AP Biology

AP Biology Practice Test: Practice Test 6

Practice Test 6 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A cell population is synchronized in metaphase. The spindle-assembly checkpoint prevents anaphase when kinetochores are unattached by generating a signal that inhibits the anaphase-promoting complex/cyclosome (APC/C), an E3 ligase that targets securin for degradation. When securin is degraded, separase becomes active and cleaves cohesin, allowing sister chromatids to separate. A drug is added that stabilizes kinetochore–microtubule attachments, reducing checkpoint signaling, but does not affect cohesin directly. Which change would most likely prevent premature anaphase in the presence of the drug?

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Question 1

Decrease APC/C activity so securin remains high and separase stays inhibited (correct answer)
Increase separase activity to accelerate cohesin cleavage and chromatid separation
Increase cyclin-CDK activity to strengthen the G1/S checkpoint signal
Prevent DNA polymerase loading to delay S phase until attachments form
Raise intracellular calcium to trigger cytokinesis before anaphase begins

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, focusing on the spindle-assembly checkpoint and APC/C inhibition. The drug stabilizes kinetochore-microtubule attachments, reducing checkpoint signaling that normally inhibits APC/C, which could lead to premature securin degradation and anaphase. Decreasing APC/C activity would maintain high securin levels, keeping separase inhibited and preventing premature chromatid separation. This counteracts the drug by mimicking sustained checkpoint inhibition of APC/C. A tempting distractor is B, which suggests increasing separase activity, but this would promote separation without addressing APC/C regulation, stemming from the misconception that direct enzyme activation overrides upstream controls. To approach similar problems, trace how perturbations affect checkpoint signals and choose options that restore balance to prevent untimely progression.

Question 2

In a tropical rainforest, a vine grows on tree trunks to reach sunlight. Measurements show that trees with heavy vine loads have reduced leaf area and slower growth compared with nearby trees without vines. The vine’s growth rate is higher on trees than on the forest floor. The vine does not penetrate tree tissues, but it shades the tree’s leaves and adds weight to branches. Which interaction best describes the relationship between the vine and the tree?

Mutualism, with vines increasing tree growth while trees provide structural support
Commensalism, with vines benefiting from support while trees are unaffected
Competition, with vines reducing tree access to light and space while using the same resources (correct answer)
Predation, with vines consuming tree leaves as their primary energy source
Parasitism, with vines living inside tree vascular tissue and extracting sugars directly

Explanation: This question assesses the skill of analyzing community ecology by classifying epiphytic relationships based on growth impacts. The vine uses the tree for support to reach light, but heavy loads reduce tree leaf area and growth by shading and adding weight, indicating competition where both vie for light and space resources. The vine's higher growth on trees versus the floor shows it benefits from the association, while harming the host without tissue penetration. This interaction logic positions competition as the dynamic, with the vine exploiting shared resources at the tree's expense. A tempting distractor is E, parasitism, which is incorrect because the vine does not extract nutrients internally, due to the misconception that all climbing plants act as internal parasites. For epiphyte studies, quantify host growth metrics with and without the associate to distinguish competition from other harms like parasitism.

Question 3

A coastal marsh has measured annual energy stored as new biomass: cordgrass 20,000 kJ/m $^2$ /yr, snails 2,000 kJ/m $^2$ /yr, crabs 200 kJ/m $^2$ /yr. These values represent energy available to the next trophic level. Which conclusion is best supported about energy flow through this food chain?

Energy transfer between trophic levels is roughly 10% at each step in this chain. (correct answer)
Crabs receive 200 kJ/m $^2$ /yr directly from cordgrass because energy bypasses snails.
Energy increases at higher trophic levels because consumers combine energy from many producers.
Snails are decomposers, so their biomass energy should exceed cordgrass biomass energy.
Energy is conserved within the chain, so total stored energy across levels must equal 20,000 kJ/m $^2$ /yr.

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 snails store 2,000 kJ/m²/yr, exactly 10% of cordgrass's 20,000 kJ/m²/yr, and crabs store 200 kJ/m²/yr, 10% of snails', demonstrating consistent trophic transfer. This pattern arises from energy losses via heat, excretion, and uneaten biomass, adhering to the 10% rule. The data illustrate unidirectional energy flow, decreasing at each step. A tempting distractor, C, is wrong because it suggests energy increases upward, a misconception confusing energy accumulation with the reality of dissipation. To analyze similar problems, verify if transfers approximate 10% by calculating ratios and assess overall flow direction.

Question 4

A toxin blocks ATP production in cells. Shortly after, an ATP-dependent proton pump in the plasma membrane stops moving H $^+$ out of the cytosol, even though the H $^+$ concentration is higher outside than inside. When ATP is restored, the pump resumes transporting H $^+$ outward. Which mechanism best explains H $^+$ movement by this pump when ATP is present?

Facilitated diffusion of H $^+$ down its electrochemical gradient through a channel
Primary active transport of H $^+$ driven directly by ATP hydrolysis (correct answer)
Simple diffusion of H $^+$ through the hydrophobic lipid interior
Secondary active transport using glucose movement to power H $^+$ export
Osmosis of H $^+$ through aquaporins with water movement

Explanation: This question assesses the skill of analyzing membrane transport mechanisms. Primary active transport is correct because the pump uses ATP hydrolysis directly to move H+ against its gradient, stopping without ATP despite favorable gradient. Resumption with ATP confirms direct energy use. Phosphorylation or ATP sites would support this. A tempting distractor is facilitated diffusion, incorrect due to the misconception that uphill movement can be passive, overlooking energy needs. To analyze transport, always evaluate if movement is down a gradient (passive) or against it (active) and check for energy requirements.

Question 5

A coastal marsh is partially drained for development, reducing flooded area by 40%. Over the next year, sediment trapped by marsh plants decreases and water turbidity in nearby channels increases. Which explanation best accounts for the increased turbidity?

Loss of marsh vegetation reduces sediment capture, allowing more particles to remain suspended in moving water. (correct answer)
Reduced marsh area increases groundwater salinity, which directly forms suspended mineral particles in the channels.
Draining marshes increases channel depth, so sunlight reaches the bottom and resuspends sediment by photosynthesis.
Development increases predator abundance, which decreases fish grazing and therefore increases suspended sediments.
Marsh drainage increases rainfall frequency, which directly causes more suspended particles in the channels.

Explanation: This question assesses the skill of analyzing disruptions in ecosystems by evaluating the effects of partially draining a coastal marsh for development. Draining reduces marsh vegetation, which normally traps sediments, leading to less sediment capture and more particles remaining suspended in water channels, increasing turbidity. This change disrupts the ecosystem's filtration services, affecting light penetration and aquatic life at a system level. Consequently, the altered hydrology exacerbates erosion and sediment resuspension, impacting overall water quality and habitat stability. A tempting distractor, like choice B, claims reduced marsh area increases groundwater salinity to form suspended particles, but this confuses salinity effects with sediment dynamics, a misconception ignoring the primary role of vegetation in physical sediment trapping. When assessing ecosystem disruptions, map out changes in physical processes like sediment flow and their links to biological components for a comprehensive view.

Question 6

A membrane-associated enzyme is assayed in solutions containing different detergent concentrations while keeping pH, temperature, and substrate constant. With 0% detergent, activity is 18 units; with 1% detergent, activity is 3 units. Detergents can insert into hydrophobic regions of proteins and disrupt hydrophobic interactions that help maintain tertiary structure and the shape of the active site. The enzyme amount added is the same in both trials. Which outcome is most likely at 1% detergent?

Activity decreases because detergent disrupts hydrophobic interactions, altering protein folding and changing active-site geometry. (correct answer)
Activity decreases because detergent removes substrate from solution by converting it into a nonreactive product.
Activity increases because detergent always improves enzyme-substrate binding by making the active site more hydrophobic.
Activity decreases because detergent prevents translation of the enzyme, reducing the number of enzyme molecules present.
Activity is unchanged because detergents affect only lipid bilayers and cannot interact with protein structure.

Explanation: This question assesses the skill of analyzing environmental impacts on enzyme function, specifically how detergents affect enzyme activity. The correct answer is choice A because the stimulus indicates that detergents disrupt hydrophobic interactions by inserting into nonpolar regions, altering protein folding and active-site geometry. This leads to decreased activity from 18 units at 0% detergent to 3 units at 1% detergent, with unchanged enzyme amount. Protein structure logic supports that such disruptions affect tertiary structure without impacting substrate directly or translation processes. A tempting distractor is choice C, which is wrong because it claims detergents always improve binding, reflecting the misconception that increasing hydrophobicity universally aids catalysis rather than potentially denaturing the enzyme. A transferable strategy for interpreting enzyme-environment questions is to evaluate how agents like detergents target specific structural elements, such as hydrophobic cores, and assess their impact on folding versus direct substrate interactions.

Question 7

Two separate populations of the same fish species live in similar lakes with no migration between them. Both populations are large. At a neutral microsatellite locus, allele $M$ starts at frequency 0.50 in each lake. After 40 generations, lake 1 has $M=0.12$ and lake 2 has $M=0.87$ , with no consistent fitness differences among genotypes detected. Which factor most likely explains the divergence in allele frequency between lakes?

Directional selection acting identically in both lakes on allele $M$
Independent genetic drift causing random allele-frequency changes in each lake (correct answer)
Gene flow between lakes equalizing allele frequencies over time
High mutation rate driving allele $M$ toward the same equilibrium in both
Nonrandom mating increasing heterozygosity and preventing divergence

Explanation: This question examines population genetics in isolated populations, specifically how genetic drift causes divergence. Despite starting with identical allele frequencies (M=0.50) and living in similar environments with no fitness differences among genotypes, the two lake populations diverged dramatically (lake 1: M=0.12, lake 2: M=0.87) over 40 generations. The absence of migration prevents gene flow, and the lack of fitness differences rules out selection. Students might incorrectly assume large populations are immune to drift, but even large populations experience some random sampling effects, and over many generations these can accumulate to produce substantial divergence. The key insight is that isolated populations will diverge through independent genetic drift even when they start identically and experience similar selective pressures.

Question 8

A researcher examines a cell in prophase I and observes synapsis: each chromosome is tightly paired with its homolog along much of its length, forming tetrads. Which event is most directly enabled by synapsis at this stage of meiosis?

Separation of sister chromatids into different cells during the first meiotic division.
Independent alignment of individual chromosomes at the metaphase plate during meiosis II.
Exchange of segments between nonsister chromatids within a homologous pair. (correct answer)
Replication of DNA to produce sister chromatids prior to chromosome condensation.
Formation of diploid nuclei after cytokinesis so that chromosome number is restored.

Explanation: This question tests understanding of synapsis and its role in enabling crossing over during prophase I. Synapsis is the tight pairing of homologous chromosomes along their length, forming tetrads where each chromosome consists of two sister chromatids. This close physical alignment directly enables crossing over—the exchange of DNA segments between nonsister chromatids of the paired homologs. Without synapsis bringing homologs together, crossing over couldn't occur because the nonsister chromatids wouldn't be positioned for exchange. Students might choose option B, confusing prophase I events with metaphase II, but independent alignment occurs later and in a different meiotic division. The strategy for understanding meiotic events is to connect each structural arrangement (like synapsis) with its functional purpose (enabling crossing over).

Question 9

In a prairie flower population, stalk height varies. Grazers preferentially eat the tallest plants, while very short plants are shaded and set fewer seeds. Plants of intermediate height produce the most seeds. After several generations, extreme heights become less common and the mean height changes little. Which pattern best illustrates the selection acting on stalk height?

Disruptive selection increasing both extremes because extremes avoid both grazing and shade
Directional selection decreasing mean height because only short plants reproduce successfully
Stabilizing selection reducing extreme heights by favoring intermediate stalk height (correct answer)
Genetic drift reducing extremes because grazers remove plants regardless of height
Plants growing to intermediate height during life and passing that change to offspring

Explanation: This question tests your ability to analyze natural selection patterns by identifying multiple selective pressures acting on a trait. Grazers eat the tallest plants while very short plants are shaded and set fewer seeds, meaning both extremes have reduced fitness while intermediate heights produce the most seeds. After several generations, extreme heights become less common and the mean changes little, demonstrating stabilizing selection that favors intermediate phenotypes. Students often choose disruptive selection (A) when they see two different selective pressures, but the key is recognizing that both pressures eliminate extremes rather than favoring them. When analyzing complex selection scenarios, determine which phenotypes have highest fitness: if intermediates survive best due to avoiding multiple threats, it's stabilizing selection.

Question 10

A scientist places cells in a solution containing 100 mM solute T; cytosolic T is 10 mM. Cells express a transporter specific for T. When the transporter is inhibited, intracellular T remains low. When the inhibitor is removed, T increases until near 100 mM, with no detectable ATP consumption. Which explanation best accounts for the increase in intracellular T after inhibitor removal?

T enters by facilitated diffusion through the transporter, moving down its concentration gradient. (correct answer)
T enters by active transport, and removing inhibitor restores ATP hydrolysis by the pump.
T enters by simple diffusion, and inhibitor removal increases bilayer permeability to ions.
T enters by exocytosis, and inhibitor removal increases vesicle release into the cell.
T enters only when internal concentration exceeds external concentration, reversing diffusion.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as T enters via facilitated diffusion down its gradient from 100 mM to 10 mM through the transporter, increasing to equilibrium after inhibitor removal without ATP consumption. Inhibitor blocks the passive pathway. Gradients drive equalization. A tempting distractor is B, which claims active pumping restored, based on energy misconception. When inhibition reversal allows gradient equalization without energy, identify facilitated diffusion.

Question 11

A ligand binds a receptor and rapidly increases intracellular cAMP. When cells are treated with a drug that inhibits phosphodiesterase (PDE), the peak cAMP level is higher and returns to baseline more slowly. Ligand binding to the receptor is unchanged. Which of the following best explains the early effect of PDE inhibition on the signaling pathway?

PDE normally breaks down cAMP, so inhibiting PDE increases cAMP accumulation and duration (correct answer)
PDE normally synthesizes cAMP from AMP, so inhibiting PDE should lower cAMP levels
PDE inhibition prevents receptor-ligand binding by altering extracellular pH around the receptor
PDE inhibition blocks cAMP transport into the nucleus, causing cytosolic cAMP to rise
PDE inhibition increases cAMP because cells detect drug exposure and amplify signaling to compensate

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on the transduction step degrading second messengers. The correct answer is A because PDE hydrolyzes cAMP, and its inhibition allows cAMP to accumulate higher and longer, as shown by altered levels post-ligand despite normal binding. This fits basic signaling principles of messenger turnover controlling signal duration. The rapid increase indicates transduction involvement. A tempting distractor is B, which is incorrect because it misattributes synthesis to PDE, a misconception of its catabolic role. In signal transduction questions, identify degradative enzymes to predict effects on messenger persistence.

Question 12

A freshwater fish population contains heritable variation in tolerance to low dissolved oxygen, influenced by alleles O (higher tolerance) and o (lower tolerance). During summer, algal blooms repeatedly reduce dissolved oxygen for several weeks, and fish with lower tolerance die at higher rates before spawning. Survivors reproduce within the same lake, and offspring oxygen tolerance resembles parental genotypes. Migration into the lake is rare. Which outcome is most likely after several bloom seasons?

The o allele increases because low oxygen causes fish to switch on genes that create lower tolerance.
The frequency of allele O increases because more O-bearing fish survive to reproduce during low oxygen. (correct answer)
Allele frequencies stay constant because the bloom affects all fish equally regardless of genotype.
All fish become highly tolerant within one generation because exposure changes their DNA permanently.
Both alleles disappear because environmental stress prevents inheritance of oxygen tolerance.

Explanation: This question tests understanding of natural selection, the process where heritable traits that improve survival and reproduction become more common in a population over generations. Repeated algal blooms reduce oxygen, causing higher mortality in fish with low-tolerance allele o before spawning, while high-tolerance fish with allele O survive and reproduce more often. This differential survival leads to an increase in the frequency of allele O, as surviving fish pass on the trait to a larger share of the next generation. At the population level, the heritable variation in oxygen tolerance drives this change under the consistent selective pressure of low oxygen. A tempting distractor is choice D, which incorrectly states that exposure permanently changes DNA in all fish, reflecting the misconception that environments directly alter genes within a generation. For natural selection questions, always identify the environmental pressure, the heritable variation it acts on, and how it leads to changes in allele frequencies through differential reproduction.

Question 13

A DNA polymer is treated with an enzyme that specifically breaks hydrogen bonds but does not break covalent bonds. After treatment, the sample contains two intact nucleotide chains of the same length as the original molecule. Which statement best describes what bonds were disrupted?

Phosphodiester bonds between adjacent nucleotides were broken, separating the backbone into monomers.
Covalent bonds between deoxyribose and phosphate were broken, shortening both strands.
Hydrogen bonds between complementary bases were broken, separating the two strands without cutting backbones. (correct answer)
Ionic bonds between phosphate groups were broken, causing bases to detach from sugars.
Glycosidic bonds between bases and sugars were broken, releasing free nitrogenous bases.

