Exam questions · Biology
Transport in Cells
- 49 exam questions
- 137 marks
- 50 quick checks
Diffusion
Just this lesson-
1 Name [1 mark]
Name the waste product that diffuses from cells into the blood plasma, to be excreted by the kidneys.
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Model answer
Urea.
Mark scheme
- Urea — 1 mark
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2 Describe [2 marks]
Describe what is meant by diffusion.
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Model answer
The spreading out of particles (of a gas or of a substance in solution), resulting in a net movement from an area of higher concentration to an area of lower concentration.
Mark scheme
- Particles spread out / move randomly — 1 mark
- Net movement from higher to lower concentration — 1 mark
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3 Name [3 marks]
Figure 1 shows particles on both sides of a cell membrane. Name the parts labelled A, B and C.
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Model answer
A is the region of higher concentration. B is the region of lower concentration. C is the cell membrane.
Mark scheme
- A: higher concentration — 1 mark
- B: lower concentration — 1 mark
- C: (cell) membrane — 1 mark
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4 Give [2 marks]
Give two substances that move into or out of cells by diffusion.
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Model answer
Any two from: oxygen, carbon dioxide, urea.
Mark scheme
- Oxygen — 1 mark
- Carbon dioxide — 1 mark
- Urea — 1 mark
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5 State [3 marks]
State three factors that affect the rate of diffusion.
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Model answer
The concentration gradient (difference in concentration); the temperature; the surface area of the membrane.
Mark scheme
- Concentration gradient — 1 mark
- Temperature — 1 mark
- Surface area (of the membrane) — 1 mark
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6 Explain [2 marks]
Explain why diffusion is faster at a higher temperature.
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Model answer
At a higher temperature the particles have more energy, so they move faster and spread out more quickly.
Mark scheme
- Particles have more energy — 1 mark
- So they move faster / spread out faster — 1 mark
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7 Calculate [2 marks]
A cell takes in 36 micrograms of a substance by diffusion in 9 minutes. Calculate the mean rate of uptake, in micrograms per minute.
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Model answer
36 ÷ 9 = 4 micrograms per minute.
Mark scheme
- 36 ÷ 9 — 1 mark
- 4 micrograms per minute (unit needed) — 1 mark
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8 Explain [3 marks]
Explain why oxygen diffuses from the air in the alveoli into the blood.
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Model answer
The concentration of oxygen is higher in the air in the alveoli than in the blood. So there is a net movement of oxygen from the alveoli into the blood, down the concentration gradient.
Mark scheme
- Higher concentration of oxygen in the alveoli — 1 mark
- Lower concentration of oxygen in the blood — 1 mark
- Net movement down the concentration gradient — 1 mark
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9 Explain [6 marks]
Explain how the concentration gradient, the temperature and the surface area of the membrane each affect the rate of diffusion.
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Model answer
The greater the concentration difference between the two sides, the faster the rate of diffusion, because more particles move down the gradient than back against it. The higher the temperature, the faster the rate of diffusion, because the particles have more energy and move faster. The larger the surface area of the membrane, the faster the rate of diffusion, because more particles can cross at the same time.
Mark scheme
- Bigger concentration difference gives faster diffusion — 1 mark
- Because more particles move down the gradient — 1 mark
- Higher temperature gives faster diffusion — 1 mark
- Because particles have more energy / move faster — 1 mark
- Larger surface area gives faster diffusion — 1 mark
- Because more particles can cross at once — 1 mark
Quick check
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1
Which statement describes diffusion?
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C: Net movement of particles from high to low concentration
Diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration.
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2
Which of these moves into cells by diffusion?
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B: Oxygen
Oxygen diffuses into respiring cells. Urea and carbon dioxide diffuse out of them.
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3
What happens to the rate of diffusion if the temperature is increased?
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A: It increases
The particles have more energy and move faster, so diffusion is faster.
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4
Which change would make diffusion across a membrane slower?
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D: A smaller membrane
A smaller surface area lets fewer particles cross at once, so the rate falls.
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5
A cell takes in 50 units of a substance in 10 minutes. What is the rate of uptake?
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B: 5 units per minute
Rate = 50 ÷ 10 = 5 units per minute.
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6
Where does urea diffuse to from the cells?
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C: Into the blood plasma
Urea diffuses from the cells into the blood plasma, and the kidneys then excrete it.
