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: an α-helix rises 0.15 nm for each amino acid; the hydrophobic interior of a membrane is roughly 3 to 4 nm thick. All proteins, cells and data sets are fictional.
Question 1 (multiple choice · foundation)
Which statement correctly describes how phospholipids are arranged in a plasma membrane?
- (A) There are two layers. The heads of one layer face the extracellular fluid, the heads of the other face the cytosol, and the tails of both layers face each other.
- (B) There are two layers. The tails face the extracellular fluid and the cytosol, and the heads of both layers face each other in the middle.
- (C) There is one layer, with heads facing the extracellular fluid and tails facing the cytosol.
- (D) There are two layers. The outer layer’s heads face the extracellular fluid, but the inner layer’s tails face the cytosol.
Answer and explanation
Answer: (A). Heads are polar and hydrophilic, so they face the watery fluid on both sides. Tails are nonpolar and hydrophobic, so they face each other and form the hydrophobic interior.
- (B) reverses the rule. Hydrophobic tails would then be in contact with water, which is unstable.
- (C) leaves hydrophobic tails exposed to the watery cytosol. A plasma membrane is a bilayer.
- (D) also exposes tails to the cytosol. The cytosol is mostly water, so the inner layer must point its heads into it.
Question 2 (multiple choice · core)
A transmembrane protein has three regions: region 1 lies in the extracellular fluid, region 2 crosses the hydrophobic interior of the membrane, and region 3 lies in the cytosol. Region 2 contains no pore. Which best describes the R groups on the outer surface of region 2?
- (A) Mostly charged R groups, which bond to the charged phosphate heads.
- (B) Mostly nonpolar R groups, which interact with the fatty acid tails.
- (C) Mostly polar R groups, which form hydrogen bonds with the fatty acid tails.
- (D) Mostly charged R groups, so that ions can travel along the outside of the protein.
Answer and explanation
Answer: (B). The part of a protein that touches the fatty acid tails must be hydrophobic. Nonpolar R groups interact well with nonpolar tails, so this arrangement is stable.
- (A) puts charged groups in the wrong place. The heads are at the surfaces, not in the middle; region 2 is surrounded by tails.
- (C) is wrong because fatty acid tails are nonpolar and cannot form hydrogen bonds with polar R groups.
- (D) confuses the outside of the protein with the lining of a pore. Ions cross through a hydrophilic pore inside a channel protein, not along its outer surface, and this region has no pore.
Question 3 (multiple choice · core)
A fictional membrane protein contains a run of 24 nonpolar amino acids that folds as an α-helix. How long is this helix?
- (A) 3.6 nm
- (B) 6.7 nm
- (C) 36 nm
- (D) 160 nm
Answer and explanation
Answer: (A). Length = 24 × 0.15 nm = 3.6 nm. This fits inside the 3 to 4 nm hydrophobic interior, so the helix could span the membrane.
- (B) divides 24 by 3.6 (the number of amino acids in one turn of a helix) and treats the result as a length. That gives a number of turns (about 6.7), not nanometres.
- (C) uses 1.5 nm per amino acid. The rise is 1.5 Å, which is 0.15 nm.
- (D) divides 24 by 0.15 instead of multiplying. A 160 nm helix would be at least 40 times as long as the 3 to 4 nm hydrophobic interior is thick.
Question 4 (diagram · core)
The diagram shows part of a membrane from an animal cell.
(a) Identify the structures labelled A to E. (b) State which side, 1 or 2, is the outside of the cell. Justify your answer from the diagram. (c) Describe one role of A and one role of B. (d) Describe one way in which this diagram is not a realistic representation of a living membrane.
Model answer and suggested Marlbridge rubric
(a) A: carbohydrate chain (so A with B is a glycoprotein). B: integral (transmembrane) protein. C: polar (phosphate) head of a phospholipid. D: hydrophobic interior made of fatty acid tails. E: peripheral protein.
(b) Side 2 is the outside. Carbohydrate chains of glycoproteins and glycolipids face the outside of the cell, and chain A points into side 2. (Side 1 is the cytosol, and E is a peripheral protein on the cytosol face.)
(c) A: cell recognition, for example identifying the cell as “self” to the immune system, or helping cells stick to each other. B (any one): a channel or carrier that lets specific hydrophilic substances cross; a receptor for a signal; an anchor to the cytoskeleton.
(d) Any one: it is not to scale; it shows a still picture although phospholipids and proteins move sideways all the time; it shows very few proteins; it shows no cholesterol, which is present in animal cell membranes; it is a flat slice of a curved, three-dimensional surface.
| Point | What earns it |
|---|---|
| 1 | At least four of A–E correctly identified |
| 1 | Side 2 chosen with the carbohydrate chain as the reason |
| 1 | A correct role for A and for B |
| 1 | A valid limitation of the diagram |
Question 5 (constructed response · core)
In a fictional experiment, purified membranes were washed in a salt solution. Protein P came off the membrane, but protein Q stayed attached. Protein Q came off only when the membranes were treated with a detergent. (A detergent molecule has a polar end and a nonpolar end; it breaks up the bilayer.) Salt solutions weaken attractions between charged groups.
(a) Classify P and Q as integral or peripheral proteins, and justify each. (b) Describe where the hydrophobic and hydrophilic regions of protein Q are likely to be. (c) Predict whether protein P has a long run of nonpolar amino acids. Explain.
