Secondary 3 Biology Practice Paper 4

TuitionGoWhere Practice Paper — Biology Secondary 3

Answer Key & Marking Scheme

Paper: Practice Paper — Cells & Biomolecules (Version 4 of 5)
Total Marks: 40

Section A: Multiple Choice & Short Answer [20 marks]

1. (a) [3 marks — 1 mark per correct structure-function pair]

Marking notes: Accept any three correctly identified organelles from the diagram with a valid function. Common mistake: students may label the smooth ER instead of RER — accept if the function matches the structure they have named.

(b) [1 mark] B) Rough endoplasmic reticulum

Explanation: Digestive enzymes are proteins. The RER is the site of protein synthesis (with attached ribosomes) and is therefore most abundant in cells that secrete large amounts of protein-based enzymes.

2. (a) [3 marks — 1 each]

• Solution X: Hypotonic
• Solution Y: Isotonic
• Solution Z: Hypertonic

(b) [2 marks] Solution X is hypotonic (has a higher water potential / lower solute concentration than the cell cytoplasm). Water molecules move into the red blood cells by osmosis (from a region of higher water potential to a region of lower water potential). The cells swell and eventually burst (haemolysis) because the cell membrane cannot withstand the increasing internal pressure.

Marking notes: Award 1 mark for identifying osmosis/water moving in, and 1 mark for explaining the bursting. "Water enters the cell" alone is insufficient without reference to osmosis or water potential gradient.

3. [4 marks — 1 each]

Marking notes: Accept "starch", "glucose/reducing sugar", "protein", and "lipid/fat". Do not award marks for vague answers like "positive result" — students must name the biomolecule.

4. (a) [1 mark] pH 7

(b) [2 marks] As pH increases from 1 to 7, the rate of enzyme activity increases gradually. The enzyme becomes more active as the pH approaches the optimum. At pH 1–3 the enzyme shows very low activity; the rate increases more steeply between pH 4 and 6, reaching a maximum at pH 7.

Marking notes: Award 1 mark for stating the trend (increases) and 1 mark for describing the pattern (gradual/steep increase, or reference to specific data points).

(c) [2 marks] Above pH 8, the enzyme becomes denatured. The alkaline conditions disrupt the hydrogen bonds and ionic bonds that maintain the enzyme's three-dimensional (tertiary) structure. The active site changes shape so that the substrate can no longer fit into it (lock-and-key / induced fit), and the enzyme can no longer catalyse the reaction.

Marking notes: Award 1 mark for "denatured" or equivalent, and 1 mark for explaining the structural change (bonds broken, active site altered). Simply stating "the enzyme stops working" without explanation scores 0.

5. (a) [3 marks] The Visking tubing is partially permeable — it allows water molecules to pass through but not sucrose molecules. The 10% sucrose solution inside the tubing has a lower water potential (higher solute concentration) than the distilled water outside. Water molecules move by osmosis from the distilled water (higher water potential) into the sucrose solution (lower water potential) through the partially permeable membrane. As more water enters the tubing, the volume of liquid inside increases and the level rises.

Marking notes: Award 1 mark for identifying the partially permeable membrane, 1 mark for the water potential gradient, and 1 mark for stating osmosis as the process. "Water moves from high to low concentration" is acceptable but "water potential" terminology is preferred.

(b) [2 marks] The liquid level would remain unchanged / there would be no net movement of water. This is because the water potential inside and outside the Visking tubing would be equal (isotonic conditions). Water molecules would still move in both directions across the membrane, but there would be no net osmosis.

Marking notes: Award 1 mark for "no change / no net movement" and 1 mark for the explanation (equal water potential / isotonic).

6. (a) [2 marks] Amylase works best at 40 °C. This is determined by the shortest time taken for starch to disappear (2 minutes), which indicates the fastest rate of reaction / highest enzyme activity at this temperature.

Marking notes: Award 1 mark for the correct temperature and 1 mark for the reasoning (shortest time = fastest rate).

