JEE Challenger
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ChemHardAdv 2013 • Paper 1

Number of Primary Structure Sequences in a Tetrapeptide

A given tetrapeptide features a −COOH-COOH terminal on alanine. Exhaustive hydrolysis of this tetrapeptide yields glycine (Gly), valine (Val), phenyl alanine (Phe), and alanine (Ala). Calculate the total number of plausible primary sequences for this tetrapeptide such that the terminal −NH2-NH_2 group is bonded to a chiral carbon center.

Official Numerical Answer4

Topics & Concepts

BiomoleculesAmino Acids and Proteins

Step-by-Step Solution

To determine the total number of plausible primary sequences (primary structures) for the given tetrapeptide, we analyze the constraints provided in the problem step-by-step:

1. Identification of the Amino Acids

Complete hydrolysis of the tetrapeptide yields four distinct amino acids:

  • Glycine (Gly\text{Gly})
  • Valine (Val\text{Val})
  • Phenylalanine (Phe\text{Phe})
  • Alanine (Ala\text{Ala})

Thus, the tetrapeptide contains exactly one residue of each of these four amino acids.

2. Fixing the C-Terminal Position

A peptide sequence is conventionally written from the N-terminal (free −NH2-\text{NH}_2 group) to the C-terminal (free −COOH-\text{COOH} group): N-terminal (Position 1)−Position 2−Position 3−C-terminal (Position 4)\text{N-terminal (Position 1)} - \text{Position 2} - \text{Position 3} - \text{C-terminal (Position 4)}

The problem states that the −COOH-\text{COOH} terminal is on Alanine (Ala\text{Ala}). Therefore, position 4 is fixed as Ala\text{Ala}: Position 1−Position 2−Position 3−Ala\text{Position 1} - \text{Position 2} - \text{Position 3} - \text{Ala}

3. Chirality Condition for the N-Terminal Amino Acid

The terminal −NH2-\text{NH}_2 group is located at Position 1. The problem specifies that this −NH2-\text{NH}_2 group must be attached to a chiral carbon center (α\alpha-carbon).

Let's evaluate the chirality of the remaining amino acids:

  1. Glycine (Gly\text{Gly}): H2N−CH2−COOH\text{H}_2\text{N}-\text{CH}_2-\text{COOH}
    • The α\alpha-carbon is attached to two hydrogen atoms (−H-\text{H}), making it achiral (optically inactive).
  2. Valine (Val\text{Val}): H2N−CH(CH(CH3)2)−COOH\text{H}_2\text{N}-\text{CH}(\text{CH}(\text{CH}_3)_2)-\text{COOH}
    • The α\alpha-carbon is attached to four different groups (−H,−NH2,−COOH,−CH(CH3)2-\text{H}, -\text{NH}_2, -\text{COOH}, -\text{CH}(\text{CH}_3)_2), making it chiral.
  3. Phenylalanine (Phe\text{Phe}): H2N−CH(CH2C6H5)−COOH\text{H}_2\text{N}-\text{CH}(\text{CH}_2\text{C}_6\text{H}_5)-\text{COOH}
    • The α\alpha-carbon is attached to four different groups (−H,−NH2,−COOH,−CH2Ph-\text{H}, -\text{NH}_2, -\text{COOH}, -\text{CH}_2\text{Ph}), making it chiral.

Therefore, the N-terminal amino acid at Position 1 cannot be Glycine. It must be either Valine (Val\text{Val}) or Phenylalanine (Phe\text{Phe}).

4. Counting the Possible Sequences

Positions 1, 2, and 3 must be filled by a permutation of the set {Gly,Val,Phe}\{\text{Gly}, \text{Val}, \text{Phe}\}.

  • Total permutations of these 3 amino acids: 3!=63! = 6

  • Invalid sequences (where Glycine is at Position 1):

    1. Gly−Val−Phe−Ala\text{Gly} - \text{Val} - \text{Phe} - \text{Ala}
    2. Gly−Phe−Val−Ala\text{Gly} - \text{Phe} - \text{Val} - \text{Ala}

    (Total invalid sequences = 2)

  • Valid sequences (where Val or Phe is at Position 1):

    • Case 1: N-terminal is Valine (Val\text{Val})

      1. Val−Gly−Phe−Ala\text{Val} - \text{Gly} - \text{Phe} - \text{Ala}
      2. Val−Phe−Gly−Ala\text{Val} - \text{Phe} - \text{Gly} - \text{Ala}

    • Case 2: N-terminal is Phenylalanine (Phe\text{Phe}) 3. Phe−Gly−Val−Ala\text{Phe} - \text{Gly} - \text{Val} - \text{Ala} 4. Phe−Val−Gly−Ala\text{Phe} - \text{Val} - \text{Gly} - \text{Ala}

Total number of valid sequences=6−2=4\text{Total number of valid sequences} = 6 - 2 = 4

Conclusion

The total number of plausible primary sequences for the given tetrapeptide is 4.

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