Practice questions · Topic 4.4

Unit 4 · Practice for the Topic 4.4 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, the caption says what they represent.

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

A veterinarian measures a healthy dog's core temperature every two hours through a day: 38.4, 38.7, 38.3, 38.6, 38.5 and 38.4 °C. The owner worries that the temperature is unsteady.

Which statement describes these readings?

A. The dog's set point moved with each reading B. Homeostasis failed, because the six readings differ C. Each reading is a new stimulus that moves the temperature away for good D. The temperature wandered around a set point near 38.5 °C and was brought back each time

Question 3

During a night's sleep a person eats nothing. Their blood glucose drifts below 90 mg/dL, and cells of the pancreas release glucagon. By morning the glucose is back near 90 mg/dL and glucagon release has fallen to its usual low level.

Why has glucagon release fallen by morning?

A. The glucose is back near its set point, so the pancreas cells detect no drop B. The pancreas has run out of glucagon overnight C. Insulin released in the early morning switched the glucagon off D. The body lowered the value it holds glucose near, to match the night-time glucose

Question 4

Loop 1: a bacterium's pathway makes an amino acid, and when the amino acid is plentiful it binds the pathway's first enzyme and slows it. Loop 2: a yeast cell's inside becomes more acidic, and pumps in its membrane move hydrogen ions out until the acidity is back to its usual level. Loop 3: a dog's core temperature rises and it pants until the temperature falls. Loop 4: after a meal, cells of a horse's pancreas release insulin and its blood glucose falls.

Which loop operates at the cellular level?

A. Loop 1, because the pathway is inside a single bacterial cell B. Loop 4, because insulin acts on liver and muscle cells C. Loop 2, because one cell detects the change and one cell responds D. Loop 3, because the dog's tongue cells lose the heat

Question 5

A mutation changes the intracellular domain of a person's insulin receptors. Insulin binds the receptors normally, but the changed intracellular domain never changes shape, so the signal is never passed on inside the cell. The person's pancreas cells and insulin are normal. The person drinks a glass of juice.

Predict the person's blood glucose two hours later, compared with a person with normal receptors.

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 the same enzyme, and within an hour thousands are active and the cell has taken itself apart.

Which statement describes this loop?

A. Each active enzyme reduces the activation that produced it: negative feedback B. The number of active enzymes is held near a set point: negative feedback C. Dismantling the cell is bad for it, so the loop is negative feedback D. Each active enzyme increases the activation that produced it: 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 statement classifies this loop, with the correct reason?

A. Cortisol reduces the release of the signal that triggered it: negative feedback B. The cortisol level rises during stress, so the loop is 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

A cell has received the signal to dismantle itself, and a few molecules of its cutting enzyme have been switched on; each active molecule normally switches on more. Researchers add a drug that blocks the active site of the cutting enzyme.

Predict what happens in the cell, and why.

A. The cell dismantles itself as usual, because the first few active enzymes are enough B. The cell dismantles itself faster, because the drug adds to the cutting C. The cell stays intact, because a blocked enzyme switches on no others and the loop never grows D. The cell dismantles itself slowly, because the loop switches to negative feedback

Question 9

Mean blood cortisol of normal mice and of mice whose brain-gland cells lack the cortisol receptor, after thirty minutes in a narrow tube; six mice in each group. Error bars represent ±2SE. Gridlines every 50 ng/mL.

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. Six normal mice and six receptor-lacking mice were each kept in a narrow tube for thirty minutes, a mild stress, and then the cortisol in their blood was measured. The graph below shows the two means; the error bars represent ±2SE.

Which claim do the data support?

A. The two groups do not differ, because both means are above 100 ng/mL B. The receptor-lacking mice failed to hold their cortisol down: the bars do not overlap C. The receptor-lacking mice have a set point of about 310 ng/mL D. The receptor-lacking mice released less signal from the brain gland, because their cortisol was high

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) Justify the claim that the pigment slows enzyme 1, using 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 and the bar with added pigment from 2.2 to 3.8 μmol/min. The two ±2SE bars do not overlap, so the difference is very unlikely to be chance: the added pigment slowed the rate at which enzyme 1 made B.

Check the box for each point your answer earns

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

Do not award the point for comparing the two means alone (12.0 against 3.0 μmol/min).

Common slip: Comparing the means alone. The claim rests on the ±2SE bars not overlapping.

(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, it changes enzyme 1's shape and slows it, so less B is made, less C, and 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. It binds a separate site and changes the enzyme's shape; that is what makes it 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 runs 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, active pepsin cutting inactive molecules into active ones, 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) Predict the effect of the drug on the number of active pepsin molecules during the meal, and justify your prediction. (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 it activates no others; every round of the loop needs that cut, so the loop never runs and most of the pepsin stays inactive.

Check the box for each point your answer earns

the number stays very low or effectively zero (at most the few the acid activates), because the drug fills the active site so an active molecule can cut nothing, so pepsin activates no inactive molecules and the loop never runs; direction and reason both needed.

Do not award the point 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 runs 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 it removes the stimulus: once every inactive pepsin molecule has been cut, there is nothing left to activate, and as the food and enzyme leave the stomach the active pepsin is gone too. Positive feedback holds no set point; it runs 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 running out (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; it is ended by an outside event or limit.

Free-response score: 0 of 4

Multiple choice checked: 0 of 9 correct.

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