Environmental Impacts on Enzyme Function

Unit 3 · Topic 3.2 end-of-topic test

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 two free-response questions, write your answer in full sentences (and show any calculation), then open the scoring guide and mark your own work against it. Formulas: mean x̄ = Σxᵢ/n; standard deviation s = √(Σ(xᵢ − x̄)²/(n − 1)); standard error SE = s/√n; an error bar of ±2SE covers the range the true mean is likely to lie in. Rounding: carry three significant figures through your working and round only the final value you report.

Question 1

Rate of the same reaction catalyzed by two enzymes, at temperatures from 10 to 90 °C. pH and substrate concentration were the same in every tube.

The graph shows the rate of the same kind of reaction catalyzed by two enzymes, one from a human and one from a bacterium that lives in a hot spring.

What is the optimal temperature of each enzyme?

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

Question 2

A sample of catalase is held at 75 °C for 15 minutes, cooled back to 37 °C and added to hydrogen peroxide. No oxygen is released. A second sample of the same catalase, kept at 37 °C throughout, releases oxygen quickly.

What happened to the heated catalase?

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

Question 3

Four equal samples of amylase were each held at one temperature for 15 minutes. The rate of each (mg of maltose, a sugar, released per minute) was measured twice: first at the holding temperature, then after every sample had been returned to 37 °C. Sample held at 4 °C: 5, then 22. Sample held at 25 °C: 14, then 22. Sample held at 37 °C: 22, then 22. Sample held at 70 °C: 0, then 0.

Which treatment denatured the amylase?

A. 4 °C B. 25 °C C. 37 °C D. 70 °C

Question 4

Tomato juice has a pH of about 4 and black coffee a pH of about 5.

How do their hydrogen ion (H⁺) concentrations compare?

A. Coffee has ten times the H⁺ concentration of tomato juice B. Coffee has a hundred times the H⁺ concentration of tomato juice C. Tomato juice has ten times the H⁺ concentration of coffee D. Tomato juice has a hundred times the H⁺ concentration of coffee

Question 5

An enzyme from the lining of the small intestine was tested at 37 °C with the same substrate concentration in solutions of pH 4, 6, 8 and 10. It broke down 1, 14, 38 and 12 mg of substrate per minute. A student expects the enzyme to work best at pH 7, because that is neutral.

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

A. pH 4; more H⁺ in the solution always speeds an enzyme up B. pH 6, nearest to neutral; every enzyme works best near pH 7 C. pH 8; different enzymes have different optimal pH values D. pH 10; fewer H⁺ in the solution always speed an enzyme up

Question 6

Two tubes held equal amounts of pepsin and of protein (20 mg/mL) at 37 °C, with no product present at the start. At pH 2 the initial rate was 3.0 mg of protein digested per minute; at pH 5 it was 0.8 mg/min. At pH 5 pepsin keeps its overall fold.

Why was the rate lower at pH 5?

A. Fewer H⁺ altered the charges on R groups at the active site, so the substrate bound less well B. Fewer H⁺ meant there was less acid to break the protein down alongside the pepsin C. At pH 5 the hydrogen bonds holding pepsin's fold broke, so the enzyme lost its shape D. Fewer H⁺ meant fewer collisions between pepsin and protein molecules each second

Question 7

An enzyme from yeast, in a cell-free solution, loses 95% of its activity while a mild chemical is present. When the chemical is washed out, its activity returns to 92% of the original. A test of how many enzyme molecules hold their normal fold shows 8% folded while the chemical is present and 90% folded after the wash.

What do these results show?

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

Question 8

An enzyme from an aquarium filter was tested at salt concentrations of 5, 15, 25 and 35 g/L. Temperature was 25 °C and pH 7.5 in every tube, with equal amounts of enzyme and substrate. The mean rates were 18, 31, 24 and 9 μmol/min.

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 0.5 to 8 mmol/L. Numbers are the measured rates in μmol/min.

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

Why does raising the substrate concentration from 4 to 8 mmol/L barely change the rate?

