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Curated PYQ Collection

Top 50 Most Repeated BIOMOLECULES PYQs | JEE MAINS

A curated collection of the most important questions from BIOMOLECULES, fully solved with step-by-step concepts to prepare for JEE MAINS.

Question #1

Practice Question

A.Monosaccharides are polymers of glucose.
B.Disaccharides consist of two monosaccharide units linked by a glycosidic bond.
C.Polysaccharides are always water soluble.
D.Monosaccharides contain a carbonyl group and at least three hydroxyl groups.

Concept Applied

A disaccharide is formed when two monosaccharide units join through a condensation reaction, creating a glycosidic bond. The other statements are eith...

Read Full Step-by-Step Solution →

Question #2

Practice Question

A.Vitamin A
B.Vitamin B2 (Riboflavin)
C.Vitamin B3 (Niacin)
D.Vitamin D

Concept Applied

Niacin (Vitamin B3) is a water‑soluble vitamin that forms the coenzyme NADâș, essential for oxidation‑reduction processes in metabolism....

Read Full Step-by-Step Solution →

Question #3

Practice Question

A.α-D-glucose and ÎČ-D-glucose
B.D-glucose and L-glucose
C.Open chain and cyclic ester
D.Aldose and ketose

Concept Applied

Mutarotation is equilibrium between anomers via open chain....

Read Full Step-by-Step Solution →

Question #4

Practice Question

A.I, III
B.I, II, III
C.II, IV
D.I, III, IV

Concept Applied

Peptide bond is planar (resonance), secondary includes α-helix; H-bonds stabilize secondary, not primary; hydrolysis breaks bond....

Read Full Step-by-Step Solution →

Question #5

Practice Question

A.C-1 attacks C-5 OH
B.C-5 OH attacks C-1 carbonyl
C.C-6 OH attacks C-1 carbonyl
D.C-2 OH attacks C-5

Concept Applied

In glucose, C-5 hydroxyl attacks C-1 aldehyde to form pyranose (6-membered) ring....

Read Full Step-by-Step Solution →

Question #6

Practice Question

A.Aldehyde carbon
B.Primary alcohol carbon
C.Ketone carbon
D.Hydroxyl carbon

Concept Applied

Cyclic hemiacetal forms when aldehyde group reacts with C5-OH; former aldehyde carbon becomes anomeric....

Read Full Step-by-Step Solution →

Question #7

Practice Question

A.Mutarotation is the change in optical rotation when anomers interconvert in solution.
B.$\alpha$-D-glucose and $\beta$-D-glucose are enantiomers.
C.Mutarotation occurs only in open-chain forms of monosaccharides.
D.Anomers differ in configuration at the C-2 carbon atom.

Concept Applied

Mutarotation is the spontaneous change in optical rotation of a sugar solution due to interconversion of anomers....

Read Full Step-by-Step Solution →

Question #8

Practice Question

A.Phosphodiester bond
B.Glycosidic bond
C.Hydrogen bond
D.Peptide bond

Concept Applied

Phosphodiester bonds link sugar and phosphate in DNA backbone....

Read Full Step-by-Step Solution →

Question #9

Practice Question

A.Active site geometry
B.Cofactor presence
C.Optimum pH
D.High temperature

Concept Applied

Lock and key model explains specificity via shape....

Read Full Step-by-Step Solution →

Question #10

Practice Question

A.Glucose
B.Fructose
C.Lactose
D.Sucrose

Concept Applied

Sucrose lacks free aldehyde or ketone group; no open-chain form → non-reducing....

Read Full Step-by-Step Solution →

Question #11

Practice Question

A.Urease
B.Hexokinase
C.Trypsin
D.Carbonic anhydrase

Concept Applied

Urease acts only on urea; others show group or bond specificity....

Read Full Step-by-Step Solution →

Question #12

Practice Question

A.The specific rotation of the solution remains constant over time.
B.Only the $\alpha$‑anomer is optically active.
C.The equilibrium mixture contains equal amounts of $\alpha$ and $\beta$ forms.
D.The observed rotation initially changes and then becomes constant.

