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AP Biology

AP Biology Practice Test: Practice Test 73

Practice Test 73 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A researcher studies a signaling pathway in animal cells where ligand M causes a rapid opening of a specific ion channel in the plasma membrane, increasing cytosolic Na+^++. The ligand binds to receptor T on the cell surface. In cells expressing a receptor T variant lacking most of its cytosolic tail, ligand binding still occurs, but the Na+^++ increase is greatly reduced. Direct application of a channel-opening drug restores Na+^++ influx in both cell types. Which of the following best explains the function of receptor T’s cytosolic tail in early signal transduction?

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Question 1

A researcher studies a signaling pathway in animal cells where ligand M causes a rapid opening of a specific ion channel in the plasma membrane, increasing cytosolic Na+^++. The ligand binds to receptor T on the cell surface. In cells expressing a receptor T variant lacking most of its cytosolic tail, ligand binding still occurs, but the Na+^++ increase is greatly reduced. Direct application of a channel-opening drug restores Na+^++ influx in both cell types. Which of the following best explains the function of receptor T’s cytosolic tail in early signal transduction?

  1. The cytosolic tail binds Na+^++ and transports it across the membrane after ligand binding.
  2. The cytosolic tail interacts with intracellular proteins that promote channel opening after receptor activation. (correct answer)
  3. The cytosolic tail is required for ligand M synthesis and secretion from the signaling cell.
  4. The cytosolic tail converts Na+^++ into a second messenger that activates the channel.
  5. The cytosolic tail enables the receptor to enter the nucleus to open ion channels from inside.

Explanation: This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is B because the receptor variant lacking the cytosolic tail binds ligand M but shows reduced Na⁺ increase, indicating the tail is essential for transducing the signal to open channels, while the channel-opening drug restores function, confirming channels are present but not activated. This suggests the cytosolic tail interacts with intracellular proteins, such as G proteins or adapters, to promote channel opening via second messengers. Basic signaling principles show that receptor cytosolic domains are crucial for coupling to effectors in pathways like those involving ion channels. A tempting distractor is A, which is wrong due to the misconception that receptors directly transport ions, ignoring that the response involves separate channels and transduction steps. For signal transduction questions, analyze receptor domain functions by comparing wild-type and mutant behaviors to determine their role in intracellular signaling interactions.

Question 2

A student compares DNA and RNA nucleotides. Both contain a pentose sugar, a phosphate group, and a nitrogenous base, but one type of nucleic acid is more chemically stable in alkaline conditions. The difference is traced to a functional group on the sugar that can participate in intramolecular reactions that break the backbone. Which feature best explains why RNA is generally less stable than DNA under basic conditions?

  1. RNA has a 2' hydroxyl group on ribose that can promote backbone cleavage (correct answer)
  2. RNA contains thymine, which is more reactive than uracil in base
  3. RNA has triple hydrogen bonds between bases that strain the backbone
  4. RNA has deoxyribose, which forms fewer phosphodiester bonds per nucleotide
  5. RNA is double-stranded, increasing repulsion between phosphates in base

Explanation: This question tests understanding of nucleic acid structure-function relationships, specifically the chemical differences between RNA and DNA that affect stability. The correct answer is A because RNA contains ribose sugar with a 2'-OH group that can act as a nucleophile, attacking the adjacent phosphodiester bond and causing backbone cleavage, especially under basic conditions. DNA lacks this 2'-OH (hence deoxyribose), making it more chemically stable. Choice B incorrectly states that RNA contains thymine when it actually contains uracil, and this base difference doesn't explain the stability difference. The strategy is to focus on the sugar component differences between RNA (ribose with 2'-OH) and DNA (deoxyribose without 2'-OH) when considering chemical stability.

Question 3

A plant leaf cell contains chloroplasts with extensive thylakoid membranes arranged in stacks. When light intensity increases, the cell’s rate of ATP formation inside chloroplasts increases. Which feature best explains why stacked thylakoid membranes support higher ATP formation rates?

  1. Greater thylakoid membrane surface area allows more electron transport complexes and ATP synthase (correct answer)
  2. Thylakoid stacks provide extra space for chromosomal DNA replication in the chloroplast
  3. Thylakoid membranes directly import glucose from the cytosol through nuclear pores
  4. Stacked thylakoids increase ATP by enabling ribosomes to attach and translate enzymes
  5. Thylakoid stacks increase ATP by allowing water to diffuse freely out of the chloroplast

Explanation: This question assesses the skill of analyzing cell structure-function relationships. The extensive stacked thylakoid membranes in chloroplasts increase surface area for embedding photosystems, electron transport chains, and ATP synthase, enhancing ATP production with higher light intensity. This structural adaptation supports efficient photophosphorylation, as noted in the stimulus. In AP Biology, thylakoid stacking optimizes light-dependent reactions. A tempting distractor is D, claiming thylakoids enable ribosome attachment, but this is incorrect due to structure-function confusion, as thylakoids are for photosynthesis, not translation. When analyzing photosynthetic efficiency, connect membrane area to reaction complex density.

Question 4

To model protocell formation, fatty acids were added to water in two treatments: (1) pure water and (2) water containing dissolved organic molecules (nucleotides and amino acids). In both treatments, fatty acids spontaneously assembled into vesicles, but only in treatment (2) did vesicles persist longer and retain a higher fraction of the dissolved organics after gentle mixing. Which conclusion is best supported regarding hypotheses about early cell-like structures?

  1. Vesicle formation requires proteins to catalyze membrane assembly in aqueous environments.
  2. Lipid vesicles can form spontaneously and can compartmentalize dissolved organics, supporting protocell models. (correct answer)
  3. The data demonstrate that DNA-based genomes existed before membranes on early Earth.
  4. Long-term vesicle stability proves that natural selection acted on fully living cells in the experiment.
  5. Compartmentalization prevents any exchange with the environment, making early metabolism impossible.

Explanation: This question requires evaluating evidence about the origins of life, specifically how experimental data supports protocell formation hypotheses. The correct answer (B) is supported because the experiment shows that fatty acids spontaneously self-assemble into vesicles in water (demonstrating no protein catalysis needed) and that these vesicles can compartmentalize dissolved organic molecules like nucleotides and amino acids, with enhanced stability when organics are present. This supports protocell models because it demonstrates that simple lipid membranes could have formed spontaneously and created isolated environments where prebiotic chemistry could occur differently than in bulk solution. Answer A represents a protein-requirement misconception—the spontaneous vesicle formation in pure water directly contradicts the need for protein catalysts in membrane assembly. When analyzing protocell experiments, look for evidence of spontaneous organization and compartmentalization capabilities, not assumptions about modern cellular complexity.

Question 5

A researcher compares two populations of secretory cells. Population 1 contains abundant rough endoplasmic reticulum (RER) and many Golgi stacks; Population 2 has relatively little RER and a small Golgi region. Both populations have similar numbers of mitochondria. The RER consists of membrane sacs studded with ribosomes, and the Golgi modifies and packages proteins into vesicles that fuse with the plasma membrane. Which outcome is most likely for Population 1 compared with Population 2?​​​

  1. Greater rate of exporting proteins to the extracellular environment via vesicle trafficking (correct answer)
  2. Greater rate of DNA replication because RER membranes provide nucleotides for the nucleus
  3. Greater rate of photosynthesis because stacked membranes increase light absorption efficiency
  4. Greater rate of cytosolic glycolysis because Golgi enzymes catalyze glucose breakdown
  5. Greater rate of chromosome movement because Golgi stacks assemble microtubules during mitosis

Explanation: This question requires analysis of cell structure-function relationships to predict cellular capabilities based on organelle abundance. The correct answer A is supported because cells with extensive rough endoplasmic reticulum (RER) and prominent Golgi apparatus are specialized for protein synthesis and secretion, as the RER's ribosomes synthesize proteins that are then modified in the Golgi and packaged into vesicles for export via exocytosis. The stimulus explicitly states that "the Golgi modifies and packages proteins into vesicles that fuse with the plasma membrane," directly describing the secretory pathway that would be enhanced in Population 1. Choice C represents a fundamental organelle confusion misconception where students attribute photosynthetic function to the Golgi apparatus rather than chloroplasts, failing to recognize that the stacked membranes mentioned refer to Golgi cisternae, not thylakoids. The key strategy for analyzing organelle abundance questions is to match specific organelle structures with their established functions in the cell, recognizing that increased abundance of secretory pathway components predicts increased secretory activity.

Question 6

In a lab, a eukaryotic cell is observed with extensive rough endoplasmic reticulum and many ribosomes attached to its surface. Vesicles frequently bud from this membrane network and move toward the cell’s periphery. The cell releases large amounts of a water-soluble protein into the surrounding medium. No unusual changes are seen in mitochondria or chloroplasts. Which feature best explains how the cell produces proteins for secretion in high quantities?

  1. Ribosomes on rough ER synthesize polypeptides into the ER lumen for vesicle transport to the membrane (correct answer)
  2. Smooth ER enzymes assemble amino acids into proteins that diffuse directly through the plasma membrane
  3. Mitochondrial cristae provide compartments where secreted proteins are folded before export
  4. Lysosomes fuse with the plasma membrane to release newly made proteins into the extracellular space
  5. The nucleolus packages completed proteins into vesicles that travel along the cytoplasm to the surface

Explanation: This question assesses the skill of analyzing cell structure and function by linking organelle features to protein secretion processes. The extensive rough endoplasmic reticulum with attached ribosomes indicates active protein synthesis for export, as ribosomes on the rough ER translate mRNA into polypeptides that enter the ER lumen, a key step in the secretory pathway. Vesicles budding from the rough ER transport these proteins to the Golgi for modification and then to the plasma membrane for secretion, explaining the high quantities of water-soluble protein released. The absence of changes in mitochondria or chloroplasts supports that energy production or photosynthesis is not directly involved, aligning with the rough ER's role in eukaryotic cells for secreting proteins like enzymes or hormones. A tempting distractor is choice B, which incorrectly suggests smooth ER assembles proteins that diffuse through the membrane, reflecting a structure-function confusion by misattributing protein synthesis to smooth ER instead of rough ER and ignoring vesicle-mediated export. To approach similar questions, identify the organelle most directly linked to the observed function and trace the cellular pathway step by step.

