IB Biology HL Enzymes & Cell Biology — C1.1, A1.1, D1.1

Complete Study Guide

Topics Covered

1. Enzymes — Structure & Function (C1.1)
2. Cell Theory & Ultrastructure (A1.1 / A1.2)
3. Membrane Structure & Transport (A1.2)
4. Cell Division (D1.1)
5. HL Extension — Enzyme Kinetics & Metabolic Pathways (HL)
6. Exam Strategy & Common Mistakes
7. Integration — Linking Enzymes to Metabolism
8. Mixed Practice — Exam Style

Videos on this page: Enzymes Overview

Aligned to IB Biology HL 2025 syllabus — C1.1, A1.1, A1.2, D1.1

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Section 1: Enzymes — Structure & Function (C1.1)

Enzymes are biological catalysts — globular proteins that speed up metabolic reactions by lowering the activation energy ($E_a$). Without enzymes, essentially no metabolic reaction would proceed fast enough to sustain life at body temperature. Every enzyme is specific to one reaction or a small family of related reactions.

1.1 How Enzymes Work

Enzymes work by binding their substrate at the active site, forming a temporary enzyme-substrate complex (ES complex). The active site provides an environment that lowers $E_a$ — by straining bonds, providing an optimal microenvironment, or positioning substrates correctly for the reaction.

$E + S \rightleftharpoons ES \rightarrow E + P$

The enzyme is unchanged after the reaction and can be reused.

1.2 Lock-and-Key vs Induced Fit

Two models explain enzyme-substrate interaction:

Lock-and-Key Model (Fischer, 1894):

• The active site is a rigid, pre-formed shape exactly complementary to the substrate
• Like a key fitting into a lock
• Explains specificity but is oversimplified

Induced-Fit Model (Koshland, 1958):

• The active site is flexible and moulds around the substrate upon binding
• Both enzyme AND substrate change conformation slightly
• More accurate — explains how some enzymes act on a range of similar substrates

1.3 Factors Affecting Enzyme Activity

Four factors control the rate of enzyme-catalysed reactions: temperature, pH, substrate concentration, and enzyme concentration.

Temperature

• As temperature increases, kinetic energy of molecules increases
• More frequent enzyme-substrate collisions → higher rate
• Optimum temperature: the temperature at which rate is maximum (typically ~37 C for human enzymes)
• Beyond the optimum: hydrogen bonds and weak interactions break → denaturation
• Active site loses its specific shape → substrate can no longer bind → rate drops to zero

pH

• Each enzyme has an optimum pH (e.g. pepsin: pH 2, trypsin: pH 8)
• Extreme pH values change the ionisation of R-groups in the active site
• Disrupts ionic bonds and hydrogen bonds → changes active site shape → denaturation
• pH changes can be reversible if small, or irreversible (denaturation) if extreme

Substrate Concentration

• At low $[\text{S}]$: rate increases proportionally — plenty of free active sites
• At high $[\text{S}]$: rate plateaus — all active sites are occupied (saturation)
• Maximum rate at saturation is called $V_{\text{max}}$
• The substrate concentration at which rate = $\frac{1}{2}V_{\text{max}}$ is the Michaelis constant ($K_m$)

1.4 Enzyme Inhibition HL

AHL — C1.1

Cells need ways to slow down or stop enzyme activity — otherwise reactions would run unchecked. Inhibitors are molecules that do this by interfering with the enzyme’s ability to bind its substrate. The critical distinction is whether the inhibitor blocks the active site directly (competitive) or attacks from a different location and changes the enzyme’s shape (non-competitive), because this determines whether you can overcome the inhibition by adding more substrate.

Inhibitors are molecules that reduce the rate of enzyme-catalysed reactions. Two main types:

Competitive Inhibition:

• Inhibitor has a shape similar to the substrate
• Binds directly to the active site — blocks the substrate from entering
• Can be overcome by increasing substrate concentration (substrates outcompete the inhibitor)
• $V_{\text{max}}$ is unchanged (at very high $[\text{S}]$, all active sites are occupied by substrate)
• $K_m$ increases (need more substrate to reach $\frac{1}{2}V_{\text{max}}$)

Non-competitive Inhibition:

• Inhibitor binds to a site other than the active site (allosteric site)
• Causes a conformational change in the enzyme that distorts the active site
• Substrate can still bind, but the reaction rate is reduced
• Cannot be overcome by increasing $[\text{S}]$
• $V_{\text{max}}$ decreases (fewer functional enzyme molecules at any given time)
• $K_m$ is unchanged (affinity for substrate is not affected)

1.5 Allosteric Regulation HL

AHL — C1.1

Some enzymes have a second binding site — separate from the active site — that acts like a volume knob for the whole pathway. When a small molecule binds there, it either speeds up or slows down the enzyme by changing its shape. This is how cells fine-tune their metabolism in real time without switching genes on or off.

