End-of-topic test: Environmental Impacts

Unit 3 · Topic 3.2a end-of-topic test

Suggested time: about 48 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question. For each multiple-choice question, pick one option. When you have answered every question, press Submit the test; the feedback then gives the reasoning for each. For the two 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. Then 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

Rate of the same kind of reaction catalyzed by an enzyme from a mouse (solid) and by an enzyme from a compost-heap bacterium (dashed), at temperatures from 10 to 90 °C. pH and substrate concentration were the same in every tube. Gridlines every 10 °C and every 2 μmol/min; the rate axis ends at 16 μmol/min.

The graph shows the rate of the same kind of reaction catalyzed by two enzymes, one from a mouse and one from a bacterium that lives in a compost heap.

What is the optimal temperature of each enzyme?

A. Mouse 37 °C; bacterium 37 °C B. Mouse 25 °C; bacterium 75 °C C. Mouse 37 °C; bacterium 75 °C D. Mouse 37 °C; bacterium 90 °C

Question 2

Sucrase speeds up the reaction sucrose → glucose + fructose. A student holds a sample of sucrase at 78 °C for 20 minutes, cools the sample back to 37 °C and adds the sample to a sucrose solution. No glucose is released. A second sample of the same sucrase, kept at 37 °C throughout, releases glucose quickly.

What happened to the heated sucrase?

A. The covalent bonds of the sucrase's amino-acid chain were broken, so the chain fell apart B. The weak interactions holding the sucrase's fold were disrupted, so the active site lost its shape C. The sucrase was used up catalyzing the reaction while it was hot D. The sucrase molecules were still moving too slowly after cooling to collide with the substrate

Question 3

Rate of the amylase (mg of maltose released per minute) while each sample was held at its temperature, and after every sample was returned to 37 °C.

Amylase speeds up the reaction starch → maltose. A student held four equal samples of amylase at 4, 25, 37 or 70 °C for 15 minutes. He measured each sample's rate twice: at its holding temperature, then after returning every sample to 37 °C. The table gives the rates in mg of maltose, a sugar, released per minute.

Which treatment denatured the amylase?

A. 4 °C B. 25 °C C. 37 °C D. 70 °C

Question 4

Plain yogurt has a pH of about 4. The water in a swimming pool has a pH of about 7.

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

A. The yogurt has three times the concentration of hydrogen ions of the pool water B. The yogurt has a hundred times the concentration of hydrogen ions of the pool water C. The yogurt has a thousand times the concentration of hydrogen ions of the pool water D. The pool water has a thousand times the concentration of hydrogen ions of the yogurt

Question 5

Rate of the caterpillar enzyme at each pH, 37 °C.

A student tests a digestive enzyme from the gut of a caterpillar at 37 °C, with the same substrate concentration, in solutions of pH 4, 6, 8 and 10. The rates are in the table. The student expects the enzyme to work best at pH 7, because pH 7 is neutral.

Of the pH values tested, which is closest to the enzyme's optimal pH, and what do the results show?

A. pH 4; a higher concentration of hydrogen ions always speeds an enzyme up B. pH 6, the value nearest to neutral; every enzyme works best near pH 7 C. pH 8; different enzymes have different optimal pH values D. pH 10; a lower concentration of hydrogen ions always speeds an enzyme up

Question 6

An amylase from a bacterium that ferments cabbage into sauerkraut has an optimal pH of 5. Amylase speeds up the reaction starch → maltose. A student gives two equal samples of the amylase the same starch solution at 30 °C, one at pH 5 and one at pH 7. At pH 5 the initial rate is 2.4 mg of starch broken down per minute; at pH 7 it is 0.6 mg/min. At pH 7 the amylase keeps its overall fold.

Why is the rate lower at pH 7?

A. The lower concentration of hydrogen ions (H⁺) alters the charges on R groups at the active site, so the starch binds less well B. The lower concentration of hydrogen ions means there is less acid to break the starch down alongside the amylase C. At pH 7 the hydrogen bonds holding the amylase's fold break, so the whole enzyme loses its shape D. The lower concentration of hydrogen ions means fewer collisions between amylase and starch molecules each second

Question 7

A researcher dissolves a purified ribonuclease, an enzyme that speeds up the reaction RNA → shorter pieces of RNA, in a strong urea solution at 25 °C. Its rate falls to 3% of normal, and a folding test (a count of how many enzyme molecules still hold their fold) finds 6% folded. The researcher then lets the urea diffuse out through a membrane. The rate returns to 95% of normal, and 93% of the molecules are folded.

What do these results show?

