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IB Biology C2.1: Chemical signalling worked solutions

Worked solutions to every practice question in C2.1: Chemical signalling of Photon Biology: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

HL[1 mark]
A white blood cell releases a small protein that diffuses a short distance and makes nearby cells divide. Which category of signalling chemical is this? (A) Hormone (B) Neurotransmitter (C) Cytokine (D) Calcium ion
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C. It is a cytokine, a signal protein that controls growth and immune responses and acts on nearby cells. A hormone is carried in the blood to distant cells, a neurotransmitter crosses a synaptic gap between a neuron and its target, and a calcium ion is not a protein.

Final answer

C

Question A2

HL[1 mark]
Insulin travels in the blood past every cell, yet only some cells respond. Why? (A) Blood reaches only some cells. (B) Only some cells carry a fitting receptor. (C) Insulin enters only some cells. (D) Only some cells make cAMP.
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B. Only cells with receptors whose binding site fits insulin can respond. A is wrong because the blood reaches all the cells, C because insulin does not enter the cell, and D because cAMP is not the second messenger for insulin.

Final answer

B

Question A3

HL[1 mark]
A marine bacterium produces light only when the culture is dense. Which statement explains this? (A) Each cell glows once it is old enough. (B) Autoinducer passes a threshold and switches on genes. (C) Light genes exist only in dense cultures. (D) A host animal makes the light.
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B. Autoinducer builds up as the population grows, and past a threshold it switches on the light genes in every cell. A and C ignore the density effect, and D is wrong because the bacteria, not a host, make the light.

Final answer

B

Question A4

HL[1 mark]
Which ligand most probably binds a receptor inside its target cell? (A) Insulin (B) Epinephrine (C) Testosterone (D) Acetylcholine
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C. Testosterone is a lipid-soluble steroid, so it crosses the plasma membrane and binds an intracellular receptor. Insulin, epinephrine and acetylcholine are water-soluble and bind receptors in the plasma membrane.

Final answer

C

Question A5

HL[1 mark]
Which is the correct order of events when epinephrine acts through a G protein-coupled receptor? (A) G protein, receptor, cAMP, adenylyl cyclase (B) receptor, cAMP, G protein, adenylyl cyclase (C) adenylyl cyclase, G protein, receptor, cAMP (D) receptor, G protein, adenylyl cyclase, cAMP
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D. Epinephrine binds the receptor, the receptor activates the G protein, the G protein activates adenylyl cyclase, and adenylyl cyclase makes cAMP. The other orders reverse at least one arrow.

Final answer

D

Question A6

HL[1 mark]
Binding of insulin to its receptor first causes which event? (A) Activation of a G protein (B) Autophosphorylation of tyrosines (C) Opening of a sodium channel (D) The receptor enters the nucleus
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B. The insulin receptor is a tyrosine kinase receptor, so binding makes the receptor phosphorylate tyrosines on itself. A describes a G protein receptor, C a ligand-gated channel and D a steroid receptor.

Final answer

B

Question A7

HL[1 mark]
Acetylcholine binds its receptor on a postsynaptic membrane. What happens next? (A) cAMP is made inside the cell. (B) Acetylcholine enters the cell. (C) A channel opens and sodium ions enter. (D) Transcription of a gene begins.
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C. The receptor is itself an ion channel; it opens and sodium ions diffuse in, changing the membrane voltage. Acetylcholine does not enter the cell, and no G protein or transcription is involved.

Final answer

C

Question A8

HL[1 mark]
Which is an example of positive feedback? (A) Autoinducer switches on its own production. (B) A signal switches on its own inhibitor. (C) A G protein hydrolyses GTP and switches off. (D) A receptor is taken into the cell.
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A. A signal that switches on its own production amplifies itself, as autoinducer does in quorum sensing. B, C and D are ways a pathway is turned down, which is negative regulation.

Final answer

A

Tutorial practice

Question B1

HL[5 marks]
Glucagon is a peptide hormone released by the pancreas. It makes liver cells release glucose from glycogen. Describe how glucagon produces this response.
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Any five of: glucagon is a peptide, so it cannot cross the plasma membrane (1); it binds a specific receptor in the plasma membrane of the liver cell (1); binding activates a G protein (1); the G protein activates adenylyl cyclase, which makes cAMP, the second messenger (1); the cascade amplifies the signal because each active molecule activates many others (1); enzymes are activated that break glycogen down to glucose (1); the glucose is released into the blood (1).

