Secondary 4 Combined Science Biology Preliminary Examination Paper 1

TuitionGoWhere Practice Paper - Combined Science Biology Secondary 4

PRELIMINARY EXAMINATION - ANSWER KEY AND MARKING SCHEME

TuitionGoWhere Secondary School (AI)

Subject: Combined Science Biology (5087/5088) Level: Secondary 4 Paper: Biology Paper 2 (Structured & Free Response) Version: 1 of 5

Section A: Structured Questions

[50 marks]

Question 1: Cell Structure and Function

[8 marks]

(a)(i) Identify cell X and cell Y. [2]

• Cell X: Sperm cell / spermatozoon [1 mark]
• Cell Y: Red blood cell / erythrocyte [1 mark]

(a)(ii) Cell X contains many mitochondria. Explain why this is necessary for its function. [2]

• Sperm cells require large amounts of energy/ATP [1 mark]
• Energy is needed for swimming/movement to reach the egg for fertilisation [1 mark]
• Mitochondria are the site of aerobic respiration where ATP is produced [accept as part of explanation]

(b)(i) Using the data in Table 1.1, explain why cardiac muscle cells have the highest number of mitochondria. [2]

• Cardiac muscle cells contract continuously/rhythmically throughout life without rest [1 mark]
• They require large amounts of energy/ATP for continuous contraction, hence they have the most mitochondria to carry out aerobic respiration [1 mark]

(b)(ii) Suggest why liver cells have more mitochondria than skin cells. [2]

• Liver cells carry out many metabolic processes (e.g., detoxification, protein synthesis, glucose regulation) [1 mark]
• These processes require more energy/ATP than skin cells, which have lower metabolic activity [1 mark]

Question 2: Movement of Substances

[10 marks]

(a) Define the term diffusion. [2]

• Diffusion is the net movement of particles/molecules [1 mark]
• From a region of higher concentration to a region of lower concentration, down a concentration gradient [1 mark]
• [Accept: passive process that does not require energy]

(b)(i) Explain why the contents of the Visking tubing turned blue-black. [2]

• Iodine molecules are small enough to diffuse through the Visking tubing membrane [1 mark]
• Iodine reacts with starch inside the tubing to form a blue-black complex [1 mark]

(b)(ii) Explain why the water outside the Visking tubing did not change colour. [2]

• Starch molecules are too large to pass through the Visking tubing membrane [1 mark]
• Therefore, starch cannot diffuse out of the tubing to react with iodine in the external solution [1 mark]

(c)(i) Name the process by which glucose is absorbed under these conditions. [1]

• Active transport [1 mark]

(c)(ii) Explain why this process requires energy. [3]

• Active transport moves substances against the concentration gradient (from low to high concentration) [1 mark]
• This requires energy in the form of ATP [1 mark]
• Carrier proteins in the cell membrane use ATP energy to transport glucose molecules across the membrane against the gradient [1 mark]

Question 3: Gas Transport and Respiration

[12 marks]

(a) State the process by which oxygen from the air in the alveoli enters the blood capillaries. [1]

• Diffusion [1 mark]

(b) Describe in detail how a molecule of oxygen present in the air breathed into the lungs reaches a muscle cell in the leg. Name the structures involved in the pathway. [6]

• Award marks for correct sequence and naming of structures:

1. Oxygen enters alveoli during inhalation [1 mark]
2. Oxygen diffuses across alveolar epithelium into blood capillaries [1 mark]
3. Oxygen binds to haemoglobin in red blood cells [1 mark]
4. Oxygenated blood travels via pulmonary vein → left atrium → left ventricle → aorta [1 mark]
5. Aorta branches to femoral artery → arterioles → capillaries in leg muscle [1 mark]
6. Oxygen diffuses from capillary into tissue fluid, then across muscle cell membrane into the cell/mitochondria [1 mark]

[Accept any 6 correct steps with named structures. Deduct marks for missing key structures like alveoli, pulmonary vein, aorta, or capillaries.]

