VCE Biology Unit 3 AOS2 — Diagnostic Test

Scientific Investigation Design

8 marks

Question 1 1 mark Easy

A student investigates whether the enzyme catalase, extracted from potato, breaks down hydrogen peroxide at different rates depending on pH. They prepare five solutions at pH 4, 6, 7, 8, and 10, add a fixed amount of catalase extract to each, and measure the volume of oxygen produced over 3 minutes. Identify the methodology used in this investigation.

Think about whether the student is observing a system in its natural state, or actively changing a variable to measure its effect.

Model Answer

Controlled experiment (also accepted: laboratory experiment, experimental methodology). The student deliberately manipulates an independent variable (pH) while controlling other variables, to measure the effect on a dependent variable (volume of oxygen produced).

1 mark: identifies the methodology as a controlled experiment (accept: laboratory experiment, experimental methodology)

Question 2 2 marks Easy

A class investigates the effect of temperature on the rate of starch digestion by amylase. Five test tubes are prepared, each containing the same concentration of starch solution and amylase. The test tubes are placed in water baths at 10°C, 20°C, 30°C, 40°C, and 50°C. Every 2 minutes, a drop from each tube is tested with iodine solution. The time taken for the blue-black colour to disappear (indicating all starch has been digested) is recorded. Identify the independent variable and the dependent variable in this investigation.

The independent variable is what the experimenter deliberately changes. The dependent variable is what is measured as a result.

Model Answer

Independent variable: temperature (°C) of the water bath. Dependent variable: time taken for the iodine solution to no longer turn blue-black (i.e., time for complete starch digestion).

1 mark: independent variable — temperature (accept: temperature of the water bath) 1 mark: dependent variable — time for complete starch digestion / time for iodine to remain yellow-brown (must indicate a time measurement)

Question 3 2 marks Medium

Two scenarios involving an investigation into the effect of light intensity on photosynthesis rate in Elodea are described below. Scenario A: The same student repeats the experiment three times in the same laboratory on the same day, using the same equipment, and obtains consistent results each time. Scenario B: Another school across the city uses the same method and obtains results consistent with the first student's results. For each scenario, state whether the results demonstrate repeatability or reproducibility. Justify your answer for Scenario B only.

Repeatability involves the same investigator; reproducibility involves different investigators or different settings.

Model Answer

Scenario A: Repeatability. Scenario B: Reproducibility. Justification for Scenario B: Reproducibility is demonstrated when independent investigators (or the same investigator in different circumstances or locations) obtain consistent results using the same method. In Scenario B, a different school obtained consistent results, confirming the findings are not specific to one laboratory or experimenter.

1 mark: correctly identifies Scenario A as repeatability AND Scenario B as reproducibility (both must be correct for the mark) 1 mark: justification for Scenario B — different investigator/school/setting obtained consistent results using the same method

Question 4 3 marks Medium

A student investigates the effect of substrate concentration on urease activity. They measure the amount of ammonia produced (in μmol/min) at six substrate concentrations. Their results are shown below.

Substrate concentration (mM)	Ammonia produced (μmol/min)
2	1.2
4	2.3
6	3.5
8	8.1
10	5.8
12	6.4

(a) Identify the outlier in the data. (1 mark) (b) State the type of error most likely responsible for this result, and give one specific reason why this type of error would produce an unusually high reading. (2 marks)

Look for the data point that doesn't fit the trend. Then consider whether the error could affect all readings equally (systematic) or just one reading unpredictably (random).

Model Answer

(a) The outlier is the reading at 8 mM substrate concentration (8.1 μmol/min), which is anomalously high compared to the expected upward trend. (b) Random error. A random error affects individual measurements unpredictably and does not follow a consistent pattern. A specific reason: the student may have accidentally added a greater volume of urease solution to the 8 mM tube than to the others, producing a falsely elevated ammonia reading for that one trial.

