Practice questions · Topic 4.3

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

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 tells you what a wrong choice assumed. For the free-response questions, write your answer in full sentences. 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. When the prey is gone, the release stops within a minute.

Which kind of response is this?

A. A change in gene expression producing new mucus proteins B. Apoptosis of the tentacle cells C. A change in cell function using vesicles the cells already had D. Growth and division of the tentacle cells

Question 3

In a rabbit, a hormone binds receptors on fur cells and a pathway makes the cells produce black pigment. A rabbit carries a mutation that locks that receptor in its active shape in every fur cell.

Predict the rabbit's phenotype.

A. Black fur, whether or not the hormone is present B. Black fur only where the hormone reaches the skin C. White fur, because the locked receptor is broken D. Fur that darkens only when extra hormone is given

Question 4

In a growing plant stem, certain cells receive a hormone and, over two days, each one cuts up its own contents and empties itself; what remains is a hollow tube of cell wall that carries water up the stem. Nothing leaks into the cells around them.

Which statement describes how these cells are emptied?

A. Water pressure bursts the cells and washes the contents away B. Enzymes from the neighboring cells digest them from outside C. The cells dry out because water flows past them D. Each cell dismantles itself with its own enzymes in response to the signal

Question 5

A kinase in a pathway is switched off when a phosphatase removes its phosphate. In a mutant cell line, a mutation swaps one amino acid at the phosphatase's active site, and the misshapen site holds the kinase only rarely.

Predict how the mutant cells' response differs after the signal stops.

A. The response never starts, because the phosphatase is needed to switch the kinase on B. The response lasts longer, because the kinase keeps its phosphate and stays active C. The response stops sooner, because the kinase is broken down D. The response is unchanged, because a mutation in the phosphatase's gene leaves the kinase untouched

Question 6

Root cells of a mutant tomato plant carry a change in a hormone receptor. When the hormone is added, binding tests find as much hormone bound to the mutant receptors as to normal ones, but the relay protein inside the cell stays unphosphorylated and the roots show no response.

Where does the mutation lie, and which measurement places it there?

A. In the ligand-binding domain: the hormone fits the site poorly, so binding is what failed B. In the intracellular domain: the hormone bound, yet the inner part stayed at rest C. In the relay protein: it stayed unphosphorylated, so it must be faulty D. In the response proteins: the roots showed no response, so they must be faulty

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. The hormone is added.

Which measurements change when the hormone is added to the mutant cells?

A. Hormone bound to the receptor, and protein A activated B. Kinase C phosphorylated, and the channel open C. Only the channel opening D. Every measurement, because the hormone still binds

Question 8

In a salivary gland cell, a pathway runs hormone → receptor → G protein → cAMP-making enzyme → cAMP → kinase → digestive enzymes secreted. A mutation locks the cAMP-making enzyme in its active shape.

Predict what these gland cells do.

A. They secrete digestive enzymes only when the hormone is present B. They secrete nothing, because the pathway is broken C. They secrete digestive enzymes continuously, with no hormone present D. They secrete more only when extra hormone is added

Question 9

A chemical from a soil fungus binds a kinase in the pathway hormone → receptor → G protein → cAMP-making enzyme → cAMP → kinase → channel opens, and holds the kinase in its inactive shape. Cells are given the hormone with the chemical present.

Which measurements change when the hormone is added?

A. Every measurement, because the hormone still binds its receptor B. The channel opens, but cAMP stays at rest C. None of the measurements, because the chemical blocks the pathway D. Hormone bound, G protein switched on and cAMP raised; the channel stays closed

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 G, makes the cells of the stem lengthen. Researchers propose that the pathway runs hormone G → receptor → kinase K → protein F → genes for wall-loosening enzymes expressed → cell lengthens. They compare a normal wheat line with a dwarf line whose stems stay short. Six pots of seedlings from each line are sprayed with hormone G every day for ten days, and six pots of each line are sprayed with water. Stem length is then measured. Two further tests are run on stem cells: a binding test finds as much hormone G bound to the dwarf line's receptors as to the normal line's; a second test finds that kinase K is present and normal in the dwarf line but is never phosphorylated after hormone G is given. 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 G 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 G?

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 G 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 G 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 G absent against its growth with hormone G present.

(c) State the null hypothesis for the effect of hormone G 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 G makes no difference to the stem length of the dwarf line; dwarf seedlings sprayed with hormone G 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 G and dwarf seedlings sprayed with water.

Accept: 'hormone G 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) Justify 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 binding test shows as much hormone G bound to the dwarf line's receptors as to the normal line's, so the ligand-binding domain works and reception succeeds. The second test shows kinase K present and normal yet never phosphorylated, so the receptor's inner part is failing to take up its active shape and pass the message to K. Binding works and the very next step fails: the fault is in the receptor's intracellular domain.

Check the box for each point your answer earns

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: the fault is in the intracellular domain.

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

Common slip: 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 G. Kinase K sits downstream of the faulty receptor, and an always-active kinase K phosphorylates protein F on its own, so the 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 G (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, six mice of each strain are injected with hormone L and six are injected with saline (salt water with no hormone), and the food each mouse eats per day is recorded. 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) Justify the claim that strain 2's fault lies in the pathway inside its brain cells rather than in a shortage of hormone L. (1 point)

A full-credit answer: Strain 2 mice have 40 ng/mL of hormone L in their blood, eight times the level in strain 1, so they are not short of the hormone; and injecting still more changed their food intake by nothing the bars can show. Hormone present in plenty with no response means the message is lost in the target cells: at the receptor or at a component downstream of it. The high blood level fits this too, since fat cells keep releasing hormone L when the pathway that would reduce appetite never runs.

Check the box for each point your answer earns

strain 2's blood carries eight times the hormone L of strain 1 (40 against 5 ng/mL), and injecting more changed nothing, so the hormone is present in plenty and the cells fail to respond to it: the fault is at the receptor or downstream of it.

Accept: 'the hormone is high yet the cells do not respond, so the fault is in the cells'. Do not award the point for the high food intake alone, or for 'strain 2 lacks the hormone'.

Common slip: Arguing from the high 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, means hormone L's message is lost, so the pathway that reduces appetite never runs. The mice eat about twice as much each day (7.8 g against 4.0 g), and repeated across 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 runs, 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 runs, 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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