Model answer and suggested Marlbridge rubric
(a) P is peripheral: it was held to the surface by attractions between charged groups (for example with phosphate heads or another protein), which the salt weakened, so it never entered the hydrophobic interior. Q is integral: it was held in place by the hydrophobic interior itself, so it came off only when the detergent broke up the bilayer.
(b) Q’s hydrophobic region (nonpolar R groups) sits among the fatty acid tails in the middle of the membrane. Its hydrophilic regions (polar and charged R groups) face the cytosol and the extracellular fluid, and line any pore inside the protein.
(c) No (or only a short one). P sits on the surface in contact with water and the heads, so its surface is mostly hydrophilic. A run of about 20 or more nonpolar amino acids would be expected only in a segment that crosses the hydrophobic interior.
| Point | What earns it |
|---|---|
| 1 | P peripheral, linked to salt breaking surface attractions |
| 1 | Q integral, linked to the detergent disrupting the bilayer |
| 1 | Q’s hydrophobic region in the tails and hydrophilic regions facing the fluids |
| 1 | P predicted to lack a long nonpolar run, with a reason |
Question 6 (constructed response · stretch)
Cells from a fictional embryo tissue normally stick together in clumps. A researcher treated one sample of these cells with an enzyme that removes carbohydrate chains from the cell surface. A control sample was given the same solution without the enzyme. After 1 hour, she counted the percentage of cells in clumps in three repeats of each.
| Sample | Repeat 1 | Repeat 2 | Repeat 3 |
|---|---|---|---|
| Control / % in clumps | 82 | 78 | 85 |
| Enzyme-treated / % in clumps | 21 | 25 | 18 |
(a) Calculate the mean for each sample and the percentage change caused by the enzyme. (b) Explain the result using the roles of membrane components. (c) Identify one weakness of the control, and propose a better one.
Model answer and suggested Marlbridge rubric
(a) Control mean = (82 + 78 + 85) ÷ 3 = 245 ÷ 3 = 81.7%. Treated mean = (21 + 25 + 18) ÷ 3 = 64 ÷ 3 = 21.3%. Percentage change = (21.3 − 81.7) ÷ 81.7 × 100 = −73.9%, a fall of about three quarters. The ranges (78–85 and 18–25) do not overlap.
(b) The carbohydrate chains of glycoproteins and glycolipids face the outside of the cell. They help cells recognise and stick to each other. Removing them leaves cells unable to recognise or bind their neighbours, so far fewer clumps form.
(c) The control did not contain the enzyme protein at all, so the result might be due to the enzyme doing something else, such as damaging other surface proteins. A better control is the same enzyme inactivated (for example by heating) in the same solution: everything is the same except the enzyme’s ability to remove carbohydrate chains. More repeats would also help.
| Point | What earns it |
|---|---|
| 1 | Both means correct |
| 1 | Percentage change about −74% (accept −73% to −74% from rounded means) |
| 1 | Carbohydrate chains of glycoproteins/glycolipids linked to recognition or adhesion and to fewer clumps |
| 1 | Valid weakness and an improved control (such as an inactivated enzyme) |
Question 7 (evaluating a model · stretch)
A student made a poster of the plasma membrane of an animal cell. The poster shows:
- a single row of phospholipids, with heads facing the extracellular fluid and tails facing the cytosol;
- every protein drawn sitting on top of the phospholipid heads;
- carbohydrate chains drawn on both faces of the membrane;
- a caption saying: “The membrane is a rigid wall. Each molecule stays in its place.”
(a) For each of the four features, state what is wrong and how to correct it. (b) Explain why feature 1 would not be stable in a living cell.
Model answer and suggested Marlbridge rubric
(a)
- A plasma membrane is a bilayer. The inner layer’s heads should face the cytosol, with the tails of both layers facing each other.
- Integral proteins should pass through the bilayer, with hydrophobic regions among the tails. Only peripheral proteins sit on one surface.
- Carbohydrate chains (of glycoproteins and glycolipids) should be on the outer face only.
- The membrane is fluid: phospholipids and many proteins move sideways within it. Some proteins are anchored, but the membrane is not a rigid wall.
(b) The cytosol is mostly water. Hydrophobic tails facing it cannot form hydrogen bonds with water molecules, so water would have to arrange itself around them in an unstable way. The tails would be pushed together, away from the water. A second layer, heads towards the cytosol, is the stable arrangement.
| Point | What earns it |
|---|---|
| 1 | Bilayer correction (feature 1) |
| 1 | Integral proteins pass through, peripheral on one face (feature 2) |
| 1 | Carbohydrate on outer face only (feature 3) and fluid, sideways movement (feature 4) |
| 1 | Instability explained: nonpolar tails cannot interact with water, so they cluster away from it |
Do not award the last point for “the tails would dissolve” or “the tails would be damaged”; the answer must use the hydrophobic nature of the tails.
How did you do?
- Q1 or Q7 wrong: re-read “Phospholipids: heads out, tails in” in the study guide.
- Q2, Q3 or Q5 wrong: re-read “Membrane proteins: R groups decide where each part sits” and rework Worked example 1.
- Q4 wrong: study Figure 1 and the steps in “Reading membrane diagrams”.
- Q6 incomplete: re-read the component table, and the control in Worked example 2.
Then tick off the topic checklist.