AP Biology
Practice Test 94 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.
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A student examines meiosis in an organism with genotype Tt for a gene on chromosome 9. The student notes that genetic variation among gametes can arise even without crossing over. In one meiosis, the two homologs of chromosome 9 align at metaphase I with the T-bearing homolog facing one pole and the t-bearing homolog facing the opposite pole, then segregate in anaphase I. Which meiotic feature most directly accounts for variation in which allele ends up in a particular gamete?
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A student examines meiosis in an organism with genotype Tt for a gene on chromosome 9. The student notes that genetic variation among gametes can arise even without crossing over. In one meiosis, the two homologs of chromosome 9 align at metaphase I with the T-bearing homolog facing one pole and the t-bearing homolog facing the opposite pole, then segregate in anaphase I. Which meiotic feature most directly accounts for variation in which allele ends up in a particular gamete?
Explanation: This question tests understanding of how meiosis generates genetic diversity through chromosome segregation. Random orientation of homologous chromosomes at metaphase I (A) is correct because each homologous pair can orient with either homolog facing either pole, and this random orientation determines which allele (T or t) ends up in each gamete after segregation in anaphase I. The question emphasizes variation without crossing over, making independent orientation the key mechanism. Separation of sister chromatids at anaphase I (B) is incorrect because sister chromatids separate during anaphase II, not anaphase I—homologous chromosomes separate in anaphase I. To identify sources of variation, distinguish between processes that shuffle existing alleles (orientation, crossing over) versus those that would require creating new genetic information.
In pea plants, purple flowers (P) are completely dominant to white flowers (p). A student crosses two heterozygous purple-flowered plants (Pp × Pp). Each parent produces gametes with one allele per gamete, and fertilization is random. Assume the trait is controlled by a single gene with two alleles and that no other genes affect flower color in this cross. Which proportion of offspring is expected to have white flowers?
Explanation: This question tests the skill of analyzing Mendelian inheritance patterns. In this cross between two heterozygous purple-flowered pea plants (Pp × Pp), each parent produces gametes with equal proportions of P and p alleles due to segregation. The Punnett square shows that the possible offspring genotypes are PP, Pp, and pp in a 1:2:1 ratio, with white flowers only appearing in the homozygous recessive pp genotype. Thus, the proportion of white-flowered offspring is 1/4, as random fertilization combines the alleles accordingly. A tempting distractor is 3/4, which might arise from the misconception of confusing the dominant phenotype ratio with the recessive one. To solve similar inheritance questions, always construct a Punnett square to visualize allele combinations and calculate probabilities based on segregation.
An animal cell contains lysosomes with hydrolytic enzymes that function best at acidic pH. The lysosomal lumen is acidic compared with the cytosol. In an experiment, lysosomal proton pumps are inhibited, raising lysosomal pH while leaving enzyme quantity unchanged; macromolecule degradation inside lysosomes decreases. Which feature best explains how compartmentalization contributes to control of this process?
Explanation: This question assesses the skill of analyzing cell compartmentalization by examining how lysosomal membranes control macromolecule degradation. The lysosome maintains an acidic lumen that optimizes hydrolytic enzyme activity, as indicated in the stimulus, aligning with the AP Biology concept that compartmentalization isolates harsh conditions like low pH to enable breakdown without harming cytosolic components. Inhibiting proton pumps raises lysosomal pH, decreasing degradation despite unchanged enzyme levels, illustrating the membrane's role in sustaining the acidic environment. This prevents premature activation of enzymes outside the lysosome, protecting cellular structures. A tempting distractor is choice C, which involves a level-of-organization error by claiming lysosomes alter whole-cell pH, whereas they target only their internal space. To approach similar questions, consider how compartments confine extreme conditions to specific functions, avoiding widespread cellular disruption.
