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AP Biology
Practice Test 3 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.
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A researcher compares the forelimb bones of a bat, whale, cat, and human. Each forelimb contains a humerus, radius, ulna, carpals, metacarpals, and phalanges arranged in the same relative order, but the bones differ in length and shape. The researcher also notes that insect wings lack these bones and are made of cuticle extensions. Which conclusion is best supported by the forelimb evidence?
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A researcher compares the forelimb bones of a bat, whale, cat, and human. Each forelimb contains a humerus, radius, ulna, carpals, metacarpals, and phalanges arranged in the same relative order, but the bones differ in length and shape. The researcher also notes that insect wings lack these bones and are made of cuticle extensions. Which conclusion is best supported by the forelimb evidence?
Explanation: This question assesses the skill of analyzing evidence of evolution, specifically identifying homologous structures as indicators of common ancestry. The consistent arrangement of forelimb bones (humerus, radius, ulna, etc.) across bat, whale, cat, and human, despite functional differences, points to inheritance from a shared vertebrate ancestor. Variations in bone length and shape reflect adaptations to diverse lifestyles, but the underlying pattern remains conserved. In contrast, insect wings lacking these bones highlight analogous rather than homologous structures for flight. A tempting distractor is choice C, which attributes similarities to need rather than ancestry, reflecting the misconception that evolution is driven by necessity instead of descent with modification. When examining structures, compare underlying patterns to distinguish homology from analogy in supporting evolutionary relationships.
A wax is composed of long-chain fatty acids ester-linked to long-chain alcohols, producing molecules that are largely nonpolar and lack charged or highly polar groups. A thin waxy coating on a leaf surface reduces water loss. Which statement best explains, at the molecular level, how the wax coating limits evaporation from the leaf?
Explanation: This question tests analysis of lipid structure–function in the context of plant water conservation. The correct answer A explains that the nonpolar hydrocarbon chains in wax repel water molecules, creating a hydrophobic barrier that reduces water's ability to diffuse outward from the leaf surface. Waxes, composed of long-chain fatty acids esterified to long-chain alcohols, are extremely hydrophobic molecules that form a water-impermeable coating because water molecules cannot dissolve in or easily pass through the nonpolar wax layer. This forces water vapor to find the few openings (stomata) where gas exchange is regulated, rather than evaporating freely across the entire leaf surface. Option B incorrectly claims wax contains phosphate groups that bind water, showing a category error—waxes are simple lipids without phosphate groups, unlike phospholipids. The key principle for understanding biological waterproofing is that extensive nonpolar surfaces create barriers to polar molecules like water, exploiting the fundamental incompatibility between polar and nonpolar substances.
Scientists tested protocell formation by adding fatty acids to warm water containing dissolved salts similar to those expected in some early Earth pools. Within minutes, microscopy showed spherical vesicles with bilayer-like boundaries. When a fluorescent dye that cannot cross lipid membranes was added outside the vesicles, fluorescence remained outside; when dye was present before vesicle formation, fluorescence was observed inside many vesicles. Which conclusion is best supported by these observations about early steps in life’s origin?
Explanation: This question assesses the skill of evaluating evidence about the origins of life on Earth. The researchers added fatty acids to warm water with dissolved salts, leading to spontaneous formation of spherical vesicles with bilayer-like boundaries, and fluorescence experiments showed that these vesicles could enclose solutes like dye, creating a compartmentalized internal environment separate from the exterior. This demonstrates a key step in protocell formation, where lipid bilayers abiotically assemble and provide isolation, which is essential in AP Biology for understanding the emergence of cellular compartments that could concentrate reactants and protect early biochemical processes. The absence of any added proteins or genomes in the setup highlights that vesicle formation can occur spontaneously under prebiotic conditions. A tempting distractor, choice B, is incorrect because it claims vesicle formation requires protein channels encoded by genomes, embodying a level-of-organization error by conflating modern cellular requirements with abiotic origins. For these question types, focus on how experimental observations support incremental abiotic steps toward life and avoid projecting advanced biological features backward in time.