Question 7

Two cube-shaped cells have the same membrane composition and live in identical solutions containing glucose. Cell X has side length 5 μm5\,\mu m5μm and cell Y has side length 15 μm15\,\mu m15μm. Both cells use facilitated diffusion through membrane transport proteins to import glucose, and both metabolize glucose throughout the cytoplasm at similar rates per unit volume. After 10 minutes, cell Y shows a larger decrease in cytoplasmic glucose concentration than cell X. Which explanation best accounts for the observed difference?

  1. Cell Y has fewer transport proteins because larger cells reduce membrane protein density.
  2. Cell Y has a lower surface area–to–volume ratio, limiting glucose entry relative to demand. (correct answer)
  3. Cell Y has a higher surface area, so glucose leaves the cell faster than it enters.
  4. Cell Y has a smaller cytoplasmic volume, so glucose concentration drops more quickly.
  5. Cell Y has a higher metabolic rate because larger cells always metabolize faster per unit volume.

Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular processes like glucose import. The correct answer is B because cell Y, being larger, has a lower surface area-to-volume ratio, which limits the membrane area available for glucose transport relative to the cytoplasmic volume that metabolizes it. Consequently, glucose entry via facilitated diffusion cannot match the demand in the larger volume, causing a faster drop in concentration. This transport efficiency logic explains why larger cells face challenges in sustaining nutrient levels through membrane-based uptake. A tempting distractor is C, which wrongly suggests that higher surface area increases glucose loss, based on the misconception that total area directly correlates with outward diffusion rates without considering the ratio. To approach similar problems, always compare surface area-to-volume ratios to evaluate how size impacts supply relative to demand.

Question 8

An enzyme from a freshwater fish is placed in buffers with different NaCl concentrations while substrate concentration is held constant. In 0.1 M NaCl, the reaction rate is 20 units; in 1.0 M NaCl, the rate is 7 units. The enzyme’s tertiary structure is maintained partly by ionic interactions among charged side chains. Increasing salt concentration can shield charges, weakening electrostatic attractions that help stabilize the active site’s shape. The temperature and pH are unchanged. Which condition would most likely explain the decreased activity at 1.0 M NaCl?

  1. High NaCl strengthens ionic bonds between side chains by providing more ions to link charges, increasing the reaction rate.
  2. High NaCl changes DNA transcription of the enzyme gene, lowering enzyme concentration and decreasing reaction rate.
  3. High NaCl shields charged side chains, weakening stabilizing electrostatic interactions and distorting the active site. (correct answer)
  4. High NaCl supplies additional substrate molecules that compete for the active site, lowering product formation.
  5. High NaCl breaks peptide bonds directly, causing immediate and complete loss of all protein primary structure.

Explanation: This question assesses the skill of analyzing environmental impacts on enzyme function, specifically how salt concentration affects enzyme activity. The correct answer is choice C because the stimulus explains that high NaCl shields charged side chains, weakening electrostatic attractions that stabilize the enzyme's tertiary structure and active site shape. This distortion reduces catalytic efficiency, as evidenced by the reaction rate dropping from 20 units in 0.1 M NaCl to 7 units in 1.0 M NaCl. Protein structure logic supports this, as ionic interactions are key to folding in many enzymes, and increased salt disrupts these without affecting covalent bonds or gene expression. A tempting distractor is choice E, which is wrong because it claims high salt breaks peptide bonds, reflecting the misconception that salinity impacts primary structure instead of noncovalent interactions in tertiary structure. A transferable strategy for interpreting enzyme-environment questions is to identify how ionic conditions influence charge-based interactions while ruling out effects on transcription or covalent modifications unless explicitly indicated.

Question 9

A researcher adds Ligand M to cells and detects rapid activation of a cytosolic kinase only when the receptor’s intracellular tail contains phosphorylatable serines. A modified receptor binds M normally but has all serines in the tail replaced with alanines, and kinase activation is not observed. Which of the following best explains the loss of early signaling?

  1. Phosphorylation of serines on the receptor tail creates binding sites needed to recruit signaling proteins. (correct answer)
  2. Alanine substitution prevents ligand M from entering the cell, eliminating receptor binding.
  3. Serine phosphorylation is required to increase receptor gene copy number, enabling kinase activation.
  4. Replacing serines with alanines increases membrane fluidity, which directly inhibits cytosolic kinases.
  5. Kinase activation requires serines to be converted into ATP, which then phosphorylates the kinase.

Explanation: This question tests your ability to analyze signal transduction by understanding how receptor phosphorylation enables downstream signaling. The stimulus shows that replacing phosphorylatable serines with alanines prevents kinase activation despite normal ligand binding, indicating that serine phosphorylation creates binding sites (phosphotyrosine or phosphoserine motifs) that recruit and activate downstream signaling proteins including kinases. This is a common mechanism where phosphorylated residues serve as docking sites for proteins with SH2 or other phospho-binding domains. Choice C incorrectly suggests phosphorylation increases gene copy number, confusing rapid post-translational modifications with slower genetic changes. When analyzing receptor signaling, remember that phosphorylation often creates specific binding sites for recruiting downstream signaling components.

Question 10

In a neuron, acetylcholine (ACh) binding to a postsynaptic receptor produces an immediate inward current measurable by patch clamp. The response occurs even when intracellular GTP is depleted, but it is abolished by a competitive ACh-binding antagonist. No change in cAMP is detected during the first 30 seconds. Which of the following best explains the receptor type and early transduction event?

  1. ACh activates a receptor tyrosine kinase that autophosphorylates and opens nearby ion channels indirectly
  2. ACh binds a ligand-gated ion channel receptor that changes conformation to allow ion flow across the membrane (correct answer)
  3. ACh diffuses into the cytosol and binds a nuclear receptor that triggers rapid ion channel synthesis
  4. ACh binds a GPCR, and GTP depletion prevents channel opening by blocking receptor–ligand binding
  5. ACh is enzymatically converted to a second messenger that directly carries charge through the membrane

Explanation: This question tests analysis of signal transduction by examining acetylcholine's rapid effect on neurons. The immediate current response measurable by patch clamp indicates direct ion flow, while the response persisting without GTP rules out GPCR involvement (which requires GTP for G protein activation). The competitive antagonist blocking the response confirms ACh must bind the receptor directly, and the lack of cAMP change eliminates second messenger pathways. These features identify a ligand-gated ion channel where ACh binding directly opens the channel for ion flow. Choice A incorrectly suggests a receptor tyrosine kinase with indirect channel opening, which would require more time and likely involve phosphorylation cascades. When analyzing neurotransmitter signaling, consider response speed: millisecond responses indicate ligand-gated channels, while slower responses suggest GPCR or enzyme-linked pathways.

Question 11

In snapdragons, flower color shows incomplete dominance: allele R produces red pigment and allele r produces no pigment. RR plants have red flowers, rr plants have white flowers, and Rr plants have pink flowers. A gardener crosses two pink-flowered plants (Rr × Rr) and grows 200 offspring under the same conditions. Which outcome best predicts the flower-color phenotypes among the offspring?

  1. About 100 red, 100 white, and 0 pink offspring
  2. About 150 red, 50 white, and 0 pink offspring
  3. About 50 red, 100 pink, and 50 white offspring (correct answer)
  4. About 200 pink offspring because pink is dominant
  5. About 100 red, 50 pink, and 50 white offspring

Explanation: This question tests your ability to analyze non-Mendelian inheritance patterns, specifically incomplete dominance in snapdragons. In incomplete dominance, heterozygotes (Rr) show a blended phenotype (pink) that is intermediate between the two homozygous phenotypes (red and white). When crossing two heterozygous pink plants (Rr × Rr), the offspring follow a 1:2:1 genotypic ratio (RR:Rr:rr), which directly translates to a 1:2:1 phenotypic ratio of red:pink:white flowers. With 200 offspring, we expect approximately 50 red (25%), 100 pink (50%), and 50 white (25%) plants. Students often incorrectly choose option A thinking incomplete dominance produces equal numbers of parental phenotypes, missing that heterozygotes have their own distinct phenotype. To solve incomplete dominance problems, remember that the phenotypic ratio matches the genotypic ratio because heterozygotes have a unique appearance.

Question 12

Two epithelial cell samples are compared. Sample 1 has many membrane proteins with attached carbohydrate chains projecting into the extracellular space, forming a dense surface coat. Sample 2 has far fewer of these carbohydrate-bearing proteins, but similar phospholipid composition. When mixed, cells from Sample 1 clump together more strongly than cells from Sample 2. Which feature best explains the increased cell-to-cell adhesion in Sample 1?

  1. Carbohydrate chains on glycoproteins enable specific extracellular interactions that increase adhesion between cells (correct answer)
  2. Extra cholesterol forms covalent bonds between adjacent cells, permanently fusing their membranes
  3. More smooth ER in Sample 1 secretes adhesive lipids directly into neighboring cell membranes
  4. Higher cytosolic ribosome density increases membrane thickness, causing cells to stick together
  5. Additional nuclear pores in Sample 1 export adhesion molecules straight into the extracellular matrix

Explanation: This question assesses the skill of analyzing cell structure and function by linking surface modifications to intercellular interactions. The carbohydrate chains on glycoproteins in Sample 1 form a glycocalyx that facilitates specific recognition and binding between cells, enhancing adhesion through extracellular matrix interactions or direct cell-cell contacts like in tissues. This dense surface coat increases clumping compared to Sample 2, despite similar phospholipid compositions, highlighting the functional role of glycosylation in cell signaling and adhesion. The membrane proteins' projections into the extracellular space enable these interactions without altering lipid bilayers. A tempting distractor is choice E, which suggests nuclear pores export adhesion molecules directly, embodying a level-of-organization error by bypassing the endomembrane system's processing pathway. In adhesion-related questions, evaluate extracellular components and their modifications to explain binding behaviors.

Question 13

Two populations of frogs occupy adjacent wetlands with no physical barrier between them. Population 1 males call at 800 Hz and females preferentially approach 800 Hz calls; population 2 males call at 1200 Hz and females preferentially approach 1200 Hz calls. When researchers move adults from both populations into the same wetland, males continue their typical calls and females rarely approach males from the other population. Genetic analyses show reduced gene flow and increasing divergence between populations. Which isolating mechanism most directly reduces interbreeding between the frog populations?

  1. Gametic isolation because sperm cannot penetrate eggs of the other population
  2. Behavioral isolation due to divergence in mating calls and female preferences (correct answer)
  3. Habitat isolation caused by use of different continents for breeding
  4. Polyploidy producing instant isolation through mismatched chromosome numbers
  5. Hybrid inviability because adults die immediately after mating occurs

Explanation: This question examines speciation through behavioral reproductive isolation. The two frog populations have evolved different mating call frequencies (800 Hz vs 1200 Hz) and corresponding female preferences, creating a prezygotic barrier to reproduction. Even when placed in the same wetland with no physical barriers, females rarely approach males from the other population because they don't recognize the different frequency calls as mating signals. This behavioral isolation maintains reproductive separation and allows continued genetic divergence between populations. Choice C incorrectly identifies habitat isolation, but the frogs occupy adjacent wetlands with no barriers, not different continents. To identify behavioral isolation, look for differences in courtship signals, mating rituals, or mate recognition that prevent interbreeding despite physical proximity.

Question 14

In a coastal snail population, shell color is controlled by two alleles. Before 2000, light shells (LL or Ll) were 70% of adults and dark shells (ll) were 30%. After a decade of increased predation by visually hunting crabs, marked-recapture data show light-shelled adults produced an average of 1.1 surviving offspring each, while dark-shelled adults produced 2.0. By 2010, the frequency of allele l increased from 0.40 to 0.62 across the population. Which pattern best illustrates the type of selection acting on shell color?

  1. Directional selection favoring darker shells, increasing the l allele frequency over generations (correct answer)
  2. Stabilizing selection maintaining intermediate shell colors and reducing phenotypic variance
  3. Disruptive selection favoring both light and dark shells while decreasing intermediate phenotypes
  4. Genetic drift causing random allele-frequency change unrelated to differences in reproductive success
  5. Balancing selection maintaining both alleles at equal frequencies because all genotypes reproduce equally

Explanation: This question assesses the skill of analyzing patterns of natural selection by interpreting data on allele frequencies and reproductive success in response to environmental pressures. The data show that dark-shelled snails had higher reproductive success (2.0 offspring) compared to light-shelled ones (1.1), leading to an increase in the l allele frequency from 0.40 to 0.62 over generations. This shift indicates directional selection favoring darker shells, as the population's phenotype moved toward the advantageous dark trait due to predation by crabs. The trend of increasing frequency of the darker allele aligns with a consistent push in one direction, rather than maintaining or splitting the distribution. A tempting distractor is choice C, disruptive selection, which is wrong because it would favor both extremes and increase variance, but here only one extreme (dark) is favored, stemming from the misconception that any change in extremes implies disruption. To identify selection types in similar problems, examine how fitness differences correlate with shifts in mean trait values and allele frequencies over time.

Question 15

Two cell types express the same receptor for ligand Q. In both, Q binds the receptor with equal affinity. In cell type 1, Q rapidly increases cAMP; in cell type 2, Q rapidly increases IP3_33​. The receptor’s extracellular domain sequence is identical in both cell types. Which of the following best explains the difference in the initial intracellular response?

  1. The receptor couples to different intracellular signaling proteins in each cell type, activating distinct effectors (correct answer)
  2. Cell type 2 lacks a plasma membrane, so Q binding must occur on internal membranes to generate IP3_33​
  3. Q changes shape after binding in cell type 2, converting into IP3_33​ as the first transduction step
  4. The receptor in cell type 1 is located in the nucleus, allowing cAMP to be produced by DNA-binding enzymes
  5. Cell type 1 produces cAMP because Q binding immediately increases transcription of adenylyl cyclase

Explanation: This question tests signal transduction analysis regarding pathway specificity. The same receptor producing different second messengers (cAMP vs IP₃) in different cell types, despite identical extracellular domains and binding affinity, indicates the receptor couples to different intracellular signaling machinery—likely different G protein subtypes (Gαs for cAMP, Gαq for IP₃) expressed in each cell type. Choice E incorrectly suggests immediate transcriptional changes cause cAMP production, impossible given the rapid timeframe of second messenger generation. When analyzing cell-type-specific responses, consider that identical receptors can activate different pathways depending on which downstream signaling proteins are expressed, allowing the same signal to produce diverse cellular responses.

Question 16

Two tide pools were sampled for algal species. Pool A contained 5 species with abundances 20, 20, 20, 20, and 20. Pool B contained 5 species with abundances 96, 1, 1, 1, and 1. Total algal individuals counted were 100 in each pool. Which conclusion is best supported about biodiversity?

  1. Pool B has higher diversity because rare species increase the total count of species
  2. Pool A has higher diversity because it has greater evenness at the same richness (correct answer)
  3. Both pools have equal diversity because they have the same number of algal species
  4. Pool B has higher richness because one species is dominant in the community
  5. Pool A has lower diversity because equal abundances reduce the number of species present

Explanation: Biodiversity analysis is the skill of evaluating species richness and evenness to compare diversity in microhabitats like tide pools. Both pools have equal richness of 5 algal species, but Pool A has greater evenness with equal abundances of 20 each, unlike Pool B's dominance with 96 in one and 1 in others. This leads to Pool A having higher diversity, as shown by the balanced distribution versus B's extreme unevenness at the same total of 100 individuals. The specific abundance numbers provide clear evidence for evenness's role. A tempting distractor is choice A, suggesting Pool B has higher diversity due to rare species increasing richness, but this misconceives that rare species boost diversity when they actually lower evenness without adding to richness. In biodiversity questions, emphasize evenness in equal-richness cases, using precise abundance data to calculate relative impacts.

Question 17

In a cell, extracellular ATP activates a purinergic receptor that opens a cation channel, allowing Na+^++ influx and depolarization. Depolarization opens voltage-gated K+^++ channels, causing K+^++ efflux that repolarizes the membrane. Repolarization reduces opening of the ATP-gated channel by lowering its driving force for Na+^++ entry. Blocking the K+^++ channels increases the duration of depolarization after ATP addition. Which outcome best illustrates the feedback’s role?

  1. Longer depolarization because K+^++ efflux provides negative feedback that opposes membrane excitation (correct answer)
  2. Shorter depolarization because K+^++ efflux provides positive feedback that sustains excitation
  3. No change because voltage-gated K+^++ channels do not affect membrane potential
  4. Longer depolarization because K+^++ channel block increases ATP binding to the receptor
  5. Shorter depolarization because K+^++ channel block increases Na+^++ channel inactivation

Explanation: This question assesses the skill of understanding feedback regulation in signal transduction pathways. The correct answer is longer depolarization because K⁺ efflux provides negative feedback that opposes membrane excitation, as the stimulus shows depolarization opens K⁺ channels for efflux, repolarizing and reducing Na⁺ driving force. This feedback limits depolarization duration. Blocking K⁺ channels prolongs depolarization, confirming negative feedback shortens the response. A tempting distractor is shorter depolarization because K⁺ efflux provides positive feedback that sustains excitation, which misidentifies opposition as reinforcement, a misconception in ion flux. A strategy is to model membrane potential changes with and without the feedback component.

Question 18

In a population of rabbits, a locus has alleles HHH and hhh. A random sample of 1,000 rabbits shows genotype counts: HH=640HH=640HH=640, Hh=320Hh=320Hh=320, and hh=40hh=40hh=40. The population is large, isolated, and no selection or mutation is detected. Which conclusion is best supported about Hardy-Weinberg equilibrium at this locus based on the observed data?

  1. The population deviates from equilibrium because p=0.64p=0.64p=0.64 and q=0.04q=0.04q=0.04.
  2. The population is in equilibrium because observed genotypes equal p2:2pq:q2p^2:2pq:q^2p2:2pq:q2 for p=0.80p=0.80p=0.80. (correct answer)
  3. The population deviates from equilibrium because HHHHHH must equal 2pq2pq2pq.
  4. The population is in equilibrium only if HHH is recessive to hhh.
  5. The population deviates from equilibrium because heterozygotes should be the most common genotype.

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium in a population. From counts HH=640, Hh=320, hh=40 in 1,000 rabbits, frequencies are 0.64, 0.32, 0.04, yielding p(H)=0.80 and q(h)=0.20. Expected HWE frequencies are p²=0.64, 2pq=0.32, and q²=0.04, matching observed values exactly. Conditions like large size, isolation, no selection, and no mutation support equilibrium. A tempting distractor is A, claiming deviation with p=0.64 and q=0.04, a misconception from equating homozygote frequencies directly to allele frequencies. To confirm HWE, compute allele frequencies from all genotypes and verify if they generate the observed ratios.

Question 19

A bacterial population in a closed flask was measured every hour. The population increased from 1,000 cells at hour 0 to 2,000 at hour 1, 4,000 at hour 2, and 8,000 at hour 3. At hour 4 it reached 8,200, and at hour 5 it decreased to 7,900. Dissolved oxygen dropped sharply after hour 3. Which explanation best accounts for the change after hour 3?