1 mark: identifies the 8 mM result (8.1 μmol/min) as the outlier 1 mark: identifies random error 1 mark: provides a specific plausible reason why a random error would produce an unexpectedly high reading (e.g. pipetting error, cross-contamination, timing error for that one trial only)

Enzyme Inhibition & Regulation

7 marks

Question 5 1 mark Easy

Methotrexate is a drug used in cancer treatment. It binds to the enzyme dihydrofolate reductase (DHFR) at the same site as the enzyme's natural substrate, preventing the substrate from binding. High concentrations of substrate can reduce the effectiveness of methotrexate. Name the type of inhibition methotrexate exerts on DHFR.

If adding more substrate can 'outcompete' the inhibitor, think about what that tells you about where the inhibitor binds.

Model Answer

Competitive inhibition. Methotrexate binds at the active site (same site as the substrate), and its effect can be reduced by increasing substrate concentration.

1 mark: competitive inhibition

Question 6 2 marks Medium

Ibuprofen inhibits the enzyme cyclooxygenase (COX), which normally converts arachidonic acid into prostaglandins. Ibuprofen binds to a site on COX that is not the active site. Explain how ibuprofen reduces prostaglandin production.

Focus on where ibuprofen binds and how binding at that site changes the enzyme's ability to act on its substrate.

Model Answer

Ibuprofen binds to an allosteric site on COX (not the active site). This binding causes a conformational change in the enzyme's shape, including the active site. The altered active site can no longer bind arachidonic acid effectively, so the reaction that converts arachidonic acid to prostaglandins cannot proceed, reducing prostaglandin production.

1 mark: ibuprofen binds to an allosteric site (accept: a site other than the active site), causing a conformational change in the enzyme's shape 1 mark: the changed shape of the active site means arachidonic acid cannot bind (or: substrate–enzyme complex cannot form), so less prostaglandin is produced

Question 7 4 marks Hard

HMG-CoA reductase is an enzyme in the cholesterol synthesis pathway. When cholesterol levels in the cell are high, cholesterol binds to an allosteric site on HMG-CoA reductase, reducing its activity. This is an example of feedback inhibition. (a) State two reasons why feedback inhibition of this enzyme is beneficial to the cell. (2 marks) (b) Describe the molecular mechanism by which cholesterol inhibits HMG-CoA reductase. (1 mark) (c) The drug atorvastatin (a statin) also inhibits HMG-CoA reductase, but it binds at the active site and competes directly with the enzyme's substrate. Compare the effect of increasing substrate concentration on feedback inhibition by cholesterol versus competitive inhibition by atorvastatin. (1 mark)

For (a), think about what the cell gains by linking the product back to control the enzyme. For (b), name the binding site and the structural change. For (c), think about whether more substrate can 'rescue' each type of inhibition.

Model Answer

(a) Any two of: 1. Prevents overproduction of cholesterol, avoiding wasteful use of cellular resources (ATP, raw materials). 2. Maintains homeostasis — cholesterol levels are kept within a functional range. 3. Allows the cell to respond rapidly to changing cholesterol demands without requiring new gene expression. (b) Cholesterol binds to an allosteric site on HMG-CoA reductase. This causes a conformational change in the enzyme's three-dimensional shape, distorting the active site so it can no longer efficiently bind the substrate and catalyse the reaction. (c) Feedback inhibition by cholesterol cannot be overcome by increasing substrate concentration (it is non-competitive — inhibitor and substrate bind at different sites). Competitive inhibition by atorvastatin can be reduced by increasing substrate concentration, because the substrate and inhibitor compete for the same active site.

1 mark: first benefit — prevents wasteful overproduction of cholesterol / conserves cellular resources (ATP, raw materials) 1 mark: second benefit — maintains homeostasis / allows dynamic regulation / rapid response without gene expression changes 1 mark: cholesterol binds allosteric site → conformational change → active site shape altered → substrate cannot bind 1 mark: feedback inhibition (non-competitive) is NOT overcome by more substrate; competitive inhibition (atorvastatin) IS reduced by increasing substrate — must contrast both

Photosynthesis

4 marks

Question 8 1 mark Easy

During the light-dependent reactions of photosynthesis, water molecules are split and electrons are transferred through the electron transport chain in the thylakoid membrane. Name the reduced coenzyme produced by the light-dependent reactions that is used to drive the Calvin cycle.

