How much of JEE Main comes from Polymers and Biomolecules?

JEEGenie estimates about 3% of JEE Main marks come from Polymers and Biomolecules, based on recent papers. The share varies from year to year.

Which formulas matter most in Polymers and Biomolecules?

The Polymers and Biomolecules sheet on this page has 13 formula entries, starting with Degree of Polymerization (DP), Carbohydrates, Proteins, Nucleic Acids, Number Average Molecular Weight (M_n). Each one comes with when to use it and the trap to avoid.

What topics does Polymers and Biomolecules cover for JEE?

Polymers and Biomolecules covers Carbohydrates, Proteins and Amino Acids, Nucleic Acids, Vitamins, Classification of Polymers, Biomolecules.

How should I revise Polymers and Biomolecules before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Polymers and Biomolecules questions on JEEGenie.

How many previous-year questions are there on Polymers and Biomolecules?

JEEGenie has 349 previous-year questions on Polymers and Biomolecules, and this page shows a few with worked explanations.

Is practice on Polymers and Biomolecules free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Polymers and Biomolecules for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 3% of JEE Main marks come from Polymers and Biomolecules. This page covers 13 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Confusing polymer with monomer; polymer is the chain, monomer is the unit.

Key points to know

Polymer. A polymer is a large molecule composed of repeating structural units (monomers) linked by covalent bonds.
Monomer. A monomer is a small molecule that can bond with other monomers to form a polymer.
Degree of Polymerization (DP). DP is the number of monomer units in a polymer chain, calculated as molecular weight of polymer divided by molecular weight of monomer.
Biomolecule. Biomolecules are organic molecules essential for life, including carbohydrates, proteins, nucleic acids, and lipids.
Carbohydrates. Carbohydrates are polyhydroxy aldehydes or ketones, classified as monosaccharides, disaccharides, or polysaccharides.
Proteins. Proteins are polymers of amino acids linked by peptide bonds, with primary to quaternary structures.
Nucleic Acids. Nucleic acids (DNA and RNA) are polymers of nucleotides, storing and transmitting genetic information.
Lipids. Lipids are hydrophobic biomolecules including fats, oils, phospholipids, and steroids, not polymers.

Formula sheet

Name	Formula	When to use
Degree of Polymerization (DP)	$D P = \frac{M_{n}}{M_{0}}$	When calculating the size or molecular weight of a polymer from its monomer.
Carbohydrates	General formula: $C_{n} (H_{2} O)_{n}$	When identifying or classifying sugars and their reactions.
Proteins	Peptide bond: $- C O - N H -$	When studying protein structure, function, or denaturation.
Nucleic Acids	Nucleotide = Nitrogenous base + Sugar + Phosphate	When studying genetics, replication, or transcription.
Number Average Molecular Weight ( $M_{n}$ )	$M_{n} = \frac{\sum N_{i} M_{i}}{\sum N_{i}}$	When calculating average molecular weight from number of chains.
Weight Average Molecular Weight ( $M_{w}$ )	$M_{w} = \frac{\sum N_{i} M_{i}^{2}}{\sum N_{i} M_{i}}$	When assessing polymer properties affected by larger chains.
Polydispersity Index (PDI)	$P D I = \frac{M_{w}}{M_{n}}$	When characterizing polymer uniformity or synthesis method.
Mutarotation of Glucose	$α$ -D-glucose: $[α]_{D} = + 112 °$ ; $β$ -D-glucose: $[α]_{D} = + 18.7 °$ ; equilibrium: $+ 52.5 °$	When explaining optical rotation changes in sugar solutions.
Peptide Bond Formation	$R - C O O H + H_{2} N - R^{'} \to R - C O - N H - R^{'} + H_{2} O$	When understanding protein synthesis or hydrolysis.
Identifying Amino Acid from Structure	General structure: $H_{2} N - C H R - C O O H$	When classifying or naming amino acids from their chemical structure.
Solving Polymer Molecular Weight Problems	$M_{n} = \frac{\sum N_{i} M_{i}}{\sum N_{i}}$ , $M_{w} = \frac{\sum N_{i} M_{i}^{2}}{\sum N_{i} M_{i}}$	When given a table of polymer fractions with number of chains and molecular weights.
Solving Carbohydrate Hydrolysis Problems	$(C_{6} H_{10} O_{5})_{n} + n H_{2} O \to n C_{6} H_{12} O_{6}$	When calculating the amount of monosaccharides from a given polysaccharide mass.
Using Saponification for Lipid Analysis	Triglyceride + 3NaOH → Glycerol + 3RCOONa	When analyzing fats or producing soap.

