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
Keywords:#essential amino acids#Glycine chiral centre#JEE Main 2026 Morning Q67#Biomolecules JEE Main 2026
More Biomolecules Previous-Year Questions — Page 2
Qjee_main_2025_07_april_morningHormones Structure
Thyroxine, the hormone has the structure given below:
The structural formula of Thyroxine shows four iodine atoms attached to the aromatic benzene rings.
The percentage of iodine in thyroxine is ______ %. (nearest integer)
(Given molar mass in mathrmg\ mol^-1$\mathrm{g\ mol}^{-1}$ of mathrmC: 12$\mathrm{C}: 12$, mathrmH: 1$\mathrm{H}: 1$, mathrmO: 16$\mathrm{O}: 16$, mathrmN: 14$\mathrm{N}: 14$, mathrmI: 127$\mathrm{I}: 127$)
Numerical Answer.Answer: 65 to 65
Solution
### Related Formula
\% mathrmI = frac4 times M_mathrmIM_textThyroxine times 100$$\% \mathrm{I} = \frac{4 \times M_{\mathrm{I}}}{M_{\text{Thyroxine}}} \times 100$$
### Core Logic
Let's first determine the molecular formula of Thyroxine from its structure shown in below :
The structural formula of Thyroxine shows four iodine atoms attached to the aromatic benzene rings.
- Formula: mathrmC_15mathrmH_11mathrmO_4mathrmNmathrmI_4$\mathrm{C}_{15}\mathrm{H}_{11}\mathrm{O}_4\mathrm{N}\mathrm{I}_4$
Now, compute the molecular mass:
- Carbon: 15 times 12 = 180$15 \times 12 = 180$
- Hydrogen: 11 times 1 = 11$11 \times 1 = 11$
- Oxygen: 4 times 16 = 64$4 \times 16 = 64$
- Nitrogen: 1 times 14 = 14$1 \times 14 = 14$
- Iodine: 4 times 127 = 508$4 \times 127 = 508$textTotal Molecular Mass = 180 + 11 + 64 + 14 + 508 = 777 text g mol^-1$$\text{Total Molecular Mass} = 180 + 11 + 64 + 14 + 508 = 777 \text{ g mol}^{-1}$$
Now, calculate the percentage of Iodine:
\% mathrmI = frac508777 times 100 approx 65.38 \% approx 65 \%$$\% \mathrm{I} = \frac{508}{777} \times 100 \approx 65.38 \% \approx 65 \%$$
Hence, the percentage of iodine in thyroxine is 65$65$.
### Pattern Recognition
Thyroxine molecular mass (777$777$) is composed mostly of Iodine (508$508$), making up approximately 508/777 approx 65\%$508/777 \approx 65\%$ of its entire mass. Always carefully account for the four phenyl-bound iodines.
### Evaluation Rubric / Model Answer
Accurate molecular weight summation leading to 65\%$65\%$ nearest integer Iodine composition.
### Chapter Mix
Class 12 Chemistry: Biomolecules
Class 12 Chemistry: Organic Chemistry - Oxygen containing compounds
Q43jee_main_2025_07_april_morningHydrolysis of Sucrose
Given below are two statements:
Statement I: textD-(+)-glucose + textD-(+)-fructose xrightarrow-mathrmH_2mathrmO textsucrose qquad textsucrose xrightarrowtextHydrolysis textD-(+)-glucose + textD-(+)-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:
A.textBoth Statement I and Statement II are true.$\text{Both Statement I and Statement II are true.}$
B.textStatement I is false but Statement II is true.$\text{Statement I is false but Statement II is true.}$
C.textStatement I is true but Statement II is false.$\text{Statement I is true but Statement II is false.}$
D.textBoth Statement I and Statement II are false.$\text{Both Statement I and Statement II are false.}$
Solution
### Core Logic
Statement I: Sucrose is formed by condensation of textD-(+)-glucose$\text{D-(+)-glucose}$ and textD-(-)-fructose$\text{D-(-)-fructose}$ (levorotatory fructose, not dextrorotatory as claimed). Hydrolysis of sucrose yields textD-(+)-glucose$\text{D-(+)-glucose}$ and textD-(-)-fructose$\text{D-(-)-fructose}$. Thus, Statement I is false.
Statement II: Hydrolysis of dextrorotatory sucrose (+66.5^circ$+66.5^{\circ}$) yields a mixture of dextrorotatory glucose (+52.5^circ$+52.5^{\circ}$) and highly levorotatory fructose (-92.4^circ$-92.4^{\circ}$). Because the overall specific rotation of the mixture becomes levorotatory (-39.9^circ$-39.9^{\circ}$), the hydrolyzed mixture is called **invert sugar**. Thus, Statement II is true.
### Pattern Recognition
Natural fructose is always levorotatory, textD-(-)-fructose$\text{D-(-)-fructose}$. Dextrorotatory fructose mentioned in Statement I is an immediate giveaway that the statement is false.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q39jee_main_2025_08_april_eveningAmino Acids
The chemical structure of the amino acid Valine is monitored across different environments:
{{Q_IMG1}}
Choose the correct option depicting the structures of products A and B under pH = 2 and pH = 10 conditions respectively.
The flowchart maps the structural modifications of Valine when treated at strongly acidic (pH = 2) versus basic (pH = 10) conditions.The flowchart maps the structural modifications of Valine when treated at strongly acidic (pH = 2) versus basic (pH = 10) conditions.
