Match List-I with List-II tracking nitrogenous bases structures:
List-I (Base)List-II (Chemical Structure Diagram)
(A) Adenine(I)
Structure of Nucleic Acids diagram for Q44 - JEE Main 2025 Evening
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(B) Cytosine(II)
Structure of Nucleic Acids diagram for Q44 - JEE Main 2025 Evening
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(C) Thymine(III)
Structure of Nucleic Acids diagram for Q44 - JEE Main 2025 Evening
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(D) Uracil(IV)
Structure of Nucleic Acids diagram for Q44 - JEE Main 2025 Evening
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
Choose the correct answer from the options given below :

Solution & Explanation

### 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) ightarrow **(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) ightarrow **(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) ightarrow **(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) ightarrow **(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:
Structure of Nucleic Acids solution diagram for Q44 - JEE Main 2025 Evening
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
Structure of Nucleic Acids solution diagram for Q44 - JEE Main 2025 Evening
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
Structure of Nucleic Acids solution diagram for Q44 - JEE Main 2025 Evening
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
Structure of Nucleic Acids solution diagram for Q44 - JEE Main 2025 Evening
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
### Pattern Recognition Quick identification keys: - Bicyclic ring = Adenine - Monocyclic ring with a -mathrmCH_3 group = Thymine - Monocyclic ring without a -mathrmCH_3 group = Uracil - Monocyclic ring with an amino group (-mathrmNH_2) = Cytosine ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

Reference Study Guides

More Biomolecules Previous-Year Questions — Page 6

Q63 jee_main_2024_30_jan_morning Carbohydrates
  • A. textSucrose
  • B. textLactose
  • C. textGlucose
  • D. textMaltose

Solution

### Core Logic Fehling's reagent is reduced by reducing sugars to give a reddish-brown precipitate of Cu_2O. Reducing sugars must have a free aldehyde/ketone group or a hemiacetal linkage that can open to form an aldehyde. Sucrose is a non-reducing sugar because its anomeric carbons (C1 of glucose and C2 of fructose) are tied up in a glycosidic linkage, leaving no free hemiacetal group. ### Step 1: Analyzing the options Lactose, glucose, and maltose are all reducing sugars and will give a positive Fehling's test. Sucrose does not. ### Pattern Recognition Sucrose = non-reducing sugar. Maltose, Lactose = reducing sugars. Monosaccharides (glucose, fructose) = always reducing. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules
Q89 jee_main_2024_31_jan_evening Vitamins and their Classification
From the vitamins A, B_1, B_6, B_12, C, D, E and K, the number vitamins that can be stored in our body is ________
Numerical Answer. Answer: 5 to 5

Solution

### Core Logic Vitamins are broadly classified into two groups based on solubility: 1) Fat-soluble vitamins: Vitamins A, D, E, and K. These are stored in the liver and adipose (fat-storing) tissues. 2) Water-soluble vitamins: B group vitamins and Vitamin C. These are readily excreted in urine and cannot be stored in the body (with the exception of Vitamin B_12, which can be stored in the liver). ### Step 1: Final List The vitamins that can be stored in the body from the given list are A, D, E, K, and B_12. Total number = 5. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules
Q78 jee_main_2024_31_jan_morning Reactions of Glucose
Match List I with List II
LIST-ILIST-II
A. Glucose/NaHCO_3/DeltaI. Gluconic acid
B. Glucose/HNO_3II. No reaction
C. Glucose/HI/DeltaIII. n-hexane
D. Glucose/Bromine waterIV. Saccharic acid
Choose the correct answer from the options given below:
  • A. textA-IV, B-I, C-III, D-II
  • B. textA-II, B-IV, C-III, D-I
  • C. textA-III, B-II, C-I, D-IV
  • D. textA-I, B-IV, C-III, D-II

Solution

### Core Logic Matching the reactions of glucose: (A) Glucose does not react with NaHCO_3, so there is no reaction. (A rightarrow II) (B) Oxidation of glucose with strong oxidizing agents like HNO_3 yields a dicarboxylic acid called saccharic acid. (B rightarrow IV) (C) Prolonged heating of glucose with HI forms n-hexane, indicating a straight chain of six carbon atoms. (C rightarrow III) (D) Oxidation with mild agents like bromine water converts glucose to gluconic acid. (D rightarrow I) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

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