Practice questions · Topic 3.2a

Unit 3 · Practice for the Topic 3.2a end-of-topic test

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Temperature and pH act through collisions, charges and the fold; substrate, enzyme and product set how often an active site is filled; a competitive inhibitor binds the active site, a noncompetitive inhibitor an allosteric site.

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 for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. 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 and third are at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a question gives mean rates from repeated trials, it also tells you that the differences between conditions were larger than the variation between trials, so you can compare the means directly.

Question 1

Amylase speeds up the reaction starch → maltose. A student adds mercury ions to a tube of purified amylase and starch at 25 °C and pH 7. The temperature and pH do not change, and the mercury ions take part in no reaction. The rate falls from 20 to 4 mg of maltose released per minute.

Which of the following explains how the mercury ions could lower the rate, and what measurement would confirm it?

A. The mercury ions lower the temperature of the solution; a second thermometer reading of the tube would confirm it B. The mercury ions are turned into product in place of the starch; measuring the mercury ions left would confirm it C. The mercury ions pull the amylase's fold apart; the fall in rate on its own is enough to confirm that the fold changed D. The mercury ions disrupt the weak interactions that hold the fold; a folding test (a count of molecules still folded) would confirm it

Question 2

Rate of the same kind of reaction catalyzed by an enzyme from a fish that lives in the icy seas around Antarctica (solid) and by an enzyme from a desert lizard (dashed), at temperatures from 0 to 60 °C. pH and substrate concentration were the same in every tube. Gridlines every 2 μmol/min and every 10 °C.

The graph shows the rate of the same kind of reaction catalyzed by two enzymes: one from a fish that lives in the icy seas around Antarctica and one from a desert lizard.

What is the optimal temperature of each enzyme, and what is happening to the Antarctic fish's enzyme at 35 °C?

A. Antarctic fish 15 °C, lizard 40 °C; at 35 °C the fish's enzyme is denatured B. Antarctic fish 15 °C, lizard 40 °C; at 35 °C the fish's enzyme is slowed by too few collisions C. Antarctic fish 35 °C, lizard 40 °C; at 35 °C the fish's enzyme is at its optimum D. Both 40 °C; at 35 °C the fish's enzyme is still warming toward its optimum

Question 3

Papain, a protease from papaya fruit, speeds up the reaction protein → shorter pieces of protein. A student heats a sample of papain to 85 °C for 10 minutes and cools it to 25 °C. Afterwards the sample's rate is zero, and the rate stays at zero however long the sample is left.

Which of the following describes the papain molecules after the heating?

A. Each chain of amino acids has been broken into separate amino acids B. Each chain is still whole, but its fold has been lost, and the active site with it C. The papain has been used up as a reactant in the reaction D. The papain molecules have stopped moving, so they no longer collide with protein

Question 4

Cola has a pH of about 3. Black tea has a pH of about 5.

How do the concentrations of hydrogen ions (H⁺) in the two drinks compare?

A. The tea has a hundred times the concentration of hydrogen ions of the cola B. The cola has two times the concentration of hydrogen ions of the tea C. The cola has ten times the concentration of hydrogen ions of the tea D. The cola has a hundred times the concentration of hydrogen ions of the tea

Question 5

Rate of each enzyme at three pH values, 30 °C.

A researcher tests an enzyme from a bacterium that lives in acidic mine water and an enzyme from a bacterium that lives in a soda lake. Both enzymes speed up the same kind of reaction. The researcher measures each enzyme's rate at 30 °C, with the same substrate concentration, at pH 3, 6 and 9. The rates are in the table.

Which of the following do the results support?

A. Every enzyme works best near pH 7, so both enzymes give their greatest rate at pH 6, the value nearest to neutral B. A lower pH always gives a higher rate, so both enzymes give their greatest rate at pH 3, the lowest value tested C. Each enzyme's optimal pH matches where it works: about pH 3 for the mine enzyme, about pH 9 for the soda-lake enzyme D. Both enzymes share the same optimal pH, because both speed up the same kind of reaction on the same substrate

Question 6

Initial rate at each substrate concentration.

A student gave a fixed amount of a purified enzyme its substrate at five concentrations, with temperature and pH the same in every tube. The initial rates are in the table.

Why does the rate rise steeply from 1 to 4 mmol/L of substrate, yet barely change from 8 to 16 mmol/L?