A cell-surface glycoprotein has a short carbohydrate chain ending in sialic acid, a monosaccharide that carries a negative charge at physiological pH. A mutant version lacks the terminal sialic acid but keeps the rest of the carbohydrate chain unchanged. In a salt solution, the mutant proteins cluster more closely on the membrane surface than the normal proteins. Which molecular-level change best explains the increased clustering? Which change best explains why mutant proteins cluster more closely on the membrane?
Explanation: This question assesses the analysis of carbohydrate structure–function. The removal of terminal sialic acid in the mutant, as described in the stimulus, eliminates the negative charge that normally causes electrostatic repulsion between glycoprotein carbohydrate chains on the membrane. This loss reduces repulsion, allowing closer clustering of proteins in salt solutions, consistent with AP Biology principles of cell surface charge influencing molecular spacing. The remaining chain structure stays polar but lacks the charge-based separation. A tempting distractor like choice C is incorrect because it claims increased branching from removal, which doesn't occur and would actually hinder clustering, representing a structure–function confusion. To address similar queries, consider how charged moieties affect intermolecular forces in biological contexts.
A peptide hormone H is added to two cell lines. Cell line 1 shows a rapid response when H is added, but cell line 2 does not. Radiolabeled H binds strongly to intact membranes from cell line 1 but not to membranes from cell line 2. Which of the following best explains the difference in responsiveness?
Which of the following best explains why only cell line 1 responds to H?
Explanation: This question assesses understanding of cell communication via signal transduction pathways. Cell line 2 lacks the specific membrane receptor for peptide hormone H, as radiolabeled H binds strongly to membranes from cell line 1 but not to those from cell line 2, explaining the difference in responsiveness. Cell line 1 shows a rapid response to H, indicating the presence of receptors initiating transduction. Since H is a peptide, it binds surface receptors, and the binding assay confirms receptor absence in cell line 2 prevents signaling. A tempting distractor is choice C, claiming H crosses the membrane to bind DNA directly, but this reflects the misconception that peptides are hydrophobic and can enter cells, whereas they require membrane receptors. To approach similar questions, use binding assay evidence to determine receptor presence and correlate it with response differences.
Two desert shrubland sites are sampled for arthropods. Site R includes 14 species, but 88% of individuals are a single ant species. Site S includes 10 species, and the most common species accounts for 20% of individuals; total individuals are equal. Which conclusion is best supported about which site likely has higher biodiversity overall?
Explanation: This question tests understanding of how extreme differences in evenness can override richness advantages in biodiversity assessment. Site S likely has higher biodiversity because its much higher evenness can result in greater overall diversity despite having fewer species. Site R has 14 species but extreme dominance with 88% being a single ant species, creating very low evenness. Site S has 10 species with the most common at only 20%, indicating relatively high evenness with no extreme dominance. The dramatic difference in evenness (88% vs 20% for top species) suggests Site S's community structure is more diverse overall. A common misconception (choice A) is that higher richness always means higher biodiversity, but extreme dominance can negate richness advantages. When comparing biodiversity, recognize that very low evenness (extreme dominance) can reduce overall diversity below that of communities with fewer but more evenly distributed species.
Biologists examine the amino acid sequence of the protein cytochrome c in four vertebrates. Compared with species M, species N differs by 2 amino acids, species O differs by 11, and species P differs by 14. The protein performs the same cellular role in all four species. Which inference is best supported by these data about common ancestry among the lineages?
Explanation: This question assesses the skill of inferring common ancestry from protein sequence differences in cytochrome c among vertebrates. The minimal differences between species M and N (2 amino acids) compared to M-O (11) and M-P (14) suggest M and N share a more recent common ancestor, as fewer changes have accumulated since their divergence. The conserved function of cytochrome c across species indicates that sequence similarities reflect inheritance from a shared lineage rather than convergence. This pattern allows inference of relative divergence times, with greater similarity pointing to closer relatedness. A tempting distractor is choice C, which assumes equal relatedness due to shared function, but this overlooks how sequence divergence quantifies ancestry and misapplies functional conservation to relatedness. To infer common ancestry from molecular data, compare sequence similarities quantitatively, remembering that fewer differences typically indicate more recent shared ancestry across lineages.
