BiologyISC Class 11 · Biomolecules

Enzymes: Active Site, Classification, Cofactors, Factors and Competitive Inhibition

Updated 17 October 2026 · Fact-checked

Enzymes are biological catalysts, mostly proteins, that lower the activation energy of reactions. Their activity depends on temperature, pH and substrate concentration and can be reduced by inhibitors. Enzymes are classified into six classes: oxidoreductases, transferases, hydrolases, lyases, isomerases and ligases.

Notice that the curve levels off at Vmax when all active sites are full, and Km is the substrate concentration that gives half of Vmax.

Illustration drawn for QuestionBuddy.

Understand Enzymes: Properties, Classification and Kinetics

Nature. Enzymes are proteins (with the exception of ribozymes, catalytic RNA) with a specific three-dimensional structure and an active site into which the substrate fits. They lower the activation energy, are specific, are needed in small amounts and are not used up in the reaction. Reaction: E + S ⇌ ES → E + P.

Cofactors. Many enzymes need non-protein components: a *prosthetic group* is tightly bound (the haem in catalase and peroxidase), a *coenzyme* is loosely bound organic cofactor (NAD, NADP, which contain the vitamin niacin) and *metal ions* (zinc in carboxypeptidase). The protein part alone is the apoenzyme; with the cofactor it is the holoenzyme.

Classification. (1) Oxidoreductases: oxidation-reduction (dehydrogenases). (2) Transferases: transfer a group between substrates. (3) Hydrolases: hydrolysis by adding water. (4) Lyases: remove groups without hydrolysis, forming double bonds. (5) Isomerases: interconvert isomers. (6) Ligases: join two compounds with ATP use.

Factors. *Temperature:* activity rises up to an optimum and then falls because the enzyme is denatured at high temperature; low temperature only makes it inactive temporarily. *pH:* each enzyme has an optimum pH (pepsin about pH 2). *Substrate concentration:* the velocity rises and reaches a maximum (Vmax) when all active sites are occupied. The Michaelis constant Km is the substrate concentration at which v = Vmax/2; v = Vmax[S]/(Km + [S]).

Inhibition. In competitive inhibition, the inhibitor resembles the substrate and competes for the active site (malonate inhibits succinic dehydrogenase; sulpha drugs). It can be overcome by increasing the substrate concentration, and Vmax stays the same while Km appears to increase. Feedback inhibition: the end product inhibits an early enzyme in the pathway.

Enzymes: Properties, Classification and Kinetics: key facts

Enzyme reaction: $E + S ⇌ E S \to E + P$; ES = enzyme-substrate complex.
Michaelis–Menten: $v = V ma x [S] / (K m + [S])$; At [S] = Km, v = Vmax/2.
Classes: $O x i d or e d u c t a ses ∣ T r an s f er a ses ∣ H y d r o l a ses ∣ L y a ses ∣ I so m er a ses ∣ L i g a ses$; The six EC classes.
Cofactors: Prosthetic group bound tightly
Coenzyme NAD, NADP
Metal ion Zn²⁺; Apoenzyme + cofactor = holoenzyme.

Prosthetic group	bound tightly
Coenzyme	NAD, NADP
Metal ion	Zn²⁺

Enzymes: Properties, Classification and Kinetics in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

A drug that blocks a bacterial enzyme

Suppose a sulpha drug has a structure similar to the substrate of a bacterial enzyme and blocks it.

Explain what kind of inhibition this is and how it could be reduced.

Try it first, then show the solutionSolution

The drug resembles the substrate and binds to the active site: competitive inhibition.
It reduces the amount of enzyme available to the real substrate.
Increasing the concentration of the substrate overcomes the effect because the substrate competes more successfully.

Answer: Competitive inhibition; overcome by raising substrate concentration.

Memory hook: A similar shape lets the inhibitor occupy the active site.

Tricks to remember Enzymes: Properties, Classification and Kinetics

Short memory aids. Say each one aloud once, then use it on a question the same day.

Optimum, not maximum
Enzymes have an optimum temperature and pH, above which they are denatured.
Use it when: Graph questions.
OTHLIL
Oxidoreductases, Transferases, Hydrolases, Lyases, Isomerases, Ligases.
Use it when: Remembering the six classes.

Enzymes: Properties, Classification and Kinetics: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Optimum pH · 1 mark · MCQBoard level

Pepsin works best at:
(a) acidic pH
(b) neutral pH
(c) alkaline pH
(d) any pH

Try it first, then show the solutionSolution

Pepsin acts in the acidic gastric juice (about pH 2).

Answer: (a) acidic pH

Example 2: Properties of enzymes · 3 marks · Short answerBoard level

List four properties of enzymes and explain what is meant by the active site.

Try it first, then show the solutionSolution

Enzymes are mostly proteins, specific, work in small amounts and are not used up. (1.5 marks)
They lower the activation energy of the reaction. (0.5 mark)
The active site is the region of the enzyme with a specific shape into which the substrate binds to form the ES complex. (1 mark)

Answer: Protein, specific, reusable, lower activation energy; active site binds substrate.

Example 3: Classification · 4 marks · Short answerHarder board level

Name the class of enzyme for each reaction: (a) glucose + ATP → glucose-6-phosphate + ADP (transfer of a phosphate), (b) hydrolysis of a peptide bond, (c) conversion of glucose-6-phosphate to fructose-6-phosphate, (d) joining two molecules using ATP.

Try it first, then show the solutionSolution

(a) Transferase. (1 mark)
(b) Hydrolase. (1 mark)
(c) Isomerase. (1 mark)
(d) Ligase. (1 mark)

Answer: Transferase, hydrolase, isomerase, ligase.

