What this diagnostic is for
Use this page before you revise Unit 3, or right after you finish it. There is at least one short question per topic, more for the largest, and each answer names the study guide to read.
These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, so your score does not predict an exam score. Treat it as a map of what to revisit.
Calculator and data. A calculator is allowed on both sections of the Biology exam. A basic one is enough here, and every number is given. All data and organisms are fictional.
How to work. Answer every question before opening any answer, and write a short reason for each choice. A right answer for the wrong reason is still a gap.
Question 1 (multiple choice · 3.1)
A fictional enzyme, renase, makes a reaction run about 10 000 times faster. Tubes with and without renase start with the same substrate and are left for a very long time. Which statement is correct?
- (A) The tube with renase ends with more product, because renase adds energy to the reaction.
- (B) Renase lowers the activation energy but not the overall energy change, so both tubes end with the same amount of product.
- (C) Renase makes an energy-requiring reaction release energy, so it can go further.
- (D) Each renase molecule is used up after one reaction, so the tube with renase stops early.
Answer and explanation
Answer: (B). An enzyme lowers the energy barrier, so the reaction reaches its end point sooner. The energies of reactants and products do not change, so the end point is the same.
- (A) is tempting because “faster” sounds like “more”, but an enzyme adds no energy.
- (C) confuses catalysis with energy coupling, such as to ATP hydrolysis.
- (D) is wrong: an enzyme is unchanged and reused.
If you missed this: read “Activation energy: the barrier every reaction must cross” in the enzymes study guide.
Question 2 (multiple choice · 3.2)
A fictional enzyme’s substrate is positively charged. A negatively charged R group in the active site holds it in place. At pH 3, this R group picks up H⁺ and loses its charge. What is the most likely effect?
- (A) The rate rises, because the extra H⁺ ions collide with the substrate more often.
- (B) The rate falls, because the acid breaks the peptide bonds of the enzyme.
- (C) The rate is unchanged, because the shape of the active site is unchanged.
- (D) The rate falls, because the substrate is no longer attracted to the active site.
Answer and explanation
Answer: (D). Binding needs a match of shape and charge. With no negative charge left in the active site, the positive substrate is not held, so fewer complexes form each second.
- (A) treats H⁺ as a reactant. H⁺ changes the enzyme; it does not join the reaction.
- (B) overstates the damage. Denaturation disrupts weak interactions; peptide bonds stay intact.
- (C) forgets that charge matters as much as shape.
If you missed this: read “pH: changing the charges” in the environmental impacts study guide.
Question 3 (multiple choice · 3.2)
Drug P is shaped like an enzyme’s substrate and binds reversibly to the active site. In a tube with the enzyme, substrate and drug P, which change would most restore the rate?
- (A) Adding much more substrate
- (B) Raising the temperature well above the optimum
- (C) Adding a molecule that binds an allosteric site
- (D) Removing half of the enzyme
Answer and explanation
Answer: (A). Drug P is a competitive inhibitor. With much more substrate, substrate molecules win the active sites more often than P does.
- (B) is tempting because heat speeds collisions, but above the optimum the enzyme denatures.
- (C) describes noncompetitive inhibition, which would lower the rate further.
- (D) removes active sites, so the rate falls.
If you missed this: read “Inhibitors” in the environmental impacts study guide.
Question 4 (multiple choice · 3.3)
A fictional poison stops all ATP production in an animal cell. What is the most likely result over the next few minutes?
- (A) Nothing changes, because energy cannot be destroyed.
- (B) The cell becomes more ordered, because it is no longer releasing heat.
- (C) Membrane pumps stop, so ion concentration gradients run down and the cell loses order.
- (D) The cell turns the heat it releases back into ATP.
Answer and explanation
Answer: (C). Order needs a constant energy input. Pumps use ATP to hold ion gradients across the membrane. Without ATP, ions diffuse down their gradients and order is lost; a large loss means death.
