Secondary 4 Pure Biology Preliminary Examination Paper 2

TuitionGoWhere Practice Paper – Pure Biology Secondary 4

PRELIMINARY EXAMINATION – Version 2 – ANSWER KEY

TuitionGoWhere Secondary School (AI)

SECTION A: Structured Questions (40 marks)

1. Fig. 1.1 – Animal cell electron micrograph

(a) Identify organelles P and Q. [2 marks]

• P: Mitochondrion / Mitochondria [1]
• Q: Ribosome / Ribosomes [1]

(b) State one function of organelle P. [1 mark]

• Site of aerobic respiration / Releases energy (ATP) from glucose / Produces ATP [1]

(c) Explain why organelle Q is described as the "site of protein synthesis". [2 marks]

• Ribosomes are where amino acids are linked together / joined to form polypeptide chains / proteins [1]
• mRNA carries the genetic code from the nucleus to the ribosome, where translation occurs [1]

(d) Suggest one reason, visible from the electron micrograph, that supports the statement that the cell is active in producing and secreting enzymes. [1 mark]

• Presence of abundant rough endoplasmic reticulum / many ribosomes / prominent Golgi apparatus / many secretory vesicles [1] (Any one visible feature)

[Total: 6 marks]

2. Potato strip osmosis investigation

(a) Calculate the percentage change in mass for 0.8 mol/dm³ sucrose solution. [2 marks]

• Percentage change = (Final mass – Initial mass) / Initial mass × 100% [1]
• = (–0.30 / 2.50) × 100% = –12.0% [1]

(b) Explain why the potato strip in 0.0 mol/dm³ sucrose solution gained mass. [3 marks]

• The 0.0 mol/dm³ solution (distilled water) has a higher water potential than the potato cells [1]
• Water moves into the potato cells by osmosis, down the water potential gradient [1]
• Water enters the vacuole, causing the cells to become turgid and the strip to gain mass [1]

(c) Estimate the concentration of sucrose solution that has the same water potential as the potato cells. Explain your answer. [2 marks]

• Approximately 0.4 mol/dm³ [1]
• At this concentration, there is no net change in mass, indicating no net movement of water / the water potential of the solution equals the water potential of the potato cells [1]

(d) State one variable that must be kept constant. [1 mark]

• Temperature / Volume of sucrose solution / Surface area of potato strips / Time of immersion / Same potato / Same variety of potato [1] (Any one)

[Total: 8 marks]

3. Root hair cell structure and function

(a) Explain two ways in which the root hair cell is adapted for absorption. [2 marks]

• Adaptation 1: Long, narrow extension / protrusion that increases surface area to volume ratio for faster absorption of water and mineral ions [1]
• Adaptation 2: Thin cell wall that reduces the diffusion distance for water and mineral ions [1] (Accept: Many mitochondria to provide energy for active transport of mineral ions / Large vacuole to maintain water potential gradient)

(b) Name the process and explain why it requires energy. [3 marks]

• Process: Active transport [1]
• Explanation: Mineral ions are absorbed against the concentration gradient (from low to high concentration) [1]
• Active transport requires energy (ATP) from respiration to move ions against the gradient via carrier proteins [1]

(c) Explain why many mitochondria are an advantage for the root hair cell. [2 marks]

• Mitochondria are the site of aerobic respiration, which produces ATP / energy [1]
• This energy is needed for active transport of mineral ions from the soil into the root hair cell [1]

[Total: 7 marks]

4. Enzymes and the lock-and-key hypothesis

(a) Explain the "lock-and-key" hypothesis of enzyme action. [3 marks]

• The active site of the enzyme has a specific three-dimensional shape that is complementary to the shape of the substrate [1]
• The substrate fits into the active site, forming an enzyme-substrate complex [1]
• This brings the substrate molecules into close proximity and lowers the activation energy, allowing the reaction to proceed; the products are then released, and the enzyme remains unchanged [1]

(b)(i) Describe the shape of the curve between 0 °C and 40 °C. [2 marks]

• As temperature increases from 0 °C to 40 °C, the rate of reaction increases [1]
• The rate increases because the kinetic energy of enzyme and substrate molecules increases, leading to more frequent and successful collisions / more enzyme-substrate complexes formed per unit time [1]