In an experiment, cells are treated with ligand Y, which binds a receptor tyrosine kinase (RTK) in the plasma membrane. Within 1 minute, the receptor becomes phosphorylated on specific tyrosines. A drug that prevents receptor dimerization is added before Y; Y still binds, but receptor phosphorylation is not detected. Which of the following best explains why phosphorylation fails to occur?
Explanation: This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is A because the rapid phosphorylation after Y binding and its absence when dimerization is prevented indicate that ligand-induced dimerization is essential for RTK activation. In basic signaling principles, RTKs dimerize upon ligand binding, allowing their kinase domains to trans-phosphorylate each other on tyrosines, initiating the signal. The evidence that Y binds but no phosphorylation occurs with the dimerization drug supports this mechanism, as monomers cannot effectively autophosphorylate. A tempting distractor is B, which is wrong because it assumes RTKs enter the nucleus for phosphorylation, reflecting the misconception that membrane receptors relocate to act directly on intracellular targets. A transferable strategy for signal transduction questions is to recall receptor-specific activation mechanisms, like dimerization for RTKs, and test them against experimental manipulations.
In a cultured animal cell, DNA replication completes during S phase, and the cell enters G2 with duplicated chromosomes. The cell is then exposed to a drug that prevents microtubules from attaching to kinetochores. When the cell enters M phase, chromosomes condense and the nuclear envelope breaks down, but kinetochores remain unattached. The spindle assembly checkpoint monitors attachment and tension before anaphase begins. After 45 minutes in mitosis, the cell still shows all sister chromatids paired at their centromeres and aligned poorly near the cell center. Which outcome is most likely if the checkpoint remains active?
Explanation: This question assesses the skill of analyzing the cell cycle, focusing on the role of checkpoints in mitosis. The stimulus describes a drug that prevents microtubule attachment to kinetochores, resulting in unattached kinetochores during M phase, which activates the spindle assembly checkpoint to delay anaphase. In AP Biology, the spindle assembly checkpoint ensures all chromosomes are properly attached and under tension before chromatid separation, preventing progression if conditions are not met. Therefore, with the checkpoint active and kinetochores unattached, the cell remains stalled in metaphase, with duplicated chromosomes aligned poorly and no chromatid separation occurring. A tempting distractor is A, which suggests separation despite unattached kinetochores, but this is incorrect due to a misunderstanding of checkpoint mechanisms, specifically a teleological misconception that the cell prioritizes completion over accuracy. To approach similar questions, identify the checkpoint involved and its specific triggers, then evaluate how disruptions affect progression through the cycle.
Some hypotheses propose that RNA served as an early genetic system because it can store information and catalyze reactions. In an experiment, a ribozyme population was provided RNA templates and nucleotides. After multiple cycles, ribozymes that copied templates more accurately and faster became more common in the population, even though no proteins were present. Which conclusion is best supported by this evidence about early life origins?
Explanation: This question requires evaluating evidence about the origins of life, specifically whether RNA molecules can undergo selection based on replication differences without proteins. The experiment demonstrates that ribozymes (catalytic RNAs) copying templates more accurately and faster become more common over multiple cycles in a protein-free system, directly supporting answer A that RNA molecules can undergo selection based on heritable differences in replication performance without requiring proteins. This provides crucial evidence for the RNA World hypothesis, showing that information storage (sequence) and catalytic function (replication activity) can reside in the same molecule type, allowing evolution to occur before the evolution of proteins or DNA. The selection for better-replicating ribozymes demonstrates natural selection operating at the molecular level based on functional differences. Answer B incorrectly claims proteins must be first self-replicators because RNA cannot store information, representing a function confusion error since RNA clearly stores sequence information in its nucleotide order. For RNA World questions, recognize that RNA's dual capacity for information storage and catalysis makes it a plausible precursor to modern DNA-protein biology.
A yeast population was grown in a closed flask with a fixed amount of sugar. Cell density rose rapidly for 10 hours, then growth slowed and cell density leveled off from hours 14–24. Measurements showed sugar concentration decreased steadily and was very low by hour 14. No cells were added or removed during the experiment. Which explanation best accounts for the slowing and leveling of population growth after hour 10?
