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. Data for every question: standard error SE = s ÷ √n, where s is the sample standard deviation; error bars are mean ± 2 SE. A calculator is assumed. Keep full calculator values until the last step. All compounds, cells and data sets are fictional unless named.
Question 1 (multiple choice · foundation)
Artificial vesicles made only of a phospholipid bilayer are placed in a solution containing four substances. Which substance will cross the bilayer most readily?
- (A) O₂
- (B) H₂O
- (C) glucose
- (D) K⁺
Answer and explanation
Answer: (A). O₂ is small and nonpolar, so it dissolves in the hydrophobic interior formed by the fatty acid tails and crosses freely.
- (B) Water does cross, because it is small and uncharged, but only in small amounts, because it is polar.
- (C) Glucose is a large polar molecule. It barely crosses a pure bilayer and needs a transport protein.
- (D) K⁺ is charged. Its charge and water shell keep it out of the hydrophobic interior, so it crosses only through channel or transport proteins, and these vesicles have none.
Question 2 (multiple choice · core)
Which statement best explains why Na⁺ ions cannot cross a phospholipid bilayer without help?
- (A) Na⁺ is charged, and it cannot leave its shell of water molecules to enter the nonpolar interior formed by the fatty acid tails.
- (B) Na⁺ is too large to fit between neighbouring phospholipids.
- (C) The phosphate groups in the heads carry a positive charge and repel Na⁺.
- (D) Cholesterol in the membrane binds Na⁺ ions and holds them.
Answer and explanation
Answer: (A). The barrier is the hydrophobic interior. A charged ion is surrounded by water; entering the nonpolar tails would mean losing those attractions with nothing to replace them, which is very unfavourable.
- (B) is a size argument. A single Na⁺ ion is smaller than an O₂ molecule, which crosses freely. Charge, not size, is the problem.
- (C) is wrong about the charge: phosphate groups are negatively charged. Even if the heads did repel one kind of ion, that could not explain why negative ions such as Cl⁻ are blocked as well. The barrier is the tails, not the heads.
- (D) invents a role for cholesterol. Cholesterol helps keep the membrane stable; it does not trap ions. Membranes without cholesterol also block ions.
Question 3 (multiple choice · core)
A plant cell and a red blood cell are each placed in distilled water. The red blood cell bursts; the plant cell does not. Which statement best explains the plant cell’s survival?
- (A) Water enters the plant cell, which swells and presses against its cell wall; the wall resists further expansion, so the cell becomes firm instead of bursting.
- (B) The cellulose wall is impermeable to water, so no water enters the plant cell.
- (C) The plant cell’s membrane has no hydrophobic interior, so water passes in and out equally.
- (D) Water leaves the plant cell, and the membrane pulls away from the wall.
Answer and explanation
Answer: (A). Water moves into both cells by osmosis. The animal cell has no wall, so it swells until it bursts (osmotic lysis). The plant cell’s wall pushes back as the cell presses on it (turgor pressure), so the cell stops swelling and becomes turgid.
- (B) is wrong because cell walls are porous; water passes through them easily.
- (C) is wrong because every plasma membrane has a hydrophobic interior.
- (D) describes what happens in a more concentrated solution, not in distilled water.
Question 4 (data analysis · core)
A fictional antibiotic stops bacteria from building their peptidoglycan wall. A student hypothesised: “The antibiotic kills bacteria by causing osmotic lysis.” She measured the percentage of cells that lysed (burst) after 2 hours, with four repeats per treatment.
| Treatment | Repeat 1 | Repeat 2 | Repeat 3 | Repeat 4 |
|---|---|---|---|---|
| No antibiotic, dilute medium | 3 | 5 | 4 | 4 |
| Antibiotic, dilute medium | 62 | 70 | 66 | 58 |
| Antibiotic, medium with high sucrose concentration | 9 | 12 | 7 | 12 |
The high-sucrose medium has a solute concentration close to that of the bacterial cytoplasm, so little water moves into the cells.