Allosteric enzymes have regulatory sites separate from the active site. Binding of an effector molecule changes the enzyme’s conformation:

• Allosteric activators stabilise the active conformation → increase activity
• Allosteric inhibitors stabilise the inactive conformation → decrease activity

End-product inhibition (negative feedback): The final product of a metabolic pathway inhibits the first enzyme in that pathway. This prevents overproduction and conserves resources.

$A \xrightarrow{E_1} B \xrightarrow{E_2} C \xrightarrow{E_3} D$

If $D$ accumulates, it inhibits $E_1$ (allosteric inhibition). When $D$ is used up, inhibition is released and the pathway resumes. This is a self-regulating system.

MCQ Practice

An enzyme has a $K_m$ of 2.5 mM. A competitive inhibitor is added. What is the most likely new $K_m$?

A. 1.0 mM

B. 2.5 mM

C. 5.0 mM ← CORRECT

D. The $K_m$ cannot be determined without knowing $V_{\text{max}}$

Why: Competitive inhibitors increase $K_m$ because more substrate is needed to outcompete the inhibitor and reach $\frac{1}{2}V_{\text{max}}$. The $V_{\text{max}}$ itself is unchanged (at infinite $[\text{S}]$, the substrate always wins). Option A describes increased affinity (wrong direction). Option B would mean no effect (that is the non-competitive pattern for $K_m$). Option D is a distractor — $K_m$ is independent of $V_{\text{max}}$.

MCQ Practice

A non-competitive inhibitor is added to an enzyme. Which graph correctly shows the effect?

A. $V_{\text{max}}$ increases, $K_m$ unchanged

B. $V_{\text{max}}$ unchanged, $K_m$ increases

C. $V_{\text{max}}$ decreases, $K_m$ unchanged ← CORRECT

D. Both $V_{\text{max}}$ and $K_m$ decrease

Why: Non-competitive inhibitors bind to an allosteric site and reduce the number of functional enzyme molecules, lowering $V_{\text{max}}$. Because the inhibitor does not compete with the substrate for the active site, $K_m$ (a measure of substrate affinity) is unaffected. Option B describes competitive inhibition. Option A is physiologically nonsensical for an inhibitor.

Quick Recall — Section 1

Try to answer without scrolling up:

1. What is the difference between competitive and non-competitive inhibition?
2. Name the four factors that affect enzyme activity.
3. What does “denaturation” mean for an enzyme?

Reveal answers

1. Competitive inhibitors bind the active site and can be overcome by increasing substrate concentration. Non-competitive inhibitors bind an allosteric site and reduce Vmax regardless of substrate concentration.
2. Temperature, pH, substrate concentration, enzyme concentration.
3. Permanent loss of the tertiary structure (and therefore function) of the enzyme, caused by disruption of hydrogen bonds and ionic bonds.

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Watch: Enzymes — Structure, Function & Inhibition

Amoeba Sisters · 9 min · Enzyme structure, active site, induced fit, cofactors, pH and temperature effects, and competitive inhibition

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Section 2: Cell Theory & Ultrastructure (A1.1 / A1.2)

2.1 Cell Theory

Cell theory has three main principles:

1. All living organisms are composed of cells (or cell products)
2. Cells are the smallest unit of life — they can carry out all functions of life
3. All cells come from pre-existing cells (no spontaneous generation)

2.2 Prokaryotic vs Eukaryotic Cells

2.3 Eukaryotic Cell Ultrastructure HL

Ultrastructure refers to the internal architecture of a cell as revealed by an electron microscope — the organelles, membranes, and structures too small to see with light. Knowing each organelle’s structure helps you understand its function, because the two are always matched: a large surface area for reactions, small volumes to build gradients, and compartments to keep incompatible processes separate.

