AP Biology

AP Biology Practice Test: Practice Test 8

Practice Test 8 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A plant population is split between two soil types that occur in adjacent patches with no physical barrier. Plants in patch X grow on serpentine soil; patch Y is normal soil. Pollinator observations show that pollinators move frequently between patches. However, pollen from patch X plants germinates poorly on patch Y stigmas, and the reverse cross also shows low pollen tube growth. Seeds are rarely produced from cross-patch hand pollinations, even though flowering times overlap. Which reproductive barrier most directly contributes to reduced gene flow between the patches?

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Question 1

Hybrid breakdown in the $F_2$ generation after viable $F_1$ offspring form
Behavioral isolation due to differences in courtship displays by individuals
Gametic isolation preventing successful fertilization between populations (correct answer)
Geographic isolation caused by a river blocking pollen movement
Polyploidy creating a new species through chromosome doubling in somatic cells

Explanation: This question tests recognition of gametic isolation as a prezygotic reproductive barrier in sympatric plant populations. Despite pollinators moving between patches and overlapping flowering times, reproduction fails at the fertilization stage—pollen from one patch germinates poorly on stigmas from the other patch, preventing successful fertilization. This gametic incompatibility maintains reproductive isolation even though the populations are adjacent and could potentially interbreed. Choice E incorrectly describes polyploidy through chromosome doubling in somatic (body) cells, which wouldn't create heritable reproductive isolation since only germline changes are passed to offspring. When analyzing plant reproductive barriers, consider the entire sequence from pollination to seed formation to identify where isolation occurs.

Question 2

A membrane receptor binds ligand K and rapidly activates a cytosolic kinase. In vitro, the kinase becomes active only when it binds a phosphorylated tyrosine motif on the receptor’s cytosolic side. A point mutation changes the tyrosine to phenylalanine; ligand binding remains normal but kinase activation is absent. Which of the following best explains the missing early transduction step?

The mutation prevents ligand K from being produced, so receptors are unoccupied.
The mutation removes a phosphorylation site needed to recruit or activate the kinase at the receptor. (correct answer)
The mutation increases receptor affinity, causing constant kinase activation even without ligand.
The mutation forces the receptor to move into the nucleus to bind DNA before signaling can begin.
The mutation converts the receptor into an ion channel, eliminating the need for kinase activation.

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on receptor-ligand interactions and early transduction events. The correct answer is B because mutating tyrosine to phenylalanine prevents kinase activation despite normal K binding, indicating the site is needed for phosphorylation to recruit or activate the kinase. In vitro data show kinase requires the phosphorylated motif, aligning with principles where receptor phosphorylation creates binding sites. The point mutation specifically disrupts this early step. A tempting distractor is A, which wrongly blames ligand production, based on the misconception that mutations affect ligand synthesis, though binding is normal. When analyzing signal transduction questions, examine mutation sites and in vitro binding to identify critical motifs in transduction.

Question 3

In a coastal snail population, shell color is controlled by two alleles. Before a storm, allele B frequency is 0.60 and allele b is 0.40. After the storm deposits dark sediment, birds more easily spot light shells; in the next generation, allele B frequency is 0.78. Which statement best explains how variation affected this change in allele frequency across generations?

Individuals changed shell color during their lifetime, then passed the new color allele to offspring.
Random mating after the storm created new alleles for dark shells, increasing allele B frequency.
Preexisting color alleles caused differential survival and reproduction, increasing the frequency of allele B. (correct answer)
The storm directly mutated most b alleles into B alleles, producing a rapid adaptive shift.
All snails benefited equally from dark sediment, so allele frequencies changed without selection.

Explanation: This question tests your ability to analyze how population variation leads to evolutionary change through natural selection. The correct answer (C) recognizes that preexisting genetic variation (alleles B and b) created differential survival when environmental conditions changed—dark-shelled snails with allele B were less visible to bird predators on dark sediment, so they survived and reproduced more than light-shelled snails with allele b. This differential reproduction increased the frequency of allele B from 0.60 to 0.78, demonstrating natural selection acting on existing variation. Answer A incorrectly suggests Lamarckian inheritance where individuals change traits during their lifetime and pass those acquired changes to offspring, which violates our understanding of genetics. When analyzing population changes, always check whether the explanation involves selection acting on preexisting variation (correct) versus individuals acquiring and passing on new traits (incorrect).

Question 4

An enzyme-catalyzed reaction is run at saturating substrate concentration (well above the level that produces a rate plateau). In Trial 1, enzyme concentration is 1×. In Trial 2, enzyme concentration is doubled while all other conditions are unchanged and no inhibitor is present. The initial rate in Trial 2 is approximately twice that of Trial 1. Which explanation best accounts for this result at the molecular level?

Doubling enzyme increases the number of active sites, increasing total ES complexes formed per unit time. (correct answer)
Doubling enzyme increases substrate concentration, raising collision frequency and driving equilibrium forward.
Doubling enzyme lowers activation energy more per molecule, so each active site catalyzes faster.
Doubling enzyme causes product inhibition, which paradoxically increases the initial reaction rate.
Doubling enzyme reduces temperature changes during mixing, preventing thermal denaturation of substrate.

Explanation: This question assesses the skill of analyzing enzyme function by examining the impact of enzyme concentration on reaction rates at saturating substrate levels. Doubling the enzyme concentration doubles the initial rate because it increases the number of available active sites, allowing more ES complexes to form per unit time and thus more product to be generated. At saturating substrate, the reaction is enzyme-limited, so additional enzyme directly proportionalizes the rate, as each enzyme molecule operates at its maximum catalytic capacity (Vmax). This is evidenced by the absence of inhibitors and unchanged conditions, confirming the rate increase is due to more catalytic units rather than alterations in kinetics or equilibrium. A tempting distractor is choice C, which claims doubling enzyme lowers activation energy more, but this reflects a teleology misconception, assuming enzymes 'work harder' individually, whereas enzymes lower activation energy consistently per molecule, and rate scales with enzyme amount. For enzyme questions varying concentrations, remember that at saturation, rate is proportional to enzyme concentration, providing a strategy to identify limiting factors.

Question 5

A cell’s cytosol contains 1 mM Ca2+, while the extracellular fluid contains 2 mM Ca2+. A specific Ca2+ transporter moves Ca2+ from the cytosol to the extracellular space, and transport stops when ATP is absent. Ca2+ continues to be exported even though cytosolic Ca2+ is lower than extracellular Ca2+. Which mechanism best explains Ca2+ export?

Facilitated diffusion of Ca2+ outward through a channel down its gradient
Simple diffusion of Ca2+ outward directly through the lipid bilayer
Primary active transport exporting Ca2+ against its gradient using ATP (correct answer)
Osmosis of Ca2+ outward through aquaporins driven by water movement
Exocytosis of Ca2+ from vesicles that fuse with the plasma membrane

Explanation: This question assesses the skill of analyzing membrane transport mechanisms by evaluating how substances move across cell membranes based on gradients and energy requirements. The correct answer is primary active transport because Ca2+ is exported against its gradient from 1 mM inside to 2 mM outside, stopping without ATP, indicating direct energy use. Continued export despite the gradient requires ATP hydrolysis to drive uphill movement. The transporter's specificity and ATP dependence fit primary active transport logic. A tempting distractor is facilitated diffusion outward, which is wrong due to the misconception that ions always follow gradients passively, ignoring ATP needs for against-gradient transport. To distinguish transport types, always check if movement is down or against the gradient and if energy like ATP is directly required.

Question 6

In a freshwater lake, a fungal pathogen infects a dominant zooplankton species. During weeks with high fungal infection, zooplankton grazing pressure decreases and phytoplankton biomass increases, while fish abundance remains stable. In weeks with low infection, zooplankton grazing increases and phytoplankton biomass decreases. Which interaction best explains the reduced zooplankton grazing during high-infection weeks?

Mutualism between fungus and zooplankton increases zooplankton feeding
Parasitism by the fungus reduces zooplankton performance and grazing (correct answer)
Competition between fungus and phytoplankton lowers zooplankton grazing
Predation by zooplankton on fungus reduces zooplankton grazing rates
Commensalism allows fungus to benefit without affecting zooplankton grazing

Explanation: This question assesses the skill of analyzing community ecology by identifying species interactions based on experimental outcomes. Parasitism by the fungus reduces zooplankton performance and grazing, as infection harms the host, leading to decreased grazing pressure and increased phytoplankton biomass during high-infection weeks. Low-infection weeks show opposite trends, with stable fish abundance ruling out predation changes. This explains the grazing reduction, as the fungus benefits at the zooplankton's expense without directly affecting phytoplankton. A tempting distractor is choice C, proposing competition with phytoplankton, but this is wrong due to the misconception that infection effects mimic resource competition, ignoring the host-parasite dynamic. To identify parasitism, correlate infection levels with host performance and cascading ecosystem effects like grazing rates.

Question 7

A membrane protein forms a pore that allows K $^+$ to cross the membrane. In an experiment, extracellular K $^+$ is 2 mM and cytosolic K $^+$ is 120 mM. When the pore is opened, net K $^+$ movement is from the cytosol to the outside until concentrations approach equilibrium. The movement continues when ATP synthesis is inhibited but stops when the pore is blocked. Which mechanism best explains K $^+$ movement through the pore?

