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Secondary 1 Science Life Sciences Quiz
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Secondary 1 Science Quiz - Life Sciences (Answer Key)
Total Marks: 40
Section A: Multiple Choice Questions (10 marks)
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B – Photosynthesis is a process carried out by plants and some microorganisms, not a universal characteristic of all living things. The characteristics of living things include: movement, respiration, sensitivity, growth, reproduction, excretion, and nutrition (MRS GREN).
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B – The cell membrane is a partially permeable membrane that controls the movement of substances in and out of the cell. The cell wall provides structural support but is fully permeable.
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B – Plant cells have a cell wall, chloroplasts, a large central vacuole, and lack centrioles. Animal cells lack a cell wall and chloroplasts. Bacterial cells lack a nucleus and membrane-bound organelles. Fungal cells have cell walls made of chitin, not cellulose.
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C – The nucleus contains genetic material (DNA) and controls cell activities. Mitochondria are the site of aerobic respiration. Chloroplasts are the site of photosynthesis. Vacuoles store substances and maintain turgor pressure.
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C – Root hair cells are specialised for absorption of water and mineral salts from the soil. Their elongated projection increases surface area for absorption.
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A – The correct hierarchy of organisation in multicellular organisms: Cell → Tissue → Organ → Organ system → Organism.
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C – Cell R lacks a cell wall, chloroplasts, and large vacuole but has a nucleus. This matches an animal cell. Cell P is a typical plant cell (has all plant structures). Cell Q is a plant cell without chloroplasts (e.g., root cell). Cell S has a cell wall but no chloroplasts or large vacuole (could be fungal or certain plant cells).
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C – Osmosis is the movement of water molecules across a partially permeable membrane from a region of higher water potential to lower water potential. It is a special type of diffusion specific to water, but the question asks what does not involve diffusion. Options A, B, and D all involve diffusion of gases or volatile molecules.
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B – In a concentrated salt solution (hypertonic), water moves out of the red blood cell by osmosis, causing the cell to shrink (crenation). Animal cells lack a cell wall, so they cannot prevent shrinkage.
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C – Active transport requires energy (from respiration) to move substances against their concentration gradient. Diffusion and osmosis are passive processes that do not require energy.
Section B: Structured Questions (18 marks)
Question 11 [5 marks]
(a) [2 marks]
- X: Nucleus (or Cell membrane / Cytoplasm / Mitochondria / Small vacuoles – depending on diagram labels)
- Y: Mitochondria (or other structure as labelled)
Marking note: Award 1 mark for each correct identification based on the diagram provided.
(b) [1 mark]
- The mitochondria is the site of aerobic respiration, where glucose is broken down in the presence of oxygen to release energy (ATP) for cellular activities.
(c) [2 marks]
- Animal cells do not have a cell wall because they need to be flexible and able to change shape for movement, phagocytosis, and formation of specialised tissues. [1]
- A rigid cell wall would prevent these functions. Plant cells have cell walls for structural support and protection since they are stationary. [1]
Common mistake: Stating "animal cells don't need support" without explaining the need for flexibility.
Question 12 [6 marks]
(a) [1 mark]
- The onion epidermal cells appear turgid (firm and swollen), with the cell membrane pressed against the cell wall and the large central vacuole filling most of the cell.
(b) [2 marks]
- The cell membrane pulls away from the cell wall (plasmolysis). [1]
- The cytoplasm and vacuole shrink, and the cell becomes flaccid. [1]
(c) [1 mark]
- Plasmolysis (or osmosis – but plasmolysis is the specific term for this observation in plant cells).
(d) [2 marks]
- The concentrated salt solution has a lower water potential (more negative) than the cell sap inside the vacuole. [1]
- Water moves out of the cell by osmosis, from a region of higher water potential (inside the cell) to a region of lower water potential (salt solution), causing the cytoplasm to shrink and pull away from the cell wall. [1]
Key concept: Water potential gradient drives osmosis. "Concentrated solution = low water potential" is essential.
Question 13 [6 marks]
(a) [1 mark]
- Palisade mesophyll (or palisade layer).
(b) [2 marks]
- Packed with many chloroplasts to absorb maximum light for photosynthesis. [1]
- Cells are elongated and closely packed vertically to reduce air spaces and maximise light absorption / increase surface area for light capture. [1]
Other acceptable answers: Thin cell walls for gas diffusion; positioned near upper epidermis to receive maximum light.
(c) [1 mark]
- Guard cells.
(d) [2 marks]
- The spongy mesophyll has loosely packed cells with large air spaces between them. [1]
- This creates a large surface area for gas exchange (diffusion of CO₂ in and O₂ out) between the air spaces and the mesophyll cells. [1]
Question 14 [5 marks]
(a) [1 mark]
- As temperature increases, the rate of diffusion increases (or time taken decreases).
(b) [1 mark]
- Rate = 1 / time = 1 / 75 s = 0.0133 s⁻¹ (or 0.013 s⁻¹, 1.33 × 10⁻² s⁻¹).
- Accept any reasonable significant figures. Unit: s⁻¹ or arbitrary units.
(c) [2 marks]
- At higher temperatures, particles (water and potassium permanganate ions) have more kinetic energy and move faster. [1]
- This increases the frequency and energy of collisions, causing the particles to spread out more quickly throughout the water. [1]
Key phrase: "Kinetic energy increases → faster movement → faster diffusion."
(d) [1 mark]
- Volume of water / amount of potassium permanganate / size of crystals / stirring (or no stirring) / pressure (any one).
Question 15 [10 marks]
(a) [2 marks]
- Independent variable: Distance from lamp (light intensity) [1]
- Dependent variable: Number of bubbles per minute (rate of photosynthesis) [1]
Marking note: "Light intensity" is acceptable for IV since distance is a proxy for light intensity.