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7
Does diffusion need energy from respiration?
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A: No
No. The particles move on their own, so the cell does not have to use energy.
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8
Why does carbon dioxide diffuse out of respiring cells?
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D: Its concentration is higher inside the cell
Respiration makes carbon dioxide, so its concentration is higher inside the cell than outside, and it moves down the gradient.
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9
What does a steeper concentration gradient do to the rate of diffusion?
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B: It makes it faster
A bigger difference in concentration means more particles move down the gradient, so diffusion is faster.
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10
Once the particles are evenly spread out, what is true?
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C: There is no net movement
The particles are still moving, but as many move one way as the other, so there is no net movement.
Surface Area to Volume Ratio and Exchange Surfaces
Just this lesson-
1 Name [1 mark]
Name the tiny air sacs in the lungs where gas exchange takes place.
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Model answer
Alveoli.
Mark scheme
- Alveoli — 1 mark
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2 Calculate [3 marks]
Figure 1 shows three cubes. Calculate the surface area to volume ratio of the cube with 2 cm sides. Show your working.
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Model answer
Surface area = 6 × 2 × 2 = 24 cm². Volume = 2 × 2 × 2 = 8 cm³. Ratio = 24 ÷ 8 = 3, so 3 : 1.
Mark scheme
- Surface area = 24 cm² — 1 mark
- Volume = 8 cm³ — 1 mark
- 24 ÷ 8 = 3 : 1 — 1 mark
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3 Explain [2 marks]
Explain why a single-celled organism can get all the oxygen it needs by diffusion across its cell membrane.
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Model answer
A single-celled organism has a large surface area to volume ratio. So oxygen can diffuse across the membrane fast enough to meet the needs of the whole cell.
Mark scheme
- Large surface area to volume ratio — 1 mark
- So enough oxygen can diffuse in to meet the needs of the cell — 1 mark
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4 Explain [3 marks]
Explain why a large multicellular organism needs exchange surfaces and a transport system.
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Model answer
A large organism has a small surface area to volume ratio. So diffusion across its outer surface would be too slow to supply the cells in the middle. Exchange surfaces increase the area for exchange, and a transport system carries substances to and from every cell.
Mark scheme
- Small surface area to volume ratio — 1 mark
- Diffusion from the outside is too slow for cells in the middle — 1 mark
- Exchange surface and transport system get substances to all cells — 1 mark
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5 State [4 marks]
State four features that make an exchange surface effective.
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Model answer
A large surface area; a thin membrane (short diffusion path); (in animals) an efficient blood supply; (in animals, for gas exchange) being ventilated.
Mark scheme
- Large surface area — 1 mark
- Thin membrane / short diffusion path — 1 mark
- Efficient blood supply — 1 mark
- Ventilation — 1 mark
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6 Describe [3 marks]
Describe how the villi in the small intestine are adapted for absorbing digested food.
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Model answer
Villi give the small intestine a large surface area. Each has a wall that is one cell thick, giving a short diffusion path. Each has a network of capillaries, which carries absorbed food away and keeps the concentration gradient steep.
Mark scheme
- Large surface area — 1 mark
- Thin wall / short diffusion path — 1 mark
- Good blood supply (capillaries) — 1 mark
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7 Describe [3 marks]
Describe how the gills of a fish are adapted for gas exchange.
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Model answer
The gills are made of many thin filaments, which give a large surface area and a short diffusion path. They have a good blood supply. Water flows over them all the time, which ventilates them and keeps the gradient steep.
Mark scheme
- Many filaments give a large surface area — 1 mark
- Thin surface and a good blood supply — 1 mark
- Water flowing over them (ventilation) — 1 mark
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8 Describe [3 marks]
Describe how a leaf is adapted for the exchange of gases.
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Model answer
The leaf is flat, which gives a large surface area, and thin, which gives a short diffusion path. It has stomata in the surface and air spaces inside, so carbon dioxide can diffuse in and oxygen can diffuse out.
Mark scheme
- Flat, with a large surface area — 1 mark
- Thin, with a short diffusion path — 1 mark
- Stomata and air spaces — 1 mark
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9 Calculate [3 marks]
A block of tissue measures 1 cm by 2 cm by 3 cm. Calculate its surface area to volume ratio, in the form x : 1.