2.4 Endosymbiotic Theory HL

AHL — A1.2

Mitochondria and chloroplasts are thought to have evolved from free-living prokaryotes that were engulfed by ancestral eukaryotic cells. Evidence:

• Double membrane — inner membrane from the prokaryote, outer from the host’s vesicle
• Own DNA — circular, like bacterial DNA
• Own ribosomes — 70S (prokaryotic size), not 80S
• Binary fission — divide independently of the host cell
• Size — similar to bacteria (~1-10 $\mu\text{m}$)

MCQ Practice

Which of the following is evidence for the endosymbiotic origin of mitochondria?

A. Mitochondria have 80S ribosomes like the cytoplasm

B. Mitochondria have circular DNA and 70S ribosomes ← CORRECT

C. Mitochondria are surrounded by a single membrane

D. Mitochondria cannot replicate independently

Why: Mitochondria have circular DNA (like bacteria), 70S ribosomes (like bacteria), and divide by binary fission (like bacteria). They have a DOUBLE membrane (not single). They CAN replicate semi-independently. 80S ribosomes are found in the eukaryotic cytoplasm, not inside mitochondria.

MCQ Practice

A student observes a cell under an electron microscope. The cell has no nucleus but contains ribosomes and a cell wall made of peptidoglycan. Which organism does this cell most likely belong to?

A. A plant

B. A fungus

C. A bacterium ← CORRECT

D. An animal

Why: No nucleus = prokaryote. Peptidoglycan cell wall = bacterium (not archaea, which have different cell wall composition). Plants have cellulose walls, fungi have chitin walls, and animals have no cell wall. The presence of ribosomes is universal and does not distinguish cell types.

Quick Recall — Section 2

Try to answer without scrolling up:

1. State the three principles of cell theory.
2. Name two organelles that have their own DNA.
3. What type of cell wall do bacteria have?

Reveal answers

1. (i) All living organisms are composed of cells, (ii) cells are the smallest unit of life, (iii) all cells come from pre-existing cells.
2. Mitochondria and chloroplasts (evidence for the endosymbiotic theory).
3. Peptidoglycan.

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Section 3: Membrane Structure & Transport (A1.2)

The cell membrane is not just a passive barrier — it is a dynamic, selective gateway that controls everything entering and leaving the cell. Understanding its structure explains why some substances cross freely while others need help, and why cells can maintain internal conditions completely different from the surrounding environment.

3.1 The Fluid Mosaic Model

The cell membrane is described by the fluid mosaic model (Singer-Nicolson, 1972):

• Fluid — phospholipids and proteins can move laterally within the bilayer
• Mosaic — proteins are scattered throughout, like tiles in a mosaic

Components:

• Phospholipid bilayer — hydrophilic heads face outward, hydrophobic tails face inward
• Integral proteins — span the entire membrane (transmembrane); involved in transport
• Peripheral proteins — attached to one surface; often enzymes or structural
• Cholesterol — embedded between phospholipids; regulates fluidity
• Glycoproteins — cell recognition and signalling
• Glycolipids — cell recognition

3.2 Transport Across Membranes

Cells need to move substances in and out constantly — nutrients in, waste out, signals received and sent. Some substances can slip through the membrane on their own; others need protein channels or pumps. Whether a cell spends ATP on this depends on whether the substance is moving with or against its concentration gradient.

Type	Direction	Energy?	Examples
Simple diffusion	High → low concentration	No (passive)	$\text{O}_2$, $\text{CO}_2$, steroid hormones
Facilitated diffusion	High → low concentration	No (passive)	Glucose (via channel/carrier proteins)
Osmosis	High water potential → low water potential	No (passive)	Water across membranes
Active transport	Low → high concentration	Yes (ATP)	$\text{Na}^+$/$\text{K}^+$ pump, mineral uptake in roots
Endocytosis	Into cell	Yes (ATP)	Phagocytosis, pinocytosis
Exocytosis	Out of cell	Yes (ATP)	Secretion of hormones, neurotransmitters

MCQ Practice

Which statement correctly describes osmosis?

A. The movement of solute molecules from high to low concentration

B. The active transport of water using ATP

C. The net movement of water from a region of higher water potential to lower water potential through a partially permeable membrane ← CORRECT

D. The diffusion of water from low solute concentration to high solute concentration without a membrane

Why: Osmosis specifically refers to water movement through a partially permeable membrane, driven by water potential differences. It is passive (no ATP). Option A describes diffusion of solutes. Option D removes the membrane requirement, which is essential to the definition of osmosis.