  1. Oxygen became limiting, increasing mortality and reducing net population growth (correct answer)
  2. The population entered continued exponential growth because doubling persisted
  3. Immigration into the flask stopped, so population size could no longer increase
  4. A new beneficial trait spread rapidly, causing the growth rate to level off
  5. Predators were introduced at hour 4, causing an immediate population crash

Explanation: This question assesses the skill of analyzing population ecology trends by explaining shifts in growth phases due to limiting factors. The bacterial population doubled consistently from 1,000 to 8,000 cells in the first three hours, showing exponential growth under unlimited conditions, but after hour 3, growth nearly halted and then declined as dissolved oxygen dropped sharply. This change indicates oxygen became a density-dependent limiting resource, increasing mortality and reducing net population growth. By hour 5, the population decreased to 7,900, consistent with resource depletion overriding reproduction. A tempting distractor is choice B, suggesting continued exponential growth because doubling persisted, but this ignores the post-hour 3 slowdown and decline, misconstruing temporary patterns as indefinite. When evaluating growth curves, identify points where rates change and link them to environmental data for causal explanations.

Question 20

A cell in an isotonic solution shows no net change in volume. When a drug blocks aquaporins, water movement across the membrane slows, but the final cell volume remains unchanged. ATP depletion does not alter the direction of water movement. Which mechanism best explains water movement across the membrane in both conditions?

  1. Osmosis, a passive process that can occur through bilayer or aquaporins (correct answer)
  2. Primary active transport pumping water to maintain equal cell volume
  3. Secondary active transport coupling water movement to Na+^++ transport
  4. Facilitated diffusion of solutes that pulls water against its gradient
  5. Endocytosis of extracellular water to balance osmotic conditions

Explanation: This question assesses the skill of analyzing membrane transport mechanisms. The correct answer is osmosis, a passive process that can occur through bilayer or aquaporins because water moves down its gradient to maintain equilibrium in isotonic conditions, with no net volume change. Aquaporin blocking slows but doesn't alter final volume, and ATP depletion doesn't change direction, confirming passivity. The process balances osmotic pressures without energy. A tempting distractor is primary active transport pumping water, but this is wrong due to the misconception that equilibrium requires active intervention; osmosis is passive. To analyze similar problems, always determine if movement is down a gradient (passive) or against (active) and check for energy dependence.

Question 21

A researcher compares permeability of a pure phospholipid bilayer to different solutes. The bilayer’s interior is hydrophobic, so solutes that are nonpolar dissolve into it more easily than polar solutes. Charge greatly reduces permeability because ions remain stabilized by water and are unfavorable in the hydrophobic core. Compare two small solutes: hydrogen sulfide (H2S), which is relatively nonpolar, and sodium ion (Na+), which is charged. No proteins are present.

Which solute would most likely cross the bilayer more readily by simple diffusion?

  1. Sodium ion (Na+), because it is smaller than H2S
  2. Hydrogen sulfide (H2S), because it is uncharged and more nonpolar (correct answer)
  3. Sodium ion (Na+), because charge increases attraction to the membrane
  4. Both, because small solutes cross regardless of polarity
  5. Neither, because all diffusion across membranes requires ATP

Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Hydrogen sulfide (H2S) crosses the bilayer more readily than sodium ion (Na+) because it is uncharged and relatively nonpolar, allowing dissolution in the hydrophobic core, while Na+ is charged and stabilized by water, facing a high energy barrier. Both are small, but charge is the dominant barrier without proteins for facilitated transport. Simple diffusion in pure bilayers strongly disfavors ions. A tempting distractor is Na+ because it is smaller (choice A), but this stems from the misconception that size trumps charge, whereas charge greatly reduces permeability. To analyze similar problems, always rank uncharged, nonpolar molecules higher than ions for permeability in phospholipid bilayers.

Question 22

In a herd of 200 cattle, ranchers allow only the 20 bulls with the greatest muscle mass to sire calves each year for 10 years; other bulls are excluded from breeding. Calves are raised similarly regardless of parent. Muscle mass varies continuously, and the trait is heritable. After 10 years, the average muscle mass in the calf population is higher than at the start. Which outcome is most likely in the population’s gene pool after this artificial selection?

  1. Alleles associated with greater muscle mass increase in frequency across generations of calves. (correct answer)
  2. Each calf develops greater muscle mass because ranchers chose muscular bulls as parents.
  3. New alleles for greater muscle mass arise because selection creates beneficial mutations.
  4. Allele frequencies stay constant because only the environment determines muscle mass.
  5. The cattle population becomes genetically identical because selection removes all variation.

Explanation: This question assesses the skill of analyzing artificial selection by examining how human-directed breeding changes trait distributions in populations over generations. The correct answer is choice A because ranchers selectively breed only the most muscular bulls, leading to calves inheriting alleles that promote greater muscle mass, and over 10 years, these alleles increase in frequency as less muscular individuals are excluded from reproduction, aligning with the AP Biology concept that artificial selection alters allele frequencies by favoring heritable traits. The stimulus specifies that muscle mass is heritable and varies continuously, indicating polygenic inheritance, where nonrandom mating increases the prevalence of favorable allele combinations in the gene pool. Since calves are raised similarly, environmental factors are controlled, confirming that the shift in average muscle mass results from genetic changes due to selection pressure. A tempting distractor is choice C, which is incorrect due to the misconception of teleology, suggesting selection creates mutations to meet needs rather than acting on existing variation. To approach similar questions, identify how selection acts on pre-existing genetic variation to shift allele frequencies without introducing new mutations or individual adaptations.

