Almost all enzymes are proteins; some nucleic acids, ribozymes, behave like enzymes. The substrate fits into the active site, and enzymes bring down the activation energy. Activity is highest at an optimum temperature and pH; low temperature preserves the enzyme in a temporarily inactive state, whereas high temperature destroys enzymatic activity because proteins are denatured by heat. With increasing substrate concentration the velocity rises at first, then reaches a maximum velocity (Vmax) not exceeded by more substrate, because the enzyme molecules are saturated; Figure 9.5 marks Km at Vmax/2. A competitive inhibitor closely resembles the substrate and competes with it for the substrate-binding site, e.g. malonate inhibits succinic dehydrogenase.
-- NCERT Class 11 Biology, Ch. 9, p. 116Enzymes
Enzymes, explained for NEET
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 116). 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.
Can you answer these Enzymes MCQs?
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 is the correct definition of Km in the Michaelis-Menten equation?
Show answer and why every option is right or wrong
Answer: B. Km is defined as the substrate concentration [S] at which v = Vmax/2 (NCERT Class 11 Biology Chapter 9, page 116). It is measured in molarity, not in rate units.
Why A is wrong: A describes Vmax, not Km. Vmax is the maximum velocity achieved at substrate saturation — it is a rate, whereas Km is a concentration (mistake: confusing Km with Vmax).
Why C is wrong: C inverts the definition — it describes a rate at a given concentration. Km is the concentration at a given rate (v = Vmax/2), not the other way around.
Why D is wrong: D refers to enzyme concentration, but the Michaelis-Menten equation relates substrate concentration to velocity. Km concerns [S], not [E].
An enzyme that catalyses the transfer of a functional group from one molecule to another belongs to which IUB class?
Show answer and why every option is right or wrong
Answer: A. Transferases catalyse group-transfer reactions (NCERT Class 11 Biology, Chapter 9). The name directly indicates the function: transfer of a functional group between donor and acceptor molecules.
Why B is wrong: B — Oxidoreductases catalyse oxidation-reduction reactions (transfer of electrons), not transfer of functional groups like amino, phosphate, or methyl groups.
Why C is wrong: C — Hydrolases catalyse hydrolysis (bond cleavage by adding water), not group transfer between two molecules.
Why D is wrong: D — Lyases catalyse elimination reactions that remove groups to form double bonds (or the reverse), without hydrolysis or oxidation. This is not group transfer.
The complete conjugated enzyme with its non-protein component is called:
Show answer and why every option is right or wrong
Answer: B. Holoenzyme = apoenzyme (protein part) + cofactor/coenzyme (non-protein part). The complete, catalytically active form is the holoenzyme (NCERT Class 11 Biology, Chapter 9).
Why A is wrong: A — Apoenzyme is the protein part ALONE, without the cofactor. It is catalytically inactive by itself.
Why C is wrong: C — Coenzyme is the organic non-protein helper molecule (e.g., NAD⁺). It is one component of the holoenzyme, not the complete enzyme.
Why D is wrong: D — A prosthetic group is a tightly bound non-protein component (e.g., haem in haemoglobin). It is part of the holoenzyme, not the whole enzyme.
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?
Show answer and why every option is right or wrong
Answer: D. A competitive inhibitor competes with substrate for the active site, raising the apparent Km (more substrate needed to reach half Vmax). However, at sufficiently high [S], substrate outcompetes the inhibitor, so Vmax is eventually reached unchanged (NCERT Class 11 Biology, Chapter 9).
Why A is wrong: A — This describes non-competitive inhibition characteristics mixed with competitive. A competitive inhibitor does not lower Vmax; excess substrate can overcome it.
Why B is wrong: B — If both were unchanged, the inhibitor would have no effect. A competitive inhibitor demonstrably increases the apparent Km.
Why C is wrong: C — Km unchanged with Vmax decreased describes non-competitive inhibition, where the inhibitor binds a site other than the active site and reduces catalytic efficiency regardless of [S].
An enzyme has a Km of 2 mM. If the substrate concentration is 2 mM, what fraction of Vmax is the reaction velocity?
Show answer and why every option is right or wrong
Answer: C. By the Michaelis-Menten equation, when [S] = Km: v = Vmax × Km / (Km + Km) = Vmax / 2. This is the defining relationship of Km (NCERT Class 11 Biology Chapter 9, page 116).
Why A is wrong: A — This would require [S] = Km/3. At [S] = Km, the equation yields exactly Vmax/2, not Vmax/4.
Why B is wrong: B — Three-quarters of Vmax occurs when [S] = 3 × Km. At [S] = Km, substitution gives v = Vmax/2.
Why D is wrong: D — v = Vmax only at infinitely high [S]. At [S] = Km, the denominator is 2Km, giving exactly half Vmax.