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Section 4: Cell Division (D1.1)

4.1 The Cell Cycle

The cell cycle consists of interphase and mitotic phase (M phase):

Interphase (longest phase, ~90% of cycle time):

• $G_1$ phase — cell grows, organelles replicate, proteins synthesised
• $S$ phase — DNA replication (each chromosome → two sister chromatids joined at centromere)
• $G_2$ phase — cell prepares for division; centrioles replicate; final checks

M phase:

• Mitosis — nuclear division (PMAT: Prophase, Metaphase, Anaphase, Telophase)
• Cytokinesis — cytoplasmic division

4.2 Mitosis — Stages

4.3 Mitosis vs Meiosis HL

Feature	Mitosis	Meiosis
Divisions	1	2 (meiosis I + meiosis II)
Daughter cells	2 identical	4 genetically different
Chromosome number	Same as parent (diploid → diploid)	Halved (diploid → haploid)
Crossing over	No	Yes (prophase I)
Independent assortment	No	Yes (metaphase I)
Purpose	Growth, repair	Production of gametes
Where	Somatic cells	Gonads (ovaries, testes)
Genetic variation	None	Significant (crossing over + independent assortment)

MCQ Practice

A cell with 8 chromosomes undergoes mitosis. How many chromosomes will be in each daughter cell?

A. 4

B. 8 ← CORRECT

C. 16

D. 2

Why: Mitosis produces daughter cells with the SAME number of chromosomes as the parent cell. The DNA is replicated in S phase, sister chromatids are separated, and each daughter cell receives a complete set. Meiosis would halve the number (to 4), but the question specifies mitosis.

MCQ Practice

During which stage of mitosis do the centromeres split?

A. Prophase

B. Metaphase

C. Anaphase ← CORRECT

D. Telophase

Why: Centromeres split during anaphase, allowing sister chromatids to be pulled to opposite poles by shortening spindle fibres. In prophase, chromosomes condense. In metaphase, they align. In telophase, nuclear envelopes reform.

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Section 5: HL Extension — Enzyme Kinetics & Metabolic Pathways HL

5.1 Metabolic Pathways

Metabolism is a web of interconnected enzyme-catalysed reactions. Pathways can be:

• Linear: $A \rightarrow B \rightarrow C \rightarrow D$
• Branched: Product $C$ can feed into two different pathways
• Cyclic: Like the Krebs cycle — the starting molecule is regenerated

Each step in a pathway is catalysed by a specific enzyme. The product of one reaction becomes the substrate for the next. This is called metabolic channelling.

5.2 Cofactors and Coenzymes

• Cofactor: Non-protein molecule required for an enzyme to function. Can be inorganic ions ($\text{Mg}^{2+}$, $\text{Fe}^{2+}$, $\text{Zn}^{2+}$) or organic molecules (coenzymes)
• Coenzyme: Organic cofactor, often derived from vitamins. Examples:
  • $\text{NAD}^+$ (from vitamin B3 / niacin) — electron carrier in respiration
  • $\text{FAD}$ (from vitamin B2 / riboflavin) — electron carrier in Krebs cycle
  • Coenzyme A (from vitamin B5) — carries acetyl groups

5.3 Immobilised Enzymes HL

AHL — C1.1

Enzymes can be fixed (immobilised) to an inert support material. Methods include:

• Gel entrapment — enzyme trapped in alginate or polyacrylamide beads
• Adsorption — enzyme bound to surface of glass beads or activated carbon
• Covalent bonding — enzyme chemically bonded to a support

Advantages of immobilisation:

• Enzyme can be reused (reduced cost)
• Product is not contaminated with enzyme
• Enzyme is more thermostable (restricted movement reduces denaturation)
• Continuous production in column reactors

Disadvantages:

• Immobilisation can reduce activity (active site may be blocked or distorted)
• Diffusion of substrate to active site may be slower

MCQ Practice

Which of the following is an advantage of using immobilised enzymes in industrial processes?