This coenzyme carries high-energy electrons from the light reactions to the stroma. It has a similar name to the coenzyme used in cellular respiration, but with one extra phosphate group.

Model Answer

NADPH (nicotinamide adenine dinucleotide phosphate, reduced form).

1 mark: NADPH

Question 9 3 marks Medium

Elodea (a water plant) is placed in a solution containing sodium bicarbonate (NaHCO₃), which releases CO₂ as it dissolves. A student counts the bubbles of oxygen produced per minute as a measure of photosynthesis rate. The student then doubles the concentration of sodium bicarbonate while keeping light intensity constant. (a) Predict what will happen to the rate of oxygen production. (1 mark) (b) Explain this prediction by referring to the Calvin cycle. (2 marks)

CO₂ is the substrate for one specific step in the Calvin cycle. Think about what enzyme fixes CO₂ and what happens to the products when more CO₂ is available.

Model Answer

(a) The rate of oxygen production will increase (up to a point, assuming light is not the limiting factor). (b) CO₂ is fixed in the Calvin cycle when it combines with ribulose-1,5-bisphosphate (RuBP) via the enzyme RuBisCO to form 3-phosphoglycerate (3-PGA). When more CO₂ is available, this fixation step occurs at a higher rate, producing more 3-PGA. More 3-PGA is reduced using NADPH and ATP from the light reactions to form G3P, increasing demand on the light-dependent reactions and driving greater water splitting and oxygen release.

1 mark: rate of oxygen production increases (accept: rate of photosynthesis increases) 1 mark: CO₂ is fixed by RuBisCO / combines with RuBP to form 3-PGA; more CO₂ → more 3-PGA produced (accept: CO₂ fixation step runs faster) 1 mark: more 3-PGA reduced to G3P using NADPH/ATP → increased demand on light reactions → more water split → more O₂ released

Biotechnology Applications

5 marks

Question 10 3 marks Medium

Sugarcane processing produces large amounts of bagasse — the fibrous plant material left after juice extraction. Bagasse is rich in cellulose (glucose polymers). Biotechnology companies are developing processes to break bagasse down into glucose, which is then fermented by yeast to produce bioethanol. (a) Name the metabolic pathway in yeast that converts glucose into pyruvate during bioethanol production. (1 mark) (b) Describe what happens to pyruvate after glycolysis in yeast cells under the conditions used for bioethanol production. Include the products formed and whether oxygen is required. (2 marks)

Bioethanol is produced anaerobically. After glycolysis, pyruvate does not enter the mitochondria — think about what yeast does to regenerate NAD⁺ and what small molecules are released.

Model Answer

(a) Glycolysis. (b) Under anaerobic conditions (no oxygen required), yeast converts pyruvate into ethanol (bioethanol) and carbon dioxide through alcoholic fermentation. Pyruvate is first decarboxylated (CO₂ is removed) to form acetaldehyde by pyruvate decarboxylase. Acetaldehyde is then reduced to ethanol by alcohol dehydrogenase, using NADH and regenerating NAD⁺ so glycolysis can continue.

1 mark: glycolysis 1 mark: pyruvate is converted to ethanol and CO₂ (alcoholic fermentation); no oxygen is required (anaerobic conditions) 1 mark: NADH is oxidised to NAD⁺ in the process (accept: acetaldehyde is an intermediate; or: the purpose is to regenerate NAD⁺ to allow glycolysis to continue)

Question 11 2 marks Medium

Scientists want to modify algae so they produce higher amounts of oil for biodiesel production. They are deciding between two methods: traditional chemical mutagenesis (exposing algae to a mutagen that causes random DNA changes) or CRISPR-Cas9 gene editing to specifically activate a gene involved in oil synthesis. (a) State one advantage of using CRISPR-Cas9 over chemical mutagenesis for this purpose. (1 mark) (b) State one environmental concern that would need to be assessed before releasing CRISPR-modified algae into outdoor cultivation ponds. (1 mark)

For (a), think about precision — what is random about mutagenesis that CRISPR avoids? For (b), consider what might happen if modified algae escaped the cultivation pond.

