Practice questions · Topic 4.4

Unit 4 · Practice for the Topic 4.4 end-of-topic test

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Feedback is the response acting back on its own trigger; negative feedback reduces the change and holds a set point; positive feedback increases the change and holds none.

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, the caption says what the bars represent. Blood glucose is in milligrams per deciliter (mg/dL).

Question 1

Minutes after birth, a newborn's core temperature falls from 37.0 °C to 36.2 °C. Blood vessels in its skin narrow, and a special fat store in its back begins to burn, releasing heat. Its temperature rises back to 37.0 °C.

Which of these is the response in this loop?

A. The fall in core temperature to 36.2 °C B. The narrowed skin vessels and the burning fat store C. The 37.0 °C the newborn's body holds near D. The cold air of the room

Question 2

reading	1	2	3	4	5	6	7	8
blood calcium (mmol/L)	2.4	2.6	2.3	2.5	2.4	2.6	2.5	2.4

Blood calcium of a healthy adult, measured every six hours for two days.

A healthy adult's blood calcium is measured every six hours for two days. The readings are in the table. A student says: “These readings show that the calcium is out of control.”

Is the student correct, and why?

A. Yes — a quantity under homeostatic control is held at exactly one value B. No — homeostasis is working, because the set point moves to match each new reading C. Yes — each reading is a new stimulus, and the body never brings the calcium back D. No — the calcium is under control, because each small drift away from about 2.5 mmol/L is brought back

Question 3

After a hard sprint, the carbon dioxide in a runner’s blood has risen well above the value her body holds it near, and she is breathing fast and deep. Five minutes after she stops, her breathing is back to normal.

Why has her fast breathing stopped?

A. The carbon dioxide in her blood is back near its set point, so her brain detects no excess B. Her lungs have breathed out all the carbon dioxide her body holds C. Her heart rate has fallen, and the slower heartbeat switched the fast breathing off D. Her body has raised the value it holds carbon dioxide near, so the excess no longer counts as a stimulus

Question 4

Four negative-feedback loops.

Four negative-feedback loops are described in the table below.

Which of the following loops operates at the cellular level?

A. The bacterium’s loop B. The horse’s loop C. The yeast cell’s loop D. The dog’s loop

Question 5

In one breed of pig, a mutation changes the intracellular domain of the insulin receptor: insulin binds the receptor normally, but the intracellular domain no longer changes shape. The pigs’ pancreas cells and insulin are normal. A pig of this breed eats a meal of grain.

Which of the following predicts the pig’s blood glucose concentration three hours later, compared with a pig whose receptors are normal, with the correct reason?

A. The same, because insulin still binds the receptor B. Higher, because liver and muscle cells are never told to take glucose up C. Lower, because bound insulin stays on the receptor longer D. The same, because the set point has not changed

Question 6

A cell has received the signal to dismantle itself. Inside it, a few molecules of a cutting enzyme are switched on. Each active cutting enzyme switches on more molecules of that enzyme, and within an hour thousands are active and the cell has taken itself apart.

Which of the following classifies this loop, with the correct reason?

A. The loop ends once every enzyme molecule is active, so it is negative feedback B. Each active enzyme switches off the enzyme that activated it, so the number is held steady: negative feedback C. Dismantling the cell is bad for it, so the loop is negative feedback D. Each active enzyme switches on more of its own kind, so the number of active enzymes climbs round after round: positive feedback

Question 7

When a person is under stress, a gland at the base of the brain releases a signal that makes the adrenal glands release the hormone cortisol. Cortisol in the blood acts on the cells of the brain gland and makes them release less of the signal. The cortisol level rises during the stress and then levels off.

Which of the following classifies this loop, with the correct reason?

A. Cortisol acts back on the brain gland, so the gland releases less signal and the cortisol levels off: negative feedback B. Cortisol acts back on the brain gland, so the gland releases more signal and the cortisol climbs during stress: positive feedback C. Cortisol helps the body cope with stress, so the loop is positive feedback D. Stress is harmful, so the loop is negative feedback

Question 8

During labor, each contraction of the uterus is stronger than the one before it.

Which of the following is the step by which one contraction leads to a stronger one?

A. The contraction squeezes oxytocin out of the muscle of the uterus into the blood B. The contraction raises the mother’s blood pressure, and the higher pressure makes the heart pump oxytocin faster C. The contraction pushes the baby’s head against the cervix, and the stretch triggers the release of more oxytocin D. The contraction pushes the baby’s head back, which relieves the stretch on the cervix

Question 9

Some mice are bred so that the cells of their brain gland lack the receptor for cortisol; their adrenal glands and their cortisol are normal. A researcher keeps six normal mice and six receptor-lacking mice in a narrow tube for thirty minutes, a mild stress, and then measures the cortisol in their blood. The receptor-lacking mice’s mean cortisol concentration is far higher than the normal mice’s, and the two ±2SE ranges lie clear of each other.