A. The enzyme works faster because it is trapped in a gel

B. The products are always purer because no reaction occurs

C. The enzyme can be recovered and reused, reducing costs ← CORRECT

D. Immobilised enzymes never denature at any temperature

Why: The primary industrial advantage of immobilisation is enzyme reuse. The enzyme is fixed in place, so it can be separated from the product easily. Option A is false — immobilisation can actually slow the reaction due to diffusion limitations. Option D is false — immobilised enzymes are more thermostable but will still denature at extreme temperatures.

MCQ Practice

Which row correctly matches the cofactor with its enzyme or process?

	Cofactor	Process
A	$\text{Zn}^{2+}$	Photolysis of water
B	$\text{NAD}^+$	Oxidation reactions in respiration
C	$\text{Ca}^{2+}$	Calvin cycle
D	$\text{Mg}^{2+}$	Electron transport chain

B is CORRECT ← CORRECT

Why: $\text{NAD}^+$ is the primary electron carrier in cell respiration, accepting hydrogen atoms during oxidation reactions (glycolysis, link reaction, Krebs cycle). $\text{Zn}^{2+}$ is a cofactor for carbonic anhydrase, not photolysis. $\text{Ca}^{2+}$ is involved in muscle contraction and signalling, not the Calvin cycle. $\text{Mg}^{2+}$ is actually a cofactor for RuBisCO and is part of chlorophyll, not the ETC.

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Section 6: Exam Strategy & Common Mistakes

Top 10 Enzyme & Cell Biology Mistakes

1. Saying enzymes “provide energy” — they lower $E_a$, not provide energy
2. Confusing denaturation with inactivation at low temperature
3. Saying non-competitive inhibitors bind to the active site
4. Forgetting that $K_m$ INCREASES with competitive inhibition
5. Saying mitochondria have 80S ribosomes — they have 70S
6. Confusing “cell wall” and “cell membrane” — plants have BOTH
7. Saying mitosis produces haploid cells — that is meiosis
8. Forgetting that DNA replication occurs in S phase, NOT in mitosis
9. Saying osmosis is “movement of water from low to high concentration” without specifying water potential and membrane
10. Saying crossing over occurs in mitosis — it occurs only in meiosis (prophase I)

MCQ Practice

Which statement about enzymes is correct?

A. Enzymes increase the activation energy of a reaction

B. Enzymes change the equilibrium position of a reaction

C. Enzymes are specific to their substrates due to the shape of the active site ← CORRECT

D. Enzymes are consumed during the reaction and must be resynthesised

Why: Enzymes are specific because the active site is complementary in shape (and charge) to the substrate. Option A is backwards — enzymes DECREASE $E_a$. Option B is false — enzymes speed up the rate of reaching equilibrium but do not change the equilibrium position itself. Option D is false — enzymes are catalysts and are recycled (not consumed).

MCQ Practice

A cell is placed in a hypertonic solution. Which of the following will occur?

A. The cell will swell and may lyse

B. Water will enter the cell by osmosis

C. Water will leave the cell by osmosis, and the cell will shrink ← CORRECT

D. There will be no net movement of water

Why: In a hypertonic solution, the solute concentration outside is higher than inside (water potential is lower outside). Water moves from higher water potential (inside the cell) to lower water potential (outside) by osmosis. The cell loses water and shrinks (crenation in animal cells, plasmolysis in plant cells). Option A describes what happens in a hypotonic solution.

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Section 7: Integration — Linking Enzymes to Metabolism

Why Enzymes Matter for Everything

Every metabolic pathway you study in IB Biology depends on enzymes:

Pathway	Key enzyme(s)	Topic
Glycolysis	Hexokinase, phosphofructokinase	C1.2
Krebs cycle	Isocitrate dehydrogenase	C1.2
Photosynthesis	RuBisCO, ATP synthase	C1.3
DNA replication	DNA helicase, DNA polymerase, ligase	D1.2
Transcription	RNA polymerase	D1.2
Translation	Peptidyl transferase (ribosomal)	D1.2
Digestion	Amylase, pepsin, lipase	D1.1

MCQ Practice

Phosphofructokinase (PFK) is inhibited by ATP and activated by AMP. This is an example of:

A. Competitive inhibition by ATP

B. Denaturation of PFK by ATP

C. Allosteric regulation by end-product inhibition ← CORRECT

D. Non-competitive inhibition that permanently inactivates PFK

Why: PFK is an allosteric enzyme. ATP (the end product of respiration) inhibits PFK by binding to an allosteric site, slowing glycolysis when energy is plentiful. AMP (indicating low energy) activates PFK. This is a classic example of end-product (feedback) inhibition — reversible, not permanent. It is NOT competitive because ATP does not bind to the active site (the active site binds fructose-6-phosphate).