Final answer

Glucagon is water-soluble and cannot cross the membrane. It binds a specific receptor in the plasma membrane. The receptor starts a cascade inside the cell through a G protein, adenylyl cyclase and cAMP as second messenger. Each step activates many molecules, so the signal is amplified. Enzymes are activated that break down glycogen to glucose, which leaves the cell.

Question B2

HL[6 marks]
Compare and contrast the way in which testosterone and insulin act on their target cells.
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One mark for a similarity: both are carried in the blood, bind a specific receptor, and act only on cells that have it (1). Paired differences, one mark per pair with both halves stated, maximum 5: testosterone is a lipid-soluble steroid that crosses the membrane / insulin is a water-soluble protein that cannot; testosterone binds an intracellular receptor in the cytoplasm or nucleus / insulin binds a transmembrane tyrosine kinase receptor; the testosterone-receptor complex binds DNA and promotes transcription / insulin binding causes autophosphorylation and a relay; insulin's relay moves vesicles with glucose transporters to the membrane / testosterone's response is new proteins from new mRNA; testosterone has no second messenger / insulin's relay passes the message on inside the cell.

Final answer

Similarity: both travel in the blood, bind a specific receptor and affect only cells with it. Differences: testosterone is lipid-soluble and enters the cell, insulin is water-soluble and stays outside. Testosterone binds an intracellular receptor, insulin a receptor in the plasma membrane. Testosterone changes transcription, insulin triggers tyrosine phosphorylation and the movement of vesicles with glucose transporters.

Question B3

HL[6 marks]
Explain how oestradiol and epinephrine each produce a response in their target cells.
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Oestradiol, any three of: it is lipid-soluble and diffuses through the plasma membrane (1); it binds a receptor in the cytoplasm or nucleus (1); the activated receptor binds a specific DNA sequence (1); transcription of the target genes is promoted (1). Epinephrine, any three of: it binds a receptor in the plasma membrane and stays outside the cell (1); the receptor activates a G protein (1); the G protein activates adenylyl cyclase, which makes cAMP from ATP (1); cAMP, the second messenger, activates enzymes that change the cell's activity (1). Up to 3 marks for each hormone.

Final answer

Oestradiol: a lipid-soluble steroid that crosses the membrane and binds an intracellular receptor. The activated receptor binds specific DNA sequences and promotes transcription of target genes. Epinephrine: stays outside, binds a G protein-coupled receptor. The G protein activates adenylyl cyclase, which makes cAMP. cAMP activates enzymes and the cell responds.

Question B4

HL[5 marks]
A newly isolated marine bacterium makes light. Cultures of different densities were grown and the light produced per cell was measured. Light per cell in model units at 10^3, 10^5, 10^7, 10^8 and 10^9 cells per mL: 0.2, 0.3, 0.9, 41 and 55. (a) Describe the effect of cell density on light per cell, quoting values. [2] (b) Explain the pattern. [3]
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(a) Light per cell is low (0.2 to 0.9) up to 10^{7} cells per mL (1); it rises from 0.9 to 41, about 46-fold (41\div0.9=45.6), between 10^{7} and 10^{8}, then levels near 55 (1). (b) Cells release a diffusible autoinducer (1); the concentration rises with density and at low density is too low to act (1); above a threshold enough autoinducer binds the receptors and the light genes are switched on in the cells together (1).

Final answer

(a) Light per cell stays below 1 unit up to 10^{7} cells per mL, then rises about 46-fold between 10^{7} and 10^{8}, and levels near 55 at 10^{9}. (b) At low density autoinducer is dilute. At high density it passes a threshold, binds its receptor and switches on the light genes in all the cells together, which is quorum sensing.

Question B5

HL[6 marks]
Liver cells were treated in four ways and the rate of glycogen breakdown was measured. A G protein blocker prevents the G protein from being activated. The cAMP analogue acts as cAMP does and is added directly to the cells. Rate of glycogen breakdown in arbitrary units: no treatment 2.0; epinephrine 31.0; epinephrine plus G protein blocker 2.4; G protein blocker plus cAMP analogue 29.5. (a) Calculate how many times epinephrine increases the rate. [1] (b) Deduce what the blocker result shows. [2] (c) Explain the last result. [3]
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(a) 31.0\div2.0=15.5 times (1). (b) With the blocker the rate (2.4) is close to the untreated 2.0 (1), so the G protein is needed between the receptor and the response (1). (c) The analogue restores the rate to 29.5 even though the G protein is blocked (1); so cAMP acts downstream of the G protein (1) and is enough to activate the enzymes that break down glycogen (1).

Final answer

(a) 31.0\div2.0=15.5. (b) Epinephrine needs an active G protein: with the blocker the rate stays at the untreated level. (c) cAMP acts after the G protein. Supplying the analogue bypasses the blocked step and still activates the enzymes, so cAMP is the second messenger downstream of the G protein.