(b)(ii) Explain why the rate of reaction decreases sharply above 50 °C. [3 marks]

• Above 50 °C, the enzyme begins to denature [1]
• The high temperature breaks the hydrogen bonds / weak bonds that maintain the specific three-dimensional shape of the enzyme's active site [1]
• The active site loses its complementary shape, so the substrate can no longer bind / fit into the active site, and the rate of reaction decreases [1]

(c) State one factor, other than temperature, that affects the rate of enzyme activity. [1 mark]

• pH / Enzyme concentration / Substrate concentration [1] (Any one)

[Total: 9 marks]

5. Red blood cells and water potential

(a) Identify the condition of the red blood cell in solution A and explain how this condition occurred. [3 marks]

• Condition: Crenation / Crenated / Shrivelled [1]
• Explanation: Solution A has a lower water potential than the cytoplasm of the red blood cell [1]
• Water moves out of the red blood cell by osmosis, down the water potential gradient, causing the cell to shrink and become crenated [1]

(b) Explain why IV fluid must have the same water potential as the patient's blood. [3 marks]

• If the IV fluid has a higher water potential than blood, water would enter the red blood cells by osmosis, causing them to swell and potentially burst (haemolysis) [1]
• If the IV fluid has a lower water potential than blood, water would leave the red blood cells by osmosis, causing them to shrink (crenation) [1]
• Both conditions would damage the red blood cells and impair their function of transporting oxygen; therefore, the IV fluid must be isotonic to prevent net water movement [1]

(c) State what would happen to a plant cell placed in a solution with a lower water potential than the cell sap. Name the process involved. [2 marks]

• Observation: The cytoplasm and vacuole shrink / the cell membrane pulls away from the cell wall / the cell becomes plasmolysed [1]
• Process: Osmosis [1]

[Total: 8 marks]

6. Biological molecules

(a) Identify the type of biological molecule shown in Fig. 6.1. [1 mark]

• Protein / Polypeptide [1]

(b) State the smaller units that make up this molecule. [1 mark]

• Amino acids [1]

[Total: 2 marks]

SECTION B: Data-Based Questions (15 marks)

7. Effect of pH on amylase activity

(a) Plot a graph of the data in Table 7.1. [4 marks]

Marking criteria:

• [1] Axes correctly labelled: x-axis "pH" and y-axis "Colorimeter reading (arbitrary units)"
• [1] Appropriate linear scales chosen, using at least half the grid
• [1] All 8 points plotted accurately (± half a small square)
• [1] Points joined with a smooth curve (not straight lines between points)

(b) Determine the optimum pH for amylase activity. [1 mark]

• Optimum pH = 7 [1]

(c) Explain the results obtained at pH 3 and pH 10. [4 marks]

• At pH 3 (strongly acidic) and pH 10 (strongly alkaline), the enzyme amylase has denatured / lost its activity [1]
• The extreme pH disrupts the hydrogen bonds and ionic bonds that maintain the specific three-dimensional shape of the enzyme's active site [1]
• The active site loses its complementary shape, so the starch substrate can no longer bind / fit into the active site [1]
• Therefore, little or no starch is broken down, resulting in a high colorimeter reading (indicating much starch remains) [1]

(d) Predict and explain the expected colorimeter reading with twice the concentration of amylase at pH 7. [3 marks]

• Prediction: The colorimeter reading would be lower than 10 arbitrary units / close to 0 arbitrary units [1]
• Explanation: At pH 7 (optimum pH), all enzyme molecules are active [1]
• With twice the enzyme concentration, there are more active sites available, so more enzyme-substrate complexes can form per unit time, leading to faster breakdown of starch [1]
• Therefore, less starch remains after 10 minutes, giving a lower colorimeter reading [1]

(e) State one way to improve the reliability of the results. [1 mark]

• Repeat the experiment at each pH and calculate the mean / average colorimeter reading [1] (Accept: Use a water bath to control temperature / Use the same batch of amylase and starch solution)

[Total: 13 marks]

8. Transpiration experiment

(a) State the purpose of the air bubble in this experiment. [1 mark]