Explanation: This question requires analyzing nucleic acids as macromolecules to distinguish between bond types in DNA structure. The enzyme broke hydrogen bonds between complementary bases, separating the two strands while leaving both sugar-phosphate backbones intact, as indicated by the presence of two chains of original length. Hydrogen bonds between A-T and G-C base pairs are the only non-covalent interactions holding the two strands together in the double helix, while phosphodiester bonds (covalent) maintain backbone integrity within each strand. Choice A incorrectly identifies phosphodiester bonds as broken, which would fragment the backbone into individual nucleotides—this represents a level-of-organization error confusing intermolecular forces (between strands) with intramolecular bonds (within strands). The strategy for DNA denaturation questions is to recognize that gentle treatments break only hydrogen bonds between strands, while harsh treatments break covalent bonds within strands.

Question 14

A researcher compares cytochrome c amino acids among four species (1–4). The number of differences from species 1 are: species 2 differs by 4, species 3 differs by 4, species 4 differs by 9. Additional comparisons show species 2 and 3 differ by 1, species 2 and 4 differ by 10, and species 3 and 4 differ by 10. Assume fewer differences indicate a more recent common ancestor among lineages. Which inference is best supported?

Species 1 and 4 are sister taxa because their difference is largest
Species 2 and 3 are most closely related to each other (correct answer)
Species 1 is equally closely related to species 2, 3, and 4
Species 4 is most closely related to species 2 because both differ from 3
Species 1 and 2 share a more recent common ancestor than species 2 and 3

Explanation: This question assesses the skill of inferring phylogenetic relatedness from protein sequence differences. Species 2 and 3 differ by only 1 amino acid in cytochrome c, the fewest among pairs, indicating they share the most recent common ancestor under the AP Biology molecular clock assumption where fewer differences reflect less divergence time. Comparisons from species 1 show 2 and 3 both differ by 4, while 4 differs by 9, and cross-pairs like 2-4=10, 3-4=10, confirm 2 and 3 are closest. This protein is conserved, making differences reliable for phylogeny. A tempting distractor is choice E (1 and 2 share more recent ancestor than 2 and 3), but 1-2=4 vs 2-3=1 shows otherwise, due to a structure-function confusion mistaking reference species differences for overall relatedness. Always compute all pairwise differences and select the minimal for the closest pair in such analyses.

Question 15

Two cells have plasma membranes with identical phospholipid composition and thickness. Cell X has many transmembrane carrier proteins specific for glucose; Cell Y has far fewer of these carriers. Both cells are placed in a solution with glucose concentration higher outside than inside. Glucose enters Cell X faster than Cell Y, even though glucose is polar. Which membrane feature best explains the higher glucose uptake rate?

Which feature best explains why glucose enters Cell X faster than Cell Y?

A higher density of specific carrier proteins increases facilitated diffusion of glucose (correct answer)
More carrier proteins increase phospholipid tail saturation, raising glucose solubility
Fewer carriers in Cell Y increase membrane fluidity, preventing glucose from entering
Glucose crosses faster in Cell X because polar molecules diffuse best through lipid cores
Carrier proteins in Cell X reduce extracellular glucose concentration, increasing diffusion force

Explanation: This question assesses the skill of analyzing plasma membrane structure and transport. Glucose enters Cell X faster because it has a higher density of specific carrier proteins that facilitate diffusion of the polar glucose molecules down their concentration gradient. Despite identical phospholipid composition, the abundance of carriers in Cell X provides more pathways for glucose, which cannot easily cross the hydrophobic bilayer without assistance. This explains the higher uptake rate even though glucose is polar and the gradient is the same for both cells. A tempting distractor is choice B, which mistakenly states more carriers increase tail saturation to raise glucose solubility, ignoring that saturation affects fluidity, not polar solute solubility in lipids. To compare transport rates, consider how the number of specific transport proteins influences facilitated diffusion of polar molecules.

Question 16

A signaling molecule Q binds a membrane receptor R on Cell Z. When excess unlabeled Q is added, binding of fluorescent Q to Cell Z decreases. A different molecule, Q2, similar in size and charge to Q, does not reduce fluorescent Q binding even at high concentration. Cell Z shows no response to Q when receptor R is removed from the membrane. Which of the following best explains these observations? Which of the following best explains these observations?