Explanation: This question tests population ecology analysis of resource-limited growth in closed systems. The yeast population exhibits classic logistic growth: rapid initial increase followed by leveling off as sugar (the limiting resource) becomes depleted. The correlation between sugar depletion at hour 14 and population plateau confirms density-dependent resource limitation is controlling growth. Choice B incorrectly suggests resources increased to cause exponential growth, but the data shows resources decreased while growth slowed. In closed systems with fixed resources, always expect logistic rather than indefinite exponential growth patterns.
A marine fish species occupies a coastline with no physical barriers. In the northern region, adults spawn in early spring; in the southern region, adults spawn in late summer. Water temperature and day length differ by latitude, and long-term surveys show little overlap in spawning dates. Genetic sampling reveals increasing allele-frequency differences with latitude. When northern and southern adults are brought into the same aquarium, they still release gametes only at their region-typical times. Which mechanism most directly reduced gene flow and promoted speciation in this system?
Explanation: This question requires analyzing speciation through temporal reproductive isolation in a continuous marine habitat. Despite no physical barriers, the fish populations spawn at completely different times of year (early spring vs. late summer), which prevents interbreeding and gene flow between regions. This temporal isolation allowed genetic divergence along the latitudinal gradient, and importantly, the spawning time differences persist even in laboratory conditions, showing they're genetically based. Choice E incorrectly invokes Lamarckian inheritance, suggesting individuals acquire traits during life and pass them genetically to offspring, which contradicts modern evolutionary theory. To identify temporal isolation, look for time-based reproductive barriers (different breeding seasons, flowering times, or daily activity patterns) that prevent populations from exchanging genes.
An RNA molecule folds into a stem-loop when complementary regions within the same strand base-pair. Which feature best explains how a single RNA strand can form double-stranded segments?
Explanation: This question examines nucleic acid structure-function analysis in RNA secondary structure formation. The correct answer B explains that single-stranded RNA can fold back on itself when complementary sequences within the same molecule form intramolecular hydrogen bonds, creating stem-loop structures. Unlike DNA, which typically exists as two separate antiparallel strands, RNA molecules can have regions that are self-complementary, allowing one part of the strand to base-pair with another part through standard Watson-Crick pairing (A-U and G-C). This intramolecular base pairing creates double-stranded regions (stems) connected by single-stranded loops. Choice C incorrectly states RNA uses thymine (RNA uses uracil instead). The strategy is to recognize that RNA's single-stranded nature allows flexibility for intramolecular interactions when complementary sequences are present.
A fluorescent ligand binds to receptors on the surface of living cells. At low ligand concentration, binding is detected and increases with concentration; at high concentration, binding reaches a maximum and no longer increases. The intracellular response also plateaus at high ligand concentration. Which of the following best explains the plateau in the context of signal reception?
Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on the reception step and dose-response. The correct answer is A because at high concentrations, all receptor sites are occupied, saturating binding and the response, as shown by the plateau in fluorescent binding and intracellular effects. This aligns with basic signaling principles of finite receptors leading to maximum responses. Low-to-high concentration data illustrate this. A tempting distractor is B, which is incorrect because it ignores diffusion at high concentrations, a misconception unrelated to saturation. In signal transduction questions, interpret dose-response curves to understand receptor-ligand dynamics like saturation.
A eukaryotic lineage contains an organelle that performs aerobic respiration. The organelle has a double membrane, and its inner membrane includes many proteins related to bacterial electron transport proteins. Antibiotics that inhibit bacterial ribosomes reduce protein synthesis inside the organelle but do not affect cytosolic protein synthesis. The organelle contains a small circular genome. Which evidence best supports the origin of this organelle by endosymbiosis?
