AP Biology

AP Biology Practice Test: Practice Test 12

Practice Test 12 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

In a rabbit population, fur thickness varies. During unusually cold winters, rabbits with intermediate fur thickness leave the most surviving offspring; very thin and very thick fur rabbits leave fewer. After six generations, the population mean remains near the original value, but the proportion of extreme phenotypes declines. Which explanation best accounts for the observed change in fur thickness distribution?

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Question 1

Disruptive selection increasing both extremes because extremes have the highest fitness
Directional selection shifting the mean toward thicker fur because cold favors thickness
Stabilizing selection reducing extremes by favoring intermediate fur thickness each generation (correct answer)
Genetic drift maintaining the mean by chance because reproductive success is equal
Natural selection acting on individuals so each rabbit adjusts fur thickness to match climate

Explanation: This question requires analyzing natural selection patterns by examining how fitness varies across a trait distribution. The data shows that rabbits with intermediate fur thickness leave the most surviving offspring during cold winters, while both very thin and very thick fur rabbits have lower fitness. After six generations, the mean remains unchanged but extreme phenotypes decline, which is the hallmark of stabilizing selection that favors intermediate phenotypes and eliminates extremes. Students often mistakenly choose directional selection (B) because they associate cold weather with thicker fur, but the data explicitly states that very thick fur rabbits also have reduced fitness. To identify selection type, examine which phenotypes have highest fitness: if intermediates survive best and extremes decline while the mean stays constant, it's stabilizing selection.

Question 2

In a large bacterial population, a new allele $R$ that confers antibiotic resistance arises once by mutation. After antibiotic treatment begins, the frequency of $R$ increases from 0.001 to 0.35 in 30 generations. The mutation rate is low and unchanged, and population size remains very large. Which process is most likely responsible for the rapid increase in $R$ frequency?

Genetic drift increasing allele $R$ frequency by random sampling
Gene flow introducing allele $R$ repeatedly from other populations
Directional selection favoring allele $R$ under antibiotic exposure (correct answer)
High mutation rates creating many new $R$ alleles each generation
Assortative mating increasing the frequency of allele $R$ in bacteria

Explanation: This question examines population genetics in the context of antibiotic resistance evolution. A rare resistance allele R increases from frequency 0.001 to 0.35 in just 30 generations after antibiotic treatment begins - an extremely rapid change. The problem states the population remains very large (ruling out drift) and mutation rates are low and unchanged (ruling out mutation pressure). This leaves directional selection as the only mechanism that can drive such rapid frequency increase: bacteria carrying allele R survive antibiotic exposure better than those without it, giving them a massive fitness advantage. Students might choose genetic drift (A), but drift cannot cause such consistent directional change in large populations, especially not this rapidly. When you see rapid allele frequency increases coinciding with environmental changes that affect survival, think directional selection.

Question 3

In rabbits, a dominant allele (R) produces round ears and a recessive allele (r) produces pointed ears. Two rabbits with round ears are crossed. Each parent could be RR or Rr. If one offspring is born with pointed ears (rr), the parents’ genotypes can be inferred using Mendelian inheritance. Which parental genotypes are consistent with producing a pointed-eared offspring?

RR × RR
RR × Rr
Rr × Rr (correct answer)
RR × rr
Rr × rr

Explanation: This question requires reverse genetics—using offspring phenotypes to deduce parental genotypes through Mendelian principles. For a pointed-eared offspring (rr) to appear, it must receive one r allele from each parent, meaning both parents must carry at least one r allele. Since both parents have round ears (dominant phenotype), they cannot be rr, so both must be Rr (heterozygous). The cross Rr × Rr produces RR (1/4), Rr (1/2), and rr (1/4), explaining how round-eared parents can have pointed-eared offspring. Students often think dominant phenotype parents cannot have recessive phenotype offspring, forgetting that heterozygotes carry hidden recessive alleles. When offspring show a recessive phenotype that neither parent displays, both parents must be heterozygous carriers.

Question 4

Two populations of frogs live in adjacent valleys with similar climate. During the breeding season, males in valley 1 produce a call with a dominant frequency near 1.2 kHz, while males in valley 2 produce a call near 2.0 kHz. Females show strong preference for local calls in playback experiments, and field surveys find few mixed pairs even where the valleys meet. Genetic data indicate reduced gene flow and increasing divergence between the populations. Which prezygotic barrier most directly contributes to reproductive isolation here?

Behavioral isolation based on divergence in mating calls and preferences (correct answer)
Hybrid inviability because embryos fail to develop after fertilization
Ecological release caused by predators disappearing from one valley
Polyploidy producing immediate reproductive isolation in amphibians
Inheritance of acquired call changes from individual males to offspring

Explanation: This question assesses the skill of analyzing speciation processes in AP Biology, focusing on behavioral barriers in adjacent populations. Behavioral isolation occurs through divergence in mating calls, with females preferring local frequencies, which reduces interbreeding between the valley frog populations. This prezygotic barrier limits gene flow, allowing genetic divergence despite the valleys meeting and similar climates. Reproductive isolation is reinforced by these preferences, as evidenced by few mixed pairs and increasing allele differences. A tempting distractor is option B, which proposes hybrid inviability, but this is postzygotic and not the primary barrier here, reflecting the misconception that isolation must involve hybrid failure rather than mating prevention. A useful strategy is to examine experimental evidence like playback tests to identify behavioral isolation in speciation scenarios.

Question 5

An organism has the genotype $AaBb$ , where the genes for traits A and B are located on different chromosomes. Which of the following represents all possible combinations of alleles in the gametes produced by this organism?

$AB, Ab, aB, ab$ (correct answer)
$Aa, Bb$
$A, a, B, b$
$AB, ab$

Explanation: Due to the independent assortment of unlinked genes, a dihybrid individual with genotype $AaBb$ will produce four types of gametes in equal proportions. Each gamete must contain one allele for each gene, resulting in the combinations $AB, Ab, aB$ , and $ab$ .

Question 6

A structural carbohydrate in plant cell walls consists of unbranched chains of glucose where each monomer is linked by $b2$ (1b4) glycosidic bonds. These bonds orient adjacent glucose units in alternating directions, allowing many chains to lie parallel and form extensive hydrogen bonds between hydroxyl groups on neighboring chains. Bundles of these chains form microfibrils with high tensile strength and limited solubility. Which statement best describes how the bond type contributes to this material property?

$b2$ (1b4) linkages produce straight chains that hydrogen-bond into strong, insoluble fibers (correct answer)
$b1$ (1b4) linkages create coils that pack into rigid sheets through ionic bonding
Glycosidic bonds join amino acids, enabling disulfide bridges that stiffen the wall
Branched chains prevent alignment, increasing solubility and decreasing tensile strength
Phosphate groups on the backbone cause repulsion that tightly cross-links adjacent chains

Explanation: This question assesses the analysis of carbohydrate structure-function relationships. The correct answer A explains that β(1→4) glycosidic bonds produce straight, unbranched chains of glucose that align parallel and form extensive hydrogen bonds between hydroxyl groups on neighboring chains, as noted in the stimulus describing the structural carbohydrate in plant cell walls forming microfibrils with high tensile strength. This bond type orients adjacent glucose units in alternating directions, preventing coiling and promoting the formation of insoluble fibers, which is a key AP Biology concept for how cellulose provides rigidity and limited solubility to cell walls. Consequently, these interactions enhance the material's tensile strength, making it suitable for structural support without dissolving in water. A tempting distractor is B, which mentions α(1→4) linkages creating rigid sheets through ionic bonding, but this is incorrect due to a structure-function confusion by misattributing alpha bond coiling and non-ionic interactions to structural properties. To approach such questions, compare glycosidic bond types and predict their impact on polymer chain alignment and intermolecular forces like hydrogen bonding.

Question 7

In a population of 800 snails, genotype counts are 320 GG, 160 Gg, and 320 gg. The population is large, and there is no mutation, migration, or selection at this locus. Which conclusion is best supported about Hardy-Weinberg equilibrium in this generation?

The population is in equilibrium because $p=q=0.50$ .
The population is in equilibrium because homozygotes are equally frequent.
The population is not in equilibrium because observed heterozygotes are fewer than $2pq$ . (correct answer)
The population is not in equilibrium because allele frequencies cannot be computed from counts.
The population is in equilibrium because heterozygotes are exactly half the population.

