Secondary 4 Combined Science Biology Preliminary Examination Paper 2

TuitionGoWhere Secondary School (AI)

Secondary 4 Combined Science Biology

Preliminary Examination — Version 2 — Answer Key & Marking Scheme

Total Marks: 65

Section A: Structured Questions (20 marks)

1. (a) Red blood cell / Erythrocyte [1]

(b) Cell P contains many mitochondria because it requires a large amount of ATP/energy for its metabolic activities / for active transport / for protein synthesis / for cell division [1]. Cell Q (red blood cell) lacks mitochondria because it does not carry out aerobic respiration / it relies on anaerobic respiration for its energy needs / it lacks a nucleus and most organelles to maximise space for haemoglobin [1]. [2]

2. (a) As temperature increases from 10°C to 40°C, the distance diffused increases / the rate of diffusion increases [1]. [1]

(b) At 50°C, the distance diffused decreased (from 14 mm to 11 mm) [0.5] because the high temperature denatured the protein structures in the agar gel / denatured the dye molecules [0.5], changing the structure of the gel and reducing the rate of diffusion / the kinetic energy of particles was still high but the structural change impeded movement [1]. [2]

3. (a) Diffusion [1]

(b) Haemoglobin [1]

(c) Actively respiring muscle cells use up oxygen for aerobic respiration, so the concentration of oxygen inside the muscle cells is low [1]. This creates a concentration gradient, and oxygen diffuses from the blood (higher concentration) into the muscle cells (lower concentration) down the concentration gradient [1]. [2]

4. (a) Iodine solution / Methylene blue [1]

(b) Onion epidermal cells are found underground / in the bulb where light does not penetrate [1]. Since chloroplasts are required for photosynthesis, and these cells are not exposed to light, they do not need chloroplasts / chloroplasts would serve no function in the absence of light [1]. [2]

5. (a) Carrier protein / Transport protein / Protein channel [1]

(b) Structure X (carrier protein) facilitates the movement of specific substances (e.g., glucose, ions, amino acids) across the cell membrane [1]. It does this by binding to the substance on one side of the membrane, undergoing a conformational change, and releasing the substance on the other side / it provides a hydrophilic channel for polar/charged substances to pass through the hydrophobic core of the membrane [1]. [2]

(c) Diffusion is a passive process that does not require energy and moves substances down a concentration gradient, whereas active transport requires energy (ATP) and moves substances against a concentration gradient. [1]

Section B: Data Interpretation Questions (20 marks)

6. (a) Heart muscle cell [0.5] because it has the highest number of mitochondria (5,000 per cell) [0.5]. Mitochondria are the site of aerobic respiration where ATP/energy is produced [0.5], so a higher number of mitochondria indicates a higher energy demand for continuous contraction of the heart muscle [0.5]. [2]

(b) A sperm cell requires a large number of mitochondria to produce ATP/energy for movement [1]. The sperm must swim a long distance to reach the egg for fertilisation, and this movement requires energy from respiration / the flagellum requires ATP to beat and propel the sperm forward [1]. [2]

(c) A mature red blood cell would have zero (0) mitochondria [1] because mature red blood cells lack a nucleus and most organelles, including mitochondria, to maximise space for haemoglobin for oxygen transport / mature red blood cells rely on anaerobic respiration for their energy needs [1]. [2]

7. (a) The potato strip in 0.0 mol/dm³ sucrose solution gained mass because the solution was hypotonic to the potato cells / the water potential of the solution was higher than the water potential inside the potato cells [1]. Water molecules moved into the potato cells by osmosis from a region of higher water potential (the solution) to a region of lower water potential (inside the cells) down the water potential gradient [1]. This caused the cells to become turgid and the mass of the potato strip to increase [1]. [3]

(b) 0.4 mol/dm³ [0.5] because at this concentration, there was no net movement of water / the percentage change in mass was 0.0% [0.5], indicating that the water potential of the sucrose solution was equal to the water potential inside the potato cells / the solution was isotonic to the potato cells [1]. [2]

(c) The potato strip would lose mass / become flaccid / decrease in mass because water would move out of the potato cells by osmosis into the hypertonic sucrose solution. [1]

8. (a) Graph requirements [4 marks]:

• Axes: x-axis labelled "pH" and y-axis labelled "Volume of oxygen produced (cm³)" [1]
• Scale: Appropriate linear scales on both axes using at least half the grid [1]
• Plotting: All 5 points plotted accurately (± half a small square) [1]
• Line: Smooth curve drawn through the points (not dot-to-dot) [1]

(b) pH 7 [1]

