A tetrapeptide "x" on complete hydrolysis produced glycine (Gly), alanine (Ala), valine (Val), leucine (Leu) in equimolar proportion each. The number of tetrapeptides (sequences) possible involving each of these amino acids is

### Related Formula textNumber of unique sequences = n!$$\text{Number of unique sequences} = n!$$ ### Core Logic A **tetrapeptide** is formed by connecting four amino acids through three peptide linkages. Since the problem specifies that complete hydrolysis of the tetrapeptide produces Gly, Ala, Val, and Leu in *equimolar proportions*, the peptide must contain exactly one molecule of each of these four distinct amino acids. ### Step 1: Calculate the Permutations The number of unique peptide sequences corresponds to the number of ways we can arrange these 4 distinct amino acids: textNumber of permutations = 4! = 4 times 3 times 2 times 1 = 24$$\text{Number of permutations} = 4! = 4 \times 3 \times 2 \times 1 = 24$$ ### Pattern Recognition Combinatorics in Chemistry: If we have n$n$ unique, non-repeating amino acids, the number of linear isomeric peptides is n!$n!$. If repetition were permitted, the number of possible peptides would be n^n$n^n$ (which would be 4^4 = 256$4^4 = 256$ in this case). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules