AP Biology

AP Biology Practice Test: Practice Test 2

Practice Test 2 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

In snapdragons for this question, red flower color (R) is completely dominant to white (r). A red-flowered plant is crossed with a white-flowered plant. The red parent is known to be heterozygous (Rr), and the white parent must be rr. Assume equal gamete production and random fertilization. Which outcome is expected among the offspring?

Which proportion of offspring from an $Rr \times rr$ cross will have red flowers?

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Question 1

Which proportion of offspring from an $Rr \times rr$ cross will have red flowers?

$\frac{1}{4}$
$\frac{1}{2}$ (correct answer)
$\frac{3}{4}$
$1$
$0$

Explanation: This question tests Mendelian inheritance analysis, focusing on phenotype proportions in a testcross-like scenario. In an Rr x rr cross, the heterozygous parent produces R and r gametes equally, while the homozygous recessive produces only r. Fertilization results in 1/2 Rr (red) and 1/2 rr (white) offspring, as the dominant R allele from the heterozygote determines red in half the cases. This 1/2 red proportion reflects independent segregation, with each gamete combination equally likely. A tempting distractor is 3/4, which might arise from confusing this with a dihybrid cross or misapplying the monohybrid 3:1 ratio, a common error in ratio recall. For inheritance questions, identify the cross type and use Punnett squares to ensure accurate allele pairing and probability calculations.

Question 2

In a cardiac muscle cell, a hormone binds a GPCR that activates Gs and increases cAMP, activating PKA. PKA phosphorylates L-type Ca $^{2+}$ channels, increasing Ca $^{2+}$ influx. The incoming Ca $^{2+}$ binds calmodulin, and the Ca $^{2+}$ -calmodulin complex binds the L-type channel to reduce its open probability. With constant hormone, Ca $^{2+}$ influx initially increases but then partially decreases. Which outcome best illustrates the feedback described in this pathway?

Blocking PKA prevents Ca $^{2+}$ influx entirely because calmodulin cannot bind the channel.
Chelating intracellular Ca $^{2+}$ reduces calmodulin binding, leading to a more sustained high Ca $^{2+}$ influx. (correct answer)
Increasing hormone concentration prevents calmodulin effects by saturating the receptor and maximizing cAMP.
Inhibiting adenylyl cyclase increases Ca $^{2+}$ influx by removing negative feedback on cAMP production.
Blocking calmodulin increases channel closure by strengthening Ca $^{2+}$ -dependent inactivation.

Explanation: This question tests understanding of feedback regulation in signal transduction pathways. Negative feedback partially decreases Ca²⁺ influx as entering Ca²⁺ binds calmodulin, which reduces channel open probability after PKA-mediated enhancement. Choice B illustrates this by chelating Ca²⁺, preventing calmodulin binding and leading to sustained high influx. This shows how feedback fine-tunes Ca²⁺ levels to avoid overload. Choice A is a tempting distractor but incorrect because blocking PKA prevents channel phosphorylation and influx entirely, not just calmodulin effects, stemming from confusing upstream activation with downstream feedback. For similar problems, distinguish initial activation from feedback modulation and predict outcomes of disrupting each.

Question 3

A plasma membrane is a phospholipid bilayer with a nonpolar interior formed by fatty acid tails. Some membrane proteins span the bilayer and can provide hydrophilic routes for polar solutes. A student compares transport of glucose into two cell types placed in the same glucose solution. Cell type 1 rapidly accumulates glucose, while cell type 2 shows minimal glucose entry over the same time. Microscopy indicates both membranes have similar phospholipid composition, but cell type 2 has fewer transmembrane proteins.

Which change would most likely increase glucose entry into cell type 2?

Decrease the number of phospholipid molecules to create gaps in the bilayer
Increase the number of transmembrane transport proteins that provide hydrophilic passageways (correct answer)
Replace fatty acid tails with additional phosphate groups to make the core nonpolar
Add more peripheral proteins to the membrane surface to attract glucose
Increase membrane carbohydrate content to convert glucose into a nonpolar solute

Explanation: This question requires analyzing plasma membrane structure and transport to determine how to increase glucose entry. The correct answer is B because glucose is a polar molecule that cannot easily cross the hydrophobic core of the phospholipid bilayer and requires transmembrane transport proteins to provide a hydrophilic pathway. The data shows that cell type 1, which has more transmembrane proteins, accumulates glucose rapidly while cell type 2, with fewer such proteins, shows minimal glucose entry despite having similar phospholipid composition. Adding more transmembrane transport proteins would create additional hydrophilic routes for glucose to cross the membrane. Answer C is incorrect because replacing fatty acid tails with phosphate groups would destroy the bilayer structure—phosphate groups are hydrophilic, not nonpolar, and the membrane needs its hydrophobic core to function. When comparing transport between cells, always check both the membrane's lipid composition and the presence of transport proteins that facilitate movement of polar solutes.

Question 4

An enzyme binds substrate S at its active site and catalyzes formation of product P. A mutation changes one amino acid in the active site, replacing a charged residue with a nonpolar residue. In assays with identical enzyme concentration and substrate concentration, the mutant enzyme shows a much lower initial reaction rate than the wild type, while the enzyme remains folded and stable at the assay temperature. Which explanation best accounts for the reduced rate in the mutant enzyme?

The mutation increases substrate concentration, which slows the reaction by causing substrate crowding.
The mutation reduces complementarity between substrate and active site, decreasing formation of enzyme–substrate complexes. (correct answer)
The mutation raises the temperature of the reaction mixture, decreasing kinetic energy and slowing catalysis.
The mutation increases enzyme concentration, which lowers rate by reducing the fraction of saturated enzyme.
The mutation shifts reaction equilibrium toward products, reducing the observed initial rate.

Explanation: This question assesses the skill of analyzing enzyme function by assessing the impact of active site mutations on catalytic activity. The mutation replaces a charged residue with a nonpolar one in the active site, reducing the complementarity between the enzyme and substrate, which decreases the formation of stable enzyme-substrate complexes and lowers the initial reaction rate. This is supported by the much lower rate in the mutant despite identical conditions and enzyme stability, indicating impaired binding or transition state stabilization without denaturation. The change disrupts specific interactions like hydrogen bonding or electrostatic forces necessary for efficient catalysis, aligning with the lock-and-key or induced-fit models. A tempting distractor is choice A, which claims the mutation increases substrate concentration causing crowding, reflecting a misconception about confusing molecular changes with concentration effects. A transferable strategy for enzyme questions is to consider how alterations in active site residues affect substrate binding affinity and catalytic efficiency separately.

Question 5

In a herd of cattle, breeders allow only the 20% with the highest milk yield to reproduce each generation. Milk yield is influenced by a gene with alleles H (higher yield) and h (lower yield). At generation 0, allele frequencies are H = 0.40 and h = 0.60. After five generations of this breeding practice, the herd’s average milk yield increases and fewer low-yield calves are born. No new animals are introduced. Which outcome is most likely in the herd’s gene pool after five generations of selection?

The frequency of allele H increases because individuals with more H alleles leave more offspring. (correct answer)
Allele h becomes more common because selection creates new h mutations in low-yield calves.
Both alleles remain at the same frequencies because selection changes only phenotypes, not genes.
Allele frequencies fluctuate randomly because breeding choices eliminate any effect of selection.
Each cow increases its own H allele proportion during life in response to frequent milking.

Explanation: This question assesses the analysis of artificial selection, where humans selectively breed organisms for desired traits, leading to changes in population genetics. The correct answer is A because breeders allow only the highest milk-yield cows to reproduce, and since milk yield is influenced by allele H for higher yield, individuals carrying more H alleles contribute disproportionately to the next generation's gene pool. Over five generations, this differential reproductive success increases the frequency of H from 0.40, as low-yield calves with more h alleles are less likely to be born due to the selection pressure. This aligns with AP Biology concepts of evolution, where artificial selection mimics natural selection by favoring heritable traits that enhance reproductive output in the controlled environment. A tempting distractor is E, which is incorrect due to the misconception of Lamarckian inheritance, suggesting individuals acquire and pass on traits during their lifetime rather than through genetic selection. A transferable strategy for this question type is to identify how selective breeding amplifies alleles linked to the desired phenotype through generational reproduction, not individual adaptation.

Question 6

A mouse population in a barn has 300 individuals. Over one month, 90 births occur, 60 deaths occur, 20 mice immigrate, and 50 mice emigrate. Which statement best predicts the population size after one month?

260 mice, because emigration exceeds immigration by 30
300 mice, because births and deaths cancel and movement is negligible
330 mice, because net change is $+90-60+20-50=0$
300 mice, because net change is $+90-60+20-50=0$ (correct answer)
360 mice, because births are added without subtracting deaths

Explanation: This question tests population ecology calculations requiring careful accounting of all demographic changes to predict future population size. Starting with 300 mice, we add births (+90), subtract deaths (-60), add immigration (+20), and subtract emigration (-50), yielding a net change of +90-60+20-50 = 0, maintaining the population at 300 mice. The calculation demonstrates that populations remain stable when gains (births + immigration) exactly balance losses (deaths + emigration), a condition that defines zero population growth. Choice A incorrectly focuses only on migration balance while ignoring the larger contributions of births and deaths to overall population change. When solving population arithmetic, use the complete demographic equation systematically, ensuring each process is included with its correct sign to avoid overlooking major population drivers.

Question 7

A breeder maintains two lines of rabbits: Line 1 is produced by mating only rabbits with very long ears, while Line 2 is produced by mating rabbits at random regardless of ear length. Ear length is influenced by alleles L (longer) and l (shorter). After 12 generations, Line 1 shows a much narrower range of ear lengths than Line 2. Which explanation best accounts for the difference between the lines?

Directional selection in Line 1 reduced allelic diversity at ear-length loci compared with the randomly bred Line 2. (correct answer)
Random mating in Line 2 caused individuals to change from allele l to L when exposed to longer-eared parents.
Line 1 gained variation because selection increases recombination rates that create new ear-length alleles.
Line 2 lost variation because random mating prevents alleles from being passed to offspring consistently.
Line 1 narrowed variation because rabbits evolved longer ears within their lifetimes before reproducing.

Explanation: This question analyzes how artificial selection affects genetic variation compared to random mating. Line 1 experiences directional selection for long ears, which increases the frequency of allele L while decreasing l, reducing allelic diversity and narrowing the phenotypic range as the population becomes more genetically uniform at ear-length loci. In contrast, Line 2 maintains both alleles through random mating, preserving genetic variation and a wider range of ear lengths. Option B incorrectly suggests individuals change alleles (Lamarckian error); option C wrongly claims selection increases variation through recombination; option D misunderstands random mating's effect on allele transmission; option E represents within-lifetime change rather than population evolution. To compare selection regimes, recognize that directional selection reduces variation while random mating maintains it.

Question 8

In a grassland, a flowering plant is visited by a native bee that collects nectar. When bees are excluded with mesh cages, the plant produces far fewer seeds, but leaf growth and survival remain similar. The bees also show higher larval survival in areas with many of these flowers compared with areas where the plant is rare. No other pollinators are observed visiting the plant during the study period. Which interaction best explains the outcomes for both species?

Commensalism, because the bee benefits while the plant is unaffected in seed production.
Mutualism, because both the bee and plant show increased reproductive success. (correct answer)
Predation, because the bee reduces plant fitness by removing nectar.
Competition, because bees and plants require the same limiting nutrient in soil.
Parasitism, because the plant reduces bee larval survival by attracting predators.

Explanation: This question requires analyzing community ecology interactions based on reciprocal fitness effects between species. The data shows that when bees are excluded, the plant produces far fewer seeds (reduced reproductive success), while bees show higher larval survival in areas with many of these flowers (increased reproductive success). This reciprocal benefit pattern where both species gain fitness advantages from their interaction defines mutualism - the bee gets nectar for larval nutrition while providing pollination services that increase plant seed production. A common misconception is choosing commensalism (A) because students focus only on the bee's benefit, missing that the plant's seed production clearly increases with bee pollination. To identify mutualism, look for evidence that both species show measurable fitness benefits from their interaction.

Question 9

A student fills dialysis tubing (permeable to water but not to starch) with 0.60 M starch solution and places it into a beaker containing 0.20 M starch solution. The initial mass of the tubing is recorded, and after 20 minutes the tubing has increased in mass. Assume temperature and pressure remain constant and no starch crosses the tubing. Which outcome is most likely responsible for the mass change?

Net water movement into the tubing because the beaker solution is hypertonic to the tubing.
Net water movement out of the tubing because the tubing solution has higher water potential.
Net water movement into the tubing because the tubing solution is hypertonic to the beaker. (correct answer)
Net starch movement into the tubing because starch diffuses down its concentration gradient.
No net water movement because both solutions contain starch, so they are isotonic.

Explanation: This question assesses understanding of tonicity and osmoregulation using dialysis tubing as a model for semipermeable membranes. The tubing's 0.60 M starch solution has higher solute concentration than the beaker's 0.20 M, making the inside hypertonic with lower water potential internally. Water moves from higher potential in the beaker to lower inside the tubing, increasing mass due to net influx. This osmosis occurs without starch diffusion, explaining the observed change. A tempting distractor is choice A, claiming influx because the beaker is hypertonic, but this reverses the tonicity comparison, a misconception arising from confusing which side has higher solutes. For a transferable strategy, visualize water potential as 'pulling' water toward higher solute areas to anticipate mass or volume changes in enclosed systems.

Question 10

Red blood cells (internal solute concentration 0.30 M) are placed in Solution Z with 0.30 M NaCl. NaCl does not cross the membrane, but water does. After 15 minutes, the average cell diameter is unchanged. Which outcome is most likely regarding net water movement in Solution Z?

Water moves into cells because the outside has more water.
Water moves out of cells because the outside is hypertonic.
No net water movement occurs because the solutions are isotonic. (correct answer)
NaCl enters cells, increasing internal solute and shrinking them.
ATP-driven pumps remove water from cells to maintain diameter.

Explanation: This question tests understanding of tonicity and osmoregulation under isotonic conditions. The red blood cells have an internal solute concentration of 0.30 M and are placed in a solution with 0.30 M NaCl, meaning the solute concentrations inside and outside the cells are equal. When solutions are isotonic, they have the same water potential, so there is no net movement of water—water moves equally in both directions across the membrane. This explains why the cell diameter remains unchanged after 15 minutes, as there is no net gain or loss of water to change cell volume. A common error is thinking that water doesn't move at all (choice C is correct but for the right reason—there's no NET movement, though water molecules still cross the membrane in both directions). When solute concentrations are equal on both sides of a membrane, predict no net water movement and no volume change.

Question 11

A student compares diffusion-limited nutrient uptake in two similarly shaped cells. Cell 1 has surface area 600 µm² and volume 200 µm³; Cell 2 has surface area 1,200 µm² and volume 800 µm³. Both have equal membrane permeability and are in the same nutrient concentration. Which prediction best describes which cell maintains higher internal nutrient concentration under steady conditions?

Cell 2, because greater total surface area always produces greater uptake regardless of volume.
Cell 1, because its surface area–to–volume ratio is higher, supporting greater uptake per unit cytoplasmic volume. (correct answer)
Cell 2, because its volume is larger and therefore nutrients diffuse faster inside the cytoplasm.
Both, because equal membrane permeability ensures equal uptake per unit volume at steady state.
Cell 1, because smaller cells have lower surface area–to–volume ratio, which increases nutrient accumulation.

Explanation: This question assesses surface area-to-volume ratio in maintaining nutrient levels. Cell 1, with its higher surface area-to-volume ratio (600/200 = 3 vs. 1200/800 = 1.5), supports greater nutrient uptake per unit volume, allowing a higher internal concentration at steady state. The smaller cell's efficient ratio better matches supply to demand. This prediction holds under equal conditions. A tempting distractor is choice A, prioritizing greater total surface area in Cell 2, but this confuses total with ratio-based efficiency, a common error. When predicting steady-state concentrations, compute surface area-to-volume ratios for transport capacity insights.

Question 12

A cell’s plasma membrane is a phospholipid bilayer with embedded proteins. The bilayer’s hydrophobic core limits diffusion of polar solutes, while certain transmembrane proteins form selective channels lined with polar amino acids. A mutation replaces several polar amino acids lining a glucose channel with nonpolar amino acids, without changing the channel’s overall size. After the change, glucose transport through the channel decreases markedly.

Which feature best explains the reduced glucose transport?

A less polar channel lining reduces favorable interactions with glucose during passage (correct answer)
A less polar channel lining increases the membrane’s surface tension, blocking diffusion
Nonpolar amino acids add charge to the channel, repelling glucose molecules away
Nonpolar amino acids convert glucose into a lipid, trapping it within the bilayer core
Nonpolar amino acids increase phospholipid head size, preventing channels from opening

Explanation: This question tests understanding of plasma membrane structure and transport selectivity. The correct answer is A because polar amino acids lining the channel create a hydrophilic environment that stabilizes polar solutes like glucose during passage, while nonpolar amino acids cannot provide these favorable interactions. The mutation replacing polar with nonpolar amino acids reduces glucose transport despite maintaining channel size, demonstrating the importance of chemical compatibility. Answer C is incorrect because nonpolar amino acids are uncharged and cannot repel molecules—only charged amino acids create electrostatic forces, reflecting a misconception about amino acid properties. When analyzing channel selectivity, match the chemical properties of the channel lining (polar/nonpolar) with the transported solute for optimal transport.

Question 13

A student places identical animal cells (internal solute concentration 0.30 M) into three beakers separated by membranes permeable only to water. Beaker 1 contains 0.10 M sucrose, Beaker 2 contains 0.30 M sucrose, and Beaker 3 contains 0.50 M sucrose. After 10 minutes, the student observes no change in the cells in Beaker 2. Assume sucrose does not cross the cell membrane and no active transport changes solute levels during the trial. Which outcome is most likely for the cells in Beaker 3 compared with their initial volume?

They swell because water moves into the cells down the solute gradient.
They shrink because water moves out of the cells toward the higher solute concentration. (correct answer)
They remain unchanged because equal solute concentrations prevent net water movement.
They lyse because sucrose diffuses into the cells, increasing internal solute concentration.
They shrink because cells actively pump water out to maintain a constant volume.