A. Nearly every active site is already occupied all the time B. The extra substrate has used up the enzyme C. The extra substrate has denatured the enzyme D. Product has built up and is turning back into substrate

Question 10

Two tubes each contain 20 μmol of substrate S in the same volume at the same temperature and pH. Tube H has twice as much enzyme as tube L. Each molecule of S becomes one molecule of product P, and both reactions run until all the S is gone.

How does tube H compare with tube L?

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

Question 11

Two sealed tubes start with the same enzyme and 10 mmol/L of substrate. In tube 1 the product is left to build up: after 20 minutes the amount of product stops rising, although 3 mmol/L of substrate remains. In tube 2 the product is removed continuously as it forms, and the substrate is used up completely. At 20 minutes a sample of the enzyme from tube 1, moved into fresh substrate with no product present, works at its full starting rate.

Why did tube 1 stop making product?

A. The enzyme molecules were used up over the 20 minutes B. All the substrate had been converted by 20 minutes C. The product that built up denatured the enzyme molecules D. As product built up, the reverse reaction matched the forward one

Question 12

An enzyme's rate (μmol/min) was measured at rising substrate concentrations with no inhibitor and with inhibitor Q present. At 1 mmol/L of substrate: 10 without Q, 4 with Q. At 2 mmol/L: 18 and 10. At 4 mmol/L: 30 and 24. At 8 mmol/L: 40 and 36.

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

Five tubes of catalase at 30 °C released oxygen at 4.2, 3.8, 4.5, 4.1 and 3.9 mL/min.

What mean rate should be reported for 30 °C?

A. 0.7 mL/min B. 4.10 mL/min C. 5.13 mL/min D. 20.5 mL/min

Question 14

Five trials with enzyme A gave rates with a mean of 8.0 mL/min and a standard deviation of 0.3 mL/min. Five trials with enzyme B gave a mean of 5.0 mL/min and a standard deviation of 0.9 mL/min.

Which statement do these values support?

A. Enzyme B's rate is higher, because its standard deviation is larger B. Enzyme A's five readings all lay within 0.3 mL/min of one another C. Enzyme B's readings were less accurate, so its mean of 5.0 should be dropped D. Enzyme B's readings varied more from one trial to the next

Question 15

Five readings of an enzyme's rate were 6.0, 6.4, 5.6, 6.2 and 5.8 μmol/min. Their mean is 6.00 μmol/min. Use the formula on the AP sheet, s = √(Σ(xᵢ − x̄)²/(n − 1)).

What is the standard deviation of these readings?

A. 0.10 μmol/min B. 0.283 μmol/min C. 0.316 μmol/min D. 0.40 μmol/min

Question 16

Nine trials of a reaction gave a mean rate of 12.0 mL/min with a standard deviation of 0.60 mL/min.

What is the standard error of this mean?

A. 0.067 mL/min B. 0.20 mL/min C. 0.60 mL/min D. 1.80 mL/min

Question 17

The student's graph: mean rate of oxygen release by catalase at two temperatures, five trials each.

A student plots the mean rate of oxygen release by catalase at two temperatures, five trials each, and draws an error bar on each mean. Her legend on the graph reads "error bars = one standard deviation". She concludes that catalase is faster at 30 °C than at 20 °C, because there is a clear gap between the two bars.

Which statement about her conclusion is correct?

A. Her bars show how spread out the readings were; the overlap test needs ±2SE bars B. Her bars show where each true mean is likely to lie, so the conclusion is sound C. A gap between one-standard-deviation bars is the overlap test itself, so no further check is needed D. Five trials is too few for any conclusion, whatever the bars show

Question 18

Mean rate of an enzyme at three pH values, five trials each, same temperature and substrate concentration. Error bars represent ±2SE.

The graph shows the mean rate of an enzyme at pH 6, 7 and 8.

What do the error bars show about the rates at pH 7 and pH 8?

A. The rates at pH 7 and pH 8 are the same B. These data do not show a difference between pH 7 and pH 8 C. pH 8 is faster, because its mean is higher D. Every reading at pH 8 lay above every reading at pH 7

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

Students measured the rate at which an amylase extracted from germinating wheat seeds breaks starch down into the sugar maltose, at four temperatures, with five trials at each temperature. The pH and the starch concentration were the same in every trial. The table gives the mean rate at each temperature and the standard error of that mean (n = 5). The students plan to plot the four means as bars with error bars representing ±2SE.

