Match List-I with List-II tracking nitrogenous bases structures:
List-I (Base)
List-II (Chemical Structure Diagram)
(A) Adenine
(I) The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(B) Cytosine
(II) The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(C) Thymine
(III) The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
(D) Uracil
(IV) The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.
Choose the correct answer from the options given below :
A.\text{(A)-(III), (B)-(IV), (C)-(II), (D)-(I)}
B.\text{(A)-(III), (B)-(I), (C)-(IV), (D)-(II)}
C.\text{(A)-(IV), (B)-(III), (C)-(II), (D)-(I)}
D.\text{(A)-(III), (B)-(IV), (C)-(I), (D)-(II)}
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$
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$
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$
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$
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:
The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.The Match List contains nitrogenous bases aligned with their corresponding biochemical molecular ring structures.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$-\mathrm{CH}_3$ group = Thymine
- Monocyclic ring without a -mathrmCH_3$-\mathrm{CH}_3$ group = Uracil
- Monocyclic ring with an amino group (-mathrmNH_2$-\mathrm{NH}_2$) = Cytosine
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
More Biomolecules Previous-Year Questions — Page 5
Q61jee_main_2024_01_february_morningNucleic Acids
If one strand of a DNA has the sequence ATGCTTCA, sequence of the bases in complementary strand is:
A. CATTAGCT
B. TACGAAGT
C. GTACTTAC
D. ATGCGACT
Solution
### Core Logic
Adenine (A) base pairs with Thymine (T) with 2 hydrogen bonds, and Cytosine (C) base pairs with Guanine (G) with 3 hydrogen bonds.
For the given DNA strand:
A rightarrow T$A \rightarrow T$T rightarrow A$T \rightarrow A$G rightarrow C$G \rightarrow C$C rightarrow G$C \rightarrow G$T rightarrow A$T \rightarrow A$T rightarrow A$T \rightarrow A$C rightarrow G$C \rightarrow G$A rightarrow T$A \rightarrow T$
### Step 1: Sequence Matching
Given sequence: A T G C T T C A
Complementary: T A C G A A G T
Nucleic Acids diagram for Q61 - JEE Main 2024 Morning
### Pattern Recognition
DNA base pairing strictly follows Chargaff's rule: A=T and G≡C. Just swap A with T, and C with G in the exact order.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q62jee_main_2024_29_january_eveningPolymers and Monomers
Match List I with List II:
List I (Bio Polymer)
List II (Monomer)
A. Starch
I. nucleotide
B. Cellulose
II. alpha$\alpha$-glucose
C. Nucleic acid
III. beta$\beta$-glucose
D. Protein
IV. alpha$\alpha$-amino acid
Choose the correct answer from the options given below :
A. A-II, B-I, C-III, D-IV
B. A-IV, B-II, C-I, D-III
C. A-I, B-III, C-IV, D-II
D. A-II, B-III, C-I, D-IV
Solution
### Related Formula
Factual knowledge of biopolymers and their fundamental repeating units (monomers).
### Core Logic
Analyzing each polymer component:
* **Starch** is a polymer composed entirely of alpha$\alpha$-glucose units.
* **Cellulose** is a linear structural polymer consisting of linear chains of beta$\beta$-glucose units.
* **Nucleic acids** (DNA/RNA) are long chains composed of repeating nucleotide units.
* **Proteins** are polypeptides made from combined alpha$\alpha$-amino acid sequences.
### Step 1: Final Match Alignment
Matching structural links properly leads cleanly to the configuration: A-II, B-III, C-I, D-IV.
### Pattern Recognition
Standard memorization trick: Plants store starch using alpha linkers, but build rigid cell walls via beta linkers.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
### Core Logic
Proteins are natural biopolymers. The fundamental building blocks (monomers) of all naturally occurring proteins are alpha$\alpha$-amino acids. In these molecules, both the amino group (-NH_2$-NH_2$) and the carboxyl group (-COOH$-COOH$) are attached to the exact same carbon atom, designated as the alpha$\alpha$-carbon.
Because they are polymers of alpha$\alpha$-amino acids linked by peptide bonds, the hydrolysis of proteins (acidic, basic, or enzymatic) will cleave these peptide bonds to yield the constituent alpha$\alpha$-amino acids.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q84jee_main_2024_30_january_eveningNucleic Acids (DNA and RNA)
The total number of correct statements, regarding the nucleic acids is
A. RNA is regarded as the reserve of genetic information.
B. DNA molecule self-duplicates during cell division
C. DNA synthesizes proteins in the cell.
D. The message for the synthesis of particular proteins is present in DNA
E. Identical DNA strands are transferred to daughter cells.
Numerical Answer.Answer: 3 to 3
Solution
### Core Logic
A. RNA is regarded as the reserve of genetic information. (False - DNA is the reserve of genetic information).
B. DNA molecule self-duplicates during cell division. (True - through replication).
C. DNA synthesizes proteins in the cell. (False - RNA molecules synthesize proteins via translation, DNA only provides the code).
D. The message for the synthesis of particular proteins is present in DNA. (True - the genetic code lies in the base sequence of DNA).
E. Identical DNA strands are transferred to daughter cells. (True - through accurate replication and cell division, genetic continuity is maintained).
### Step 1: Final Conclusion
The true statements are B, D, and E.
Total number of correct statements is 3.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
More Biomolecules Questions — jee_main_2025_24_jan_evening
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