Explanation: This question assesses understanding of tonicity and osmoregulation, which involve how cells maintain water balance in different solute environments. The cells in Beaker 3 are exposed to a 0.50 M sucrose solution, which has a higher solute concentration than the internal 0.30 M, resulting in lower water potential outside the cells. Water potential drives water movement from areas of higher potential (inside the cells) to lower potential (outside), causing net water efflux and cell shrinkage. This hypertonic condition explains why the cells shrink compared to their initial volume, as observed similarly with no change in the isotonic Beaker 2. A tempting distractor is choice A, which suggests swelling due to water influx, but this confuses hypertonic with hypotonic environments, a common misconception where tonicity directions are reversed. To approach similar problems, always compare solute concentrations to determine tonicity and predict water movement based on water potential gradients.

Question 14

A fish is transferred from well-aerated water to water with low dissolved oxygen for 15 minutes. The fish increases gill ventilation rate and spends more time near the surface. When returned to well-aerated water, ventilation rate decreases. Which response best explains the fish’s short-term response to the oxygen change?

Chemoreceptors detect low oxygen and increase ventilation and surface activity to raise oxygen uptake. (correct answer)
The fish grows new gill filaments within minutes, permanently increasing surface area for exchange.
The fish switches to photosynthesis near the surface, producing oxygen internally to meet demand.
The fish reduces diffusion by thickening gill membranes, preventing oxygen loss to the water.
The fish moves to the surface to increase future mating opportunities, not to change respiration.

Explanation: This question assesses the skill of analyzing how organisms respond to changes in their external environment. In low-oxygen water, chemoreceptors detect the decrease and trigger increased gill ventilation and surface activity to enhance oxygen diffusion into the blood. When returned to well-aerated water, these behaviors reverse as oxygen levels normalize, indicating a temporary adjustment. This short-term physiological and behavioral response maintains adequate oxygen uptake without permanent modifications to the respiratory system. A tempting distractor is choice B, which claims the fish grows new gill filaments quickly, but this misconceptions confuses rapid behavioral changes with long-term developmental growth. A transferable strategy is to identify sensory detection and immediate adjustments in ventilation or positioning as short-term responses to gas levels, separating them from evolutionary or growth-based adaptations.

Question 15

A mammalian gene has a CpG-rich promoter. When promoter cytosines are methylated, a methyl-binding protein recruits chromatin remodeling factors that increase nucleosome packing, and mRNA levels are low. In a treated cell population, promoter methylation increases while transcription decreases. Which change would most likely increase transcription without altering the gene’s coding sequence?

Increase DNA methyltransferase activity to add additional methyl groups
Introduce a mutation that removes CpG sites from the promoter region (correct answer)
Add a translation inhibitor to reduce ribosome movement along mRNA
Increase availability of RNA nucleotides to speed RNA chain elongation
Duplicate the gene’s exons to increase the number of possible proteins

Explanation: This question investigates transcriptional regulation via epigenetic mechanisms in mammals, particularly DNA methylation's role in gene silencing. Methylation of CpG sites in the promoter recruits proteins that compact chromatin, reducing transcription, and increased methylation correlates with decreased mRNA. Mutating to remove CpG sites prevents methylation, avoiding recruitment of repressive factors and allowing more open chromatin for higher transcription without changing the coding sequence. This eliminates the sites targeted for silencing, bypassing the methylation-dependent repression. A tempting distractor is option A, increasing DNA methyltransferase activity, but this would enhance methylation and further repress transcription, stemming from the misconception that more methylation activates rather than silences genes. To address epigenetics problems, identify modifiable elements like CpG sites and consider alterations that disrupt repressive marks.

Question 16

In smooth muscle cells, ligand Z binds a GPCR that activates Gs, increasing cAMP and activating PKA. PKA phosphorylates myosin light-chain kinase (MLCK), decreasing MLCK activity and reducing contraction. A drug that blocks PKA prevents MLCK phosphorylation and contraction remains high after Z addition. Which manipulation would most likely restore Z-dependent reduction in contraction while PKA remains blocked?

