Identify the correct statements.
A. Arginine and Tryptophan are essential amino acids.
B. Histidine does not contain heterocyclic ring in its structure.
C. Proline is a six membered cyclic ring amino acid.
D. Glycine does not have chiral centre.
E. Cysteine has characteristic feature of side chain as mathrmMeS-CH_2-CH_2-$\mathrm{MeS-CH_2-CH_2-}$.
Choose the correct answer from the options given below:
A.textC and E Only$\text{C and E Only}$
B.textB and E Only$\text{B and E Only}$
C.textC and D Only$\text{C and D Only}$
D.textA and D Only$\text{A and D Only}$
Solution & Explanation
### Core Logic
A. Arginine and Tryptophan are indeed essential amino acids (cannot be synthesized by the human body). (Correct)
B. Histidine contains an imidazole ring, which is a heterocyclic ring. So statement B is incorrect.
C. Proline contains a pyrrolidine ring, which is a five-membered cyclic ring. So statement C is incorrect.
D. Glycine is mathrmNH_2-CH_2-COOH$\mathrm{NH_2-CH_2-COOH}$. Because its alpha carbon has two identical hydrogen atoms attached, it lacks a chiral center. (Correct)
E. Cysteine's side chain is a thiol group (-mathrmCH_2-SH$-\mathrm{CH_2-SH}$), not a thioether (mathrmMeS-CH_2-CH_2-$\mathrm{MeS-CH_2-CH_2-}$ which belongs to Methionine). So statement E is incorrect.
Correct statements: A and D.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
### 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
Q29jee_main_2025_04_april_eveningAmino Acids and Peptides
A dipeptide, "x" on complete hydrolysis gives "y" and "z". "y" on treatment with aq. HNO_2$HNO_{2}$ produces lactic acid. On the other hand "z" on heating gives the following cyclic molecule.
The image shows a heterocyclic six-membered cyclic molecule containing amide groups, formed by heating amino acid residue z.
Based on the information given, the dipeptide X is:
A. valine-glycine
B. alanine-glycine
C. valine-leucine
D. alanine-alanine
Solution
### Related Formula
textDipeptide X xrightarrowtextHydrolysis textAmino Acid y + textAmino Acid z$$\text{Dipeptide } X \xrightarrow{\text{Hydrolysis}} \text{Amino Acid } y + \text{Amino Acid } z$$
### Core Logic
- Since **y** reacts with nitrous acid (HNO_2$HNO_2$) to give lactic acid (CH_3-CH(OH)-COOH$CH_3-CH(OH)-COOH$), **y** must be alanine (CH_3-CH(NH_2)-COOH$CH_3-CH(NH_2)-COOH$).
- When glycine (NH_2-CH_2-COOH$NH_2-CH_2-COOH$) is heated, two molecules undergo intermolecular cyclization to produce a six-membered diketopiperazine ring as shown in the problem diagram. Therefore, **z** is glycine.
Hence, combining residue **y** (alanine) and **z** (glycine), the dipeptide X is **alanine-glycine**.
### Step 1: Stepwise Degradation Overview
Reaction scheme:
1. Alanine-Glycine linkage rightarrow$\rightarrow$ Alanine + Glycine
2. textAlanine + HNO_2 rightarrow textLactic acid + N_2uparrow + H_2O$\text{Alanine} + HNO_2 \rightarrow \text{Lactic acid} + N_2\uparrow + H_2O$
3. 2 times textGlycine xrightarrowDelta textCyclic diketopiperazine + 2H_2O$2 \times \text{Glycine} \xrightarrow{\Delta} \text{Cyclic diketopiperazine} + 2H_2O$
### Pattern Recognition
Lactic acid generation from alpha-amino acids via nitrous acid deamination is a definitive chemical fingerprint for alanine. The unsubstituted cyclic diketopiperazine product confirms glycine as the second component.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q39jee_main_2025_04_april_morningProteins and Amino Acids
Identify the pair of reactants that upon reaction, with elimination of HCl$HCl$ will give rise to the dipeptide Gly-Ala.
A.mathrmNH_2-CH_2-COCl text and mathrmNH_2-CH(CH_3)-COOH$\mathrm{NH_2-CH_2-COCl} \text{ and } \mathrm{NH_2-CH(CH_3)-COOH}$
B.mathrmNH_2-CH_2-COCl text and mathrmNH_3-CH(CH_3)-COCl$\mathrm{NH_2-CH_2-COCl} \text{ and } \mathrm{NH_3-CH(CH_3)-COCl}$
C.mathrmNH_2-CH_2-COOH text and mathrmNH_2-CH(CH_3)-COCl$\mathrm{NH_2-CH_2-COOH} \text{ and } \mathrm{NH_2-CH(CH_3)-COCl}$
D.mathrmNH_2-CH_2-COOH text and mathrmNH_2-CH(CH_3)-COOH$\mathrm{NH_2-CH_2-COOH} \text{ and } \mathrm{NH_2-CH(CH_3)-COOH}$
Solution
### Core Logic
A dipeptide sequence is parsed strictly from the **N-terminus** to the **C-terminus**. Therefore, in Gly-Ala:
* **Glycine (Gly)** must supply its carbonyl end for coupling.
