ACT Science: Research Summaries (Drill 3)

Question 1. According to Table 2, at which pH was the catalase reaction rate the highest?

• A) pH 5
• B) pH 6
• C) pH 7 ✓
• D) pH 8

Explanation: Table 2 shows the highest reaction rate, 4.1 mL O₂/min, at pH 7. The rate increases from pH 4 to pH 7 and decreases from pH 7 to pH 10, making pH 7 the optimum pH for this enzyme. This is consistent with the biology of catalase, which functions in the near-neutral cellular environment of most organisms. At pH values far from the optimum, the shape of the enzyme's active site is disrupted by changes in ionization, reducing its ability to bind and process the substrate.

Question 2. Which variable was held constant in all three experiments?

• A) H₂O₂ concentration
• B) Catalase concentration
• C) pH
• D) Temperature ✓

Explanation: The passage states that "all experiments were performed at 25°C unless otherwise stated," and no experiment states otherwise. Temperature was therefore held constant at 25°C in all three experiments. H₂O₂ concentration was the independent variable in Experiment 1 (varied), held constant in Experiments 2 and 3. Catalase concentration was held constant in Experiments 1 and 2, but was the independent variable in Experiment 3. pH was the independent variable in Experiment 2, and was held constant in Experiments 1 and 3. Only temperature was constant throughout all three.

Question 3. Both Experiments 1 and 3 show reaction rate increasing as concentration increases, but with different behavior at high concentrations. Which of the following best describes this difference?

• A) Experiment 1 plateaus at a lower rate than Experiment 3 at high concentrations
• B) Experiment 1 plateaus quickly; Experiment 3 continues increasing beyond the Experiment 1 plateau ✓
• C) Experiment 3 plateaus at a lower rate than Experiment 1
• D) Both experiments plateau at the same maximum rate

Explanation: In Experiment 1, the rate reaches 5.0 mL O₂/min at 5% H₂O₂ and essentially stops increasing, the enzyme molecules are fully saturated with substrate so adding more substrate has no additional effect. In Experiment 3, the rate is already at 4.1 mL O₂/min at 0.5% catalase and continues rising to 8.4 mL O₂/min at 4% catalase before approaching its plateau. The plateau in Experiment 3 occurs at a much higher rate than Experiment 1's plateau because more enzyme molecules means more active sites available to process substrate simultaneously. This illustrates that when enzyme is the limiting factor, adding more enzyme continues to increase the rate; when substrate is the limiting factor, adding more substrate does not.

Question 4. A student wants to maximize the reaction rate using 3% H₂O₂ and pH 7. Based on Experiment 3, approximately what catalase concentration would she need to roughly double the rate observed at 0.5% catalase?

• A) 0.5%
• B) 1.0% ✓
• C) 2.0%
• D) 4.0%

Explanation: At 0.5% catalase, the rate is 4.1 mL O₂/min (Table 3). Double that rate would be approximately 8.2 mL O₂/min. Reading Table 3, a rate of 8.2 mL O₂/min occurs at 2.0% catalase. At 1.0% catalase the rate is 6.8 mL O₂/min, not yet double, and at 4.0% the rate is 8.4 mL O₂/min, only slightly above double. The concentration that most closely achieves double the 0.5% rate is therefore 2.0% catalase.

Question 5. In Experiment 3, the students included a trial with 0.0% catalase. What was the most likely purpose of this trial?

• A) To show that the reaction rate decreases when catalase is removed
• B) To serve as a negative control, confirming that catalase is required for the reaction to proceed at a measurable rate ✓
• C) To determine the maximum possible reaction rate
• D) To measure how fast H₂O₂ naturally breaks down without any influence from the experiment

Explanation: The 0.0% catalase trial is a negative control, a baseline condition in which the key variable (the enzyme) is absent. It produced a rate of 0.0 mL O₂/min, confirming that without catalase, the decomposition of H₂O₂ does not proceed at a measurable rate under these conditions. This is important because it validates that any oxygen production in the other trials was due to catalase activity, not some other factor. Option A is backward, 0% catalase is the starting point of the experiment, not a removal of catalase from an ongoing reaction. Option D is partly right in spirit but "natural breakdown" implies a spontaneous rate; at 0% the measured rate is zero, not a baseline spontaneous rate.