Practice questions · Topic 4.3

Unit 4 · Practice for the Topic 4.3 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Topic 4.3 summary: pathways broken on and off

Video coming soon

A pathway can be broken on as well as off: a look-alike switches it on, a blocker switches it off; what still happens locates the change; the change predicts what the organism shows.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a graph carries error bars, they represent ±2SE, the range the true mean is likely to lie in.

Question 1

A silk moth caterpillar's silk glands begin making silk protein about a day after a hormone reaches them. Before the hormone arrives, the gland cells contain no silk protein.

Why does the silk protein appear a day after the hormone, rather than within seconds?

A. The hormone needs about a day to diffuse from the blood into the gland cells B. The cells have to express the silk genes and build the protein, which takes hours C. The receptor takes a day to change shape once the hormone binds D. cAMP has to build up for a day before the silk is released

Question 2

When prey touches a sea anemone's tentacle, a signal reaches the tentacle cells, and within three seconds they release sticky mucus they had stored in vesicles. A researcher adds a chemical that stops all gene expression to the tentacle cells an hour before prey touches them.

Predict the mucus release when the prey touches.

A. No release B. Release only after several hours C. Release within three seconds, as before D. A smaller release than before

Question 3

In a rabbit, a hormone released only in cold weather binds receptors on fur cells, and a pathway with one kinase makes the cells produce black pigment, so a normal rabbit's fur darkens in winter. In a normal rabbit the kinase carries a phosphate, and is switched on, only while the hormone is bound. A researcher measures the kinase in the fur cells of a mutant rabbit kept away from the hormone: the kinase carries its phosphate in summer and in winter.

Which of the following best predicts the mutant rabbit's phenotype?

A. Black fur all year round B. Black fur only in warm weather C. White fur all year round D. Darker fur than normal, but only in cold weather

Question 4

In a growing plant stem, certain cells receive a hormone. Over two days each of these cells empties, and what remains is a hollow tube of cell wall that carries water up the stem. The cells around the tubes stay healthy. A researcher applies a chemical that enters these cells and blocks protein-cutting enzymes; with the chemical present the cells stay full, and no tube forms.

Which of the following describes how these cells are emptied?

A. Water pressure bursts the cells from outside, and the flow of water washes their contents away B. Enzymes released by the neighboring cells digest the contents from outside, and the flow clears them C. The cells dry out as water flows past them, and the dried contents are carried away D. Enzymes inside each cell, switched on by the hormone, break down its contents and clear them away

Question 5

Four changes in a liver cell are described.

Which of the following changes is a mutation?

A. The cell makes twice as much of a receptor after a hormone arrives B. One base in the cell's gene for a receptor is swapped for a different base C. A receptor changes shape while its ligand is bound to it D. A kinase adds a phosphate group to a relay protein

Question 6

Root cells of a mutant tomato plant carry a hormone receptor whose intracellular domain has been changed by a mutation; the ligand-binding domain is normal. A researcher adds the hormone and measures three things: whether hormone is found bound to the receptors, whether the relay protein inside the cell is phosphorylated, and whether the roots respond.

Which pattern of results do you predict?

A. Hormone bound: no; relay protein phosphorylated: no; response: none B. Hormone bound: yes; relay protein phosphorylated: no; response: none C. Hormone bound: yes; relay protein phosphorylated: yes; response: none D. Hormone bound: yes; relay protein phosphorylated: yes; response: weaker than normal

Question 7

The pathway in the snail's gill cells from the hormone to the channel. The dashed box is the component missing in the mutant line.

The model below shows a pathway in a snail's gill cells: a hormone binds a receptor, the receptor activates protein A, protein A phosphorylates kinase B, kinase B phosphorylates kinase C, and kinase C opens a channel. In a mutant line, kinase B is missing. A researcher adds the hormone.

Which pattern of measurements do you predict in the mutant cells?