Concept Applied

Mutarotation is the change in optical rotation as $\alpha$ and $\beta$ anomers interconvert until an equilibrium mixture (typically not 1:1) is reache...

Read Full Step-by-Step Solution →

Question #13

Practice Question

A.Alanine
B.Glutamine
C.Valine
D.Serine

Concept Applied

Valine is essential (cannot be synthesized); others are non-essential amino acids....

Read Full Step-by-Step Solution →

Question #14

Practice Question

A.Lock and key model
B.High molecular weight
C.Low concentration
D.Zymogen form

Concept Applied

Specificity arises from active site geometry matching substrate (lock and key)....

Read Full Step-by-Step Solution →

Question #15

Practice Question

A.A-T
B.G-C
C.A-U
D.C-T

Concept Applied

G-C pair has 3 H-bonds; A-T has 2 in DNA....

Read Full Step-by-Step Solution →

Question #16

Practice Question

A.Primary structure
B.Secondary structure
C.Tertiary structure
D.Quaternary structure

Concept Applied

Peptide bonds link amino acids in sequence, forming primary structure....

Read Full Step-by-Step Solution →

Question #17

Practice Question

A.Monosaccharides are polymers of glucose.
B.Disaccharides consist of two monosaccharide units linked by a glycosidic bond.
C.Polysaccharides are always water soluble.
D.Monosaccharides contain a carbonyl group and at least three hydroxyl groups.

Concept Applied

A disaccharide is formed when two monosaccharide units join through a condensation reaction, creating a glycosidic bond. The other statements are eith...

Read Full Step-by-Step Solution →

Question #18

Practice Question

A.Vitamin A
B.Vitamin B2 (Riboflavin)
C.Vitamin B3 (Niacin)
D.Vitamin D

Concept Applied

Niacin (Vitamin B3) is a water‑soluble vitamin that forms the coenzyme NADâș, essential for oxidation‑reduction processes in metabolism....

Read Full Step-by-Step Solution →

Question #19

Practice Question

A.α-D-glucose and ÎČ-D-glucose
B.D-glucose and L-glucose
C.Open chain and cyclic ester
D.Aldose and ketose

Concept Applied

Mutarotation is equilibrium between anomers via open chain....

Read Full Step-by-Step Solution →

Question #20

Practice Question

A.I, III
B.I, II, III
C.II, IV
D.I, III, IV

Concept Applied

Peptide bond is planar (resonance), secondary includes α-helix; H-bonds stabilize secondary, not primary; hydrolysis breaks bond....

Read Full Step-by-Step Solution →

Question #21

Practice Question

A.C-1 attacks C-5 OH
B.C-5 OH attacks C-1 carbonyl
C.C-6 OH attacks C-1 carbonyl
D.C-2 OH attacks C-5

Concept Applied

In glucose, C-5 hydroxyl attacks C-1 aldehyde to form pyranose (6-membered) ring....

Read Full Step-by-Step Solution →

Question #22

Practice Question

A.Aldehyde carbon
B.Primary alcohol carbon
C.Ketone carbon
D.Hydroxyl carbon

Concept Applied

Cyclic hemiacetal forms when aldehyde group reacts with C5-OH; former aldehyde carbon becomes anomeric....

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.Mutarotation is the change in optical rotation when anomers interconvert in solution.
B.$\alpha$-D-glucose and $\beta$-D-glucose are enantiomers.
C.Mutarotation occurs only in open-chain forms of monosaccharides.
D.Anomers differ in configuration at the C-2 carbon atom.

Concept Applied

Mutarotation is the spontaneous change in optical rotation of a sugar solution due to interconversion of anomers....

Read Full Step-by-Step Solution →

Question #24

Practice Question

A.Phosphodiester bond
B.Glycosidic bond
C.Hydrogen bond
D.Peptide bond

Concept Applied

Phosphodiester bonds link sugar and phosphate in DNA backbone....