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Mixed Practice — Exam Style

1. [Enzyme Kinetics] At very high substrate concentrations, the rate of an enzyme-catalysed reaction plateaus. The best explanation is:

   A. The enzyme is denatured by excess substrate

   B. All active sites are occupied (enzymes are saturated), so adding more substrate has no effect on rate

   C. The substrate begins to inhibit the enzyme competitively at high concentrations

   D. High substrate concentrations lower the pH, reducing enzyme activity

2. [Cell Membrane Structure] The fluid mosaic model describes the plasma membrane as:

   A. A rigid bilayer of phospholipids with proteins embedded at fixed positions

   B. A fluid phospholipid bilayer in which proteins can move laterally, with the hydrophilic heads facing inward and hydrophobic tails facing outward

   C. A fluid phospholipid bilayer with hydrophilic heads facing the aqueous environment and hydrophobic tails facing the interior of the bilayer; proteins are embedded or associated with this structure

   D. A single layer of glycoproteins embedded in a lipid monolayer

3. [Inhibition Types] An inhibitor reduces enzyme activity. Increasing substrate concentration fully restores the original maximum reaction rate ($V_{\text{max}}$). This inhibitor is best described as:

   A. Non-competitive

   B. Irreversible

   C. Allosteric (activating)

   D. Competitive

4. [Cell Division — Mitosis] During which phase of mitosis do sister chromatids separate and move to opposite poles?

   A. Prophase

   B. Metaphase

   C. Anaphase

   D. Telophase

5. [Transport Mechanisms] A large polar molecule cannot cross the phospholipid bilayer by simple diffusion. The most likely mechanism by which it crosses the membrane is:

   A. Osmosis through aquaporins

   B. Facilitated diffusion through a specific channel or carrier protein, moving down its concentration gradient without ATP

   C. Active transport using a pump protein, requiring ATP

   D. Endocytosis, regardless of the molecule’s size

6. [Enzyme Kinetics — Distractor] A student heats an enzyme solution to 80°C for 10 minutes and then returns it to 37°C. They observe no enzyme activity. The most accurate explanation is:

   A. The enzyme is still inhibited by the high temperature even after cooling

   B. The enzyme has been permanently denatured — its tertiary structure has been disrupted, changing the shape of the active site so that substrate can no longer bind

   C. The enzyme has been competitively inhibited by heat

   D. The enzyme has been reversibly inhibited; activity will return after several hours at 37°C

7. [Cell Membrane — Transport] Sodium ions ($\text{Na}^+$) are maintained at a higher concentration outside the cell than inside. The sodium-potassium pump moves $\text{Na}^+$ out of the cell against this gradient. This is an example of:

   A. Facilitated diffusion — no energy required because ions use a protein channel

   B. Active transport — energy (ATP) is required to move ions against their concentration gradient

   C. Osmosis — water follows the sodium ions

   D. Passive transport — the gradient drives the movement

8. [Allosteric Regulation — Distractor] Phosphofructokinase (PFK) is inhibited by high ATP concentrations. A student claims this is competitive inhibition because ATP is also a substrate for PFK. Evaluate this claim:

   A. Correct — ATP binds to the active site as both a substrate and an inhibitor

   B. Incorrect — ATP as an inhibitor binds to an allosteric (regulatory) site, not the active site; this is allosteric/non-competitive inhibition, not competitive inhibition

   C. Correct — increasing substrate (fructose-6-phosphate) concentration would not restore activity

   D. Incorrect — PFK is not regulated by ATP at all

9. [Cell Division — Meiosis vs Mitosis] Which of the following is unique to meiosis and does NOT occur in mitosis?