Explanation: This question requires analyzing the origins of cell compartmentalization to identify endosymbiotic evidence. The correct answer A is supported by the specific antibiotic sensitivity pattern: antibiotics that target bacterial ribosomes inhibit the organelle but not the cytosol, directly demonstrating the organelle contains bacterial-type ribosomes distinct from eukaryotic ones. Combined with circular DNA (bacterial genome organization) and bacterial-related electron transport proteins, this provides strong molecular evidence for bacterial ancestry through endosymbiosis. Answer C incorrectly attributes electron transport proteins to membrane infolding, a structure-function confusion—infolded membranes would contain the host's proteins, not bacterial-type proteins. To confirm endosymbiotic origins, look for molecular markers that specifically link to bacterial ancestry, particularly differential antibiotic sensitivity that reveals the presence of bacterial-type translation machinery.
In pea plants, purple flowers (P) are completely dominant to white flowers (p). A student crosses two purple-flowered plants and is told both parents are heterozygous (Pp). Each parent produces gametes carrying either P or p with equal probability. Assume fertilization is random and the gene is autosomal. Which proportion of offspring is expected to have white flowers?
Which proportion of offspring from a Pp×Pp cross will have white flowers?
Explanation: This question tests Mendelian inheritance analysis, focusing on predicting offspring phenotypes from a monohybrid cross. In a Pp x Pp cross, each parent produces gametes with P or p alleles in equal proportions due to segregation during meiosis. The Punnett square shows that the possible genotypes are PP, Pp, and pp, with pp occurring in 1/4 of the offspring, resulting in white flowers since p is recessive. This 1/4 proportion arises because both parents must contribute a p allele, each with 1/2 probability, so (1/2) x (1/2) = 1/4. A tempting distractor is 1/2, which might come from mistakenly thinking white flowers require only one p allele, confusing dominance with codominance. For inheritance questions, always construct a Punnett square to visualize allele combinations and calculate probabilities based on independent segregation.
In an isolated mitochondrion preparation supplied with abundant NADH and ADP + Pi, researchers add cyanide, which blocks electron transfer to oxygen at the end of the electron transport chain. Protons had been pumped into the intermembrane space as electrons moved through complexes, creating a gradient used by ATP synthase. After cyanide addition, electron flow through the chain stops and NADH can no longer be oxidized to NAD+. Which outcome is most likely to occur immediately in this preparation?
Explanation: This question assesses the skill of analyzing cellular respiration, specifically how inhibitors disrupt the electron transport chain and ATP synthesis. Cyanide blocks the final step of electron transfer to oxygen, halting electron flow through the chain and preventing proton pumping into the intermembrane space. Without ongoing proton pumping, the existing gradient dissipates as protons leak back or are used without replenishment, leading to decreased ATP production by ATP synthase. The stimulus notes that NADH can no longer be oxidized, which aligns with stopped electron flow, further supporting that oxidative phosphorylation ceases. A tempting distractor is choice B, which incorrectly suggests increased ATP via substrate-level phosphorylation due to electron accumulation, stemming from the misconception that blocked chains shift to alternative ATP pathways in mitochondria. A transferable strategy for respiration questions is to trace the flow of electrons and protons, identifying how disruptions affect the proton gradient and ATP yield.
In a rabbit population, fur thickness varies and is heritable. A new parasite attaches more successfully to rabbits with very thick fur, reducing their reproductive success compared with rabbits with thinner fur. Both fur types continue to appear among newborns. Which explanation best accounts for how allele frequencies will change over 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 parasite attaches more easily to rabbits with thick fur, reducing their reproductive success, while thinner-furred rabbits reproduce more successfully. As a result, thinner-furred rabbits contribute more offspring, leading to an increase in alleles for thinner fur across the population over generations. This change reflects selection against thick fur due to the parasitic pressure, with variation continuing among newborns. A tempting distractor is choice D, which incorrectly proposes that individuals evolve thinner fur during their lifetimes and pass it on, representing the misconception of Lamarckian evolution instead of selection on genetic variation. For natural selection questions, always identify the selective pressure, the heritable trait, and how it affects reproductive success at the population level.