Open the Polymers and Biomolecules formula page

What JEE asks

Polymer. JEE often asks to identify polymers from given monomers or vice versa, e.g., nylon-6,6 from hexamethylenediamine and adipic acid.
Monomer. JEE tests identification of monomers from polymer structures, e.g., glucose for cellulose.
Degree of Polymerization (DP). JEE may ask to find DP given polymer and monomer molecular weights, e.g., for polyethylene.
Biomolecule. JEE often asks to classify given compounds as biomolecules or identify their functions.
Carbohydrates. JEE tests mutarotation, reducing nature, and hydrolysis of carbohydrates.
Proteins. JEE often asks about levels of structure, denaturation, and essential amino acids.
Nucleic Acids. JEE tests base pairing, structure differences, and functions of DNA vs RNA.
Lipids. JEE tests saponification, fatty acid types, and lipid functions.
Number Average Molecular Weight ( $M_{n}$ ). JEE may ask to compute $M_{n}$ from given data of polymer fractions.
Weight Average Molecular Weight ( $M_{w}$ ). JEE tests calculation of $M_{w}$ and polydispersity index.
Polydispersity Index (PDI). JEE may ask to interpret PDI values, e.g., living polymerization gives PDI close to 1.
Mutarotation of Glucose. JEE tests the mechanism and equilibrium of mutarotation, often with numerical values.
Peptide Bond Formation. JEE tests the direction of peptide bond formation and hydrolysis conditions.
Identifying Polymerization Type. JEE gives polymer structures and asks to identify type, e.g., Teflon is addition, polyester is condensation.
Distinguishing Reducing and Non-reducing Sugars. JEE tests with Tollens' test or Fehling's test, e.g., glucose gives positive, sucrose negative.
Determining Monomer from Polymer Structure. JEE often shows polymer like polyvinyl chloride (PVC) and asks for monomer (vinyl chloride).
Identifying Amino Acid from Structure. JEE tests identification of amino acids like glycine, alanine, or cysteine from structures.
Solving Polymer Molecular Weight Problems. JEE provides data like: 10 chains of 1000 g/mol, 20 chains of 2000 g/mol; find $M_{n}$ , $M_{w}$ , PDI.
Analyzing Protein Structure Levels. JEE tests which levels are disrupted by denaturation (secondary, tertiary and quaternary; primary structure stays intact).
Identifying Nucleic Acid Components. JEE tests base ratios (Chargaff's rules) and structure differences.
Solving Carbohydrate Hydrolysis Problems. JEE may ask: How many grams of glucose from 100 g of starch?
Using Saponification for Lipid Analysis. JEE tests saponification value and its relation to fatty acid chain length.
Interpreting Polymer Classification by Structure. JEE tests classification of polymers like polyethylene (linear) vs Bakelite (cross-linked).