(b) [3 marks] At 60 °C, the enzyme is beginning to denature — the high temperature is disrupting the bonds that hold the enzyme's tertiary structure together. The active site is partially altered, so the substrate does not fit as well, and the reaction is slower (15 minutes). At 70 °C, the enzyme is completely denatured — the active site has been permanently changed in shape, so the enzyme can no longer catalyse the reaction at all (starch not broken down after 30 minutes).

Marking notes: Award 1 mark for "denaturation" at 60 °C, 1 mark for explaining partial loss of function, and 1 mark for complete denaturation at 70 °C. Students who only say "the enzyme is destroyed" without mentioning structure/bonds lose 1 mark.

(c) [1 mark] Any one of the following: volume of starch solution / volume of amylase / concentration of starch solution / concentration of amylase / time allowed before testing for starch.

Marking notes: Do not accept "temperature" — this is the independent variable being tested.

7. (a) [2 marks — 1 per correct pair]

Other acceptable differences: Cell A has a large central vacuole (both may have this, so it is less ideal); Cell B has more mitochondria; Cell A is found in the leaf, Cell B is found in the root.

Marking notes: The differences must be structural, not functional. "Cell A photosynthesises" is not a structural difference and scores 0.

(b) [2 marks] The long, thin extension greatly increases the surface area of the root hair cell. This increases the surface area-to-volume ratio, which allows more efficient absorption of water and mineral ions from the soil by diffusion and active transport.

Marking notes: Award 1 mark for "increases surface area" and 1 mark for linking this to absorption efficiency. Simply saying "it absorbs more water" without mentioning surface area scores 1 mark only.

8. (a) [1 mark] Carbohydrate (or: polysaccharide)

(b) [1 mark] Glycosidic bond

(c) [3 marks] Starch is first broken down by salivary amylase (in the mouth) and then by pancreatic amylase (in the small intestine) into maltose. Maltose is then broken down by the enzyme maltase (on the membrane of small intestine epithelial cells) into glucose. Glucose is the final product that is absorbed into the blood.

Marking notes: Award 1 mark for naming amylase, 1 mark for maltose as an intermediate product, and 1 mark for maltase → glucose. Students who only say "starch → glucose" without naming enzymes or intermediate products score a maximum of 1 mark.

9. (a) [2 marks — ½ mark each, all four must be correct for full marks]

Explanation: Radioactive amino acids are first incorporated into proteins at the ribosomes (1). The proteins are then transported to the rough endoplasmic reticulum (2) for folding and modification. They move to the Golgi body (3) for further processing and packaging into secretory vesicles (4), which transport the proteins to the cell membrane for secretion.

(b) [2 marks] The Golgi body receives proteins from the rough endoplasmic reticulum, modifies them (e.g., adds carbohydrate groups to form glycoproteins), packages them into secretory vesicles, and directs the vesicles to the cell membrane for exocytosis / secretion out of the cell.

Marking notes: Award 1 mark for "modifies/packages proteins" and 1 mark for "forms secretory vesicles / directs secretion". Simply saying "it transports proteins" is too vague for full marks.

10. (a) [4 marks — 1 each]

• W: Carbohydrate (contains C, H, O only; ratio of H:O is typically 2:1)
• X: Protein (contains C, H, O, N; no phosphorus)
• Y: Nucleic acid (DNA or RNA) (contains C, H, O, N, and P)
• Z: Lipid (fat/oil) (contains C, H, O only; typically has a much higher proportion of H relative to O compared to carbohydrates)

Marking notes: For Z, accept "lipid" or "fat". Some students may argue Z could also be a carbohydrate — accept "lipid" as the intended answer based on typical exam conventions where the H:O ratio differs. If a student writes "carbohydrate" for Z, award the mark only if they have not already used "carbohydrate" for W (i.e., each answer must be unique).

Section B: Structured Response [20 marks]

11. (a) [3 marks] As the enzyme concentration increases, the rate of reaction increases. At 0% enzyme, there is no reaction. At 1% enzyme, the rate is low (reaching 15 cm³ in 5 min). At 2% enzyme, the rate is higher (30 cm³ in 5 min). At 4% enzyme, the rate is highest (55 cm³ in 5 min). This shows a direct / positive relationship between enzyme concentration and rate of reaction — doubling the enzyme concentration approximately doubles the rate of reaction.