- (A) misuses the first law. Energy is not destroyed, but it spreads out as heat the cell cannot use.
- (B) reverses the second law: order is kept only while energy flows through the cell.
- (D) is impossible for a cell: heat cannot be used to make ATP.
If you missed this: read “Order, energy and the laws of thermodynamics” in the cellular energy study guide.
Question 5 (multiple choice · 3.4)
Thylakoids are separated from their chloroplasts, leaving no stroma, and placed in bright light with water, ADP, Pᵢ, NADP⁺ and CO₂. Which products form?
- (A) O₂, ATP and NADPH, but no sugar
- (B) O₂ and sugar, but no ATP
- (C) ATP only; no O₂ forms, because O₂ comes from CO₂
- (D) Nothing, because the light reactions need the Calvin cycle enzymes
Answer and explanation
Answer: (A). The light reactions happen in the thylakoid membranes: water is split (releasing O₂), H⁺ flows through ATP synthase (making ATP) and NADP⁺ is reduced to NADPH. Fixing CO₂ into sugar needs the Calvin cycle, in the stroma.
- (B) forgets that the Calvin cycle needs the stroma.
- (C) repeats a common error: the O₂ comes from water.
- (D) is tempting because the stages depend on each other, but the light reactions stop only when the ADP and NADP⁺ run out.
If you missed this: read “Inside the chloroplast” in the photosynthesis study guide.
Question 6 (multiple choice · 3.4)
Which sequence correctly traces the electrons that reduce NADP⁺ in the light reactions?
- (A) water → photosystem I → electron transport chain → photosystem II → NADP⁺
- (B) CO₂ → photosystem II → electron transport chain → photosystem I → NADP⁺
- (C) ATP → photosystem I → photosystem II → NADP⁺
- (D) water → photosystem II → electron transport chain → photosystem I → NADP⁺
Answer and explanation
Answer: (D). Light boosts electrons out of photosystem II; water is split to replace them. They pass along the chain to photosystem I, are boosted again and reduce NADP⁺ to NADPH.
- (A) swaps the photosystems. Photosystem II comes first in the path, even though its number is higher.
- (B) names the wrong source. CO₂ is used in the Calvin cycle.
- (C) treats ATP as an electron source. ATP is a product of the light reactions.
If you missed this: read “The light reactions” in the photosynthesis study guide.
Question 7 (multiple choice · 3.5)
Animal cells are given glucose in which every carbon atom is labelled. Where in the cell is labelled CO₂ first released?
- (A) The cytosol, during glycolysis
- (B) The mitochondrial matrix, during pyruvate oxidation and the Krebs cycle
- (C) The intermembrane space, when protons are pumped
- (D) The inner membrane, when oxygen accepts electrons
Answer and explanation
Answer: (B). Glycolysis releases no carbon. Pyruvate moves into the mitochondrion, and in the matrix its carbon leaves as CO₂ in pyruvate oxidation and the Krebs cycle.
- (A) is tempting because glycolysis comes first, but it makes no CO₂.
- (C) and (D) are electron transport: oxygen forms water, and no carbon is involved.
If you missed this: read “The four stages and where they happen” in the cellular respiration study guide.
Question 8 (multiple choice · 3.5)
In an aerobic cell, the electron transport chain is stopped. Soon after, the Krebs cycle stops too, even though it uses no oxygen directly. Why?
- (A) The Krebs cycle needs ATP from the electron transport chain as its only fuel.
- (B) The Krebs cycle needs the CO₂ made by the electron transport chain.
- (C) NAD⁺ and FAD are no longer regenerated from NADH and FADH₂, so the Krebs cycle has no electron acceptors.
- (D) The proton gradient is needed to carry pyruvate through the Krebs cycle.
Answer and explanation
Answer: (C). Each stage’s products are the next stage’s inputs. The Krebs cycle passes electrons to NAD⁺ and FAD. If the ETC cannot take electrons from NADH and FADH₂, NAD⁺ and FAD run out and the cycle stalls.