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.Active site geometry
B.Cofactor presence
C.Optimum pH
D.High temperature

Concept Applied

Lock and key model explains specificity via shape....

Read Full Step-by-Step Solution →

Question #26

Practice Question

A.Glucose
B.Fructose
C.Lactose
D.Sucrose

Concept Applied

Sucrose lacks free aldehyde or ketone group; no open-chain form → non-reducing....

Read Full Step-by-Step Solution →

Question #27

Practice Question

A.Urease
B.Hexokinase
C.Trypsin
D.Carbonic anhydrase

Concept Applied

Urease acts only on urea; others show group or bond specificity....

Read Full Step-by-Step Solution →

Question #28

Practice Question

A.The specific rotation of the solution remains constant over time.
B.Only the $\alpha$‑anomer is optically active.
C.The equilibrium mixture contains equal amounts of $\alpha$ and $\beta$ forms.
D.The observed rotation initially changes and then becomes constant.

Concept Applied

Mutarotation is the change in optical rotation as $\alpha$ and $\beta$ anomers interconvert until an equilibrium mixture (typically not 1:1) is reache...

Read Full Step-by-Step Solution →

Question #29

Practice Question

A.Alanine
B.Glutamine
C.Valine
D.Serine

Concept Applied

Valine is essential (cannot be synthesized); others are non-essential amino acids....

Read Full Step-by-Step Solution →

Question #30

Practice Question

A.Lock and key model
B.High molecular weight
C.Low concentration
D.Zymogen form

Concept Applied

Specificity arises from active site geometry matching substrate (lock and key)....

Read Full Step-by-Step Solution →

Question #31

Practice Question

A.A-T
B.G-C
C.A-U
D.C-T

Concept Applied

G-C pair has 3 H-bonds; A-T has 2 in DNA....

Read Full Step-by-Step Solution →

Question #32

Practice Question

A.Primary structure
B.Secondary structure
C.Tertiary structure
D.Quaternary structure

Concept Applied

Peptide bonds link amino acids in sequence, forming primary structure....

Read Full Step-by-Step Solution →

Question #33

Practice Question

A.Monosaccharides are polymers of glucose.
B.Disaccharides consist of two monosaccharide units linked by a glycosidic bond.
C.Polysaccharides are always water soluble.
D.Monosaccharides contain a carbonyl group and at least three hydroxyl groups.

Concept Applied

A disaccharide is formed when two monosaccharide units join through a condensation reaction, creating a glycosidic bond. The other statements are eith...

Read Full Step-by-Step Solution →

Question #34

Practice Question

A.Vitamin A
B.Vitamin B2 (Riboflavin)
C.Vitamin B3 (Niacin)
D.Vitamin D

Concept Applied

Niacin (Vitamin B3) is a water‑soluble vitamin that forms the coenzyme NADâș, essential for oxidation‑reduction processes in metabolism....

Read Full Step-by-Step Solution →

Question #35

Practice Question

A.α-D-glucose and ÎČ-D-glucose
B.D-glucose and L-glucose
C.Open chain and cyclic ester
D.Aldose and ketose

Concept Applied

Mutarotation is equilibrium between anomers via open chain....

Read Full Step-by-Step Solution →

Question #36

Practice Question

A.I, III
B.I, II, III
C.II, IV
D.I, III, IV

Concept Applied

Peptide bond is planar (resonance), secondary includes α-helix; H-bonds stabilize secondary, not primary; hydrolysis breaks bond....

Read Full Step-by-Step Solution →

Question #37

Practice Question

A.C-1 attacks C-5 OH
B.C-5 OH attacks C-1 carbonyl
C.C-6 OH attacks C-1 carbonyl
D.C-2 OH attacks C-5

Concept Applied

In glucose, C-5 hydroxyl attacks C-1 aldehyde to form pyranose (6-membered) ring....

