JEE Challenger
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ChemMedAdv 2012 • Paper 1

Positively Charged Peptides at Neutral pH

Consider nine peptides whose substituents R1\text{R}_1 and R2\text{R}_2 are specified in the given table. Determine how many of these peptides carry a net positive charge at pH=7.0\text{pH} = 7.0.

H3N⊕−CH∣H−CO−NH−CH∣R1−CO−NH−CH∣R2−CO−NH−CH∣H−COO⊖\overset{\oplus}{\text{H}_3\text{N}}-\underset{\underset{\text{H}}{\vert}}{\text{CH}}-\text{CO}-\text{NH}-\underset{\underset{\text{R}_1}{\vert}}{\text{CH}}-\text{CO}-\text{NH}-\underset{\underset{\text{R}_2}{\vert}}{\text{CH}}-\text{CO}-\text{NH}-\underset{\underset{\text{H}}{\vert}}{\text{CH}}-\text{COO}^{\ominus}

PeptideR1R2IHHIIHCH3IIICH2COOHHIVCH2CONH2(CH2)4NH2VCH2CONH2CH2CONH2VI(CH2)4NH2(CH2)4NH2VIICH2COOHCH2CONH2VIIICH2OH(CH2)4NH2IX(CH2)4NH2CH3\begin{array}{|c|c|c|} \hline \text{Peptide} & \text{R}_1 & \text{R}_2 \\ \hline \text{I} & \text{H} & \text{H} \\ \text{II} & \text{H} & \text{CH}_3 \\ \text{III} & \text{CH}_2\text{COOH} & \text{H} \\ \text{IV} & \text{CH}_2\text{CONH}_2 & (\text{CH}_2)_4\text{NH}_2 \\ \text{V} & \text{CH}_2\text{CONH}_2 & \text{CH}_2\text{CONH}_2 \\ \text{VI} & (\text{CH}_2)_4\text{NH}_2 & (\text{CH}_2)_4\text{NH}_2 \\ \text{VII} & \text{CH}_2\text{COOH} & \text{CH}_2\text{CONH}_2 \\ \text{VIII} & \text{CH}_2\text{OH} & (\text{CH}_2)_4\text{NH}_2 \\ \text{IX} & (\text{CH}_2)_4\text{NH}_2 & \text{CH}_3 \\ \hline \end{array}
Official Numerical Answer4

Topics & Concepts

BiomoleculesAmino Acids and Proteins

Step-by-Step Solution

To determine the number of peptides carrying a net positive charge at pH=7.0\text{pH} = 7.0, we analyze the charge contribution from the backbone terminal groups and the variable side-chain substituents R1\text{R}_1 and R2\text{R}_2.

1. Charge on the Peptide Backbone at pH=7.0\text{pH} = 7.0

The backbone of the given tetrapeptide consists of:

  • An N-terminal amino group (−NH3⊕-\text{NH}_3^\oplus), which remains protonated at neutral pH\text{pH} (charge = +1+1).
  • A C-terminal carboxylate group (−COO⊖-\text{COO}^\ominus), which remains ionized at neutral pH\text{pH} (charge = −1-1).

Thus, the net charge contribution from the backbone terminals is: Chargebackbone=(+1)+(−1)=0\text{Charge}_{\text{backbone}} = (+1) + (-1) = 0

2. Charge Contributions of the Side Chains (R1\text{R}_1 and R2\text{R}_2) at pH=7.0\text{pH} = 7.0

  • −H,−CH3,−CH2OH,−CH2CONH2-\text{H}, -\text{CH}_3, -\text{CH}_2\text{OH}, -\text{CH}_2\text{CONH}_2: Neutral groups   ⟹  Charge=0\implies \text{Charge} = 0
  • −(CH2)4NH2-(\text{CH}_2)_4\text{NH}_2: Basic side chain (Lysine-like), protonated to −(CH2)4NH3⊕-(\text{CH}_2)_4\text{NH}_3^\oplus at pH=7.0  ⟹  Charge=+1\text{pH} = 7.0 \implies \text{Charge} = +1
  • −CH2COOH-\text{CH}_2\text{COOH}: Acidic side chain (Aspartate-like), deprotonated to −CH2COO⊖-\text{CH}_2\text{COO}^\ominus at pH=7.0  ⟹  Charge=−1\text{pH} = 7.0 \implies \text{Charge} = -1

