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O Level Biology Practice Paper 5
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TuitionGoWhere Practice Paper - Biology O-Level
ANSWER KEY AND MARKING SCHEME
Paper: PRACTICE - Cells & Biomolecules
Version: 5 of 5
Total Marks: 60
Section A: Multiple Choice (10 marks)
| Question | Answer | Mark |
|---|---|---|
| 1 | C | 1 |
| 2 | B | 1 |
| 3 | C | 1 |
| 4 | B | 1 |
| 5 | B | 1 |
| 6 | B | 1 |
| 7 | C | 1 |
| 8 | C | 1 |
| 9 | B | 1 |
| 10 | C | 1 |
Marking notes:
- Q1: Nucleus is present in both plant and animal cells. Cell wall, chloroplasts, and large central vacuole are plant-only features.
- Q4: Rough ER has ribosomes attached for protein synthesis and transport. Smooth ER is for lipid synthesis and detoxification.
- Q5: In concentrated salt solution (hypertonic), water leaves the red blood cell by osmosis, causing it to shrink (crenation).
- Q9: Root hair cells have cell walls but no chloroplasts (underground). Palisade and guard cells contain chloroplasts. Cheek cells lack cell walls.
Section B: Structured Questions (30 marks)
Question 11 (5 marks)
(a) Structure X: Chloroplast [1]
Function: Carries out photosynthesis / absorbs light energy to produce glucose/food for the plant [1]
(b) Structure Y: Cell membrane [1]
(c) The plant cell has a cell wall made of cellulose [1] which is rigid and provides structural support, giving the cell a fixed, regular shape [1]. Animal cells lack a cell wall and only have a flexible cell membrane, so they have an irregular shape.
Question 12 (6 marks)
(a) As temperature increases from 10°C to 40°C, the time taken for starch breakdown decreases, indicating that enzyme activity increases [1]. The fastest rate of reaction occurs at 40°C (only 3 minutes) [1]. Above 40°C, enzyme activity decreases sharply, and at 60°C, the enzyme is completely inactive (no breakdown after 30 minutes) [1].
(b) At 60°C, the high temperature causes the enzyme to denature [1]. The shape of the active site is permanently changed, so the substrate (starch) can no longer fit into the active site, and no enzyme-substrate complex can form [1].
(c) Benedict's test [1]
Question 13 (6 marks)
(a) Diagram labels:
- Phospholipid bilayer correctly indicated [1]
- Protein channel correctly indicated [1]
(b) The cell membrane is composed of a phospholipid bilayer [1]. The hydrophobic (water-repelling) tails face inwards, while the hydrophilic (water-attracting) heads face outwards [1]. This arrangement allows small, non-polar molecules (such as oxygen and carbon dioxide) to pass through freely by diffusion, while larger or charged molecules require protein channels or carriers, making the membrane partially permeable [1].
(c) Substances moving across the membrane:
(i) Diffusion: Oxygen / Carbon dioxide / Water (any one correct) [1]
(ii) Active transport: Mineral ions / Glucose / Amino acids (any one correct) [1]
Question 14 (7 marks)
(a) Percentage change in mass = (Final mass - Initial mass) / Initial mass × 100%
= (4.3 - 5.0) / 5.0 × 100% [1]
= -0.7 / 5.0 × 100%
= -14% [1]
(b) The 0.0 mol/dm³ solution is distilled water, which has a higher water potential than the potato cells [1]. Water moves from a region of higher water potential (the solution) to a region of lower water potential (the potato cells) by osmosis [1]. This causes water to enter the potato cells, increasing their mass [1].
(c) At 0.4 mol/dm³ [1]. At this concentration, there was no change in mass, indicating that there was no net movement of water into or out of the potato cells. This means the water potential of the potato cells was equal to the water potential of the surrounding solution [1].
Question 15 (3 marks)
(a) Two features present in a bacterial cell but not in an animal cell:
Feature 1: Cell wall (made of peptidoglycan) [1]
Feature 2: Plasmids / Circular DNA / Flagella / Capsule (any one correct) [1]
(b) One feature present in an animal cell but not in a bacterial cell:
Nucleus (membrane-bound) / Mitochondria / Membrane-bound organelles (any one correct) [1]
Section C: Free Response Questions (20 marks)
Question 16 (8 marks)
(a) The lock-and-key hypothesis states that the active site of an enzyme has a specific three-dimensional shape that is complementary to the shape of its specific substrate [1]. The substrate fits into the active site like a key fits into a lock [1]. When the substrate binds to the active site, an enzyme-substrate complex is formed [1]. This lowers the activation energy required for the reaction, allowing the reaction to proceed more quickly. The products are then released, and the enzyme remains unchanged and can be reused [1].
(b) Each enzyme has an optimum pH at which it works most efficiently [1]. At pH values above or below the optimum, enzyme activity decreases [1]. This is because changes in pH alter the ionic bonds and hydrogen bonds that maintain the specific three-dimensional shape of the enzyme's active site [1]. At extreme pH values, the enzyme becomes denatured — the shape of the active site is permanently changed, and the substrate can no longer bind, so the enzyme loses its catalytic function [1].
Question 17 (9 marks)
(a) Reagent: Biuret solution (sodium hydroxide followed by copper sulfate solution) [1]
Positive result: The solution changes from blue to purple/violet [1]
(b) Importance of proteins:
Reason 1: Growth and repair of body tissues — Proteins are used to build new cells and repair damaged tissues, as they are the main structural component of cells [2].
Reason 2: Enzymes are proteins — Enzymes are biological catalysts that speed up metabolic reactions in the body, and all enzymes are proteins [2].
(Also accept: Formation of hormones, antibodies for immunity, haemoglobin for oxygen transport, etc.)
(c) Table completion:
| Large molecule | Subunit(s) |
|---|---|
| Starch | Glucose [1] |
| Protein | Amino acids [1] |
| Lipid | Glycerol and fatty acids [1] |
Question 18 (3 marks)
(a) Graph requirements:
- Axes correctly labelled: x-axis = pH, y-axis = Volume of oxygen produced (cm³) [1]
- Appropriate scales used on both axes [1]
- All five points plotted accurately and connected with a smooth curve showing a peak at pH 7 [1]
Expected graph shape: The curve should rise from pH 3 to a maximum at pH 7, then decrease through pH 9 to pH 11, showing the optimum pH of catalase is around pH 7 (neutral).
Mark Allocation Summary
| Section | Questions | Marks |
|---|---|---|
| A: Multiple Choice | 1–10 | 10 |
| B: Structured Questions | 11–15 | 30 |
| C: Free Response | 16–18 | 20 |
| Total | 60 |
END OF ANSWER KEY