Primary active transport by a K $^+$ pump that hydrolyzes ATP each cycle
Facilitated diffusion through an ion channel down the concentration gradient (correct answer)
Simple diffusion of K $^+$ through the lipid bilayer without proteins
Secondary active transport of K $^+$ coupled to glucose cotransport
Endocytosis of extracellular fluid containing K $^+$ into vesicles

Explanation: This question assesses the skill of analyzing membrane transport mechanisms. The correct answer is facilitated diffusion through an ion channel down the concentration gradient because K+ moves passively from high cytosolic to low extracellular concentration via a pore-forming protein, without energy input. The continuation of movement despite ATP inhibition confirms it's passive, driven by the concentration gradient toward equilibrium. Blocking the pore stops the process, highlighting the role of the channel in facilitating ion passage that wouldn't occur easily through the bilayer alone. A tempting distractor is simple diffusion through the lipid bilayer, but this is wrong due to the misconception that charged ions like K+ can freely cross hydrophobic lipids without proteins; ions require channels or carriers. To analyze similar problems, always determine if movement is down a gradient (passive) or against (active) and check for energy dependence.

Question 8

In an experiment, a membrane vesicle contains 100 mM sucrose and is placed in 100 mM sucrose solution. The membrane is impermeable to sucrose but permeable to water. Then 50 mM NaCl is added only to the external solution; Na+ and Cl− cannot cross the membrane. The vesicle’s volume decreases over several minutes, and ATP is absent. Which mechanism best explains the volume change?

Water leaves the vesicle by osmosis because external solute concentration is higher (correct answer)
Sucrose leaves the vesicle by facilitated diffusion, lowering internal osmolarity
NaCl enters the vesicle by simple diffusion, increasing internal osmolarity
Water is pumped out by primary active transport using ATP hydrolysis
The vesicle loses volume by exocytosis of water-containing internal vesicles

Explanation: This question tests the skill of analyzing membrane transport mechanisms. The correct answer A is osmosis because water leaves the vesicle as the external solution becomes hypertonic after adding 50 mM NaCl (increasing osmolarity), creating a water potential gradient outward, with no energy needed since ATP is absent. The membrane's permeability to water but not solutes drives shrinkage to equalize potentials. Initial equal sucrose concentrations ensure the change is due to added impermeable NaCl. A tempting distractor is B, sucrose leaving, but this is incorrect because the membrane is impermeable to sucrose, based on the misconception that solutes move osmotically like water. A transferable strategy is to calculate effective osmolarities and predict water flow direction in response to impermeable solute differences.

Question 9

In a eukaryotic cell line, gene Y transcription requires an activator that binds an enhancer 2 kb upstream. The activator recruits a histone acetyltransferase (HAT), increasing acetylation of nearby histone tails and correlating with higher gene Y mRNA. A drug that inhibits the HAT is added; the activator still binds DNA, but histone acetylation near the enhancer decreases. Which change would most likely restore high transcription of gene Y in the presence of the drug?

Overexpress a histone deacetylase to increase chromatin compaction
Increase tRNA abundance to accelerate elongation during translation
Mutate the gene Y start codon to improve ribosome recognition
Recruit a different coactivator with HAT activity to the enhancer-bound activator (correct answer)
Delete the introns from gene Y to reduce mRNA processing time

Explanation: This question evaluates knowledge of transcriptional regulation in eukaryotes, particularly how chromatin modifications influence gene activation. The activator normally recruits a histone acetyltransferase (HAT) to acetylate histones, loosening chromatin and promoting transcription, but the drug inhibits this HAT, reducing acetylation and transcription. Recruiting a different coactivator with HAT activity bypasses the inhibited HAT, restoring acetylation near the enhancer and allowing high gene Y transcription despite the drug. This works because the new coactivator provides the necessary enzymatic function to modify chromatin without relying on the original inhibited HAT. A tempting distractor is option A, overexpressing a histone deacetylase, but this would further compact chromatin and decrease transcription, based on the misconception that deacetylation aids activation rather than repression. For such questions, consider how epigenetic modifiers like HATs and HDACs alter chromatin accessibility and test interventions that compensate for specific defects.

Question 10

A patient is treated with a full course of a specific antibiotic for a severe bacterial infection. The patient's symptoms improve dramatically. However, several weeks later, the symptoms return. The same antibiotic is prescribed again, but this time it has little effect on the infection.

Which of the following best explains, from an evolutionary perspective, why the second antibiotic treatment was ineffective?

The initial antibiotic treatment acted as a selective pressure, allowing a few naturally resistant bacteria to survive and reproduce, leading to a population that is now largely resistant. (correct answer)
The bacteria that survived the first treatment developed an immunity to the antibiotic and passed this acquired trait on to their offspring, making the new population immune.
The patient's immune system was weakened by the initial infection and antibiotic use, preventing it from effectively combating the bacterial population during the second treatment.
The antibiotic caused specific mutations to arise in the bacterial DNA that conferred resistance, allowing the population to adapt directly to the presence of the drug.