Question B6

HL[6 marks]
Muscle cells were incubated in four conditions and glucose uptake was measured (model data). Glucose uptake in arbitrary units: no insulin 1.8; insulin 9.6; insulin plus tyrosine kinase inhibitor 2.1; insulin plus inhibitor of vesicle fusion 2.3. (a) Calculate the percentage increase in uptake caused by insulin. [1] (b) Explain what the two inhibitor results show. [2] (c) Outline how insulin binding leads to the rise in uptake. [3]
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(a) (9.6-1.8)\div1.8\times100=433\% (1). (b) The kinase inhibitor lowers uptake to 2.1, close to the no-insulin value, so tyrosine kinase activity is needed (1); blocking vesicle fusion lowers it to 2.3, so vesicles have to fuse with the plasma membrane for uptake to rise (1). (c) Insulin binds the receptor on the outside of the cell (1); the receptor phosphorylates tyrosines and starts a relay (1); vesicles carrying glucose transporters move to the plasma membrane and fuse with it, so uptake rises (1).

Final answer

(a) (9.6-1.8)\div1.8\times100=433\%. (b) Uptake falls to 2.1 with the inhibitor, so tyrosine kinase activity is needed. Uptake also falls to 2.3 when vesicle fusion is blocked, so the vesicles must reach the membrane. (c) Insulin binds the receptor; the receptor phosphorylates tyrosines; a relay moves vesicles with glucose transporters to the plasma membrane; more transporters means more uptake.

Question B7

HL[5 marks]
Two receptors, R1 and R2, were tested in cultured cells. Both respond to a ligand, but the cells give different results. R1: delay 2 ms; response lost without external sodium; unchanged with the G protein blocker; cAMP no change. R2: delay 800 ms; response unchanged without external sodium; lost with the G protein blocker; cAMP rises 6-fold. (a) Identify the receptor with a built-in ion channel, using two pieces of evidence. [3] (b) Explain why the response of the other receptor is slower. [2]
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(a) R1 (1). Any two of: the response begins after 2 ms, much faster than R2 (1); it needs sodium ions outside the cell (1); the G protein blocker has no effect (1); cAMP does not change (1). (b) R2 works through a chain of steps, G protein, then adenylyl cyclase, then cAMP, then enzyme activation (1); each step takes time, while binding at R1 opens a channel directly (1).

Final answer

(a) R1 has the built-in channel: the response starts after 2 ms, needs external sodium and is unaffected by the G protein blocker, and cAMP does not change. (b) R2 acts through several steps: G protein, adenylyl cyclase, cAMP and enzyme activation, so the response is slower than the direct opening of a channel.

Question B8

HL[5 marks]
Outline the action of oestradiol on cells of the hypothalamus and of progesterone on cells of the endometrium.
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Any five of: oestradiol is lipid-soluble and enters the cell (1); it binds a nuclear/intracellular receptor in GnRH-secreting hypothalamic cells (1); the activated receptor binds DNA and promotes transcription (1); progesterone enters endometrial cells and binds its receptor (1); the receptor changes gene expression in the endometrium (1); this regulates development and function of the endometrium, for implantation and the maintenance of pregnancy (1).

Final answer

Both are lipid-soluble steroids that diffuse into the cell and bind intracellular receptors. The activated receptor promotes transcription. Oestradiol acts on GnRH-secreting cells of the hypothalamus. Progesterone acts on endometrial cells and regulates the development and function of the lining, for implantation and pregnancy.

Question B9

HL[3 marks]
Outline the role of calcium ions as a signal in a cell.
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Channels open and Ca^{2+} enters the cytoplasm (1); the rise in concentration above a low resting level is the signal, and calcium binds proteins to trigger a response such as vesicles fusing with the membrane (1); calcium is removed again by pumps, which ends the signal (1).

Final answer

A signal opens calcium channels, so Ca^{2+} enters the cytoplasm and its concentration rises sharply from a very low resting level. Calcium binds proteins and triggers a response such as vesicle fusion. It is then pumped out to restore the low level.

Question B10

HL[4 marks]
Explain, with reference to amino acids, how a transmembrane receptor is held in the plasma membrane and how it binds its ligand.
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Hydrophobic amino acids sit in the region in contact with the phospholipid tails and hold the protein in the bilayer (1); hydrophilic amino acids lie on the parts in contact with the water inside and outside the cell (1); the binding site on the outer face has a shape and charges that fit the ligand (1); so only that ligand binds and the receptor changes shape (1).