Q binds specifically to receptor R, and unlabeled Q competitively occupies the same binding site (correct answer)
Q2 blocks signaling by entering the nucleus and preventing Q synthesis
Q binding is nonspecific, so any similar molecule should reduce fluorescent Q binding equally
Removing receptor R increases Q binding by exposing more phospholipids for attachment
Fluorescent Q binds irreversibly, so unlabeled Q decreases fluorescence by quenching the dye

Explanation: This question assesses understanding of cell communication via signal transduction pathways, emphasizing specific ligand-receptor interactions. Excess unlabeled Q reducing fluorescent Q binding, while Q2 does not, demonstrates competitive inhibition at the same specific site on receptor R, with unlabeled Q occupying sites and preventing fluorescent binding. The lack of response without receptor R confirms its necessity for Q signaling, ruling out nonspecific attachment to membrane lipids. Q2's ineffectiveness despite similarity in size and charge highlights binding specificity rather than general molecular properties or nuclear interference. A tempting distractor is choice C, reflecting the misconception that binding is nonspecific and any similar molecule competes equally, but specificity arises from precise ligand-receptor matches. In binding studies, use competition with labeled and unlabeled ligands to verify specificity and receptor dependence.

Question 17

An unknown macromolecule is composed of a long chain of repeating units, each containing an amino group (–NH2), a carboxyl group (–COOH), and a variable R group. Adjacent units are connected by covalent bonds formed when the carboxyl group of one unit reacts with the amino group of the next, releasing water. The resulting polymer can fold into specific shapes based on interactions among R groups, influencing catalytic activity and binding. Which feature best explains why this molecule is classified as a protein?

It contains nucleotides with phosphate groups that form a sugar-phosphate backbone.
It consists of fatty acids ester-linked to glycerol, creating a hydrophobic molecule.
It is a polymer of amino acids joined by peptide bonds formed through dehydration synthesis. (correct answer)
It is a polymer of monosaccharides linked by glycosidic bonds for short-term energy.
It is a chain of repeating phosphate groups that store energy in their bonds.

Explanation: This question requires analyzing macromolecule structure-function relationships to identify protein characteristics. The molecule contains amino acids (units with –NH2, –COOH, and variable R groups) connected by peptide bonds formed through dehydration synthesis between carboxyl and amino groups, which definitively identifies it as a protein. The ability to fold into specific shapes based on R group interactions and influence catalytic activity further confirms protein identity, as these are hallmark protein functions. Option D incorrectly suggests carbohydrate classification, representing a monomer confusion error since carbohydrates are built from monosaccharides with hydroxyl groups, not amino acids with amino and carboxyl groups. To identify macromolecules, match the monomer structure (amino group + carboxyl group + R group = amino acid) to the correct category and verify the bond type matches.

Question 18

In a freshwater fish population, body coloration ranges from very pale to very dark. Predators more easily detect intermediate-colored fish against patchy substrates, while very pale fish blend with sand and very dark fish blend with rocks. Over ten generations, the frequency of intermediate-color phenotypes decreases, while both pale and dark phenotypes increase and the distribution becomes bimodal. Which pattern best illustrates the selection acting on body coloration?

Stabilizing selection favoring intermediates, narrowing variation around a single peak
Directional selection favoring darker fish, shifting the entire distribution toward one extreme
Genetic drift producing random changes in coloration unrelated to predator detection
Disruptive selection favoring both extremes, reducing intermediates and increasing bimodality (correct answer)
Fish choose habitats that change their color during life, altering allele frequencies directly

Explanation: This question tests your ability to analyze natural selection patterns by examining changes in phenotype frequency distributions. The scenario describes intermediate-colored fish being more easily detected by predators, while extreme phenotypes (pale and dark) have camouflage advantages, leading to decreased intermediate frequencies and increased extreme frequencies with a bimodal distribution. This perfectly describes disruptive selection, where extreme phenotypes have higher fitness than intermediates, creating two peaks in the distribution. Choice A incorrectly suggests stabilizing selection, but that would favor intermediates and create a single narrow peak, not the observed bimodal pattern. When you see selection against intermediates producing a bimodal distribution with increased extreme frequencies, identify it as disruptive selection.

Question 19

Two bacterial cells communicate using a secreted molecule AHL that accumulates in the environment. In low-density culture, no response occurs. At high density, many cells respond simultaneously. If AHL is removed by frequent medium replacement, the response does not occur even at high density. Which of the following best explains this signaling behavior?

Which of the following best explains why the response requires high cell density?

