Question

Partial hydrolysis of a pentapeptide yields all possible tripeptides and dipeptides. The dipeptides that are obtained upon hydrolysis are given below:

• Val–Ala
• Gln–His
• Phe–Val
• Ala–Gln

The total number of tripeptides obtained that contain ‘Ala’ as one of the amino acids is 2 (in integer).

Hint:

To count tripeptides in a linear peptide chain, use a sliding window of 3 residues. Make sure to cross-check against the sequence constraints derived from known dipeptides.

Answer 1

Let’s first reconstruct the pentapeptide sequence consistent with the given dipeptides:

• Val–Ala → Val precedes Ala
• Ala–Gln → Ala precedes Gln
• Gln–His → Gln precedes His
• Phe–Val → Phe precedes Val

Therefore, the likely pentapeptide is:

\[ \text{Phe} - \text{Val} - \text{Ala} - \text{Gln} - \text{His} \]

Now, generate all possible tripeptides from this sequence (sliding window of 3):

• Phe–Val–Ala ✅ contains Ala
• Val–Ala–Gln ✅ contains Ala
• Ala–Gln–His ✅ contains Ala

Only the valid tripeptides from the linear sequence that contain Ala are:

\[ \boxed{ \begin{aligned} \text{1. } & \text{Phe–Val–Ala} \\ \text{2. } & \text{Val–Ala–Gln} \\ \text{3. } & \text{Ala–Gln–His} \end{aligned} } \]

\[ \boxed{3} \]