AP Biology Unit 1: Chemistry of Life — Worked Examples

Water Potential and Solute Concentration in Potato Cores

Hard

A student places potato cores of equal mass into five beakers of sucrose solution (0.0 M to 0.8 M). After 24 hours, the student measures the percent change in mass. The data are shown in the table below. Based on these data, what is the approximate molar concentration of solute inside the potato cells?

0.2 M, because the cell gains the most water below this concentration
0.4 M, because the cell neither gains nor loses mass at this concentration ✓ Correct
0.6 M, because the cell begins to plasmolyze at this concentration
0.0 M, because pure water has the highest water potential

Solution

At 0.4 M sucrose, the potato core shows 0% change in mass, meaning no net water movement occurred. This indicates the water potential of the cell equals the water potential of the surrounding solution. Since $\\Psi = \\Psi_s + \\Psi_p$ and the solution has no pressure component, the solute concentration inside the cell must be approximately 0.4 M. Choice A (0.2 M) confuses the point of maximum water gain with equilibrium. Choice C (0.6 M) confuses plasmolysis onset with isotonicity. Choice D (0.0 M) confuses external water potential with internal solute concentration.

Competitive vs. Noncompetitive Inhibition from Kinetic Data

Hard

Researchers studying a metabolic enzyme measured the reaction rate at various substrate concentrations with and without an inhibitor. The data are shown below. Based on the data, what type of inhibition is occurring?

Competitive inhibition, because the inhibitor reduces the reaction rate at all substrate concentrations
Competitive inhibition, because increasing substrate concentration would eventually overcome the inhibitor
Noncompetitive inhibition, because the $V_{max}$ is reduced but the apparent $K_m$ is unchanged ✓ Correct
Noncompetitive inhibition, because the inhibitor binds to the active site permanently

Solution

The inhibitor reduces the rate to exactly half at every substrate concentration, meaning $V_{max}$ is halved while $K_m$ remains unchanged. This is the hallmark of noncompetitive inhibition, where the inhibitor binds to an allosteric site and reduces the number of functional enzyme molecules. Choice A incorrectly states the reasoning for competitive inhibition (competitive also reduces rate, but the ratio changes with [S]). Choice B describes competitive inhibition correctly but does not match the data — the inhibitor's effect is never overcome. Choice D confuses noncompetitive inhibition with irreversible inhibition.