3.2 Environmental Impacts on Enzyme Function

1. AIts peptide bonds are broken, so it has been split up into its separate amino acids.
   A student who thinks denaturation breaks a protein into amino acids picks this. Denaturation disrupts the bonds and interactions that hold the folded shape; the peptide bonds that join the amino acids stay intact.
2. BIt has been killed by the acid, in the same way that a strong acid kills living cells.
   A student who thinks enzymes are living things picks this. An enzyme is a protein molecule; it is not alive and cannot be killed. It lost its activity because it lost its folded shape.
3. CIt is inhibited but not denatured, as only high temperature can denature enzymes.
   A student who thinks only heat denatures enzymes picks this. A change in pH, like a change in temperature or chemical environment, can disrupt an enzyme's structure and denature it.
4. DIts folded shape is disrupted, although its amino acids are still joined in sequence. Correct
   A large change in pH disrupts the hydrogen bonds and ionic interactions that hold the enzyme's three-dimensional shape, so the active site loses its shape and the enzyme can no longer catalyze the reaction. Denaturation does not break the peptide bonds, so the sequence of amino acids is unchanged.

CED 3.2.A.1.i · Read this in Fix

1. AActivity returned to exactly the untreated level, as the two sets of error bars overlap.
   A student who thinks overlapping error bars prove two means are equal picks this. Overlap means a difference has not been shown, not that the means are the same; the sample means here differ (95 and 100).
2. BActivity returned in part, as the mean after removal is below the untreated mean.
   A student who treats any difference between means as real picks this. The mean after removal is lower, but the ±2 SE bars overlap widely, so the difference could be due to chance variation between samples.
3. CActivity returned, so urea cannot have caused the loss of activity by denaturing it.
   A student who thinks denaturation can never be reversed picks this. Some denatured enzymes refold and regain activity once the denaturing condition is removed, so recovery does not rule out denaturation.
4. DActivity returned to near the untreated level, as the bars show no significant difference. Correct
   Activity fell to about 4% in urea and rose to about 95% after the urea was removed. The ±2 SE bars for the untreated enzyme (100 ± 8) and the urea-removed enzyme (95 ± 10) overlap widely, so the data show no significant difference between them: the loss of activity in urea was reversed.

Working No test statistic is calculated; the decision rests on the ±2 SE error bars. Untreated: mean 100, bar 92–108 (SE = 4). In urea: mean 4, bar 1–7 (SE = 1.5). Urea removed: mean 95, bar 85–105 (SE = 5). In urea vs untreated: the bars do not overlap (7 < 92), so the loss of activity in urea is likely a real difference. Urea removed vs untreated: the bars overlap widely (both cover 92–105), so the 5-point difference between the means (95 vs 100) is not significant. Conclusion: after the urea is removed, activity returns to near the untreated level, so the denaturation in urea was reversed. Overlap does not show the means are exactly equal, a difference within overlapping bars is not shown to be real, and recovery of activity is consistent with, not evidence against, denaturation.

CED 3.2.A.2 · Read this in Fix

1. A28%
   A student who reports the change in rate, 80 − 52 = 28 μmol/min, as the percent change picks this. The change must be divided by the original rate: 28/80 × 100 = 35%.
2. B54%
   A student who divides the change by the new value picks this: (28/52) × 100 = 54%. Percent change is calculated relative to the original value, 80 μmol/min.
3. C35% Correct
   Percent decrease = (80 − 52)/80 × 100 = 35%. With more product present at the start, the reaction proceeds less efficiently in the forward direction: the relative concentrations of substrate and product determine how efficiently it proceeds.
4. D65%
   A student who gives the new rate as a percentage of the original picks this: (52/80) × 100 = 65%. The new rate is 65% of the original, which is a decrease of 35%.

Working From the table, the initial rate is 80 μmol/min with 0 mM product added and 52 μmol/min with 20 mM added. Percent change = (original − new)/original × 100 = (80 − 52)/80 × 100 = 35% decrease. Distractors: reporting the change in rate, 28 μmol/min, as a percentage gives 28%; dividing the change by the new rate gives (28/52) × 100 = 54%; giving the new rate as a percentage of the original gives (52/80) × 100 = 65%.