Marking notes: Award 1 mark for stating the relationship (positive/direct), 1 mark for referencing specific data from the graph, and 1 mark for explaining that more enzyme means more active sites available for substrate binding.

(b) [2 marks] The curves flatten because the substrate (protein) is being used up / is running out. As the substrate concentration decreases, fewer enzyme-substrate complexes can be formed per unit time, so the rate of reaction decreases. Eventually, all the substrate may be used up and the reaction stops.

Marking notes: Award 1 mark for "substrate is used up / limiting" and 1 mark for explaining the effect on the rate. Simply saying "the reaction stops" without explanation scores 1 mark.

(c) [3 marks] The rate of reaction would be very low / no reaction would occur. Pepsin has an optimum pH of approximately pH 2 (it works in the acidic conditions of the stomach). At pH 8 (alkaline conditions), pepsin would be denatured. The alkaline environment would disrupt the hydrogen and ionic bonds maintaining the enzyme's tertiary structure, changing the shape of the active site so that the protein substrate can no longer bind to it.

Marking notes: Award 1 mark for predicting low/no activity, 1 mark for stating pepsin's optimum pH is ~2, and 1 mark for explaining denaturation at pH 8. Students who only say "the enzyme is denatured" without explaining why (structural change) score a maximum of 2 marks.

(d) [2 marks — 1 each] Any two of the following:

• Temperature (kept at 37 °C)
• pH of the solution (kept at pH 2)
• Volume / concentration of protein substrate
• Total volume of the reaction mixture
• Time interval for measuring the product

Marking notes: Do not accept "enzyme concentration" — this is the independent variable. Do not accept "volume of gas produced" — this is the dependent variable.

12. (a) [1 mark] Plasmolysis

(b) [4 marks] The concentrated salt solution has a lower water potential (higher solute concentration) than the cell sap in the vacuole. Water molecules move out of the cell by osmosis, from the vacuole (higher water potential) through the cell membrane and cell wall (which is fully permeable) to the external solution (lower water potential). As water leaves, the vacuole shrinks and the cell membrane pulls away from the cell wall. The cell wall remains in its original shape because it is rigid and fully permeable.

Marking notes: Award 1 mark for each correct use of the four required terms in a meaningful context:

• Cell wall: 1 mark for stating it remains rigid / does not shrink / is fully permeable
• Cell membrane: 1 mark for stating it pulls away from the cell wall / shrinks
• Vacuole: 1 mark for stating it shrinks / loses water
• Water potential: 1 mark for correctly describing the water potential gradient (water moves from high to low water potential)

Students who use all four terms but incorrectly describe the process (e.g., saying the cell wall shrinks) lose the mark for that term.

(c) [2 marks] Animal cells do not have a cell wall. When placed in a concentrated salt solution, water leaves the cell by osmosis, and the cell shrinks and becomes crenated (wrinkled). Unlike plant cells, animal cells do not undergo plasmolysis because there is no rigid cell wall to pull away from — the entire cell simply shrinks.

Marking notes: Award 1 mark for "no cell wall" and 1 mark for describing the result (cell shrinks / crenates).

(d) [3 marks] The cell would become deplasmolysed / return to its original turgid state. Distilled water has a higher water potential than the cell sap. Water would move into the cell by osmosis, from the distilled water (higher water potential) through the cell membrane into the vacuole (lower water potential). As water enters, the vacuole would expand and the cell membrane would be pushed back against the cell wall. The cell wall prevents the cell from bursting by exerting an inward pressure (wall pressure) that balances the inward osmotic pressure.

Marking notes: Award 1 mark for predicting the cell returns to normal / deplasmolysis, 1 mark for explaining water enters by osmosis (with water potential gradient), and 1 mark for mentioning the cell wall prevents bursting. Students who only say "the cell swells" without explaining osmosis or the role of the cell wall score a maximum of 1 mark.

End of Answer Key

Total: 40 marks