:
- N-terminus (–NH₂): 1
- C-terminus (–COOH

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