Step 1: Start with the easiest unique structure.
Structure (II) contains NH2, COOH and an indole ring, so it can only be tryptophan. So C - II. Options 1 and 2 say C - III, so both are wrong.
Step 2: Use the number of acid groups.
Saccharic acid is the dicarboxylic acid of glucose, HOOC-(CHOH)4-COOH. Structure (I) is the only one with COOH at both ends. So D - I. Options 3 and 4 both say D - I, so both stay.
Step 3: Separate options 3 and 4 with fructose.
Fructose has a ketone group. Structure (IV) is the only one with C=O inside the chain. So B - IV. Option 3 gives B - II, which is the amino acid, so option 3 fails.
Step 4: Check the last pair.
Gluconic acid has COOH on one end and CH2OH on the other. That is structure (III), so A - III. Option 4 agrees with every pair.
Final Answer:
The answer is option (4): A - III, B - IV, C - II, D - I.
\[ \boxed{\text{(A)-(III), (B)-(IV), (C)-(II), (D)-(I)}} \]