Researchers measure water movement across a membrane containing aquaporins. The extracellular solution has lower solute concentration than the cytosol, and water enters the cell rapidly. When aquaporins are chemically blocked, the rate of water entry decreases, but the direction of net water movement remains the same. Cellular ATP concentration is unchanged during the assay. Which explanation best accounts for the reduced water entry rate when aquaporins are blocked?
Explanation: This question tests understanding of facilitated diffusion of water through aquaporin channels. Water moves from the hypotonic solution (lower solute concentration outside) to the hypertonic cytosol (higher solute concentration inside) through aquaporins, following its concentration gradient without ATP input. When aquaporins are blocked, water entry slows but doesn't stop completely because some water can still cross the lipid bilayer directly, though much more slowly. Choice B incorrectly suggests ATP-driven pumping, contradicting the observation that ATP levels remain unchanged. Remember that facilitated diffusion increases the rate of movement down a gradient but doesn't change the direction of net transport.
A small bird species colonized two nearby islands separated by 30 km of open ocean. Banding data show that adults almost never move between islands. Over many generations, allele frequencies have diverged, and song recordings show consistent differences: males from island 1 sing at a higher frequency than males from island 2. In aviary trials, females prefer the song typical of their own island, reducing cross-island mating. Which process most directly initiated the divergence that later led to behavioral reproductive isolation?
Explanation: This question examines how allopatric speciation initiates through geographic isolation and subsequent behavioral divergence. The 30 km of ocean between islands prevents gene flow (adults almost never move between islands), creating the geographic separation necessary for populations to diverge independently through genetic drift and/or local selection. This divergence led to different song frequencies, which now act as a behavioral reproductive barrier since females prefer their own island's song type. Choice C incorrectly suggests Lamarckian inheritance where learned song changes are passed genetically, rather than recognizing that genetic divergence underlies the behavioral differences. To analyze speciation scenarios, first identify the initial isolating mechanism (here, geographic), then trace how it enables subsequent reproductive barriers to evolve.
Cells with internal solute concentration 0.28 M are placed into a solution containing 0.18 M of a nonpenetrating solute. The membrane is permeable to water but not to the solute. After 5 minutes, the cells are larger than at the start. Which statement best explains the direction of net water movement?
Explanation: This question tests tonicity and osmoregulation concepts by examining water movement when cells are placed in a hypotonic solution. The cells have an internal solute concentration of 0.28 M while the external solution contains only 0.18 M solute, making the external solution hypotonic relative to the cells. Water moves from regions of higher water potential (the hypotonic external solution with lower solute) to regions of lower water potential (inside the cells with higher solute). This net influx of water causes the cells to swell and become larger, as observed after 5 minutes. A common misconception is that water moves out because the external solution has lower solute (choice A), but this reverses the direction—water moves TOWARD higher solute concentration, not away from it. Remember that hypotonic solutions cause cells to gain water and swell because water moves into the cells.
A researcher compares two spherical cells made of the same membrane material. Cell 1 has a diameter of 10μm and cell 2 has a diameter of 30μm. Both rely on diffusion across the plasma membrane for oxygen uptake and produce oxygen-consuming reactions throughout the cytoplasm at similar rates per unit volume. When placed in the same oxygen concentration, cell 2 shows a lower internal oxygen concentration than cell 1 after several minutes. Which explanation best accounts for the difference in internal oxygen concentration between the cells?
Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular processes like oxygen diffusion. The correct answer is A because cell 2, being larger, has a lower surface area-to-volume ratio, meaning there is less membrane surface available for oxygen to diffuse in relative to the cytoplasmic volume that consumes it. As a result, oxygen entry cannot keep pace with consumption throughout the larger volume, leading to a lower internal concentration. This transport efficiency logic highlights why larger cells struggle to maintain adequate supply via diffusion alone. A tempting distractor is B, which incorrectly assumes that higher total surface area reduces oxygen entry, stemming from the misconception that absolute size overrides the importance of the ratio. To approach similar problems, always calculate the surface area-to-volume ratio to compare transport limitations in cells of different sizes.
A steroid-like signaling molecule S is experimentally tethered to a large bead, preventing membrane crossing. In normal cells, free S rapidly decreases cytosolic enzyme F activity by activating a membrane receptor M that stimulates PI3K, producing PIP3. PIP3 recruits kinase Akt to the membrane, where it becomes active and phosphorylates F, decreasing F activity. When bead-tethered S is added, F activity still decreases similarly to free S. Which conclusion is best supported by the data?
Explanation: This question assesses the skill of analyzing a signal transduction pathway. The bead-tethered S still decreases F activity, indicating it can activate the membrane receptor M without entering the cell, leading to PI3K activation, PIP₃ production, Akt recruitment, and F phosphorylation. Normally, free S crosses the membrane, but the tethered version's effect shows signaling occurs via a cell-surface receptor. This supports that S does not require intracellular access for its effect. Choice B is tempting but wrong because it assumes S must bind an intracellular receptor for direct phosphorylation, which is a misconception ignoring the evidence of surface-mediated signaling. A transferable strategy is to use modified ligands like tethered versions to distinguish between surface and intracellular receptor mechanisms in steroid-like signaling.
A membrane carrier transports amino acid Y. Outside concentration is 5 mM and inside is 5 mM at the start. When outside is increased to 50 mM, Y enters the cell, and the rate rises then plateaus as outside concentration continues to increase. ATP depletion does not change the initial influx rate, but a competitive inhibitor reduces influx at all concentrations. Which mechanism best explains Y entry?
Explanation: This question tests the skill of analyzing membrane transport mechanisms. The correct answer A is facilitated diffusion because amino acid Y moves from high external (50 mM) to equal or lower internal concentration, down its gradient, without energy as ATP depletion does not affect influx. The carrier protein binds Y, showing saturation as rate plateaus at high concentrations, and competitive inhibition confirms protein mediation. Equal starting concentrations ensure gradient-driven entry upon increase. A tempting distractor is C, primary active transport, but this is wrong because no ATP is required, arising from the misconception that saturation always indicates active processes. A transferable strategy is to analyze kinetics like saturation and energy needs to distinguish facilitated diffusion from active transport.
Genetically identical tadpoles were raised in aquaria containing either predator chemical cues (no actual predation) or no cues. Tadpoles exposed to cues developed deeper tail fins and different swimming behavior, but sequencing found no DNA differences between groups. Which explanation best accounts for the cue-induced phenotype differences?
Explanation: This question assesses understanding of environmental effects on phenotype, where external factors influence traits without altering the underlying DNA sequence. The correct answer, B, is right because predator cues can trigger developmental gene regulation, altering expression of growth-related genes to produce deeper tail fins and adaptive behaviors. In tadpoles, chemical signals activate signaling pathways that modify transcription during development, leading to phenotypic plasticity without DNA changes. Since the tadpoles are genetically identical and sequencing shows no differences, the variation is an environmental induction of gene expression. A tempting distractor is A, which is wrong because it assumes cues cause mutations, a misconception that confuses plasticity with mutagenesis. To approach similar questions, always check if phenotypic differences in identical genotypes under varying conditions point to gene expression changes rather than genetic mutations or evolution.
A fruit fly population feeds on two host plants growing intermixed in an orchard. Flies that develop on host plant 1 usually mate on that plant; flies from host plant 2 usually mate on that plant. Mark–recapture data show adults often fly between plants, but most matings occur on the natal host. Over time, allele frequencies differ between host-associated groups, and lab choice tests show preference for mating on the natal host plant. Which mechanism most directly promotes divergence in this system?