A snail population shows genetic variation in shell thickness. A crab predator is introduced and can crush thin shells more easily than thick shells. After several years, surveys show an increased proportion of thick-shelled snails, and breeding experiments indicate shell thickness is heritable. Which statement best explains the role of variation in the population’s evolution?
Explanation: This question assesses the skill of analyzing variations in populations by examining how genetic diversity in shell thickness evolves under predatory pressure. The correct answer, B, highlights that variation in shell thickness created survival differences, with thicker-shelled snails more resistant to crabs, resulting in natural selection that increased the frequency of thick-shell alleles over time. Breeding experiments confirming heritability underscore that the change was evolutionary, driven by differential reproduction of preexisting variants. The increased proportion of thick-shelled snails after years illustrates adaptive evolution without the need for new genetic inputs. A tempting distractor is A, which represents the misconception of induced acquired traits, implying attacks caused shell thickening that was inherited, but selection requires heritable variation from the outset. A transferable approach is to evaluate evolutionary scenarios by checking for heritable variation and selective advantages that alter population genetics.
In a diploid cell (2n=4), homologous chromosome pair 1 carries alleles A and a, and pair 2 carries alleles B and b. During metaphase I, the two homologous pairs align independently at the metaphase plate, so either the maternal or paternal homolog of each pair can face the same pole. No crossing over occurs in this cell. After meiosis I and II, four haploid gametes form. Considering only the arrangement of whole chromosomes into gametes, the cell can produce gametes with different combinations of the A/a chromosome and the B/b chromosome. Which meiotic process best explains the production of gametes AB, Ab, aB, and ab from this one cell?
Explanation: This question assesses understanding of how meiosis generates genetic diversity through processes like independent assortment. The cell has two homologous pairs (2n=4) with alleles A/a and B/b, and no crossing over occurs, so variation arises solely from how whole chromosomes are distributed. During metaphase I, the random alignment of each homologous pair means the maternal or paternal homolog can face either pole independently, leading to different combinations in the gametes after meiosis II. This independent assortment produces the four gamete types: AB, Ab, aB, and ab, as each gamete receives one chromosome from each pair in varying maternal-paternal mixes. A tempting distractor is choice D, random fertilization, which is wrong because it refers to combining gametes from different parents, not variation within one cell's meiosis, reflecting a misconception about when diversity is generated. To approach similar questions, identify whether the variation comes from within meiosis or post-meiosis events like fertilization.
A word-described phylogeny for four plant species is: (A,B) form a clade; that clade is sister to C; the clade containing A, B, and C is sister to D. No timing information is given. Based on this branching pattern, which statement is best supported?
Explanation: This question assesses the skill of inferring phylogenetic relatedness from a described branching pattern in plants. The topology shows ((A,B),C) as a clade sister to D, meaning C shares a more recent common ancestor with A (at the node uniting (A,B,C)) than with D (at the root). A and B are sisters, with C as their outgroup within the larger clade excluding D, and no timing data alters this hierarchical inference. This supports that divergence between C and A occurred after the split from D. A tempting distractor is D, claiming A equally related to C and D due to A's basal position, a misconception of equating branching order with equal genetic distance, a level-of-organization error in tree reading. A transferable strategy for this question type is to identify the most recent shared node for pairs and compare node depths to determine relative closeness without assuming branch lengths.
Researchers measure early signaling in cells exposed to ligand M. M binds a membrane receptor, and within 10 seconds IP3 levels rise sharply. When phospholipase C (PLC) is inhibited, M still binds the receptor but IP3 does not increase. Which of the following best explains the role of PLC in early transduction here?
Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on receptor-ligand interactions and early transduction events. The correct answer is A because inhibiting PLC prevents IP3 increase despite M binding, indicating PLC cleaves membrane phospholipids to produce IP3 after receptor activation. This follows basic GPCR-PLC pathway principles where activated receptors stimulate PLC to generate IP3 as a second messenger within 10 seconds. The rapid rise and specific inhibition confirm PLC's enzymatic role in early transduction. A tempting distractor is C, which wrongly posits PLC degrades IP3, based on the misconception that PLC is a terminator like phosphodiesterase, though inhibition prevents IP3 production. When analyzing signal transduction questions, identify enzyme inhibitors' effects to map second messenger generation pathways.