Explanation: This question tests detecting Hardy-Weinberg equilibrium violations by comparing observed and expected heterozygote frequencies. From counts (320 GG, 160 Gg, 320 gg), allele frequencies are p(G) = (640 + 160)/1600 = 0.50 and q(g) = 0.50. Under Hardy-Weinberg, expected Gg frequency = 2pq = 2(0.50)(0.50) = 0.50, meaning 400 heterozygotes expected. With only 160 observed (0.20 frequency vs 0.50 expected), there's a significant heterozygote deficit, indicating the population is not in equilibrium (option C). Option E incorrectly focuses on heterozygotes being "half the population" without checking if this matches 2pq expectations. Remember: even with equal allele frequencies, Hardy-Weinberg predicts specific genotype ratios—always calculate and compare to observations.

Question 8

In endothelial cells, nitric oxide (NO) diffuses into the cytosol and binds soluble guanylyl cyclase (sGC), increasing cGMP. cGMP activates protein kinase G (PKG), which phosphorylates myosin light-chain phosphatase, increasing its activity and reducing myosin phosphorylation, relaxing the cell. A cGMP phosphodiesterase (PDE) converts cGMP to GMP. When PDE is overexpressed, NO produces a smaller relaxation response despite normal sGC activation. Which change would most likely restore relaxation in PDE-overexpressing cells?

Add a PDE inhibitor to slow cGMP breakdown and sustain PKG activation (correct answer)
Block sGC so NO cannot bind and initiate the pathway
Chelate intracellular $\mathrm{Ca^{2+}}$ to prevent NO diffusion into the cytosol
Inhibit myosin light-chain phosphatase to increase relaxation signaling
Increase actin polymerization to enhance the mechanical response to NO

Explanation: This question assesses the skill of analyzing a signal transduction pathway. Adding a PDE inhibitor would slow cGMP breakdown, sustaining PKG activation and restoring relaxation in PDE-overexpressing cells despite normal sGC activation by NO. In the pathway, NO activates sGC to produce cGMP, activating PKG to enhance phosphatase activity and reduce myosin phosphorylation for relaxation. Overexpressed PDE reduces cGMP, but inhibiting it counters this. Choice B is tempting but wrong because blocking sGC would prevent cGMP synthesis entirely, which is a misconception about compensating for degradation rather than blocking initiation. A transferable strategy is to target degradative enzymes like PDEs to modulate second messenger levels in gaseous signaling pathways like NO.

Question 9

In a plant cell assay, peptide P binds a membrane receptor kinase and triggers phosphorylation of a cytosolic relay protein K. Phosphorylated K activates an NADPH oxidase in the plasma membrane, producing reactive oxygen species (ROS). ROS oxidize cysteines on a nearby protein phosphatase (PP), decreasing PP activity. PP normally dephosphorylates K, limiting signal duration. When P is added, K phosphorylation rises quickly and then plateaus. A mutant line expresses PP that cannot be oxidized by ROS but retains normal catalytic activity. Compared with wild-type cells, what is most likely in the mutant after P addition?

Higher plateau of K phosphorylation because PP is permanently inhibited by mutation
Lower K phosphorylation because PP remains active and continues dephosphorylating K (correct answer)
No ROS production because PP oxidation is required to activate NADPH oxidase
Greater receptor kinase activity because PP directly phosphorylates the receptor
Delayed response because oxidation must occur before P can bind the receptor

Explanation: This question examines how feedback regulation affects signal transduction dynamics. In wild-type cells, ROS production leads to phosphatase (PP) oxidation and inactivation, reducing K dephosphorylation and allowing K phosphorylation to plateau at a higher level. In the mutant, PP cannot be oxidized by ROS, so it remains fully active throughout signaling. This continued PP activity constantly dephosphorylates K even as the receptor kinase phosphorylates it, resulting in lower steady-state K phosphorylation compared to wild-type cells where PP is partially inactivated. Choice A incorrectly assumes the mutation permanently inactivates PP, when actually it prevents ROS-mediated inactivation while maintaining normal catalytic function. When analyzing feedback loops, consider how preventing regulatory modifications affects the balance between activating and deactivating processes.

