AP Biology
Practice Test 54 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.
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A transporter moves Cl− from the cytosol to the extracellular space. Cytosolic Cl− is 5 mM and extracellular Cl− is 120 mM, yet Cl− continues to be exported. Export stops when ATP is depleted. The transporter is not affected by disrupting Na+ gradients. Which mechanism best explains Cl− export?
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A transporter moves Cl− from the cytosol to the extracellular space. Cytosolic Cl− is 5 mM and extracellular Cl− is 120 mM, yet Cl− continues to be exported. Export stops when ATP is depleted. The transporter is not affected by disrupting Na+ gradients. Which mechanism best explains Cl− export?
Explanation: This question assesses the skill of analyzing membrane transport mechanisms. The correct answer is primary active transport exporting Cl- against gradient using ATP because Cl- moves from low cytosolic to high extracellular concentration, directly powered by ATP hydrolysis. Export stops with ATP depletion, confirming direct energy dependence, and Na+ gradient disruption has no effect. The transporter maintains the gradient actively. A tempting distractor is facilitated diffusion through a channel, but this is incorrect due to the misconception that against-gradient movement can be passive; active transport is needed. To analyze similar problems, always determine if movement is down a gradient (passive) or against (active) and check for energy dependence.
A protein contains several proline residues within an α-helix region. Proline’s rigid ring structure restricts rotation of the polypeptide backbone and can introduce kinks that change secondary structure. A new mutation inserts an additional proline into the middle of a helix that helps position residues forming the active site. The amino acid sequence is otherwise unchanged. Enzyme activity decreases markedly. Which feature best explains the decrease in activity?
Explanation: This question tests analysis of protein structure-function by examining how proline insertion affects secondary structure and enzyme activity. The correct answer A identifies that proline's rigid ring structure and inability to form standard α-helix hydrogen bonds disrupts helix geometry, introducing kinks that shift the positions of residues involved in forming the active site. The stimulus establishes that the helix helps position active site residues and that proline restricts backbone rotation, so inserting proline into the helix alters the precise three-dimensional arrangement of catalytic residues, reducing their ability to bind and process substrate effectively. Option B incorrectly suggests proline forms phosphodiester bonds (a bond type error), confusing protein and nucleic acid chemistry - proline is an amino acid that forms peptide bonds, not phosphodiester bonds found in DNA/RNA. The key strategy is to understand how specific amino acids (proline) have unique structural constraints that affect secondary structure formation and propagate to alter functional sites.
A signaling molecule Q is produced by gland G and circulates throughout the body. Target cells in tissue R respond, but cells in tissue S do not. Both tissues receive equal blood flow, and Q concentration is the same in capillaries supplying R and S. Which of the following best explains why only tissue R responds?
Which of the following best explains the specificity of Q’s effect?
Explanation: This question assesses understanding of cell communication via signal transduction pathways. Cells in tissue R express the receptor for signaling molecule Q, while tissue S does not, explaining why only R responds despite equal Q concentrations in capillaries supplying both. Q circulates body-wide from gland G, indicating endocrine signaling, but response is tissue-specific due to receptor expression. Equal blood flow ensures Q delivery is not the issue, so receptor presence determines specificity. A tempting distractor is choice C, suggesting proximity enables better diffusion, but this reflects the misconception that endocrine signals rely on diffusion rather than circulation, ignoring bloodstream delivery. To approach similar questions, focus on receptor distribution to explain tissue-specific responses in endocrine signaling.
A gene encodes a membrane protein with a signal peptide at the N-terminus. A mutation changes the start codon from ATG to ACG in the coding strand, with the rest of the coding sequence unchanged. Assume no alternative start codon is used. Which outcome is most likely in the cell?
Explanation: This question examines the critical role of the start codon in translation initiation. ATG (coding for methionine) is the universal start codon recognized by the ribosome-tRNA complex to begin translation. When mutated to ACG, the ribosome cannot recognize this as a start signal, preventing translation initiation at the normal position. Without a functional start codon, the ribosome either fails to initiate translation entirely or initiates at a downstream ATG (if present), producing a different or nonfunctional protein. A common error is thinking any codon can serve as a start codon (choice B), but translation initiation requires specific recognition of ATG. When analyzing start codon mutations, consider that translation may fail completely or initiate at an alternative downstream ATG.
