Edexcel A Level Biology: Gas Exchange Surfaces, Membranes and Osmosis — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Gas Exchange Surfaces, Membranes and Osmosis study guide | Gas Exchange Surfaces, Membranes and Osmosis revision notes

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Section A

1. State Fick’s Law of Diffusion, naming all three variables it relates. [3]

2. Explain how the structure of the mammalian lung is adapted for a rapid rate of gas exchange, linking two named features to Fick’s Law. [4]

Section B

3. A patient develops a respiratory condition that causes fluid to build up in the alveoli, increasing the diffusion distance for oxygen.

(a) Using Fick’s Law, explain the effect on the rate of gas exchange. [3] (b) Suggest why this could impair oxygen supply to the body’s tissues. [2]

4. Describe the fluid mosaic model of cell membrane structure. [4]

5. In Core Practical 3, students investigate the effect of ethanol concentration on the permeability of beetroot cell membranes by measuring pigment leakage.

(a) Predict and explain the effect of increasing ethanol concentration on pigment leakage. [3] (b) State one other variable, besides ethanol concentration, that this practical can investigate. [1]

6. Define osmosis precisely, and explain why it is incorrect to describe it simply as “diffusion of water”. [3]

7. A plant cell is placed in a solution with a lower water potential than the cell’s cytoplasm.

(a) State the direction in which water moves, referring to water potential. [2] (b) Describe and explain what happens to the plant cell as a result. [3]

8. Complete a comparison of diffusion, facilitated diffusion and active transport in terms of energy requirement and direction of movement relative to the concentration gradient. [4]

9. Distinguish between endocytosis and exocytosis, and explain why both are described as forms of “bulk transport”. [3]

10. Explain how carrier proteins differ from channel proteins in membrane transport. [3]

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Answers

1. The rate of diffusion is proportional to (surface area × concentration difference) ÷ diffusion distance [1]. The three variables are surface area, concentration difference, and diffusion distance [1] [1].

2. Any two, e.g.: alveoli provide a large surface area [1], increasing the rate of diffusion (surface area is in the numerator) [1]; thin alveolar walls (one cell thick) give a short diffusion distance [1], increasing the rate (diffusion distance is in the denominator, so a smaller value increases the rate) [1].

3. (a) Fluid build-up increases the diffusion distance [1]. Since diffusion distance is in the denominator of Fick’s Law [1], an increase reduces the rate of diffusion, assuming surface area and concentration difference are unchanged [1]. (b) A reduced rate of oxygen diffusion into the blood means less oxygen is available for aerobic respiration in tissues, which could impair normal cell function [1] [1].

4. A fluid phospholipid bilayer [1] with a varied (“mosaic”) arrangement of embedded proteins [1], cholesterol [1] and other components, forming a dynamic, scientifically modelled (not fixed) structure [1].

5. (a) Pigment leakage increases as ethanol concentration increases [1], because ethanol disrupts the phospholipid bilayer [1], making the membrane more permeable and allowing more pigment (betalain) to leak out of the vacuole [1]. (b) Temperature [1] (also accept: exposure time).

6. Osmosis is the movement of free water molecules through a partially permeable membrane, from a region of higher to lower water potential [1] [1]. Calling it “diffusion of water” is imprecise because osmosis specifically requires a partially permeable membrane and refers only to water, whereas diffusion is a general term that can apply to any particle with or without a membrane [1].

7. (a) Water moves out of the cell [1], from the region of higher water potential (inside the cell) to lower water potential (the solution) [1]. (b) The cell loses water and its cytoplasm shrinks and pulls away from the cell wall [1] — this is called plasmolysis [1], and the cell wall itself remains rigid and does not collapse, unlike in an animal cell [1].

8. Diffusion: no ATP required; moves down the concentration gradient [1] [1]. Facilitated diffusion: no ATP required (uses a carrier/channel protein); moves down the gradient [1] [1]. Active transport: requires ATP; moves against the concentration gradient [1] [1] (any four of these six points).

9. Endocytosis is the bulk uptake of material into a cell by the membrane folding inwards to form a vesicle [1]; exocytosis is the bulk release of material from a cell by a vesicle fusing with the membrane [1]. Both are “bulk transport” because they move large quantities or large particles that cannot pass individually through carrier or channel proteins, and both require ATP [1].

10. Carrier proteins bind to the specific molecule being transported and change shape to move it across the membrane, and can be involved in both facilitated diffusion and active transport [1] [1]. Channel proteins form a fixed pore or channel through which a specific ion or molecule passes, without changing shape, and are only involved in facilitated diffusion [1].

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Where marks are usually lost

• Stating Fick’s Law in words without naming all three variables explicitly.
• Linking a named structure (e.g. alveoli) to “gas exchange” generally rather than to the specific Fick’s Law variable it improves.
• Describing osmosis as “diffusion of water” without mentioning the partially permeable membrane or water potential gradient.
• Confusing plasmolysis (a plant cell in a low water potential solution) with crenation (the equivalent process in an animal cell without a cell wall).
• Saying facilitated diffusion “requires energy” — it does not; only active transport, endocytosis and exocytosis require ATP.
• Describing carrier and channel proteins as interchangeable, rather than distinguishing the shape-change mechanism of carriers from the fixed-pore mechanism of channels.

Approaching “explain using Fick’s Law” questions

Whenever a question describes a change to a gas exchange surface — thickening, damage, disease, or a structural difference between species — the mark scheme rewards identifying which of the three named variables (surface area, concentration difference, diffusion distance) has changed, stating whether that variable has increased or decreased, and only then concluding what happens to the rate. Jumping straight to “the rate decreases” without naming the variable and its position in the equation (numerator or denominator) is the single most common way marks are lost on this outcome, even when the final conclusion is correct.