Use a MLCK variant that mimics the phosphorylated (low-activity) state (correct answer)
Increase cAMP breakdown to reduce PKA inhibition and allow MLCK phosphorylation
Block Gs activation so cAMP does not rise and contraction decreases through another route
Increase intracellular Ca $^{2+}$ to activate PKA directly and phosphorylate MLCK
Add ligand Z at higher concentration so it can phosphorylate MLCK without kinases

Explanation: This question assesses the skill of analyzing signal transduction pathways by restoring effects despite PKA blockage in smooth muscle signaling. PKA blockage prevents MLCK phosphorylation and reduction in contraction after Z, as PKA normally inhibits MLCK via phosphorylation. Using a MLCK variant mimicking the phosphorylated (low-activity) state directly reduces MLCK activity, bypassing PKA and restoring Z-dependent contraction decrease. This phosphomimetic simulates the downstream effect. A tempting distractor is choice B, increasing cAMP breakdown, but this would reduce PKA activity further, not restore inhibition, confusing activation with inhibition. For kinase blocks, use variants that mimic the modified state of their substrates.

Question 17

An enzyme is tested with two molecules: Substrate X and Substrate Y. With equal enzyme and substrate concentrations, the initial rate is high for X but near zero for Y. Both molecules are stable under assay conditions, and no inhibitor is added. The enzyme’s activity returns when X is reintroduced after exposure to Y. Which explanation best accounts for the difference in reaction rates?

The enzyme’s active site is complementary to X, so Y forms few productive enzyme–substrate complexes. (correct answer)
Y increases enzyme concentration, which reduces reaction rate by lowering substrate collision frequency.
Y permanently denatures the enzyme, but X later refolds it into an active conformation.
The enzyme catalyzes Y only after the reaction reaches equilibrium, so initial rate appears near zero.
Y is converted into X spontaneously, and the enzyme only catalyzes the conversion after saturation.

Explanation: This question assesses the skill of analyzing enzyme function by comparing substrate specificity in reaction rates. The high rate with Substrate X occurs because the enzyme's active site is complementary in shape and chemistry, allowing efficient ES complex formation and catalysis. In contrast, the near-zero rate with Y indicates poor fit or interaction, resulting in few productive complexes, despite equal concentrations and stable conditions. The return of activity upon reintroducing X confirms no permanent alteration, emphasizing the enzyme's selectivity based on molecular recognition rather than denaturation or other factors. A tempting distractor is choice C, claiming Y denatures the enzyme but X refolds it, but this reflects a misconception about substrate roles, as substrates don't typically refold enzymes, and the stimulus notes stability and reversibility. For enzyme questions on specificity, focus on active site complementarity as the key to distinguishing effective substrates from ineffective ones.

Question 18

In an experiment, ligand Z binds to a receptor and causes rapid opening of an ion channel, producing a measurable influx of Na $^+$ . A competitive antagonist is added that occupies the receptor’s ligand-binding site without activating the receptor. After antagonist treatment, Z no longer triggers Na $^+$ influx. Which of the following best explains the antagonist’s effect on reception?

The antagonist prevents Z from binding the receptor, so the receptor does not undergo the activation step needed to open the channel. (correct answer)
The antagonist increases Z binding affinity, causing sustained channel opening and Na $^+$ influx.
The antagonist blocks Na $^+$ synthesis, reducing the extracellular Na $^+$ available for influx.
The antagonist enters the nucleus and suppresses ion channel gene transcription immediately.
The antagonist converts Z into Na $^+$ , eliminating the ligand and the gradient.

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on receptor-ligand interactions and early transduction events. The correct answer is A because the antagonist occupies the binding site without activating the receptor, preventing Z from binding and thus blocking the activation needed for channel opening and Na+ influx. This is supported by the loss of influx after treatment and aligns with principles of ligand-gated channels where binding induces conformational changes for gating. The competitive nature ensures it specifically disrupts reception. A tempting distractor is B, which wrongly suggests increased binding causes sustained opening, based on the misconception that antagonists activate rather than block, confusing agonist and antagonist roles. When analyzing signal transduction questions, distinguish agonists from antagonists by their effects on activation versus inhibition.

Question 19

In a salt marsh, periwinkle snails graze on cordgrass leaves. Researchers placed cages that excluded snails from some plots for 6 weeks. Exclusion plots developed thicker cordgrass stems and greater leaf cover than uncaged plots. In uncaged plots, more bare sediment was visible and cordgrass leaves showed grazing scars. No other herbivores were observed, and soil salinity and flooding frequency were similar across plots. Which interaction best explains the difference between caged and uncaged plots?

