The dipeptide $(I I I)$ is:

V a l

\to

L e u

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L e u

\to

V a l

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G l y

\to

A l a

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A l a

\to

G l y

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Solution

The correct option is C $V a l$ $\to$ $L e u$
(1) DBP method or Leucine aminopeptidase enzyme releases $N$ -terminal amino acid.
$T e t r a p e p t i d e (I) ⟶ A l a n i n e (N - t e r m i n a l) + T r i p e p t i d e$
$A l a (N - t e r m i n a l) ⟶ T r i p e p t i d e$
(2) Hydrazinolys or carboxypeptidase enzyme releases C-terminal amino acid.
$T e t r a p e p t i d e (I) ⟶ D i p e p t i d e + G l y c i n e (C - t e r m i n a l)$
$A l a (N - t e r m i n a l) ⟶ D i p e p t i d e ⟶ G l y (C - t e r m i n a l)$
(3) Dipeptide $(I I I)$ now contains only two amino acids, valine and Leucine. Edman method releases N-terminal amino acid.
$D i p e p t i d e (I I I) ⟶ V a l i n e (N - t e r m i n a l) ⟶ L e u c i n e (C - t e r m i n a l)$
Hence, option A is correct.

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The dipeptide (III) is:

The dipeptide $(I I I)$ is: