Enzymes

8 MCQs1 revision card9-step worked example
Source: NCERT Plant KingdomPYQ coverage: NEET 2020, 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Enzymes are biological catalysts — proteins (with a few RNA exceptions called ribozymes) that accelerate biochemical reactions without being consumed. NEET questions on enzymes revolve around a single high-frequency confusion: misinterpreting what Km actually represents.

The trap you need to fix. Most aspirants, when asked about Km, describe it as "the rate at half-maximum velocity." That is wrong. Km is a substrate concentration (measured in molarity, M), not a rate. Specifically, Km is the substrate concentration [S] at which the reaction velocity v equals exactly half of Vmax. Lower Km means higher enzyme–substrate affinity — the enzyme reaches half its maximum speed at a lower substrate concentration.

Core concept (NCERT Class 11 Biology, Chapter 9, page 200). Enzyme activity depends on temperature, pH, and substrate concentration. The Michaelis-Menten equation captures the substrate-velocity relationship:

v = Vmax × [S] / (Km + [S])

At very low [S] (much less than Km), v rises nearly linearly with [S]. At very high [S] (much greater than Km), v plateaus at Vmax because all active sites are saturated.

Classification basics for NEET recall. Enzymes are classified into six IUB classes: oxidoreductases, transferases, hydrolases, lyases, isomerases, and ligases. Each name tells you the reaction type — hydrolases catalyse hydrolysis, transferases transfer functional groups, and so on. The suffix "-ase" is appended to the substrate name (e.g., lipase acts on lipids, sucrase on sucrose).

Cofactors vs. coenzymes. Non-protein helpers that enzymes need: cofactors are inorganic ions (Zn²⁺, Mg²⁺); coenzymes are organic molecules (NAD⁺, FAD). The protein part alone (without its helper) is the apoenzyme; the complete active form is the holoenzyme.

Watch-out for NEET. Competitive inhibitors raise the apparent Km (enzyme needs more substrate to reach half Vmax) but do not change Vmax. Non-competitive inhibitors lower Vmax but leave Km unchanged. This distinction is a frequent distractor source.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

Which of the following is the correct definition of Km in the Michaelis-Menten equation?

MCQ 2Easy RecallPractice

An enzyme that catalyses the transfer of a functional group from one molecule to another belongs to which IUB class?

MCQ 3Easy RecallPractice

The complete conjugated enzyme with its non-protein component is called:

MCQ 4Direct ApplicationPractice

A competitive inhibitor is added to an enzyme-catalysed reaction at saturating substrate concentration. What happens to Km and Vmax compared to the uninhibited reaction?

MCQ 5Direct ApplicationPractice

An enzyme has a Km of 2 mM. If the substrate concentration is 2 mM, what fraction of Vmax is the reaction velocity?

MCQ 6Direct ApplicationPractice

Enzyme X has Km = 0.5 mM for substrate P. Enzyme Y has Km = 5 mM for the same substrate. Which statement is correct?

MCQ 7Concept TrapPractice

A student claims that Km has units of mol·L⁻¹·s⁻¹. Why is this claim incorrect?

MCQ 8CalculationPractice

For an enzyme following Michaelis-Menten kinetics, Vmax = 100 µmol/min and Km = 4 mM. What is the initial velocity when [S] = 12 mM?

Quick recall before you leave

Worked Example

  1. 1

    Given

    An enzyme has Vmax = 200 µmol/min and Km = 5 mM. The substrate concentration is 15 mM.

  2. 2

    Required

    Find the initial reaction velocity v.

  3. 3

    Concept

    The Michaelis-Menten equation relates substrate concentration to initial velocity for a single-substrate enzyme with no cooperative or allosteric effects.

  4. 4

    Formula

    v = Vmax × [S] / (Km + [S])

  5. 5

    Substitution

    v = 200 × 15 / (5 + 15)

  6. 6

    Calculation

    v = 3000 / 20 = 150 µmol/min Note: All numerical values in this problem (200, 15, 5) are exact values given in the problem statement. They do not limit significant figures in the answer.

  7. 7

    Final answer

    v = 150 µmol/min At [S] = 3 × Km, the enzyme operates at 75% of Vmax. This confirms the enzyme is NOT yet saturated — Vmax (200 µmol/min) would require much higher [S].

  8. 8

    Common trap

    The frequent mistake is interpreting Km as a rate ("the rate at half Vmax") instead of as a concentration. If a question asks "what does Km = 5 mM mean?", the answer is: "the substrate concentration at which v = Vmax/2 is 5 mM." The units (mM = millimoles per litre) confirm it is a concentration, not a rate.

  9. 9

    Similar NEET-style question

    "An enzyme with Km = 2 mM is incubated with substrate at 10 mM. If Vmax = 60 µmol/min, what is the initial velocity?" Apply: v = 60 × 10 / (2 + 10) = 600 / 12 = 50 µmol/min. ---

Before solving, remember these

Key Fact

Enzymes — Michaelis-Menten

Enzymes: globular proteins (RNA = ribozyme exception); active site; lower activation energy. Rate vs [S] — hyperbolic; Vmax at saturation; Km = [S] at Vmax/2. Inhibition: competitive (Km up, Vmax same), non-competitive (Km same, Vmax down).

-- NCERT Class 11 Biology, Ch. 9, p. 200

Formulas

Michaelis-Menten equation

Initial reaction velocity v of an enzyme as a function of substrate concentration. Km = [S] at v = Vmax/2.

SymbolQuantitySI Unit
vrate-
Vmaxmax rate-
KmMichaelis constantM
[S]substrateM

Valid when

  • Single substrate
  • No allosteric/cooperative effects

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Mitosis vs meiosis stage characteristics swapped

Category: Similar Terms

Synapsis, crossing over, chiasmata are meiosis-only (prophase I). Equational mitosis has none.

When it triggers

Question describes a phase event and asks which division it belongs to.

How to avoid

Synapsis/CO/chiasmata = meiosis I. Sister chromatid separation = mitosis or meiosis II. Homologue separation = meiosis I.

Trap

70S vs 80S ribosome confusion

Category: Similar Terms

Prokaryotes 70S; eukaryotic cytosol 80S; mitochondria + chloroplast 70S (endosymbiotic).

When it triggers

Question gives organelle/cell type and asks ribosome type.

How to avoid

Cytosolic of eukaryote = 80S; everywhere else (bacteria, mito, chloroplast) = 70S.

Past Year Questions

83 questions from NEET 2020, 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

Given below are two statements : One is labelled as Assertion (A) and other is labelled as Reason (R). Assertion (A) : Cells of the tapetum possess dense cytoplasm and generally have more than one nucleus. Reason (R) : Presence of more than one nucleus in the tapetum increases the efficiency of nourishing the developing microspore mother cells. In light of the above statements, choose the most appropriate answer from the options given below:

1A is false but R is true
2Both A and R are true and R is the correct explanation of A
3Both A and R are true but R is NOT the correct explanation of A
4A is true but R is false
NTA Answer: Option 4(final)
NEET 2025

Histones are enriched with -

1Phenylalanine & Arginine
2Lysine & Arginine
3Leucine & Lysine
4Phenylalanine & Leucine
NTA Answer: Option 2(final)
NEET 2025

Given below are two statements : one is labelled as Assertion (A), and the other is labelled as Reason (R). Assertion (A) : The primary function of the Golgi apparatus is to package the materials made by the endoplasmic reticulum and deliver it to intracellular targets and outside the cell. Reason (R) : Vesicles containing materials made by the endoplasmic reticulum fuse with the cis face of the Golgi apparatus, and they are modified and released from the trans face of the Golgi apparatus. In the light of the above statements, choose the correct answer from the options given below :

1A is false but R is true
2Both A and R are true and R is the correct explanation of A
3Both A and R are true but R is not the correct explanation of A
4A is true but R is false
NTA Answer: Option 3(final)
NEET 2025

From the statements given below choose the correct option : A. The eukaryotic ribosomes are 80S and prokaryotic ribosomes are 70S. B. Each ribosome has two sub-units. C. The two sub-units of 80S ribosome are 60S and 40S while that of 70S are 50S and 30S. D. The two sub-units of 80S ribosome are 60S and 20S and that of 70S are 50S and 20S. E. The two sub-units of 80S are 60S and 30S and that of 70S are 50S and 30S.

1B, D, E are true
2A, B, C are true
3A, B, D are true
4A, B, E are true
NTA Answer: Option 2(final)
NEET 2025

Given below are two statements : Statement I : Transfer RNAs and ribosomal RNA do not interact with mRNA. Statement II : RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence. In the light of the above statements, choose the most appropriate answer from the options given below :

1Statement I is incorrect but statement II is correct
2Both statement I and statement II are correct
3Both statement I and statement II are incorrect
4Statement I is correct but statement II is incorrect
NTA Answer: Option 1(final)
NEET 2024

What is the fate of a piece of DNA carrying only gene of interest which is transferred into an alien organism? A. The piece of DNA would be able to multiply itself independently in the progeny cells of the organism. B. It may get integrated into the genome of the recipient. C. It may multiply and be inherited along with the host DNA. D. The alien piece of DNA is not an integral part of chromosome. E. It shows ability to replicate. Choose the correct answer from the options given below:

1A and B only
2D and E only
3B and C only
4A and E only
NTA Answer: Option 3(final)
NEET 2024

Given below are two statements: Statement I : Chromosomes become gradually visible under light microscope during leptotene stage. Statement II : The beginning of diplotene stage is recognized by dissolution of synaptonemal complex. In the light of the above statements, choose the correct answer from the options given below:

1Both Statement I and Statement II are true
2Both Statement I and Statement II are false
3Statement I is true but Statement II is false
4Statement I is false but Statement II is true
NTA Answer: Option 1(final)
NEET 2024

The DNA present in chloroplast is:

1Linear, double stranded
2Circular, double stranded
3Linear, single stranded
4Circular, single stranded
NTA Answer: Option 2(final)
NEET 2024

Read the following statements and choose the set of correct statements: In the members of Phaeophyceae, A. Asexual reproduction occurs usually by biflagellate zoospores. B. Sexual reproduction is by oogamous method only. C. Stored food is in the form of carbohydrates which is either mannitol or laminarin. D. The major pigments found are chlorophyll a, c and carotenoids and xanthophyll. E. Vegetative cells have a cellulosic wall, usually covered on the outside by gelatinous coating of algin. Choose the correct answer from the options given below:

1A, B, C and D only
2B, C, D and E only
3A, C, D and E only
4A, B, C and E only
NTA Answer: Option 3(final)
NEET 2024

Given below are two statements: Statement I: Mitochondria and chloroplasts both double membranes bound organelles. Statement II: Inner membrane of mitochondria is relatively less permeable, as compared chloroplast. In the light of the above statements, choose the mis appropriate answer from the options given below

1Both Statement I and Statement II are correct.
2Both Statement I and Statement II are incorrect.
3Statement I is correct but Statement Il is incorrect.
4Statement I is incorrect but Statement Il is correct.
NTA Answer: Option 3(final)
NEET 2023

Given below are two statements: Statement I: A protein is imagined as a line, the left end represented by first amino acid (C-terminal) and the right end represented by last amino acid (N-terminal). Statement II: Adult human haemoglobin, consists of 4 subunits (two subunits of - 35- type and two subunits of type.) In the light of the above statements, choose the correct answer from the options given below:

1Both Statement I and Statement II are false.
2Statement I is true but Statement II is false.
3Statement I is false but Statement II is true.
4Both Statement I and Statement II are true
NTA Answer: Option 3(final)
NEET 2023

Given below are two statements : Statement I : Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat. Statement II : When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor. In the light of the above statements, choose the correct answer from the options given below :

1Both Statement I and Statement II are false.
2Statement I is true but Statement II is false.
3Statement I is false but Statement II is true.
4Both Statement I and Statement II are true.
NTA Answer: Option 4(final)
NEET 2023

Given below are two statements: Statement I : During G phase of cell cycle, the cell is metabolically inactive. 0 Statement II : The centrosome undergoes duplication during S phase of interphase. In the light of the above statements, choose the most appropriate answer from the options given below:

1Both Statement I and Statement II are incorrect.
2Statement I is correct but Statement II is incorrect.
3Statement I is incorrect but Statement II is correct.
4Both Statement I and Statement II are correct
NTA Answer: Option 3(final)
NEET 2022

Given below are two statements : Statement I : The primary CO acceptor in C plants is phosphoenolpyruvate and is found in the mesophyll cells. 2 4 Statement II : Mesophyll cells of C plants lack RuBisCo enzyme. In the light of the above statements, choose the correct 4 answer from the options given below:

1Statement I is incorrect but Statement II is correct
2Both Statement I and Statement II are correct
3Both Statement I and Statement II are incorrect
4Statement I is correct but Statement II is incorrect
NTA Answer: Option 2(final)
NEET 2022

Match List-I with List-II List-I List-II (a) Manganese (i) Activates the enzyme catalase (b) Magnesium (ii) Required for pollen germination (c) Boron (iii) Activates enzymes of respiration (d) Iron (iv) Functions in splitting of water during photosynthesis Choose the correct answer from the options given below :

1(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
2(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
3(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
4(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
NTA Answer: Option 3(final)
NEET 2022

Read the following statements on lipids and find out correct set of statements: (a) Lecithin found in the plasma membrane is a glycolipid (b) Saturated fatty acids possess one or more c = c bonds (c) Gingely oil has lower melting point, hence remains as oil in winter (d) Lipids are generally insoluble in water but soluble in some organic solvents (e) When fatty acid is esterified with glycerol, monoglycerides are formed Choose the correct answer from the option given below:

1(a), (b) and (d) only
2(a), (b) and (c) only
3(a), (d) and (e) only
4(c), (d) and (e) only
NTA Answer: Option 4(final)
NEET 2022

Match List-I with List-II. List-I List-II (a) Metacentric chromosome (i) Centromere situated close to the end forming one extremely short and one very long arms (b) Acrocentric chromosome (ii) Centromere at the terminal end (c) Submetacentric (iii) Centromere in the middle forming two equal arms of chromosomes (d) Telocentric chromosome (iv) Centromere slightly away from the middle forming one shorter arm and one longer arm Choose the correct answer from the options given below :

1(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
2(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
3(a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
4(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
NTA Answer: Option 2(final)
NEET 2022

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Mendel’s law of Independent assortment does not hold good for the genes that are located closely on the same chromosome. Reason (R) : Closely located genes assort independently. In the light of the above statements, choose the correct answer from the options given below:

1(A) is not correct but (R) is correct
2Both (A) and (R) are correct and (R) is the correct explanation of (A)
3Both (A) and (R) are correct but (R) is not the correct explanation of (A)
4(A) is correct but (R) is not correct
NTA Answer: Option 4(final)
NEET 2022

Regarding Meiosis, which of the statements is incorrect?

1Four haploid cells are formed at the end of Meiosis-II
2There are two stages in Meiosis, Meiosis-I and II
3DNA replication occurs in S phase of Meiosis-II
4Pairing of homologous chromosomes and recombination occurs in Meiosis-I
NTA Answer: Option 3(final)
NEET 2022

Which of the following statements are true for spermatogenesis but do not hold true for Oogenesis? (a) It results in the formation of haploid gametes (b) Differentiation of gamete occurs after the completion of meiosis (c) Meiosis occurs continuously in a mitotically dividing stem cell population (d) It is controlled by the Luteinising hormone (LH) and Follicle Stimulating Hormone (FSH) secreted by the anterior pituitary (e) It is initiated at puberty Choose the most appropriate answer from the options given below:

1(b), (c) and (e) only
2(c) and (e) only
3(b) and (c) only
4(b), (d) and (e) only
NTA Answer: Option 1(final)
NEET 2022

Which of the following is a correct match for disease and its symptoms?

1Muscular dystrophy – An auto immune disorder causing progressive degeneration of skeletal muscle
2Arthritis – Inflammed joints
3Tetany – High Ca2+ level causing rapid spasms.
4Myasthenia gravis – Genetic disorder resulting in weakening and paralysis of skeletal muscle
NTA Answer: Option 2(final)
NEET 2021

Identify the correct statement. (b)G 2 phase (ii) Inactive phase (c) Quiescent (iii) Interval between mitosis

1Split gene arrangement is characteristic of stage and initiation of prokaryotes DNA replication (d)G phase (iv) DNA replication 1
2In capping, methyl guanosine triphosphate is added to the 3' end of hnRNA (a) (b) (c) (d)
3RNA polymerase binds with Rho factor to (1) (ii) (iv) (iii) (i) terminate the process of transcription in bacteria (2) (iii) (ii) (i) (iv)
4The coding strand in a transcription unit is (3) (iv) (ii) (iii) (i) copied to an mRNA (4) (iv) (i) (ii) (iii)
NTA Answer: Option 3(final)
NEET 2020

∑ȧ¿U Áfl÷ÊÁ¡Ã „Ê ⁄U„Ë ∑§Ê Á‡Ê∑§Êÿ ∑§Ê Á‡Ê∑§Ê ø∑˝§áÊ ‚ ’Ê„⁄U 95. Some dividing cells exit the cell cycle and enter ÁŸ∑§‹ ¡ÊÃË „Ò •ÊÒ⁄U ∑§ÊÁÿ∑§ ÁŸÁc∑˝§ÿÃÊ •flSÕÊ ◊ ¬˝fl ‡Ê ∑§⁄U vegetative inactive stage. This is called quiescent stage (G ). This process occurs at the end of : ¡ÊÃË „Ò– ß‚ ‡ÊÊ Ã •flSÕÊ (G ) ∑§„Ê ¡ÊÃÊ „Ò– ÿ„ ¬˝Á∑˝§ÿÊ 0 0 Á∑§‚∑ § •ãà ◊ „Ê ÃË „Ò?