Herbivory, with snails reducing cordgrass biomass through grazing (correct answer)
Mutualism, with snails increasing cordgrass growth by cleaning leaf surfaces
Commensalism, with cordgrass unaffected while snails benefit from shelter
Interspecific competition, with snails and cordgrass competing for sediment nutrients
Parasitism, with cordgrass extracting energy from snails during feeding

Explanation: This question assesses the skill of analyzing community ecology by identifying species interactions based on experimental exclusion evidence. The caged plots, which exclude snails, show thicker cordgrass stems and greater leaf cover, indicating that snails are reducing plant biomass through grazing, a form of herbivory where one species consumes plant tissue, negatively impacting the plant's growth and survival. In uncaged plots, visible grazing scars and more bare sediment further support that snails are herbivores directly feeding on the cordgrass, leading to decreased plant health. This interaction logic highlights herbivory as the driving force, as the absence of snails allows cordgrass to thrive without consumption pressure. A tempting distractor is B, mutualism, which is wrong because the plants are harmed rather than benefited, stemming from the misconception that any close association between species must be mutually positive. To analyze similar interactions, always compare outcomes in presence versus absence experiments to determine if a species causes harm, benefit, or no effect.

Question 20

In cultured liver cells, the peptide hormone H binds a specific protein on the outer surface of the plasma membrane. Within 10 seconds of adding H, cytosolic cAMP increases. When cells are pretreated with a drug that prevents the receptor from activating a nearby G protein, H still binds the receptor but cAMP does not increase. No changes in total receptor amount are detected during the experiment. Which of the following best explains how the signal is initiated after H binds its receptor?

H enters the cell and directly activates adenylyl cyclase in the cytosol.
Ligand binding enables the receptor to activate a G protein that stimulates cAMP production. (correct answer)
H binding causes receptor degradation, which releases cAMP from membrane vesicles.
The receptor phosphorylates DNA-associated proteins, increasing cAMP as a byproduct.
H binding opens a nuclear pore, allowing cAMP to diffuse from the nucleus.

Explanation: This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is B because the rapid increase in cAMP after H binding and the prevention of this increase by a drug blocking G protein activation indicate that the receptor is a G protein-coupled receptor (GPCR). In basic signaling principles, ligand binding to a GPCR activates an associated G protein, which then stimulates adenylyl cyclase to produce cAMP from ATP. The evidence that H binds the receptor without cAMP increase when G protein activation is blocked, and no receptor degradation occurs, supports that the signal initiation relies on G protein-mediated transduction rather than direct receptor actions. A tempting distractor is C, which is wrong because it assumes receptor degradation releases cAMP, reflecting the misconception that signaling involves receptor breakdown rather than conformational changes activating intracellular messengers. A transferable strategy for signal transduction questions is to identify the receptor type and immediate downstream effectors based on timing and inhibitor effects.

Question 21

After a severe drought, a grassland’s flowering plant cover drops by 60% and insect pollinator counts drop by 50% the following month. Which consequence is most likely for the ecosystem over the next growing season?

Primary productivity and seed set decrease further because fewer pollination events reduce plant recruitment. (correct answer)
Primary productivity increases because reduced plant cover concentrates soil nutrients into fewer individuals.
Pollinator counts rise sharply because fewer flowers reduce competition for nectar among insects.
The grassland becomes more stable because drought eliminates fluctuations in plant and insect populations.
Decomposer activity stops because drought removes the need for nutrient recycling in the soil.

Explanation: This question assesses the skill of analyzing disruptions in ecosystems by examining the cascading effects of drought on plant-pollinator interactions in a grassland. The severe drought reduces flowering plant cover and insect pollinator counts, leading to fewer pollination events that are critical for seed production and plant recruitment. As a result, primary productivity declines further because the ecosystem's ability to regenerate plant biomass is impaired at a system level, creating a feedback loop of diminished resources for both plants and pollinators. This interconnected decline affects the overall energy flow and stability of the grassland ecosystem over the next growing season. A tempting distractor is choice B, which suggests primary productivity increases due to nutrient concentration, but this overlooks the misconception that reduced plant numbers inherently limit total productivity rather than enhance it. To analyze similar disruptions, always trace mutualistic relationships and their impacts on reproduction and energy cycles across seasons.

Question 22

In yeast, pheromone F binds a GPCR on the plasma membrane. Shortly after binding, the G protein’s beta-gamma subunits dissociate from the alpha subunit and interact with a downstream effector at the membrane. If a mutation prevents dissociation of the subunits, F binding still occurs but the effector is not activated. Which of the following best explains the earliest blocked step?

The mutation blocks receptor synthesis, so F cannot bind at the cell surface.
Subunit dissociation is required to expose interaction surfaces that activate the membrane-associated effector. (correct answer)
The mutation increases effector activity because intact G proteins are always active.
The effector must enter the nucleus before it can bind the G protein subunits.
F must be phosphorylated into a second messenger to cause subunit dissociation.