Final answer

Hydrophobic R groups lie in the belt that touches the fatty-acid tails, so they hold the protein in the bilayer. Hydrophilic R groups lie where the protein meets water on either side. The binding site on the outer face has a shape and charge pattern that fits one ligand, so only that ligand binds.

Advanced practice

Question C1

HL[8 marks]
A bacterium releases a toxin only in dense cultures. A mutant cannot make the autoinducer, and tubes were set up at two densities. Toxin was measured after 6 hours (model data). Toxin in arbitrary units at 10^6 and 10^9 cells per mL: wild type 2 and 88; mutant 2 and 4; mutant with added autoinducer 80 and 84. (a) Describe the pattern in the data. [2] (b) Explain the wild-type result. [3] (c) Deduce what the last row shows about the trigger. [2] (d) Suggest how a drug that blocks the receptor could treat an infection. [1]
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(a) Toxin is 88 for the wild type at high density but 2 at low density (1); the mutant stays low at both densities, 2 and 4 (1). (b) The wild type releases autoinducer, which accumulates as density rises (1); above a threshold it binds receptors and switches on the toxin gene (1); the mutant cannot make autoinducer, so the gene stays off (1). (c) Adding autoinducer restores toxin even at low density: 80 (1), so it is the concentration of autoinducer, and not density itself, that triggers the response (1). (d) A drug that blocks the receptor stops cells reading the signal (1), so the toxin is not made even when autoinducer is high (1).

Final answer

(a) Toxin is high only for the wild type at high density; the mutant makes little at either density. (b) The autoinducer builds up with density and at a threshold switches on the toxin gene; the mutant cannot make it. (c) The mutant makes a lot of toxin at low density once autoinducer is added, so it is the autoinducer concentration that matters, not density as such. (d) A drug that blocks the receptor stops the response to the autoinducer, so toxin is not made even in a dense infection.

Question C2

HL[7 marks]
In a model of a G protein pathway, one active receptor activates 15 G proteins. Each active G protein activates one adenylyl cyclase, and each cyclase makes 40 cAMP per second. The G protein switches itself off when it breaks down its bound GTP. cAMP is itself broken down by an enzyme that the pathway activates. (a) Calculate the number of cAMP molecules made per second per active receptor. [2] (b) Predict the effect of a mutation that stops the G protein breaking down GTP. [2] (c) Name the type of feedback shown by the cAMP-breaking enzyme and predict the effect of blocking it. [3]
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(a) 15\times1\times40=600 cAMP per second (2: one for the multiplication, one for the value). (b) The G protein stays active (1); adenylyl cyclase keeps making cAMP, so the response is prolonged / not switched off (1). (c) Negative feedback (1); the enzyme lowers the level of cAMP, so the pathway is turned down (1); with the enzyme blocked, cAMP would keep rising or stay high (1).

Final answer

(a) 15\times40=600 cAMP per second. (b) A G protein that cannot break down GTP stays active, so cyclase keeps making cAMP and the response is prolonged. (c) Negative feedback acting on the cAMP level; if the enzyme is blocked cAMP rises further and stays high.

Question C3

HL[7 marks]
A gland releases hormone H. H stimulates cells that release inhibitor I, and I lowers the release of H. After a pulse of H, the blood level was recorded every 10 minutes (model data). Level of H in units at 0, 10, 20, 30, 40 and 50 minutes: 100, 62, 40, 31, 28, 28. (a) Name the type of feedback and the step it acts on. [2] (b) Calculate the percentage fall in H during the first 10 minutes. [1] (c) Predict what happens to the level of H if the inhibitor I is blocked after 20 minutes. [2] (d) A second system gives 3, 6, 12, 24 units at 0, 1, 2 and 3 minutes. State the next value, and why it cannot rise for ever. [2]
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(a) Negative feedback (1); acting on the release of H by the gland (1). (b) (100-62)\div100\times100=38\% (1). (c) With I blocked, nothing lowers the release of H (1), so the level stays high or rises instead of settling at 28 (1). (d) A second system doubles every minute: 3, 6, 12, 24; the next value is 48 (1); it must level off eventually because a substrate or receptor becomes limiting (1).

Final answer

(a) Negative feedback, acting on the gland's release of H. (b) 38%. (c) With I blocked nothing lowers the release of H, so the level stays high or rises instead of settling at 28. (d) 48, because each value is twice the one before; it cannot continue because a substrate or receptor runs out.