Explanation: This is a classic example of natural selection. The antibiotic kills susceptible bacteria, but individuals with pre-existing resistance survive and reproduce. This differential survival and reproduction lead to an increase in the frequency of the resistance allele in the population over time. Choice B describes an incorrect, Lamarckian view of acquired characteristics. Choice C offers a physiological explanation for the patient's condition but not an evolutionary explanation for the bacteria's resistance. Choice D is incorrect because mutations are random events, not directed responses to environmental pressures like antibiotics.

Question 11

A researcher monitors NADH and NAD+ levels in a cell suspension. When oxygen is removed, NADH levels rise and NAD+ levels fall within minutes, even though glucose is still available. ATP production decreases compared with oxygen-present conditions. No inhibitors are added. The observation is limited to changes in electron carrier oxidation state and ATP output under different terminal electron acceptor availability. This scenario focuses on how availability of a terminal electron acceptor affects electron flow, NADH oxidation, and the cell’s capacity to generate ATP via oxidative phosphorylation.

Without oxygen as a terminal electron acceptor, electron transport slows and NADH is less readily oxidized to NAD+ (correct answer)
Without oxygen, NAD+ is rapidly converted into NADH by ATP synthase during proton pumping
Without oxygen, NADH donates phosphate to ADP, causing NADH accumulation and ATP decrease
Without oxygen, ATP increases because electrons remain at higher energy in NADH
Without oxygen, NADH levels rise because NADH is generated directly from oxygen reduction

Explanation: This question assesses the skill of analyzing cellular energy transformations, specifically how terminal electron acceptor availability affects NADH/NAD+ balance and ATP production. The correct answer is A because oxygen absence prevents electron transport chain progression, slowing NADH oxidation to NAD+ and causing NADH accumulation, which reduces ATP via impaired oxidative phosphorylation. Evidence shows NADH rises and NAD+ falls upon oxygen removal despite glucose availability, with decreased ATP, indicating blocked electron flow halts regeneration of NAD+ for glycolysis and TCA cycle. This is based on energy principles where terminal acceptors maintain electron flow, enabling NADH oxidation and proton gradient formation for ATP. A tempting distractor is D, which is incorrect by claiming ATP increases with trapped energy in NADH, arising from the misconception that reduced carriers store usable energy without oxidation. A transferable strategy for cellular energy questions is to monitor redox carrier states to predict impacts on metabolic flux and energy output under varying conditions.

Question 12

Two groups of genetically identical fruit fly larvae are raised on diets that differ only in protein content. Adults from the high-protein diet are larger and have higher expression of insulin-like signaling genes in fat body cells; DNA sequencing shows no differences in those genes. If high-protein larvae are switched to low protein early in development, adult size decreases. Which explanation best accounts for the diet-dependent differences in adult body size?

Diet protein levels altered transcriptional regulation in developing larvae, changing growth-related gene expression without DNA changes. (correct answer)
High protein caused point mutations in insulin-like genes that increased signaling and permanently increased adult size.
Low protein removed insulin-like genes from chromosomes, reducing gene dosage and producing smaller adults.
High protein increased allele frequency for large size in the population, so larger adults became more common.
Low protein prevented meiosis from occurring, so adults inherited fewer chromosomes and became smaller.

Explanation: This question tests understanding of environmental effects on phenotype through diet-dependent body size regulation. The correct answer A explains that diet protein levels altered transcriptional regulation in developing larvae, changing growth-related gene expression without DNA changes - this accounts for the larger size and higher insulin-like gene expression in high-protein flies while DNA sequences remained identical. The fact that switching diets early in development changed adult size confirms this is an environmentally regulated developmental response. Answer B incorrectly suggests high protein caused point mutations, which contradicts the finding of identical DNA sequences between groups - this represents the misconception that all trait differences require genetic mutations. The strategy is to recognize that environmental factors during development can alter gene expression patterns and phenotypes without changing the underlying DNA sequence.

Question 13

A scientist examines two cell sizes of the same species. Both have the same membrane thickness and the same permeability to CO2. Under identical conditions, larger cells show higher internal CO2 levels than smaller cells. Which explanation best accounts for the higher CO2 levels in larger cells?

Larger cells have increased membrane permeability because membrane thickness decreases as cell size increases.
Larger cells have decreased total surface area, so CO2 accumulates due to reduced diffusion across the membrane.
Larger cells have lower surface area–to–volume ratio, limiting CO2 efflux relative to the amount of producing cytoplasm. (correct answer)
Larger cells have greater volume, which lowers CO2 concentration by diluting CO2 and reducing accumulation.
Larger cells have higher surface area–to–volume ratio, causing CO2 to diffuse out more slowly per unit area.