Question 10

An mRNA contains the sequence 5′-AUG GAA CUA UAG-3′, where UAG is a stop codon. A researcher introduces a modified tRNA whose anticodon base-pairs with UAG and is charged with an amino acid. In normal cells, release factors bind stop codons and promote polypeptide release. Assume the modified tRNA competes successfully with the release factor at UAG. Which outcome is most likely for translation of this mRNA?

Translation terminates normally because stop codons cannot base-pair with any anticodon.
Translation continues past UAG, adding an amino acid at that position before proceeding. (correct answer)
Transcription of the mRNA stops at UAG because stop codons terminate RNA polymerase.
The ribosome moves 3′ to 5′ to avoid the stop codon and maintain termination.
The first codon is skipped because a modified tRNA prevents initiator tRNA binding at AUG.

Explanation: This question assesses the skill of analyzing translation processes in protein synthesis. The correct answer is B because the modified tRNA base-pairs with the UAG stop codon and adds its charged amino acid, competing with the release factor and allowing translation to continue past UAG. Normally, stop codons like UAG trigger release factors for termination, but the suppressor tRNA enables read-through by incorporating an amino acid instead. This results in elongation proceeding beyond the intended stop, potentially extending the polypeptide if more codons follow. A tempting distractor is A, which incorrectly asserts termination occurs normally, based on the misconception that stop codons cannot pair with any anticodon even if a modified tRNA is present. A transferable strategy is to consider how alterations like suppressor tRNAs can override normal stop codon function, evaluating competition with release factors.

Question 11

A researcher studies the restriction point in late G1, where sustained CDK signaling is required to commit a cell to DNA synthesis. In the experiment, cyclin is present and binds CDK, but a competitive inhibitor occupies the CDK active site, preventing phosphorylation of downstream targets. The inhibitor does not affect cyclin binding or CDK abundance. Cells remain in G1 despite adequate nutrients. Which change would most likely allow the cells to enter S phase while the inhibitor is present?

Add more cyclin so additional cyclin–CDK complexes outcompete the inhibitor at the active site
Increase phosphatase activity to remove inhibitory phosphate and activate inhibited CDK
Introduce a CDK allele with reduced inhibitor binding but retained catalytic activity (correct answer)
Block APC/C so securin persists, enabling earlier initiation of DNA replication
Destabilize microtubules to silence the spindle checkpoint and promote S phase entry

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, focusing on CDK regulation at the G1 restriction point. The competitive inhibitor occupies the CDK active site, preventing substrate phosphorylation despite cyclin binding and adequate nutrients, causing G1 arrest. Introducing a CDK allele with reduced inhibitor binding but retained catalytic activity allows the mutant CDK to phosphorylate targets, committing cells to S phase. This bypasses the inhibition without altering cyclin or inhibitor levels. A tempting distractor is A, which suggests adding more cyclin, but excess cyclin would still bind inhibited CDK, stemming from the misconception that increasing ligand overcomes active-site competition. To approach similar problems, distinguish between types of inhibition and select genetic changes that specifically evade the block.

Question 12

A lab compares two cell types from the same individual and finds that a specific microRNA is abundant in cell type Q but rare in cell type R. Both cell types contain similar amounts of mRNA transcribed from Gene W, yet Gene W protein levels are low in Q and high in R. The DNA sequence of Gene W is identical in both cell types. Which explanation best accounts for the difference in Gene W protein levels?

Cell type Q has fewer copies of Gene W DNA, so less protein can be produced.
A mutation in cell type R created a new promoter that increases transcription of Gene W mRNA.
MicroRNA in cell type Q inhibits translation or promotes degradation of Gene W mRNA, reducing protein. (correct answer)
Cell type R lacks ribosomes, so Gene W protein accumulates without being synthesized.
Gene W protein is produced only when DNA is condensed, which occurs more in cell type R.