In a large population of frogs, allele frequencies at a locus are p(F)=0.80 and q(f)=0.20. The population is isolated and mates randomly. A parasite reduces reproductive success of ff individuals relative to FF and Ff, while no mutation is detected. Which outcome is most consistent with this violation of Hardy-Weinberg conditions over multiple generations?
Explanation: This question assesses the skill of predicting outcomes from violations of Hardy-Weinberg equilibrium. With selection against ff homozygotes reducing their reproductive success, the frequency of allele f (q=0.20) is expected to decrease over generations. Since ff has lower fitness, f alleles are removed more often, shifting frequencies despite random mating and isolation. This directional selection violates HWE, leading to declining q. A tempting distractor is D, suggesting f increases because ff is recessive, a misconception ignoring that selection acts on phenotypes regardless of dominance. When selection is present, model allele frequency changes based on relative fitness of genotypes.
A signaling ligand G binds to a receptor on the plasma membrane and quickly increases cytosolic cGMP. Cells pretreated with a drug that locks heterotrimeric G proteins in the GDP-bound state show no cGMP increase after ligand addition, even though ligand binding is unchanged. When a membrane-permeable cGMP analog is added, the intracellular response occurs despite the drug. Which of the following best explains the early transduction step normally required for cGMP production?
Explanation: This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is B because locking G proteins in GDP-bound state prevents cGMP increase despite ligand binding, indicating that receptor activation promotes GTP exchange on G proteins, which then activate guanylyl cyclase to produce cGMP. The permeable cGMP analog restoring the response shows the block is upstream of cGMP production. This fits basic signaling principles of GPCRs where ligand binding activates G proteins by facilitating GDP-GTP exchange, enabling effector activation. A tempting distractor is C, which is wrong due to the misconception that rapid second messenger increases involve transcription, whereas cGMP production occurs enzymatically within seconds. For signal transduction questions, use drugs affecting G protein states to identify their role in linking receptors to second messenger-generating enzymes.
In smooth muscle cells, ligand L binds receptor tyrosine kinase (RTK) and causes receptor dimerization and autophosphorylation on cytosolic tyrosines. An adaptor protein binds the phosphotyrosines and recruits a Ras-GEF, which converts Ras-GDP to Ras-GTP. Ras-GTP activates a kinase cascade: Raf phosphorylates MEK, and MEK phosphorylates ERK. Activated ERK phosphorylates a cytosolic enzyme, increasing its catalytic rate within minutes. A mutant cell line expresses Ras that can bind GDP but cannot exchange GDP for GTP; all other proteins are normal and L binding to RTK is unchanged. Which outcome is most likely when mutant cells are treated with ligand L?
Explanation: This question requires analyzing how a mutation in a signal transduction pathway affects downstream signaling events. The RTK pathway normally proceeds: ligand binding → RTK dimerization/autophosphorylation → adaptor protein recruitment → Ras-GEF activation → Ras-GTP formation → Raf/MEK/ERK cascade → enzyme phosphorylation. The mutant Ras cannot exchange GDP for GTP, blocking its activation despite normal RTK function upstream. Without active Ras-GTP, the Raf/MEK/ERK cascade cannot be initiated, preventing ERK activation and subsequent enzyme phosphorylation. Choice A incorrectly assumes ERK activation depends only on RTK autophosphorylation, ignoring the requirement for Ras-GTP in activating the downstream kinase cascade. To analyze pathway disruptions, identify where the block occurs and predict effects on all downstream components.
In an experiment, intact chloroplasts are illuminated while the pH of the stroma is artificially lowered to match the pH of the thylakoid lumen, reducing the H+ gradient across the thylakoid membrane. Electron flow from water to NADP+ continues, and oxygen is produced. However, the reduced gradient decreases the driving force for ATP synthase. CO2 concentration is kept constant and high. Which outcome is most likely for the ratio of NADPH to ATP produced during illumination compared with controls?