3. Analysis of Each Peptide

  1. Peptide I: R1=−H (0),  R2=−H (0)\text{R}_1 = -\text{H}\,(0),\;\text{R}_2 = -\text{H}\,(0) Net Charge=0+0+0=0(Neutral)\text{Net Charge} = 0 + 0 + 0 = 0 \quad (\text{Neutral})

  2. Peptide II: R1=−H (0),  R2=−CH3 (0)\text{R}_1 = -\text{H}\,(0),\;\text{R}_2 = -\text{CH}_3\,(0) Net Charge=0+0+0=0(Neutral)\text{Net Charge} = 0 + 0 + 0 = 0 \quad (\text{Neutral})

  3. Peptide III: R1=−CH2COOH (−1),  R2=−H (0)\text{R}_1 = -\text{CH}_2\text{COOH}\,(-1),\;\text{R}_2 = -\text{H}\,(0) Net Charge=0−1+0=−1(Negatively charged)\text{Net Charge} = 0 - 1 + 0 = -1 \quad (\text{Negatively charged})

  4. Peptide IV: R1=−CH2CONH2 (0),  R2=−(CH2)4NH2 (+1)\text{R}_1 = -\text{CH}_2\text{CONH}_2\,(0),\;\text{R}_2 = -(\text{CH}_2)_4\text{NH}_2\,(+1) Net Charge=0+0+1=+1(Positively charged)\text{Net Charge} = 0 + 0 + 1 = +1 \quad (\mathbf{Positively\ charged})

  5. Peptide V: R1=−CH2CONH2 (0),  R2=−CH2CONH2 (0)\text{R}_1 = -\text{CH}_2\text{CONH}_2\,(0),\;\text{R}_2 = -\text{CH}_2\text{CONH}_2\,(0) Net Charge=0+0+0=0(Neutral)\text{Net Charge} = 0 + 0 + 0 = 0 \quad (\text{Neutral})

  6. Peptide VI: R1=−(CH2)4NH2 (+1),  R2=−(CH2)4NH2 (+1)\text{R}_1 = -(\text{CH}_2)_4\text{NH}_2\,(+1),\;\text{R}_2 = -(\text{CH}_2)_4\text{NH}_2\,(+1) Net Charge=0+1+1=+2(Positively charged)\text{Net Charge} = 0 + 1 + 1 = +2 \quad (\mathbf{Positively\ charged})

  7. Peptide VII: R1=−CH2COOH (−1),  R2=−CH2CONH2 (0)\text{R}_1 = -\text{CH}_2\text{COOH}\,(-1),\;\text{R}_2 = -\text{CH}_2\text{CONH}_2\,(0) Net Charge=0−1+0=−1(Negatively charged)\text{Net Charge} = 0 - 1 + 0 = -1 \quad (\text{Negatively charged})

  8. Peptide VIII: R1=−CH2OH (0),  R2=−(CH2)4NH2 (+1)\text{R}_1 = -\text{CH}_2\text{OH}\,(0),\;\text{R}_2 = -(\text{CH}_2)_4\text{NH}_2\,(+1) Net Charge=0+0+1=+1(Positively charged)\text{Net Charge} = 0 + 0 + 1 = +1 \quad (\mathbf{Positively\ charged})

  9. Peptide IX: R1=−(CH2)4NH2 (+1),  R2=−CH3 (0)\text{R}_1 = -(\text{CH}_2)_4\text{NH}_2\,(+1),\;\text{R}_2 = -\text{CH}_3\,(0) Net Charge=0+1+0=+1(Positively charged)\text{Net Charge} = 0 + 1 + 0 = +1 \quad (\mathbf{Positively\ charged})

Conclusion

The peptides that carry a net positive charge at pH=7.0\text{pH} = 7.0 are IV, VI, VIII, and IX.

Total number of positively charged peptides = 4

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