### Core Logic
To assign D or L configurations to ketohexoses like fructose, locate the configuration at the highest-numbered asymmetric carbon atom (carbon-5).
* In D-fructose, the -textOH$-\text{OH}$ group on C-5 is situated on the \right side in the Fischer projection.
* In L-fructose, the configuration of every stereocenter is inverted compared to D-fructose, meaning the -textOH$-\text{OH}$ group on C-5 is situated on the \left hand side. L-Fructose configuration diagram for Q31 - JEE Main 2025 MorningL-Fructose configuration diagram for Q31 - JEE Main 2025 MorningL-Fructose configuration diagram for Q31 - JEE Main 2025 Morning
### Pattern Recognition
Shortcut: L-sugar means the stereocenter farthest from the carbonyl group (C5 for hexoses) has its -textOH$-\text{OH}$ pointing \left. Ensure it is the perfect non-superimposable mirror image of standard D-fructose.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Keywords:#Structure of L-fructose#Biomolecules carbohydrates Fischer projection#Biomolecules JEE Main 2025#JEE Main 2025 Morning Q31
More Biomolecules Previous-Year Questions — Page 4
Q362025Carbohydrates - Nucleic Acids component
The carbohydrates "Ribose" present in DNA, is
A. A pentose sugar
B. present in pyranose from
C. in "D" configuration
D. a reducing sugar, when free
E. in alpha$\alpha$ -anomeric form
Choose the correct answer from the options given below:
A. A, C and D Only
B. A, B and E Only
C. B, D and E Only
D. A, D and E Only
Solution
### Core Logic
The specific carbohydrate residue present across backbone units in DNA molecules is beta$\beta$-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. Carbohydrates - Nucleic Acids component diagram for Q36 - JEE Main 2025 Morning
### Pattern Recognition
Ribose in nucleic acids occurs primarily as a five-membered furanose ring configuration system.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q282025Carbohydrates and Glycosidic Linkages
Match List-I with List-II
List-I (Saccharides)
List-II (Glycosidic-linkages found)
(A) Sucrose
(I) alphatext 1-4$\alpha\text{ 1-4}$
(B) Maltose
(II) alphatext 1-4 and alphatext 1-6$\alpha\text{ 1-4 and }\alpha\text{ 1-6}$
### Related Formula
Glycosidic linkages define the connectivity between monosaccharide units in disaccharides and polysaccharides.
### Core Logic
Analyzing each saccharide configuration:
- **Sucrose**: Formed by alpha$\alpha$-D-glucose and beta$\beta$-D-fructose via a alpha 1 - beta 2$\alpha 1 - \beta 2$ glycosidic linkage.
- **Maltose**: Composed of two alpha$\alpha$-D-glucose units connected by a alpha 1-4$\alpha 1-4$ glycosidic linkage.
- **Lactose**: Composed of beta$\beta$-D-galactose and beta$\beta$-D-glucose via a beta 1-4$\beta 1-4$ glycosidic linkage.
- **Amylopectin**: A branched polymer of glucose with linear alpha 1-4$\alpha 1-4$ linkages and branching at alpha 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 (alpha 1 - beta 2$\alpha 1 - \beta 2$).
- Lactose has a beta$\beta$-linkage (beta 1-4$\beta 1-4$).
- Amylopectin represents branched starch (alpha 1-4$\alpha 1-4$ and alpha 1-6$\alpha 1-6$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q322025Hydrolysis of Carbohydrates
Identify correct conversion during acidic hydrolysis from the following:
(A) starch gives galactose.
(B) cane sugar gives equal amount of glucose and fructose.
(C) milk sugar gives glucose and galactose.
(D) amylopectin gives glucose and fructose.
(E) amylose gives only glucose.