A. The enzyme was only slowed by the urea; its fold was never lost B. The urea destroyed the enzyme, and new enzyme molecules were made C. The urea sat in the active site, blocking the RNA while it was there D. The enzyme was denatured and then refolded once the urea was removed

Question 8

Mean rate at each salt concentration.

A student tests an enzyme from an aquarium filter at four salt concentrations. She keeps every tube at 25 °C and pH 7.5, with equal amounts of enzyme and substrate. The mean rates are in the table.

Which additional measurement would show whether the enzyme's shape itself changed at 35 g/L?

A. The rate at 35 g/L when the substrate concentration is doubled B. How many enzyme molecules still hold their normal fold at each salt level C. The amount of product formed at 35 g/L after 60 minutes rather than 10 D. The pH of each tube, measured a second time to confirm it

Question 9

Initial rate of an enzyme-catalyzed reaction with a fixed amount of enzyme, at substrate concentrations from 1 to 12 mmol/L. Numbers are the measured rates in μmol/min. Gridlines every 2 mmol/L and every 10 μmol/min.

The graph shows the initial rate of a reaction with a fixed amount of enzyme as a student raises the substrate concentration.

Why does the rate rise steeply from 1 to 3 mmol/L of substrate, yet barely change from 6 to 12 mmol/L?

A. From 1 to 3 mmol/L each added substrate molecule finds empty active sites; by 6 mmol/L nearly every active site is already occupied all the time B. From 1 to 3 mmol/L the enzyme is still being made; by 6 mmol/L the extra substrate has used the enzyme up C. From 1 to 3 mmol/L the extra substrate lowers the activation energy; by 6 mmol/L the extra substrate has denatured the enzyme D. From 1 to 3 mmol/L no product has built up yet; by 6 mmol/L product is being turned back into substrate as fast as it forms

Question 10

Amylase speeds up the reaction starch → maltose. Two tubes each hold 8.0 mg of starch in the same volume at the same temperature and pH. Tube 1 receives 0.10 mg of amylase; tube 2 receives 0.30 mg of the same amylase. Both tubes are left until all the starch is gone.

Which of the following describes tube 2, compared with tube 1?

A. Same initial rate; same final amount of maltose B. Same initial rate; more maltose in the end C. Faster initial rate; same final amount of maltose D. Faster initial rate; more maltose in the end

Question 11

Glucose isomerase speeds up the reaction glucose → fructose. Two tubes start with the same amount of the enzyme and 10 mmol/L of glucose. In tube 1 the fructose builds up: after 20 minutes the amount of fructose stops rising, with 3 mmol/L of glucose left. In tube 2 the fructose is removed as it forms, and the glucose is used up. At 20 minutes, enzyme taken from tube 1 converts fresh glucose at its full initial rate.

Why has tube 1 stopped making fructose with glucose left, while tube 2 has used up all of its glucose?

A. Tube 1's enzyme is used up after 20 minutes; tube 2 has more enzyme, so its enzyme lasts until the glucose is gone B. Tube 1's fructose denatures its enzyme; in tube 2 the fructose is removed before it can denature the enzyme C. All of tube 1's glucose had been converted by 20 minutes; tube 2 only looks different because its fructose was removed D. In tube 1 fructose is turned back into glucose as fast as glucose is converted; in tube 2 no fructose stays to be turned back

Question 12

Rate with and without inhibitor Q.

A student measures an enzyme's rate at rising substrate concentrations, with no inhibitor and with inhibitor Q present. The rates, in μmol/min, are in the table.

What kind of inhibitor is Q, and what in the data shows it?

A. Competitive: with more substrate, the inhibited rate catches up with the uninhibited rate B. Competitive: even with more substrate, the inhibited rate stays well below the uninhibited rate C. Noncompetitive: with more substrate, the inhibited rate catches up with the uninhibited rate D. Noncompetitive: even with more substrate, the inhibited rate stays well below the uninhibited rate

Question 13

A cheesemaker added chymosin, the enzyme used to set milk into curd for cheese, to two portions of the same milk. At 10 °C the milk took 40 minutes to set; at 30 °C it took 8 minutes. The enzyme kept its shape at both temperatures.

Why does the milk set faster at 30 °C?

A. The warmth lowers the activation energy of the reaction, so fewer collisions are needed B. The molecules of enzyme and milk protein move faster, so they collide more often and with more energy C. The active site opens wider at 30 °C, so the milk protein fits into it more easily D. The warm milk holds more protein for the enzyme to act on, so product forms sooner

Question 14

Rate of a reaction catalyzed by an enzyme from a mushroom, at temperatures from 0 to 60 °C, with the same substrate concentration in every tube. Gridlines every 10 °C and every 2 μmol/min.