Enzyme X has Km = 0.5 mM for substrate P. Enzyme Y has Km = 5 mM for the same substrate. Which statement is correct?
Show answer and why every option is right or wrong
Answer: D. Lower Km means the enzyme reaches half its maximum velocity at a lower substrate concentration — it binds substrate more effectively. Enzyme X (Km = 0.5 mM) therefore has higher affinity than Enzyme Y (Km = 5 mM) (NCERT Class 11 Biology, Chapter 9).
Why A is wrong: A — This reverses the Km–affinity relationship. A higher Km means the enzyme needs MORE substrate to reach half Vmax, indicating LOWER affinity, not higher.
Why B is wrong: B — Km is the standard measure for comparing enzyme-substrate affinity. Lower Km = higher affinity is a foundational enzymology concept.
Why C is wrong: C — The Km values differ by 10-fold, so affinity is clearly different. Equal affinity would require equal Km.
A student claims that Km has units of mol·L⁻¹·s⁻¹. Why is this claim incorrect?
Show answer and why every option is right or wrong
Answer: C. Km represents the substrate concentration at which v = Vmax/2. As a concentration, its SI-compatible unit is mol·L⁻¹ (M), not a rate unit. The student confused Km (a concentration) with a rate or rate constant (NCERT Class 11 Biology Chapter 9, page 116).
Why A is wrong: A — Km is NOT dimensionless. It represents a substrate concentration and carries concentration units (M). Dimensionless constants have no units by definition.
Why B is wrong: B — s⁻¹ is the unit of a first-order rate constant (like kcat). Km is not a rate constant; it is a concentration threshold derived from the steady-state treatment.
Why D is wrong: D — Joules per mole is the unit of energy (e.g., activation energy). Km measures substrate concentration, not energy.
For an enzyme following Michaelis-Menten kinetics, Vmax = 100 µmol/min and Km = 4 mM. What is the initial velocity when [S] = 12 mM?
Show answer and why every option is right or wrong
Answer: A. v = Vmax × [S] / (Km + [S]) = 100 × 12 / (4 + 12) = 1200 / 16 = 75 µmol/min (NCERT Class 11 Biology Chapter 9, page 116; Michaelis-Menten equation applied).
Why B is wrong: B — 50 µmol/min occurs only when [S] = Km = 4 mM (half Vmax). Here [S] = 12 mM, which is three times Km, giving a higher velocity.
Why C is wrong: C — 25 µmol/min would result from [S] = Km/3 ≈ 1.33 mM. With [S] = 12 mM, the numerator is 1200 and denominator is 16, giving 75.
Why D is wrong: D — v = Vmax (100 µmol/min) requires [S] → ∞. At [S] = 12 mM (3 × Km), the enzyme is not fully saturated; v = 75 µmol/min.
Free NEET study resources
Get a structured 30-day study plan and a complete formula booklet — delivered to your inbox instantly.
Enzymes: quick recall before you leave
How do you solve a Enzymes question? A worked example
- 1
Given
An enzyme has Vmax = 200 µmol/min and Km = 5 mM. The substrate concentration is 15 mM.
- 2
Required
Find the initial reaction velocity v.
- 3
Concept
The Michaelis-Menten equation relates substrate concentration to initial velocity for a single-substrate enzyme with no cooperative or allosteric effects.
- 4
Formula
v = Vmax × [S] / (Km + [S])
- 5
Substitution
v = 200 × 15 / (5 + 15)
- 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
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
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
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.
---
What to remember before solving Enzymes questions
Which Enzymes formulas do you need for NEET?
Michaelis-Menten equation
Initial reaction velocity v of an enzyme as a function of substrate concentration. Km = [S] at v = Vmax/2.
| Symbol | Quantity | SI Unit |
|---|---|---|
| v | rate | - |
| Vmax | max rate | - |
| Km | Michaelis constant | M |
| [S] | substrate | M |
Valid when
- Single substrate
- No allosteric/cooperative effects
Where do students lose marks on Enzymes?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Root cause: term confusion
Correction
Km is [S] at v = Vmax/2 — measured in M, not /s. Lower Km = higher affinity.
More in Cell: Structure and Function: 13 exam traps and mistakes · 1 question pattern from its other lessons.
Enzymes questions from past NEET papers
11 questions from NEET 2021, 2023, 2024, 2025. Answers verified against NTA official keys.
The protein portion of an enzyme is called:
Which one of the following enzymes contains ‘Haem’ as the prosthetic group?
The cofactor of the enzyme carboxypeptidase is :
Inhibition of Succinic dehydrogenase enzyme by malonate is a classical example of:
Melonate inhibits the growth of pathogenic bacteria by inhibiting the activity of
All 69 past-paper questions from Cell: Structure and Function →
Sources
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.
Test yourself on this topic with real past-paper questions:
Practice this topic →