Explanation: This question analyzes sympatric speciation through ecological (habitat) isolation linked to host plant specialization. Despite living in the same orchard with no geographic barriers, flies show strong host fidelity—they preferentially mate on their natal host plant, creating assortative mating that reduces gene flow between host-associated groups. This ecological specialization allows genetic divergence even though adults can fly between plants, demonstrating how reproductive isolation can evolve without physical separation. Choice D incorrectly suggests individuals change their genotype to match their host, implying genetic changes within an individual's lifetime rather than population-level evolution. To identify sympatric speciation, look for reproductive barriers evolving within a continuous population through mechanisms like host specialization or polyploidy.
In a fish population, some individuals have a heritable allele that produces a larger tail fin, while others have smaller fins. A new predator is introduced that more easily catches fish with larger fins, and small-finned fish survive and reproduce more often. Tail-fin size variation persists among juveniles. Which outcome is most likely after many generations?
Explanation: This question tests understanding of natural selection, the process where heritable traits that enhance survival and reproduction become more common in a population over generations. The new predator catches large-finned fish more easily, allowing small-finned fish to survive and reproduce more often. Consequently, small-finned fish pass on more alleles for smaller fins, increasing their frequency in the population over generations. This occurs through differential survival and reproduction favoring the small-fin trait, with variation persisting among juveniles. A tempting distractor is choice B, which incorrectly assumes fish learn to avoid predators and pass this behavior genetically, confusing learned behaviors with heritable traits. For natural selection questions, always identify the selective pressure, the heritable trait, and how it affects reproductive success at the population level.
A student tests three lipid types in water: (1) triglycerides with three fatty acid tails, (2) phospholipids with two tails and a charged phosphate head, and (3) steroids with four fused hydrocarbon rings and few polar groups. After mixing, only one type consistently forms a stable, sheet-like boundary between water and water (a bilayer). Which property best explains why that lipid type forms bilayers?
Explanation: This question assesses the analysis of lipid structure–function relationships. Phospholipids form stable bilayers because they are amphipathic, with a hydrophilic phosphate head that interacts with water and two hydrophobic tails that avoid it, driving self-assembly into sheet-like structures to minimize hydrophobic exposure, as described in choice C. Triglycerides lack a significant polar head and are mostly hydrophobic, forming droplets, while steroids' fused rings do not promote bilayer formation without amphipathic balance. This behavior exemplifies the hydrophobic effect in lipid organization, a fundamental AP Biology mechanism. A tempting distractor is choice B, which incorrectly claims triglycerides are amphipathic and form bilayers, representing a structure–function confusion by overlooking the absence of a polar head in triglycerides. When approaching these questions, classify lipids by their amphipathic properties and predict their aqueous behavior based on polarity distribution.
In a lab, two single-stranded nucleic acids are mixed and allowed to hybridize. Strand X is DNA with sequence 5'-ATGCCG-3'. Strand Y is RNA with sequence 5'-CGGCAU-3'. In the mixture, complementary bases can form hydrogen bonds: A pairs with T in DNA and with U in RNA; C pairs with G. Hybridization requires antiparallel alignment of the strands. Which statement best describes the most likely duplex that forms in the mixture?
Explanation: This question assesses the analysis of nucleic acids as macromolecules. A duplex forms with strands X and Y aligned antiparallel, as per choice B, because their sequences are complementary, allowing A-U and C-G hydrogen bonding in a DNA-RNA hybrid. Antiparallel alignment is required for proper base pairing geometry in hybrids, similar to DNA-DNA duplexes. In AP Biology, nucleic acids can form stable hybrids if bases complement each other, with orientation ensuring optimal hydrogen bonding and backbone alignment. A tempting distractor is choice A, which is incorrect due to a level-of-organization error by assuming DNA and RNA bases cannot pair, ignoring their compatible hydrogen-bonding patterns. In similar scenarios, check sequence complementarity and recall that antiparallel orientation is key for hybridization across nucleic acid types.