Explanation: This question tests understanding of how the proton gradient affects ATP versus NADPH production. When the stromal pH is artificially lowered to match the lumen pH, the proton gradient across the thylakoid membrane is reduced or eliminated. ATP synthase requires this gradient to drive ATP production, so ATP synthesis decreases significantly. However, NADPH production depends on electron flow to NADP+, not on the proton gradient, so it continues relatively normally as long as electrons are flowing. This causes the NADPH:ATP ratio to increase - more NADPH is produced relative to ATP compared to normal conditions. Choice B incorrectly claims NADPH formation requires ATP synthase, but NADPH is produced by NADP+ reductase using electrons, not by ATP synthase. To analyze changes in ATP:NADPH ratios, consider what drives each process: ATP needs the proton gradient, while NADPH needs electron flow.
An enzyme-catalyzed reaction S → P is measured at saturating substrate (100 mM S). At 0.5 µM enzyme, the initial rate is 30 µM/min; at 1.0 µM enzyme, the initial rate is 60 µM/min; at 2.0 µM enzyme, the initial rate is 120 µM/min. All other conditions are constant. Which explanation best accounts for the effect of changing enzyme concentration on reaction rate?
Explanation: This question requires analysis of enzyme function to understand how enzyme concentration affects reaction rate. The data shows a direct linear relationship: doubling enzyme concentration doubles the reaction rate (0.5 µM enzyme gives 30 µM/min, 1.0 µM gives 60 µM/min, 2.0 µM gives 120 µM/min), which occurs because more enzyme molecules provide more active sites for simultaneous catalysis. Since substrate is saturating (100 mM), each enzyme molecule can work at maximum capacity, and the total rate is simply proportional to the number of enzyme molecules present. Choice D incorrectly suggests that enzyme molecules compete for substrate, which reflects a teleological misconception—enzymes don't "compete" but rather each independently binds and processes substrate molecules. The key strategy is to recognize that at saturating substrate concentrations, reaction rate is directly proportional to enzyme concentration because rate depends on the total number of active sites available.
In a developing embryo, scientists isolate two differentiated cell types that share the same genome. In Cell Type 1, a specific enhancer region is heavily methylated and the nearby gene’s mRNA is low. In Cell Type 2, the same enhancer is unmethylated and the gene’s mRNA is high. Which explanation best accounts for the difference in mRNA levels?
Explanation: This question tests gene expression and cell specialization through epigenetic mechanisms. The correct answer (C) explains that differential DNA methylation alters transcription machinery access to regulatory DNA—heavy methylation in Cell Type 1 blocks transcription factor binding, reducing mRNA levels, while unmethylated DNA in Cell Type 2 allows active transcription. Option A incorrectly suggests Cell Type 2 has extra gene copies, contradicting their shared genome. Option B wrongly claims methylation changes codons affecting translation, but methylation doesn't alter DNA sequence. The key concept is that epigenetic modifications like methylation regulate gene expression without changing DNA sequence, creating stable cell-type differences.
In cultured liver cells, adding peptide hormone P causes a rapid increase in cytosolic Ca2+ within 10 seconds. P cannot cross the plasma membrane. When cells are treated with a drug that blocks GTP binding to G proteins, adding P no longer increases cytosolic Ca2+, but P still binds to a specific protein on the cell surface. Which of the following best explains how the signal is initiated in untreated cells?
Explanation: This question assesses the ability to analyze signal transduction pathways, emphasizing G protein-coupled receptor mechanisms in rapid calcium signaling. The correct answer is B because peptide hormone P, unable to cross the membrane, binds a surface receptor and requires G protein GTP binding for Ca2+ release, as the inhibitor blocks the response while binding persists. Basic signaling principles indicate that GPCRs activate G proteins to stimulate phospholipase C, producing IP3 that releases Ca2+ from the endoplasmic reticulum, aligning with the rapid 10-second response and G protein dependency. The surface binding and impermeability of P confirm a membrane receptor-initiated cascade rather than direct intracellular action. A tempting distractor is A, which is wrong because it suggests P diffuses through the membrane, ignoring its hydrophilic nature and the evidence of receptor binding, a misconception about ligand permeability. A transferable strategy for signal transduction questions is to evaluate ligand properties like hydrophobicity and inhibitor effects to infer receptor type and pathway involvement.