CED 3.2.B.1 · Read this in Fix

1. AThe higher temperature lowers the activation energy of the reaction being catalyzed.
   A student who thinks heating lowers the activation energy picks this. The model shows a change in molecular speed, not in activation energy; lowering the activation energy is what the enzyme does.
2. BThe enzyme is now closer to 37 °C, the temperature at which all enzymes work best.
   A student who thinks every enzyme works best at 37 °C picks this. Optimal temperatures differ between enzymes, and the model explains the higher rate by faster molecular movement, not by closeness to a particular temperature.
3. CThe enzyme molecule is more alive when it is warm, so it works harder on the substrate.
   A student who thinks enzymes are living things picks this. The enzyme is a molecule; warming changes how fast it and the substrate move, not how alive it is.
4. DThe molecules move faster, so enzyme and substrate collide more often. Correct
   The arrows are longer at 30 °C, showing faster-moving molecules. Faster movement makes collisions between enzyme and substrate more frequent, so substrate enters the active site more often and the rate increases, up to the optimal temperature.

CED 3.2.B.2 · Read this in Fix

1. AIt binds reversibly to the active site, so substrate cannot bind while it is there. Correct
   A competitive inhibitor binds reversibly to the active site. While it occupies the site, substrate cannot bind there, so fewer enzyme–substrate complexes form; because binding is reversible, a high substrate concentration can outcompete it.
2. BIt binds to an allosteric site, so the active site changes shape and fits poorly.
   A student who has swapped the two kinds of inhibitor picks this. Binding to an allosteric site and changing the active site's shape describes a noncompetitive inhibitor.
3. CIt unfolds the whole enzyme, so the active site and the rest of it lose their shape.
   A student who thinks inhibitors denature enzymes picks this. An inhibitor binds to a specific site; unfolding of the whole enzyme is denaturation, caused by temperature, pH or chemical conditions.
4. DIt binds to the substrate, so less of the substrate is free to reach the enzyme.
   A student who thinks an inhibitor competes with the enzyme for the substrate picks this. A competitive inhibitor binds to the enzyme, at the active site, competing with the substrate for that site.

CED 3.2.B.3 · Read this in Fix

1. AIt binds the substrate as readily as before, as the active site's shape is not changed.
   A student who thinks binding depends only on the active site's shape picks this. The new R group is about the same size, but its charge changes the chemistry of the site, and charge compatibility matters as well as shape.
2. BIt binds the substrate more readily, as mutations arise to improve enzymes a cell needs.
   A student who thinks mutations arise because a cell needs them picks this. Mutations are not directed toward improvement, and nothing about a charged group facing a nonpolar region would strengthen binding.
3. CIt binds the substrate less readily, as the site's chemistry no longer suits it. Correct
   The active site must be compatible with the substrate in charge as well as in shape. A negatively charged R group in place of a nonpolar one does not interact favorably with the substrate's nonpolar region, so the substrate binds less readily and the enzyme catalyzes the reaction less efficiently.
4. DIt unfolds and loses all its activity, as any change in its amino acids destroys it.
   A student who thinks every mutation destroys an enzyme picks this. A single substitution in the active site changes how well the substrate binds; there is no reason to expect the whole enzyme to unfold.

CED 3.2.A.1 · Read this in Fix

1. AA substitution outside the active site can greatly reduce the enzyme's activity. Correct
   Mutant 3's substitution is far from the active site, yet its activity is only about 35% of the wild type's. Losing a hydrogen bond that holds part of the active site in position changes the active site's shape, so a change elsewhere in the molecule can reduce activity.
2. BOnly a substitution located inside the active site can change how active the enzyme is.
   A student who thinks only the active site matters picks this. Mutant 3 has its substitution outside the active site and has lost about two-thirds of its activity, because the active site's shape depends on bonds elsewhere in the folded chain.
3. CEach substitution, wherever it is, leaves the enzyme with far less activity.
   A student who thinks every mutation ruins an enzyme picks this. Mutant 2's activity, about 97% of the wild type's, is barely changed: replacing a surface amino acid with a similar one did not alter the active site.
4. DOne amino acid is too small a part of the enzyme for a change in it to matter much.
   A student who thinks one amino acid is too small to matter picks this. Mutant 1 has lost almost all of its activity (about 8% of the wild type's) and mutant 3 most of it, each from a single substitution.