Mean rate of maltose release by wheat amylase at four temperatures, five trials each. The standard error of each mean is given; n = 5.

(a) Describe the relationship between temperature and the mean rate shown in the table, using values from the table. (1 point)

A full-credit answer: The mean rate rises with temperature from 20 to 40 °C, from 2.20 to 4.40 to 7.20 mg of maltose per minute, and then levels off: at 50 °C the mean is 7.40 mg/min, close to the 40 °C value.

Check the box for each point your answer earns

the mean rate rises with temperature from 20 to 40 °C (2.20 to 4.40 to 7.20 mg/min) and is about the same from 40 to 50 °C (7.20 and 7.40 mg/min).

Accept "rises steeply, then levels off" with at least two values quoted. Do not award the point for a trend with no values, or for a description that has the rate falling between 40 and 50 °C.

Common slip: Writing "the rate increases with temperature" and stopping. A describe-the-data point needs the values and the whole pattern, including the leveling off from 40 to 50 °C.

(b) Calculate the range covered by a ±2SE error bar for 40 °C and for 50 °C. (1 point)

A full-credit answer: For 40 °C the bar runs from 6.94 to 7.46 mg/min, and for 50 °C from 7.01 to 7.79 mg/min.

Write down the values in the question:

40 °C: x̄ = 7.20 mg/min, SE = 0.131 mg/min
50 °C: x̄ = 7.40 mg/min, SE = 0.195 mg/min

Write down the equation:

error bar runs from x̄ − 2SE to x̄ + 2SE

Substitute in the values, and calculate:

40 °C: 2SE = 2 × 0.131 = 0.262
bar: 7.20 − 0.262 = 6.94 to 7.20 + 0.262 = 7.46 mg/min
50 °C: 2SE = 2 × 0.195 = 0.390
bar: 7.40 − 0.390 = 7.01 to 7.40 + 0.390 = 7.79 mg/min

Check the box for each point your answer earns

both ranges: 40 °C, 2SE = 0.262, so the bar runs from 6.94 to 7.46 mg/min; 50 °C, 2SE = 0.390, so the bar runs from 7.01 to 7.79 mg/min (accept each limit within ±0.01).

Accept the ranges written as 7.20 ± 0.26 and 7.40 ± 0.39. Do not award the point for bars of ±1SE, or for bars built from the standard deviation.

Common slip: Adding and subtracting one SE instead of two, or using the standard deviation. The bar the students planned is ±2SE, the range the true mean is likely to lie in.

(c) One student claims that the enzyme works faster at 50 °C than at 40 °C. Use the two error bars to evaluate the claim, and state whether the null hypothesis of no difference between 40 and 50 °C is rejected. (1 point)

A full-credit answer: The 40 °C bar (6.94 to 7.46) and the 50 °C bar (7.01 to 7.79) overlap, so the gap between the means, 7.40 against 7.20, could be chance. These data do not show a difference between the two temperatures, so the null hypothesis of no difference is not rejected. That is not the same as showing the two rates are equal.

Check the box for each point your answer earns

the two ±2SE bars overlap (6.94 to 7.46 shares its range with 7.01 to 7.79), so the difference between the means could be chance; the data do not show a difference, and the null hypothesis is not rejected.

Accept "the claim is not supported by these data". Do not award the point for "the rates are the same" (overlap shows no difference, not equality), or for a decision made from the two means alone.

Common slip: Reading overlapping bars as "the rates are the same". Overlap means the data have not shown a difference; the true means may still differ.

(d) The students propose repeating the experiment at 75 °C. Predict how the mean rate at 75 °C will compare with the mean at 50 °C, and justify your prediction in terms of the enzyme's structure. (1 point)

A full-credit answer: The mean rate at 75 °C will be far lower than at 50 °C, close to zero. Warming speeds collisions, but above the optimum the heat disrupts the hydrogen bonds and other weak interactions that hold the enzyme's fold, so the active site loses its shape, the starch no longer fits, and the denatured enzyme can no longer catalyze the reaction; that loss far outweighs the extra collisions.

