These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. No constants are needed. Rates are in the units given in each question. A calculator is assumed. Keep full calculator values until the last step, then round to the precision of the data. All data sets are fictional.
Question 1 (multiple choice · foundation)
A sodium ion (Na⁺) is smaller than a water molecule, yet it crosses a phospholipid bilayer far less easily. Which statement best explains this?
- (A) Na⁺ is too large to fit between the phospholipid molecules.
- (B) Na⁺ is charged and surrounded by water molecules, so it is repelled by the nonpolar hydrocarbon tails in the middle of the bilayer.
- (C) Na⁺ is attracted to the phosphate heads and becomes permanently stuck to the membrane surface.
- (D) Na⁺ can only cross a membrane using energy from ATP.
Answer and explanation
Answer: (B). A charged ion holds a shell of water molecules. Moving it into the nonpolar core of the bilayer would cost a great deal of energy, so it stays out. It crosses through channel proteins instead.
- (A) The question tells you Na⁺ is smaller than water, which does cross the bilayer slowly. Size is not the reason.
- (C) Ions interact with the polar heads, but they are not permanently bound, and this does not explain the barrier.
- (D) Na⁺ can cross passively through Na⁺ channels, down its gradient. ATP is only needed to pump it against its gradient (Topic 2.8).
Question 2 (multiple choice · foundation)
Which statement about facilitated diffusion is correct?
- (A) It moves substances down their concentration gradient through channel or carrier proteins, without energy input from the cell.
- (B) It can move substances against their concentration gradient if enough carrier proteins are present.
- (C) It uses ATP to change the shape of carrier proteins.
- (D) It is only used to move water.
Answer and explanation
Answer: (A). Facilitated diffusion is passive: the protein provides a hydrophilic route, and the gradient drives the movement.
- (B) More carriers make movement faster, but only down the gradient. Movement against a gradient needs energy.
- (C) Carriers change shape when they bind their substance; no ATP is used in facilitated diffusion.
- (D) Water uses aquaporins, but ions and large polar molecules such as glucose also move by facilitated diffusion.
Question 3 (multiple choice · core)
Fictional cells take up a sugar only through carrier proteins. At a very high outside sugar concentration, the rate of uptake has levelled off at 8.0 units. A second line of the same cells has 50% more of the same carrier protein in each membrane. What rate would you predict for the second line at the same very high concentration?
- (A) 8.0 units
- (B) 12.0 units
- (C) 16.0 units
- (D) 5.3 units
Answer and explanation
Answer: (B). At saturation every carrier is busy, so the rate depends on how many carriers there are. 50% more carriers gives 1.5 × 8.0 = 12.0 units.
- (A) assumes the rate is fixed by the sugar concentration. At saturation, extra sugar does not help, but extra carriers do.
- (C) doubles the rate, which would need twice as many carriers, not 1.5 times as many.
- (D) divides by 1.5 instead of multiplying (8.0 ÷ 1.5 ≈ 5.3). More carriers cannot slow uptake.
Question 4 (data analysis · core)
Fictional cells are placed in a very dilute solution, so water enters them by osmosis. The table shows how fast the cell volume increases at the start.
| Cells | Rate of volume increase / % per second |
|---|---|
| Normal cells | 2.4 |
| Normal cells with an aquaporin blocker | 0.30 |
| Artificial vesicles of pure phospholipid (no proteins) | 0.25 |
(a) Calculate the percentage decrease in rate caused by the blocker. (b) Explain why the normal cells swell so much faster than the blocked cells. (c) Explain why the rate with the blocker is similar to the rate for artificial vesicles, and why neither is zero. (d) A mutant cell line has twice as many aquaporins as normal cells. Predict its rate, showing your reasoning.
Worked solution and suggested Marlbridge rubric
(a) (2.4 − 0.30) ÷ 2.4 × 100 = 2.1 ÷ 2.4 × 100 = 87.5% (about 88%).
(b) Normal cells have aquaporins, water channels that let large numbers of water molecules cross quickly. The blocker closes off this route, so water can only cross more slowly.
(c) Once aquaporins are blocked, the only route left is the phospholipid bilayer itself, which is also the only route in the artificial vesicles. Water is small and uncharged, so it crosses the bilayer slowly but not at zero rate.
(d) Split the normal rate into two parts. Through the bilayer: about 0.30. Through aquaporins: 2.4 − 0.30 = 2.1. Doubling aquaporins doubles only the second part: 0.30 + 2 × 2.1 = 4.5% per second. (Simply doubling 2.4 gives 4.8, which wrongly doubles the bilayer route too.)
| Point | What earns it |
|---|---|
| 1 | 87.5% (accept 87% or 88%) |
| 1 | Aquaporins named as water channels that allow fast water movement |
| 1 | Bilayer named as the remaining route, with water’s small size and lack of charge |
| 1 | 4.5% per second, with only the aquaporin share doubled |
Question 5 (constructed response · stretch)
An artificial vesicle contains 150 mmol L⁻¹ potassium chloride (K⁺ and Cl⁻ ions). It is placed in a solution of 5 mmol L⁻¹ potassium chloride. The vesicle membrane contains only open K⁺ channels.
(a) Predict the direction of net K⁺ movement, and explain why. (b) Explain why the Cl⁻ ions do not move with the K⁺. (c) Describe how the charge across the membrane changes as K⁺ moves. (d) Explain why net K⁺ movement stops before the concentrations inside and outside become equal.