In a lab, a student places a drop of water on clean glass and observes it spreads into a thin film rather than forming a tall bead. Water molecules are polar because oxygen is more electronegative than hydrogen, creating partial negative charge near oxygen and partial positive charge near hydrogens. Adjacent water molecules form hydrogen bonds when the partially positive hydrogen of one molecule is attracted to the partially negative oxygen of another. Water can also hydrogen-bond to polar groups on the glass surface. These intermolecular attractions influence how strongly water molecules stick to each other and to other materials. Which feature best explains why the water drop spreads on glass? Which feature best explains the spreading of water into a thin film on glass?
Explanation: This question assesses the analysis of water structure and hydrogen bonding. The correct answer, choice C, explains that adhesive hydrogen bonds between water and polar sites on glass cause the drop to spread by competing with the cohesive forces among water molecules. The stimulus describes water's polarity due to oxygen's electronegativity, enabling hydrogen bonds with polar groups on glass, which allows adhesion to overcome some cohesion and flatten the drop into a thin film. This aligns with the AP Biology concept that water's hydrogen bonding leads to properties like adhesion and cohesion, influencing how liquids interact with surfaces. A tempting distractor is choice D, which is incorrect due to a structure-function confusion by claiming water is nonpolar and molecules avoid each other, ignoring water's actual polarity and attractive hydrogen bonds. To approach similar questions, identify how polarity and hydrogen bonding dictate interactions between water and other materials by balancing adhesive and cohesive forces.
A membrane contains a cotransporter that moves Na+ and an amino acid into the cell in the same direction. Amino acid uptake occurs even when its intracellular concentration is higher than extracellular. Uptake stops if the Na+ gradient is eliminated, even though ATP is still present. ATP is not used by the cotransporter itself, but another membrane protein uses ATP to maintain low intracellular Na+. Which process best explains amino acid uptake?
Explanation: This question assesses the skill of analyzing membrane transport mechanisms. Secondary active transport is correct because the cotransporter uses the Na+ electrochemical gradient (maintained by ATP elsewhere) to drive amino acid uptake against its gradient, without directly using ATP. Uptake stops when the Na+ gradient is eliminated, showing dependence on this stored energy rather than direct hydrolysis. Amino acid accumulation above extracellular levels confirms active transport powered indirectly. A tempting distractor is primary active transport, incorrect due to the misconception that all uphill transport directly uses ATP, overlooking coupled ion gradients. To analyze transport, always evaluate if movement is down a gradient (passive) or against it (active) and check for energy requirements.
A cell is placed in a solution with 9 mM molecule U outside and 3 mM inside. The membrane has U carriers that bind and release U without ATP hydrolysis. If the number of carriers is reduced by half, the initial rate of U uptake decreases. Which explanation best accounts for the reduced uptake rate?
Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as facilitated diffusion relies on carrier proteins to transport molecule U down its concentration gradient from 9 mM outside to 3 mM inside without requiring ATP. Reducing the number of carriers by half decreases the overall transport capacity, leading to a slower initial uptake rate because fewer proteins are available to bind and shuttle U across the membrane. The process is passive and driven solely by the concentration gradient, with carriers facilitating the movement without energy input. A tempting distractor is B, which incorrectly assumes ATP is involved, reflecting the misconception that all protein-mediated transport requires energy like active transport. To distinguish facilitated diffusion from other transport types, always check if movement is down the gradient and ATP-independent.
A researcher measures ATP levels in muscle cells before and after adding a compound that specifically inhibits the Na+/K+ ATPase. Within minutes, total cellular ATP increases slightly while the Na+ gradient across the plasma membrane decreases. No changes occur in oxygen availability or substrate supply, and mitochondria remain functional. The inhibitor does not affect ion channels directly. This scenario highlights how ATP hydrolysis can be coupled to endergonic transport and how blocking a major ATP-consuming process changes ATP abundance and gradient maintenance at the membrane.