Explanation: This question explores gene expression and cell specialization through post-transcriptional regulation by microRNAs. The correct answer C explains that microRNA in cell type Q binds to Gene W mRNA, either blocking its translation or promoting its degradation, resulting in low protein levels despite normal mRNA amounts. MicroRNAs are crucial regulators that allow cells to fine-tune protein production after transcription has occurred, adding another layer of specialization control. Answer A incorrectly suggests fewer DNA copies in cell type Q, misunderstanding that protein differences here result from post-transcriptional regulation, not genomic variation. When both cell types have similar mRNA but different protein levels, look for translational or post-translational regulatory mechanisms like microRNAs or translation factors.

Question 13

A scientist prepares a cDNA library from pancreatic cells by isolating mRNA and using reverse transcriptase to synthesize DNA copies. The resulting cDNAs are cloned into plasmids and transformed into bacteria, creating many bacterial colonies. The researcher later screens the colonies to find clones containing the insulin cDNA. Compared with a genomic DNA library made from the same organism, which feature would be expected in the insulin clone from the cDNA library?

It lacks introns because the template mRNA was processed by splicing before reverse transcription. (correct answer)
It includes upstream promoter sequences because mRNA contains regulatory DNA elements.
It contains all intergenic regions flanking the insulin gene because reverse transcriptase copies whole chromosomes.
It includes histone proteins bound to the insulin sequence because plasmids package DNA into nucleosomes.
It has a higher mutation rate because reverse transcriptase proofreads more than DNA polymerase.

Explanation: This question assesses biotechnology analysis by comparing features of cDNA and genomic DNA libraries for gene cloning. The correct answer is that the insulin cDNA lacks introns because the template mRNA was processed by splicing in eukaryotic cells before reverse transcription, resulting in a continuous coding sequence. Reverse transcriptase copies the mature mRNA into DNA, excluding non-coding introns that are present in genomic DNA. This makes cDNA libraries useful for expressing functional proteins in bacteria, as they lack splicing machinery. A tempting distractor is choice B, which is wrong because mRNA does not contain promoters, which are DNA elements upstream of genes, based on the misconception that transcription copies regulatory sequences into RNA. When choosing library types, use cDNA for expression studies and genomic for regulatory analysis, a strategy guiding molecular biology resource selection.

Question 14

A cell imports phosphate (Pi) using a cotransporter that moves Pi into the cell only when H $^+$ also moves in. Pi accumulates inside the cell above its extracellular concentration. When the H $^+$ gradient is dissipated, Pi uptake stops even though ATP is present in the cytosol. Which mechanism best explains Pi uptake?

Simple diffusion of phosphate through the bilayer down its concentration gradient
Facilitated diffusion of phosphate through a channel down its gradient
Primary active transport of phosphate by direct ATP hydrolysis at cotransporter
Secondary active transport driven by H $^+$ moving down its gradient (correct answer)
Endocytosis internalizing phosphate in vesicles that then fuse with cytosol

Explanation: This question assesses the skill of analyzing membrane transport mechanisms. The correct answer is secondary active transport driven by H+ moving down its gradient because Pi accumulates against its gradient, powered by H+ influx through the cotransporter, not direct ATP. Uptake stops when H+ gradient dissipates, despite ATP presence, confirming secondary dependence on the ion gradient. The cotransporter requires both ions for movement. A tempting distractor is primary active transport by ATP hydrolysis, but this is wrong due to the misconception that all uphill transport directly uses ATP; H+ dependence indicates secondary. To analyze similar problems, always determine if movement is down a gradient (passive) or against (active) and check for energy dependence.

Question 15

A disaccharide is formed by linking one glucose and one fructose through a glycosidic bond that uses both anomeric carbons. In solution, the disaccharide does not open to expose a free aldehyde or ketone group. In a lab assay, this sugar fails to reduce Cu $^{2+}$ to Cu $^{+}$ , unlike maltose. Which statement best describes the structural reason for the negative test result?