A.textA = textH_3textN^+text-CH(CH(CH_3text)_2text)-COOH text and textB = textH_2textN-CH(CH(CH_3text)_2text)-COO^-$\text{A} = \text{H}_3\text{N}^+\text{-CH(CH(CH}_3\text{)}_2\text{)-COOH} \text{ and } \text{B} = \text{H}_2\text{N-CH(CH(CH}_3\text{)}_2\text{)-COO}^-$
B.textA = textH_2textN-CH(CH(CH_3text)_2text)-COO^- text and textB = textH_3textN^+text-CH(CH(CH_3text)_2text)-COOH$\text{A} = \text{H}_2\text{N-CH(CH(CH}_3\text{)}_2\text{)-COO}^- \text{ and } \text{B} = \text{H}_3\text{N}^+\text{-CH(CH(CH}_3\text{)}_2\text{)-COOH}$
C.textA = textH_2textN-CH(CH(CH_3text)_2text)-COO^- text and textB = textH_3textN^+text-CH(CH(CH_3text)_2text)-COO^-$\text{A} = \text{H}_2\text{N-CH(CH(CH}_3\text{)}_2\text{)-COO}^- \text{ and } \text{B} = \text{H}_3\text{N}^+\text{-CH(CH(CH}_3\text{)}_2\text{)-COO}^-$
D.textA = textH_3textN^+text-CH(CH(CH_3text)_2text)-COO^- text and textB = textH_3textN^+text-CH(CH(CH_3text)_2text)-COOH$\text{A} = \text{H}_3\text{N}^+\text{-CH(CH(CH}_3\text{)}_2\text{)-COO}^- \text{ and } \text{B} = \text{H}_3\text{N}^+\text{-CH(CH(CH}_3\text{)}_2\text{)-COOH}$
Solution
### Core Logic
Amino acids exhibit amphoteric behavior due to the simultaneously present basic amino (-textNH_2$-\text{NH}_2$) and acidic carboxyl (-textCOOH$-\text{COOH}$) functional groups:
1. **At pH = 2 (Highly Acidic Medium)**:
The abundant concentration of hydronium ions (H^+$H^+$) protonates the carboxylate ion back into its non-ionized acid state (-textCOO^- rightarrow -textCOOH$-\text{COO}^- \rightarrow -\text{COOH}$), while the amine group remains securely protonated as an ammonium ion (-textNH_3^+$-\text{NH}_3^+$). Hence, product **A** exists purely as a **cation**.
2. **At pH = 10 (Highly Basic Medium)**:
The high concentration of hydroxide ions (OH^-$OH^-$) abstracts protons from the system, deprotonating the carboxyl group into a carboxylate anion (-textCOO^-$-\text{COO}^-$) and neutralizing the ammonium group back into a free amine fraction (-textNH_2$-\text{NH}_2$). Hence, product **B** exists purely as an **anion**. The flowchart maps the structural modifications of Valine when treated at strongly acidic (pH = 2) versus basic (pH = 10) conditions.
### Pattern Recognition
Acidic environments (low pH) force positive overall charges onto amino structures (cation form). Basic environments (high pH) drive a net negative structure (anion form). This simple rule makes picking Option (1) immediate.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q29jee_main_2025_29_jan_eveningAmino Acids and Proteins
Identify the essential amino acids from below:
(A) Valine
(B) Proline
(C) Lysine
(D) Threonine
(E) Tyrosine
Choose the correct answer from the options given below:
A. (A), (C) and (D) only
B. (A), (C) and (E) only
C. (B), (C) and (E) only
D. (C), (D) and (E) only
Solution
### Core Logic
Essential amino acids cannot be synthesized by the body and must be obtained from diet. From the given choices:
* Valine (Essential)
* Proline (Non-essential)
* Lysine (Essential)
* Threonine (Essential)
* Tyrosine (Non-essential)
Hence, (A), (C), and (D) are the essential amino acids.
### Pattern Recognition
Mnemonic for essential amino acids: TV TILL PM MALL (Threonine, Valine, Tryptophan, Isoleucine, Leucine, Lysine, Phenylalanine, Methionine, Arginine, Histidine).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
Q45jee_main_2025_28_jan_morningReactions of Glucose and Starch
Given below are two statements :
Statement I : D-glucose pentaacetate reacts with 2, 4-dinitrophenylhydrazine.
Statement II : Starch, on heating with concentrated sulfuric acid at 100^circmathrmC$100^{\circ}\mathrm{C}$ and 2-3 atmosphere pressure produces glucose.
In the light of the above statements, choose the correct answer from the options given below
A.textBoth Statement I and Statement II are false$\text{Both Statement I and Statement II are false}$
B.textStatement I is false but Statement II is true.$\text{Statement I is false but Statement II is true.}$
C.textStatement I is true but Statement II is false.$\text{Statement I is true but Statement II is false.}$
D.textBoth Statement I and Statement II are true.$\text{Both Statement I and Statement II are true.}$
Solution
### Core Logic
Statement I is false because glucose pentaacetate fixes the cyclic hemiacetal system structure securely into an unreactive ester configuration. As a result, it cannot revert to an open-chain form containing a free aldehyde group, meaning it does not react with carbonyl reagents like 2,4-DNP.
Statement II is true because starch, a polysaccharide composed of glucose monomer blocks, undergoes acid-catalyzed hydrolysis to yield glucose when heated under pressure.
### Pattern Recognition
Sees: Pentacetate reactivity vs polysaccharide hydrolysis.
Shortcut: Acetylation locks the cyclic structure of glucose, preventing reactions that require an open-chain carbonyl group (like 2,4-DNP).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Biomolecules
More Biomolecules Questions — jee_main_2026_21_jan_morning
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