Common traps

Confusing polymer with monomer; polymer is the chain, monomer is the unit. Polymer
Mistaking a dimer or trimer for a monomer; monomers are single units. Monomer
Using weight-average molecular weight instead of number-average; DP uses $M_{n}$ . Degree of Polymerization (DP)
Excluding lipids as biomolecules; they are crucial for membranes and energy storage. Biomolecule
Assuming all carbohydrates follow $C_{n} (H_{2} O)_{n}$ ; deoxy sugars like deoxyribose do not. Carbohydrates
Confusing peptide bond with disulfide bond; peptide bonds are between amino acids. Proteins
Confusing DNA and RNA; RNA has ribose and uracil, DNA has deoxyribose and thymine. Nucleic Acids
Classifying lipids as polymers; they are not made of monomers. Lipids
Using $M_{n}$ when $M_{w}$ is needed; $M_{n}$ is lower than $M_{w}$ for polydisperse samples. Number Average Molecular Weight ( $M_{n}$ )
Confusing $M_{w}$ with $M_{n}$ ; $M_{w}$ uses $M_{i}^{2}$ in numerator. Weight Average Molecular Weight ( $M_{w}$ )
Assuming PDI is always >1; it can be 1 for perfectly uniform polymers. Polydispersity Index (PDI)
Thinking mutarotation is only for glucose; it occurs for all reducing sugars. Mutarotation of Glucose
Forgetting that water is lost when the peptide bond forms (a condensation), and that hydrolysis needs water to break it again. Peptide Bond Formation
Classifying nylon as addition polymer; it's condensation (releases water). Identifying Polymerization Type

Topics in Polymers and Biomolecules

Carbohydrates
Proteins and Amino Acids
Nucleic Acids
Vitamins
Classification of Polymers
Biomolecules

Previous-year questions on Polymers and Biomolecules

JEEGenie has 349 previous-year questions on Polymers and Biomolecules. A few worked examples follow.

Year	Questions
2026	19
2025	21
2024	25
2023	36
2022	42
2021	43
2020	19
2019	41
2018	9
2017	5
2016	9
2015	8
2014	7
2013	6
2012	13
2011	5
2010	5
2009	4
2008	3
2007	3
2006	2
2005	6
2004	6
2003	5
2002	3
2001	1
2000	3

Question 1 medium

A D-aldotetrose on oxidation with concentrated

{H N O}_{3}

resulted in optically inactive dicarboxylic acid. The structure of the D-aldotetrose is :

Fischer projection, CHO at top and $C H_{2} O H$ at bottom. C-2: HO on the LEFT. C-3: HO on the LEFT. (Both OH left - an L-sugar.)
Fischer projection, CHO top / $C H_{2} O H$ bottom. C-2: OH on the RIGHT. C-3: HO on the LEFT. (OH right then left.)
Fischer projection, CHO top / $C H_{2} O H$ bottom. C-2: OH on the RIGHT. C-3: OH on the RIGHT. (Both OH on the right = D-erythrose; conc. $H N O_{3}$ oxidises it to meso-tartaric acid, which has an internal mirror plane and is therefore optically INACTIVE.)
Fischer projection, CHO top / $C H_{2} O H$ bottom. C-2: HO on the LEFT. C-3: OH on the RIGHT. (D-threose; its aldaric acid is optically active.)

Answer: C

Explanation. 1. A D-aldotetrose has the general structure:

C H O - C H O H - C H O H - C H_{2} O H

. 2. Oxidation with concentrated

H N O_{3}

converts both the aldehyde group (C-1) and the primary alcohol group (C-4) into carboxylic acids, giving an aldaric acid:

H O O C - C H O H - C H O H - C O O H

. 3. The product is optically inactive. This means the aldaric acid must be *meso*-tartaric acid, which has an internal plane of symmetry. 4. For *meso*-tartaric acid, the two chiral centers (C-2 and C-3) must have opposite configurations: one is

R

, the other is

S

. 5. In the parent aldotetrose, the configurations at C-2 and C-3 are the same as in the aldaric acid (oxidation does not change stereochemistry at these carbons). 6. Therefore, the D-aldotetrose must have opposite configurations at C-2 and C-3. Among the D-aldotetroses, this is **D-erythrose**, not D-threose. 7. The Fischer projection of D-erythrose is:

C H O

on top,

O H

on right at C-2,

O H

on left at C-3,

C H_{2} O H

at bottom. 8. This matches option **C**.