Model Answer

(a) CRISPR-Cas9 is targeted — it edits a specific, known gene — whereas chemical mutagenesis causes random, unpredictable mutations throughout the genome, which may disrupt other essential genes or produce unintended effects. (Accept any one of: greater precision/specificity; no off-target mutations in unrelated genes; predictable outcome.) (b) Any one of: risk of gene transfer to wild algal populations (via horizontal gene transfer or hybridisation) that could spread the modification; potential competitive advantage allowing modified algae to outcompete native species if they escaped; unintended ecological effects on organisms that consume or interact with algae in the local ecosystem.

1 mark: CRISPR is targeted/specific (edits a defined gene) whereas mutagenesis is random (may cause unintended mutations elsewhere) — must imply greater precision 1 mark: any valid environmental concern — gene transfer to wild populations / disruption of native species / ecological impact on food webs / unknown effects on aquatic organisms

Cellular Respiration

4 marks

Question 12 1 mark Easy

During aerobic cellular respiration, pyruvate from glycolysis is converted to acetyl-CoA in the mitochondrial matrix. Acetyl-CoA then enters a cyclic metabolic pathway where it is completely oxidised, and carbon dioxide and reduced coenzymes (NADH and FADH₂) are produced. Name this metabolic pathway.

This cycle is named after the scientist who discovered it, and it involves citric acid as one of its intermediates.

Model Answer

The Krebs cycle (also accepted: citric acid cycle / tricarboxylic acid cycle / TCA cycle).

1 mark: Krebs cycle (accept: citric acid cycle, TCA cycle, tricarboxylic acid cycle)

Question 13 3 marks Hard

The electron transport chain (ETC) in the inner mitochondrial membrane is the final stage of aerobic cellular respiration. (a) Explain the role of oxygen in the electron transport chain. In your answer, refer to electrons and the proton gradient. (2 marks) (b) Explain why water is produced as a byproduct of the electron transport chain. (1 mark)

For (a): what happens to electrons at the END of the chain? What does this have to do with the proton gradient being maintained? For (b): what two things combine at the final electron acceptor?

Model Answer

(a) Oxygen acts as the final electron acceptor at the end of the electron transport chain. Electrons passed along the chain from NADH and FADH₂ are eventually accepted by oxygen at Complex IV. Without oxygen accepting these electrons, the chain would stall — electrons could not continue to flow, NADH and FADH₂ could not be reoxidised, and the protein complexes could no longer pump protons (H⁺) into the intermembrane space. The proton gradient that drives ATP synthase (and therefore ATP production via chemiosmosis) would collapse. (b) When oxygen accepts electrons at Complex IV (cytochrome c oxidase), it combines with those electrons and with protons (H⁺) drawn from the matrix. This produces water: 2H⁺ + 2e⁻ + ½O₂ → H₂O.

1 mark: oxygen is the final electron acceptor at the end of the ETC; without it, electrons cannot flow along the chain (and NADH/FADH₂ cannot be reoxidised) 1 mark: if electrons cannot flow, protons cannot be pumped into the intermembrane space, so the proton gradient and ATP synthesis via chemiosmosis is lost 1 mark: water is produced when oxygen accepts electrons AND combines with protons (H⁺) from the matrix at Complex IV — must mention both electrons and protons combining with oxygen

Scientific Investigation Design

Enzyme Inhibition & Regulation

Photosynthesis

Biotechnology Applications

Cellular Respiration

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