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Model answer
Surface area = 2 × (1×2 + 1×3 + 2×3) = 2 × 11 = 22 cm². Volume = 1 × 2 × 3 = 6 cm³. Ratio = 22 ÷ 6 = 3.7, so 3.7 : 1.
Mark scheme
- Surface area = 22 cm² — 1 mark
- Volume = 6 cm³ — 1 mark
- 22 ÷ 6 = 3.7 : 1 — 1 mark
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10 Explain [6 marks]
Use ideas about surface area to volume ratio to explain why a single-celled organism does not need an exchange organ, but a large animal does.
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Model answer
A single-celled organism has a large surface area to volume ratio. Every part of the cell is close to the surface, so oxygen and food can diffuse in, and waste can diffuse out, fast enough to meet its needs. A large animal has a much smaller surface area to volume ratio. The cells in the middle are a long way from the surface, so diffusion across the outside would be too slow. It therefore has exchange organs such as lungs, which have a large surface area, thin walls and a good blood supply, and a transport system, the blood, to carry oxygen to the cells.
Mark scheme
- Single cell: large surface area to volume ratio — 1 mark
- So diffusion across its surface is fast enough — 1 mark
- Large animal: small surface area to volume ratio — 1 mark
- Cells in the middle are too far from the surface for diffusion alone — 1 mark
- Exchange surface with a large area, thin walls, blood supply — 1 mark
- Transport system carries substances to the cells — 1 mark
Quick check
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1
What is the surface area of a cube with sides of 1 cm?
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B: 6 cm²
A cube has 6 faces, each 1 cm × 1 cm, so the surface area is 6 cm².
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2
As an object gets bigger, what happens to its surface area to volume ratio?
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C: It decreases
Volume increases faster than surface area, so the ratio falls.
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3
A cube has a surface area of 24 cm² and a volume of 8 cm³. What is its ratio?
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A: 3 : 1
24 ÷ 8 = 3, so the ratio is 3 : 1.
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4
Why do large animals need a transport system?
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D: Diffusion alone is too slow to reach all their cells
They have a small surface area to volume ratio, so diffusion alone is too slow to supply all the cells. The blood carries substances around.
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5
Which is NOT a feature of an effective exchange surface?
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C: A thick membrane
Exchange surfaces are thin, to give a short diffusion path. A thick membrane would slow diffusion down.
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6
What are villi?
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B: Folds in the small intestine lining
Villi are finger-like folds in the lining of the small intestine that give a large surface area for absorption.
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7
How are the gills of a fish ventilated?
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A: Water flows over them
Water flows over the gills all the time, which brings fresh oxygen and keeps the gradient steep.
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8
Why do root hair cells have a long thin extension?
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D: To increase the surface area for taking up water and ions
The extension increases the surface area, so more water and mineral ions can be taken up from the soil.
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9
Where does gas exchange take place in the lungs?
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B: In the alveoli
Oxygen and carbon dioxide are exchanged across the thin walls of the alveoli.
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10
One cube has sides of 4 cm. Another has sides of 2 cm. Which statement is correct?
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C: The 2 cm cube has the larger ratio
The 2 cm cube has the larger ratio (3 : 1 compared with 1.5 : 1), because the smaller an object, the larger its surface area compared with its volume.
Osmosis
Just this lesson-
1 Name [1 mark]
Name the type of membrane through which osmosis takes place.
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Model answer
A partially permeable membrane.
Mark scheme
- Partially permeable (membrane) — 1 mark
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2 State [2 marks]
State what is meant by osmosis.
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Model answer
The diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane.
Mark scheme
- Diffusion of water from a dilute to a concentrated solution — 1 mark
- Through a partially permeable membrane — 1 mark
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3 Name [3 marks]
Figure 1 shows two solutions separated by a membrane. Name the parts labelled A, B and C.
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Model answer
A is the dilute solution. B is the concentrated solution. C is the partially permeable membrane.
Mark scheme
- A: dilute solution — 1 mark
- B: concentrated solution — 1 mark
- C: partially permeable membrane — 1 mark
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4 Explain [2 marks]
In Figure 1, the sugar molecules do not move across the membrane. Explain why.
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Model answer
The membrane is only partially permeable. The sugar molecules are too large to pass through its tiny holes, but the small water molecules can.
