Test your understanding of Lessons 11–15: cellular respiration, cell requirements, enzyme structure and function, and SA:V ratio.
⏱ 35 min10 MC + 4 SA22 marksLessons 11–15
Lessons Covered
11
Cellular Respiration
12
Cell Requirements
13
Enzyme Structure
14
Enzyme Factors
15
SA:V Ratio
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Multiple Choice — 10 marks
1. Which is the correct sequence of stages in aerobic cellular respiration?
A Krebs cycle → Glycolysis → Electron transport chain
B Glycolysis → Electron transport chain → Krebs cycle
C Glycolysis → Krebs cycle → Electron transport chain
D Electron transport chain → Glycolysis → Krebs cycle
2. During anaerobic respiration in human muscle cells, pyruvate is converted to:
A Ethanol and CO₂
B Lactic acid
C Acetyl-CoA
D Water and CO₂
3. The electron transport chain produces most of the ATP in aerobic respiration. Where does this process occur?
A In the cytoplasm
B In the mitochondrial matrix
C On the inner mitochondrial membrane (cristae)
D In the nucleus
4. Which of the following is a metabolic waste product of protein catabolism that must be excreted by the kidneys?
A Carbon dioxide
B Water
C Urea
D Lactic acid
5. Carbon dioxide is classified as a metabolic waste product because it:
A Cannot be used in any biological reaction
B Is toxic at any concentration and must be removed immediately
C Is produced during cellular respiration and would accumulate to lower blood pH if not excreted
D Is only produced during anaerobic respiration in muscle cells
6. The lock-and-key model and the induced fit model of enzyme action both agree that:
A The active site is completely rigid and does not change shape when the substrate binds
B Multiple substrates can bind simultaneously to the same active site
C Enzyme specificity results from the complementary shape between the active site and the substrate
D All enzymes work most effectively at pH 7 and 37°C
7. An enzyme's specific 3D active site geometry is determined primarily by its:
A Primary structure (amino acid sequence) alone, regardless of folding
B Tertiary structure — the 3D folding of the polypeptide maintained by non-covalent bonds
C The number of disulfide bonds in the enzyme
D The lipid composition of the surrounding membrane
8. An enzyme is exposed to pH 2 (well below its optimum of 7.4). What happens to enzyme activity?
A Activity increases because more H⁺ ions provide energy for catalysis
B Activity is unchanged because enzymes are proteins and resist pH change
C Activity decreases because H⁺ ions disrupt ionic bonds and hydrogen bonds in the active site, altering its shape
D Activity increases temporarily then the enzyme denatures after exactly one hour
9. A student graphs enzyme activity against temperature for an enzyme with an optimum of 37°C. Which best describes the shape of the graph?
A A straight line increasing from 0°C to 37°C then dropping vertically to zero
B Activity gradually increases from 0°C to the optimum at 37°C, reaches a peak, then drops sharply as denaturation occurs above the optimum
C A flat plateau because the enzyme maintains constant activity across all temperatures
D A perfectly symmetric bell curve equally sloped on both sides of 37°C
10. A spherical cell increases in radius from 1 μm to 3 μm. What happens to its surface area to volume (SA:V) ratio?
A The SA:V ratio increases by a factor of 3
B The SA:V ratio stays the same because both surface area and volume increase
C The SA:V ratio decreases because volume increases proportionally faster than surface area
D The SA:V ratio increases because larger cells have more surface area in total
Short Answer — 12 marks
1. Compare aerobic and anaerobic respiration in human muscle cells. In your answer, refer to oxygen requirements, ATP yield, location within the cell, and end products. (3 marks)
Any 3 correct comparison points — 1 mark each: O₂ (aerobic requires; anaerobic does not); ATP yield (~36–38 vs 2 per glucose); location (aerobic: cytoplasm + mitochondria; anaerobic: cytoplasm only); end products (aerobic: CO₂ + H₂O; anaerobic: lactic acid)
Oxygen: Aerobic respiration requires oxygen as the final electron acceptor in the electron transport chain. Anaerobic respiration does not require oxygen — it regenerates NAD⁺ by converting pyruvate to lactic acid instead.