(a) Calculate the mean, standard error and mean ± 2 SE for the two antibiotic treatments. (b) State a null hypothesis for comparing the two antibiotic treatments, and decide whether to reject it. (c) Explain how the results support the student’s hypothesis.
Worked solution and suggested Marlbridge rubric
(a) Antibiotic, dilute medium. Mean = 256 ÷ 4 = 64.0%. Deviations: −2, 6, 2, −6; squares: 4, 36, 4, 36; sum = 80. s = √(80 ÷ 3) = 5.16%. SE = 5.16 ÷ √4 = 2.58%. Interval: 64.0 ± 5.2, from 58.8% to 69.2%.
Antibiotic, high sucrose. Mean = 40 ÷ 4 = 10.0%. Deviations: −1, 2, −3, 2; squares: 1, 4, 9, 4; sum = 18. s = √(18 ÷ 3) = 2.45%. SE = 2.45 ÷ 2 = 1.22%. Interval: 10.0 ± 2.4, from 7.6% to 12.4%.
(b) H₀: the sucrose concentration of the medium has no effect on the percentage of antibiotic-treated cells that lyse. The intervals (58.8–69.2% and 7.6–12.4%) do not overlap, so reject H₀.
(c) In dilute medium, water enters the cells by osmosis. Without a proper wall to resist swelling, most cells burst. When the medium matches the cytoplasm, little water enters, so even cells with weak walls survive (10%, much nearer the 4.0% with no antibiotic than the 64% in dilute medium). This shows the antibiotic kills mainly by allowing osmotic lysis, not by poisoning the cells directly.
| Point | What earns it |
|---|---|
| 1 | Both means and both SEs correct |
| 1 | Both ± 2 SE intervals correct |
| 1 | Correct null hypothesis and “reject” justified by non-overlapping intervals |
| 1 | Explanation links weak wall + water entry to lysis, using the sucrose result to rule out direct toxicity |
Common error: SE = s ÷ n (5.16 ÷ 4 = 1.29) instead of s ÷ √n. The decision here would not change, but the interval would be wrong.
Question 5 (constructed response · core)
A fictional medicine, compound M, can exist in two forms. Form 1 is small and uncharged with no polar groups exposed. Form 2 has the same size but carries a positive charge.
(a) Predict which form can enter cells by crossing the phospholipid bilayer directly. Explain. (b) A researcher finds that form 2 still enters liver cells quickly, but does not enter artificial vesicles made of pure phospholipid. Propose an explanation. (c) Describe how the researcher could test your explanation.
Model answer and suggested Marlbridge rubric
(a) Form 1. It is small and nonpolar, so it can dissolve in and pass through the hydrophobic interior formed by the fatty acid tails. Form 2 is charged, so it is held in water and cannot enter the nonpolar interior.
(b) Liver cells have an embedded channel or transport protein that carries form 2 across, bypassing the hydrophobic interior. Pure phospholipid vesicles have no proteins, so form 2 cannot cross them.
(c) Any valid test, for example: add the suspected transport protein to vesicles and show that form 2 now enters, compared with protein-free vesicles; or block or remove the protein in liver cells and show that form 2 entry falls, compared with untreated cells. Keep other conditions the same and repeat each treatment.
| Point | What earns it |
|---|---|
| 1 | Form 1 chosen, linked to being small and nonpolar |
| 1 | Form 2 blocked because its charge keeps it out of the hydrophobic interior |
| 1 | Transport or channel protein proposed for liver cells, absent from vesicles |
| 1 | Test with a suitable comparison (with versus without the protein) |
Question 6 (constructed response · stretch)
In a fictional study, plant cells were exposed to labelled, uncharged molecules of different diameters. Researchers measured the percentage of each that reached the outer surface of the plasma membrane within 10 minutes, in cells with their walls intact and in cells whose walls had been removed (kept in a solution that prevents bursting).
| Molecule diameter / nm | 1 | 3 | 5 | 10 |
|---|---|---|---|---|
| Wall intact / % reaching membrane | 100 | 85 | 20 | 0 |
| Wall removed / % reaching membrane | 100 | 100 | 100 | 100 |
(a) Describe the effect of the cell wall, using values from the table. (b) Explain what these data show about the cell wall as a permeability barrier. (c) A student claims: “These data show that the cell wall controls what enters the cytoplasm.” Evaluate this claim.
