Covering Lessons 12–15: Cell requirements and waste removal, enzymes (structure, action, and environmental factors), and SA:V ratio and exchange surfaces.
What's Covered
Multiple Choice — 10 marks
1. Which of the following correctly identifies TWO forms of energy that are suitable for cells?
2. Lactic acid produced during anaerobic respiration in muscle cells is eventually transported to the liver. What happens to it there?
3. Which of the following best explains why amylase cannot digest proteins?
4. A student boils an enzyme solution for 5 minutes, then cools it to 37°C and tests it with its substrate. No reaction occurs. The student then adds fresh (unboiled) enzyme to the same mixture. The reaction proceeds. What does this confirm?
5. An enzyme has an optimum temperature of 40°C. Which of the following correctly explains why the rate of reaction is lower at 25°C than at 40°C?
6. A student measures how long it takes for amylase to digest starch at different pH values and records the time in seconds. To convert this to a rate of reaction, the student should:
7. An enzyme reaction has reached Vmax. Which of the following would increase the reaction rate beyond Vmax?
8. A cube has a side length of 4 cm. What is its SA:V ratio?
9. The small intestine has villi and microvilli that dramatically increase its internal surface area. Which variable in Fick's Law does this primarily optimise?
10. A patient has severe liver disease. Blood tests show elevated urea AND elevated lactic acid. Enzyme tests show many liver enzymes are non-functional. Using concepts from L12–L15, which of the following best explains all three findings?
Short Answer — 12 marks
1. A cell requires a continuous supply of matter. Explain why ions such as Ca²⁺ and K⁺ are essential cell requirements, and describe what would happen to enzyme function if intracellular pH dropped significantly due to CO₂ accumulation. (3 marks)
1 mark: Ca²⁺ — essential for muscle contraction and nerve signalling; K⁺ — resting membrane potential / cardiac rhythm; 1 mark: CO₂ + H₂O → H₂CO₃ → H⁺ + HCO₃⁻, lowering intracellular pH; 1 mark: lower pH disrupts ionic bonds and H-bonds in enzyme tertiary structure, alters active site shape, reduces substrate binding and enzyme activity
2. Compare the lock-and-key and induced fit models of enzyme action, then explain why the rate of an enzyme-catalysed reaction first increases then decreases as temperature rises above the optimum. (3 marks)
1 mark: lock-and-key rigid active site; induced fit — active site flexible, changes shape on substrate binding; 1 mark: increase to optimum — more kinetic energy → more frequent E-S collisions → more complexes per unit time; 1 mark: above optimum — denaturation disrupts non-covalent bonds maintaining tertiary structure → active site shape lost → substrate cannot bind
3. A student investigates the effect of pH on catalase activity (H₂O₂ → H₂O + O₂) but does not use a buffer solution. Explain why failing to control pH is a problem for this investigation, and describe how this links to why the alveolus must maintain stable pH for gas exchange enzymes to function. (3 marks)
1 mark: without buffer, pH may vary between or within trials — pH is a confounding variable that affects enzyme activity independently of the IV; results are invalid; 1 mark: pH affects active site shape via ionic bond disruption — uncontrolled pH means observed rate changes could be due to pH, not the IV; 1 mark: alveolar cells contain carbonic anhydrase and other enzymes — if CO₂ is not removed efficiently, pH falls, these enzymes denature or lose activity, impairing gas exchange further (positive feedback)
4. A flatworm (phylum Platyhelminthes) is a multicellular organism that exchanges gases entirely through its body surface with no specialised respiratory organs. Using SA:V ratio, Fick's Law, and cell requirements, explain why this strategy works for a flatworm but would be impossible for a large animal like a dog. (3 marks)
1 mark: flatworm is flat and thin — all cells within short diffusion distance of body surface; high SA:V relative to body mass; 1 mark: Fick's Law — thin body = short diffusion distance (↓ membrane thickness); flat shape maximises SA:V; moist skin maintains solubility; 1 mark: a dog's large volume means most cells are far from the body surface; SA:V too low for diffusion to supply O₂/remove CO₂ for all cells; requires specialised lungs + circulatory system to meet cell requirements
Answers
SA1: Calcium ions (Ca²⁺) are an essential cell requirement because they trigger muscle contraction — Ca²⁺ released from the sarcoplasmic reticulum binds to troponin, exposing myosin binding sites on actin and initiating the contraction cycle. Insufficient Ca²⁺ causes muscle spasms and tetany. Potassium ions (K⁺) are essential for maintaining the resting membrane potential in excitable cells (neurons and cardiac muscle) — K⁺ gradients across the membrane are established by the Na⁺/K⁺ pump; disruption causes cardiac arrhythmia and impaired nerve signalling. If CO₂ accumulates inside a cell, it reacts with water to form carbonic acid (CO₂ + H₂O → H₂CO₃), which partially dissociates to release H⁺ ions, lowering intracellular pH. This acidic environment disrupts the ionic bonds and hydrogen bonds that maintain the tertiary structure of enzymes. The active site geometry changes, substrate binding decreases, enzyme-substrate complex formation is impaired, and overall catalytic activity falls — potentially halting critical metabolic pathways.