C

From NTA/IIT papers: JEE Main 2026 · 4 Apr Shift 2

Question 2 easy

A tetrapeptide, "

x

" on complete hydrolysis produced glycine (Gly), alanine (Ala), valine (Val), leucine (Leu) in equimolar proportion each. The number of tetrapeptides (sequences) possible involving each of these amino acids is :

Answer: B

Explanation. The tetrapeptide contains one each of Gly, Ala, Val, Leu. The number of distinct sequences is the number of permutations of 4 distinct amino acids:

4! = 4 \times 3 \times 2 \times 1 = 24

. Thus, 24 tetrapeptide sequences are possible. The correct answer is option B.

From NTA/IIT papers: JEE Main 2025 · 2 Apr Shift 2

Question 3 hard

Aspartame, an artificial sweetener, is a dipeptide aspartyl phenylalanine methyl ester. The structure of aspartame is [The four options are chemical structures; structures of phenylalanine (Ph-CH

_{2}

-CH(NH

_{2}

)-COOH) and aspartic acid (HOOC-CH

_{2}

-CH(NH

_{2}

)-COOH) are given in the question.]

H $_{2}$ N-CH(COOH)-CH $_{2}$ -C(=O)-NH-CH(CH $_{2}$ Ph)-COOMe (aspartyl side-chain / $β$ -carboxyl forms the amide bond; $α$ -COOH free)
HOOC-CH $_{2}$ -CH(NH $_{2}$ )-C(=O)-NH-CH(CH $_{2}$ Ph)-COOMe (aspartic acid $α$ -carboxyl forms the amide bond, $β$ -COOH free; phenylalanine as methyl ester)
H $_{2}$ N-CH(CH $_{2}$ Ph)-C(=O)-NH-CH(CH $_{2}$ COOH)-COOMe (phenylalanyl-aspartic acid methyl ester; order reversed)
MeOOC-CH $_{2}$ -CH(NH $_{2}$ )-C(=O)-NH-CH(CH $_{2}$ Ph)-COOH (aspartic acid $β$ -carboxyl as methyl ester; phenylalanine COOH free)

Answer: B

Explanation. Aspartame is the methyl ester of the dipeptide **aspartylphenylalanine**. The N-terminal amino acid is aspartic acid, and the C-terminal amino acid is phenylalanine, whose carboxyl group is esterified with methanol. 1. **Identify the peptide bond formation:** In aspartame, the amide bond is formed between the

α

-COOH of aspartic acid and the

α

-NH

_{2}

of phenylalanine. The

β

-COOH of aspartic acid remains free. 2. **Write the aspartic acid fragment:** Aspartic acid:

H O O C - C H_{2} - C H (N H_{2}) - C O O H

. When its

α

-COOH forms the amide bond, the fragment becomes:

H O O C - C H_{2} - C H (N H_{2}) - C (= O) -

. 3. **Write the phenylalanine fragment as methyl ester:** Phenylalanine:

P h - C H_{2} - C H (N H_{2}) - C O O H

. As the methyl ester, it is:

- N H - C H (C H_{2} P h) - C O O C H_{3}

. 4. **Combine the two fragments:**

H O O C - C H_{2} - C H (N H_{2}) - C (= O) - N H - C H (C H_{2} P h) - C O O C H_{3}

. 5. **Match with the given options:** This structure exactly matches **Option B**.

B

From NTA/IIT papers: JEE Advanced 2024

Question 4 easy

Sulphur (S) containing amino acids from the following are: (a) isoleucine (b) cysteine (c) lysine (d) methionine (e) glutamic acid

b, d
a, d
b, c, e
a, b, c

Answer: A

Explanation. Cysteine and methionine are the only sulfur-containing amino acids among the options. Isoleucine, lysine, and glutamic acid do not contain sulfur in their side chains.

From NTA/IIT papers: JEE Main 2023 · 8 Apr Shift 1

Question 5 hard

Treatment of D-glucose with aqueous

N a O H

results in a mixture of monosaccharides, which are (each option shows a set of three Fischer projections). The correct option is the set comprising D-glucose, D-mannose and D-fructose (Lobry de Bruyn-van Ekenstein rearrangement).