Explanation: This question assesses the skill of analyzing cellular energy transformations, specifically how inhibiting ATP-consuming pumps alters cellular ATP levels and ion gradients. The correct answer is A because the Na+/K+ ATPase uses ATP hydrolysis to pump ions against their gradients, so inhibiting it reduces ATP consumption, leading to a slight increase in ATP while the Na+ gradient dissipates. Evidence from the scenario shows ATP rises shortly after inhibition without changes in oxygen or substrates, indicating the pump is a major ATP sink, and blocking it conserves ATP normally spent on active transport. This follows energy principles where endergonic processes like ion pumping are directly coupled to ATP hydrolysis, and inhibition shifts the balance toward ATP accumulation. A tempting distractor is B, which is incorrect because it suggests direct phosphorylation by the Na+ gradient without enzymes, stemming from the misconception that gradients alone can synthesize ATP without coupling mechanisms. A transferable strategy for cellular energy questions is to identify major ATP-consuming processes and predict how their inhibition affects overall energy balance and coupled functions.
A student examines red blood cells (RBCs) and typical epithelial cells. RBCs lack a nucleus and most membrane-bound organelles, while epithelial cells contain a nucleus, endoplasmic reticulum, and mitochondria. When both cell types are placed in a solution containing a fluorescent lipid that inserts into membranes, epithelial cells show labeled internal membranes in addition to the plasma membrane, but RBCs show labeling only at the cell surface. Which feature best explains the difference in fluorescence patterns?
Explanation: This question assesses the analysis of cell structure and function, specifically differences in membrane-bound organelles between cell types. The correct answer is A because the stimulus indicates epithelial cells show internal fluorescence besides the plasma membrane, while RBCs label only at the surface, reflecting AP Biology principles where epithelial cells possess organelles like ER and mitochondria providing additional bilayers for lipid insertion. RBCs lack these organelles, limiting labeling to the plasma membrane. The fluorescent lipid inserts into all available membranes, highlighting the presence of internal structures. A tempting distractor is B, which embodies a structure-function confusion by incorrectly attributing a cell wall to RBCs, which are animal cells without walls. For these questions, compare organelle inventories across cell types and predict labeling patterns based on membrane availability.
A bacterial population uses a secreted molecule Q to coordinate behavior. Q binds to a receptor protein embedded in the plasma membrane. Within seconds of adding Q, a cytosolic response protein becomes phosphorylated. If the receptor’s extracellular binding site is blocked by an antibody, phosphorylation does not occur. However, if purified active kinase (the enzyme that phosphorylates the response protein) is introduced into permeabilized cells, phosphorylation occurs without Q. Which of the following best explains how signal Q is received and initiates the intracellular response?
Explanation: This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is B because Q binds a membrane receptor and initiates rapid phosphorylation, but blocking the extracellular site prevents it, indicating receptor activation is required, while introducing active kinase bypasses Q, showing the receptor triggers a kinase or cascade. This is consistent with bacterial two-component systems where ligand binding activates a receptor histidine kinase, leading to phosphorylation of response regulators. Basic signaling principles emphasize that extracellular signals like Q, which coordinate behavior, act through membrane receptors to transduce signals intracellularly without the ligand entering. A tempting distractor is C, which is wrong due to the misconception that rapid responses involve transcription, whereas phosphorylation occurs within seconds and does not require new gene expression. For signal transduction questions, assess the speed of the response and use bypass experiments to pinpoint where the signal initiates intracellular changes like phosphorylation.
In mice, black fur (B) is completely dominant to brown fur (b). Two black mice are crossed, and genetic testing shows each parent is heterozygous (Bb). Each parent produces gametes carrying either B or b with equal probability. The offspring fur color depends on whether the genotype includes at least one B allele. Which outcome is most likely for the offspring phenotypes?
Explanation: This problem requires analyzing a cross between two heterozygous individuals using Mendelian genetics. When crossing Bb × Bb, each parent produces B and b gametes in equal proportions (50% each). The Punnett square yields: BB (25%), Bb (50%), and bb (25%), following the classic 1:2:1 genotypic ratio. Since black fur (B) is completely dominant, both BB and Bb mice will have black fur, resulting in 75% black offspring (3/4). The remaining 25% will be brown (bb). Students often mistakenly think heterozygous crosses produce 50% of each phenotype, but this only occurs in testcrosses. Remember that for monohybrid crosses between heterozygotes, the phenotypic ratio is always 3:1 dominant to recessive.
