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. Model values are given in each question where you need them. A calculator is assumed. Keep full calculator values until the last step, then round to the precision of the data. All data sets, and all ligands, receptors and kinases named with a letter, are fictional.
Question 1 (multiple choice · foundation)
Which sequence correctly shows the events in a G protein-coupled receptor pathway that uses cAMP?
- (A) Ligand binds receptor → G protein swaps GDP for GTP → adenylyl cyclase makes cAMP from ATP → protein kinase is activated → target proteins are phosphorylated
- (B) Ligand enters the cell → ligand binds the G protein → G protein makes cAMP → cAMP phosphorylates target proteins → response
- (C) Ligand binds receptor → adenylyl cyclase phosphorylates the G protein → protein kinase makes cAMP → cAMP binds the receptor → response
- (D) Ligand binds receptor → cAMP is made outside the cell → cAMP enters the cell and binds the G protein → adenylyl cyclase is activated → response
Answer and explanation
Answer: (A). The bound receptor changes shape and activates the G protein, which activates adenylyl cyclase. This enzyme makes cAMP, the second messenger, which activates a protein kinase that phosphorylates targets.
- (B) is wrong because the ligand of a GPCR stays outside the cell, and cAMP activates a kinase rather than phosphorylating targets itself.
- (C) swaps the jobs of the enzymes: adenylyl cyclase makes cAMP, and kinases phosphorylate proteins.
- (D) is wrong because cAMP is made inside the cell, after the G protein has activated adenylyl cyclase.
Question 2 (multiple choice · core)
Testosterone is a small, nonpolar steroid hormone. Which statement best describes its receptor and the type of response?
- (A) The receptor is in the cytoplasm or nucleus; the hormone–receptor complex changes the expression of particular genes.
- (B) The receptor is a G protein-coupled receptor on the cell surface, because all hormones carried in the blood bind surface receptors.
- (C) The receptor is a ligand-gated channel that lets testosterone into the cell.
- (D) The receptor is on the cell surface, because nonpolar molecules cannot pass through the phospholipid bilayer.
Answer and explanation
Answer: (A). Small nonpolar molecules diffuse through the hydrophobic core of the membrane, so their receptors are inside the cell. The hormone–receptor complex acts as a transcription factor, so the response is a change in gene expression.
- (B) is wrong: many blood-borne hormones are steroids or thyroid hormones with internal receptors.
- (C) is wrong: testosterone does not need a channel; ligand-gated channels let ions through, not the ligand.
- (D) reverses the rule: nonpolar molecules cross the bilayer easily; charged and large polar molecules cannot.
Question 3 (multiple choice · core)
At the junction between a motor neuron and a muscle cell, acetylcholine binds a ligand-gated channel in the muscle cell membrane. Which statement is correct?
- (A) Binding opens the channel, Na⁺ ions flow into the muscle cell down their concentration gradient, and the change in charge across the membrane triggers a fast response.
- (B) Acetylcholine passes through the open channel into the muscle cell, where it acts as a second messenger.
- (C) Binding makes the channel pump Na⁺ out of the cell, using energy from ATP.
- (D) Binding activates a G protein, which opens the channel after cAMP is made.
Answer and explanation
Answer: (A). The receptor is itself a channel. Binding changes its shape and opens it, so ions flow by diffusion. With no relay chain, the response takes milliseconds.
- (B) is wrong: the channel lets ions through; the ligand binds outside and is not a second messenger.
- (C) describes an active-transport pump, not a channel. Ions flow through channels passively.
- (D) describes a GPCR pathway. A ligand-gated channel opens directly.
Question 4 (graph · core)
Cells are treated with a fictional growth factor at time 0. The percentage of three kinases (P, Q and R) in the phosphorylated (active) form is measured:
| Time / min | 0 | 0.5 | 1 | 2 | 3 | 4 | 6 |
|---|---|---|---|---|---|---|---|
| P / % | 0 | 60 | 90 | 95 | 95 | 95 | 95 |
| Q / % | 0 | 10 | 45 | 85 | 92 | 93 | 93 |
| R / % | 0 | 0 | 8 | 40 | 75 | 88 | 90 |
(a) Estimate, by linear interpolation, the time at which each kinase reaches 50% phosphorylated. (b) Propose the order of the kinases in the cascade, and justify it. (c) Describe one further experiment that would test your proposed order. (d) Explain why a cascade of kinases, rather than a single kinase, can produce a larger response.
Worked solution and suggested Marlbridge rubric
(a) P: between 0 (0%) and 0.5 min (60%): 0 + (50 ÷ 60) × 0.5 = 0.42 min. Q: between 1 min (45%) and 2 min (85%): 1 + (5 ÷ 40) × 1 = 1.1 min. R: between 2 min (40%) and 3 min (75%): 2 + (10 ÷ 35) × 1 = 2.3 min.
(b) P → Q → R. Each kinase becomes active after the one before it; R shows no phosphorylation at 0.5 min, when P is already 60% active. This fits P phosphorylating Q, and Q phosphorylating R. (Timing alone is only correlation; it does not prove which kinase acts on which.)
(c) For example: in cells that lack Q (or where Q cannot be phosphorylated), add the growth factor and measure P and R. If the order is right, P is still phosphorylated but R is not. Alternatively, mix purified active Q with R and test whether R becomes phosphorylated.
(d) Each active kinase is an enzyme that can phosphorylate many molecules of the next kinase, so the number of active molecules multiplies at each step. A cascade also gives several points at which the signal can be regulated.
| Point | What earns it |
|---|---|
| 1 | All three times correct (about 0.4, 1.1 and 2.3 min) |
| 1 | Order P → Q → R justified by the timing |
| 1 | A valid experiment with the expected result if the order is right |
| 1 | Amplification explained by each kinase acting on many molecules |
Question 5 (constructed response · core)
Read this description of a fictional plant pathway:
Peptide hormone F binds the extracellular domain of receptor FR in the plasma membrane of root cells. The intracellular domain of FR changes shape and activates kinase K1. K1 phosphorylates kinase K2. K2 moves into the nucleus and phosphorylates transcription factor T, which switches on genes for root-hair growth.