• To measure the rate of water uptake / transpiration by tracking the distance moved per unit time [1]

(b) State one precaution when setting up this apparatus. [1 mark]

• Ensure the apparatus is airtight / sealed with Vaseline to prevent leakage / Cut the stem under water to prevent air bubbles entering the xylem / Ensure no air bubbles are trapped in the capillary tube [1] (Any one)

[Total: 2 marks]

SECTION C: Free Response Question (10 marks)

9. DNA structure, function, and genetic engineering

(a) Describe the structure of a DNA molecule. [4 marks]

• DNA is a double helix composed of two polynucleotide strands wound around each other [1]
• Each nucleotide consists of a deoxyribose sugar, a phosphate group, and a nitrogenous base (adenine, thymine, cytosine, or guanine) [1]
• The sugar-phosphate backbones form the outside of the helix, while the nitrogenous bases pair on the inside via hydrogen bonds [1]
• Complementary base pairing occurs: adenine (A) pairs with thymine (T) via two hydrogen bonds; cytosine (C) pairs with guanine (G) via three hydrogen bonds [1]

(b) Explain how the structure of DNA allows it to carry the genetic code and replicate accurately. [4 marks]

• Carrying the genetic code: The sequence of nitrogenous bases along the DNA strand forms the genetic code [1]; each sequence of three bases (a codon) codes for one specific amino acid, determining the sequence of amino acids in a polypeptide / protein [1]
• Accurate replication: During DNA replication, the double helix unwinds and the hydrogen bonds between complementary base pairs break [1]; each strand acts as a template, and free nucleotides pair with exposed bases according to complementary base pairing rules (A with T, C with G), ensuring that two identical DNA molecules are produced [1]

(c) Discuss one application of genetic engineering in medicine and one ethical concern. [2 marks]

• Application: Production of human insulin using genetically modified bacteria – the human insulin gene is inserted into bacterial plasmids, and the bacteria produce human insulin, which is harvested and used to treat diabetes [1]
• Ethical concern: Concerns about "playing God" / altering the genetic makeup of organisms / potential unforeseen long-term effects on ecosystems if genetically modified organisms are released into the environment / issues of informed consent and genetic privacy [1] (Any one valid ethical concern)

[Total: 10 marks]

10. Transport processes and kidney function

(a) Describe the processes of diffusion, osmosis, and active transport. [6 marks]

• Diffusion: The net movement of particles (molecules or ions) from a region of higher concentration to a region of lower concentration, down a concentration gradient [1]; it is a passive process that does not require energy [1]
• Osmosis: The net movement of water molecules from a region of higher water potential to a region of lower water potential, through a partially permeable membrane [1]; it is a passive process that does not require energy [1]
• Active transport: The movement of molecules or ions from a region of lower concentration to a region of higher concentration, against a concentration gradient [1]; it requires energy (ATP) from respiration and involves carrier proteins in the cell membrane [1]

(b) Explain the importance of one of these processes in the functioning of the human kidney. [4 marks]

Example using diffusion:

• In the nephron, during ultrafiltration in the Bowman's capsule, small molecules such as water, glucose, amino acids, urea, and mineral ions are forced out of the glomerular blood capillaries into the Bowman's capsule due to high hydrostatic pressure [1]
• During selective reabsorption in the proximal convoluted tubule, useful substances such as glucose and amino acids are reabsorbed back into the blood by active transport and diffusion [1]
• Diffusion is important because it allows urea and other waste products to remain in the filtrate and be excreted as urine, while water is reabsorbed by osmosis in the collecting duct [1]
• This ensures that toxic nitrogenous waste (urea) is removed from the body while maintaining water and solute balance (osmoregulation) [1]

Example using osmosis:

• Osmosis is important in the collecting duct of the nephron, where water is reabsorbed from the filtrate back into the blood [1]; the medulla has a low water potential, so water moves out of the collecting duct by osmosis, down the water potential gradient [1]; this is controlled by antidiuretic hormone (ADH), which regulates the permeability of the collecting duct walls [1]; this process is essential for osmoregulation, allowing the body to conserve water when dehydrated or excrete excess water when overhydrated [1]

[Total: 10 marks]

END OF ANSWER KEY

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