Drill 3 ·
AP Biology: Unit 1, pH, Buffers & Biological Molecules (Drill 3) is a practice drill. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice analyzing pH, acid-base chemistry, and buffer systems in biological contexts with this AP Biology drill. You will interpret pH scale relationships, evaluate how buffers maintain homeostasis, and explain how pH changes affect enzyme function and protein structure.
| Enzyme | Optimal pH | Location |
|---|---|---|
| Pepsin | 2.0 | Stomach |
| Salivary amylase | 6.8–7.0 | Mouth |
| Trypsin | 7.5–8.5 | Small intestine |
| Urease | 7.0 | Soil bacteria |
Question 1. A solution has a pH of 5. A second solution has a pH of 3. How does the H+ concentration in the pH 3 solution compare to the pH 5 solution?
Explanation: Correct: (C) The pH scale is logarithmic base 10. Each unit decrease in pH corresponds to a 10-fold increase in H+ concentration. Moving from pH 5 to pH 3 is a decrease of 2 pH units. Therefore the H+ concentration increases by a factor of 10^2 = 100. The pH 3 solution has 100 times the H+ concentration of the pH 5 solution.
Question 2. Based on the table, which enzyme would be most active in the stomach immediately after a meal, and why?
Explanation: Correct: (A) The stomach maintains a highly acidic environment (pH approximately 1.5–2.0) due to secretion of hydrochloric acid. Pepsin has an optimal pH of 2.0, meaning its active site geometry functions most efficiently at this very low pH. The other enzymes have optimal pH values far from 2.0 and would be denatured or function at minimal efficiency in the acidic stomach. Trypsin (C) functions in the small intestine at pH 7.5–8.5 and is irreversibly denatured by strong acid.
Question 3. During intense exercise, CO2 production increases dramatically. Using the bicarbonate buffer equilibrium shown in the passage, predict the effect on blood pH if the respiratory system does not increase ventilation rate to compensate.
Explanation: Correct: (B) The bicarbonate equilibrium is: CO2 + H2O ⇌ H2CO3 ⇌ H+ + HCO3-. When CO2 concentration rises, Le Chatelier's principle predicts the equilibrium shifts right to consume excess CO2. This produces more carbonic acid, which dissociates to release more H+ ions. The increase in H+ lowers blood pH. This is called respiratory acidosis. The respiratory system normally compensates by increasing breathing rate to expel more CO2 through the lungs. Option A incorrectly identifies bicarbonate as a base in this context; while HCO3- can act as a base, it is produced alongside H+, so pH drops.
Question 4. A researcher adds a small amount of strong acid to a buffered solution (pH 7.4) and to an unbuffered solution (pH 7.4). The buffered solution's pH drops from 7.4 to 7.3, while the unbuffered solution's pH drops from 7.4 to 4.9. Which of the following best explains the difference?
Explanation: Correct: (D) A buffer consists of a weak acid and its conjugate base in equilibrium. When H+ ions are added, the conjugate base reacts with the excess H+ to form the weak acid, consuming the added H+ and preventing a large pH decrease. In the unbuffered solution, there is no such mechanism and the added H+ ions remain free, dramatically lowering the pH. Dilution (A) alone cannot account for such a small pH change in the buffered solution. Salt ions (B) do not have this acid-absorbing mechanism.
Question 5. A student is studying an enzyme that normally functions optimally at pH 7.4. When the solution is acidified to pH 5, enzyme activity drops to nearly zero. Which of the following provides the best molecular explanation for this loss of activity?
Explanation: Correct: (B) Enzyme function depends on the precise three-dimensional shape of the active site, maintained by interactions among amino acid R groups. Many R groups contain ionizable side chains (e.g., carboxyl, amino, imidazole groups) whose charge depends on surrounding pH. When pH drops from the optimum, these R groups become protonated in ways that disrupt the interactions maintaining active site geometry. The substrate can no longer bind properly. Peptide bonds (A) are covalent bonds that are extremely stable under physiological pH conditions and are not hydrolyzed at pH 5. Option D describes competitive inhibition, which is a different mechanism.