A. Hormone bound: yes; protein A active: yes; kinase C phosphorylated: no; channel: closed B. Hormone bound: yes; protein A active: yes; kinase C phosphorylated: yes; channel: closed C. Hormone bound: yes; protein A active: no; kinase C phosphorylated: no; channel: closed D. Hormone bound: yes; protein A active: yes; kinase C phosphorylated: yes; channel: open

Question 8

In a cell of a lizard's salt gland, a signal passes along this pathway: hormone → receptor → G protein → cAMP-making enzyme → cAMP → kinase → salt secreted. A researcher measures cAMP in normal salt-gland cells and in mutant salt-gland cells, with no hormone present: normal cells 2 units, mutant cells 60 units. With the hormone present, normal cells reach 58 units.

Which of the following best predicts what the mutant salt-gland cells do?

A. They secrete salt only while the hormone is present B. They secrete no salt C. They secrete salt continuously, with no hormone present D. They secrete more salt only when extra hormone is added

Question 9

A chemical from a soil fungus stops cells from responding to a hormone whose signal passes along this pathway: hormone → receptor → G protein → cAMP-making enzyme → cAMP → kinase → channel opens. Researchers know the chemical holds one component in its inactive shape. That component is either the G protein or the kinase.

Which single measurement, made with the hormone and the chemical both present, tells the two possibilities apart?

A. Whether the hormone is found bound to the receptor B. Whether the channel opens C. Whether the kinase is present in the cells D. Whether cAMP rises

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Scientific Investigation · 5 points

In wheat, a plant hormone, hormone B, makes the cells of the stem lengthen. Researchers propose that the signal passes along this pathway: hormone B → receptor → kinase K → protein F → genes for cell-wall-loosening enzymes expressed → cell lengthens. They compare a normal wheat line with a dwarf line whose stems stay short. The researchers spray six pots of each line with hormone B daily for ten days and six pots with water, then measure stem length. Two further tests on stem cells: a binding test finds as much hormone B bound to dwarf receptors as to normal ones; a kinase test finds kinase K present and normal in the dwarf line but never phosphorylated after hormone B. The graph below shows the mean stem lengths; the error bars represent ±2SE. The researchers plan a fifth group: dwarf seedlings carrying an always-active kinase K, sprayed with water.

Mean stem length of wheat seedlings after ten days of daily spraying, six pots per group. Error bars represent ±2SE. Gridlines every 5 cm.

(a) Identify the dependent variable in this experiment. (1 point)

Hint: Which quantity did the researchers measure at the end of the ten days, and what unit is on the graph's axis?

A full-credit answer: The dependent variable is the stem length of the seedlings after ten days, in centimeters.

Check the box for each point your answer earns

stem length (cm) after ten days as the dependent variable.

Do not award the point for the wheat line or the spray (independent variables), or for 'growth' with no measurement named.

Common slip: Naming hormone B or the wheat line as the dependent variable. Those are what the researchers changed; stem length is what they measured.

(b) Explain why seedlings of each line were also sprayed with water. (1 point)

Hint: Think about what the hormone-sprayed pots are being compared against. What would you need to know about each line before you could credit any extra growth to hormone B?

A full-credit answer: The water-sprayed pots are the control for each line.
They show how long each line's stems grow in ten days with no added hormone, so any extra length in the hormone-sprayed pots of that line can be credited to hormone B rather than to a difference the lines have anyway.

Check the box for each point your answer earns

the water-sprayed pots give each line's stem length with hormone B absent (a control), so the effect of the hormone can be measured against it within each line.

Accept: 'a baseline for each line' with what is compared against it. Do not award the point for 'it is the control' alone, or for 'to check that water does nothing'.

Common slip: Writing 'they are the control' and stopping. Say what the control lets you compare: each line's growth with hormone B absent against its growth with hormone B present.

(c) State the null hypothesis for the effect of hormone B on the dwarf line's stem length. (1 point)

Hint: A null hypothesis is a statement of no difference. Which two groups are being compared here, and which measurement?