Check the box for each point your answer earns

the mean rate at 75 °C will be much lower than at 50 °C (near zero), because above the enzyme's optimum heat disrupts the hydrogen bonds and other weak interactions that hold the protein's fold, so the active site loses its shape and the substrate no longer fits (denaturation), and this loss outweighs the extra collisions that warming brings.

Accept "denatured" only with what it does to the active site or to substrate binding. Do not award the point for "faster, because molecules move faster", or for "lower" with no structural reason.

Common slip: Predicting a higher rate because hotter molecules collide more often. That is true only below the optimum; the leveling off from 40 to 50 °C is already the sign that denaturation has begun to cancel the gain.

Free-response score: 0 of 4

Free response 2 · Conceptual Analysis · 4 points

A liver enzyme converts a toxin, T, into a harmless product. Researchers measured the enzyme's rate (nmol of product per minute) 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. At 2 μM of T: 48 with no inhibitor, 24 with D, 24 with P. At 20 μM of T: 92 with no inhibitor, 84 with D, 46 with P. With no inhibitor present, the rate at 200 μM of T is 96 nmol/min and rises no further when more T is added.

(a) Describe what an enzyme inhibitor is and what it does to the reaction the enzyme catalyzes. (1 point)

A full-credit answer: An enzyme inhibitor is a molecule that binds to an enzyme and lowers its activity, so the reaction the enzyme catalyzes runs more slowly: here, both D and P halve the rate at 2 μM of T.

Check the box for each point your answer earns

an inhibitor is a molecule that binds to an enzyme and lowers its activity, so the rate of the reaction the enzyme catalyzes falls.

Accept "slows the reaction" for the effect. Do not award the point for "destroys the enzyme" or "uses up the substrate".

Common slip: Saying the inhibitor "stops the reaction" or "kills the enzyme". An inhibitor lowers the enzyme's activity; the point needs binding to the enzyme and a lower 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 it binds reversibly in the active site and blocks T only while it sits there. When T is ten times more concentrated, T molecules reach each empty active site far more often than D does, so D is outcompeted and the rate rises from 24 toward the uninhibited 92 (84 nmol/min). D is a competitive inhibitor.

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 "ran out". 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) Predict how the rate with P present at 200 μM of T will compare with the uninhibited rate of 96 nmol/min at that concentration. (1 point)

A full-credit answer: The rate with P present will stay far below 96 nmol/min, at about half, roughly 46 to 50. Without P the enzyme is already saturated at 96, every active site busy, and P's effect did not shrink when T rose from 2 to 20 μM (24 against 48, then 46 against 92), so adding still more T will not remove it.

Check the box for each point your answer earns

the rate with P present will still be far below 96, about half (roughly 46 to 50 nmol/min); the uninhibited enzyme is saturated at 96 (nearly every active site occupied all the time), and P's effect has not shrunk as T rose from 2 to 20 μM, so more T will not remove it.

Accept "about half the uninhibited rate" without a number. Do not award the point for a prediction that the rate rises to about 96.

Common slip: Predicting that enough T will bring the rate back to 96, as it did for D. The data already show that ten times more T left P's effect unchanged.

(d) Justify your prediction using where P binds on the enzyme and what its binding does to the active site. (1 point)

A full-credit answer: P does not resemble T, so it does not sit in the active site. It binds at an allosteric site, a site elsewhere on the enzyme, and its binding changes the enzyme's shape so the active site works poorly. Because P and T are not competing for the same site, more T cannot push P off, and the rate stays about half whatever the T concentration. P is a noncompetitive inhibitor.

Check the box for each point your answer earns

P binds at an allosteric site, a site on the enzyme other than the active site, and its binding changes the enzyme's shape so that the active site works poorly; because P does not compete with T for the active site, raising the T concentration does not displace it, so the rate stays low however much T is added (P is a noncompetitive inhibitor).

Accept "binds somewhere other than the active site and bends the enzyme" for the mechanism. Do not award the point for P "blocking the active site" or for the classification alone with no mechanism.

Common slip: Writing "noncompetitive" and stopping. The point needs the mechanism: binding away from the active site, a changed shape, and why more T therefore does not help.

Free-response score: 0 of 4

Multiple choice checked: 0 of 18 correct.

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