Explanation: This question tests surface area-to-volume ratio roles in CO2 homeostasis. Larger cells exhibit a lower surface area-to-volume ratio, restricting CO2 efflux relative to the producing cytoplasmic volume, causing higher internal levels despite equal permeability. The mismatch between production and export efficiency leads to accumulation in bigger cells. This principle limits cell size in diffusion-reliant organisms. A tempting distractor is choice B, claiming decreased total surface area in larger cells, but total area increases, highlighting a misconception in how area scales with size. For gas exchange problems, analyze surface area-to-volume ratios to understand accumulation trends.

Question 14

A mitotic cell reaches metaphase with all chromosomes aligned at the metaphase plate. However, one chromosome is attached to microtubules from only one spindle pole (monotelic attachment), while all others are bi-oriented. The spindle checkpoint detects the improper attachment and prevents the onset of anaphase. Which outcome is most likely if the improper attachment persists?

The cell enters anaphase on schedule, and sister chromatids separate simultaneously
The cell remains in metaphase with sister chromatids still paired at centromeres (correct answer)
The cell returns to S phase to re-replicate DNA until attachments are correct
The cell completes telophase first, then aligns chromosomes at the equator
The cell undergoes synapsis of homologs and forms tetrads at the equator

Explanation: This question requires analysis of the cell cycle, specifically how the spindle checkpoint responds to improper chromosome attachments during metaphase. The spindle checkpoint monitors whether all kinetochores are properly bi-oriented (attached to opposite poles), and even one improperly attached chromosome prevents anaphase onset. With persistent monotelic attachment, the cell remains in metaphase with sister chromatids still paired at centromeres (answer B), as the checkpoint continuously blocks separase activation and cohesin cleavage until all attachments are corrected. Answer A incorrectly suggests normal anaphase progression, representing a quantitative error where students think most correct attachments are sufficient when actually all chromosomes must be properly attached. The strategy is to remember that the spindle checkpoint operates on an all-or-nothing principle—even one error blocks progression.

Question 15

Two marine food chains begin with the same phytoplankton production of 10,000 kJ/m $^2$ /yr. Chain 1: phytoplankton → krill → penguins. Chain 2: phytoplankton → krill → small fish → seals. Assume ~10% of energy stored at one level becomes new biomass at the next level. Which outcome is most likely when comparing top-level biomass production between the two chains?

Seals will have higher biomass production because longer food chains concentrate energy at the top.
Penguins will have higher biomass production because fewer trophic transfers reduce cumulative energy loss. (correct answer)
Both tops will have equal biomass production because they start with identical phytoplankton production.
Seals will have higher biomass production because energy increases with each consumer level.
Both tops will have near-zero biomass production because energy is recycled back to producers each step.

Explanation: This question assesses the skill of analyzing energy flow through ecosystems by comparing biomass production in marine food chains with different lengths. With ~10% efficiency per transfer, the shorter chain to penguins accumulates less cumulative loss, yielding higher top-level biomass (100 kJ/m²/yr) compared to the longer chain to seals (10 kJ/m²/yr). This occurs because each additional trophic level introduces more energy dissipation through respiration and waste, reducing available energy for top consumers. Therefore, choice B is correct in predicting higher penguin biomass due to fewer transfers. A tempting distractor is choice A, which is wrong because it assumes longer chains concentrate energy, a misconception that reverses the actual pattern of energy dilution up the chain. To compare food chains, multiply the initial production by the efficiency raised to the power of the number of transfers for accurate top-level estimates.

Question 16

In a coastal grass population, blade waxiness varies continuously. After a decade of frequent salt spray, plants with higher waxiness produce more seeds per plant than less waxy plants. Over four generations, the mean waxiness increases from 4.1 to 6.3 (arbitrary units), while the variance stays similar. Which pattern best illustrates the type of natural selection acting on waxiness in this population?

Directional selection shifting the trait mean toward higher waxiness over generations (correct answer)
Stabilizing selection reducing variance by favoring intermediate waxiness each generation
Disruptive selection favoring both low and high waxiness and reducing intermediates
Genetic drift causing random changes in waxiness because selection is not occurring
Individuals increasing waxiness during life and passing that acquired change to offspring

Explanation: This question tests your ability to analyze natural selection patterns by examining changes in trait distribution over time. The data shows that plants with higher waxiness produce more seeds (higher fitness) in response to salt spray, and the mean waxiness increases from 4.1 to 6.3 over four generations while variance remains similar. This pattern indicates directional selection, where selection consistently favors individuals at one end of the trait distribution, causing the population mean to shift in that direction over generations. Students often incorrectly choose stabilizing selection (B) because they focus on the unchanged variance, but stabilizing selection would maintain the original mean while reducing variance by eliminating extremes. When analyzing selection patterns, look for three key indicators: changes in mean (directional), changes in variance (stabilizing reduces it), and changes in distribution shape (disruptive creates bimodality).