(a) Classify F, FR, K1, K2 and T as reception, transduction or response components. (b) Explain why the receptor for F is in the plasma membrane rather than in the cytoplasm. (c) Explain how binding of F outside the cell starts events inside the cell. (d) Suggest how the root cell could switch the pathway off once F is removed.
Model answer and suggested Marlbridge rubric
(a) Reception: F binding to FR (F is the ligand; FR the receptor). Transduction: K1 and K2 (a phosphorylation cascade). Response: T switching on genes, leading to root-hair growth. (Accept T as the last transduction component, with gene expression as the response.)
(b) F is a peptide: large and water-soluble, so it cannot cross the hydrophobic interior of the bilayer. Its receptor must have a ligand-binding domain outside the cell.
(c) Binding changes the shape of FR. Because the protein spans the membrane, its intracellular domain changes shape too, and in its new shape it activates K1. The message crosses the membrane as a change in protein shape, not as the hormone.
(d) Protein phosphatases remove the phosphate groups from K1, K2 and T, returning them to their inactive form. When F leaves the receptor, FR returns to its inactive shape and stops activating K1.
| Point | What earns it |
|---|---|
| 1 | All five components correctly classified |
| 1 | Peptide cannot cross the bilayer, so receptor faces outside |
| 1 | Shape change of the intracellular domain links outside to inside |
| 1 | Phosphatases remove phosphates or receptor returns to inactive shape |
Question 6 (constructed response · stretch)
Fictional receptor A binds ligand La and makes cells secrete a protein. Fictional receptor B binds ligand Lb and switches on growth genes. Researchers build receptor C: the ligand-binding domain of A joined to the transmembrane and intracellular domains of B. They put C into cells that have neither A nor B, and measure growth-gene expression (relative units):
| Cells and treatment | No ligand | + La | + Lb |
|---|---|---|---|
| Cells with receptor C | 1.0 | 7.5 | 1.1 |
(a) Predict, before using the data, which ligand will activate C and what response it will cause. Explain your prediction. (b) Calculate the fold change in growth-gene expression with La, and state whether the data support your prediction. (c) Explain what this experiment shows about the role of each receptor domain.
Model answer and suggested Marlbridge rubric
(a) La should activate C, because C has A’s ligand-binding domain, which recognises La. The response should be growth-gene expression, because the intracellular domain, which starts transduction, comes from B. No secretion is expected.
(b) 7.5 ÷ 1.0 = 7.5-fold (+650%). With Lb, expression is 1.1, close to the no-ligand value. The data support the prediction.
(c) The ligand-binding domain decides which signal the receptor responds to. The intracellular domain decides which pathway is started, and so the response. The two domains do separate jobs and can be swapped.
| Point | What earns it |
|---|---|
| 1 | La predicted, linked to A’s ligand-binding domain |
| 1 | Growth genes predicted, linked to B’s intracellular domain |
| 1 | 7.5-fold calculated and Lb shown to have little effect |
| 1 | Clear statement of the separate roles of the two domains |
Accept “secretion will not happen” as part of the prediction; do not award the second point for “both responses”.
Question 7 (calculation · stretch)
In a model liver-cell pathway, one ligand-bound receptor activates 15 G proteins. Each G protein activates one adenylyl cyclase, which makes 200 cAMP. Every 4 cAMP activate one kinase. Each kinase activates 100 target enzymes, and each enzyme releases 10 molecules of product per second. The cell responds only if at least 2.0 × 10⁶ product molecules are released per second.
(a) Calculate the product released per second from one bound receptor. (b) Calculate the minimum number of bound receptors needed for a response. (c) Without the cascade, one bound receptor would activate one enzyme directly (10 product s⁻¹). How many bound receptors would then be needed? Comment on the answer.
Worked solution
(a) 15 × 200 = 3000 cAMP; 3000 ÷ 4 = 750 kinases; 750 × 100 = 75 000 enzymes; 75 000 × 10 = 7.5 × 10⁵ product s⁻¹.
(b) 2.0 × 10⁶ ÷ (7.5 × 10⁵) = 2.67. Receptors come in whole numbers, so at least 3. Check: 3 give 2.25 × 10⁶ (enough); 2 give 1.5 × 10⁶ (not enough).
(c) 2.0 × 10⁶ ÷ 10 = 200 000 bound receptors, about 67 000 times the 3 needed with the cascade (the cascade multiplies the output of each receptor 75 000-fold). A cell may not have that many receptors, and the hormone would have to be far more concentrated. The cascade lets a few hormone molecules trigger a full response.
Suggested mark points (4): 1 for 7.5 × 10⁵; 1 for 3 receptors with rounding up justified; 1 for 200 000; 1 for a comment linking amplification to sensitivity to low hormone levels.
Common errors: leaving out the ÷ 4 gives 3 × 10⁶ per receptor and an answer of 1 receptor; rounding 2.67 down to 2 gives too little product.
How did you do?
- Q1 or Q3 wrong: re-read “Three kinds of receptor to know” in the study guide, with Figure 1.
- Q2 or Q5(b) wrong: re-read “Where is the receptor?”.
- Q4 or Q5 incomplete: re-read “Protein phosphorylation” and “Amplification”.
- Q6 wrong: re-read the paragraph on receptor domains.
- Q7 wrong: rework Worked example 1, dividing at the cAMP step.
Then tick off the topic checklist.