Question

A tetrapeptide (X) has –COOH group on glycine. Upon complete hydrolysis of (X), glycine, valine, phenyl alanine and alanine are obtained. For this tetrapeptide, number of possible sequence with –NH2 group attached to a chiral centre is

• A. 4
• B. 6
• C. 8
• D. 16

Answer 1

Answer: (B)

6

Answer 2

A tetrapeptide is a peptide composed of four amino acid residues. Let the sequence of the tetrapeptide be represented as $AA_1-AA_2-AA_3-AA_4$, where $AA_1$ is the N-terminal amino acid and $AA_4$ is the C-terminal amino acid.

The problem states that the tetrapeptide (X) has a –COOH group on glycine. This means that glycine is the C-terminal amino acid ($AA_4$). So, the sequence is $AA_1-AA_2-AA_3-Gly$.

Upon complete hydrolysis of (X), glycine, valine, phenylalanine, and alanine are obtained. This tells us that the four amino acids that make up the tetrapeptide are Glycine (Gly), Valine (Val), Phenylalanine (Phe), and Alanine (Ala). Since Glycine is fixed as the C-terminal amino acid ($AA_4$), the remaining three amino acids ($AA_1$, $AA_2$, and $AA_3$) must be Valine, Phenylalanine, and Alanine in some order. Thus, the set of amino acids {$AA_1, AA_2, AA_3$} is {Val, Phe, Ala}.

The question asks for the number of possible sequences where the –NH2 group is attached to a chiral center. The free –NH2 group in a peptide is the amino group of the N-terminal amino acid ($AA_1$). This amino group is attached to the alpha-carbon of the N-terminal amino acid. Therefore, the condition means that the alpha-carbon of the N-terminal amino acid ($AA_1$) must be a chiral center.

Let's examine the chirality of the alpha-carbons of the given amino acids:

• Glycine (Gly): Has the structure $NH_2-CH_2-COOH$. The alpha-carbon is bonded to two hydrogen atoms, so it is achiral.
• Alanine (Ala): Has the structure $NH_2-CH(CH_3)-COOH$. The alpha-carbon is bonded to a hydrogen atom, a methyl group, a carboxyl group, and an amino group. These are four different groups, so the alpha-carbon is chiral.
• Valine (Val): Has the structure $NH_2-CH(CH(CH_3)_2)-COOH$. The alpha-carbon is bonded to a hydrogen atom, an isopropyl group, a carboxyl group, and an amino group. These are four different groups, so the alpha-carbon is chiral.
• Phenylalanine (Phe): Has the structure $NH_2-CH(CH_2C_6H_5)-COOH$. The alpha-carbon is bonded to a hydrogen atom, a benzyl group, a carboxyl group, and an amino group. These are four different groups, so the alpha-carbon is chiral.

The amino acids with a chiral alpha-carbon among {Gly, Val, Phe, Ala} are Val, Phe, and Ala. Glycine has an achiral alpha-carbon.

The N-terminal amino acid ($AA_1$) must have a chiral alpha-carbon. Therefore, $AA_1$ cannot be Glycine. As established earlier, the set {$AA_1, AA_2, AA_3$} is {Val, Phe, Ala}. This means $AA_1$ must be one of Val, Phe, or Ala. Since all of Val, Phe, and Ala have chiral alpha-carbons, any permutation of these three amino acids for the first three positions ($AA_1$, $AA_2$, $AA_3$) will result in a tetrapeptide whose N-terminal amino acid ($AA_1$) has a chiral alpha-carbon.

The number of ways to arrange the three distinct amino acids {Val, Phe, Ala} in the first three positions ($AA_1$, $AA_2$, $AA_3$) is the number of permutations of 3 items, which is 3!. $3! = 3 \times 2 \times 1 = 6$.

The possible sequences are:

1. Val-Phe-Ala-Gly (N-terminal is Val, chiral)
2. Val-Ala-Phe-Gly (N-terminal is Val, chiral)
3. Phe-Val-Ala-Gly (N-terminal is Phe, chiral)
4. Phe-Ala-Val-Gly (N-terminal is Phe, chiral)
5. Ala-Val-Phe-Gly (N-terminal is Ala, chiral)
6. Ala-Phe-Val-Gly (N-terminal is Ala, chiral)

All 6 possible sequences have an N-terminal amino acid with a chiral alpha-carbon. Therefore, there are 6 possible sequences that satisfy the given conditions.