The graph shows the rate of a reaction catalyzed by an enzyme from a mushroom, measured at temperatures from 0 to 60 °C with the same substrate concentration in every tube.

Which of the following explains the whole shape of the curve?

A. Collisions grow more frequent up to 30 °C; above it the substrate is used up before the rate can be measured B. The activation energy falls up to 30 °C and rises again above it, so fewer collisions succeed C. Collisions grow more frequent up to 30 °C; above it the molecules move too fast to stay in the active site D. Collisions grow more frequent up to 30 °C; above it denaturation removes working enzymes faster than the extra collisions help

Question 15

A researcher tests a purified enzyme from red blood cells in two cell-free mixtures with the same substrate concentration, pH and temperature. On its own the enzyme makes 60 nmol of product per minute; with a small amount of a new drug dissolved in the mixture it makes 15 nmol/min. The drug is the only difference between the two mixtures.

Which of the following explains these results?

A. The drug bound to the enzyme and lowered its rate B. The drug reacted with the substrate and used it up before the test began C. The drug was turned into product in place of the substrate D. The drug cooled the mixture, so the molecules collided less often

Question 16

A researcher adds a molecule shaped almost exactly like the substrate of a kidney enzyme to a mixture of the enzyme and 2 μM of its substrate. The rate falls from 40 to 16 nmol/min. The researcher then raises the substrate concentration to 200 μM, with the same amount of the look-alike molecule, enzyme, temperature and pH.

Which of the following predicts the rate at 200 μM of substrate with the look-alike present, compared with the uninhibited rate at 200 μM?

A. The rate stays at about 40% of the uninhibited rate, because the look-alike holds a fixed share of the enzyme molecules B. The rate rises to close to the uninhibited rate, because substrate now reaches each empty active site far more often than the look-alike C. The rate falls further below the uninhibited rate, because the extra substrate crowds the look-alike molecules into the active sites D. The rate rises above the uninhibited rate, because the look-alike and the substrate are both turned into product

Question 17

A compound from the leaves of the foxglove plant halves the rate of an enzyme in heart-muscle cells at every substrate concentration tested, from 1 μM to 1,000 μM, with equal enzyme in every tube. The compound looks nothing like the substrate.

Where does the foxglove compound bind, and what does its binding do?

A. In the active site; it blocks the substrate while it sits there B. In the active site; it is turned into product more slowly than the substrate is C. At a site away from the active site; it cuts the peptide bonds of the chain D. At a site away from the active site; it changes the enzyme's shape so the active site works poorly

Question 18

Rate of the same kind of reaction catalyzed by an enzyme from a mold (solid) and by an enzyme from a soil bacterium (dashed), at pH 1 to 12. Temperature and substrate concentration were the same in every tube. Gridlines every pH unit and every 2 μmol/min; the rate axis ends at 12 μmol/min.

The graph shows the rate of the same kind of reaction catalyzed by two enzymes: one from a mold that grows on fruit and one from a soil bacterium. Temperature and substrate concentration were the same in every tube. The mold enzyme's optimal pH is 4. At pH 9, five units above its optimal pH, its rate is close to zero.

Why is the mold enzyme's rate close to zero at pH 9?

A. The lower concentration of hydrogen ions (H⁺) means that enzyme and substrate molecules collide far less often each second B. The changed concentration of hydrogen ions (H⁺) disrupts the weak interactions that hold the fold, so the active site loses its shape C. The charges on R groups at the active site have changed while the fold holds, so the substrate binds only a little less well D. The lower concentration of hydrogen ions (H⁺) breaks the covalent peptide bonds of the chain, so the chain falls into pieces

Question 19

Lactase speeds up the reaction lactose → glucose + galactose. A student adds a small amount of dishwashing detergent to a tube of lactase and milk at 37 °C and pH 7. The detergent takes part in no reaction, and the temperature and pH do not change. The rate falls from 8 to 2 mg of glucose released per minute.

Which of the following explains the fall in rate?