- (A) is tempting because the ETC powers most ATP synthesis, but the Krebs cycle runs on carbon from pyruvate.
- (B) reverses the flow: the Krebs cycle releases CO₂.
- (D) misuses the gradient, which drives ATP synthase.
If you missed this: read “Stage 4: electron transport and oxidative phosphorylation” in the cellular respiration study guide.
Question 9 (short answer · 3.1)
A student tests whether the amount of a fictional fungal enzyme affects how fast it releases glucose from a polysaccharide. Every tube has the same polysaccharide solution at 30 °C and the same total volume. She measures the glucose released in the first 2.0 minutes.
| Tube | Enzyme added | Glucose released in 2.0 min / µmol |
|---|---|---|
| 1 | none | 0.1 |
| 2 | 0.5 mL | 6.0 |
| 3 | 1.0 mL | 12.2 |
| 4 | 2.0 mL | 24.0 |
| 5 | 2.0 mL, boiled and cooled | 0.1 |
(a) Identify the independent and dependent variables. (b) Calculate the initial rate for tubes 2 and 4. (c) State the purpose of tube 1 and of tube 5. (d) Explain why doubling the enzyme doubles the rate here.
Answer and explanation
(a) Independent: volume of enzyme. Dependent: rate of glucose release (glucose released in 2.0 minutes).
(b) Tube 2: 6.0 ÷ 2.0 = 3.0 µmol min⁻¹. Tube 4: 24.0 ÷ 2.0 = 12.0 µmol min⁻¹.
(c) Tube 1, a negative control with no enzyme, shows almost no glucose is released without it. Tube 5, a negative control with denatured enzyme, shows activity is lost when the protein’s shape is destroyed, so a protein catalyst is responsible.
(d) Substrate is plentiful, so the number of active sites limits the rate. Twice the enzyme gives twice the active sites, so twice as many enzyme–substrate complexes form each minute.
If you missed this: read “Worked example 2: choosing and justifying controls” in the enzymes study guide.
Question 10 (short answer · 3.5)
Mitochondria from the flight muscle and from the fat tissue of a fictional insect are compared.
| Mitochondria from | Inner membrane area / µm² | Maximum ATP synthesis / units |
|---|---|---|
| Flight muscle | 42 | 6.3 |
| Fat tissue | 14 | 2.0 |
(a) Calculate the ATP synthesis per µm² of inner membrane for each. (b) Use your answer to (a) to explain why the flight muscle mitochondria make more ATP. (c) Name the structures that give the flight muscle mitochondria their large inner membrane area.
Answer and explanation
(a) Flight muscle: 6.3 ÷ 42 = 0.15 units per µm². Fat tissue: 2.0 ÷ 14 = 0.14 units per µm² (0.143).
(b) ATP synthesis per µm² is almost the same, so the difference comes from the amount of inner membrane (3.0 times as much, for about 3.2 times the ATP). The ETC and ATP synthase sit in the inner membrane, so more membrane holds more of them.
(c) The cristae (folds of the inner membrane).
If you missed this: read “Stage 4: electron transport and oxidative phosphorylation” in the cellular respiration study guide.
Your next step
A short answer missed in any part counts as missed.
| Topic | Question(s) | If you missed it, read |
|---|---|---|
| 3.1 Enzymes | 1, 9 | Study guide 3.1 |
| 3.2 Environmental impacts on enzyme function | 2, 3 | Study guide 3.2 |
| 3.3 Cellular energy | 4, 8 | Study guide 3.3 |
| 3.4 Photosynthesis | 5, 6 | Study guide 3.4 |
| 3.5 Cellular respiration | 7, 8, 10 | Study guide 3.5 |
How to use your result
- Wrong answers first. Read the named section, then do that topic’s practice questions.
- Lucky guesses count as misses if your reason did not match the explanation.
- No misses? Go to the mixed unit review, which joins the topics in exam-style questions.
- Retry in a week to check that your reasons are complete.