A biologist monitored two fenced grassland plots of equal size. Plot 1 began with 20 rabbits, and Plot 2 began with 80 rabbits. Over three months, rainfall and temperature were similar in both plots. By the end, Plot 2 had more individuals with bite wounds and a lower proportion of pregnant females, while Plot 1 showed few wounds and higher pregnancy rates. No new predators entered the fences. Which factor is most likely density dependent in regulating the rabbit populations?
Explanation: This question examines density-dependent population regulation through behavioral changes at different population densities. The correct answer is B because intraspecific aggression increases with population density - when more rabbits occupy the same space, they compete more intensely for resources and territory, leading to more bite wounds and reduced reproduction. Plot 2 with 80 rabbits shows classic signs of density stress: more wounds and lower pregnancy rates compared to Plot 1 with only 20 rabbits. Answer A (lightning-caused fire) is incorrect because it represents a density-independent factor that would affect both plots similarly regardless of rabbit density - students often confuse any negative effect with density dependence. The key strategy is to identify factors whose impact scales with population size: competition, aggression, and disease transmission increase with density, while weather events affect populations regardless of their size.
A mitotic checkpoint protein (M) binds APC/C and inhibits it when M is phosphorylated. A phosphatase removes the phosphate from M after kinetochores attach, releasing APC/C. In mutant cells, M cannot be dephosphorylated, but kinetochore attachment occurs normally. Which outcome is most likely in the mutant cells?
Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, particularly the spindle assembly checkpoint involving APC/C inhibition. Normally, phosphorylated M inhibits APC/C until dephosphorylation after kinetochore attachment releases this inhibition, allowing securin degradation and anaphase. In mutant cells where M cannot be dephosphorylated despite normal attachment, M remains phosphorylated and persistently inhibits APC/C, stabilizing securin and delaying anaphase onset, as described in choice A. This prevents the progression signal, leading to a checkpoint arrest. A tempting distractor is choice E, which claims normal timing because dephosphorylation is irrelevant, but this misconceives the checkpoint's reliance on phosphorylation status for signal transduction. A transferable strategy is to trace how mutations affect signal transduction steps in checkpoints to predict downstream effects on cell cycle progression.
A membrane protein has a transmembrane region enriched in amino acids with nonpolar side chains composed mostly of C and H, and an external region enriched in polar side chains containing O and N. The phospholipid bilayer interior is largely hydrocarbon, while the surrounding cytosol is water-based. Nonpolar side chains interact favorably with the bilayer’s hydrocarbon tails via dispersion forces, whereas polar side chains can form hydrogen bonds with water. Which statement best explains why the C- and H-rich segment spans the membrane?
Explanation: This question assesses the analysis of elements of life and chemical properties. The correct answer is A because nonpolar side chains rich in C and H interact favorably with the bilayer's hydrophobic hydrocarbon interior via dispersion forces, stabilizing membrane insertion, as the stimulus describes the transmembrane region enriched in such chains versus polar external regions. This matches AP Biology concepts of membrane protein structure, where hydrophobic matching dictates transmembrane segments in nonpolar environments. Polar side chains with O and N prefer water via hydrogen bonds, avoiding the bilayer core, highlighting C and H's role in hydrophobicity. A tempting distractor is C, claiming polar side chains are excluded from water, reflecting a structure-function confusion by inverting hydrophilic and hydrophobic behaviors. When solving, evaluate how elemental composition determines regional polarity and membrane positioning.
A stream is sampled upstream and downstream of a wastewater outfall. Upstream, macroinvertebrates include 18 taxa with fairly similar abundances. Downstream, the sample includes 7 taxa, and 85% of individuals are one worm taxon; total individuals collected are equal at both sites. Water temperature is similar. Which conclusion is best supported about the community-level consequences downstream?