Question 17

A lab measures oxygen consumption and ATP synthesis in mitochondria supplied with ADP + Pi and a steady source of NADH. When ADP becomes depleted, oxygen consumption drops sharply even though oxygen is still present. Electron transport proteins remain intact, and no inhibitors are added. The inner membrane remains impermeable to protons except through ATP synthase. Because ATP synthase lacks its substrate, protons accumulate outside the matrix, strengthening the gradient and opposing further proton pumping by the electron transport chain. Which explanation best accounts for the decreased oxygen consumption when ADP is depleted?

Without ADP, glycolysis stops in the cytosol and no longer supplies oxygen to mitochondria
Without ADP, the proton gradient builds up and slows electron transport, reducing oxygen reduction at the end of the chain (correct answer)
Without ADP, oxygen cannot diffuse into mitochondria because ATP synthase normally transports oxygen
Without ADP, the citric acid cycle stops producing CO2, and oxygen consumption requires CO2 release
Without ADP, fermentation begins in mitochondria and replaces oxygen as the terminal electron acceptor

Explanation: This question assesses the skill of analyzing cellular respiration, specifically respiratory control by ADP availability. Without ADP, ATP synthase cannot phosphorylate it, causing protons to accumulate in the intermembrane space and build a steep gradient that opposes further pumping, slowing electron transport and oxygen consumption as in choice B. The stimulus indicates the membrane remains impermeable except through ATP synthase, leading to gradient buildup and reduced electron flow. Oxygen is present, but consumption drops due to inhibited chain activity. A tempting distractor is choice A, which is incorrect because glycolysis supplies pyruvate and NADH, not oxygen, to mitochondria, stemming from the misconception that ADP directly influences cytosolic oxygen delivery. For respiration questions, consider how feedback from proton gradients regulates electron transport and oxygen use in response to energy demand.

Question 18

In a receptor signaling pathway, ligand L activates receptor R, which activates kinase K. K phosphorylates inhibitor I, causing I to release enzyme E. Free E then catalyzes conversion of substrate to product. In cells expressing a mutant I that cannot be phosphorylated, L binding to R is normal but product formation does not increase. Which step is most directly disrupted by the mutation?

Release of enzyme E from inhibitor I due to phosphorylation-dependent conformational change (correct answer)
Ligand L binding to receptor R at the extracellular surface
Activation of kinase K by receptor R through membrane depolarization
Catalysis by enzyme E because phosphorylation of E is required for activity
Transport of product out of the cell by vesicles to complete the response

Explanation: This question assesses the skill of analyzing a signal transduction pathway. The mutation disrupts release of enzyme E from inhibitor I, as phosphorylation of I by K is required for conformational change and E release, leading to no product formation despite normal L binding. In the pathway, L activates R to K, which phosphorylates I to free E for catalysis, and non-phosphorylatable I blocks this. This step is directly affected. Choice D is tempting but wrong because it assumes phosphorylation of E is needed, which is a misconception as E is released, not phosphorylated. A transferable strategy is to analyze mutations in regulatory proteins to uncover inhibition-release mechanisms in signaling.

Question 19

A researcher isolates two membrane fractions from the same eukaryotic cell. Fraction 1 contains proteins embedded in a lipid bilayer and is enriched in cholesterol. Fraction 2 contains proteins embedded in a lipid bilayer but has little cholesterol. At low temperatures, proteins in fraction 1 remain more laterally mobile than proteins in fraction 2. Both fractions have similar phospholipid types and protein densities. Which feature best explains the difference in lateral mobility at low temperature?

Cholesterol reduces tight packing of phospholipids at low temperature, helping maintain membrane fluidity (correct answer)
Cholesterol forms covalent cross-links between proteins, increasing their diffusion rate in the bilayer
Cholesterol replaces phospholipids and creates a single-layer membrane that allows faster movement
Cholesterol pumps protons across the membrane, generating energy that moves proteins laterally
Cholesterol binds ribosomes to the membrane, preventing proteins from aggregating at low temperature

Explanation: This question assesses the analysis of cell structure and function, specifically how cholesterol modulates membrane fluidity. The correct answer is A because the stimulus indicates that fraction 1, enriched in cholesterol, maintains protein mobility at low temperatures compared to fraction 2 with little cholesterol, reflecting AP Biology principles where cholesterol prevents tight phospholipid packing and phase transitions to a gel state. Both fractions have similar phospholipids and proteins, isolating cholesterol's role in fluidity preservation. This allows embedded proteins to diffuse laterally even in cold conditions. A tempting distractor is B, which involves a structure-function confusion by suggesting cholesterol cross-links proteins rather than interacting with lipids to affect bilayer properties. For such questions, compare membrane compositions and relate molecular interactions to physical properties like mobility under varying conditions.