ATP yield: Aerobic respiration produces approximately 36–38 ATP molecules per glucose, through glycolysis (net 2 ATP), the Krebs cycle (2 ATP), and the electron transport chain (~32–34 ATP). Anaerobic respiration produces only 2 ATP per glucose (glycolysis only).
Location: Aerobic respiration occurs across multiple compartments: glycolysis in the cytoplasm, the Krebs cycle in the mitochondrial matrix, and the electron transport chain on the inner mitochondrial membrane (cristae). Anaerobic respiration occurs entirely in the cytoplasm.
End products: Aerobic respiration produces carbon dioxide (CO₂) and water (H₂O), which are excreted. Anaerobic respiration in muscle cells produces lactic acid (lactate), which accumulates in muscle tissue, lowering pH and contributing to muscle fatigue. Lactic acid is later transported to the liver where it can be converted back to glucose (Cori cycle).
2. Explain the induced fit model of enzyme action. In your answer, describe how it differs from the lock-and-key model and explain what happens to enzyme function during denaturation. (3 marks)
1 mark: induced fit — active site flexes and changes shape when substrate approaches, forming a more precise fit; 1 mark: differs from lock-and-key — lock-and-key proposes a rigid, pre-formed active site; induced fit proposes a flexible active site that moulds around the substrate; 1 mark: denaturation — non-covalent bonds maintaining tertiary structure disrupted → active site shape permanently lost → cannot bind substrate → enzyme inactivated
The induced fit model (Koshland, 1958) proposes that the active site of an enzyme is not rigid but flexible. When a substrate approaches, the active site changes shape to more precisely fit and accommodate the substrate — forming a closer, more effective enzyme-substrate complex that stabilises the transition state and lowers activation energy more efficiently. This differs from the lock-and-key model (Fischer, 1894), which proposes a rigid, pre-formed active site with a shape exactly complementary to the substrate — like a key fitting a lock — with no change in active site geometry upon binding. The induced fit model better explains substrate specificity: only substrates that can induce the correct conformational change will bind productively. During denaturation, high temperatures or extreme pH disrupt the non-covalent bonds (hydrogen bonds, ionic bonds, van der Waals forces) that maintain the enzyme's tertiary (3D) structure. The polypeptide chain unfolds and the precise geometry of the active site is permanently destroyed. The substrate can no longer bind, no enzyme-substrate complex forms, and the enzyme is inactivated. Unlike inhibition (which is often reversible), denaturation is largely irreversible — the enzyme cannot refold to its original active configuration.
3. Explain how increasing substrate concentration affects the rate of an enzyme-catalysed reaction. Include reference to enzyme saturation and Vmax in your answer. (3 marks)
1 mark: as substrate concentration increases, rate initially increases — more substrate collisions with free active sites; 1 mark: rate levels off at Vmax — all active sites saturated, every enzyme is processing a substrate at maximum speed; 1 mark: adding more substrate above Vmax has no effect — no free active sites available; only adding more enzyme increases rate further
As substrate concentration increases from zero, the rate of the enzyme-catalysed reaction initially increases proportionally. This is because more substrate molecules are available to collide with and bind to free enzyme active sites, forming enzyme-substrate complexes and producing product. When substrate concentration is low, many active sites are unoccupied, so each additional substrate molecule added finds a free active site and increases the reaction rate. As substrate concentration continues to increase, the rate of increase slows. Eventually, the rate reaches a plateau — the maximum reaction rate (Vmax). At Vmax, all active sites are occupied at all times: every enzyme molecule is continuously processing a substrate molecule. The enzyme preparation is saturated — there are no free active sites available. Adding more substrate above this point has no further effect on the rate because incoming substrate molecules must wait for an active site to become free before they can bind. The only way to increase the rate above Vmax is to add more enzyme — increasing the total number of active sites available.