Mark scheme
- Membrane is partially permeable — 1 mark
- Sugar molecules are too big to pass through — 1 mark
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5 Describe [2 marks]
A plant cell is placed in pure water. Describe what happens to the cell.
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Model answer
Water enters the cell by osmosis, and the cell becomes turgid (swollen and firm).
Mark scheme
- Water enters by osmosis — 1 mark
- Cell becomes turgid / swollen and firm — 1 mark
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6 Explain [3 marks]
Explain why a red blood cell placed in pure water may burst, but a plant cell does not.
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Model answer
Water enters both cells by osmosis. A red blood cell has no cell wall, so nothing stops it swelling, and it may burst. A plant cell has a strong cell wall, which stops it bursting.
Mark scheme
- Water enters by osmosis — 1 mark
- Red blood cell has no cell wall, so it may burst — 1 mark
- Plant cell wall prevents bursting — 1 mark
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7 Explain [3 marks]
A plant that has not been watered for several days wilts. Use ideas about osmosis to explain why.
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Model answer
The soil water becomes more concentrated, so water leaves the plant cells by osmosis. The cells lose water and go flaccid, so they can no longer hold the leaves and stems up and the plant wilts.
Mark scheme
- Water leaves the cells by osmosis — 1 mark
- Cells become flaccid / lose firmness — 1 mark
- So stems and leaves are no longer supported (wilting) — 1 mark
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8 Explain [2 marks]
Solution P has a concentration of 0.2 mol/dm³ and solution Q has a concentration of 0.6 mol/dm³. They are separated by a partially permeable membrane. State the direction in which water moves and explain why.
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Model answer
Water moves from P to Q. P is the more dilute solution, and water moves from a dilute solution to a concentrated solution by osmosis.
Mark scheme
- From P to Q (into the more concentrated solution) — 1 mark
- Water moves from dilute to concentrated by osmosis — 1 mark
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9 Explain [4 marks]
A student puts a red blood cell into a very concentrated salt solution. Describe and explain what happens to the cell.
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Model answer
The salt solution is more concentrated than the cytoplasm of the red blood cell. Water leaves the cell by osmosis, moving from the more dilute cytoplasm to the more concentrated solution. The cell loses water, so it shrinks and shrivels.
Mark scheme
- Solution is more concentrated than the cell — 1 mark
- Water leaves the cell by osmosis — 1 mark
- From dilute (cell) to concentrated (outside) — 1 mark
- Cell shrinks / shrivels — 1 mark
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10 Describe and explain [6 marks]
Describe and explain what happens to plant cells when they are placed in (a) a dilute solution and (b) a concentrated solution.
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Model answer
In a dilute solution, the solution outside is more dilute than the cell sap, so water enters the cells by osmosis. The vacuole swells and the cell becomes turgid. The cell wall stops it bursting. In a concentrated solution, the solution outside is more concentrated than the cell sap, so water leaves the cells by osmosis. The vacuole shrinks and the cell becomes flaccid. If a lot of water is lost, the membrane pulls away from the cell wall.
Mark scheme
- Dilute: water enters by osmosis — 1 mark
- Cell becomes turgid / swollen and firm — 1 mark
- Cell wall prevents bursting — 1 mark
- Concentrated: water leaves by osmosis — 1 mark
- Cell becomes flaccid / soft — 1 mark
- Reason: outside solution is more concentrated than the cell sap — 1 mark
Quick check
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1
Osmosis is the movement of which substance?
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A: Water
Osmosis is the diffusion of water through a partially permeable membrane.
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2
In which direction does water move by osmosis?
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C: From a dilute solution to a concentrated solution
Water moves from a dilute solution to a concentrated solution through a partially permeable membrane.
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3
A plant cell is placed in a very dilute solution. What happens?
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B: It becomes turgid
Water enters by osmosis and the cell becomes turgid. The cell wall stops it bursting.
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4
What does flaccid mean?
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D: Soft and limp from losing water
A flaccid cell has lost water and is soft and limp.
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5
Why does a red blood cell burst in pure water, but a plant cell does not?
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C: A red blood cell has no cell wall
A red blood cell has no cell wall, so it cannot resist the pressure as water enters. A plant cell has a strong cell wall.
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6
Two solutions of 0.1 and 0.5 mol/dm³ are separated by a partially permeable membrane. Which way does water move?