High density increases AHL accumulation, allowing receptor binding once a threshold concentration is reached (correct answer)
High density causes AHL to become membrane-bound, enabling contact-dependent signaling
High density forces AHL to bind DNA directly, bypassing receptors
Medium replacement increases receptor number by stimulating transcription immediately
AHL cannot diffuse, so only crowded cells can physically push it into neighbors

Explanation: This question assesses understanding of cell communication via signal transduction pathways. At high density, AHL accumulates to a threshold concentration allowing receptor binding and response, as removing AHL by medium replacement prevents response even at high density. In low-density culture, AHL does not accumulate sufficiently, but high density enables quorum sensing via increased local concentration. This density-dependent behavior relies on AHL diffusion and accumulation for transduction. A tempting distractor is choice B, claiming high density makes AHL membrane-bound, but this reflects the misconception that quorum sensing involves contact-dependent signaling, whereas it uses diffusible molecules. To approach similar questions, evaluate how signal accumulation thresholds drive density-dependent responses in bacterial communication.

Question 20

Two solutions contain nonpenetrating solutes: Solution 1 is 0.12 M and Solution 2 is 0.30 M. Identical cells have internal solute concentration 0.20 M, and membranes allow water movement only. A cell placed in Solution 1 increases in volume. Which outcome is most likely for a cell placed in Solution 2?

It increases in volume because Solution 2 has higher solute concentration.
It decreases in volume because Solution 2 is hypertonic to the cell. (correct answer)
It remains unchanged because both solutions are hypotonic to the cell.
It bursts because solute enters the cell and draws in water.
It remains unchanged because water movement requires ATP.

Explanation: This question tests tonicity and osmoregulation by using comparative observations to predict cell behavior. The cell with 0.20 M internal solute swells in Solution 1 (0.12 M), confirming that Solution 1 is hypotonic and water enters the cell. Solution 2 contains 0.30 M solute, which is higher than the cell's internal concentration of 0.20 M, making Solution 2 hypertonic to the cell. In hypertonic solutions, water moves from inside the cell (higher water potential, lower solute) to outside the cell (lower water potential, higher solute), causing the cell to shrink or decrease in volume. A misconception is that both solutions being on opposite sides of the cell's concentration would produce no change (choice C), but each solution must be evaluated independently. Use the cell's internal concentration as a reference point—solutions below it are hypotonic (cells swell) and solutions above it are hypertonic (cells shrink).

Question 21

A diploid cell with $2n=4$ completes DNA replication before meiosis. At metaphase I, both homologous pairs align. In anaphase I, homologous chromosomes separate normally. After meiosis I, each daughter cell contains two chromosomes, each still composed of two sister chromatids. Without another round of DNA replication, meiosis II occurs. Which change in chromosome structure occurs during anaphase II?

Homologous chromosomes separate, reducing chromosome number from $2n$ to $n$ .
Sister chromatids separate when centromeres divide, forming single-chromatid chromosomes. (correct answer)
Tetrads form as homologous chromosomes synapse and exchange segments.
Chromosomes replicate again, restoring two sister chromatids per chromosome.
Homologous chromosomes pair at the metaphase plate to ensure identical gametes.

Explanation: This question focuses on the specific events of anaphase II in meiosis. After meiosis I, each daughter cell has the haploid number of chromosomes (n=2), but each chromosome still consists of two sister chromatids joined at the centromere. During anaphase II, centromeres divide and sister chromatids separate, converting each double-chromatid chromosome into two single-chromatid chromosomes that move to opposite poles. This is the same mechanism as mitotic anaphase but occurs in haploid cells rather than diploid cells. Students choosing A confuse this with meiosis I events, incorrectly thinking homologs separate again in meiosis II. To distinguish meiosis I from meiosis II events, remember that homolog separation (meiosis I) reduces chromosome number while sister chromatid separation (meiosis II) changes chromosome structure from double to single chromatids.

Question 22

A cell is placed in a solution containing a polar solute that cannot cross the lipid bilayer. When a specific membrane protein is added to the bilayer, the solute begins moving down its concentration gradient. Which feature best explains the change?