A researcher scored four taxa for derived characters relative to an outgroup. Character 1 is present in A, B, C, and D. Character 2 is present only in C and D. Character 3 is present only in D. Character 4 is present only in A and B. Assume each derived character evolved once and was not lost. Which inference about most recent common ancestry is best supported by these data?
Explanation: This question assesses the skill of inferring phylogenetic relatedness using shared derived characters in cladistic analysis. The presence of character 2 only in C and D indicates they share a more recent common ancestor than either does with A, as this synapomorphy defines their clade relative to the outgroup. Character 1 is shared by all, representing an ancestral trait for the ingroup, while character 3 is unique to D and character 4 groups A and B separately. These patterns support a tree where C and D form a subclade branching after the split from A and B. A tempting distractor is D, suggesting equal relatedness due to shared character 1, which reflects a misconception of treating plesiomorphies as evidence of close relatedness, a teleological error assuming shared ancestry implies equal divergence. A transferable strategy for this question type is to identify the most exclusive shared derived characters to determine the most recent common ancestors, assuming parsimony in trait evolution.
In a culture of human somatic cells, one cell completes S phase normally (DNA replicated) and enters mitosis. A drug prevents microtubules from attaching to kinetochores, so chromosomes remain unattached at metaphase. DNA replication does not occur again during this arrest. After 2 hours, microscopy shows condensed chromosomes aligned near the cell center but no sister chromatid separation. The cell’s cyclin-dependent kinase activity remains characteristic of M phase. Which outcome is most likely if the spindle checkpoint is not satisfied?
Explanation: This question requires analysis of the cell cycle, focusing on the mitotic spindle checkpoint. The drug prevents microtubule attachment to kinetochores, leaving chromosomes unattached at metaphase, which activates the spindle assembly checkpoint to halt progression to anaphase. In AP Biology, this checkpoint ensures all chromosomes are properly aligned and attached before sister chromatid separation, preventing aneuploidy. Since the checkpoint is not satisfied, the cell remains arrested in metaphase with sister chromatids still joined at centromeres, as indicated by the condensed chromosomes aligned near the center and sustained M-phase cyclin-dependent kinase activity. A tempting distractor is choice E, which suggests sister chromatids separate despite unattached kinetochores, but this is incorrect due to a structure-function confusion where the role of kinetochore attachment in triggering anaphase is overlooked. To approach similar questions, always identify the specific checkpoint involved and recall its criteria for cell cycle progression.
In a river ecosystem, measured annual energy stored as new biomass is: algae 10,000 kJ/m2/yr, mayfly larvae 800 kJ/m2/yr, trout 80 kJ/m2/yr. Assume each value represents energy available to the next trophic level. Which conclusion is best supported about trophic efficiency across the two transfers?
Explanation: This question assesses the skill of analyzing energy flow through ecosystems by examining biomass energy storage across trophic levels. The correct answer, A, is supported because mayflies store 800 kJ/m²/yr, which is 8% of algae's 10,000 kJ/m²/yr, and trout store 80 kJ/m²/yr, 10% of mayflies', showing slight variation in efficiency. This reflects energy dissipation through respiration and unassimilated food in aquatic systems. The data emphasize that efficiencies are approximate and can differ by level. A tempting distractor, C, is wrong because it misinterprets numerical patterns as high efficiency, a misconception disregarding actual ratios. To analyze similar problems, compute distinct efficiencies for each transfer and compare them for ecosystem insights.
In a kelp forest, sea urchins graze on kelp holdfasts. Sea otters prey on sea urchins. In areas where otters are common, kelp cover is high and urchin density is low. In nearby areas with few otters, urchin density is high and kelp cover is reduced. Wave exposure and water temperature are similar between areas. Which community interaction best explains the difference in kelp cover between the two areas?