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A: From 0.1 into 0.5
Water moves from the more dilute solution (0.1) to the more concentrated one (0.5).
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7
A cell is in a solution of the same concentration as its contents. What happens to the water?
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B: There is no net movement
There is no net movement of water. Water molecules still move both ways, but equally.
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8
Why does a plant wilt when it is short of water?
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D: Its cells become flaccid
Its cells lose water by osmosis and become flaccid, so they no longer hold the plant up.
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9
What does a partially permeable membrane do?
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A: Lets some substances through but not others
It lets small molecules such as water through but stops larger molecules.
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10
A student says: "Water moves by osmosis to where there is more water." What is wrong with this?
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C: It is the wrong way round: water moves from more water to less
Water moves from where there is more water (a dilute solution) to where there is less (a concentrated solution).
Investigating Osmosis in Plant Tissue
Just this lesson-
1 State [2 marks]
State two variables that should be controlled in the investigation.
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Model answer
Any two from: the length (and width) of the potato cylinders; the volume of the solutions; the time the cylinders are left in the solutions; the temperature; the type of potato.
Mark scheme
- Size / length of cylinder — 1 mark
- Volume of solution — 1 mark
- Time in solution — 1 mark
- Temperature / type of potato — 1 mark
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2 Explain [2 marks]
Explain why the potato cylinders are blotted dry before their final mass is measured.
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Model answer
To remove solution from the surface of the cylinder. Any left on the surface would add mass, so the result would not be due only to osmosis.
Mark scheme
- To remove liquid from the surface — 1 mark
- Because it would add extra mass / give an inaccurate result — 1 mark
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3 Calculate [2 marks]
A potato cylinder has a mass of 2.50 g at the start and 2.85 g at the end. Calculate the percentage change in mass.
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Model answer
(2.85 − 2.50) ÷ 2.50 × 100 = +14%.
Mark scheme
- 0.35 ÷ 2.50 — 1 mark
- +14 (%) — 1 mark
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4 Calculate [3 marks]
A potato cylinder has a mass of 3.20 g at the start and 2.72 g at the end. Calculate the percentage change in mass.
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Model answer
(2.72 − 3.20) ÷ 3.20 × 100 = −15%.
Mark scheme
- Change = −0.48 g — 1 mark
- −0.48 ÷ 3.20 × 100 — 1 mark
- −15 (%) — 1 mark
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5 Explain [3 marks]
Figure 1 shows the results of an investigation into the effect of sugar concentration on potato cylinders. Estimate the concentration at which there is no change in mass, and explain what this tells you.
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Model answer
About 0.33 mol/dm³ (accept 0.30 to 0.35), where the line crosses 0% change. At this concentration the solution is the same as the concentration of the potato cells, so there is no net movement of water.
Mark scheme
- 0.30 to 0.35 mol/dm³ — 1 mark
- Same concentration as the potato cells — 1 mark
- So no net movement of water by osmosis — 1 mark
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6 Use the graph [2 marks]
Use Figure 1 to give the percentage change in mass in the solution of concentration 0.6 mol/dm³, and say what this shows about the movement of water.
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Model answer
−12% (accept −11 to −13). A loss of mass shows that water moved out of the potato by osmosis.
Mark scheme
- −12% — 1 mark
- Water left the potato (by osmosis) — 1 mark
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7 Explain [2 marks]
The potato cylinders in the most concentrated solutions lost mass. Explain why.
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Model answer
The solution was more concentrated than the cell sap in the potato. Water moved out of the potato cells into the solution by osmosis, so the mass fell.
Mark scheme
- Solution more concentrated than the potato cell sap — 1 mark
- Water left the cells by osmosis — 1 mark
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8 Explain [2 marks]
A student calculates the percentage change in mass for each cylinder instead of just the change in mass. Explain why.
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Model answer
The cylinders do not all start with exactly the same mass, so a percentage of the starting mass allows a fair comparison.
Mark scheme
- Cylinders have different starting masses — 1 mark
- Percentage change allows a fair comparison — 1 mark
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9 Suggest [2 marks]
Suggest two ways in which the student could make the results more reliable.
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Model answer
Repeat the investigation with several cylinders in each solution and calculate a mean; check for anomalous results; keep the control variables the same each time.