Which of the following explains the higher cortisol concentration in the receptor-lacking mice?

A. Their adrenal glands released cortisol on their own, without waiting for any signal from the brain gland B. Cortisol could not act on the brain gland, so the signal kept coming and the adrenal glands kept releasing cortisol C. Their set point for cortisol is higher, so their loop held the cortisol near that higher value D. Cortisol that cannot bind its receptor is not broken down, so the cortisol built up in the blood over the thirty minutes

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

Cells of a red alga make a red pigment in three enzyme steps: enzyme 1 turns A into B, enzyme 2 turns B into C, and enzyme 3 turns C into the pigment. When the pigment is plentiful it binds enzyme 1 at a site away from the active site. Students broke open alga cells to make an extract that contains all three enzymes, and measured how fast enzyme 1 made B. Six tubes received extract and A; six tubes received extract, A and a high concentration of the pigment. Everything else, including the temperature, was the same in every tube. The graph below shows the mean rate at which B formed in each set of tubes; the error bars represent ±2SE.

Mean rate at which enzyme 1 made B in alga extract, with and without added pigment; six tubes in each set, same temperature and same amount of A. Error bars represent ±2SE. Gridlines every 1 μmol/min.

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

Hint: Which one thing did the students choose to make different between the two sets of tubes, and which quantity did they measure?

A full-credit answer: The independent variable is whether the pigment was added to the tube.
The dependent variable is the rate at which enzyme 1 made B, in μmol/min.

Check the box for each point your answer earns

independent variable, whether pigment was added to the tube (present or absent); dependent variable, the rate at which B formed, in μmol/min.

Do not award the point if the variables are reversed, or if the temperature, the extract or the substance A is named as the independent variable.

Common slip: Naming the extract or the substance A as the independent variable. Every tube had the same extract and the same A. The students changed only whether the pigment was present.

(b) State the null hypothesis for this experiment. (1 point)

Hint: A null hypothesis names the factor the students changed and the quantity they measured, and it predicts no difference.

A full-credit answer: There is no difference in the rate at which B forms between tubes with added pigment and tubes with none.

Check the box for each point your answer earns

there is no difference in the rate at which B forms between tubes with added pigment and tubes with none.

Do not award the point for a prediction of a difference in either direction.

Common slip: Writing the expected result, 'the pigment slows the pathway', as the null. The null predicts no difference.

(c) Support the claim that the pigment slows enzyme 1, using evidence from the error bars. (1 point)

Hint: The caption tells you what each bar represents. Compare the ends of the two bars before you compare the means, and recall what the overlap rule says about bars of that kind.

A full-credit answer: The bar for extract and A runs from 11.0 to 13.0 μmol/min.
The bar with added pigment runs from 2.2 to 3.8 μmol/min.
The two ±2SE bars do not overlap, so the difference is very unlikely to be chance.
So the added pigment slowed the rate at which enzyme 1 made B.

Check the box for each point your answer earns

the evidence (the ±2SE bars, 11.0 to 13.0 μmol/min with extract and A, 2.2 to 3.8 μmol/min with added pigment, do not overlap) AND the reasoning (so the lower rate with pigment is very unlikely to be chance, so the added pigment slowed enzyme 1).

Do not award the point for comparing the two means alone (12.0 against 3.0 μmol/min), or for the ranges with no link to the claim.

Common slip: Comparing the means alone, or quoting the bars with no link. Supporting the claim needs the non-overlap and what it shows about enzyme 1.

(d) Explain how the pigment slows its own production in a living alga cell. (1 point)

Hint: What does a molecule bound at a site away from the active site do to an enzyme? Then follow the pathway from enzyme 1 through B and C to the pigment.

A full-credit answer: When the pigment is plentiful, it binds enzyme 1 at a site away from the active site.
Bound there, the pigment changes enzyme 1's shape and slows enzyme 1.
So less B is made, then less C, then less pigment.
The end product reduces its own production: negative feedback at the molecular level.
As the cell uses the pigment up, the pigment leaves enzyme 1 and the pathway speeds up again.

Check the box for each point your answer earns

the pigment binds enzyme 1 at a site away from its active site (an allosteric site), which changes the enzyme's shape and slows it, so less B is made, then less C and less pigment: the end product reduces its own production (negative feedback at the molecular level); when the cell uses pigment up, the pigment leaves enzyme 1 and the pathway speeds up again.

Accept 'binds an allosteric site on enzyme 1, changes its shape, slows it, so less pigment is made'. Do not award the point for 'the pigment competes with A for the active site', or for 'the pigment slows enzyme 1' with no shape change and no link to less pigment.

Common slip: Saying the pigment blocks the active site. The pigment binds a separate site and changes the enzyme's shape. Binding a separate site is what makes this feedback by the end product rather than competition with A.

