1. Given amino acids in the tetrapeptide

On complete hydrolysis, the tetrapeptide gives:

• Glycine (Gly)
• Valine (Val)
• Phenylalanine (Phe)
• Alanine (Ala)

Also given:

• The terminal $-COOH$ group is on alanine.

So, alanine must be the C-terminal amino acid.

Hence the peptide has the form $\boxed{\_\; -\; \_\; -\; \_\; -\; Ala}$ where the first three positions are occupied by Gly, Val, and Phe in some order.

2. Total possible sequences with Ala fixed at C-terminus

The remaining three amino acids Gly, Val, and Phe can be arranged in $3! = 6$ ways.

These are:

1. Gly–Val–Phe–Ala
2. Gly–Phe–Val–Ala
3. Val–Gly–Phe–Ala
4. Val–Phe–Gly–Ala
5. Phe–Gly–Val–Ala
6. Phe–Val–Gly–Ala

3. Condition: $-NH_2$ group attached to a chiral center

The free $-NH_2$ group is present at the N-terminus of the peptide.

So the first amino acid must have its amino-bearing carbon as a chiral center.

Among the given amino acids:

• Glycine: achiral, since its $\alpha$-carbon has two H atoms
• Valine: chiral
• Phenylalanine: chiral
• Alanine: chiral

But Ala is fixed at the C-terminus, so the N-terminal amino acid can only be one of Gly, Val, or Phe.

To satisfy the condition, the N-terminal residue must not be Gly.

So valid sequences are only those beginning with Val or Phe.

From the 6 sequences above, exclude those starting with Gly:

• Gly–Val–Phe–Ala ❌
• Gly–Phe–Val–Ala ❌

Valid ones are:

• Val–Gly–Phe–Ala ✅
• Val–Phe–Gly–Ala ✅
• Phe–Gly–Val–Ala ✅
• Phe–Val–Gly–Ala ✅

Thus, number of possible sequences is $\boxed{4}$

4. Comparison with stored answer

Stored correct answer = $4$

Our derived answer = $4$

So the answer agrees with the stored correct answer.