Mark scheme
- Repeat and calculate a mean — 1 mark
- Spot and deal with any anomalous results / control variables carefully — 1 mark
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10 Describe [6 marks]
Describe how you would carry out an investigation into the effect of sugar concentration on the mass of potato cylinders.
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Model answer
Cut cylinders of potato of the same length with a cork borer. Measure the mass of each cylinder. Put the same volume of different concentrations of sugar solution into test tubes, including pure water. Put one cylinder in each tube and leave them for the same length of time at the same temperature. Take the cylinders out, blot them dry and measure the mass of each again. Calculate the percentage change in mass for each, and repeat the investigation to calculate a mean.
Mark scheme
- Cut identical cylinders of potato — 1 mark
- Measure the starting mass — 1 mark
- Same volume of different concentrations of solution — 1 mark
- Same time (and temperature) in the solutions — 1 mark
- Blot dry and measure the final mass — 1 mark
- Calculate the percentage change / repeat and take a mean — 1 mark
Quick check
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1
What is the independent variable in the potato investigation?
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B: The concentration of the solution
The independent variable is the one you change: the concentration of the solution.
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2
What is the dependent variable?
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A: The percentage change in mass
The dependent variable is what you measure: the percentage change in mass of the potato.
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3
A potato cylinder gains mass in a solution. What does this show?
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C: The solution was more dilute than the cell sap
The potato gained water by osmosis, so the solution was more dilute than the potato cell sap.
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4
A cylinder goes from 2.00 g to 1.80 g. What is the percentage change in mass?
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D: −10%
The change is −0.20 g. Dividing by the start mass gives −0.1, so the change is −10%.
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5
Why is the surface of each cylinder blotted dry?
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B: To remove liquid that would add extra mass
Solution left on the surface would add mass and affect the result.
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6
On the graph, where does the line cross zero change?
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A: At the concentration of the potato cell sap
This is the concentration at which the solution matches the potato cell sap, so there is no net movement of water.
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7
Which would make the results more reliable?
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C: Repeating and calculating a mean
Repeating the test and calculating a mean reduces the effect of random errors and anomalies.
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8
Which of these is a control variable?
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D: The volume of the solution
The volume of the solution has to be the same in every tube. The concentration is what is changed, and the mass is what is measured.
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9
Why is pure water used as one of the solutions?
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A: It is the most dilute solution and gives the zero concentration
Pure water is the most dilute solution, so it shows the biggest gain in mass, and gives a zero point on the concentration scale.
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10
A student finds the line crosses zero at 0.3 mol/dm³ and wants a more accurate value. What should they do?
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B: Use extra solutions with concentrations close to 0.3
Using more solutions close to 0.3, such as 0.25, 0.30 and 0.35 mol/dm³, narrows down where the line crosses zero.
Active Transport
Just this lesson-
1 State [1 mark]
State what is needed for active transport.
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Model answer
Energy (from respiration).
Mark scheme
- Energy from respiration — 1 mark
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2 Describe [2 marks]
Describe what is meant by active transport.
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Model answer
The movement of substances from a more dilute solution to a more concentrated solution (against a concentration gradient), using energy from respiration.
Mark scheme
- Movement from dilute to concentrated / against a concentration gradient — 1 mark
- Needs energy from respiration — 1 mark
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3 Name [4 marks]
Figure 1 shows ions being taken into a cell. A is a carrier protein. Name the structure B, name the process shown, and explain how you know it is that process.
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Model answer
B is a mitochondrion. The process is active transport. The ions move from a lower to a higher concentration, which is against the concentration gradient, and the mitochondrion supplies energy from respiration, which diffusion would not need.
Mark scheme
- B: mitochondrion — 1 mark
- Active transport — 1 mark
- Ions move from lower to higher concentration / against the gradient — 1 mark
- Needs energy from respiration — 1 mark
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4 Give [1 mark]
Give one example of active transport in a plant.
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Model answer
Mineral ions being taken into root hair cells from the soil.
Mark scheme
- Mineral ions into root hair cells — 1 mark
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5 Explain [3 marks]
Explain why root hair cells take up mineral ions by active transport and not by diffusion.
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Model answer
The concentration of mineral ions in the soil solution is lower than in the root hair cell. Diffusion only moves substances from a higher to a lower concentration. Active transport can move ions against the concentration gradient, using energy.