SA2: Lock-and-key model: the active site is rigid and fixed in shape, perfectly complementary to the substrate, which slots in without any conformational change in either molecule. Induced fit model: the active site is flexible — when the substrate approaches, the enzyme changes shape slightly to wrap more precisely around the substrate, optimising the fit for catalysis. The induced fit model is more accurate and accounts for enzyme flexibility observed in X-ray crystallography studies. Regarding temperature: as temperature increases toward the optimum (~40°C), molecules gain kinetic energy and move faster. Enzyme and substrate molecules collide more frequently per unit time, meaning more enzyme-substrate complexes form per second and the reaction rate increases. Above the optimum, however, the excess thermal energy disrupts the non-covalent bonds — especially hydrogen bonds and ionic interactions — that maintain the enzyme's tertiary structure. The polypeptide chain unfolds (denaturation), the precise geometry of the active site is lost, and the substrate can no longer bind. Since denaturation is irreversible, the rate cannot recover when temperature is lowered, and it continues to fall toward zero as more enzyme molecules denature.
SA3: Failing to use a buffer solution means pH is not controlled between trials. Since pH is a known factor affecting enzyme activity — it disrupts the ionic bonds and hydrogen bonds maintaining active site shape — any observed differences in catalase activity across conditions could be due to pH variation rather than the intended independent variable. This introduces pH as a confounding variable and compromises the validity of the investigation; conclusions drawn from the data would be unreliable. In the alveolus, carbonic anhydrase and other metabolic enzymes function within narrow pH ranges. If CO₂ is not removed efficiently from alveolar cells — for example, due to impaired ventilation or thickened membranes — CO₂ reacts with water to form carbonic acid, lowering intracellular pH. This disrupts the active sites of alveolar enzymes, reducing their function. Impaired enzyme function in alveolar cells reduces their capacity to maintain the exchange surface (e.g. producing surfactant, repairing membranes), which further impairs gas exchange — a positive feedback loop where poor CO₂ removal leads to pH changes that further impair the cells responsible for gas exchange.
SA4: A flatworm's body plan — extremely flat and thin (often less than 1 mm thick) — means that no cell in its body is more than a fraction of a millimetre from the external environment. Its high SA:V ratio relative to its body mass ensures there is ample surface area for the volume of cells to be supplied. Applying Fick's Law: the flat, thin body minimises the diffusion distance (equivalent to membrane thickness in Fick's Law), maximising the rate of O₂ and CO₂ exchange. The moist, permeable skin maintains the solubility of gases at the exchange surface. The organism's small overall mass means metabolic demand is low, and diffusion can meet it. For a dog, this strategy is impossible because its large body volume means that the vast majority of cells are metres away from the skin surface — far beyond the effective range of diffusion (which operates efficiently only over distances of tens of micrometres). Its SA:V ratio is extremely low, meaning even though the total skin surface area is large, it is negligible relative to the enormous volume of tissue that must receive O₂ and expel CO₂ every second. A dog's metabolic rate also demands far more gas exchange than any surface diffusion strategy could provide. It therefore requires specialised internal exchange surfaces (lungs with alveoli, maximising SA within a compact space), a circulatory system to transport gases between exchange surfaces and all body cells, and ventilation to maintain concentration gradients at the exchange surface.
Which topics do you need to revisit before the Module Review? Note them here.