A. From 1 to 4 mmol/L doubling the substrate doubles the rate; from 8 to 16 mmol/L the extra substrate has begun to denature the enzyme B. From 1 to 4 mmol/L added substrate molecules find empty active sites; by 8 mmol/L nearly every active site is already occupied all the time C. From 1 to 4 mmol/L the enzyme is still being made; by 8 mmol/L the extra substrate has used the enzyme up D. From 1 to 4 mmol/L no product has built up yet; by 8 mmol/L product is being turned back into substrate as fast as it forms

Question 7

A weedkiller lowers the rate of a plant enzyme. In leaf extracts with equal amounts of the enzyme, the weedkiller cuts the rate to about half at every substrate concentration tested, from 2 μM to 2,000 μM.

Why does raising the substrate concentration leave the weedkiller's effect unchanged?

A. The substrate and the weedkiller compete for the active site, and the weedkiller binds it more tightly B. The weedkiller binds at an allosteric site, so more substrate cannot displace it C. More substrate denatures the enzyme, canceling any gain from the extra collisions D. The weedkiller has permanently denatured every enzyme molecule in the extract

Question 8

In a closed tube, an enzyme speeds up the reaction fumarate → malate (two small molecules of the cell's energy pathway). After 40 minutes the amount of malate has stopped rising, though 33% of the fumarate remains. A sample of the enzyme taken from the tube converts fresh fumarate at its full initial rate.

Which change to the tube would keep the reaction going forward until all of the fumarate is converted?

A. Removing malate from the tube as fast as it forms B. Adding more of the enzyme to the tube C. Warming the tube to speed the molecules up D. Sealing the tube more tightly so that nothing escapes

Question 9

Initial rate of a tapeworm enzyme at substrate concentrations from 1 to 16 mmol/L, with and without a fixed amount of a worm-killing drug; equal enzyme, temperature and pH in every tube. Gridlines every 2 mmol/L and every 10 μmol/min.

The graph shows the initial rate of an enzyme from a tapeworm (a gut parasite) at rising substrate concentrations, with and without a fixed amount of a new worm-killing drug. Enzyme amount, temperature and pH were the same in every tube.

Which kind of inhibitor is the drug, and what in the graph shows it?

A. Competitive; the inhibited rate stays about half at every concentration B. Noncompetitive; the inhibited rate nearly reaches the uninhibited rate at high substrate concentration C. Competitive; the inhibited rate nearly reaches the uninhibited rate at high substrate concentration D. Noncompetitive; the inhibited rate stays about half at every concentration

Question 10

A results table lists an enzyme's rate as 30 nmol/min and its substrate concentration as 2 μM (micromoles per liter).

Which of the following states what 30 nmol/min means?

A. 30 millionths of a mole of product formed each minute B. 30 nanometers moved by each product molecule every minute C. 30 billionths of a mole of product in each liter of solution D. 30 billionths of a mole of product formed each minute

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 · Analyze Data · 5 points

Amylase in saliva begins breaking starch down into the sugar maltose in the mouth: it speeds up the reaction starch → maltose. A student mixed a fixed amount of her own saliva with a starch solution at five temperatures, three trials at each, and measured the rate of maltose release. The table gives the mean rate at each temperature; the differences between temperatures were far larger than the variation between her three trials at any one temperature. She then brought the 65 °C tube down to 37 °C and measured 0.1 mg/min again, and brought the 10 °C tube up to 37 °C and measured 5.9 mg/min. The amylase's optimal pH is about 7.

Mean rate at which the student's salivary amylase released maltose from starch at five temperatures, three trials each, with the same starch concentration and pH in every tube.

(a) Identify the optimal temperature of the amylase among those tested. (1 point)

Hint: The optimal temperature is the temperature at which the enzyme's rate is greatest. Compare the rows before you answer.

A full-credit answer: The optimal temperature is 37 °C, because the rate of maltose release is greatest there: 6.0 mg/min.

Check the box for each point your answer earns

37 °C, identified as the temperature at which the rate is greatest (6.0 mg/min).

Accept 37 °C alone. Do not award the point for 65 °C (the hottest tube) or for 25 °C (the biggest jump from the reading before).

Common slip: Choosing the highest temperature tested, or the temperature with the biggest jump from the reading before it. The optimal temperature is where the rate itself is greatest.

(b) Explain why the rate at 37 °C is higher than the rate at 25 °C. (1 point)

Hint: Think about how an amylase molecule and a starch molecule have to collide before anything happens, and what warming changes about that.

A full-credit answer: At 37 °C the molecules move faster than at 25 °C.
So the amylase molecules and the starch molecules collide more often and with more energy.
More of those collisions carry at least the activation energy, so more collisions succeed each second.
Therefore the rate rises from 3.4 mg/min to 6.0 mg/min.