(e) A mutant alga has an enzyme 1 that lacks the site the pigment binds. Predict how the amount of pigment in the mutant's cells compares with a normal alga's, and justify your prediction. (1 point)

Hint: Name the part of the loop the mutant lacks, then follow the loop to the pigment and say which way the amount moves.

A full-credit answer: The mutant's cells hold more pigment.
The pigment has no site to bind on enzyme 1, so nothing slows the pathway as the pigment builds up.
Enzyme 1 keeps making B at full speed.
So more C and more pigment are made, and the pigment climbs well above the level a normal cell holds.

Check the box for each point your answer earns

more pigment (higher), because the site the pigment binds is missing, so the pigment can no longer slow enzyme 1 and the pathway makes pigment at full speed however much pigment has built up; the response that would reduce the pigment's own production is gone.

Do not award the point for 'less pigment', or for 'more pigment' with no reference to the missing binding site removing the slowing of enzyme 1.

Common slip: Predicting less pigment because 'the loop is broken'. Say which part is missing and follow the loop: the missing part is the brake on enzyme 1, so the pigment rises.

Free-response score: 0 of 5

Free response 2 · Analyze Model · 4 points

Cells lining the stomach release pepsin, an enzyme that digests protein, in an inactive form that has an extra stretch of amino acids folded over its active site. Stomach acid removes that stretch from a few molecules, making them active. Each active pepsin molecule can then cut the extra stretch off other inactive molecules, making them active too. The model below shows the loop. A drug that binds the active site of pepsin and stays there is swallowed with a meal; the step it blocks is marked X.

Model of the loop that activates pepsin in the stomach. The box marked X is the step the drug blocks.

(a) Describe how the number of active pepsin molecules changes over the first minutes after the acid makes the first few active, using the model. (1 point)

A full-credit answer: The number rises faster and faster.
The few molecules the acid activated each cut several inactive molecules into active ones.
Those new active molecules cut more.
So each round activates more than the round before, until nearly every pepsin molecule is active.

Check the box for each point your answer earns

it rises, and rises faster and faster, because each newly active molecule cuts and activates more, so every round activates more than the last.

Do not award the point for 'it rises' alone, or for 'it is held at a set level'.

Common slip: Describing a steady rise. The returning arrow means each round feeds the next, so the rise speeds up.

(b) Identify the kind of feedback in the model, and explain how the model shows it. (1 point)

A full-credit answer: Positive feedback.
The stimulus is active pepsin appearing.
The response is active pepsin cutting inactive molecules into active ones.
The response increases the number of active pepsin molecules, the very change that triggered it.
The arrow returning from the last box to the cutting step shows the response feeding its own trigger.

Check the box for each point your answer earns

positive feedback, because the response, more pepsin molecules becoming active, increases the very change that triggered it, the number of active pepsin molecules; shown by the arrow returning from the 'number of active molecules rises' box to the cutting step.

Do not award the point for 'positive feedback' alone, or for 'positive because digestion is useful'.

Common slip: Naming the kind without saying what the response does to its trigger. 'Positive' is earned by the returning arrow.

(c) Determine the effect of the drug on the number of active pepsin molecules during the meal, and justify your decision. (1 point)

A full-credit answer: The number of active pepsin molecules stays very low, near the few that the acid activates directly.
With its active site filled by the drug, an active pepsin molecule can cut nothing.
So the molecule activates no others.
Every round of the loop needs that cut.
So no round of the loop happens, and most of the pepsin stays inactive.

Check the box for each point your answer earns

the decision (the number stays very low, at most the few the acid activates) AND the reasoning it rests on (the drug fills the active site, so an active molecule can cut nothing, so pepsin activates no inactive molecules and no round of the loop happens).

Do not award the point for the decision alone, for 'the drug blocks pepsin' with no link to the number of active molecules, or for 'no pepsin is made'.

Common slip: Describing the blocked component and stopping. Follow the loop: no cutting, so no new active molecules, so the number stays low.

(d) In a healthy stomach the loop repeats for a while and then the number of active molecules stops rising. Explain what ends the loop. (1 point)

A full-credit answer: The loop ends when something outside the loop removes the stimulus.
Once every inactive pepsin molecule has been cut, there is nothing left to activate.
As the food and enzyme leave the stomach, the active pepsin is gone too.
Positive feedback holds no set point.
A positive-feedback loop repeats until an outside limit ends it.

Check the box for each point your answer earns

the loop ends when an outside limit removes the stimulus, here the supply of inactive pepsin being used up (every molecule has been activated), or the food and enzyme leaving the stomach; a positive-feedback loop has no set point and ends only when something outside it removes the stimulus.

Do not award the point for 'the number reaches its set point' or for 'active pepsin switches itself off'.

Common slip: Saying the number 'reaches its set point'. Positive feedback holds no set point. An outside event or limit ends the loop.

Free-response score: 0 of 4

Multiple choice checked: 0 of 9 correct.

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