Both anomeric carbons are involved in the glycosidic bond, leaving no free anomeric carbon to act as a reducing end (correct answer)
The disaccharide contains only $b2$ linkages, which prevents any interaction with metal ions
The disaccharide is nonpolar, so it cannot dissolve well enough to participate in redox reactions
The glycosidic bond is a peptide bond, which lacks electrons needed for reduction of copper ions
The disaccharide contains phosphate groups that keep it permanently in a closed-ring form

Explanation: This question requires analyzing carbohydrate structure–function relationships to understand reducing sugar behavior. The disaccharide formed between glucose and fructose uses both anomeric carbons in the glycosidic bond, meaning neither sugar retains a free anomeric carbon that could undergo ring-opening to expose a reactive aldehyde or ketone group. Without this free carbonyl group, the disaccharide cannot donate electrons to reduce Cu2+ to Cu+ in Benedict's or Fehling's test, explaining the negative result. Option D incorrectly states the glycosidic bond is a peptide bond, confusing carbohydrate bonds (between sugars) with protein bonds (between amino acids)—this represents a biomolecule class error. When predicting reducing sugar behavior, check whether at least one anomeric carbon remains free after glycosidic bond formation to allow ring-opening and carbonyl group exposure.

Question 16

Three ponds are sampled for zooplankton. Pond A has 4 species with 25 individuals each. Pond B has 8 species with 12–13 individuals each. Pond C has 8 species, but one species has 70 individuals and the remaining 7 species share 30 individuals. Each pond sample totals 100 individuals. Which pond has the highest biodiversity based on richness and evenness together?

Pond A, because it has the most even distribution of individuals among species.
Pond B, because it combines higher richness with relatively high evenness. (correct answer)
Pond C, because it has the highest richness and a clearly dominant species.
Pond A, because it has the lowest richness and therefore the lowest competition.
All ponds, because each sample includes exactly 100 individuals.

Explanation: This question requires analyzing biodiversity by integrating both species richness and evenness to determine overall diversity. Pond B has the highest biodiversity because it combines higher species richness (8 species) with relatively high evenness (12-13 individuals per species). Pond A has perfect evenness but only 4 species, limiting its overall diversity due to low richness. Pond C has high richness (8 species) but very low evenness with one species having 70 individuals while the other 7 share only 30 individuals. The data shows Pond B achieves the best balance of both biodiversity components. A common misconception (choice C) is that highest richness automatically means highest biodiversity, ignoring how extreme dominance reduces overall diversity. When comparing biodiversity across communities, evaluate both richness and evenness together, recognizing that the highest diversity occurs when both components are reasonably high, not when one is maximized at the expense of the other.

Question 17

A river fish species occupies both upstream and downstream habitats with no physical barrier between them. Upstream water is clear and fast; downstream water is turbid and slow. Over many generations, females upstream mate mostly with males showing bright flank coloration, while females downstream mate mostly with males showing dull coloration that is less visible in turbid water. Mark-recapture studies show fish move between regions, but genetic data reveal strong allele-frequency differences at loci linked to coloration and mate preference, with fewer mixed-genotype offspring than expected under random mating. Which mechanism most directly explains divergence despite ongoing gene flow?

Allopatric speciation caused by complete geographic separation of subpopulations
Sympatric divergence driven by assortative mating and disruptive selection (correct answer)
Genetic drift alone producing identical allele frequencies in both regions
Inheritance of acquired traits from individuals adjusting coloration to water
Postzygotic isolation from sterile hybrids that never survive to adulthood

Explanation: This question assesses the skill of analyzing speciation processes by evaluating divergence mechanisms in connected habitats. Sympatric divergence occurs through assortative mating, where females prefer males with coloration suited to local water conditions, combined with disruptive selection favoring extremes in clear versus turbid environments. This reduces gene flow despite movement, leading to reproductive isolation via prezygotic behavioral barriers and allele-frequency differences at relevant loci. Fewer mixed-genotype offspring than expected confirm isolation without a physical barrier. A tempting distractor is choice A, allopatric speciation, but this is wrong as no complete geographic separation exists, highlighting the misconception that divergence requires barriers rather than selection in sympatry. To approach similar problems, assess gene flow and selection pressures to differentiate sympatric from allopatric speciation.

Question 18

A phospholipid bilayer lacking proteins is exposed to equal concentrations of iodine (I $_2$ , 254 Da, nonpolar) and lysine (146 Da, charged at neutral pH). Which molecule would most likely cross the membrane more readily by simple diffusion?