CED 3.2.A.1 · Read this in Fix

1. AThe bonds joining its active-site amino acids into the chain have been broken by urea.
   A student who thinks denaturation breaks the chain into pieces picks this. The model shows every bond between neighboring amino acids still present after treatment; only the folding has been lost.
2. BIts active-site amino acids are still in the chain but are no longer held close together. Correct
   In the model, the three active-site amino acids are still linked into the same chain in the same order, but after treatment the chain has unfolded and they are far apart. With them no longer held together, there is no active site of the right shape and charge for the substrate.
3. CIts active site has kept its shape, and only the rest of the chain has unfolded.
   A student who thinks the active site is a separate part, unaffected by changes elsewhere, picks this. The model shows the active-site amino acids spread apart along the unfolded chain, so the active site itself has lost its shape.
4. DUrea molecules have filled its active site, so the substrate can no longer enter it.
   A student who thinks any loss of activity comes from something blocking the active site picks this. The model shows no urea in the active site; it shows the enzyme's folded structure lost, which is denaturation.

CED 3.2.A.1.i · Read this in Fix

1. ATube 3, which contains no enzyme, so the substrate is left on its own
   A student who thinks the control is always the tube without enzyme picks this. Tube 3 shows whether the substrate reacts without enzyme; it differs from tube 2 in having no enzyme at all, so it cannot isolate the effect of heating the enzyme.
2. BTube 4, which received the same heat treatment as the enzyme in tube 2
   A student who thinks the control should be treated in the same way as the experimental tube picks this. Tube 4's enzyme was heated too, so it cannot show what heating does; tube 4 also lacks substrate.
3. CTube 1, which differs from tube 2 only in its enzyme being unheated Correct
   Tube 1 contains the same enzyme and substrate under the same conditions as tube 2, but its enzyme was not heated. Heating is the only difference, so a lower rate of product formation in tube 2 than in tube 1 can be attributed to the heat treatment.
4. DNo other tube, as tube 2's result alone shows whether heat denatures it
   A student who thinks a treated sample alone shows a treatment's effect picks this. Without the result for unheated enzyme under the same conditions, low activity in tube 2 could not be attributed to heating.

Working No calculation. Tube 2 (heated enzyme + substrate) must be compared with a tube that differs from it only in the factor being tested, heating: tube 1 (unheated enzyme + substrate, same temperature and pH during the assay). Tube 3 lacks enzyme, tube 4 lacks substrate and was also heated, and a single tube gives no comparison.

CED 3.2.A.1.i · Read this in Fix

1. ABoth enzymes reach their highest rate at about 37 °C, near human body temperature.
   A student who thinks every enzyme works best at body temperature picks this. At 37 °C each enzyme works at only about 20–25% of its maximum rate; their optima are about 20 °C and 70 °C.
2. BEach enzyme's rate rises steadily as the temperature rises across the whole range of the graph.
   A student who thinks higher temperatures always speed up enzyme-catalyzed reactions picks this. Each curve rises to a peak and then falls, as rising temperature eventually disrupts the enzyme's structure.
3. CEach enzyme catalyzes the reaction only at the single temperature where its curve peaks.
   A student who thinks an enzyme works only at its optimum picks this. Each enzyme is active over a wide range of temperatures; enzyme A, for example, works at 40% of its maximum rate at 0 °C. The peak marks the highest rate, not the only one.
4. DEach enzyme's rate peaks at a different temperature and is lower on either side of it. Correct
   Enzyme A's rate is highest at about 20 °C and enzyme B's at about 70 °C; for each enzyme, the rate is lower at temperatures below and above its peak. Each enzyme has its own optimal temperature, and temperatures outside its optimal range reduce its efficiency.

CED 3.2.A.1.ii · Read this in Fix

1. AEnzyme and substrate molecules move so fast that they pass each other before they are able to bind.
   A student who explains the fall above the optimum by molecular speed picks this. Faster-moving molecules collide more often, which by itself would raise the rate; the fall comes from changes to the enzyme's structure.
2. BSome hydrogen bonds that hold the enzyme's shape are disrupted, so the active site fits less well. Correct
   Above an enzyme's optimal temperature, increased molecular motion disrupts some of the hydrogen bonds that hold its three-dimensional shape. The active site's shape changes, so the enzyme catalyzes the reaction less efficiently, and this outweighs the effect of more frequent collisions.
3. CSome peptide bonds between the enzyme's amino acids break, splitting its chain into pieces.
   A student who thinks heat damage breaks a protein's chain picks this. Temperatures just above the optimum disrupt the weaker bonds that hold the folded shape, not the peptide bonds that join the amino acids.
4. DSome of the enzyme molecules have been killed by the heat, so fewer are left alive to react.
   A student who thinks enzymes are living things picks this. Enzyme molecules are not alive; heat reduces their activity by disrupting their shape.

