Enzymes are biological catalysts — proteins (with rare RNA exceptions called ribozymes) that accelerate biochemical reactions without being consumed. NCERT Class 12 Chemistry (pre-2023 edition), Chapter 14, page 425 describes enzymes as biocatalysts, almost all of them globular proteins, that are very specific and let the body's reactions run under very mild conditions.
The NEET-critical properties:
Specificity — Each enzyme acts on one substrate or a narrow class. Lock-and-key model (Fischer, 1894) vs. induced-fit model (Koshland, 1958). NEET questions test whether you can distinguish these two models.
Nomenclature — Enzymes named by substrate + suffix "-ase" (maltase hydrolyses maltose; lipase hydrolyses lipids). Six IUBMB classes: oxidoreductases, transferases, hydrolases, lyases, isomerases, ligases.
Optimum conditions — Each enzyme has a pH and temperature optimum. Beyond these, activity drops due to denaturation of the protein's tertiary structure. Pepsin (stomach, pH ~2) vs. trypsin (intestine, pH ~8) is a high-frequency recall pair.
Cofactors and coenzymes — Many enzymes need non-protein helpers. Metal ions (Zn²⁺, Mg²⁺) = cofactors. Organic molecules (NAD⁺, FAD) = coenzymes. The protein part alone = apoenzyme; apoenzyme + cofactor = holoenzyme.
Enzyme inhibition — Competitive inhibitors resemble substrate and occupy the active site. Non-competitive inhibitors bind elsewhere and alter enzyme shape. NEET tests whether a given inhibitor is reversible or irreversible.
Watch-out: Students confuse "enzyme denaturation" with "enzyme inhibition." Denaturation is irreversible structural unfolding (temperature/pH extreme). Competitive inhibition is reversible and overcome by excess substrate.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which of the following correctly defines a holoenzyme?
Answer: D. D is correct. A holoenzyme is the complete, catalytically active enzyme formed by the apoenzyme (protein part) combined with its required cofactor or coenzyme (the apoenzyme and its co-factors are in NCERT Class 11 Biology Chapter 9, page 118; NCERT does not use the word holoenzyme).
Why A is wrong: A describes the apoenzyme, not the holoenzyme — the apoenzyme is inactive without its cofactor.
Why B is wrong: B describes the enzyme-substrate complex (ES complex), which is a transient intermediate in catalysis, not a term for the enzyme itself.
Why C is wrong: C is incorrect because enzymes that need no cofactor are called simple enzymes, not holoenzymes. The term 'holoenzyme' specifically implies cofactor requirement.
Enzymes that catalyse the transfer of a functional group from one substrate to another belong to which IUBMB class?
Answer: C. C is correct. Transferases catalyse the transfer of a functional group (amino, phosphate, methyl, etc.) between substrates. Example: transaminases transfer amino groups (NCERT Class 12 Chemistry (pre-2023 edition), Chapter 14).
Why A is wrong: A is wrong — oxidoreductases catalyse oxidation-reduction reactions (transfer of electrons/hydrogen), not functional group transfer between substrates.
Why B is wrong: B is wrong — hydrolases catalyse bond cleavage by addition of water (hydrolysis), such as lipases and proteases.
Why D is wrong: D is wrong — lyases catalyse elimination reactions forming double bonds (or the reverse, addition to double bonds) without hydrolysis or oxidation.
Which enzyme is active at approximately pH 2 in the human digestive system?
Answer: D. D is correct. Pepsin is a gastric protease with an optimum pH of approximately 1.5–2, suited to the highly acidic stomach environment (NCERT Class 12 Chemistry (pre-2023 edition), Chapter 14).
Why A is wrong: A is wrong — trypsin operates in the alkaline environment of the small intestine (optimum pH ~7.5–8). Students who confuse gastric vs. intestinal proteases select this.
Why B is wrong: B is wrong — pancreatic lipase functions in the duodenum at pH ~7–8, not in the acidic stomach.
Why C is wrong: C is wrong — salivary amylase operates in the mouth at near-neutral pH (~6.8). It is inactivated in the stomach's acidic conditions.
A competitive inhibitor of an enzyme works by:
Answer: B. B is correct. Competitive inhibitors structurally resemble the substrate and compete for the active site. Their effect is reversed by increasing substrate concentration, which outcompetes the inhibitor for binding (NCERT Class 12 Chemistry (pre-2023 edition), Chapter 14).
Why A is wrong: A describes non-competitive or allosteric inhibition, not competitive. Additionally, 'permanently altering' conflates irreversible inhibition with allosteric regulation — these are distinct concepts.
Why C is wrong: C describes hydrolysis (breaking peptide bonds), not inhibition. Enzyme inhibition does not involve breaking covalent bonds of the enzyme's backbone.
Why D is wrong: D is wrong — no inhibitor lowers activation energy below the uninhibited enzyme's value. Inhibitors raise the effective barrier or reduce the available active enzyme, never enhance catalysis.
An enzyme requires Zn²⁺ ions for catalytic activity. The Zn²⁺ ion is best described as:
Answer: B. B is correct. Metal ions such as Zn²⁺, Mg²⁺, and Fe²⁺ that are required for enzyme activity are classified as cofactors. Coenzymes are specifically organic non-protein molecules (e.g., NAD⁺, FAD) (NCERT Class 12 Chemistry (pre-2023 edition), Chapter 14).