Model answer and suggested Marlbridge rubric
(a) K⁺ moves out of the vesicle, from 150 to 5 mmol L⁻¹, down its concentration gradient, through the K⁺ channels. This is facilitated diffusion.
(b) Cl⁻ is charged, so it cannot cross the hydrophobic core of the bilayer, and the K⁺ channels are specific: they do not let Cl⁻ through. There is no route for Cl⁻.
(c) Each K⁺ that leaves carries a positive charge out and leaves a Cl⁻ behind. The inside becomes negative compared with the outside. The membrane becomes polarized.
(d) As the inside becomes more negative, it attracts positive K⁺ ions back. The charge difference pulls K⁺ in while the concentration gradient pushes it out. When the two balance, there is no net movement, even though K⁺ is still more concentrated inside. (Only a tiny number of ions needs to move to build up this charge difference.)
| Point | What earns it |
|---|---|
| 1 | K⁺ moves out, down its gradient, through channels |
| 1 | Cl⁻ blocked: charged, so excluded by the bilayer, and channels are specific |
| 1 | Inside becomes negative; membrane polarized |
| 1 | Electrical attraction of K⁺ back into the negative inside balances the concentration gradient |
Do not award (d) for “the channels close”: the question states they stay open.
Question 6 (prediction · core)
A fictional liver cell takes in glucose through carrier proteins. Outside, glucose is at 5.0 mmol L⁻¹. Inside, glucose stays at about 0.5 mmol L⁻¹, because an enzyme converts glucose into another molecule as soon as it enters.
(a) State the concentration difference that drives glucose into the cell, and explain why glucose needs a carrier protein. (b) A drug blocks the enzyme that converts glucose. Predict what happens to the inside glucose concentration and to the rate of uptake over time. Explain. (c) A different drug blocks the carrier proteins. Predict the effect on glucose uptake, and one effect on the cell.
Model answer and suggested Marlbridge rubric
(a) 5.0 − 0.5 = 4.5 mmol L⁻¹ (outside is 10 times higher). Glucose is large and polar, so it cannot cross the hydrophobic core of the bilayer at a useful rate.
(b) Glucose that enters is no longer used up, so the inside concentration rises towards 5.0 mmol L⁻¹. The gradient becomes smaller, so the rate of uptake falls. When inside and outside are equal, net uptake stops. Facilitated diffusion cannot raise the inside above the outside.
(c) Glucose uptake would almost stop, because there is no other route through the membrane. The cell would have less glucose for respiration, so it would make less ATP.
| Point | What earns it |
|---|---|
| 1 | 4.5 mmol L⁻¹ and glucose large and polar |
| 1 | Inside concentration rises, gradient falls, so uptake slows |
| 1 | Net uptake stops at equal concentrations (cannot exceed outside) |
| 1 | Blocking carriers stops uptake, with a consequence such as less respiration or ATP |
Question 7 (data analysis · stretch)
Two substances, A and B, are taken up by the same fictional cells. Neither uptake is affected by blocking ATP production. Rates are in µmol per minute per gram of cells.
| Outside concentration / mmol L⁻¹ | 2 | 4 | 8 | 16 |
|---|---|---|---|---|
| Rate for A | 1.0 | 2.0 | 4.0 | 8.0 |
| Rate for B | 3.0 | 4.8 | 6.9 | 8.7 |
(a) Calculate rate ÷ concentration for each substance at 2 and at 16 mmol L⁻¹. (b) Identify which substance moves by facilitated diffusion, and justify your answer from (a). (c) Suggest what kind of molecule A might be, and what kind B might be. (d) A drug blocks the protein used by B. Predict the effect on the uptake of A and of B.
Worked solution and suggested Marlbridge rubric
(a) A: 1.0 ÷ 2 = 0.50 and 8.0 ÷ 16 = 0.50. B: 3.0 ÷ 2 = 1.5 and 8.7 ÷ 16 = 0.54.
(b) B moves by facilitated diffusion. For A, the rate is proportional to concentration (the ratio stays at 0.50), so it does not level off: the pattern of simple diffusion. For B, the ratio falls from 1.5 to 0.54: an 8-fold rise in concentration gives less than a 3-fold rise in rate (8.7 ÷ 3.0 ≈ 2.9). B’s rate is levelling off, which shows that a limited number of proteins is becoming saturated. Since ATP is not needed, it is facilitated diffusion, not active transport.
(c) A: a small nonpolar molecule that crosses the bilayer directly (for example a dissolved gas). B: an ion or a large polar molecule (for example glucose or an amino acid) that needs a carrier or channel.
(d) Uptake of B would fall to near zero, because it has no other route. Uptake of A would be unchanged, because it does not use the protein.
| Point | What earns it |
|---|---|
| 1 | All four ratios correct |
| 1 | B identified, with the falling ratio or levelling off as evidence of saturation |
| 1 | Passive (no ATP) distinguished from active transport |
| 1 | Sensible molecule types for both A and B |
| 1 | Correct predictions for both A and B, with reasons |
How did you do?
- Q1 or Q2 wrong: re-read “Structure decides the route” and “Channel proteins and carrier proteins” in the study guide.
- Q3 or Q7 wrong: rework Worked example 1 and look again at Figure 2.
- Q4 or Q6 incomplete: rework Worked example 2; follow the effect from the protein to the cell.
- Q5 incomplete: re-read “Ions, channels and polarized membranes”.
Then tick off the topic checklist.