AP Biology · Topic 3.2
Enzyme Catalysis Practice
Part of Cellular Energetics.(ENE-1.M)
Practice questions
17
Sample questions
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Sample 1difficulty 2/5
A student plots free-energy curves for the same reaction with and without an enzyme catalyst.
Which statement best describes the effect of the enzyme shown?
- A
The enzyme converts an endergonic reaction to exergonic
- Bcheck_circle
The enzyme lowers the activation energy but does not change ΔG
- C
The enzyme raises activation energy and lowers ΔG
- D
The enzyme makes the reaction more exergonic
Why
Enzymes are catalysts - they reduce the activation energy barrier (Ea) so reactions occur faster, but the start and end energy levels (and therefore ΔG) are unchanged.
- A
Sample 2difficulty 2/5
A student investigates catalase, an enzyme that breaks down hydrogen peroxide (H2O2) into water and oxygen. Filter paper disks are soaked in liver extract (catalase source) and dropped into test tubes containing 3% H2O2 at five different temperatures: 5, 20, 35, 50, and 65 degrees Celsius. The student records the time (in seconds) for each disk to rise to the surface, propelled by O2 bubbles. Each temperature is tested in triplicate, and the mean rise time is converted to reaction rate (1/time). The pH is held constant at 7.0 and substrate concentration is held at 3% throughout.
Which factor is the independent variable in this experiment?
- A
pH of the buffer
- Bcheck_circle
Temperature of the H2O2 solution
- C
Time for the disk to rise
- D
Concentration of H2O2
Why
The student deliberately varies temperature across five levels (5-65 C); this is the independent variable. Rise time (or its reciprocal, rate) is the dependent variable, while pH and substrate concentration are controlled.
- A
Sample 3difficulty 2/5
A student measured initial reaction rate of an enzyme at increasing substrate concentrations and produced the curve shown.
The graph shows enzyme reaction rate vs substrate concentration. What does Km represent?
- A
Rate at which enzyme denatures under increased substrate
- B
Substrate concentration at which enzyme activity stops
- C
Maximum reaction rate when enzyme is fully saturated
- Dcheck_circle
Substrate concentration at half Vmax, indicating enzyme-substrate affinity
Why
Km (Michaelis constant) is the substrate concentration at which the reaction rate equals 1/2 Vmax. A lower Km indicates higher affinity between enzyme and substrate.
- A
Sample 4difficulty 2/5
Students measured the initial rate of catalase activity by recording the volume of O2 produced per second when the enzyme was exposed to varying concentrations of hydrogen peroxide. All reactions were run at 25 degrees C and pH 7. The table below shows mean rates from triplicate trials.
Which conclusion is best supported by the data?
- Acheck_circle
The enzyme approaches saturation between 20 and 40 mM substrate.
- B
Rate is directly proportional to substrate at all tested concentrations.
- C
The enzyme is denatured at 40 mM substrate.
- D
The enzyme has greater affinity for product than for substrate.
Why
The rate plateaus between 20 mM (4.0) and 40 mM (4.1), indicating active sites are saturated. Linear proportionality fails at high [S]; denaturation would lower the rate.
- A
Sample 5difficulty 2/5
The rate plateau at high substrate concentration occurs because
- A
All cofactors become tightly bound to substrate
- B
Allosteric inhibitors fully occupy regulatory sites
- C
Active sites change shape at high concentrations
- Dcheck_circle
All enzyme active sites become saturated with substrate
Why
Once every active site is occupied, additional substrate cannot further increase the reaction rate (V_max).
- A