Answer:
(i) Glucose does not have open chain structure and hence it does not have a free \[-CHO\] group. Actually \[CHO\] group combines with \[{{C}_{5}}-OH\] to form an hemiacetal. Thus, glucose largely exists in the cyclic hemiacetal form, with a small amount (< 0.5%) of the open chain form. Since the concentration of the open chain form is low and its reactions with 2, 4-DNP reagent and Schiff's reagent are reversible, therefore, formation of 2, 4-DNP derivative or restoration of pink colour of the Schiff's reagent cannot disturb the equilibrium to generate more of the open chain form from the cyclic hemiacetal forms and hence does not give 2, 4-DNP test/Schiff's test.
(ii) \[\alpha \]-Amino acids contain both an acidic carboxyl group and a basic amino group. These two groups neutralize each other. As a result, \[\alpha \]-amino acids largely exist as dipolar ions or zwitterions.
\[N{{H}_{2}}-\underset{\begin{matrix} | \\ R \\ \end{matrix}}{\mathop{C}}\,H-COOH\rightleftharpoons \underset{\text{Zwitterion}}{\mathop{\overset{+}{\mathop{N}}\,{{H}_{3}}-\underset{\begin{matrix} | \\ R \\ \end{matrix}}{\mathop{C}}\,H-CO{{O}^{-}}}}\,\]
Due to dipole-dipole interactions, the molecules of \[\alpha \]-amino acids are held together by strong forces of attraction and hence amino acids have high melting points. Further, due to strong interactions between the +ve and -ve poles of \[\alpha \]-amino acids with polar water molecules, \[\alpha \]-amino acids are moderately soluble in water.
(iii) Secondary structure of proteins refers to the conformation or the shape which the polypeptide chains assume as a result of H-bonding. There are two types of secondary structures of proteins. These are \[\alpha \]-helix and \[\beta \]-pleated sheet structure.
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