1G phase 2 (1) G ¬˝ÊflSÕÊ
2M phase 2 (2) M ¬˝ÊflSÕÊ
3G 1 phase
4S phase (3) G ¬˝ÊflSÕÊ 1 (4) S ¬˝ÊflSÕÊ
NTA Answer: Option 3(final)
NEET 2020

©Ÿ ¬ŒÊÕÊ Z ∑§Ê ¬„øÊÁŸ∞, Á¡Ÿ∑§Ë ‚ ⁄UøŸÊ•Ê ◊ ∑˝§◊‡Ê— 126. Identify the substances having glycosidic bond and peptide bond, respectively in their structure : Ç‹Êß∑§Ê ‚ÊßÁ«U∑§ ’ œ •ÊÒ⁄U ¬ å≈UÊß«U ’ œ ¬Êÿ ¡Êà „Ò —

1Inulin, insulin (1) ߟÈÁ‹Ÿ, ß ‚ÈÁ‹Ÿ
2Chitin, cholesterol (2) ∑§ÊßÁ≈UŸ, ∑§Ê ‹ S≈U⁄UÊÚ‹
3Glycerol, trypsin (3) ÁÇ‹‚⁄UÊÚ‹, Á≈˛UÁå‚Ÿ
4Cellulose, lecithin (4) ‚ ‹È‹Ê ¡, ‹ Á‚ÁÕŸ
NTA Answer: Option 1(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Cell theory, organelles, biomolecules, enzymes, cell cycle, mitosis/meiosis stages

RecallMedium

Common distractors

enzyme subunit terminology swap

Apoenzyme, holoenzyme, cofactor, coenzyme, and prosthetic group cluster tightly. Students who have not anchored each term structurally reach for the nearest-sounding label (e.g., selecting 'coenzyme' or 'cofactor' when the answer is 'apoenzyme' for the protein part alone).

enzyme inhibition type confusion

When malonate (structural analog of succinate) inhibits succinate dehydrogenase, students with uncertain vocabulary fall for 'feedback inhibition' because they associate inhibition with a product feedback loop, even though malonate is a substrate analog competing for the active site.

meiosis stage event misassignment

The five Prophase I sub-stages have rhyming names and adjacent events. Students confuse recombination nodule appearance (Pachytene, when crossing-over occurs) with synapsis initiation (Zygotene), or place centromere division in Anaphase I instead of Anaphase II.

endomembrane membership error

Mitochondria and chloroplasts are membrane-bound organelles, so students assume they belong to the endomembrane system. The actual criterion (vesicle-mediated ER-Golgi secretory network) is not recalled under exam pressure; peroxisomes are also frequently misclassified.

metabolic pathway compartment swap

Because glycolysis feeds the Krebs cycle (which is in the mitochondrial matrix), students assume glycolysis itself also occurs in the mitochondria. The distinction cytosol=glycolysis, matrix=Krebs, inner membrane=ETS is lost under time pressure.

assertion reason unchecked reason

A/R questions with a true Assertion invite accepting the Reason without independent verification. A plausible-sounding R alongside a true A leads to selecting 'both true, R explains A' (option 2) when R is factually incorrect or logically irrelevant to A — a direct negative-mark risk.

organelle substructure recall error

Match-list questions on organelle sub-structures (axoneme/cilia, cartwheel/centriole, crista/mitochondria, satellite/chromosome) test fine-grained recall. Students who hold only a coarse concept of each organelle confuse structurally related components that share the same cell region.

mitosis meiosis phase conflation

Mitosis and meiosis share named phases but have fundamentally different events. Students conflate Anaphase I (homologous chromosomes separate, centromeres intact) with Anaphase II (centromeres split, sister chromatids separate), or attribute spindle fiber attachment to the wrong division entirely.

prokaryotic eukaryotic ribosome confusion

Students know eukaryotes use 80S ribosomes but forget that organelles of endosymbiotic origin (mitochondria, chloroplasts) retain 70S ribosomes inside eukaryotic cells. They also confuse prokaryotic (70S = 50S + 30S) with eukaryotic (80S = 60S + 40S) subunit composition.

Sources

NCERT refs: Class 11 Biology Chapter 9, p.200

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