A. The detergent breaks the covalent peptide bonds of the lactase's chain, so the chain falls into separate amino acids B. The detergent is turned into product by the lactase, so less lactose is converted into glucose each minute C. The detergent disrupts the weak interactions that hold the lactase's fold, so the active site loses its shape D. The detergent slows the lactase and lactose molecules down, so they collide less often each second

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 · 4 points

Detergent makers add a lipase from a soil bacterium to laundry detergents to break down the fats in stains. Lipase speeds up the reaction fat → fatty acids + glycerol. A researcher measured the rate at which the purified lipase released fatty acids from a fat at five temperatures, with the same fat concentration and the same pH in every tube and five trials at each temperature. The table gives the mean rate at each temperature; the differences between temperatures were far larger than the variation between the five trials at any one temperature. After the 60 °C trials, the researcher cooled one 60 °C tube to 40 °C and measured that tube again: 0.5 mg/min. The researcher will warm a separate tube, held at 5 °C, to 40 °C and measure that tube.

Mean rate at which the purified lipase released fatty acids at five temperatures, five trials each, with the same fat concentration and pH in every tube.

(a) Identify the optimal temperature of the lipase among those tested, and describe how the rate changes on either side of it, using values from the table. (1 point)

A full-credit answer: The optimal temperature is 40 °C, because the rate is greatest there: 9.6 mg/min.
Below 40 °C the rate rises with temperature, from 3.0 mg/min at 20 °C to 5.8 mg/min at 30 °C. Above 40 °C the rate falls, to 6.2 mg/min at 50 °C and 0.4 mg/min at 60 °C.

Check the box for each point your answer earns

40 °C (9.6 mg/min, the greatest rate), with the rate rising toward it (3.0 and 5.8 mg/min at 20 and 30 °C) and falling above it (6.2 and 0.4 mg/min at 50 and 60 °C).

Accept "about 40 °C" with at least two values quoted for the rise or the fall. Do not award the point for 60 °C as the optimal temperature, or for the optimum with no description of the pattern.

Common slip: Writing "the rate increases with temperature" and stopping, or picking 60 °C because it is the hottest tube. The optimal temperature is where the rate itself is greatest. A describe-the-data point needs values on both sides of the optimal temperature.

(b) Explain why the rate is higher at 40 °C than at 20 °C. (1 point)

A full-credit answer: Warming a solution makes its molecules move faster.
So the lipase molecules and the fat 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 3.0 mg/min at 20 °C to 9.6 mg/min at 40 °C.

Check the box for each point your answer earns

warming makes the molecules move faster, so enzyme and fat molecules 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 between enzyme and substrate, 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" or that heat "gives the enzyme energy". 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 60 °C is close to zero. (1 point)

A full-credit answer: Above the optimal temperature, the heat disrupts the hydrogen bonds and other weak interactions that hold the lipase's fold.
So the active site loses its shape.
The fat no longer fits the active site, so the enzyme can no longer catalyze the reaction.
The lipase is denatured.
Denaturation removes working enzymes faster than the extra collisions can help, so the rate falls to 0.4 mg/min.

Check the box for each point your answer earns

at 60 °C the heat disrupts the hydrogen bonds and other weak interactions that hold the lipase's fold, so the active site loses its shape and the fat no longer fits (the enzyme is denatured), and this loss outweighs the extra collisions.

Accept "denatured" only with what it does to the active site or to substrate binding. Do not award the point for "the peptide bonds break" or for "denatured" with nothing after it.

Common slip: Writing "the enzyme denatures" with nothing after it, 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) Predict the rate of the 5 °C tube after it is warmed to 40 °C, and justify your prediction using the result for the cooled 60 °C tube. (1 point)

A full-credit answer: The 5 °C tube will release fatty acids at about 9.6 mg/min once it is at 40 °C.
Cold slows an enzyme because its molecules move more slowly and collide less often.
The cold does not disturb the fold.
So the rate returns as soon as the tube warms.
The cooled 60 °C tube shows the other case.
Its rate stayed at 0.5 mg/min at 40 °C, because a fold pulled apart by strong heating does not re-form.

Check the box for each point your answer earns

about 9.6 mg/min (accept 9 to 10 mg/min), because cold only slows an enzyme, its fold is kept and the rate returns on warming; by contrast the 60 °C tube stayed at 0.5 mg/min at 40 °C because a fold lost to strong heating does not re-form.

Accept "the full 40 °C rate" for the prediction, and accept the justification framed either way: cold leaves the fold intact so the rate returns, with the cooled 60 °C tube as the contrasting case; or the 60 °C result first, with the 5 °C tube as the opposite case. Do not award the point for a rate near 0.5 mg/min, or for the right rate with no reference to the cooled 60 °C tube.

Common slip: Predicting a rate near 0.5 mg/min, as if the cold tube had been damaged like the hot one. Recovery on warming is the mark of an enzyme slowed by cold. A rate that stays low after the return is the mark of an enzyme denatured by heat.