CED 3.2.A.1.ii · Read this in Fix

1. AIts activity decreases, as the new pH disrupts bonds that hold the enzyme's shape. Correct
   pH 7.0 is outside this enzyme's optimal range. The change disrupts hydrogen bonds and ionic interactions that hold the enzyme's shape, so the active site changes and the enzyme catalyzes the reaction less efficiently.
2. BIts activity decreases, as the lower pH breaks the peptide bonds joining its amino acids.
   A student who thinks that disrupting an enzyme's structure breaks the bonds between its amino acids picks this. A fall from pH 8.5 to 7.0 disrupts hydrogen bonds and ionic interactions that hold the enzyme's folded shape; the peptide bonds that join its amino acids stay intact.
3. CIts activity is unchanged, as the enzyme reshapes itself to suit the new pH of the lake.
   A student who thinks enzymes change to meet an organism's needs picks this. An enzyme molecule does not adjust its shape to suit new conditions; the new pH alters its structure in a way that reduces its activity.
4. DIts activity stops, as an enzyme cannot catalyze a reaction at a pH other than its optimum.
   A student who thinks an enzyme works only at its optimum picks this. Enzymes catalyze reactions over a range of pH values, with efficiency falling as the pH moves away from the optimum; an enzyme does not work at its optimum alone.

CED 3.2.A.1.ii · Read this in Fix

1. AEnzyme kept at 50 °C, then returned to 25 °C, regains much of its activity. Correct
   Reversible denaturation means the enzyme regains activity when the condition that disrupted its shape is removed. Activity returning after the heated enzyme is cooled shows that its chain has folded back into its functional shape.
2. BEnzyme cooled to 5 °C loses activity and regains it when warmed to 25 °C.
   A student who thinks low temperatures denature enzymes picks this. Cooling slows molecular motion and reduces collisions, so this observation concerns the effect of low temperature, not reversal of denaturation at 50 °C.
3. CEnzyme heated to 50 °C splits into amino acids, which rejoin into enzyme on cooling.
   A student who thinks denaturation breaks an enzyme into amino acids picks this. Denaturation leaves the peptide bonds intact; this observation would describe an enzyme being broken down and rebuilt, not a denatured enzyme refolding.
4. DBacteria heated to 50 °C survive and then resume growing after being cooled to 25 °C.
   A student who thinks of enzymes as living things picks this. The survival and growth of the bacteria concern whole cells, which can make new enzyme molecules; they do not show whether the heated enzyme molecules regained activity.

Working No calculation. Reversible denaturation means activity returns when the denaturing condition (here 50 °C) is removed, so the supporting observation is heated enzyme regaining activity after cooling to 25 °C. The other observations concern low temperature (5 °C), breakdown and rebuilding of the chain, or the survival of whole cells.

CED 3.2.A.2 · Read this in Fix

1. ANet conversion of CO₂ and H₂O to H₂CO₃, as the enzyme catalyzes only that one direction.
   A student who thinks an enzyme catalyzes its reaction in one direction only picks this. Carbonic anhydrase catalyzes both directions; with CO₂ scarce, the net reaction runs toward CO₂.
2. BNet conversion of H₂CO₃ to CO₂ and H₂O, as the loss of CO₂ slows down the forward reaction. Correct
   The enzyme catalyzes the reaction in both directions, and the relative concentrations of substrates and products set the net direction. As CO₂ diffuses out, its concentration falls, so the forward reaction slows while the reverse reaction continues; the net reaction converts H₂CO₃ to CO₂ and H₂O, and the CO₂ formed diffuses out to be exhaled.
3. CNo reaction, as the enzyme molecules were used up making H₂CO₃ in the body's tissues.
   A student who thinks enzymes are used up in reactions picks this. Carbonic anhydrase is released unchanged after each reaction and catalyzes the reverse reaction in the lungs.
4. DConversion of all the H₂CO₃ to CO₂ and H₂O, as the enzyme drives the reaction to completion.
   A student who thinks an enzyme drives a reaction to completion picks this. An enzyme speeds the approach to equilibrium but does not convert all of the H₂CO₃; both directions continue, and the net direction depends on the concentrations.

