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AP Biology

AP Biology Practice Test: Practice Test 100

Practice Test 100 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 3

A protein-free phospholipid bilayer separates two solutions. Equal concentrations of H+^++ (1 Da, charged), glucose (180 Da, polar uncharged), and CO2_22​ (44 Da, nonpolar) are added to one side. Which solute would most likely show the greatest net movement across the bilayer over a short time?

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Question 1

A protein-free phospholipid bilayer separates two solutions. Equal concentrations of H+^++ (1 Da, charged), glucose (180 Da, polar uncharged), and CO2_22​ (44 Da, nonpolar) are added to one side. Which solute would most likely show the greatest net movement across the bilayer over a short time?

  1. H+^++, because it is the smallest and therefore crosses membranes most easily
  2. Glucose, because it is uncharged and therefore freely permeable
  3. CO2_22​, because nonpolar molecules cross the hydrophobic interior readily (correct answer)
  4. Glucose, because larger molecules maintain gradients longer and thus diffuse faster
  5. All three equally, because diffusion depends only on initial concentration

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. CO₂ shows the greatest net movement because it is nonpolar, crossing the hydrophobic interior readily unlike charged H⁺ or polar glucose. H⁺ is impermeable due to charge, and glucose is slow due to size and polarity. The stimulus adds equal concentrations to one side with no proteins, focusing on short-term diffusion. A tempting distractor is choice A (H⁺), due to tiniest size, but this reflects the misconception that charge doesn't block tiny ions. For transferable strategy, always identify nonpolar solutes for fastest net flux, considering barriers for charged and polar ones.

Question 2

A secreted ligand Z triggers a response only when presented in pulses every 5 minutes; constant exposure to the same average concentration produces little response. Binding assays show that receptors bind Z in both conditions. Which of the following best explains why pulsing changes signaling effectiveness?

Which of the following best explains why pulses of Z produce a stronger response than constant Z?

  1. Constant Z likely causes receptor desensitization or internalization, reducing signaling over time (correct answer)
  2. Pulses increase Z solubility, allowing it to cross the membrane without receptors
  3. Constant Z prevents diffusion, so Z cannot reach the receptor-binding site
  4. Pulses convert Z into a steroid, enabling intracellular receptor binding
  5. Constant Z increases cell wall thickness, blocking receptor access to the ligand

Explanation: This question assesses understanding of cell communication via signal transduction pathways. Constant exposure to ligand Z likely causes receptor desensitization or internalization, reducing signaling over time, while pulses allow recovery and stronger responses despite the same average concentration. Binding occurs in both conditions, but pulsing maintains effectiveness by preventing adaptation. This explains why constant Z produces little response compared to pulses. A tempting distractor is choice B, claiming pulses increase solubility, but this reflects the misconception that delivery mode alters molecular properties, whereas adaptation affects receptor responsiveness. To approach similar questions, consider how signal presentation timing influences receptor dynamics and adaptation.

Question 3

In cultured liver cells, the peptide hormone H binds a specific protein on the outer surface of the plasma membrane. Within 5 seconds of adding H, intracellular cAMP increases threefold. When cells are treated with a drug that prevents H from binding its receptor, cAMP levels remain at baseline. A separate drug that blocks adenylyl cyclase also prevents the cAMP increase even though H still binds the receptor. Which of the following best explains how the signal is initiated at the membrane?

  1. H diffuses through the lipid bilayer and directly activates adenylyl cyclase in the cytosol.
  2. H binding changes the receptor’s conformation, enabling activation of a membrane-associated protein that stimulates adenylyl cyclase. (correct answer)
  3. H binding triggers transcription of the adenylyl cyclase gene, increasing enzyme abundance within seconds.
  4. H binding increases membrane fluidity, allowing cAMP to enter the cell from the extracellular solution.
  5. H is converted into cAMP at the cell surface, and cAMP then binds the receptor to amplify the signal.

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on receptor-ligand interactions and early transduction events. The correct answer is B because H binds a membrane receptor, leading to a rapid cAMP increase within 5 seconds, and drugs blocking binding or adenylyl cyclase prevent this, indicating receptor conformational change activates a membrane-associated protein like a G protein to stimulate adenylyl cyclase. This matches basic GPCR signaling principles where ligand binding induces shape changes that enable G protein activation and subsequent effector enzyme stimulation. The specificity of the drugs—binding blocker prevents cAMP rise despite intact cyclase, and cyclase blocker allows binding but not cAMP—supports transduction via protein relay at the membrane. A tempting distractor is A, which incorrectly claims H diffuses through the bilayer to activate cyclase directly, arising from the misconception that all signals are intracellular without membrane reception, ignoring the receptor's role. When analyzing signal transduction questions, evaluate drug effects on binding versus downstream outputs to determine if signaling involves direct diffusion or receptor-mediated relays.