* **Alanine (Ala)** must provide its free amine end.
To drive peptide bond formation via the explicit elimination of HCl$HCl$, the carboxyl group of glycine must be pre-activated as an acyl chloride variant: mathrmNH_2-CH_2-COCl$\mathrm{NH_2-CH_2-COCl}$.
This reacts smoothly with the unsubstituted amine terminus of alanine, mathrmNH_2-CH(CH_3)-COOH$\mathrm{NH_2-CH(CH_3)-COOH}$, liberating HCl$HCl$ to form the amide bridge linkage: mathrmNH_2-CH_2-CONH-CH(CH_3)-COOH$\mathrm{NH_2-CH_2-CONH-CH(CH_3)-COOH}$.
### Pattern Recognition
peptide naming structure sequence convention dictates: textFirst name = textN-terminus acyl donor$\text{First name} = \text{N-terminus acyl donor}$, textSecond name = textC-terminus amine acceptor$\text{Second name} = \text{C-terminus amine acceptor}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q50jee_main_2025_04_april_morningNucleic Acids
The total number of hydrogen bonds of a DNA-double Helix strand whose one strand has the following sequence of bases is:
5' - G - G - C - A - A - A - T - C - G - G - C - T - A - 3'$$5' - G - G - C - A - A - A - T - C - G - G - C - T - A - 3'$$
Numerical Answer.Answer: 33 to 33
Solution
### Related Formula
textG-C base pairing implies 3 text Hydrogen bonds$$\text{G-C base pairing} \implies 3 \text{ Hydrogen bonds}$$textA-T base pairing implies 2 text Hydrogen bonds$$\text{A-T base pairing} \implies 2 \text{ Hydrogen bonds}$$
### Core Logic
Let's audit the nucleotide base distribution across the given single strand structure:
5' - G - G - C - A - A - A - T - C - G - G - C - T - A - 3'$$5' - G - G - C - A - A - A - T - C - G - G - C - T - A - 3'$$
* **Count the Guanine (G) and Cytosine (C) bases:**
Bases present: G_1, G_2, C_3, C_8, G_9, G_10, C_11$G_1, G_2, C_3, C_8, G_9, G_{10}, C_{11}$implies$\implies$ Total of 7 bases.
Each G-C interaction forms 3 hydrogen bonds:
textBonds_G-C = 7 times 3 = 21$$\text{Bonds}_{G-C} = 7 \times 3 = 21$$
* **Count the Adenine (A) and Thymine (T) bases:**
Bases present: A_4, A_5, A_6, T_7, T_12, A_13$A_4, A_5, A_6, T_7, T_{12}, A_{13}$implies$\implies$ Total of 6 bases.
Each A-T interaction forms 2 hydrogen bonds:
textBonds_A-T = 6 times 2 = 12$$\text{Bonds}_{A-T} = 6 \times 2 = 12$$
Summing them up yields the total hydrogen bonds in the helix:
textTotal Htext-bonds = 21 + 12 = 33$$\text{Total } H\text{-bonds} = 21 + 12 = 33$$
### Pattern Recognition
Quick check optimization: Total Bonds = 3 times (\#G + \#C) + 2 times (\#A + \#T)$3 \times (\#G + \#C) + 2 \times (\#A + \#T)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q26jee_main_2025_07_april_eveningProteins and Amino Acids
Given below are two statements:
Statement (I): On hydrolysis, oligo peptides give rise to fewer number of alpha$\alpha$-amino acids while proteins give rise to a large number of beta$\beta$-amino acids.
Statement (II): Natural proteins are denatured by acids which convert the water soluble form of fibrous proteins to their water insoluble form.
In the light of the above statements, choose the most appropriate answer from the options given below:
A.textBoth statement I and statement II are correct$\text{Both statement I and statement II are correct}$
B.textStatement I is incorrect but Statement II is correct$\text{Statement I is incorrect but Statement II is correct}$
C.textBoth statement I and statement II are incorrect$\text{Both statement I and statement II are incorrect}$
D.textStatement I is correct but Statement II is incorrect$\text{Statement I is correct but Statement II is incorrect}$
Solution
### Related Formula
textProtein xrightarrowtextHydrolysis textPeptides xrightarrowtextHydrolysis alphatext-amino acids$$\text{Protein} \xrightarrow{\text{Hydrolysis}} \text{Peptides} \xrightarrow{\text{Hydrolysis}} \alpha\text{-amino acids}$$
### Core Logic
Statement (I) is incorrect because the complete hydrolysis of both oligopeptides and proteins yields alpha$\alpha$-amino acids, not beta$\beta$-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 alpha$\alpha$-amino acids, so any statement mentioning beta$\beta$-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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
More Biomolecules Questions — jee_main_2026_21_jan_morning
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
)
)
)