CED 3.2.B.1 · Read this in Fix

1. AValid: changing both factors together can make differences between tubes easier to see.
   A student who thinks changing several variables at once is acceptable picks this. Larger differences are of no use if they cannot be traced to one cause; here they could come from the H₂O₂ concentration or from the amount of enzyme.
2. BInvalid: a difference in rate could be caused by the amount of extract, not the H₂O₂ level. Correct
   The amount of extract (enzyme) should be a controlled variable: more enzyme provides more active sites and can raise the rate on its own. Because it changes along with the H₂O₂ concentration, any difference in rate cannot be attributed to the H₂O₂ concentration alone.
3. CValid: the extra extract replaces enzyme used up by the higher concentrations of H₂O₂.
   A student who thinks enzymes are used up in reactions picks this. Enzyme molecules are not consumed, so nothing needs replacing; adding more extract to some tubes only introduces a second variable.
4. DInvalid only because there is no control tube containing neither extract nor H₂O₂.
   A student who thinks an experiment's validity rests on having an empty control tube picks this. A blank tube would not remove the problem: the amount of extract differs between tubes, so its effect is mixed with that of the H₂O₂ concentration.

CED 3.2.B.1 · Read this in Fix

1. AAt 4 °C, the enzyme has been denatured by the cold, so its active site has lost its shape.
   A student who thinks low temperatures denature enzymes picks this. Cooling to 4 °C slows the movement of enzyme and substrate molecules; it does not denature the enzyme, and its activity rises again as it is warmed toward its optimal temperature.
2. BAt 4 °C, the reaction's activation energy is higher than it is at 35 °C.
   A student who thinks temperature changes the activation energy picks this. The enzyme lowers the activation energy; temperature changes how fast molecules move and how often they collide.
3. CAt 4 °C, the enzyme and substrate molecules move slowly and collide less often. Correct
   Lower temperature reduces the average speed of molecules, so enzyme and substrate collide less often and fewer enzyme–substrate complexes form each second. Below the optimal temperature, collision frequency, not denaturation, limits the rate.
4. DAt 4 °C, the enzyme is far from 37 °C, the optimal temperature for every enzyme.
   A student who thinks every enzyme works best at 37 °C picks this. Enzymes differ in their optimal temperatures, and the low activity at 4 °C is explained by slow molecular movement, not by distance from a particular temperature.

Working No calculation. Below the optimal temperature, molecules move more slowly and enzyme–substrate collisions are less frequent (EK 3.2.B.2), so the rate at 4 °C is low; cooling does not denature the enzyme, and the activation energy is set by the enzyme, not by the temperature.

CED 3.2.B.2 · Read this in Fix

1. AX binds in the active site and occupies it, so the substrate cannot bind there at all.
   A student who thinks every inhibitor blocks the active site picks this. The model shows the active site empty and distorted, with X bound at a separate allosteric site.
2. BX unfolds the whole enzyme chain, so the enzyme loses all of its structure and shape.
   A student who thinks inhibitors denature enzymes picks this. In panel 2 the enzyme keeps its overall shape; only the active site is altered, by X bound at the allosteric site.
3. CX binds at a second site and changes the active site's shape, so substrate fits poorly. Correct
   In panel 2, X is bound at the allosteric site, away from the active site, and the active site has changed shape so that the substrate no longer fits. This is how a noncompetitive inhibitor changes the enzyme's activity.
4. DX binds at a second site, and its effect disappears once more substrate is added.
   A student who thinks extra substrate overcomes any inhibitor picks this. X does not occupy the active site, so substrate cannot displace it; while X is bound, the active site stays distorted however much substrate is present.

CED 3.2.B.3 · Read this in Fix

1. AQ binds the active site, as it lowers the maximum rate; P binds an allosteric site.
   A student who has swapped the two kinds of inhibitor picks this. An inhibitor whose effect extra substrate does not overcome, like Q, binds away from the active site; one that substrate outcompetes, like P, binds at the active site.
2. BP and Q both bind the active site, as enough extra substrate can overcome either one of them.
   A student who thinks extra substrate overcomes any inhibitor picks this. Q's curve levels off at about half the uninhibited rate; across the highest concentrations shown, more substrate brings it no closer to the uninhibited curve.
3. CNeither binds the enzyme; each lowers the rate by binding to part of the substrate supply.
   A student who thinks inhibitors act on the substrate picks this. An inhibitor that tied up some substrate would be overcome at high substrate concentrations, but Q's effect persists there; inhibitors act by binding to the enzyme.
4. DP binds the active site, as more substrate outcompetes it; Q binds an allosteric site. Correct
   With P, the rate is low at low substrate concentrations but approaches the uninhibited rate at high concentrations: substrate outcompetes P for the active site. With Q, the rate levels off at about half the uninhibited rate however much substrate is added, consistent with binding at an allosteric site that changes the enzyme's activity.

CED 3.2.B.3 · Read this in Fix