A full-credit answer: Null hypothesis: hormone B makes no difference to the stem length of the dwarf line; dwarf seedlings sprayed with hormone B and dwarf seedlings sprayed with water will have the same mean stem length after ten days.

Check the box for each point your answer earns

a statement of no difference in stem length between dwarf seedlings sprayed with hormone B and dwarf seedlings sprayed with water.

Accept: 'hormone B has no effect on the dwarf line's stem length'. Do not award the point for a prediction of a difference in either direction, or for a null hypothesis about the normal line.

Common slip: Writing the expected result ('the dwarf plants stay short') as the null hypothesis. The null hypothesis is the statement of no difference between the two treatments.

(d) Evaluate the claim that the dwarf line's fault lies in the receptor's intracellular domain, using both tests. (1 point)

Hint: Each test checks one link of the pathway. Which link does the binding test check, and which link does kinase K's phosphorylation check?

A full-credit answer: The claim is supported.
The binding test shows as much hormone B bound to the dwarf line’s receptors as to the normal line’s.
So the ligand-binding domain works, and reception succeeds.
The kinase test shows kinase K present and normal, yet never phosphorylated.
So the receptor’s inner part fails to take up its active shape and pass the message to kinase K.
Binding works and the very next step fails, so the fault is in the receptor’s intracellular domain.

Check the box for each point your answer earns

the judgement (the claim is supported) AND the ground (equal binding shows the ligand-binding domain works and reception succeeds, and kinase K, though present and normal, is never phosphorylated, so the receptor's inner part never passes the message on).

Accept 'supported: binding is normal but the receptor never activates kinase K'. Do not award the point for the judgement alone, for the short stems alone (that shows the pathway failed, not where), or for placing the fault in kinase K, which the kinase test rules out.

Common slip: Giving the judgement without the ground, or citing only the short stems. Short stems show the pathway failed somewhere; the two tests place the failure between binding and kinase K.

(e) Predict the stem length of the planned fifth group compared with the four groups on the graph, and justify your prediction. (1 point)

Hint: Find kinase K's place in the pathway relative to the fault you located in part (d). Are the components after it intact in the dwarf line?

A full-credit answer: The fifth group's stems will be long, about 32 cm, close to the normal line sprayed with hormone B.
Kinase K sits downstream of the faulty receptor, and an always-active kinase K phosphorylates protein F on its own, so the cell-wall-loosening genes are expressed and the cells lengthen whether or not the receptor ever passes a message on.

Check the box for each point your answer earns

long stems, about the length of the normal line sprayed with hormone B (about 32 cm), because kinase K is downstream of the faulty receptor and an always-active K drives protein F and the genes with no input from the receptor.

Accept: 'stems as long as the normal line's with hormone'. Do not award the point for 'short, because no hormone was sprayed' or 'short, because the receptor is faulty'.

Common slip: Predicting short stems because the pots get only water. An always-active kinase needs no message from the receptor; it acts on protein F by itself.

Free-response score: 0 of 5

Free response 2 · Scientific Investigation · 4 points

Fat cells release a hormone, hormone L, into the blood. Hormone L binds receptors on certain brain cells, and a pathway in those cells reduces appetite, so the animal eats less. Researchers study two strains of mice. Every day for two weeks, the researchers inject six mice of each strain with hormone L and six with saline (salt water with no hormone), and record the food each mouse eats per day. Blood tests show hormone L at 5 ng/mL in strain 1 mice and at 40 ng/mL in strain 2 mice. The graph below shows the mean food eaten per day; the error bars represent ±2SE.

Mean food eaten per day by mice of the two strains injected daily with saline or with hormone L, six mice per group. Error bars represent ±2SE. Gridlines every 1 g.

(a) Explain why six mice of each strain were injected with saline. (1 point)

A full-credit answer: The saline mice are the control for each strain.
They show how much each strain eats with no added hormone, and they receive the same daily injection, so any fall in food eaten by the hormone-injected mice of that strain can be credited to hormone L rather than to the injection or to a difference between the strains.