Free-response score: 0 of 4

Free response 2 · Conceptual Analysis · 4 points

An enzyme from a soil bacterium converts a toxin, T, into a harmless product. The enzyme's rate (nmol of product per minute) was measured at two concentrations of T: with no other molecule present, with drug D present, and with pesticide P present. D has almost the same shape as T; P has no resemblance to T. Temperature, pH and the amount of enzyme were the same in every test. The rates are in the table. With no inhibitor present, the rate at 200 μM of T is 95 nmol/min and rises no further with more T.

Rate of the enzyme at two concentrations of T.

(a) Explain how the results at 2 μM of T show that D and P are both enzyme inhibitors. (1 point)

A full-credit answer: With D present the rate falls from 40 to 20 nmol/min, and with P present it also falls to 20 nmol/min.
The enzyme, the amount of T, the temperature and the pH are the same in every test.
So the only change is the added molecule.
Therefore D and P each bind to the enzyme and lower 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 D or with P present the rate falls (from 40 to 20 nmol/min) while the enzyme, T, temperature and pH are unchanged, so each added molecule must bind to the enzyme and lower its rate, which is what an enzyme inhibitor does.

Accept the evidence (the rate falls with only the added molecule changed) with the conclusion that the molecule binds the enzyme and lowers its rate (its activity). Do not award the point for "the rate falls" with no link to binding the enzyme, or for "D and P destroy the enzyme" or "use up the substrate".

Common slip: Writing "the rate falls, so they are inhibitors" with no link to the enzyme. The point needs the evidence (only the added molecule changed) and the conclusion that each molecule binds the enzyme and lowers its rate.

(b) Explain why raising the concentration of T from 2 to 20 μM almost removed the effect of D. (1 point)

A full-credit answer: D is shaped like T.
So D binds reversibly in the active site, and D blocks T only while D sits there.
When T is ten times more concentrated, T molecules reach each empty active site far more often than D molecules do.
So D is outcompeted.
The rate rises from 20 nmol/min toward the uninhibited 90 nmol/min, reaching 81 nmol/min.

Check the box for each point your answer earns

D resembles T and binds reversibly at the active site, blocking T only while it sits there; with ten times as much T, molecules of T reach the empty active sites far more often than D does, so D is outcompeted and the rate rises toward the uninhibited value (D is a competitive inhibitor).

Accept an answer in terms of the two molecules competing for the same site. Do not award the point for "D was used up" or "D was washed away".

Common slip: Saying the extra T "knocked D out" or that D "was used up". D comes and goes on its own. The point is that with more T, the empty active site is far more likely to be filled by T than by D.

(c) Make a claim about how the rate with P present at 200 μM of T will compare with the uninhibited rate of 95 nmol/min at that concentration. (1 point)

A full-credit answer: The rate with P present will stay at about half the uninhibited rate, roughly 45 to 48 nmol/min.
It will not climb back toward 95 nmol/min.

Check the box for each point your answer earns

a correct, specific claim: the rate with P present will still be far below 95 nmol/min, about half (roughly 45 to 48 nmol/min).

Make a claim earns the point for the assertion; the reasoning is scored in part (d). Accept "about half the uninhibited rate" without a number. Do not award the point for a claim that the rate rises to about 95.

Common slip: Claiming that enough T will bring the rate back to 95 nmol/min, as it did for D. The data already show that ten times more T left P's effect unchanged.

(d) Support your claim by comparing how P and D each interact with the enzyme. (1 point)

A full-credit answer: D resembles T, so D sits in the active site, and plentiful T outcompetes D.
P does not resemble T, so P binds at an allosteric site.
P's binding changes the enzyme's shape, so the active site works poorly.
P and T do not compete for the same site, so more T cannot push P off.
So at 200 μM the rate with P stays at about half, as it did from 2 to 20 μM.

Check the box for each point your answer earns

the evidence (P's effect did not shrink when T rose from 2 to 20 μM, unlike D's) AND the reasoning that links it to the claim: P binds at an allosteric site and bends the enzyme so the active site works poorly, so T does not compete with P and more T cannot push it off, whereas D sits in the active site and is outcompeted.

Support a claim needs the evidence and the link to the claim. Accept "binds somewhere other than the active site and bends the enzyme" for the mechanism, with D placed in the active site for the contrast. Do not award the point for P "blocking the active site", or for the label "noncompetitive" alone with no mechanism.

Common slip: Writing "noncompetitive" and stopping. Supporting the claim needs the mechanism and the data: binding away from the active site, a changed shape, and the effect that did not shrink as T rose.

Free-response score: 0 of 4

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