Mark scheme
- Lower concentration of ions in the soil than in the cell — 1 mark
- Diffusion would not move ions into the cell / only goes down the gradient — 1 mark
- Active transport moves them against the gradient using energy — 1 mark
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6 Explain [2 marks]
Sugar is absorbed from the gut into the blood by active transport. Explain why the sugar must be absorbed.
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Model answer
Sugar molecules are used for cell respiration, so the body needs to absorb them.
Mark scheme
- Sugar is used for respiration — 1 mark
- To release energy for the body's cells — 1 mark
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7 Explain [3 marks]
Root hair cells contain large numbers of mitochondria. Explain why.
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Model answer
Mitochondria release energy by respiration. Root hair cells take up mineral ions by active transport, which needs energy. So they need many mitochondria to supply it.
Mark scheme
- Mitochondria release energy / respiration — 1 mark
- Active transport needs energy — 1 mark
- Root hair cells take in ions by active transport — 1 mark
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8 Explain [3 marks]
A farmer notices that plants in waterlogged soil take up fewer mineral ions, although the soil contains plenty. Use ideas about active transport to explain why.
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Model answer
Waterlogged soil contains little oxygen, so the root cells respire less. They release less energy, and active transport of mineral ions needs energy. So fewer ions are taken up.
Mark scheme
- Little oxygen in the waterlogged soil, so less respiration — 1 mark
- Less energy released — 1 mark
- Active transport of ions needs energy, so less uptake — 1 mark
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9 Compare [3 marks]
Give three differences between diffusion and active transport.
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Model answer
Diffusion moves substances from a higher to a lower concentration, but active transport moves them from a lower to a higher concentration. Diffusion does not need energy, but active transport needs energy from respiration. Diffusion goes down the concentration gradient, but active transport goes against it.
Mark scheme
- Direction: high to low (diffusion) compared with low to high (active transport) — 1 mark
- Diffusion needs no energy, active transport does — 1 mark
- Down the gradient compared with against the gradient — 1 mark
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10 Compare [6 marks]
Compare diffusion, osmosis and active transport.
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Model answer
Diffusion is the net movement of particles from a higher to a lower concentration, down the concentration gradient, and needs no energy. Osmosis is the diffusion of water only, from a dilute solution to a concentrated solution through a partially permeable membrane, and needs no energy. Active transport moves substances from a more dilute to a more concentrated solution, against a concentration gradient, and needs energy from respiration.
Mark scheme
- Diffusion: particles from high to low concentration — 1 mark
- Diffusion needs no energy — 1 mark
- Osmosis: water only — 1 mark
- Osmosis: dilute to concentrated through a partially permeable membrane — 1 mark
- Active transport: dilute to concentrated (against the gradient) — 1 mark
- Active transport needs energy from respiration — 1 mark
Quick check
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1
What does active transport need?
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C: Energy from respiration
Active transport needs energy from respiration, because it moves substances against a concentration gradient.
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2
In which direction does active transport move substances?
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A: From low concentration to high concentration
Active transport moves substances from a more dilute to a more concentrated solution, against the concentration gradient.
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3
Which structures supply the energy for active transport?
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D: Mitochondria
Mitochondria release energy from respiration, so cells that do a lot of active transport have many of them.
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4
Which of these is an example of active transport?
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B: Mineral ions moving into root hair cells
Root hair cells take in mineral ions from the very dilute soil solution by active transport.
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5
Why do plants need mineral ions?
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A: For healthy growth
Plants need ions for healthy growth, for example nitrate to make proteins and magnesium to make chlorophyll.
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6
Sugar concentration is lower in the gut than in the blood. How does sugar move into the blood?
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C: By active transport
It has to move against the concentration gradient, so it is absorbed by active transport.
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7
Which process needs no energy from the cell?
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D: Diffusion and osmosis
Diffusion and osmosis are passive processes, which move things down a gradient without energy from the cell.
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8
What happens to active transport if respiration in a cell stops?
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B: It stops or slows down
Active transport needs energy from respiration, so it stops or slows down.
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9
Which statement about osmosis is correct?
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A: It needs no energy and moves only water
Osmosis is the diffusion of water from a dilute solution to a concentrated solution through a partially permeable membrane. It needs no energy.
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10
A cell takes in a substance from a place where it is at a lower concentration. Which is the most likely process?
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C: Active transport
Moving a substance from a lower to a higher concentration is against the gradient, so it must be active transport.