Question 1: Define the term 'surface area to volume ratio' and explain its significance in cell biology.

Answer: Surface area to volume ratio is the ratio of a cell's surface area to its volume. It is significant because it affects the efficiency of nutrient uptake and waste removal; a higher ratio allows for better exchange with the environment.

Question 2: Why do smaller cells generally have a higher surface area to volume ratio compared to larger cells?

Answer: Smaller cells have a higher surface area to volume ratio because as size increases, volume grows faster than surface area, reducing the ratio and making exchange less efficient.

Question 3: Explain how the surface area to volume ratio influences the size of cells in multicellular organisms.

Answer: In multicellular organisms, cells tend to be small to maintain a high surface area to volume ratio, which facilitates efficient exchange of nutrients, gases, and waste products. Larger cells would have a lower ratio, hindering efficient transport and leading to the evolution of specialised structures and division of labour among cells.

Question 1: A spherical cell has a radius of 5 micrometres. Calculate its surface area, volume, and surface area to volume ratio. (Use π ≈ 3.14)

Answer: Surface area = 4πr² = 4 × 3.14 × 5² = 4 × 3.14 × 25 = 314 μm²; Volume = 4/3 π r³ = 4/3 × 3.14 × 5³ = 4/3 × 3.14 × 125 ≈ 523.33 μm³; Ratio = 314 / 523.33 ≈ 0.6 μm⁻¹

Question 2: A cube-shaped cell measures 10 micrometres on each side. Calculate its surface area, volume, and ratio. (Use π ≈ 3.14 for calculations involving spheres only)

Answer: Surface area = 6 × side² = 6 × 10² = 600 μm²; Volume = side³ = 10³ = 1000 μm³; Ratio = 600 / 1000 = 0.6 μm⁻¹

Question 1: Design a simple experiment to investigate how cell size affects the rate of nutrient uptake. Describe the variables involved and safety precautions you would take.

Answer: An experiment could involve using model cells of different sizes (e.g., beads or small spheres) submerged in a nutrient solution with a dye. The independent variable is cell size; the dependent variable is the rate of dye absorption. Control variables include solution concentration and temperature. Safety precautions include wearing gloves and goggles, handling chemicals carefully, and disposing of waste properly.

Question 2: A cell's surface area is 200 μm² and its volume is 400 μm³. If the cell grows so that its surface area doubles and volume triples, what is the new surface area to volume ratio? Comment on the potential impact on cell efficiency.

Answer: New surface area = 2 × 200 = 400 μm²; New volume = 3 × 400 = 1200 μm³; Ratio = 400 / 1200 ≈ 0.33 μm⁻¹. The ratio decreases, indicating reduced efficiency in exchange processes.

Question 3: Discuss how the surface area to volume ratio limits the size of single-celled organisms and how multicellularity helps overcome this limitation.

Answer: Single-celled organisms are limited in size because a low surface area to volume ratio reduces the efficiency of nutrient uptake and waste removal, which can compromise survival. Multicellularity allows organisms to have many small cells, each with a high surface area to volume ratio, thus maintaining efficient exchange across cell membranes while supporting larger overall body sizes.

Question 1: Describe how the concept of surface area to volume ratio is utilised in the design of artificial organs or medical implants.

Answer: Designs focus on increasing surface area (e.g., porous structures) to improve exchange of nutrients and waste, ensuring efficiency and compatibility with the body's functions, similar to how small cells maintain high surface area to volume ratios.

Question 2: In industry, why is surface area to volume ratio important in the production of catalysts or chemical reactions?

Answer: A high surface area to volume ratio allows catalysts to have more active sites for chemical reactions, increasing efficiency and rate of reactions in industrial processes.