Question 20

Five species (A–E) were analyzed for a DNA marker. Pairwise differences (number of nucleotide substitutions) are: A–B:2, A–C:7, A–D:8, A–E:9; B–C:7, B–D:8, B–E:9; C–D:3, C–E:4; D–E:2. Assume fewer substitutions indicate a more recent common ancestor. Which inference is best supported by the data?

A and E are sister taxa because A differs most from E
C and D are sister taxa because they differ by 3 substitutions
D and E are sister taxa because they differ by 2 substitutions (correct answer)
A and B are not closely related because they differ at 2 sites
B is ancestral to A because their sequences are most similar

Explanation: This question assesses the skill of inferring phylogenetic relatedness from nucleotide substitution data. The pairwise differences show D and E differ by only 2 substitutions, one of the smallest values, supporting they are sister taxa under the assumption that fewer substitutions indicate a more recent common ancestor. Comparatively, C differs from D by 3 and from E by 4, while A and B differ by 2 but are distant from the C-D-E group with differences of 7-9, suggesting two separate close pairs. This pattern is consistent with a tree where (D,E) form a clade sister to C. A tempting distractor is B (C and D as sisters), which might tempt due to their 3-substitution difference, reflecting a misconception of not comparing all pairs holistically, a structure-function confusion in molecular distance interpretation. A transferable strategy for this question type is to identify clusters with minimal internal differences and build a mental tree starting from the smallest pairwise values.

Question 21

Scientists tested whether protocells could grow by incorporating additional lipids. Fatty-acid vesicles were placed in solution, and extra fatty acids were added gradually. Over time, vesicle diameters increased and some vesicles divided into smaller vesicles when gently agitated. A control with no added fatty acids showed little size change. Which claim is best supported by these observations about protocells?

Simple vesicles can grow by acquiring lipids from the environment and can undergo physical division. (correct answer)
Vesicle division requires mitotic spindle fibers and regulated checkpoints.
The data demonstrate that vesicles evolved to divide in order to increase complexity.
The results prove that vesicles can copy DNA genomes with high accuracy.
Fatty acids cannot form abiotically and therefore must have been imported by meteorites.

Explanation: This question assesses the skill of evaluating evidence about the origins of life on Earth. Adding extra fatty acids to vesicles caused them to increase in diameter over time and divide into smaller vesicles with gentle agitation, while controls without added fatty acids showed little change, demonstrating that simple lipid structures can grow by incorporating environmental lipids and undergo physical division abiotically. This illustrates AP Biology principles of protocell evolution, where membrane growth and fission provide a mechanism for reproduction-like processes before genetic systems. The spontaneous nature without biological machinery highlights early steps toward cellular replication. A tempting distractor, choice B, is incorrect because it insists on mitotic spindle fibers and checkpoints for division, representing a level-of-organization error by applying eukaryotic cell cycle regulation to prebiotic entities. For similar questions, examine evidence for abiotic growth and division mechanisms and avoid conflating them with modern cellular processes.

Question 22

A photosynthetic protist contains an organelle where light reactions occur. The organelle has two surrounding membranes, and inside it are membrane stacks resembling thylakoids. The organelle contains circular DNA with sequences most similar to cyanobacterial genes, and its ribosomes are inhibited by chloramphenicol. During cell division, the organelle divides by binary fission, independent of mitosis. Which explanation best accounts for the origin of this organelle?

Infolding of the nuclear envelope created a photosynthetic compartment that later acquired bacterial genes.
Endosymbiosis of a cyanobacterium produced a double-membraned organelle with bacterial DNA and fission. (correct answer)
Spontaneous formation of thylakoid stacks occurred from cytosolic proteins assembling into membranes.
Fusion of multiple vesicles formed a photosynthetic organelle whose DNA originated from the host nucleus.
Engulfment of a mitochondrion converted it into a photosynthetic organelle through gene loss.

Explanation: This question assesses the analysis of the origins of cell compartmentalization. The correct answer, choice B, is indicated by the organelle's double membrane, thylakoid-like stacks, circular DNA similar to cyanobacteria, and ribosomes inhibited by chloramphenicol, all pointing to endosymbiosis of a cyanobacterium as per the AP Biology endosymbiotic theory for chloroplast origins. The independent binary fission during cell division further supports that the organelle retains prokaryotic replication traits from the engulfed symbiont. These features collectively demonstrate how photosynthetic capability was acquired through symbiosis, with the double membrane arising from the host's phagocytic membrane and the bacterium's own membrane. A tempting distractor, choice A, is incorrect due to teleology, as it implies purposeful infolding of the nuclear envelope to create photosynthesis, ignoring the lack of evidence for nuclear membrane involvement and the prokaryotic genetic similarities. To approach similar questions, look for prokaryotic traits like circular DNA and independent division to confirm endosymbiotic rather than host-derived origins.