Example 4: Michaelis–Menten kinetics · 4 marks · Short answerHarder board level

An enzyme has Vmax = 100 µmol min⁻¹ and Km = 2 mM. Find the rate at [S] = 2 mM and at [S] = 18 mM, and the [S] needed for 80% of Vmax.

Try it first, then show the solutionSolution

At [S] = Km, v = Vmax/2 = 50 µmol min⁻¹. (1 mark)
At [S] = 18 mM: v = 100 × 18/(2 + 18) = 90 µmol min⁻¹. (1.5 marks)
For 0.8 Vmax: 0.8 = [S]/(2 + [S]) → [S] = 0.8 × 2/0.2 = 8 mM. (1.5 marks)

Answer: 50; 90 µmol min⁻¹; 8 mM.

Example 5: Competitive inhibition · 5 marks · Long answerHOTS / competency

Succinic dehydrogenase is inhibited by malonate. (i) Name the type of inhibition and explain it. (ii) State the effect on Vmax and the apparent Km. (iii) Explain why increasing substrate concentration reduces the inhibition. (iv) Contrast it with the effect of high temperature.

Try it first, then show the solutionSolution

Competitive inhibition: malonate resembles succinate and competes for the active site. (1 mark)
Vmax is unchanged; the apparent Km increases (more substrate is needed for half Vmax). (1.5 marks)
A high substrate concentration outcompetes the inhibitor at the active site. (1 mark)
High temperature denatures the enzyme irreversibly, changing the active site shape; inhibition is reversible. (1.5 marks)

Answer: Competitive; Vmax same and Km higher; excess substrate relieves it; denaturation is permanent.

Where students lose marks in Enzymes: Properties, Classification and Kinetics

The slips examiners see most often, and what to write instead.

Mistake: Saying enzymes are used up
Writing that enzymes are consumed in the reaction.
Do this: Enzymes are not changed at the end of the reaction.
Usually costs: 1 mark.
Mistake: Confusing denaturation with inactivation
Saying cold denatures enzymes.
Do this: Cold slows enzymes temporarily; heat denatures them.
Usually costs: 1 mark.
Mistake: Wrong effect on Vmax
Saying a competitive inhibitor lowers Vmax.
Do this: Vmax is unchanged; Km increases.
Usually costs: 1 to 2 marks.

Exam tips

Draw the graphs of rate against temperature, pH and substrate concentration.
Name the six classes with one example reaction each.
Substitute carefully in the Michaelis–Menten equation.
Distinguish between competitive and non-competitive inhibition.

Practice questions on Enzymes: Properties, Classification and Kinetics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The non-protein, loosely bound organic part of an enzyme is the:
(a) prosthetic group
(b) coenzyme
(c) apoenzyme
(d) active site
Hint
NAD is an example.
Answer
(b) coenzyme
Q2 · 2 marks · Short answerBoard level
Why does an enzyme not work at a very high temperature?
Hint
Think about the protein structure.
Answer
High temperature denatures the protein, destroying the shape of the active site, so the substrate no longer fits.
Q3 · 3 marks · Short answerHarder board level
Differentiate between coenzyme and prosthetic group, with an example each.
Hint
Compare the binding strength.
Answer
A coenzyme is a loosely bound organic cofactor that can be separated by dialysis (NAD). A prosthetic group is tightly bound to the enzyme (haem in catalase).
Q4 · 3 marks · Short answerHarder board level
What is the effect of increasing substrate concentration on the velocity of an enzyme reaction? Explain the saturation.
Hint
Active sites become fully occupied.
Answer
The velocity rises hyperbolically with [S] at first and then levels off at Vmax, because all the active sites are occupied and extra substrate cannot increase the rate.
Q5 · 3 marks · Assertion-reasonHarder board level
Assertion (A): Enzymes lower the activation energy. Reason (R): They change the equilibrium of the reaction. Choose: (a) both true and R explains A (b) both true, R does not explain A (c) A true, R false (d) A false, R true.
Hint
Catalysts do not shift equilibrium.
Answer
(c) A is true, R is false: enzymes speed up the approach to equilibrium but do not change its position.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a student studies an enzyme at pH 7 and measures the rate at three temperatures: 10 °C (slow), 37 °C (maximum) and 70 °C (almost none).
(i) Explain the low rate at 10 °C. (1 mark)
(ii) State the term for 37 °C. (1 mark)
(iii) Explain the loss at 70 °C. (1 mark)
(iv) State whether cooling back to 37 °C would restore activity from 70 °C. (1 mark)
Hint
Low temperature is reversible; high is not.
Answer
(i) Low kinetic energy gives fewer collisions between enzyme and substrate; the enzyme is not denatured. (ii) The optimum temperature. (iii) Denaturation of the protein. (iv) No, denaturation is generally irreversible.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Optimum pH of pepsin?Show answer
About 2.
2.Enzyme class that joins molecules?Show answer
Ligases.
3.Km is the [S] at?Show answer
The substrate concentration at which the velocity is half of Vmax (v = Vmax/2).
4.Example of competitive inhibition?Show answer
Malonate on succinic dehydrogenase.

Enzymes: Properties, Classification and Kinetics: frequently asked questions

What is a ribozyme?

An RNA molecule with catalytic activity, showing that not all biological catalysts are proteins.

Why are enzymes specific?

The shape of the active site fits only particular substrates, like a lock and key (or induced fit).

What is feedback inhibition?

The end product of a metabolic pathway inhibits an early enzyme, preventing over-production.