Check the box for each point your answer earns

warming makes the molecules move faster, so amylase and starch collide more often (or with more energy), so more collisions carry at least the activation energy and more succeed each second.

Accept faster molecules and more frequent collisions, with the rate rising as the result. Do not award the point for "heat gives the enzyme energy" or "heat lowers the activation energy".

Common slip: Saying only that "heat speeds up reactions". The point needs the molecules moving faster and colliding more often, with more of the collisions succeeding.

(c) Explain, in terms of the enzyme's structure, why the rate at 65 °C is close to zero. (1 point)

Hint: Think about what holds the amylase's fold together, and what the active site needs in order to hold starch.

A full-credit answer: At 65 °C the heat disrupts the hydrogen bonds and other weak interactions that hold the amylase's fold.
So the active site loses its shape.
The starch no longer fits the active site, so the enzyme can no longer catalyze the reaction: the amylase is denatured.
Denaturation removes working enzymes faster than the extra collisions can help, so the rate is close to zero.

Check the box for each point your answer earns

the heat disrupts the hydrogen bonds (or other weak interactions) holding the amylase's fold, so the active site loses its shape and the starch no longer fits: the enzyme is denatured.

Accept "denatured" together with the active site losing its shape. Do not award the point for "the peptide bonds break" or for "denatured" with nothing after it.

Common slip: Writing "the enzyme denatures" and stopping, or saying the heat breaks the chain into amino acids. Name what is disrupted, the weak interactions holding the fold. Then follow it to the active site and the substrate.

(d) Evaluate the claim that the 10 °C sample was slowed rather than denatured, using the results after both tubes were brought to 37 °C. (1 point)

Hint: Compare what each tube did when it was brought to 37 °C. What does a rate that returns tell you about the fold?

A full-credit answer: The claim is supported.
When the 10 °C tube was brought to 37 °C, its rate came back to 5.9 mg/min, almost the full 6.0 mg/min.
So its fold was never lost: the cold had only slowed its molecules.
The 65 °C tube stayed at 0.1 mg/min at 37 °C.
So its fold was lost, because heating that strong pulls the fold apart and it does not re-form.
Therefore the 10 °C sample was slowed, not denatured.

Check the box for each point your answer earns

the judgement (the claim is supported) AND the ground for it: at 37 °C the 10 °C tube released maltose at 5.9 mg/min, close to the full 6.0 mg/min, so its fold was never lost and the cold had only slowed it, whereas the 65 °C tube stayed at 0.1 mg/min, so its fold was lost and did not re-form.

Evaluate needs the judgement and the ground for it. Do not award the point for 'supported' with no data, or for quoting the 10 °C rate alone with no comparison to the 65 °C tube.

Common slip: Giving the judgement without the ground, or the data without the judgement. Evaluate needs both: the claim is supported, because the 10 °C tube recovered and the 65 °C tube did not.

(e) Predict what happens to the amylase's rate when the saliva is swallowed into the stomach, where the pH is about 2, and justify your prediction. (1 point)

Hint: Is pH 2 close to this enzyme's optimal pH or far from it? Which stage of the pH effect applies at that distance?

A full-credit answer: In the stomach the rate falls to close to zero.
pH 2 is five units below the amylase's optimal pH of about 7, so the concentration of hydrogen ions is a hundred thousand times higher.
The changed concentration of hydrogen ions alters the charges on the R groups.
It also disrupts the hydrogen bonds and other weak interactions that hold the fold.
So the active site loses its shape, and the starch no longer fits.
The amylase is denatured.

Check the box for each point your answer earns

the rate falls to close to zero, because pH 2 is five units below the optimal pH: the far higher concentration of hydrogen ions (H⁺) alters the charges on the R groups and disrupts the weak interactions that hold the fold, so the active site loses its shape and the starch no longer fits (the amylase is denatured).

Accept "denatured" with the active site losing its shape, or the two-stage account (charges in the active site changed, then the fold lost). Do not award the point for "the rate rises because there are more hydrogen ions to react" or for "lower" with no reason.

Accept: the far higher concentration of hydrogen ions changes the charges on the R groups in the active site, so the starch no longer binds and the rate falls to close to zero, with or without the fold being named as lost.

Common slip: Predicting a faster rate because there are more hydrogen ions. Hydrogen ions do not take part in the reaction. They change the charges on the R groups and, this far from the optimal pH, pull the fold apart.