   A. Chromosome condensation in prophase

   B. Spindle fibre formation and attachment to centromeres

   C. Homologous chromosomes pairing (synapsis) and crossing over in prophase I

   D. Sister chromatids separating at the centromere

10. [Enzyme Kinetics — Michaelis Constant] The Michaelis constant $K_m$ is defined as the substrate concentration at which reaction rate is half of $V_{\text{max}}$. A low $K_m$ indicates:

    A. The enzyme has a low affinity for its substrate — it requires a high substrate concentration to function

    B. The enzyme has a high affinity for its substrate — it reaches half-maximal rate at a low substrate concentration

    C. The enzyme has a low maximum velocity and is not efficient

    D. The enzyme is more likely to be allosteric

Show Answers

1. B — Enzyme saturation. At high $[\text{S}]$, all active sites are occupied and the enzyme is working at $V_{\text{max}}$. Adding more substrate cannot increase rate because there are no free active sites. A is incorrect — substrate does not denature enzymes. C is incorrect — substrate competing with itself is not a standard concept.

2. C — The fluid mosaic model: a phospholipid bilayer with hydrophilic (polar) heads facing aqueous environments (both extracellular and intracellular), hydrophobic tails facing inward, and proteins embedded to varying depths or spanning the bilayer. B reverses the orientation of the heads and tails.

3. D — Competitive inhibition. The defining feature of competitive inhibition is that increasing substrate concentration can overcome the inhibitor and fully restore $V_{\text{max}}$. Non-competitive inhibition (A) reduces $V_{\text{max}}$ regardless of substrate concentration.

4. C — Anaphase. During anaphase, the spindle fibres shorten and pull sister chromatids to opposite poles after the centromere splits. Metaphase (B) is when chromosomes align at the metaphase plate. Telophase (D) is when nuclear envelopes reform.

5. B — Facilitated diffusion. Large polar molecules use specific channel or carrier proteins and move down their concentration gradient (no ATP required). C (active transport) would be needed only if movement is against the gradient. A (osmosis) applies to water movement only.

6. B — Irreversible denaturation. At 80°C, hydrogen bonds and other non-covalent interactions maintaining the tertiary structure break down, permanently misshaping the active site. This is not reversible on returning to 37°C. D is incorrect — denaturation by heat is generally permanent. A is incorrect — temperature does not inhibit like a molecule.

7. B — Active transport. The Na$^+$/K$^+$ pump moves Na$^+$ against its concentration gradient, which requires ATP hydrolysis. A is incorrect — facilitated diffusion follows the gradient. D is incorrect — passive transport always follows the gradient.

8. B — This is allosteric (non-competitive) inhibition. When ATP is at high concentration, it binds to a regulatory allosteric site on PFK, changing its conformation and reducing activity. The active site is where fructose-6-phosphate binds. A is incorrect — ATP as inhibitor does not bind to the active site. A student who confuses “ATP is used by the enzyme” with “ATP competes at the active site” will choose A.

9. C — Synapsis (homologous chromosome pairing) and crossing over (chiasmata formation) are unique to prophase I of meiosis. A, B, and D all occur in both mitosis and meiosis.

10. B — A low $K_m$ means the enzyme achieves half-maximal rate at a low substrate concentration, indicating high affinity for the substrate. A describes high $K_m$ (low affinity). $K_m$ is independent of $V_{\text{max}}$ (C), and allostery is a separate concept (D).

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IB Biology HL — Enzymes & Cell Biology — Complete Study Guide — 2025 Syllabus — Good luck!

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May 2026 Prediction Questions

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Virtual Lab Alignment: Labster Simulations

Labster Simulation	IB HL Topic	What It Covers
Enzyme Kinetics: Unravel the Effect of Substrate Concentration	B4/C3: Michaelis-Menten kinetics	$V_{max}$, competitive vs non-competitive inhibition
Enzyme Kinetics: Investigate the Effect of Temperature and pH	B4/C3: Lock-and-key vs induced fit	Denaturation, optimum conditions
Protein Folding: Alphafold’s Revolution	B2/C3: Protein structure	Primary to quaternary structure, active site, allosteric regulation
Lactase: The Science Behind Lactose-Free Products	B4/C3: Applied enzyme activity	Industrial enzymes, real-world IB context

How to use these simulations for IB exam prep:

• Enzyme Kinetics simulations generate rate vs. substrate concentration graphs — practice interpreting these for Paper 3 data analysis
• The Lactase simulation is explicitly cited by Labster as aligned to AP Biology and IB Biology — it is a strong IA context too
• The Protein Folding simulation builds conceptual understanding of active site shape and why denaturation is irreversible