Model answer and suggested Marlbridge rubric
(a) With the wall intact, small molecules pass freely (100% at 1 nm) but fewer larger ones get through: 85% at 3 nm, 20% at 5 nm and 0% at 10 nm. The largest fall is between 3 and 5 nm (65 percentage points). Without the wall, all sizes reach the membrane.
(b) The wall is porous. Its pores let small molecules through but hold back larger ones, so it acts as a barrier for some substances, mainly by size. The wall-removed cells are the comparison that shows the wall itself causes the blocking.
(c) The claim goes too far. The data only show which molecules reach the outside of the plasma membrane. Whether a molecule then enters the cytoplasm depends on the membrane: its hydrophobic interior and its channel and transport proteins. A 1 nm molecule passes the wall, but if it were charged or large and polar it would still need a protein to cross the membrane. So the wall is a coarse filter; the membrane does the selecting.
| Point | What earns it |
|---|---|
| 1 | Trend described with at least two quoted values and a comparison with wall-removed cells |
| 1 | Wall as a size-based barrier to some substances, with the wall-removed cells as the comparison |
| 1 | Notes the data measure arrival at the membrane, not entry into the cytoplasm |
| 1 | Explains that the membrane’s hydrophobic interior and proteins decide entry |
Question 7 (calculation · stretch)
The table gives fictional relative rates at which five substances cross a pure phospholipid bilayer (arbitrary units). The values are invented, but their order follows real measurements.
| Substance | CO₂ | H₂O | glycerol | glucose | K⁺ |
|---|---|---|---|---|---|
| Relative rate | 1 × 10² | 1 × 10⁻¹ | 1 × 10⁻⁴ | 1 × 10⁻⁶ | 1 × 10⁻¹⁰ |
(a) How many times faster does CO₂ cross than H₂O? How many times faster does H₂O cross than glucose? (b) Glycerol and glucose are both polar and uncharged. Suggest why glycerol crosses 100 times faster. (c) Explain why K⁺ has the lowest rate, and state how cells move K⁺ across their membranes.
Worked solution
(a) CO₂ ÷ H₂O = 10² ÷ 10⁻¹ = 10³ = 1000 times. H₂O ÷ glucose = 10⁻¹ ÷ 10⁻⁶ = 10⁵ = 100 000 times. (Subtract the powers of ten when dividing.)
(b) Glycerol is a smaller molecule (three carbons, against six in glucose) with fewer polar –OH groups. It has fewer hydrogen bonds with water to give up, and less polar surface to push through the nonpolar tails, so it crosses faster.
(c) K⁺ is an ion. Its full charge and water shell keep it out of the hydrophobic interior, so it hardly crosses at all: 10⁻⁶ ÷ 10⁻¹⁰ = 10 000 times slower than even glucose. Cells move K⁺ through embedded channel and transport proteins.
Suggested mark points (4): 1 for 1000; 1 for 100 000; 1 for linking glycerol’s smaller size and fewer polar groups to faster crossing; 1 for charge blocking K⁺ and proteins as the route.
Common errors: adding powers instead of subtracting (10² ÷ 10⁻¹ is not 10¹); saying K⁺ is too big.
How did you do?
- Q1, Q2 or Q7 wrong: re-read “Why the hydrophobic interior decides” and “Who crosses, and how” in the study guide, and look again at Figure 1.
- Q3 or Q6 wrong: re-read “Cell walls: an extra layer for some cells”.
- Q4 calculations wrong: rework Worked example 2; check you divided s by √n.
- Q5 incomplete: rework Worked example 1, especially the comparison between vesicles with and without proteins.
Then tick off the topic checklist.