Explanation: This question assesses the skill of analyzing community ecology by identifying multi-level effects in marine food webs. Sea otters prey on urchins, reducing urchin density and allowing kelp to flourish, exemplifying a trophic cascade where top predator control of herbivores indirectly benefits primary producers. Areas with otters show high kelp cover and low urchins, while otter-scarce areas have the opposite, with similar abiotic conditions confirming the cascade's role. This interaction logic traces the indirect positive effect on kelp through predator-herbivore dynamics. A tempting distractor is C, mutualism, which is wrong because otters and kelp do not directly exchange benefits, stemming from the misconception that all indirect positives indicate direct symbiosis. To uncover cascades, compare ecosystems with varying predator densities to map abundance changes across trophic levels.
Ecologists compared three prairie sites after grazing. Site 1 had 14 plant species; Site 2 had 10; Site 3 had 6. Sampling area and season were the same, and species abundances were similarly even within each site. Which conclusion is best supported about biodiversity among the sites?
Explanation: This question assesses biodiversity comparison when only species richness data is provided with similar evenness. Site 1 has the highest biodiversity because it contains the most plant species (14), compared to Site 2 (10) and Site 3 (6), representing the greatest species richness. Since the question states that species abundances were similarly even within each site, evenness doesn't differentiate the sites, making species richness the sole determinant of biodiversity differences. The misconception in choice A is that fewer species somehow enhances biodiversity through reduced competition, when actually more species indicates higher biodiversity. When evenness is similar across communities, species richness alone determines biodiversity rankings—the community with the most species has the highest biodiversity.
A marine alga secretes a polysaccharide made of repeating galactose units with many sulfate (–SO3–) groups that remain negatively charged in seawater. The polymer is highly hydrophilic and forms a viscous gel because water molecules align around the charged groups. When the alga is exposed to strong wave action, the gel layer remains attached to the cell surface and resists being washed away. Which molecular feature best explains the gel’s ability to retain water and adhere as a protective coating?
Explanation: This question requires analyzing carbohydrate structure–function relationships to understand how molecular features determine polymer properties. The marine alga's polysaccharide contains sulfate groups (–SO3–) that remain negatively charged in seawater, creating strong ion-dipole interactions with water molecules that form extensive hydration shells around each charged group. These water molecules become organized and bound to the polymer, creating a viscous gel that resists mechanical disruption because the electrostatic attractions between charged sulfates and polar water molecules are stronger than the shearing forces from waves. Option B incorrectly suggests nonpolar hydrocarbon chains would help retain water, when actually hydrophobic groups would exclude water and prevent gel formation—this represents a polarity misconception. When analyzing polysaccharide properties, identify charged or polar groups that can interact with water through electrostatic or hydrogen-bonding interactions to predict hydration and gel-forming behavior.
A lipid sample is treated with a reagent that selectively breaks ester linkages. Triglycerides and phospholipids contain ester bonds between glycerol and fatty acids, whereas steroids lack fatty acid tails and do not contain glycerol–fatty acid ester linkages. After treatment, one class of lipids remains structurally intact while others are cleaved into smaller components. Which lipid type is most likely to remain intact after ester-bond cleavage?
Explanation: This question assesses the analysis of lipid structure–function. Steroids remain intact after ester-bond cleavage, as per choice A, because their fused-ring core lacks the glycerol–fatty acid ester linkages present in triglycerides and phospholipids, making them resistant to the reagent that targets those bonds. Triglycerides and phospholipids are cleaved into smaller components like glycerol and free fatty acids due to their ester bonds. This reflects the AP Biology distinction in lipid classes, where steroids are not ester-linked like glycerolipids. Choice B is a tempting distractor, suggesting phospholipids are protected by phosphate groups, which embodies a misconception of level-of-organization by attributing protective roles without chemical basis. To solve similar questions, identify the specific bonds in each lipid type and predict susceptibility to targeted reagents.