4. Explain why large multicellular organisms require specialised exchange surfaces and transport systems to meet their metabolic needs. Refer to SA:V ratio and Fick's Law in your answer. (3 marks)
1 mark: as size increases, SA:V ratio decreases (volume grows as cube, SA as square) → body surface too small relative to metabolic demand; 1 mark: Fick's Law — diffusion rate ∝ SA/thickness; low SA:V and large diffusion distances mean diffusion alone cannot supply all cells; 1 mark: specialised exchange surfaces (lungs, villi) solve the SA problem; transport systems (circulatory system) bridge the gap between exchange surfaces and remote tissues
As the size of an organism increases, its SA:V ratio decreases. Volume grows proportionally to the cube of linear dimensions, while surface area grows only as the square — so for every doubling of linear size, volume increases 8-fold but surface area only 4-fold. For a large multicellular organism, the outer body surface area is negligible relative to the vast volume of metabolically active cells. Most cells are far from the body surface — far beyond the effective range of diffusion. According to Fick's Law, the rate of diffusion is proportional to surface area and concentration gradient, and inversely proportional to membrane thickness. With a very low SA:V ratio, the body surface provides insufficient area to supply O₂, nutrients, and remove CO₂ and wastes from the entire volume of tissue. Additionally, the large diffusion distances from the body surface to internal cells mean passive diffusion would be far too slow to meet metabolic demands. Specialised exchange surfaces solve the surface area problem: organs such as lungs (alveoli), small intestine (villi and microvilli), and gills provide enormously folded internal surfaces with high SA:V ratios, thin walls, and maintained concentration gradients — maximising diffusion rates for specific substances. Transport systems (the circulatory system) are also required to bridge the gap between exchange surfaces and remote tissues — rapidly distributing substances to all cells regardless of their distance from the exchange organ.
Answers
Q1 — C: The correct sequence is: Glycolysis (cytoplasm) → Krebs cycle (mitochondrial matrix) → Electron transport chain (inner mitochondrial membrane). Glycolysis converts glucose to pyruvate; the Krebs cycle oxidises acetyl-CoA releasing CO₂ and reducing electron carriers; the ETC uses these carriers to drive ATP synthesis via oxidative phosphorylation.
Q2 — B: In human muscle cells (and other animal cells), anaerobic respiration converts pyruvate to lactic acid. This regenerates NAD⁺ from NADH, allowing glycolysis to continue producing 2 ATP per glucose even without oxygen. Ethanol and CO₂ (A) is the product in yeast (fermentation). Acetyl-CoA (C) is produced in aerobic respiration before the Krebs cycle. Water and CO₂ (D) are the products of aerobic respiration.
Q3 — C: The electron transport chain is embedded in the inner mitochondrial membrane (cristae). The cristae's folded structure increases the surface area for the ETC, maximising ATP production. Glycolysis (A) occurs in the cytoplasm. The Krebs cycle (B) occurs in the mitochondrial matrix. The nucleus (D) is not involved in respiration.
Q4 — C: Urea is produced in the liver by deamination of amino acids — the amino group (–NH₂) is removed and converted to ammonia, then to the less toxic urea, which is excreted by the kidneys. CO₂ (A) and water (B) are products of cellular respiration excreted by lungs/kidneys. Lactic acid (D) is a product of anaerobic respiration, not protein catabolism.
Q5 — C: CO₂ dissolves in blood plasma forming carbonic acid (H₂CO₃ → H⁺ + HCO₃⁻). Accumulation lowers blood pH (acidosis), disrupting enzyme function and other physiological processes. It is not toxic at all concentrations (B) — CO₂ is actually a respiratory stimulus and has physiological roles. It is produced during aerobic respiration, not only anaerobic (D).
Q6 — C: Both models agree that enzyme specificity arises from the complementary shape between the active site and the substrate — only the correctly shaped substrate can bind productively. They differ on whether the active site is rigid (lock-and-key) or flexible (induced fit). Options A, B, and D are incorrect — the lock-and-key model does propose a rigid active site (A), but this is a point of difference, not agreement.