Free-response score: 0 of 5

Free response 2 · Conceptual Analysis · 4 points

An enzyme from a yogurt bacterium digests milk protein: it speeds up the reaction milk protein → amino acids. A researcher measured its rate (mg of protein digested per minute) at 37 °C with the same protein concentration at five pH values; the rates are in the table. A fold test showed the enzyme's overall fold intact at pH 5 and at pH 6. The researcher then moved a sample to pH 8: its rate fell to 1 mg/min and the fold test found only 15% of its molecules folded normally. Returned to pH 5 for 30 minutes, the same sample digested protein at 38 mg/min and 92% of its molecules were folded normally.

Rate of protein digestion at each pH, 37 °C.

(a) Determine the enzyme's optimal pH from the data, and describe how the rate changes on either side of it, using values. (1 point)

A full-credit answer: The optimal pH is 5, because the rate is greatest there: 42 mg/min.
On either side of pH 5 the rate falls.
Below pH 5 the rate falls to 30 mg/min at pH 4 and 8 mg/min at pH 3.
Above pH 5 the rate falls to 20 mg/min at pH 6 and 4 mg/min at pH 7.

Check the box for each point your answer earns

the decision (the optimal pH is 5) AND the observation it rests on (42 mg/min is the greatest rate in the table), with the fall on both sides described with values: 30 and 8 mg/min at pH 4 and 3; 20 and 4 mg/min at pH 6 and 7.

Determine needs the decision and the observation it rests on. Accept 'about pH 5' with at least two values quoted for the fall. Do not award the point for pH 7 because it is neutral, or for an optimum with no values.

Common slip: Deciding on pH 7 because pH 7 is neutral, or naming pH 5 with no value to rest it on. Determine needs the decision and the reading it rests on: pH 5 gives 42 mg/min, the greatest rate.

(b) Explain why the rate at pH 6 is lower than at pH 5, given that the fold test shows the enzyme's overall fold intact at both. (1 point)

A full-credit answer: At pH 6 the concentration of hydrogen ions is 10% of the concentration at pH 5.
The changed concentration of hydrogen ions alters the charges on the R groups in and around the active site.
So the milk protein's charges no longer match the pocket as well, and the milk protein binds less well.
Therefore the rate falls to 20 mg/min.
The fold as a whole is intact, so the enzyme is slowed by mismatched charges, not denatured.

Check the box for each point your answer earns

at pH 6 the concentration of hydrogen ions (H⁺) is lower than at pH 5, which alters the charges on the R groups in and around the active site, so the substrate's charges no longer match the pocket as well and it binds less well; the fold as a whole still holds, so this is not denaturation.

Accept 'the changed concentration of hydrogen ions changes the charges in the active site so the protein binds less well'. Do not award the point for 'the enzyme is denatured at pH 6' (the fold test rules that out) or for 'fewer collisions'.

Common slip: Saying the enzyme is denatured at pH 6. The fold test shows the fold intact. Near the optimum it is the charges in the active site that change, not the shape of the whole protein.

(c) Explain what the results at pH 8 and after the return to pH 5 show about this enzyme. (1 point)

A full-credit answer: At pH 8 the enzyme was denatured: only 15% of its molecules held their fold, and the rate fell to 1 mg/min.
Far from its optimal pH, the changed concentration of hydrogen ions disrupted the weak interactions holding the fold.
Back at pH 5, 92% of the molecules were folded again and the rate returned to 38 mg/min.
So the denaturation was reversible: the pH was restored before the unfolded chains tangled.

Check the box for each point your answer earns

at pH 8 the enzyme was denatured (only 15% folded, rate 1 mg/min), because far from the optimal pH the weak interactions holding the fold are disrupted and the active site loses its shape; the return of the fold (92%) and the rate (38 mg/min) at pH 5 shows that this denaturation was reversible: the disrupting condition was removed before the unfolded chains tangled, so the molecules refolded.

Accept 'denatured at pH 8, but reversibly'. Do not award the point for 'the enzyme was destroyed and replaced' (no cell is present to make new enzyme) or for 'the enzyme was only slowed at pH 8' (the fold test shows the fold lost).

Common slip: Stopping at 'the enzyme was denatured at pH 8'. The point needs the second half: the fold and the rate came back at pH 5, so this denaturation was reversible.

(d) At pH 5, the amount of enzyme is doubled while the tube keeps the same 200 mg of milk protein. Predict how the initial rate and the final amount of protein digested will compare with the original tube, and justify your prediction. (1 point)

A full-credit answer: The initial rate will be about twice as high, roughly 84 mg/min.
Doubling the enzyme doubles the number of active sites the milk protein can bind to, so the protein is converted twice as fast at the start.
The final amount digested will be the same, 200 mg, because 200 mg is all the protein there is.
More enzyme gets to the end sooner, not further.