Lysine, because it is smaller than I $_2$ and therefore diffuses faster
I $_2$ , because nonpolar molecules can dissolve in the hydrophobic bilayer core (correct answer)
Lysine, because charged molecules are stabilized by water inside the bilayer
Both equally, because both have similar molecular masses
Neither, because diffusion across membranes occurs only for gases

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. I₂ crosses more readily because it is nonpolar, dissolving in the hydrophobic core unlike the charged lysine. Despite I₂'s larger mass, nonpolarity overcomes size barriers. The stimulus provides masses and properties with no proteins, underscoring polarity's dominance. A tempting distractor is choice A, favoring lysine's smaller size, but this ignores the misconception that charge allows stabilization in bilayers. For transferable strategy, always rank nonpolar over charged solutes, even if larger, for bilayer permeability.

Question 19

A researcher isolates nuclei from eukaryotic cells and measures RNA lengths from a single gene. The primary transcript averages 2,400 nucleotides and contains one intron of 900 nucleotides flanked by two exons. After RNA processing, the most abundant RNA from this gene averages 1,500 nucleotides and contains the same exon sequences as the primary transcript. No changes are detected in the DNA sequence of the gene. Which explanation best accounts for the 900-nucleotide decrease in RNA length after processing?

RNA polymerase II skipped the intron region during transcription initiation.
The intron was excised during splicing, and exons were ligated together. (correct answer)
The intron sequence was edited into exon sequence by polyadenylation enzymes.
The 5′ cap was removed, shortening the RNA by 900 nucleotides.
The intron was degraded from the DNA template after transcription ended.

Explanation: This question tests understanding of transcription and RNA processing, particularly the mechanism of intron removal. The primary transcript is 2,400 nucleotides long and contains a 900-nucleotide intron, while the processed RNA is 1,500 nucleotides (exactly 900 nucleotides shorter). This 900-nucleotide decrease precisely matches the intron size, indicating that the intron was excised during splicing and the flanking exons were ligated together by the spliceosome. The DNA sequence remains unchanged, confirming this is an RNA processing event, not a DNA alteration. Choice A incorrectly suggests RNA polymerase can skip DNA regions during transcription, but RNA polymerase must transcribe the template continuously. To solve RNA length problems, calculate the difference between primary and mature RNA lengths and compare this to known intron sizes.

Question 20

A sealed artificial phospholipid bilayer separates two solutions. Five solutes are added separately at equal concentration: O $_2$ (32 Da, nonpolar), ethanol (46 Da, polar uncharged), glucose (180 Da, polar uncharged), Na $^+$ (23 Da, charged), and a 12–amino acid peptide (~1300 Da, polar). No transport proteins are present. Which solute would most likely cross the bilayer fastest by simple diffusion?

O $_2$ (32 Da, nonpolar) (correct answer)
Ethanol (46 Da, polar uncharged)
Glucose (180 Da, polar uncharged)
Na $^+$ (23 Da, charged)
12–amino acid peptide (~1300 Da, polar)

Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. O₂ crosses the bilayer fastest because it is small and nonpolar, allowing it to easily dissolve in and diffuse through the hydrophobic core of the membrane without needing transport proteins. In contrast, ethanol and glucose are polar uncharged, which hinders their entry into the nonpolar interior, while Na⁺ is charged and strongly repelled, and the large peptide faces both size and polarity barriers. The stimulus details emphasize equal concentrations and no proteins, highlighting that simple diffusion favors nonpolar solutes like O₂ over others. A tempting distractor is Na⁺ (choice D), as its small size might suggest fast diffusion, but this ignores the misconception that charge prevents ions from crossing the hydrophobic barrier effectively. For transferable strategy, always prioritize nonpolarity over size for diffusion through pure lipid bilayers, then consider molecular weight for polar uncharged solutes.

Question 21

A vesicle preparation contains membranes with an ion transporter that moves $\text{Ca}^{2+}$ from the cytosolic side to the lumen. When the luminal $\text{Ca}^{2+}$ concentration is already higher than the cytosolic $\text{Ca}^{2+}$ concentration, net $\text{Ca}^{2+}$ movement into the lumen still occurs if ATP is present. If ATP is removed, net movement stops even though the concentration gradient remains the same. Adding a channel blocker has no effect, but adding an inhibitor that binds the transporter’s ATP-binding site stops transport. Which process is most likely occurring?

