Question 1
Which process needs energy from respiration?
- diffusion
- osmosis
- active transport
- diffusion and osmosis
Show mark scheme for question 1
Answer: C (1)
Question 2
Describe what is meant by osmosis.
Show mark scheme for question 2
- (diffusion / movement) of water (1)
- from a dilute solution to a more concentrated solution (1) allow from a high to a low water concentration
- through a partially permeable membrane (1)
Question 3
Root hair cells take up mineral ions from the soil by active transport. Give two differences between active transport and diffusion, and state one reason why root hair cells need active transport.
Show mark scheme for question 3
- active transport moves substances against the concentration gradient / from low to high concentration, diffusion moves them down the gradient (1)
- active transport needs energy from respiration, diffusion does not (1)
- mineral ions are at a higher concentration inside the cell than in the soil, so diffusion alone could not take them in (1)
Question 4
A cell is modelled as a cube with sides of 4 cm. (a) Calculate the surface area of the cube. (b) Calculate the volume of the cube. (c) Calculate the surface area : volume ratio. (d) A smaller cube with sides of 2 cm has a ratio of 3 : 1. State how the surface area : volume ratio changes as the cube gets larger.
Show mark scheme for question 4
- (a) 6 × 4 × 4 = 96 cm² (1)
- (b) 4 × 4 × 4 = 64 cm³ (1)
- (c) 96 : 64 = 1.5 : 1 (1) allow 3 : 2
- (d) it gets smaller / decreases (1)
Question 5
Explain how two factors, other than surface area to volume ratio, affect the rate at which oxygen diffuses into a cell.
Show mark scheme for question 5
- steeper / bigger concentration gradient (1)
- faster diffusion because more particles move down the gradient than back (1)
- higher temperature (1)
- faster diffusion because particles have more kinetic energy / move faster (1)
- allow shorter distance (1) for faster diffusion because particles travel a shorter way (1)
- Max 4
Question 6
A student fills a piece of Visking tubing with a mixture of starch and glucose solution and places it in a beaker of water. After 30 minutes the water in the beaker contains glucose but no starch. Explain these results.
Show mark scheme for question 6
- Visking tubing is partially permeable / has tiny pores (1)
- glucose molecules are small (1)
- so they pass through the pores / diffuse out down a concentration gradient (1)
- starch molecules are too large to pass through the pores (1)
Question 7
A potato cylinder has a mass of 4.00 g. It is placed in a sugar solution for one hour. Its mass is then 4.40 g. Calculate the percentage change in mass and explain what this shows about the solution.
Show mark scheme for question 7
- (4.40 − 4.00) ÷ 4.00 × 100 (1)
- = +10% / 10% increase (1)
- the solution is more dilute than the potato cell contents, so water moved into the potato by osmosis (1)