Check the box for each point your answer earns

the initial rate roughly doubles (about 84 mg/min), because there are twice as many active sites, but the final amount of protein digested is the same, 200 mg, because it is set by the substrate available, so the richer tube finishes sooner.

Accept 'faster at first, same total in the end' with both reasons. Do not award the point for 'more protein is digested in the end' or for 'the rate is the same because the substrate is the same'.

Common slip: Predicting that more enzyme digests more protein in the end. The final amount is set by the substrate available. The enzyme changes only how quickly that amount is reached.

Free-response score: 0 of 4

Free response 3 · Conceptual Analysis · 3 points

An enzyme from the gut of a locust speeds up the reaction sucrose → glucose + fructose. A molecule found in the leaves the locust eats, L, has almost the same shape as sucrose. A researcher measured the enzyme's rate (nmol of glucose per minute) at 30 °C with 5 μM of sucrose, once with no L and once with L present. The enzyme, the temperature and the pH were the same in both tests. The rates are in the table. With no L, the rate at 500 μM of sucrose is 90 nmol/min and rises no further with more sucrose.

Rate of the locust enzyme at 5 μM of sucrose, with and without L.

(a) Explain how the results at 5 μM of sucrose show that L is an enzyme inhibitor. (1 point)

A full-credit answer: With L present the rate falls from 30 to 10 nmol/min.
The enzyme, the sucrose, the temperature and the pH are the same in both tests.
So the only change is the added L.
Therefore L binds to the enzyme and lowers its rate.
A molecule that binds to an enzyme and lowers its rate is an enzyme inhibitor.

Check the box for each point your answer earns

with L present the rate falls from 30 to 10 nmol/min while the enzyme, the sucrose concentration, the temperature and the pH are unchanged, so L must bind to the enzyme and lower its rate, which is what an enzyme inhibitor does.

Accept the evidence (only L was added and the rate fell) with the conclusion that L binds the enzyme and lowers its rate (its activity). Do not award the point for "the rate falls" alone, or for "L destroys the enzyme" or "L uses up the sucrose".

Common slip: Writing "the rate falls, so L is an inhibitor" with no link to the enzyme. The point needs the evidence (only L changed) and the conclusion that L binds the enzyme and lowers its rate.

(b) Explain, using L's shape, why fewer sucrose molecules are converted each second when L is present. (1 point)

A full-credit answer: L has almost the same shape as sucrose.
So L fits into the active site and sits there for a while.
While L sits in the active site, no sucrose molecule can enter it.
So fewer sucrose molecules are converted each second, and the rate falls.
L is a competitive inhibitor.

Check the box for each point your answer earns

because L resembles sucrose, L fits into the active site and sits there for a while; while L is in the active site sucrose cannot enter, so fewer sucrose molecules are converted each second (L is a competitive inhibitor).

Accept "L competes with sucrose for the active site". Do not award the point for "L changes the enzyme's shape" or "L binds elsewhere on the enzyme" (that is a noncompetitive inhibitor), or for "L is turned into product".

Common slip: Saying L "changes the enzyme's shape". A molecule shaped like the substrate acts by sitting in the active site itself. The point needs L in the active site, blocking sucrose while it is there.

(c) Predict the rate with L present at 500 μM of sucrose, and justify your prediction. (1 point)

A full-credit answer: With L present, the rate at 500 μM of sucrose will be close to 90 nmol/min.
L leaves the active site after a while, so the active site is often empty.
At 500 μM there are a hundred times as many sucrose molecules as at 5 μM.
So a sucrose molecule reaches each empty active site far more often than L does.
L is outcompeted, and the rate rises close to the uninhibited rate.

Check the box for each point your answer earns

a rate close to 90 nmol/min (accept 80 to 90 nmol/min), because L comes and goes from the active site, and with a hundred times as much sucrose, sucrose molecules reach each empty active site far more often than L does, so L is outcompeted.

Accept "close to the uninhibited rate" with the competition reasoning. Do not award the point for a rate near 30 nmol/min (a third, as at 5 μM) or for "L is used up" or "L is washed away".

Common slip: Predicting about 30 nmol/min, as at 5 μM. L does not hold a fixed share of the enzyme. With far more sucrose, the empty active site is filled by sucrose almost every time.

Free-response score: 0 of 3

Multiple choice checked: 0 of 10 correct.

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