Check the box for each point your answer earns

the saline mice give each strain's food intake with the added hormone absent (a control), so the effect of injected hormone L can be measured against it within each strain, with the injection itself matched.

Accept: 'a baseline for each strain, so the hormone's effect can be seen'. Do not award the point for 'it is the control' alone, or for 'to see if saline changes appetite'.

Common slip: Writing 'they are the control' and stopping. Say what the control lets you compare: each strain's intake with hormone L absent against its intake with hormone L added.

(b) Using the graph, describe the effect of the hormone L injections on the food eaten by each strain. (1 point)

A full-credit answer: In strain 1, hormone L cut the food eaten from 4.0 g per day to 2.6 g per day, and the two ±2SE bars (3.7 to 4.3 and 2.3 to 2.9) do not overlap, so the fall is a real effect.
In strain 2, food eaten was 7.8 g per day with saline and 7.7 g per day with hormone L; the bars overlap, so the injections made no difference that the data can show.

Check the box for each point your answer earns

strain 1 ate less with hormone L (4.0 g falling to 2.6 g per day, bars not overlapping), while strain 2 ate the same with or without it (7.8 g against 7.7 g, bars overlapping, so no difference is shown).

Accept: values read to within 0.1 g. Do not award the point for a description that gives no direction for strain 1, or that reads the overlapping strain 2 bars as a real difference.

Common slip: Reading the strain 2 means as a small fall. The bars overlap, so the data show no difference; overlap is read before any difference in means is claimed.

(c) Support the claim that strain 2's fault lies in the pathway inside its brain cells rather than in a shortage of hormone L, using evidence from the description and the graph. (1 point)

A full-credit answer: Strain 2 mice have 40 ng/mL of hormone L in their blood, eight times strain 1’s: no shortage.
Injecting still more hormone L changed their food intake by nothing the bars can show.
Hormone present in plenty with no response shows the message is lost in the target cells.
So the fault is at the receptor or a component downstream of it.

Check the box for each point your answer earns

the evidence (strain 2's blood carries 40 ng/mL of hormone L, eight times strain 1's 5 ng/mL, and injecting more changed food intake by nothing the bars can show) AND the reasoning (hormone present in plenty with no response means the message is lost in the target cells, at the receptor or downstream of it).

Accept 'the hormone concentration is high yet the cells do not respond, so the fault is in the cells'. Do not award the point for the large food intake alone, for the 40 ng/mL with no link to the claim, or for 'strain 2 lacks the hormone'.

Common slip: Arguing from the large food intake alone. That shows the appetite pathway is failing; it is the high blood level of hormone L, with no response to injections, that rules out a shortage of the hormone.

(d) Predict the phenotype of strain 2 mice compared with strain 1 mice, and explain how a change in one protein produces it. (1 point)

A full-credit answer: Strain 2 mice will be heavier and fatter than strain 1 mice.
One changed protein in their brain cells, the receptor or a relay component, stops hormone L’s message.
So the pathway that reduces appetite never passes the signal on.
So the mice eat about twice as much each day, 7.8 g against 4.0 g.
Repeated at every meal, that molecular fault shows on the whole animal as extra body fat: the phenotype.

Check the box for each point your answer earns

strain 2 mice are heavier (fatter), because a changed receptor or relay protein in the brain cells stops hormone L's message, so the pathway that reduces appetite never passes the signal on, the mice eat about twice as much every day, and the extra food is stored as fat.

Accept: 'heavier, because the cells never receive the message to eat less'. Do not award the point for 'they eat more' with no link to the failed pathway, or for a prediction that they are lighter.

Common slip: Stopping at 'strain 2 eats more'. The point needs the chain: a changed protein blocks the message, the appetite pathway never passes the signal on, the mice eat more, and the whole animal is heavier.

Free-response score: 0 of 4

Multiple choice checked: 0 of 9 correct.

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