Choose the correct answer from the options given below :
A. (C), (D) and (E) only
B. (A), (B) and (C) only
C. (B), (C) and (E) only
D. (B), (C) and (D) only
Solution
### Related Formula
Acidic or enzymatic hydrolysis cleaves the glycosidic linkages to yield constituent monosaccharides:
textPolysaccharide/Disaccharide xrightarrowH^+ / H_2O textMonosaccharides$\text{Polysaccharide/Disaccharide} \xrightarrow{H^+ / H_2O} \text{Monosaccharides}$
### Core Logic
Evaluating each conversion statement:
- (A) Starch xrightarrowH^+$\xrightarrow{H^+}$ only Glucose (not galactose) rightarrow$\rightarrow$ **Incorrect**
- (B) Cane sugar (Sucrose) xrightarrowH^+$\xrightarrow{H^+}$50\%$50\%$ Glucose + 50\%$50\%$ Fructose rightarrow$\rightarrow$ **Correct**
- (C) Milk sugar (Lactose) xrightarrowH^+$\xrightarrow{H^+}$ Glucose + Galactose rightarrow$\rightarrow$ **Correct**
- (D) Amylopectin xrightarrowH^+$\xrightarrow{H^+}$ only Glucose (not fructose) rightarrow$\rightarrow$ **Incorrect**
- (E) Amylose xrightarrowH^+$\xrightarrow{H^+}$ only Glucose rightarrow$\rightarrow$ **Correct**
### Step 1: Finding the Matching Options
Statements (B), (C), and (E) are strictly correct according to carbohydrate biochemistry properties.
### Pattern Recognition
Amylose and amylopectin are both structural components of starch, meaning their hydrolysis yields *only* D-glucose units. Fructose is obtained from sucrose, while galactose comes exclusively from lactose.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q432025Carbohydrates - Polysaccharides
Match List - I with List - II.
List - I (Carbohydrate)
List - II (Linkage Source)
(A) Amylose
(I) beta -C_1 - C_4$\beta -C_{1} - C_{4}$ plant
(B) Cellulose
(II) alpha -mathrmC_1 - mathrmC_4$\alpha -\mathrm{C}_1 - \mathrm{C}_4$ , animal
(IV) alpha -mathrmC_1 - mathrmC_4$\alpha -\mathrm{C}_1 - \mathrm{C}_4$ plant
Choose the correct answer form the options given below:
The structural visual index links common natural polysaccharides to glycosidic configurations.The structural visual index links common natural polysaccharides to glycosidic configurations.The structural visual index links common natural polysaccharides to glycosidic configurations.
A. \text{(A)-(III), (B)-(II), (C)-(I), (D)-(IV)}
B. \text{(A)-(IV), (B)-(I), (C)-(II), (D)-(III)}
C. \text{(A)-(II), (B)-(III), (C)-(I), (D)-(IV)}
D. \text{(A)-(IV), (B)-(I), (C)-(III), (D)-(II)}
Solution
### Related Formula
Polysaccharides are defined by specific monomer units bound through distinct alpha$\alpha$ or beta$\beta$ glycosidic linkages.
### Core Logic
Reviewing biological carbohydrate structural configurations based on standard biochemical definitions :
* (A) Amylose: Linear unbranched chain polymer of glucose connected via alpha -mathrmC_1 - mathrmC_4$\alpha -\mathrm{C}_1 - \mathrm{C}_4$ paths in plants
ightarrow$
ightarrow$ (IV) .
* (B) Cellulose: Linear polymer containing glucose linkages connected exclusively via beta -C_1 - C_4$\beta -C_{1} - C_{4}$ paths in plants
ightarrow$
ightarrow$ (I) .
* (C) Glycogen: Animal storage polysaccharide with highly branched links alpha -mathrmC_1 - mathrmC_4$\alpha -\mathrm{C}_1 - \mathrm{C}_4$ and alpha -mathrmC_1 - mathrmC_6$\alpha -\mathrm{C}_1 - \mathrm{C}_6$ pathways
ightarrow$
ightarrow$ (II) .
* (D) Amylopectin: Branched plant starch components showing mixed linear alpha -mathrmC_1 - mathrmC_4$\alpha -\mathrm{C}_1 - \mathrm{C}_4$ and branched alpha -mathrmC_1 - mathrmC_6$\alpha -\mathrm{C}_1 - \mathrm{C}_6$ lines
ightarrow$
ightarrow$ (III) .
This maps precisely to option (2).
### Pattern Recognition
Cellulose represents the primary standard framework containing beta$\beta$-linkages exclusively; standard starch components like amylose utilize alpha$\alpha$-configurations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
More Biomolecules Questions — jee_main_2025_03_april_morning
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