Read Full Step-by-Step Solution →

Question #38

Practice Question

A.Aldehyde carbon
B.Primary alcohol carbon
C.Ketone carbon
D.Hydroxyl carbon

Concept Applied

Cyclic hemiacetal forms when aldehyde group reacts with C5-OH; former aldehyde carbon becomes anomeric....

Read Full Step-by-Step Solution →

Question #39

Practice Question

A.Mutarotation is the change in optical rotation when anomers interconvert in solution.
B.$\alpha$-D-glucose and $\beta$-D-glucose are enantiomers.
C.Mutarotation occurs only in open-chain forms of monosaccharides.
D.Anomers differ in configuration at the C-2 carbon atom.

Concept Applied

Mutarotation is the spontaneous change in optical rotation of a sugar solution due to interconversion of anomers....

Read Full Step-by-Step Solution →

Question #40

Practice Question

A.Phosphodiester bond
B.Glycosidic bond
C.Hydrogen bond
D.Peptide bond

Concept Applied

Phosphodiester bonds link sugar and phosphate in DNA backbone....

Read Full Step-by-Step Solution →

Question #41

Practice Question

A.Active site geometry
B.Cofactor presence
C.Optimum pH
D.High temperature

Concept Applied

Lock and key model explains specificity via shape....

Read Full Step-by-Step Solution →

Question #42

Practice Question

A.Glucose
B.Fructose
C.Lactose
D.Sucrose

Concept Applied

Sucrose lacks free aldehyde or ketone group; no open-chain form → non-reducing....

Read Full Step-by-Step Solution →

Question #43

Practice Question

A.Urease
B.Hexokinase
C.Trypsin
D.Carbonic anhydrase

Concept Applied

Urease acts only on urea; others show group or bond specificity....

Read Full Step-by-Step Solution →

Question #44

Practice Question

A.The specific rotation of the solution remains constant over time.
B.Only the $\alpha$‑anomer is optically active.
C.The equilibrium mixture contains equal amounts of $\alpha$ and $\beta$ forms.
D.The observed rotation initially changes and then becomes constant.

Concept Applied

Mutarotation is the change in optical rotation as $\alpha$ and $\beta$ anomers interconvert until an equilibrium mixture (typically not 1:1) is reache...

Read Full Step-by-Step Solution →

Question #45

Practice Question

A.Alanine
B.Glutamine
C.Valine
D.Serine

Concept Applied

Valine is essential (cannot be synthesized); others are non-essential amino acids....

Read Full Step-by-Step Solution →

Question #46

Practice Question

A.Lock and key model
B.High molecular weight
C.Low concentration
D.Zymogen form

Concept Applied

Specificity arises from active site geometry matching substrate (lock and key)....

Read Full Step-by-Step Solution →

Question #47

Practice Question

A.A-T
B.G-C
C.A-U
D.C-T

Concept Applied

G-C pair has 3 H-bonds; A-T has 2 in DNA....

Read Full Step-by-Step Solution →

Question #48

Practice Question

A.Primary structure
B.Secondary structure
C.Tertiary structure
D.Quaternary structure

Concept Applied

Peptide bonds link amino acids in sequence, forming primary structure....

Read Full Step-by-Step Solution →

Question #49

Practice Question

A.Monosaccharides are polymers of glucose.
B.Disaccharides consist of two monosaccharide units linked by a glycosidic bond.
C.Polysaccharides are always water soluble.
D.Monosaccharides contain a carbonyl group and at least three hydroxyl groups.

Concept Applied

A disaccharide is formed when two monosaccharide units join through a condensation reaction, creating a glycosidic bond. The other statements are eith...

Read Full Step-by-Step Solution →

Question #50

Practice Question

A.Vitamin A
B.Vitamin B2 (Riboflavin)
C.Vitamin B3 (Niacin)
D.Vitamin D

Concept Applied

Niacin (Vitamin B3) is a water‑soluble vitamin that forms the coenzyme NADâș, essential for oxidation‑reduction processes in metabolism....

Read Full Step-by-Step Solution →
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