(c) At pH 11, the volume of oxygen produced decreased significantly (to 1 cm³) because the pH is far from the optimum (pH 7) [1]. The high pH (alkaline conditions) disrupts the ionic and hydrogen bonds that maintain the specific three-dimensional shape / tertiary structure of the catalase enzyme [1]. This causes the active site of the enzyme to change shape / become denatured, so the substrate (hydrogen peroxide) can no longer fit into the active site / the enzyme-substrate complex cannot form, and the rate of reaction decreases [1]. [3]

Section C: Extended Response Questions (25 marks)

9. Pathway of oxygen from air to leg muscle cell [6 marks]:

1. Oxygen is inhaled into the lungs and enters the alveoli (air sacs) [1].
2. Oxygen diffuses across the alveolar epithelium / alveolar wall into the blood capillaries surrounding the alveoli, down a concentration gradient [1].
3. In the blood, oxygen binds to haemoglobin in red blood cells to form oxyhaemoglobin [1].
4. The oxygenated blood travels from the lungs via the pulmonary vein to the left atrium of the heart, then to the left ventricle [1].
5. The left ventricle pumps the oxygenated blood into the aorta, which branches into arteries (e.g., femoral artery) that carry blood to the leg [1].
6. In the leg muscle, oxygen diffuses from the blood capillaries into the tissue fluid, then across the cell membrane of the muscle cell, and finally into the mitochondria where it is used for aerobic respiration [1].

Accept any 6 correct and sequential steps with named structures. Award marks for correct sequence and appropriate terminology.

10. (a) Lock and key hypothesis [3 marks]:

• The substrate molecule has a specific shape that is complementary to the shape of the active site of the enzyme, like a key fitting into a lock [1].
• The substrate binds to the active site to form an enzyme-substrate complex [1].
• The reaction takes place at the active site, and the product(s) are released, leaving the enzyme unchanged and free to catalyse another reaction [1].

(b) Effect of temperature on enzyme activity [4 marks]:

• At low temperatures, the enzyme and substrate molecules have low kinetic energy, so they move slowly and collide less frequently, resulting in a low rate of reaction [1].
• As temperature increases to the optimum, the kinetic energy of molecules increases, leading to more frequent successful collisions between enzyme and substrate, so the rate of reaction increases [1].
• At the optimum temperature, the rate of reaction is at its maximum because the enzyme is functioning at its most efficient [1].
• Above the optimum temperature, the high temperature disrupts the hydrogen and ionic bonds that maintain the specific three-dimensional shape / tertiary structure of the enzyme. The active site changes shape (the enzyme is denatured), so the substrate can no longer fit, and the rate of reaction decreases sharply / the enzyme loses its catalytic function [1].

11. (a) Surface area to volume ratio for 4 cm cube [2 marks]:

• Surface area = 6 × (side length)² = 6 × 4² = 6 × 16 = 96 cm² [1]
• Volume = (side length)³ = 4³ = 64 cm³ [0.5]
• Surface area : Volume ratio = 96 : 64 = 1.5 : 1 [0.5]

(b) As the surface area to volume ratio decreases, the time taken for the cubes to turn colourless increases / there is an inverse relationship [1]. A larger surface area to volume ratio results in faster diffusion / shorter time for the acid to diffuse throughout the cube [1]. [2]

(c) Single-celled organisms have a large surface area to volume ratio [1], which means that the distance from the cell surface to the centre of the cell is very short / diffusion is sufficient to supply oxygen and nutrients and remove waste products quickly enough [1]. Therefore, they do not require a specialised transport system (e.g., blood circulatory system) because diffusion alone can meet their metabolic needs [1]. [3]

(d) The small intestine has villi / microvilli that increase the surface area for absorption [1]. (Accept: The small intestine is very long / has a thin epithelial lining / has a rich blood capillary network.)

12. (a) Active transport [3 marks]:

• Active transport is the movement of molecules or ions against a concentration gradient / from a region of lower concentration to a region of higher concentration [1].
• The process requires energy in the form of ATP (produced during respiration) [1].
• It involves carrier proteins in the cell membrane that bind to the specific molecule, change shape using energy from ATP, and transport the molecule across the membrane [1].

(b) Glucose is absorbed by active transport rather than diffusion in the small intestine because the concentration of glucose in the intestinal epithelial cells is often higher than in the lumen of the small intestine / glucose needs to be absorbed against a concentration gradient [1]. Diffusion can only move substances from a region of higher concentration to a region of lower concentration, so active transport is required to absorb all available glucose from the digested food [1]. [2]

(c) The sports drink contains sodium ions because sodium ions are co-transported with glucose during absorption in the small intestine [1]. The movement of sodium ions down their concentration gradient provides the energy / driving force for glucose to be absorbed against its concentration gradient via co-transport proteins / symport [1]. [2]

END OF ANSWER KEY

Marking notes: Award marks for correct scientific terminology, logical sequencing, and accurate explanations. Partial marks may be awarded for partially correct responses where appropriate. Spelling errors should not be penalised unless they change the meaning of a scientific term.