IB DP Biology: Chemical signalling: ligands, receptors and signal transduction (HL) (DP Biology) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise topic C2.1, Chemical signalling, statements C2.1.1 to C2.1.14, from the IB Biology guide, first assessment 2025. All of C2.1 is HL only, so every question is HL. Data values are supplied in the questions.

Learn the content in the study guide and revision notes. Course pages: IB DP Biology hub, printable checklist and free 10-minute diagnostics.

Questions

1. (HL only) State the term for a signalling chemical that binds to a receptor, and explain why only some cells respond to a hormone carried in the blood. [2]

2. (HL only) A culture of the marine bacterium Vibrio fischeri was sampled at different cell densities.

Density (million cells per cm³)	0.05	0.5	5	50	500
Light output (arbitrary units)	0.1	1	10	4000	90 000

(a) Calculate the light output per million cells at 500 million cells per cm³. [1] (b) State the density range over which light output per cell begins to rise. [1] (c) Explain these results in terms of quorum sensing. [3]

3. (HL only) Distinguish between hormones and neurotransmitters in the way they reach their target cells, and state one role of calcium ions as a signal inside cells. [4]

4. (HL only) (a) State the chemical group of each of: testosterone, insulin, epinephrine. [3] (b) Suggest one reason why signalling chemicals belong to many chemical groups. [1] (c) State the function of cytokines. [1]

5. (HL only) Explain why the insulin receptor is a transmembrane protein while the oestradiol receptor is found inside cells, including the distribution of amino acids in each. [5]

6. (HL only) A neuron has a resting potential of −70 mV. When acetylcholine binds to its receptors, the potential changes to −58 mV. The threshold for an action potential is −50 mV. (a) Calculate the change in membrane potential. [1] (b) Explain how acetylcholine causes this change. [3] (c) Deduce whether an action potential occurs, and suggest what would be needed to trigger one. [2]

7. (HL only) Outline how binding of epinephrine to its receptor leads to a response inside a liver cell, describing the roles of the G protein and the second messenger. [6]

8. (HL only) Liver cell preparations were treated and glucose released was measured (µmol per g in 10 minutes).

Treatment	Glucose released
Control	1.8
Epinephrine	14.4
Epinephrine plus an adenylyl cyclase inhibitor	2.1
A cell-permeable cAMP analogue, no epinephrine	13.2

(a) Calculate how many times greater glucose release is with epinephrine than in the control. [1] (b) Calculate the percentage fall caused by the inhibitor compared with epinephrine alone, to 3 significant figures. [2] (c) Deduce what the data suggest about cAMP. [2]

9. (HL only) Muscle cells have 4.0 glucose transporter molecules per µm² of plasma membrane without insulin and 31.0 with insulin. (a) Calculate the percentage increase. [2] (b) Explain how insulin causes this change. [3] (c) Predict the effect of a tyrosine kinase inhibitor on glucose uptake. [1]

10. (HL only) A gene responds to a steroid hormone. Relative mRNA concentration was measured.

Time after adding hormone (min)	0	30	90	240	720
Relative mRNA	1	1	6	22	35

(a) Calculate the mean rate of increase between 90 and 240 minutes, to 2 significant figures. [2] (b) Explain why mRNA does not increase in the first 30 minutes. [3] (c) State the target cells of progesterone that are covered in C2.1. [1]

11. (HL only) After a meal, blood glucose rose from 5.0 to 8.4 mmol per dm³ and then returned to near 5.4 mmol per dm³ as insulin secretion rose. (a) Calculate the percentage rise in glucose. [2] (b) Explain this as negative feedback. [4] (c) Outline one example of positive feedback and state how it differs from your answer to (b). [3]

Answers

1. Ligand [1]. Only target cells have the specific receptor with a binding site complementary to the hormone [1]. [2] Examiner insight: “Specific receptor” earns credit only when linked to cells with or without it; “the hormone goes only to those cells” does not answer why.

2. (a) 90 000/500 = 180 (units per million cells) [1]. (b) Between 5 and 50 million cells per cm³, where the value goes from 2 to 80 [1]. (c) Each cell releases autoinducer [1]. At low density it diffuses away and stays below a threshold [1]. At high density the concentration is high enough to bind receptors and switch on bioluminescence genes in the population together [1]. [5] Examiner insight: Refer to light per cell in (c); total light would rise with numbers even without quorum sensing.