(c)(i) Explain why the oxygen concentration decreases from the alveoli (80 units) to the muscle cell (5 units). [3]

• Oxygen is used by muscle cells for aerobic respiration [1 mark]
• Oxygen is consumed in mitochondria to produce ATP/energy [1 mark]
• Continuous consumption maintains a low oxygen concentration in muscle cells, creating a concentration gradient for continued diffusion from blood to cells [1 mark]

(c)(ii) Using the data, explain why oxygen moves from the blood leaving the lungs into the muscle tissue fluid. [2]

• Oxygen concentration in blood leaving lungs is 75 units, while in muscle tissue fluid it is 20 units [1 mark]
• Oxygen diffuses down the concentration gradient from a region of higher concentration (blood, 75 units) to a region of lower concentration (tissue fluid, 20 units) [1 mark]

Question 4: Enzymes and Biomolecules

[10 marks]

(a)(i) Explain why the rate of reaction increases from 0°C to 40°C. [3]

• As temperature increases, kinetic energy of enzyme and substrate molecules increases [1 mark]
• Molecules move faster and collide more frequently [1 mark]
• More successful enzyme-substrate complexes form per unit time, increasing the rate of reaction [1 mark]

(a)(ii) Explain why the rate of reaction decreases after 40°C and reaches zero at 60°C. [3]

• Above the optimum temperature (40°C), the enzyme begins to denature [1 mark]
• High temperatures break the hydrogen bonds and other bonds maintaining the enzyme's three-dimensional shape/active site [1 mark]
• The active site loses its specific shape and can no longer bind to the substrate; at 60°C, all enzyme molecules are denatured, so the rate is zero [1 mark]

(b)(i) Identify the optimum pH for amylase activity. [1]

• pH 7 [1 mark]

(b)(ii) Explain why the time taken to digest starch increases at pH 2 and pH 10. [3]

• Amylase has an optimum pH of 7; at pH 2 (acidic) and pH 10 (alkaline), the pH is far from the optimum [1 mark]
• Extreme pH values cause the enzyme to denature by disrupting the bonds maintaining the enzyme's three-dimensional shape [1 mark]
• The active site changes shape and can no longer bind to starch effectively, so the rate of digestion decreases and takes longer [1 mark]

Question 5: Plant Biology and Photosynthesis

[10 marks]

(a) Write the word equation for photosynthesis. [2]

• Carbon dioxide + water → glucose + oxygen [1 mark for correct reactants, 1 mark for correct products]
• [Must include "light energy" and "chlorophyll" written above/below the arrow for full marks, or accept if stated in words]

(b)(i) Describe the relationship between distance from light and the rate of photosynthesis. [2]

• As the distance from the light source increases, the rate of photosynthesis decreases [1 mark]
• The number of oxygen bubbles produced per minute decreases from 45 at 10 cm to 10 at 50 cm [1 mark for using data]

(b)(ii) Explain why the rate of photosynthesis decreases as the distance from the light source increases. [3]

• Light intensity decreases as distance from the light source increases (inverse square law) [1 mark]
• Light is a limiting factor for photosynthesis; less light energy is available for the light-dependent reactions [1 mark]
• With less light energy, less ATP and reduced NADP are produced, so the rate of the light-independent reactions (Calvin cycle) also decreases, reducing overall photosynthesis rate [1 mark]

(c)(i) Describe and explain the distribution of chloroplasts in the palisade layer and the spongy layer. [3]

• Palisade mesophyll cells contain many more chloroplasts than spongy mesophyll cells [1 mark]
• Palisade cells are located near the upper epidermis and receive the most direct sunlight; more chloroplasts maximise light absorption for photosynthesis [1 mark]
• Spongy mesophyll cells receive less light (filtered through palisade layer); fewer chloroplasts are needed; the loose arrangement and air spaces facilitate gas exchange (CO₂ in, O₂ out) [1 mark]

Section B: Free Response Question

[15 marks]

Question 6: Transport in Humans and Plants

[15 marks]

(a) Describe the pathway taken by a glucose molecule from the small intestine to a muscle cell in the arm. Include the names of all blood vessels and organs involved in the transport pathway. [8]

Marking Scheme:

• Glucose absorbed from small intestine into blood capillaries of villi [1 mark]
• Travels via hepatic portal vein to the liver [1 mark]
• In the liver, excess glucose may be converted to glycogen for storage; remaining glucose enters hepatic vein [1 mark]
• Hepatic vein carries blood to inferior vena cava → right atrium of heart [1 mark]
• Right atrium → right ventricle → pulmonary artery → lungs (for oxygenation, though glucose is not involved in gas exchange) [1 mark]
• Pulmonary vein returns blood to left atrium → left ventricle → aorta [1 mark]
• Aorta branches to subclavian artery → brachial artery → arterioles → capillaries in arm muscle [1 mark]
• Glucose diffuses from capillaries into tissue fluid, then into muscle cells (or via facilitated diffusion/active transport) [1 mark]

[Accept any 8 correct sequential steps with named structures. Deduct marks for incorrect vessel names or sequence errors.]

(b) Explain how the structure of a red blood cell is adapted for its function in oxygen transport. [4]

• Biconcave disc shape increases surface area to volume ratio for faster diffusion of oxygen [1 mark]
• Contains haemoglobin, which binds to oxygen to form oxyhaemoglobin for efficient oxygen transport [1 mark]
• Lacks a nucleus, creating more space for haemoglobin molecules [1 mark]
• Small and flexible, allowing it to squeeze through narrow capillaries [1 mark]

(c) State three differences between the transport systems in plants (xylem and phloem) and the circulatory system in humans. [3]

• Any three of the following (1 mark each):
  1. Plants have two separate transport tissues (xylem and phloem); humans have a single circulatory system with blood vessels
  2. Plants do not have a pump/heart; humans have a heart that pumps blood
  3. Transport in xylem is unidirectional (upwards); transport in phloem is bidirectional; human blood circulates in a closed loop
  4. Xylem transports water and mineral ions; phloem transports sucrose and amino acids; human blood transports oxygen, nutrients, hormones, waste products
  5. Xylem vessels are dead cells with no cytoplasm; human blood vessels are lined with living endothelial cells

OR

Question 7: Homeostasis and Hormonal Control

[15 marks]

(a) Describe and explain the role of hormones in maintaining a relatively constant blood glucose concentration in humans. Include the names of the hormones, the glands that produce them, and their target organs. [8]

Marking Scheme:

• Blood glucose concentration is maintained by negative feedback [1 mark]
• When blood glucose rises (e.g., after a meal): [1 mark for context]
  • Pancreas (islets of Langerhans, β cells) detects high glucose and releases insulin [1 mark]
  • Insulin travels in blood to target organs: liver and muscles [1 mark]
  • Insulin stimulates liver and muscle cells to increase glucose uptake and convert glucose to glycogen for storage [1 mark]
  • Blood glucose concentration decreases back to normal [1 mark]
• When blood glucose falls (e.g., during fasting/exercise): [1 mark for context]
  • Pancreas (α cells) detects low glucose and releases glucagon [1 mark]
  • Glucagon stimulates liver to break down glycogen into glucose and release it into blood [1 mark]
  • Blood glucose concentration increases back to normal [1 mark]

[Accept any 8 correct points covering both hormones, their sources, targets, and effects.]

(b) Explain why it is important for the body to maintain a constant internal temperature. [4]

• Enzymes in the body function optimally at around 37°C [1 mark]
• If temperature rises too high, enzymes denature; their active sites change shape and can no longer catalyse reactions [1 mark]
• If temperature falls too low, enzyme activity slows down due to reduced kinetic energy and fewer successful collisions [1 mark]
• Maintaining constant temperature ensures metabolic reactions proceed at an appropriate rate for normal body function [1 mark]

(c) A person with Type 1 diabetes does not produce enough insulin. Explain the effects this would have on blood glucose concentration after a meal and suggest how this condition can be managed. [3]

• After a meal, blood glucose concentration rises but remains high because without insulin, glucose cannot be taken up efficiently by liver and muscle cells [1 mark]
• Glucose is not converted to glycogen, so blood glucose remains elevated (hyperglycaemia) [1 mark]
• Management: Regular insulin injections (or insulin pump) to replace the missing hormone; monitoring blood glucose levels; controlling carbohydrate intake through diet [1 mark for any one management strategy]

END OF ANSWER KEY

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