Question 23

In yeast, gene Y has an enhancer located 1 kb upstream. An activator binds the enhancer and recruits a mediator complex that helps RNA polymerase II assemble at the promoter. In nutrient-rich conditions, the activator is phosphorylated and localized in the nucleus; under starvation, the activator remains unphosphorylated in the cytosol. Starved cells show reduced gene Y pre-mRNA levels compared with nutrient-rich cells. Which change would most likely restore high transcription of gene Y during starvation?

A mutation that adds a nuclear localization signal to the activator protein (correct answer)
A mutation that removes the poly(A) signal from the gene Y transcript
A mutation that increases tRNA abundance for codons used in gene Y
A mutation that changes one amino acid in gene Y without affecting the promoter
An increase in proteasome activity that degrades RNA polymerase II

Explanation: This question assesses understanding of transcriptional regulation, specifically how activator localization and phosphorylation influence eukaryotic gene expression. The activator must be in the nucleus to bind the enhancer and recruit the mediator complex for RNA polymerase II assembly, but during starvation, it remains in the cytosol. Adding a nuclear localization signal would force the activator into the nucleus even without phosphorylation, restoring its ability to enhance transcription under starvation conditions. This bypasses the phosphorylation-dependent localization, directly enabling transcriptional activation. A tempting distractor is choice B, which removes the poly(A) signal, but this is wrong because it impairs mRNA stability and export, reflecting the misconception that post-transcriptional modifications control initiation rates. To approach similar problems, focus on whether the intervention restores the presence or activity of transcriptional regulators at the promoter.

Question 24

A cell is experimentally synchronized so that it exits mitosis and begins G1 at time 0. DNA content is measured every 2 hours and reported as relative units: at 0 h, 2C; at 2 h, 2C; at 4 h, 3C; at 6 h, 4C; at 8 h, 4C. Microscopy at 8 h shows duplicated chromatids but no spindle apparatus. The cell has not divided. Based on the pattern, the cell has completed S phase by 6 h and is in G2 at 8 h. Which observation would best indicate entry into M phase next?

Chromosomes condense and a mitotic spindle begins forming while DNA remains 4C. (correct answer)
DNA content decreases from 4C to 2C while the nuclear envelope remains intact.
Homologous chromosomes pair and align as tetrads while DNA remains 2C.
DNA content increases from 4C to 8C with no change in chromosome condensation.
Two daughter nuclei form before sister chromatids separate at the centromeres.

Explanation: This question requires analysis of the cell cycle, tracking DNA content changes to identify phase transitions. From post-mitosis 2C in G1, DNA increases to 4C by 6 hours via S phase, remaining 4C at 8 hours in G2 with no spindle, indicating preparation for M phase. In AP Biology, entry into M phase involves chromosome condensation and mitotic spindle formation while DNA stays at 4C, marking prophase. This is the next expected observation after G2. A tempting distractor is choice B, suggesting DNA decrease to 2C, but this is incorrect due to a teleology misconception assuming reduction without division. Use time-course data to predict progression and hallmark features of each phase in such problems.

Question 25

In cultured mammalian cells, entry into mitosis correlates with a sharp rise in M-cyclin bound to Cdk1. The complex is inactive when Cdk1 carries an inhibitory phosphate. A checkpoint phosphatase removes this phosphate, rapidly increasing kinase activity and promoting phosphorylation of mitotic target proteins. In an experiment, cells are treated with a small molecule that specifically inhibits the checkpoint phosphatase but does not affect cyclin levels or Cdk1 abundance. After treatment, cells accumulate with duplicated DNA and intact nuclear envelopes.

Which change would most likely restore progression into mitosis under these conditions?

Increase inhibitory phosphorylation of Cdk1 to strengthen the G2 checkpoint signal
Activate a kinase that adds inhibitory phosphate groups to Cdk1
Add a compound that prevents inhibitory phosphorylation on Cdk1 (correct answer)
Block spindle microtubule attachment to delay anaphase until alignment occurs
Increase production of M-cyclin by stimulating gene expression in G2 phase

Explanation: Signaling-based regulation of the cell cycle is a key skill in understanding how checkpoints and molecular interactions control progression through phases like mitosis. In this experiment, the small molecule inhibits the checkpoint phosphatase, preventing removal of the inhibitory phosphate from Cdk1, which keeps the M-cyclin–Cdk1 complex inactive and halts mitotic entry, leading to accumulation in G2 with duplicated DNA and intact nuclear envelopes. Adding a compound that prevents inhibitory phosphorylation on Cdk1 would bypass the need for the phosphatase, allowing Cdk1 activation and restoring mitotic progression. This directly counters the treatment's effect by ensuring the complex becomes active without phosphatase intervention. A tempting distractor is choice E, which suggests increasing M-cyclin production, but this misinterprets the issue as a cyclin shortage rather than phosphorylation status, as cyclin levels are unchanged. To approach similar problems, always identify the specific blocked step in the pathway and look for options that directly circumvent that block.

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