Question C4

HL[6 marks]
Researchers tested two ligands, L and M, on cultured cells. Each ligand was also attached to a large plastic bead that cannot enter a cell. Ligand L switched on gene X after 45 minutes, but L on a bead did nothing. Ligand M raised the cAMP level within 20 seconds, and M on a bead did the same. (a) For each ligand, deduce the position of its receptor and the type of route, with evidence. [4] (b) State which ligand is more likely to be lipid-soluble. [1] (c) Suggest why the response to L is much slower than the response to M. [1]
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(a) L: the bead stops it entering and it does nothing, so its receptor is inside the cell (1); it changes transcription of gene X, consistent with an activated receptor binding DNA (1). M: still works on a bead that cannot enter, so its receptor is in the plasma membrane (1); the rise in cAMP suggests a G protein receptor (1). (b) L is lipid-soluble, small enough to cross the bilayer, like a steroid hormone (1). (c) The change in gene X takes 45 minutes because transcription and translation are needed, while M acts within seconds through second messengers (1).

Final answer

(a) L enters the cell and binds an intracellular receptor that promotes transcription. M binds a receptor in the plasma membrane and works through a G protein and cAMP. (b) L is the steroid-like, lipid-soluble ligand. (c) The slow change in gene X reflects transcription and translation.

Connected questions

Question X1

Connected questions · HL[6 marks]
C2.1 + D2.2 + D1.2. The oestradiol receptor, once activated, binds a specific DNA sequence and promotes transcription of the genes beside it. (a) Explain why only certain genes respond to the hormone. [2] (b) Outline how the transcribed gene leads to a new protein. [3] (c) Suggest why a steroid response is slower than a G protein response. [1]
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(a) The activated receptor recognises a specific DNA base sequence (1); only genes with that sequence are switched on in that cell (1). (b) RNA polymerase transcribes the gene into mRNA (1); the mRNA leaves the nucleus and is translated at ribosomes, where tRNA anticodons pair with mRNA codons (1); amino acids join to make the new protein that carries out the response (1). (c) The response needs transcription and translation, which take longer than activating enzymes that are already present (1).

Final answer

(a) Only genes with that DNA sequence beside them are switched on. (b) mRNA is made, leaves the nucleus, and is translated at ribosomes: tRNA anticodons pair with codons and amino acids join into a polypeptide, a new protein. (c) A G protein response needs only existing proteins; a steroid response needs transcription and translation.

Question X2

Connected questions · HL[6 marks]
C2.1 + D3.3 + B2.1. After a meal the blood glucose level rises. Insulin is released and muscle cells take up glucose through transporter proteins. (a) Name the type of feedback controlling blood glucose and explain it. [2] (b) State how glucose crosses the plasma membrane through the transporter. [2] (c) Outline how insulin increases the number of transporters in the membrane. [2]
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(a) Negative feedback (1); the rise in glucose causes insulin release, which lowers glucose, and the fall reduces insulin release (1). (b) Facilitated diffusion (1); glucose moves down its concentration gradient through the transporter without ATP (1). (c) Insulin binds the receptor and a relay follows (1); vesicles with transporters fuse with the plasma membrane, increasing the number of transporters there (1).

Final answer

(a) Negative feedback: high glucose causes insulin release, which lowers glucose; the lower glucose reduces insulin release. (b) Glucose moves down its gradient through the transporter by facilitated diffusion, with no ATP. (c) Insulin binding causes vesicles carrying transporters to fuse with the plasma membrane.

Question X3

Connected questions · HL[6 marks]
C2.1 + C2.2 + C1.1. Acetylcholinesterase breaks down acetylcholine in the synaptic gap. A poison inhibits the enzyme. (a) State the role of acetylcholinesterase. [1] (b) Predict, with reasons, the effect of the poison on the postsynaptic membrane. [3] (c) Explain why acetylcholinesterase breaks down acetylcholine but not other molecules. [2]
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(a) It breaks down acetylcholine, so the channel closes and the resting potential returns (1). (b) Acetylcholine is not removed, so it keeps binding the receptors (1); the sodium channels stay open or reopen, so sodium keeps entering (1); the postsynaptic membrane stays depolarised and the response is prolonged (1). (c) The active site has a shape and charges that fit acetylcholine (1), so only that substrate binds and is broken down (1).

Final answer

(a) It hydrolyses acetylcholine. (b) With the enzyme inhibited, acetylcholine stays in the gap and keeps binding its receptors, so channels stay open longer, sodium keeps entering and the postsynaptic membrane stays depolarised. (c) The enzyme's active site has a shape complementary to acetylcholine only.

From the book

Photon Biology: Study Guide for the IB Diploma (SL & HL)

Chemical signalling from the core ideas to full exam questions, with question types, worked examples and graded practice.

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