Q7 — B: The active site's 3D geometry is determined by the enzyme's tertiary structure — the specific 3D folding of the polypeptide chain maintained by non-covalent bonds (hydrogen bonds, ionic interactions, van der Waals forces). While the primary structure (A) determines the sequence of amino acids, the tertiary structure determines the 3D shape of the active site. Disulfide bonds (C) contribute to stabilising some proteins but are not the primary determinant. Membrane lipids (D) have no role in determining active site geometry.
Q8 — C: At pH 2, the high concentration of H⁺ ions protonates basic amino acid side chains in the active site, disrupting ionic bonds and altering the charge distribution. This changes the geometry of the active site — it no longer fits the substrate precisely — reducing or eliminating enzyme activity. This effect may be reversible if pH is restored. Option A is incorrect — H⁺ ions do not provide energy. Option B is incorrect — enzymes are sensitive to pH.
Q9 — B: The activity-temperature graph is asymmetric: as temperature rises from 0°C, increasing kinetic energy increases substrate-enzyme collision frequency — so activity gradually increases to the optimum (37°C). Above the optimum, additional thermal energy disrupts non-covalent bonds in the enzyme's tertiary structure (denaturation) — activity drops sharply. The graph peaks at the optimum and is steep on the high-temperature side, not symmetric (D).
Q10 — C: At radius 1 μm: SA = 4π(1)² = ~12.6 μm², V = (4/3)π(1)³ = ~4.2 μm³, SA:V ≈ 3:1. At radius 3 μm: SA = 4π(9) = ~113 μm², V = (4/3)π(27) = ~113 μm³, SA:V ≈ 1:1. The SA:V ratio decreased from 3:1 to 1:1 as size tripled — volume increased 27-fold while surface area increased only 9-fold. Total SA does increase (D) but SA relative to volume decreases (C).
SA1: Aerobic requires O₂; anaerobic does not. ATP yield: aerobic ~36–38 ATP per glucose (glycolysis 2 ATP, Krebs 2 ATP, ETC ~32–34 ATP); anaerobic only 2 ATP (glycolysis only). Location: aerobic — glycolysis in cytoplasm, Krebs cycle in mitochondrial matrix, ETC on inner mitochondrial membrane (cristae); anaerobic — entirely in cytoplasm. End products: aerobic produces CO₂ and H₂O (excreted); anaerobic in muscle produces lactic acid, which accumulates and lowers muscle pH, contributing to fatigue.
SA2: The induced fit model proposes the active site is flexible — when substrate approaches, the active site changes shape to more precisely fit the substrate, forming a closer enzyme-substrate complex. This differs from the lock-and-key model, which proposes a rigid, pre-formed active site exactly matching the substrate with no shape change upon binding. During denaturation, high temperature or extreme pH disrupts the non-covalent bonds (H-bonds, ionic bonds, van der Waals) maintaining the enzyme's tertiary structure. The polypeptide unfolds and the active site's precise 3D shape is permanently destroyed — the substrate cannot bind, no reaction occurs. Unlike inhibition, denaturation is largely irreversible.
SA3: As substrate concentration increases from zero, reaction rate increases proportionally — more substrate molecules find and bind to free active sites. As concentration increases further, the rate increase slows and eventually plateaus at Vmax. At Vmax, all active sites are occupied continuously — the enzyme is saturated. Adding more substrate has no effect because no free active sites are available. The only way to increase rate above Vmax is to add more enzyme, increasing the total number of active sites.
SA4: As organism size increases, SA:V ratio decreases — volume grows as the cube of linear dimensions while surface area grows as the square. For large organisms, the body surface is negligible relative to cell volume, and most cells are far from the surface. Fick's Law: diffusion rate ∝ SA × concentration gradient / thickness. Low SA:V means insufficient surface area and excessive diffusion distances — passive diffusion cannot supply all cells. Specialised exchange surfaces (lungs, villi) solve the SA problem by providing large, folded internal surfaces with high SA:V. Circulatory systems bridge the gap between exchange surfaces and remote tissues, rapidly delivering substances to all cells.
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