AP Biology: Enzyme Catalysis — Inhibition Experiments Practice Questions
Test yourself on AP Biology: Enzyme Catalysis — Inhibition Experiments with 12 original AP Exams practice questions. Pick an answer to see instant feedback and a full explanation.
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1. A student adds a candidate enzyme inhibitor dissolved in ethanol to one reaction. Which comparison best isolates the effect of the candidate compound?
Explanation. The solvent is introduced together with the compound, so it must be matched in the comparison. Keeping enzyme, substrate, volume, temperature, pH, and ethanol concentration equal makes the candidate compound the intended difference. Changing several factors would leave alternative explanations for a rate change.
2. With a fixed amount of enzyme, a reversible competitive inhibitor reduces the rate at low substrate concentration. In the simple competitive model, what happens as substrate concentration becomes sufficiently high while inhibitor concentration stays fixed?
Explanation. In this model, substrate and inhibitor binding are mutually exclusive. Sufficient substrate reduces inhibition, leaving the limiting maximum rate unchanged. A finite substrate increase need not restore the rate fully.
3. In a stated model of pure noncompetitive inhibition, the inhibitor binds free enzyme and enzyme-substrate complex equally well and prevents catalysis when bound. Which rate-curve change is predicted at a fixed inhibitor concentration?
Explanation. Pure noncompetitive inhibition lowers the maximum rate. Increasing substrate does not eliminate inhibition in this model. This prediction should not be generalized to every compound that binds outside an active site.
4. A vehicle control has an initial rate of 0.80 micromoles per minute. Under matched conditions, a compound-treated reaction has a rate of 0.30 micromoles per minute. Using 100 times (control rate minus treated rate) divided by control rate, what is the percent inhibition?
Explanation. The rate decrease is 0.80 minus 0.30, or 0.50 micromoles per minute. Dividing by 0.80 gives 0.625, so inhibition is 62.5%. The remaining activity is 37.5%; that is a different quantity from the requested percent inhibition.
5. A fluorescent product assay shows a lower signal when compound X is present. Which control most directly tests whether X changes the detection signal rather than product formation?
Explanation. Holding the amount of product fixed separates detection from production. If X changes fluorescence even with equal product concentrations, a lower reaction signal cannot automatically be interpreted as less product. That result would call for a corrected measurement or another detection method.
6. After an enzyme preparation is exposed to a compound, the compound is removed. The enzyme's rate returns to the vehicle-control rate when final enzyme amount, buffer, substrate, and temperature are matched. Which conclusion is best supported?
Explanation. Recovery after compound removal demonstrates that the observed activity loss was reversible in this preparation. The observation does not identify the compound's binding site. No measurement of gene expression is described, and complete permanent destruction conflicts with the recovered activity.
7. At one substrate concentration, an enzyme reaction is 40% slower with compound Y than with vehicle alone. Which statement is justified by that single comparison?
Explanation. A single rate comparison describes an effect under one set of conditions. It provides no direct binding-site or structural measurement, no recovery test, and no results for other enzymes. Additional controlled measurements are needed to distinguish possible explanations.
8. For a reaction containing compound Z, the signal rises at 0.21 absorbance units per minute. A matched no-enzyme blank containing Z rises at 0.06 units per minute. If the background is additive, what is the blank-corrected signal slope?
Explanation. Subtract the matched blank from the total slope: 0.21 minus 0.06 equals 0.15 absorbance units per minute. Adding the signals or dividing them does not remove an additive background. Product calibration would still be needed to convert this optical slope into a chemical production rate.
9. A treated reaction contains 0, 2, 4, and 5 micromoles of product at 0, 1, 2, and 3 minutes. Independent checks establish that the first two minutes form the initial linear interval. What initial rate should be calculated from that interval?
Explanation. During the specified initial interval, product increases by 4 micromoles over 2 minutes. The initial rate is therefore 2 micromoles per minute. The slower increase between minutes 2 and 3 lies outside that interval and should not replace the initial slope.
10. A student wants to estimate how reproducible a compound's effect is across independently prepared reactions. Which plan best addresses that goal?
Explanation. Independent preparations allow variation between reactions to be observed. Reporting their rates and spread gives a clearer picture of reproducibility than repeated readings of one tube. Selecting only the largest effect would discard relevant results and bias the description.
11. At low substrate concentration, control and inhibitor-treated rates are 10 and 4 units. At high substrate concentration, they are 40 and 32 units. What happens to percent inhibition, calculated relative to the control at each substrate concentration?
Explanation. At low substrate, the fractional decrease is (10 minus 4) divided by 10, or 60%. At high substrate, it is (40 minus 32) divided by 40, or 20%. Each treatment is compared with its matching control; the absolute differences of 6 and 8 units are not percentages.
12. A compound-treated reaction has pH 5 and a rate of 3 units, while the vehicle control has pH 7 and a rate of 9 units. Which follow-up best tests whether the compound lowers activity independently of this pH difference?
Explanation. Compound exposure and pH differ in the original comparison, so either could contribute to the rate difference. Matching and verifying pH removes that confounding difference in the follow-up. Enzyme amount, substrate, solvent, temperature, and timing should also be matched.
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FAQ
Can one lower rate identify an inhibitor's mechanism?
No. One comparison establishes a measured difference under those conditions. Controls and further measurements are needed before making claims about binding sites, reversibility, or structural changes.
How do I calculate percent inhibition?
Use 100 times (control rate minus treated rate) divided by control rate, after appropriate background correction. Compare rates measured under matching conditions, and distinguish inhibition from the percentage of activity remaining.
Are these experimental data taken from a real exam or laboratory study?
No. The numerical examples are invented for independent study practice. They are designed to teach controlled comparisons and data interpretation, not to report measurements from a particular enzyme or research study.