Ch 2 · 4.2 Movement into and out of cells · AS & A Level Biology 9700
Topic 2 of 2 in Cell membranes and transport — AS & A Level Biology 9700 — 78 lessons.
Introduction
Cells must exchange gases, nutrients and waste with their surroundings fast enough to keep pace with metabolism, yet the cell surface membrane is a barrier by design. Cambridge examiners expect four named routes of exchange to be distinguished precisely by energy requirement, direction relative to the gradient, and the proteins involved, plus fluent use of water potential to explain osmosis, and the standard practicals with agar, Visking tubing and plant tissue. (Syllabus section 4.2.)
Key ideas — Diffusion, facilitated diffusion and active transport
Diffusion is the net movement of molecules or ions from a region of higher concentration to a region of lower concentration, down a concentration gradient, as a result of random motion; it needs no metabolic energy. Small non-polar molecules such as O_2 and CO_2 diffuse directly through the phospholipid bilayer. Facilitated diffusion moves polar molecules and ions down their concentration gradient through specific channel proteins (water-filled pores for ions and small polar molecules) or carrier proteins (which bind the substance and change shape), and also needs no ATP. Active transport moves substances against their concentration gradient using carrier proteins that hydrolyse ATP to change shape, for example the sodium–potassium pump.
🧑🏫 Teacher's explanation
Imagine a crowded shopping mall. If everyone is rushing from a packed corridor into an empty hallway, that is like diffusion—it happens naturally because people move from where it is crowded to where it is not. If people need to use a specific revolving door to move through a wall, that is facilitated diffusion; the door is the protein that helps them through, but they are still moving from a crowd to an empty space.
📖 Definitions
🔍 Worked example — Easy
A cell has a glucose concentration of 2.0 mol dm^-3 inside and 0.5 mol dm^-3 outside. Calculate the percentage increase in concentration if the external concentration increases to 0.8 mol dm^-3.
🔍 Worked example — Standard
A cube of agar containing phenolphthalein indicator is placed in a solution of sodium hydroxide. The cube has a side length of 1 cm. After 10 minutes, the NaOH has diffused 0.2 mm into the cube from all faces. Calculate the volume of the cube that has changed color (the 'reacted' volume).
🔍 Worked example — Exam-style
A cell membrane uses active transport to pump 3 sodium ions out for every 1 potassium ion pumped in. If the cell uses 12 ATP molecules to move 4 ions in total, how many sodium ions were moved?
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Surface area to volume ratio
As an object gets bigger, its volume grows faster than its surface area, so the surface area to volume ratio (SA:V) falls. Check with cubes of side L (surface area 6L^2, volume L^3):
🧑🏫 Teacher's explanation
Imagine you are trying to cool down a hot potato. A small, thin slice of potato cools down almost instantly because most of its surface is touching the air. However, a whole, large potato stays hot in the middle for a long time because the heat has to travel through a huge volume to reach the center, and there isn't enough surface area for the heat to escape quickly.
📖 Definitions
🔍 Worked example — Easy
A cubical agar block has a side length of 2 cm. Calculate its surface area to volume ratio.
🔍 Worked example — Standard
A cylindrical cell has a radius r = 0.5 mm and a length h = 2 mm. Calculate the surface area to volume ratio.
🔍 Worked example — Exam-style
An agar block is used to model diffusion. Block A has a side length of 1 cm. Block B has a side length of 4 cm. Calculate the ratio of their SA:V values.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the surface area and volume of the original cube. [2] (b) Explain why the center of the cube remains clear of dye. [2] [4 marks]
Key ideas — Investigating diffusion with agar and Visking tubing
Agar blocks (SA:V): make agar containing a pH indicator, such as phenolphthalein with dilute sodium hydroxide (pink). Cut cubes of side 0.5, 1 and 2 cm and put them in dilute hydrochloric acid. As acid diffuses in, the agar goes colourless. Record the time for each cube to lose its colour completely: the smallest cube, with the largest SA:V, clears first. Keep the acid's concentration, volume and temperature the same.
🧑🏫 Teacher's explanation
Imagine you are trying to cool down a hot potato. If you cut the potato into tiny cubes, it cools down much faster than if you leave it as one large whole potato. This is because the 'surface area' (the skin/exposed parts) is much larger relative to the 'volume' (the inside) when the pieces are small. In biology, cells rely on this principle to move nutrients in and waste out efficiently.
📖 Definitions
🔍 Worked example — Easy
A cube of agar has a side length of 2 cm. Calculate its surface area and its volume.
🔍 Worked example — Standard
Calculate the surface area to volume ratio (SA:V) for the 2 cm cube from the previous example.
🔍 Worked example — Exam-style
An agar block with a side length of 0.5 cm is placed in acid. If the acid diffuses 0.2 cm into the block from all sides, calculate the volume of the block that remains pink (unaffected by acid).
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the surface area and volume of the original cube. (b) Calculate the volume of the cube that has been neutralized by the acid (the volume that is no longer pink). [4 marks]
Key ideas — Osmosis and water potential
Osmosis is 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. Water potential (psi, kPa) is the tendency of water to move out of a solution. Pure water has the highest water potential, 0 kPa; dissolving solutes lowers it, so all solutions have negative values. A more concentrated solution has a more negative water potential.
🧑🏫 Teacher's explanation
Imagine a crowded party in a small room. If one room is packed with people and the next room is almost empty, people will naturally tend to spread out toward the empty room to find more space. In biology, water molecules act like those people. They always want to move from where they are 'crowded' (high water potential) to where they are 'less crowded' (low water potential).