A researcher isolates nuclear RNA from eukaryotic cells and treats it with an enzyme that specifically removes the 5′ cap but does not affect the RNA backbone elsewhere. The researcher then repeats the treatment on cytosolic RNA from the same cells. Both samples contain transcripts produced by RNA polymerase II. The 5′ cap is normally added co-transcriptionally after the nascent RNA reaches about 20–30 nucleotides in length. Which explanation best accounts for why most cytosolic RNA molecules resist this decapping treatment compared with nuclear RNA molecules?
Explanation: This question tests understanding of transcription and RNA processing, specifically the protection of mature mRNAs in the cytoplasm. The 5' cap is added co-transcriptionally in the nucleus to all RNA polymerase II transcripts, so both nuclear and cytoplasmic RNAs initially have caps. However, cytoplasmic mRNAs that are being translated have cap-binding proteins (like eIF4E) bound to their 5' caps, which would physically block access of the decapping enzyme to its substrate. Nuclear RNAs, especially those still undergoing processing, have less stable cap-binding protein associations, making their caps more accessible to the decapping enzyme. Choice B incorrectly claims that capping occurs after nuclear export, but capping actually happens co-transcriptionally when the RNA is only 20-30 nucleotides long. To analyze RNA modifications, consider not just when modifications are added but also what proteins interact with those modifications in different cellular compartments.
A plant population contains a rare mutation that causes some individuals to produce unreduced (2n) gametes. When 2n gametes fuse with normal n gametes, resulting offspring have 3n and are largely sterile. When two 2n gametes fuse, resulting offspring have 4n and can produce fertile 2n gametes with each other but not with the original n plants. Over several generations, the 4n plants increase in frequency in one area of the population. Which process most directly generated reproductive isolation between the 4n plants and the original plants?
Explanation: This question examines speciation through polyploidy, a common mechanism in plants. The mutation causing unreduced (2n) gametes leads to polyploid offspring - when two 2n gametes fuse, they create 4n individuals. These tetraploid (4n) plants can reproduce with each other but not with the original diploid (2n) plants due to chromosome number mismatch during meiosis. This creates instant reproductive isolation without geographic separation or gradual divergence. The 3n offspring from 2n × n crosses are largely sterile, further preventing gene flow. Choice E incorrectly suggests genetic drift involves intentional avoidance, but drift is random change in allele frequencies, not directed behavior. To recognize polyploid speciation, look for chromosome number changes that create immediate reproductive barriers through meiotic incompatibility.
A scientist studies transcription of a protein-coding gene in the nucleus. After transcription begins, the nascent RNA is rapidly modified at its 5′ end by addition of a 7-methylguanosine cap through an unusual 5′–5′ triphosphate linkage. In an experimental condition, the capping enzyme is inhibited, but RNA polymerase II still initiates and elongates normally. When RNA is extracted, uncapped transcripts are detected but are much less abundant than capped transcripts from control cells. Which explanation best accounts for the lower abundance of uncapped transcripts?
Explanation: This question tests understanding of transcription and RNA processing, specifically the protective role of the 5' cap. The correct answer is A because the 7-methylguanosine cap protects RNA from degradation by 5'→3' exonucleases in the nucleus, so uncapped transcripts are rapidly degraded. The cap is added co-transcriptionally through an unusual 5'-5' triphosphate linkage shortly after transcription initiation, making uncapped RNAs vulnerable. Answer B incorrectly claims capping must occur before transcription, but capping happens after initiation when the nascent RNA is about 20-30 nucleotides long—this reflects confusion about the temporal order of transcription and processing events. To understand RNA stability, remember that both the 5' cap and 3' poly(A) tail protect mRNA from exonuclease degradation.
A cell’s plasma membrane is the only site for exchange of a particular solute with the environment. Two cells have the same membrane permeability and are in the same solution. Cell U has volume V and surface area SA; Cell V has volume 8V and surface area 4SA (same shape). Both cells consume the solute at the same rate per unit volume. Cell V shows a lower internal solute concentration. Which explanation best accounts for Cell V’s lower concentration?
Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular transport efficiency. Cell V, with volume 8V and surface area 4SA, has a lower surface area-to-volume ratio (4SA/8V = 0.5 SA/V vs. original SA/V), reducing solute influx per unit volume. Since consumption is the same per volume, the diminished supply leads to lower internal concentrations in Cell V. The transport efficiency logic illustrates that maintaining shape scales SA slower than V, limiting exchange relative to demand. A tempting distractor is choice C, which assumes higher total SA guarantees higher concentration, ignoring the per-volume mismatch misconception. To approach similar problems, calculate ratios from given multiples and assess impacts on transport vs. consumption.
A student observes that a thin stream of water can momentarily bridge a small gap between two closely spaced glass surfaces. Water molecules are polar: oxygen has a partial negative charge and hydrogens have partial positive charges. Hydrogen bonds form between neighboring water molecules, creating cohesion, and water molecules can also form hydrogen bonds with polar groups on glass, producing adhesion. Together, these interactions can hold water in a continuous shape across a short distance, resisting separation. Which feature best explains the ability of water to bridge the gap between the glass surfaces?
Explanation: This question assesses the analysis of water structure and hydrogen bonding in cohesion and adhesion phenomena. The correct answer is choice A because the stimulus explains that water's polarity allows hydrogen bonds for cohesion among water molecules and adhesion to polar glass, combining to maintain a continuous bridge across the gap by resisting disruptive forces. This relates to AP Biology concepts like capillary action, where hydrogen bonding supports water's ability to form stable structures in narrow spaces. Additionally, the interplay of cohesion and adhesion enables the water to span short distances without breaking. A tempting distractor is choice C, which is incorrect due to a level-of-organization error by misapplying hydrophobic interactions to polar water instead of nonpolar substances. To approach similar questions, distinguish between cohesive and adhesive forces enabled by hydrogen bonding in different surfaces.
In a lab, a cell type shows abundant rough ER membranes studded with ribosomes and a large Golgi apparatus. When these cells are treated with a drug that prevents vesicles from fusing with the Golgi, newly made proteins accumulate in small transport vesicles near the ER, and very little protein appears outside the cell. Which feature best explains why secretion decreases when Golgi fusion is blocked?
Explanation: This question assesses the skill of analyzing cell structure-function relationships. The abundant rough ER studded with ribosomes indicates active protein synthesis for secretion, and the large Golgi apparatus suggests its role in processing these proteins, as seen when the drug blocks vesicle fusion to the Golgi, causing proteins to accumulate in transport vesicles near the ER. This aligns with the endomembrane system's secretory pathway in AP Biology, where proteins synthesized in the rough ER are transported via vesicles to the Golgi for modification and sorting into secretory vesicles that fuse with the plasma membrane for exocytosis. Blocking fusion prevents this processing, reducing secretion as proteins cannot reach the cell exterior. A tempting distractor is choice B, which is incorrect due to structure-function confusion, as ribosomes are not located inside the nucleus for translating secreted proteins, and nuclear pores export mRNA, not proteins. To approach similar questions, map the sequence of organelles involved in a process and identify how disruptions affect the pathway.
Two closely related frog populations occupy the same forest. Males call from similar locations, and females approach calling males. Playback experiments show that females from population 1 approach only the call frequency pattern typical of population 1, while females from population 2 approach only the pattern typical of population 2. When researchers place males and females together in an enclosure, mating occurs mostly within populations even though breeding seasons overlap and hybrids, when produced, develop into viable and fertile adults. Genetic analyses show increasing divergence in allele frequencies over time. Which prezygotic barrier most directly maintains reproductive isolation?
Explanation: This question tests identification of behavioral isolation through mate choice based on acoustic signals. Female frogs show strong preferences for their own population's call frequency pattern, approaching only males with matching calls despite physical proximity and overlapping breeding seasons. This prezygotic barrier operates through female choice before mating attempts, with "mating occurs mostly within populations" due to call recognition preferences. The populations maintain genetic divergence through assortative mating based on acoustic communication. Choice B incorrectly suggests postzygotic isolation, but the passage states hybrids are "viable and fertile," ruling out hybrid inviability or sterility. To recognize behavioral isolation, look for differences in courtship signals (visual, acoustic, chemical) that influence mate choice and reduce interpopulation mating despite opportunity for contact.