JEE Main 2020 · 3 September · evening · Numerical answer · +4 / −0
The number of > C=O groups present in a tripeptide Asp–Glu–Lys is ____.
Answer and solution
Answer 5
Solution
Chemistry ma’am · 4 steps
Count the carbonyls in three places: the peptide bonds, the C-terminal COOH and the acidic side chains.
- Peptide bonds. A tripeptide has 3 residues joined by 2 peptide bonds, -CO-NH⁻. Each one has one >C=O, so that gives 2.
- Free ends. Lys is the last residue, so it keeps a free -COOH at the C-terminus, which adds 1. The N-terminus on Asp is a free -NH₂, with no carbonyl.
Residue Side chain >C=O in side chain Asp -CH₂COOH 1 Glu -CH₂CH₂COOH 1 Lys -(CH₂)₄NH₂ 0 - Side chains. Asp and Glu are acidic amino acids, and each carries one extra -COOH in its side chain. Lys is basic and has an amino group, not a carbonyl.
- So. Total >C=O groups = 2 + 1 + 2 = 5.
The common slip: 3. Leaving out the side-chain -COOH groups of Asp and Glu gives 3, not 5.
The idea to keep. In any peptide, count the carbonyls as (number of peptide bonds) + 1 for the C-terminal COOH + 1 for every Asp or Glu side chain. Basic residues like Lys add none.
Answer from the NTA final answer key. Solution by PrepWiser.