Question 23

In a fish population, a gene has alleles A and a that influence tolerance to low oxygen. When a lake becomes eutrophic, low-oxygen events become frequent. Fish with genotype aa have the highest survival and produce more offspring than AA fish. After multiple generations, allele a frequency increases. Which statement best explains the role of variation in this change?

  1. Allele a increased because low oxygen caused new a alleles to appear in offspring at high rates.
  2. Allele a increased because differential survival and reproduction among genotypes shifted allele frequencies. (correct answer)
  3. Allele a increased because individual fish altered their genotype when oxygen levels dropped.
  4. Allele a increased because all genotypes had equal fitness and allele frequencies changed directionally.
  5. Allele a increased because selection always increases heterozygote frequency, regardless of environment.

Explanation: This question examines how genetic variation at a single locus enables evolutionary responses to environmental stress. The correct answer (B) identifies that fish with different genotypes (AA, Aa, aa) had different survival rates during low-oxygen events, with aa fish having the highest survival and reproductive success. This genotype-based differential survival and reproduction shifted allele frequencies in the population, increasing the frequency of allele a over multiple generations. Answer A incorrectly suggests that low oxygen caused new mutations to allele a at high rates, confusing the action of selection (which changes frequencies of existing alleles) with mutation (which creates new alleles rarely). To analyze allele frequency changes correctly, focus on how environmental conditions create fitness differences among existing genotypes rather than inducing specific mutations.

Question 24

Two human cell lines are observed under a microscope during migration across a culture dish. In line X, actin filaments rapidly assemble at the leading edge, forming broad lamellipodia, and the cell changes shape frequently. In line Y, actin filaments are present but are bundled into stable stress fibers, and the leading edge forms fewer protrusions. When both lines are treated with a compound that prevents actin polymerization, neither line forms lamellipodia and migration slows markedly. Which feature best explains the role of actin in cell migration?

  1. Actin polymerization at the membrane generates protrusive force that extends the leading edge (correct answer)
  2. Actin filaments replicate DNA at the leading edge to supply nuclei for new protrusions
  3. Actin bundles form the cell wall that anchors the cell as it pushes forward
  4. Actin depolymerization increases ATP levels, providing energy to move the entire cell
  5. Actin filaments open membrane channels so water influx inflates lamellipodia

Explanation: This question assesses the analysis of cell structure and function, specifically the role of actin filaments in cell motility and shape changes. The correct answer is A because the stimulus describes rapid actin assembly at the leading edge forming lamellipodia in line X, enabling frequent shape changes and migration, which aligns with AP Biology concepts of actin polymerization generating protrusive forces for cell extension. Treatment preventing polymerization halts lamellipodia formation and slows migration in both lines, confirming actin's dynamic role in pushing the membrane forward. In line Y, stable actin bundles limit protrusions, further illustrating how polymerization state influences motility. A tempting distractor is E, which represents a structure-function confusion by attributing water influx to actin instead of its mechanical force generation. To tackle similar questions, focus on cytoskeletal dynamics and test predictions by considering effects of inhibitors on movement.

Question 25

A researcher compares two cube-shaped cells with identical membrane protein density. Cell M has side length 4 units; Cell N has side length 8 units. A waste product leaves the cell by facilitated diffusion through membrane transporters, and the outside concentration is kept near zero. Waste is produced at the same rate per unit cytoplasmic volume in both cells. Which explanation best accounts for why Cell N is more likely to accumulate waste internally?​​​

  1. Cell N has a lower surface area–to–volume ratio, reducing transporter area relative to waste-producing cytoplasmic volume. (correct answer)
  2. Cell N has a larger surface area, so waste export per unit volume is greater than in Cell M.
  3. Cell N has a larger volume, which decreases the amount of waste produced per unit volume compared with Cell M.
  4. Cell N has more transporters total, which lowers internal waste concentration by increasing export beyond production.
  5. Cell N has thicker cytoplasm, which increases the concentration gradient across the membrane and speeds waste export.

Explanation: This question tests understanding of how surface area-to-volume ratio affects waste export efficiency. Cell N, with twice the side length of Cell M, has half the surface area-to-volume ratio (for cubes, SA/V = 6/s where s is side length). With identical transporter density per membrane area, Cell N has proportionally fewer transporters available to export waste from each unit of cytoplasmic volume. Since waste production per unit volume is the same in both cells, Cell N's reduced export capacity per volume makes it more likely to accumulate waste internally. Answer D incorrectly suggests that more total transporters prevent accumulation, ignoring that the larger volume produces proportionally more total waste. The critical concept is that cells must balance production rates with export capacity, and larger cells have less export capacity per unit of producing volume.