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 preventing subunit dissociation blocks effector activation despite F binding, indicating dissociation exposes surfaces for effector interaction. This follows GPCR principles where ligand binding triggers subunit separation for signaling. The mutation's specificity identifies the step. A tempting distractor is C, which wrongly claims intact G proteins are active, based on the misconception that dissociation inhibits rather than enables, confusing activation states. When analyzing signal transduction questions, mutate key proteins to pinpoint dynamic changes like dissociation in transduction.

Question 23

In a classroom investigation, students shine a bright light on a fish in a tank. Within seconds, the fish swims to the darker side and remains there while the light is on. When the light is turned off, the fish resumes swimming throughout the tank. The water temperature and oxygen level remain constant during the test. Which response best explains the fish’s movement when the light is turned on?

The fish shows a rapid behavioral response to light intensity, moving away from bright light (negative phototaxis). (correct answer)
The fish develops new eye structures during the test, allowing it to permanently avoid bright areas.
The fish moves because it wants to protect the students from glare by staying in darker water.
Bright light causes the fish to stop neural signaling, forcing random drifting to the tank edge.
The fish’s population becomes genetically resistant to light within seconds of exposure.

Explanation: This question assesses the skill of analyzing organisms' responses to environmental stimuli in AP Biology. The fish's swift movement to the darker side under bright light and resumption of normal swimming when off demonstrate negative phototaxis, a behavioral response to avoid potentially harmful light intensity. This is supported by the seconds-long reaction time, constant temperature and oxygen, and reversibility, indicating an innate sensory-motor reflex. The lack of environmental changes rules out other factors influencing the behavior. A tempting distractor is choice B, which incorrectly claims new eye structure development, based on the misconception that brief stimuli cause permanent anatomical evolution. To assess taxis behaviors, connect stimulus detection to immediate actions while avoiding attributions of permanence or intention.

Question 24

An investigator compares diffusion of a nutrient into two spherical cells with the same membrane permeability. Cell S has radius 3 µm; Cell T has radius 9 µm. Both are placed in the same nutrient concentration, and both metabolize the nutrient at the same rate per unit volume. After reaching steady conditions, Cell T has a lower nutrient concentration in its central cytoplasm than Cell S. Which explanation best accounts for the lower central concentration in Cell T?

Cell T has a lower surface area–to–volume ratio, reducing nutrient entry per unit volume of cytoplasm. (correct answer)
Cell T has a higher surface area–to–volume ratio, which decreases membrane transport into the cell interior.
Cell T has more total membrane area, so its central cytoplasm should have the highest nutrient concentration.
Cell T has a smaller internal volume, so nutrients are diluted less and concentration drops at the center.
Cell T has a lower external nutrient concentration because larger cells reduce solute concentration in the medium.

Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular transport efficiency. Cell T's larger radius results in a lower surface area-to-volume ratio, reducing nutrient influx per unit volume relative to metabolic use. This limitation causes a steeper internal gradient, with lower concentrations at the center of the larger cell. The transport efficiency logic explains that diffusion supply depends on surface area, while consumption is volumetric, leading to central depletion in bigger cells. A tempting distractor is choice C, which suggests more total membrane improves central supply, but overlooks that lower ratio hampers per-volume efficiency. To approach similar problems, consider how size affects internal diffusion gradients and surface-limited supply.

Question 25

In an insect population, wing color is determined by a single gene with two alleles: $W_D$ (dark) and $W_L$ (light). After a new bird predator becomes common, mark–recapture data over 10 generations show that dark-winged insects have higher survival and produce more offspring than light-winged insects. The frequency of $W_D$ increases from 0.48 to 0.86, while $W_L$ decreases accordingly. Which pattern best illustrates the evolutionary mechanism changing allele frequencies in this population?

Genetic drift, because allele frequencies change over generations regardless of fitness differences
Directional selection, because one allele increases due to higher reproductive success of its carriers (correct answer)
Stabilizing selection, because intermediate phenotypes would be favored in a constant environment
Disruptive selection, because both light and dark phenotypes should increase relative to intermediates
Gene flow, because immigration of dark-winged insects is the primary cause of the allele increase

Explanation: This question assesses the skill of analyzing patterns of natural selection by interpreting population-level changes in traits and allele frequencies. With a new bird predator, dark-winged insects have higher survival and offspring production, causing the dark allele frequency to rise from 0.48 to 0.86 over generations. This consistent advantage shifts the population toward darker wings, without evidence of immigration or random changes. The pattern reflects directional selection, favoring one allele due to fitness differences in the changed environment. A tempting distractor is genetic drift, which is wrong as it implies random changes without fitness ties, based on the misconception that any allele shift is non-adaptive. To differentiate mechanisms, always check for consistent fitness advantages linked to traits when allele frequencies change predictably.

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