📖 Definitions
🔍 Worked example — Easy
A plant cell is placed in a solution with a water potential of -500 kPa. The water potential inside the cell is -200 kPa. State the direction of the net movement of water.
🔍 Worked example — Standard
Calculate the percentage decrease in surface area to volume ratio (SA:V) when the side length of a cube is doubled.
🔍 Worked example — Exam-style
A student uses a Visking tubing bag containing 10 mol/dm^3 sucrose solution. The bag is placed in distilled water (psi = 0 kPa). Describe the effect on the volume of the solution inside the bag.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Key ideas — Effects on animal cells and plant cells
The difference comes from the cell wall, which resists expansion. This is why animal cells must live in fluid of the right water potential, and why plants need turgid cells for support.
🧑🏫 Teacher's explanation
Imagine a crowded room where people are trying to move from a packed area to an empty area. This is like molecules moving from high to low concentration. In biology, this movement is driven by the 'water potential' (psi), which is essentially the tendency of water to move from a less negative to a more negative value.
📖 Definitions
🔍 Worked example — Easy
A red blood cell is placed in a solution with a water potential of -500 kPa. The water potential inside the cell is -450 kPa. Predict the effect on the cell.
🔍 Worked example — Standard
A cube-shaped agar block has a side length of 1 cm. Calculate its surface area to volume ratio (SA:V).
🔍 Worked example — Exam-style
A plant cell is placed in a solution with a water potential of -800 kPa. The cell's internal water potential is -400 kPa. Describe the state of the cell after 30 minutes.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
\n(a) Predict the effect on the red blood cell. [1]\n(b) Predict the effect on the plant cell. [1] [2 marks]
Key ideas — Estimating the water potential of a plant tissue
Cut identical cylinders of potato (same length and diameter, skin removed), blot and weigh each. Place them in a range of sucrose solutions (e.g. 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 mol dm⁻³) for the same time at the same temperature, then blot and reweigh.
🧑🏫 Teacher's explanation
Imagine you have two sponges: one is soaking wet, and the other is bone dry. If you press them together, water will move from the wet one to the dry one until they are equally damp. In biology, cells act like these sponges. They contain solutes (like sugars and salts) that 'pull' on water.
📖 Definitions
🔍 Worked example — Easy
A potato cylinder has an initial mass of 0.45 g. After being placed in a sucrose solution, its final mass is 0.48 g. Calculate the percentage change in mass.
🔍 Worked example — Standard
A student plots a graph of percentage change in mass (y-axis) against sucrose concentration (x-axis). The line of best fit crosses the x-axis (0% change) at a sucrose concentration of 0.45 mol dm^-3. If a calibration table shows that 0.4 mol dm^-3 = -0.85 MPa and 0.5 mol dm^-3 = -1.05 MPa, estimate the water potential of the potato tissue.
🔍 Worked example — Exam-style
A potato cylinder starts at 2.00 g. In 0.2 mol dm^-3 sucrose, it becomes 2.10 g. In 0.8 mol dm^-3 sucrose, it becomes 1.80 g. Determine if the water potential of the potato is higher or lower than the 0.2 mol dm^-3 solution.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Calculate the percentage change in mass for the 0.4 mol dm⁻³ solution. [1] (b) Describe how the student can use these results to estimate the water potential of the potato tissue. [3] [4 marks]
Key ideas — Endocytosis and exocytosis
Large molecules and particles cannot cross the membrane through proteins, so they are moved in bulk within membrane-bound vesicles. In endocytosis the cell surface membrane folds inward around material outside the cell, pinching off to form a vesicle inside the cytoplasm, as in phagocytosis of bacteria by a white blood cell. In exocytosis a vesicle, often from the Golgi body, moves to the cell surface membrane and fuses with it, releasing its contents outside the cell, as in the secretion of digestive enzymes. Both processes require ATP to move vesicles along the cytoskeleton.
🧑🏫 Teacher's explanation
Imagine you are trying to move a large pile of laundry through a narrow doorway. If the laundry is small (like individual socks), you can pass them through one by one easily. But if you have a massive, bulky duvet, you can't fit it through the door piece by piece; instead, you wrap the duvet in a large plastic bag and carry the whole bag through the door at once. This 'bulk transport' is exactly what cells do when molecules are too large for transport proteins.
📖 Definitions
🔍 Worked example — Easy
A white blood cell (phagocyte) encounters a bacterium. Describe the process by which the bacterium is taken into the cell.
🔍 Worked example — Standard
Compare the energy requirements and mechanisms of facilitated diffusion and exocytosis.
🔍 Worked example — Exam-style
A cell secretes digestive enzymes into the extracellular matrix. Explain the mechanism involved.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
Worked example
Question. Potato cylinders of equal mass were placed in sucrose solutions. In 0.4\ mol dm^-3 sucrose a cylinder of initial mass 5.20\ g had a final mass of 4.94\ g. (a) Calculate the percentage change in mass. [2] (b) State, with a reason, whether the water potential of the sucrose solution was higher or lower than that of the potato cells. [2]
🔍 Worked example — Easy
A cube of agar with a side length of 2 cm is placed in a dye solution. Calculate the surface area to volume ratio (SA:V) of the cube.
🔍 Worked example — Standard
A plant tissue sample has an initial mass of 4.50 g. After being placed in a sucrose solution, its mass decreases to 4.15 g. Calculate the percentage change in mass.
🔍 Worked example — Exam-style
A potato cylinder is placed in a solution. The initial mass is 6.00 g. The final mass is 6.30 g. (a) Calculate the percentage change in mass. (b) Interpret the result in terms of water potential.
⚠️ Common errors
Common mistakes
Practice
Answers