Why A is wrong: A is wrong — coenzymes are organic molecules (vitamins or vitamin derivatives like NAD⁺, coenzyme A). Metal ions are inorganic and classified as cofactors, not coenzymes.
Why C is wrong: C is wrong — while some metal ions are tightly bound (prosthetic groups), the question describes a required metal ion generically. The broadest correct classification is 'cofactor.' Additionally, prosthetic groups are defined by tight/covalent binding, which is not stated here.
Why D is wrong: D is wrong — the apoenzyme is the protein portion of the enzyme, not the metal ion helper.
The enzyme sucrase (invertase) hydrolyses sucrose but does not act on maltose or lactose. This property of enzymes is called:
Answer: A. A is correct. Substrate specificity means each enzyme catalyses reactions involving only its specific substrate(s). Sucrase acts only on sucrose due to the complementary shape of its active site (NCERT Class 12 Chemistry (pre-2023 edition), Chapter 14).
Why B is wrong: B is wrong — pH sensitivity describes how enzyme activity varies with hydrogen ion concentration. It does not explain why an enzyme acts on one substrate but not another.
Why C is wrong: C is wrong — reversibility refers to some enzymatic reactions being bidirectional (reaching equilibrium), not to selectivity for a particular substrate.
Why D is wrong: D is wrong — competitive inhibition is a mechanism of enzyme regulation (inhibitor competing for the active site), not a property describing why an enzyme selects one substrate over others.
When an enzyme is heated to 80°C, it loses catalytic activity irreversibly. Which level(s) of protein structure are disrupted?
Answer: C. C is correct. Thermal denaturation disrupts hydrogen bonds, hydrophobic interactions, and ionic bonds that maintain secondary, tertiary, and quaternary structures. Peptide bonds (primary structure) remain intact because they are covalent and require hydrolysis (acid/base/enzyme) to break (NCERT Class 12 Chemistry (pre-2023 edition), Chapter 14).
Why A is wrong: A is wrong — peptide bonds are covalent and are not broken by heating alone. This is a common confusion: students equate loss of function with loss of primary structure. Hydrolysis breaks primary structure, not heat.
Why B is wrong: B is wrong — denaturation is not limited to quaternary structure. Secondary (α-helices, β-sheets maintained by H-bonds) and tertiary (overall 3D folding) structures are also disrupted by heat.
Why D is wrong: D is wrong — primary structure (the amino acid sequence held by peptide bonds) survives denaturation. Claiming all four levels are disrupted conflates denaturation with complete degradation.
A linear polypeptide chain contains 150 amino acid residues. During its synthesis, how many water molecules were released in forming all the peptide bonds?
Answer: A. A is correct. Each peptide bond is formed by a condensation reaction releasing one water molecule. For N amino acid residues in a linear polypeptide, the number of peptide bonds = N − 1 = 150 − 1 = 149. Therefore, 149 water molecules are released (NCERT Class 12 Chemistry (pre-2023 edition), Chapter 14).
Why B is wrong: B is wrong — 150 would mean each residue forms one bond, but the first residue has no peptide bond to its left. Students who forget the 'N − 1' rule count one bond per residue and get N instead of N − 1.
Why C is wrong: C is wrong — 151 would mean N + 1 bonds, which has no chemical basis. No additional bond forms beyond those connecting adjacent residues.
Why D is wrong: D is wrong — 148 = N − 2 has no basis in peptide bond formation. This could result from incorrectly subtracting for both terminal residues, but only one subtraction is needed (the first residue lacks a preceding partner).
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Given
An enzyme catalyses the reaction: Maltose + H₂O → 2 Glucose. You are asked to classify this enzyme by IUBMB class and name it.
Required
(a) IUBMB class of this enzyme. (b) Systematic naming principle applied.
Concept
Enzyme classification depends on the type of reaction catalysed. Hydrolases catalyse hydrolysis (bond cleavage using water). Naming convention: substrate name + "-ase" suffix.
Formula/Rule
IUBMB Class 3 = Hydrolases (catalyse hydrolytic cleavage of C–O, C–N, C–C, and other bonds with addition of water). Name = maltase (acts on maltose).
Substitution
Reaction involves: substrate (maltose) + water → products (glucose units). Bond broken: glycosidic bond in maltose. Method: hydrolysis.
Calculation
Classification: reaction uses water to cleave a bond → hydrolase (Class 3). Systematic name: maltose glycohydrolase. Common name: maltase.
Final answer
The enzyme belongs to IUBMB Class 3 (Hydrolases) and is named maltase.
Note: This is a classification question — no numerical calculation is involved. The integers (Class 3, "2 Glucose") are exact counting numbers.
Common trap
Students confuse hydrolases with lyases. Key distinction: hydrolases use water to break bonds; lyases remove groups to form double bonds (or add groups to double bonds) WITHOUT hydrolysis or oxidation.
Similar NEET-style question
"The enzyme that catalyses the hydrolysis of a peptide bond in a protein is classified as: (a) Oxidoreductase (b) Transferase (c) Hydrolase (d) Isomerase." Answer: (c) Hydrolase — peptidases/proteases cleave peptide bonds using water.
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Biological catalysts (mostly proteins) that lower activation energy of biochemical reactions. Specific (lock-and-key model). Substrate binds at active site. Examples: pepsin, amylase, urease.
-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 14, p. 425More in Biomolecules: 4 exam traps and mistakes · 3 formulas · 1 question pattern from its other lessons.
1 question from NEET 2022. Answers verified against NTA official keys.
The incorrect statement regarding enzymes is
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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