The added transmembrane protein provides a hydrophilic pathway for facilitated diffusion of the polar solute (correct answer)
The added protein converts the polar solute into a nonpolar solute that dissolves in the bilayer
The added protein increases the number of phospholipids, raising osmotic pressure and pulling solute inward
The added protein forms covalent bonds with the solute, dragging it across against its gradient
The added protein disrupts the cell wall, allowing the solute to diffuse through cellulose pores

Explanation: This question assesses the skill of analyzing cell structure-function relationships. The added transmembrane protein forms a hydrophilic channel or carrier, enabling facilitated diffusion of the polar solute down its gradient, which couldn't cross the hydrophobic bilayer alone. This illustrates membrane transport in AP Biology, where specific proteins provide pathways for polar molecules, increasing permeability without energy input. The protein's integration allows passive movement once added. A tempting distractor is choice B, which is incorrect due to structure-function confusion, as proteins facilitate passage but do not chemically convert solutes; that's enzymatic function elsewhere. To approach similar questions, identify barriers to diffusion and how proteins alter membrane selectivity for specific solutes.

Question 23

A fluorescent reporter indicates cyclin E–CDK2 activity. In control cells, activity rises in late G1 and peaks at the G1/S transition. When cells are treated with a phosphatase inhibitor, CDK2 remains phosphorylated at an inhibitory site, and reporter signal stays low despite normal cyclin E binding. Which change would most likely restore reporter signal in the presence of the phosphatase inhibitor?

Add a CDK2 variant that lacks the inhibitory phosphorylation site (correct answer)
Activate APC/C to degrade cyclin E earlier in G1
Increase kinetochore detachment to delay anaphase onset
Inhibit separase so cohesin is not cleaved during anaphase
Upregulate cyclin E gene expression to raise reporter signal indirectly

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically inhibitory phosphorylation regulating cyclin E–CDK2 at the G1/S transition. The phosphatase inhibitor keeps CDK2 phosphorylated and inactive, reducing reporter signal despite cyclin E binding. Adding a CDK2 variant lacking the inhibitory site, as in choice A, would prevent phosphorylation and restore activity. This counters the inhibitor's effect on the phosphorylation state. A tempting distractor is choice E, which proposes upregulating cyclin E expression, but this misconceives that higher cyclin levels activate a phosphorylatable CDK2. A transferable strategy is to modify target sites to evade regulatory modifications in signaling pathways.

Question 24

In a signaling assay, a CDK inhibitor binds and inactivates multiple cyclin–CDK complexes. The inhibitor is normally sequestered by a phosphorylated docking protein during late G1, reducing inhibitor availability and increasing CDK activity. A mutation prevents phosphorylation of the docking protein, so the inhibitor remains free. Which outcome is most likely in mutant cells?

Reduced cyclin–CDK activity and delayed progression from G1 into S phase (correct answer)
Increased APC/C activity and earlier anaphase onset in mitosis
Increased CDK1 activation and earlier entry into mitosis from G2
Repeated initiation of DNA replication because inhibitor binding promotes origin firing
Normal cycling because cells can compensate by increasing cyclin gene expression

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically inhibitor sequestration by phosphorylated docking proteins in G1. The mutation prevents docking protein phosphorylation, keeping the inhibitor free to inactivate CDKs. This reduces cyclin–CDK activity and delays G1 to S progression, as in choice A. The outcome aligns with increased inhibitor availability in the assay. A tempting distractor is choice C, which predicts earlier mitosis, but this misconceives that free inhibitor activates rather than inhibits CDKs. A transferable strategy is to trace mutations affecting sequestration to their impact on inhibitor efficacy.

Question 25

A receptor tyrosine kinase (RTK) binds ligand V on the cell surface. In cells expressing a mutant RTK that can bind V but cannot dimerize, V binding occurs normally but signaling does not proceed. Which of the following best explains why signaling fails?

Which of the following best explains why V binding does not trigger signaling in the dimerization-defective RTK?

RTK dimerization is required to activate the receptor’s intracellular signaling capability after ligand binding (correct answer)
RTKs function only inside the nucleus, so dimerization at the membrane is irrelevant
Dimerization prevents ligand binding, so binding assays must be incorrect
RTKs signal by diffusing through gap junctions into neighboring cells
Mutant RTKs convert V into an antagonist that blocks all receptors in the medium

Explanation: This question assesses understanding of cell communication via signal transduction pathways. RTK dimerization is required to activate the receptor’s intracellular signaling capability after ligand V binding, as the mutant RTK binds V but cannot dimerize, preventing transduction. Normal RTKs dimerize upon binding to enable kinase activity and pathway initiation. The mutation specifically blocks dimerization, explaining failed signaling despite binding. A tempting distractor is choice C, claiming dimerization prevents binding, but this reflects the misconception that dimerization inhibits receptors, contradicting binding assay evidence. To approach similar questions, examine receptor activation mechanisms like dimerization in kinase signaling.

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