3. Hormones travel in the blood to distant cells [1]; neurotransmitters diffuse across a synaptic gap to a nearby cell [1]. Hormones are slower and longer-lasting [1]; neurotransmitters fast and brief. Calcium ions trigger exocytosis or muscle contraction [1]. [4] Examiner insight: A comparison needs both sides of each point; giving only the hormone route does not distinguish.

4. (a) Testosterone is a steroid [1]; insulin is a protein [1]; epinephrine is an amine [1]. (b) Different chemicals vary in solubility, stability or speed, suiting different jobs [1]. (c) They regulate immune responses [1]. [5] Examiner insight: A reason must connect a chemical property to a function, not just “more variety”.

5. Insulin is a protein and hydrophilic, so it cannot cross the bilayer [1]. Its receptor spans the membrane, binding it outside [1], with hydrophobic amino acids in the part within the bilayer [1] and hydrophilic ones outside and inside [1]. Oestradiol is lipid-soluble and enters, so its receptor is a soluble protein inside the cell, mostly hydrophilic on the surface, with a pocket for the ligand [1]. [5] Examiner insight: Place the hydrophobic residues in the membrane-spanning region by name; “mixed amino acids” gains nothing.

6. (a) −58 − (−70) = +12 mV [1]. (b) Acetylcholine binds the receptor [1]. The channel in the receptor opens [1] and positive ions, mainly sodium, diffuse in [1]. (c) No, −58 mV is 8 mV more negative than the −50 mV threshold [1]. More receptors opening, from more acetylcholine, would be needed [1]. [6] Examiner insight: Compare the final potential with the threshold using the values given; state the sign of the change.

7. Epinephrine binds to a receptor in the plasma membrane [1]. The receptor changes shape [1]. The G protein swaps GDP for GTP and is activated [1]. It activates adenylyl cyclase [1], which converts ATP to cAMP, the second messenger [1]. cAMP starts a cascade that breaks down glycogen and releases glucose [1]. [6] Examiner insight: Keep the steps in order and name the enzyme; “a cascade occurs” alone describes nothing.

8. (a) 14.4/1.8 = 8 times [1]. (b) (14.4 − 2.1)/14.4 × 100 [1] = 85.4% [1]. (c) The analogue raises glucose release without epinephrine [1], and blocking cAMP production prevents the epinephrine effect, so cAMP carries the signal [1]. [5] Examiner insight: Support a conclusion with both the analogue and the inhibitor results; one alone leaves the link incomplete.

9. (a) (31.0 − 4.0)/4.0 × 100 [1] = 675% [1]. (b) Insulin binds the receptor [1], activating tyrosine kinase, which phosphorylates tyrosine inside the cell [1]. A sequence of reactions brings vesicles carrying glucose transporter proteins to the plasma membrane [1]. (c) Uptake would fall [1]. [6] Examiner insight: Use the order binding, phosphorylation, vesicle movement; state that vesicles fuse with the membrane.

10. (a) (22 − 6)/(240 − 90) [1] = 0.11 per min [1]. (b) The steroid must cross the membrane and bind its receptor [1]. The receptor binds DNA sequences [1], and transcription must occur before mRNA appears [1]. (c) Cells of the endometrium [1]. [6] Examiner insight: Quote units for a rate and link the delay to the steps from the hormone to transcription.

11. (a) (8.4 − 5.0)/5.0 × 100 [1] = 68% [1]. (b) Glucose rises, which is a change from the set point [1]. Insulin is released and binds to receptors on target cells [1], causing glucose uptake [1]. Glucose falls and insulin secretion decreases, reversing the change [1]. (c) Stretching of the cervix triggers oxytocin release [1], which strengthens contractions, causing more stretching [1]. Positive feedback amplifies a change instead of reversing it [1]. [9] Examiner insight: Show that the response opposes the change; for positive feedback, state the loop and how it escalates.

Where marks are usually lost

• Saying “specific” without linking the binding site shape to the ligand.
• Treating the response as caused by the hormone entering a transmembrane receptor.
• Forgetting that the G protein needs GTP.
• Giving cAMP as the first messenger.
• Mixing fold change and percentage change.
• Omitting units from rates.
• Missing the DNA-sequence and transcription steps for steroid receptors.
• Naming a feedback example without the change that is reversed or amplified.

Next steps

• Revisit the revision notes and the study guide.
• Tick off the topic on the printable checklist and return to the IB DP Biology hub.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

IB Diploma Programme Biology guide, first assessment 2025, published by the International Baccalaureate Organization. Topic C2.1, Chemical signalling (additional higher level).