Explanation: This question tests biodiversity analysis in the context of environmental disturbance from wastewater pollution. Downstream biodiversity is lower because both species richness and evenness have decreased compared to upstream conditions. The upstream site has 18 taxa with fairly similar abundances (high richness and high evenness), while downstream has only 7 taxa (reduced richness) with 85% being one worm taxon (very low evenness). This pattern indicates severe biodiversity loss, as both components of biodiversity have declined. The dominance of pollution-tolerant worms downstream is a classic indicator of degraded water quality reducing community diversity. A common misconception (choice C) is that equal numbers of individuals collected means unchanged biodiversity, but biodiversity depends on how individuals are distributed among different taxa, not total abundance. When assessing biodiversity changes, examine both richness (number of different groups) and evenness (dominance patterns) to determine if diversity has increased, decreased, or remained stable.
In a coastal grass population, individuals vary in leaf wax thickness. Wax thickness is heritable and determined largely by alleles at one gene. After several unusually dry growing seasons, plants with thicker wax lose less water and produce more seeds than plants with thinner wax. No new individuals immigrate during this period. Which outcome is most likely in the population after several generations of these dry seasons?
Explanation: This question tests understanding of natural selection in a plant population experiencing drought conditions. The correct answer is A because natural selection occurs when individuals with advantageous heritable traits (thicker leaf wax) have higher reproductive success (produce more seeds) than others in the population. Over generations, this differential reproduction causes the frequency of alleles for thicker wax to increase in the population, as these alleles are passed on more frequently to offspring. The stimulus clearly states that plants with thicker wax produce more seeds during drought, providing the necessary conditions for natural selection. Answer B incorrectly suggests that individual plants can change their traits within their lifetime (acquired characteristics), which confuses natural selection with phenotypic plasticity. To solve natural selection problems, identify the heritable trait, determine which variants have higher reproductive success, and predict that alleles for the advantageous trait will increase in frequency over generations.
In a temperate pond, larval dragonflies consume mosquito larvae. When dragonflies are experimentally removed from several enclosures, mosquito larvae density increases and grazing on algae by mosquito larvae increases, causing algae biomass to decrease compared with control enclosures. No other species are added or removed, and abiotic conditions are similar across enclosures. Which interaction best explains the initial decrease in mosquito larvae in control enclosures?
Explanation: This question assesses the skill of analyzing community ecology by identifying species interactions based on experimental outcomes. Predation by dragonflies on mosquito larvae directly reduces mosquito abundance in control enclosures, as dragonflies consume the larvae, limiting their population. When dragonflies are removed, mosquito larvae increase, leading to higher grazing on algae and decreased algae biomass, confirming the predatory relationship. This interaction explains the initial decrease in mosquito larvae, as no other species or abiotic changes account for the difference. A tempting distractor is choice C, which suggests competition for algae, but this is wrong due to the misconception that consumption of one species by another equates to resource competition rather than predation. To identify interactions in community ecology, always examine experimental manipulations and their direct effects on population densities.
In a cell-free system, activated receptor fragments phosphorylate adaptor protein Ad on tyrosines. Ad recruits enzyme PI3K, which converts membrane lipid PIP2 to PIP3. PIP3 recruits kinase Akt to the membrane, where Akt becomes phosphorylated and then phosphorylates cytosolic target X. When a lipid phosphatase that converts PIP3 back to PIP2 is added, Akt phosphorylation decreases and X phosphorylation drops. Which change would most likely counteract the effect of the lipid phosphatase?
Explanation: This question assesses the skill of analyzing signal transduction pathways by countering phosphatase effects in PI3K-Akt signaling. The lipid phosphatase converts PIP3 to PIP2, reducing PIP3, Akt membrane recruitment, phosphorylation, and X phosphorylation. Increasing PI3K activity raises PIP3 levels, opposing the phosphatase and restoring Akt activation and X phosphorylation. This boosts production to overcome degradation, maintaining the lipid signal. A tempting distractor is choice E, adding PIP2 to increase PIP3 via phosphatase, but phosphatases degrade, not produce PIP3, confusing reaction direction. For balancing enzymes, enhance the opposing activity to shift equilibrium toward the desired product.