Set of three aldohexose Fischer projections: (i) CHO / HO-H / HO-H / H-OH / H-OH / $C H_{2} O H$ ; (ii) CHO / H-OH / HO-H / HO-H / H-OH / $C H_{2} O H$ ; (iii) CHO / HO-H / HO-H / HO-H / H-OH / $C H_{2} O H$ .
Set: (i) a 2-ketose $C H_{2} O H$ / C=O / HO-H / H-OH / H-OH / $C H_{2} O H$ ; (ii) CHO / HO-H / H-OH / H-OH / H-OH / $C H_{2} O H$ ; (iii) CHO / HO-H / HO-H / HO-H / H-OH / $C H_{2} O H$ .
Set: (i) D-glucose CHO / H-OH / HO-H / H-OH / H-OH / $C H_{2} O H$ ; (ii) D-fructose $C H_{2} O H$ / C=O / HO-H / H-OH / H-OH / $C H_{2} O H$ ; (iii) D-mannose CHO / HO-H / HO-H / H-OH / H-OH / $C H_{2} O H$ .
Set: (i) CHO / H-OH / HO-H / HO-H / H-OH / $C H_{2} O H$ ; (ii) CHO / H-OH / HO-H / H-OH / H-OH / $C H_{2} O H$ ; (iii) CHO / HO-H / HO-H / H-OH / H-OH / $C H_{2} O H$ .

Answer: C

Explanation. D-Glucose in aqueous NaOH undergoes the Lobry de Bruyn–van Ekenstein rearrangement. 1. **Enolization:** The open-chain form of D-glucose (an aldohexose) is converted to its 1,2-enediol by base-catalyzed proton abstraction at C-2. 2. **Isomerization:** This enediol is a common intermediate that can revert back to an aldose or tautomerize to a ketose. 3. **Products from the enediol:** - Re-protonation at C-1 gives back D-glucose. - Re-protonation at C-2 from the opposite face gives D-mannose (the C-2 epimer of glucose). - Keto-enol tautomerization of the enediol gives D-fructose (a 2-ketohexose). 4. **Stereochemistry check:** The configurations at C-3, C-4, and C-5 remain unchanged during this rearrangement. Therefore, all three products have the same stereochemistry at C-3, C-4, and C-5 as D-glucose. 5. **Fischer projections:** - **D-Glucose:** CHO / H-OH / HO-H / H-OH / H-OH / CH

_{2}

OH - **D-Mannose:** CHO / HO-H / HO-H / H-OH / H-OH / CH

_{2}

OH - **D-Fructose:** CH

_{2}

OH / C=O / HO-H / H-OH / H-OH / CH

_{2}

OH 6. **Matching with options:** Option C lists exactly these three structures: D-glucose, D-fructose, and D-mannose. **Correct answer: C**

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Polymers and Biomolecules?: JEEGenie estimates about 3% of JEE Main marks come from Polymers and Biomolecules, based on recent papers. The share varies from year to year.
Which formulas matter most in Polymers and Biomolecules?: The Polymers and Biomolecules sheet on this page has 13 formula entries, starting with Degree of Polymerization (DP), Carbohydrates, Proteins, Nucleic Acids, Number Average Molecular Weight (M_n). Each one comes with when to use it and the trap to avoid.
What topics does Polymers and Biomolecules cover for JEE?: Polymers and Biomolecules covers Carbohydrates, Proteins and Amino Acids, Nucleic Acids, Vitamins, Classification of Polymers, Biomolecules.
How should I revise Polymers and Biomolecules before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Polymers and Biomolecules questions on JEEGenie.
How many previous-year questions are there on Polymers and Biomolecules?: JEEGenie has 349 previous-year questions on Polymers and Biomolecules, and this page shows a few with worked explanations.
Is practice on Polymers and Biomolecules free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.