(b) [3 marks] Graph marking criteria:
- Axes labelled correctly with units: X-axis "Distance from lamp (cm)", Y-axis "Number of bubbles per minute" [1]
- Appropriate scales covering all data points (e.g., X: 0–50 cm, Y: 0–50 bubbles/min) [1]
- All 5 points plotted accurately and smooth curve drawn (not straight lines) [1]
Expected graph shape: Curve decreasing steeply at first, then levelling off (inverse relationship).
(c) [2 marks]
- As distance from the lamp increases (light intensity decreases), the rate of photosynthesis decreases. [1]
- The relationship is non-linear: the rate decreases rapidly at first (short distances), then more gradually at greater distances. [1]
Alternative phrasing: "Rate of photosynthesis is directly proportional to light intensity at low intensities, but the relationship curves as light intensity increases." For Sec 1, descriptive answer is sufficient.
(d) [2 marks]
- The number of bubbles per minute would increase. [1]
- Sodium hydrogen carbonate provides more carbon dioxide (a raw material for photosynthesis), which increases the rate of photosynthesis when CO₂ was previously a limiting factor. [1]
Concept: CO₂ concentration as a limiting factor.
(e) [1 mark]
- Bubbles may be of different sizes (volume not constant), so counting bubbles does not give an accurate measure of gas volume. [1]
- Other acceptable answers: Bubbles may be missed / difficult to count at high rates / gas may dissolve in water / some oxygen used by plant for respiration.
Section C: Free Response / Data-Based Questions (12 marks)
Question 16 [7 marks]
(a) [3 marks]
- (i) Producer: Grass [1]
- (ii) Primary consumer: Grasshopper / Rabbit / Deer / Mouse (any one) [1]
- (iii) Tertiary consumer: Snake / Hawk / Fox / Lion / Owl (any one that feeds on secondary consumers) [1]
(b) [1 mark]
- Grass → Grasshopper → Frog → Snake (or Grass → Rabbit → Fox → [tertiary], Grass → Deer → Lion, Grass → Mouse → Owl → [tertiary])
- Must have 4 trophic levels: Producer → Primary → Secondary → Tertiary consumer.
(c) [2 marks]
- Snakes are a food source (prey) for hawks. [1]
- If snake population decreases, hawks have less food available, leading to starvation, reduced reproduction, and population decline. [1]
(d) [1 mark]
- Decomposers break down dead organisms and waste materials, releasing nutrients back into the environment for reuse by producers (plants).
Question 17 [6 marks]
(a) [1 mark]
- pH 7 (shortest time taken = fastest rate).
(b) [1 mark]
- Rate = 1 / time = 1 / 4 min = 0.25 min⁻¹ (or 0.25 arbitrary units).
- Accept 1/4 or 0.25 with unit min⁻¹ or arbitrary units.
(c) [2 marks]
- At pH 3 (acidic) and pH 11 (alkaline), the extreme pH denatures the enzyme by altering the shape of its active site. [1]
- The substrate (starch) can no longer bind effectively, so the rate of reaction decreases. [1]
Key concept: Enzymes have an optimum pH; deviation causes denaturation (loss of 3D shape).
(d) [2 marks]
- No digestion of starch would occur (time taken would be infinite / no change). [1]
- Boiling denatures the enzyme – high temperature destroys the tertiary structure of the protein, permanently altering the active site so it cannot function. [1]
Question 18 [7 marks]
(a) [2 marks]
- P: Stomach [1]
- Q: Small intestine (accept duodenum or ileum) [1]
(b) [1 mark]
- Bile emulsifies fats (breaks large fat globules into smaller droplets) to increase the surface area for lipase action.
(c) [2 marks]
- The small intestine has villi and microvilli which greatly increase the surface area for absorption. [1]
- It has a rich blood supply (capillaries) and lacteals to transport absorbed nutrients away quickly, maintaining a steep concentration gradient. [1]
Other acceptable: Thin epithelium (one cell thick) for short diffusion distance; length of small intestine (~6 m) provides time for absorption.
(d) [2 marks]
- Enzyme: Pepsin [1]
- Optimum pH: pH 1.5 – 2 (acidic) [1]
Question 19 [6 marks]
(a) [1 mark]
- Percentage decrease = ((21 – 16) / 21) × 100% = (5 / 21) × 100% = 23.8% (accept 24% or 23.81%).
(b) [1 mark]
- Nitrogen is inert / not used or produced by the body during respiration; it is simply breathed in and out unchanged.
(c) [2 marks]
- During inhalation, the diaphragm contracts and flattens, and the external intercostal muscles contract, pulling the ribs up and out. [1]
- This increases the volume of the thorax, causing the pressure inside to decrease below atmospheric pressure, so air flows in. [1]
(d) [2 marks]
- Name: Diaphragm [1]
- Action: Contracts and flattens (moves downwards) [1]
Question 20 [5 marks]
(a) [1 mark]
- To exclude air (oxygen) and create anaerobic conditions for the yeast.
(b) [1 mark]
- Glucose → Ethanol + Carbon dioxide (+ Energy)
- Accept: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂ (+ energy)
(c) [1 mark]
- Yeast produces carbon dioxide during anaerobic respiration; CO₂ bubbles through the limewater and reacts with calcium hydroxide to form calcium carbonate (white precipitate), turning it cloudy.
(d) [1 mark]
- Anaerobic respiration in yeast releases much less energy (only 2 ATP per glucose) compared to aerobic respiration in humans (36–38 ATP per glucose).
(e) [1 mark]
- Brewing (beer/wine production) / Baking (bread making) / Biofuel production (any one).