Given below are two statements: Statement I: D-(+)-glucose + D-(+)-fructose -H₂O sucrose sucrose Hydrolysis D-(+)-glucose + D-(+)-fructose$\text{D-(+)-glucose} + \text{D-(+)-fructose} \xrightarrow{-\mathrm{H}_2\mathrm{O}} \text{sucrose} \qquad \text{sucrose} \xrightarrow{\text{Hydrolysis}} \text{D-(+)-glucose} + \text{D-(+)-fructose}$ Statement II: Invert sugar is formed during sucrose hydrolysis. In the light of given statements, choose the correct answer from the options given below:

Related Formula

Core Logic

Let's audit the nucleotide base distribution across the given single strand structure:

• Count the Guanine (G) and Cytosine (C) bases:

Bases present: G₁, G₂, C₃, C₈, G₉, G₁₀, C₁₁$G_1, G_2, C_3, C_8, G_9, G_{10}, C_{11}$ $\implies$ Total of 7 bases. Each G-C interaction forms 3 hydrogen bonds:

• Count the Adenine (A) and Thymine (T) bases:

Bases present: A₄, A₅, A₆, T₇, T₁₂, A₁₃$A_4, A_5, A_6, T_7, T_{12}, A_{13}$ $\implies$ Total of 6 bases. Each A-T interaction forms 2 hydrogen bonds:

Summing them up yields the total hydrogen bonds in the helix:

Pattern Recognition

Quick check optimization: Total Bonds = 3 × (#G + #C) + 2 × (#A + #T)$3 \times (\#G + \#C) + 2 \times (\#A + \#T)$.

Chapter Mix

Class 12 Chemistry: Biomolecules

Related Formula

Core Logic

Statement (I) is incorrect because the complete hydrolysis of both oligopeptides and proteins yields α$\alpha$-amino acids, not β$\beta$-amino acids.

Statement (II) is incorrect because fibrous proteins are inherently water-insoluble structural materials. Denaturation typically disrupts the tertiary and secondary structures of water-soluble globular proteins, making them insoluble.

Step 1: Final Conclusion

Since both Statement I and Statement II are false, option (3) is the correct choice.

Pattern Recognition

All naturally occurring proteins are polymers of α$\alpha$-amino acids, so any statement mentioning β$\beta$-amino acids as direct translation products can be confidently ruled out. Fibrous proteins (like keratin or collagen) are structural and always insoluble, unlike globular proteins.

Chapter Mix

Class 12 Chemistry: Biomolecules

Core Logic

Let's identify the chemical structures of the nitrogenous bases used in nucleic acids:

* (A) Adenine: A purine derivative featuring a characteristic fused bicyclic ring system with an amino substituent at position 6 (6-aminopurine) arrow$\rightarrow$ (III). * (B) Cytosine: A pyrimidine monocyclic derivative with an amino group at position 4 and a carbonyl group at position 2 (4-amino-2-oxo-pyrimidine) arrow$\rightarrow$ (IV). * (C) Thymine: Found in DNA, this pyrimidine derivative features a methyl substituent at position 5 along with carbonyl groups at positions 2 and 4 (5-methyl-2,4-dioxo-pyrimidine) arrow$\rightarrow$ (II). * (D) Uracil: Found in RNA, this pyrimidine derivative lacks the methyl group found in thymine, featuring just carbonyl groups at positions 2 and 4 (2,4-dioxo-pyrimidine) arrow$\rightarrow$ (I).

Matching these structures gives the sequence: (A)-(III), (B)-(IV), (C)-(II), (D)-(I).

Step-by-Step Structural Validation

The biochemical structures correspond to the following configurations:

Pattern Recognition

Quick identification keys:

• Bicyclic ring = Adenine
• Monocyclic ring with a -CH₃$-\mathrm{CH}_3$ group = Thymine
• Monocyclic ring without a -CH₃$-\mathrm{CH}_3$ group = Uracil
• Monocyclic ring with an amino group (-NH₂$-\mathrm{NH}_2$) = Cytosine

Chapter Mix

Class 12 Chemistry: Biomolecules

Core Logic

The specific carbohydrate residue present across backbone units in DNA molecules is β$\beta$-2-deoxy-D-ribose. Its spatial parameters satisfy the following conditions:

• It is a 5-carbon pentose sugars skeleton structure (A).
• It maintains a canonical D configuration pathway sequence (C).
• When localized in free, open unlinked molecular solutions, it standardly functions as a reducing carbohydrate assembly agent (D).
• It exists predominantly in a furanose cycle form inside DNA, rather than a pyranose ring.

  

Pattern Recognition

Ribose in nucleic acids occurs primarily as a five-membered furanose ring configuration system.

Chapter Mix

Class 12 Chemistry: Biomolecules

Related Formula

Glycosidic linkages define the connectivity between monosaccharide units in disaccharides and polysaccharides.

Core Logic

Analyzing each saccharide configuration:

• Sucrose: Formed by α$\alpha$-D-glucose and β$\beta$-D-fructose via a α 1 - β 2$\alpha 1 - \beta 2$ glycosidic linkage.
• Maltose: Composed of two α$\alpha$-D-glucose units connected by a α 1-4$\alpha 1-4$ glycosidic linkage.
• Lactose: Composed of β$\beta$-D-galactose and β$\beta$-D-glucose via a β 1-4$\beta 1-4$ glycosidic linkage.
• Amylopectin: A branched polymer of glucose with linear α 1-4$\alpha 1-4$ linkages and branching at α 1-6$\alpha 1-6$ positions.

Step 1: Final Mapping

Matching pairs lead to: (A)-(III), (B)-(I), (C)-(IV), (D)-(II).

Pattern Recognition

Remember shortcuts for common linkages:

• Sucrose is a non-reducing sugar involving the anomeric carbons of both units (α 1 - β 2$\alpha 1 - \beta 2$).
• Lactose has a β$\beta$-linkage (β 1-4$\beta 1-4$).
• Amylopectin represents branched starch (α 1-4$\alpha 1-4$ and α 1-6$\alpha 1-6$).

Chapter Mix

Class 12 Chemistry: Biomolecules