3.2.1 Surface area to volume ratio exam questions

Question 1

Describe the relationship between size and surface area to volume ratio of organisms. (1 mark)

Answer 1

As size increases, the surface area to volume ratio decreases. Smaller organisms have a larger surface area to volume ratio. (1 mark)

Question 2

Explain why a large surface area to volume ratio is advantageous for single-celled organisms. (2 marks)

Answer 2

Allows efficient diffusion of substances directly across the cell membrane. (1 mark)

Meets the metabolic demands of the organism. (1 mark)

Question 3

Explain three ways in which an insect’s tracheal system is adapted for efficient gas exchange. (3 marks)

Answer 3

Thin walls of tracheoles provide a short diffusion distance to cells. (1 mark)

Highly branched tracheoles increase surface area for gas exchange. (1 mark)

Ventilation maintains a concentration gradient for oxygen and carbon dioxide. (1 mark)

Question 4

Explain how the presence of gills adapts damselfly larvae to their way of life as active hunters. (2 marks)

Answer 4

High metabolic rate requires more oxygen. (1 mark)

Gills increase the surface area for oxygen uptake from water. (1 mark)

Question 5

Describe how the counter-current flow mechanism in fish gills allows efficient oxygen uptake. (2 marks)

Answer 5

Blood and water flow in opposite directions. (1 mark)

Maintains a concentration gradient along the length of the gill lamellae. (1 mark)

Question 6

Describe how gas exchange occurs in single-celled organisms and why this method is not suitable for larger multicellular organisms. (3 marks)

Answer 6

Diffusion occurs directly across the cell surface membrane. (1 mark)

Large organisms have a smaller surface area to volume ratio. (1 mark)

Diffusion would be too slow to meet the metabolic demands of large organisms. (1 mark)

Question 7

Complete the table showing the surface area, volume, and surface area to volume ratio for blocks of different shapes. (2 marks)
Provided Data:
Shape A: Side length = 1 cm.
Shape B: Side length = 2 cm.
Shape C: Side length = 3 cm.

Answer 7

Shape C: Surface area = 96 cm²; Surface area to volume ratio = 1.5:1. (1 mark)

Shape D: Surface area = 136 cm²; Surface area to volume ratio = 2.13:1. (1 mark)

Question 8

Why does increasing the surface area of an exchange surface improve the rate of diffusion? (1 mark)

Answer 8

Increases the amount of substance that can diffuse across the surface per unit time. (1 mark)

Question 9

Explain why the surface area to mass ratio was used instead of the surface area to volume ratio in a study on frog eggs, tadpoles, and adults. (1 mark)

Answer 9

Measuring mass is easier, quicker, and more accurate for irregularly shaped organisms compared to measuring volume. (1 mark)

Question 10

Explain why oxygen uptake is a measure of metabolic rate in organisms. (1 mark)

Answer 10

Oxygen is used in respiration, which is a metabolic process that provides energy (ATP). (1 mark)

Question 11

Use the equation Surface area = 4πr² to calculate the mean diameter of a spherical frog egg with a surface area of 9.73 mm². (2 marks)

Answer 11

Radius (r) = 0.87 mm. (1 mark)

Diameter = 2 × r = 1.75 mm. (1 mark)

Question 12

Calculate the surface area to volume ratio for a cube with a side length of 2 cm. (2 marks)

Answer 12

Surface area = 6 × (side length)² = 24 cm². (1 mark)

Volume = side length³ = 8 cm³. (1 mark)

Surface area to volume ratio = 24:8 = 3:1. (1 mark)

Question 13

Explain the relationship between surface area to volume ratio and the need for specialized gas exchange surfaces in larger animals. (2 marks)

Answer 13

Larger animals have a smaller surface area to volume ratio. (1 mark)

Specialized surfaces, like lungs or gills, increase the area available for gas exchange to meet metabolic demands. (1 mark)

Question 14

Why do small organisms with a large surface area to volume ratio lose heat more quickly? (2 marks)

Answer 14

Large surface area relative to their volume increases the rate of heat loss. (1 mark)

Faster heat loss requires higher metabolic activity to maintain body temperature. (1 mark)

Question 15

Using provided data, calculate the mean surface area of one side of a damselfly larva’s gill lamella and the percentage error. (3 marks)

Provided Data:
Surface area measurements (mm²): 9.5, 10.0, 10.1, 9.8, 9.7.
True value for surface area: 10.5 mm².

Answer 15

Mean surface area = 9.85 mm². (1 mark)

Percentage error = 19%. (1 mark)