Biomolecules JEE Advanced previous year questions with solutions

5 solved JEE Advanced questions on Biomolecules, free to read — no sign-in needed. The full chapter has 25 questions; sign in to attempt the remaining 20 in the exam simulator.

1. Q1JEE Advanced Adv 2026 (Paper 2)
   The correct statement(s) regarding sugars is(are) Given: Specific rotations of L-($-$)-glucose and L-($+$)-fructose are $-52.5^\circ$ and $+92.5^\circ$, respectively.
   1. A.On treatment with $\text{HNO}_3$, gluconic acid is oxidized to saccharic acid, whereas glucose is not oxidized to saccharic acid.
   2. B.Fructose gives a positive Fehling's test because it isomerises to glucose and another aldohexose in the presence of Fehling's reagent.
   3. C.Invert sugar is an equimolar mixture of D-glucose and D-fructose formed after hydrolysis of the corresponding disaccharide.
   4. D.Specific rotation of invert sugar is $-40^\circ$.
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   Answer: B,C

   Statement (A) is incorrect: Nitric acid ($\text{HNO}_3$) is a strong oxidizing agent that oxidizes both glucose and gluconic acid to saccharic acid. Statement (B) is correct: Fehling's reagent provides an alkaline medium. Under these conditions, fructose (a ketose) undergoes the Lobry de Bruyn-van Ekenstein rearrangement to form an equilibrium mixture of fructose, glucose, and mannose. Glucose and mannose are aldohexoses that readily reduce Fehling's solution, giving a positive test. Statement (C) is correct: Hydrolysis of sucrose (a disaccharide) yields an equimolar mixture of D-($+$)-glucose and D-($-$)-fructose. This mixture is known as invert sugar because the sign of specific rotation changes from positive (for sucrose) to negative (for the mixture). Statement (D) is incorrect: The specific rotation of L-($-$)-glucose is $-52.5^\circ$, so for D-($+$)-glucose it is $+52.5^\circ$. The specific rotation of L-($+$)-fructose is $+92.5^\circ$, so for D-($-$)-fructose it is $-92.5^\circ$. The specific rotation of invert sugar (an equimolar mixture of D-glucose and D-fructose, which have the same molar mass) is the average of their specific rotations: $[\alpha]_{\text{mix}} = \dfrac{+52.5^\circ + (-92.5^\circ)}{2} = -20^\circ$. Answer: Fructose gives a positive Fehling's test because it isomerises to glucose and another aldohexose in the presence of Fehling's reagent.; Invert sugar is an equimolar mixture of D-glucose and D-fructose formed after hydrolysis of the corresponding disaccharide.
2. Q2JEE Advanced Adv 2025 (Paper 2)
   A linear octasaccharide (molar mass $=1024 \mathrm{~g} \mathrm{~mol}^{-1}$ ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is $58.26 \%(\mathrm{w} / \mathrm{w})$ of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is $\qquad$ Use : Molar mass $\left(\right.$ in $\left.\mathrm{g} \mathrm{mol}^{-1}\right)$ : ribose $=150,2$-deoxyribose $=134$, glucose $=180$; Atomic mass (in amu): $\mathrm{H}=1, \mathrm{O}=16$
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   Answer: 2

   $\underset{\mathrm{M} . \mathrm{M} .=1024}{\text { Octasaccharide }}+\underset{\mathrm{M} . \mathrm{M} .=126}{7 \mathrm{H}_2 \mathrm{O}} \longrightarrow \begin{aligned} & \text { Ribose }+ \text { 2deoxyribose }+ \text { glucose } \\ & \text { Total mass }=1024+126=1150\end{aligned}$ $\begin{aligned} & 58.26=\frac{134 \times n}{1150} \times 100 \\ & \frac{66.999}{100}=134 n \quad n=4.99=5 \end{aligned}$ 5 units of 2-Deoxyribose $1150=(5 \times 150)+(\mathrm{x} \times 150)+(\mathrm{y} \times 180)$ $\begin{aligned} & 1150=\underbrace{750}_{5 \text { unit }}+\underbrace{150 x}_{300}+\underbrace{180 y}_{180} \\ & \qquad \quad \qquad \quad \text{2 unit} \quad \text{1 unit}\end{aligned}$ n = 2.00
3. Q3JEE Advanced Adv 2019 (Paper 1)
   Which of the following statement(s) is(are) true?
   1. A.Oxidation of glucose with bromine water gives glutamic acid
   2. B.The two six-membered cyclic hemiacetal forms of $D-\left(+\right)-$ glucose are called anomers
   3. C.Hydrolysis of sucrose gives dextrorotatory glucose and levorotatory fructose
   4. D.Monosaccharides cannot be hydrolysed to give polyhydroxy aldehydes and ketones
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   Answer: B,C,D

   $\rightarrow$ Glucose $\rightarrow _{[\text{0}]}^{{\text{Br}}_{\text{2}}{\text{-H}}_{\text{2}}\text{O}}$ Glycolic acid $\rightarrow \propto -\Delta -$ Glucose and $\text{β}-\text{D}$ Glucose $\rightarrow$ Hydrolysis of sucrose given dextrorotatory glucose and levorotatory fructose, $\rightarrow$ Monosaccharide cannot be hydrolyzed to give polyhydroxy aldehyde and ketones.
4. Q4JEE Advanced Adv 2013 (Paper 1)
   A tetrapeptide has -COOH group on alanine. This produces glycine (Gly), Valine (Val), Phenyl alanine (Phe) and Alanine (Ala) on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with -NH2 group attached to a chiral center is
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   Answer: 4

   Following combinations are possible for tetrapeptide: $\begin{matrix}\text{Val}-\text{Phe}-\text{Gly}-\text{Ala} \\ \text{Val}-\text{Gly}-\text{Phe}-\text{Ala} \\ \text{Phe}-\text{Gly}-\text{Val}-\text{Gly} \\ \text{Phe}-\text{Val}-\text{Gly}-\text{Ala}\end{matrix}$ In all above sequences, C-terminal is alanine. Glycine is optically inactive amino acid, it can not be N-terminal. Hence, only above combinations are possible.
5. Q5JEE Advanced Adv 2007 (Paper 2)
   Statement I Glucose gives a reddish-brown precipitate with Fehling's solution. Statement II Reaction of glucose with Fehling's solution gives $\mathrm{CuO}$ and gluconic acid.
   1. A.Statement I is true, Statement II is true, Statement II is a correct explanation for Statement I
   2. B.Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I
   3. C.Statement I is true, Statement II is false
   4. D.Statement I is false, Statement II is true
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   Answer: (C)

   Glucose gives a reddish brown precipitate with Fehling's solution. Solution consists of $\mathrm{Cu}^{2+}$ which gets reduced to $\mathrm{Cu}^{+}\left(\mathrm{Cu}_2 \mathrm{O}\right)$.