Facilitated diffusion of $\text{Ca}^{2+}$ through a gated channel down its gradient
Simple diffusion of $\text{Ca}^{2+}$ through the lipid bilayer down its gradient
Osmosis of water across the membrane driven by $\text{Ca}^{2+}$ concentration
Primary active transport of $\text{Ca}^{2+}$ by an ATP-powered pump against gradient (correct answer)
Secondary active transport of $\text{Ca}^{2+}$ using ATP to build a cotransport gradient

Explanation: This question assesses the skill of analyzing mechanisms of membrane transport. The transporter moves Ca²⁺ into the lumen against its concentration gradient, as luminal concentration is higher yet net movement continues, requiring direct energy input from ATP hydrolysis in primary active transport. The cessation of transport upon ATP removal, despite the persistent gradient, and the effect of an inhibitor binding the transporter’s ATP site confirm that ATP powers the pump directly. No effect from a channel blocker indicates it's not a passive channel but an active pump overcoming the unfavorable gradient. A tempting distractor is choice A, facilitated diffusion through a channel, which is wrong because it assumes passive downhill movement, misconstruing the against-gradient transport and ATP dependence as gradient-driven. When evaluating active versus passive transport, focus on direction relative to gradients, direct ATP involvement, and responses to inhibitors targeting energy sources or proteins.

Question 22

A lab compares meiosis in two cells from the same individual. In cell 1, no chiasmata are observed for chromosome pair 3. In cell 2, a chiasma is observed between the homologs of chromosome pair 3 before metaphase I. After meiosis II, cell 2 produces some gametes with chromatids containing a segment originally on the other homolog, while cell 1 does not. Assume all other chromosomes behave normally. Which process best explains why cell 2 produces additional chromosomal combinations in its gametes?

Point mutations occur only in cell 2 during interphase, changing allele combinations by altering DNA sequences
Crossing over in cell 2 during prophase I creates recombinant chromatids by exchanging homologous segments (correct answer)
Independent assortment occurs only in cell 2, while cell 1 lacks metaphase I alignment of homologs
Fertilization differences cause cell 2 gametes to contain mixed homolog segments before gamete formation
Sister chromatids separate at anaphase I only in cell 2, producing recombinant chromatids by segregation

Explanation: Crossing over contributes to meiotic genetic diversity by generating recombinant chromatids in gametes. The chiasma in cell 2 during prophase I leads to segment exchange between homologs, producing gametes with mixed segments from the other homolog, unlike cell 1 without chiasmata. This recombination creates additional chromosomal combinations specifically in cell 2, while other chromosomes behave normally. The comparison shows that the presence of crossing over accounts for the extra variation. A tempting distractor is choice C, which suggests independent assortment occurs only in cell 2, reflecting the misconception that assortment varies between cells rather than being a consistent meiotic feature. To compare meiotic processes, note how the presence or absence of crossing over affects recombinant outcomes in otherwise identical cells.

Question 23

In a lab model of early cells, repeated osmotic swelling causes the plasma membrane to invaginate and pinch off, forming internal vesicles. After many cycles, microscopy shows a stable network of internal membranes that remains continuous with the plasma membrane in some regions, and the internal compartments contain the same lipid composition as the cell surface. No internal compartment contains its own DNA, and all ribosomes observed are free in the cytosol. Which explanation best accounts for the origin of these internal membranes?

Infolding of the plasma membrane produced internal membranes with matching lipids and occasional continuity. (correct answer)
Endosymbiosis of an aerobic bacterium formed internal membranes that remained connected to the surface membrane.
Horizontal gene transfer created internal compartments by adding bacterial DNA to the cytosol.
Spontaneous assembly of ribosomes into sheets generated internal membranes independent of the plasma membrane.
Fusion of two cells created internal membranes by trapping extracellular fluid without membrane invagination.

Explanation: This question assesses the analysis of the origins of cell compartmentalization. The correct answer, choice A, is supported by the stimulus describing internal membranes formed through invagination and pinching off of the plasma membrane, resulting in vesicles with matching lipid composition and occasional continuity, which aligns with the infolding hypothesis for the endomembrane system's origin in eukaryotic cells. The absence of internal DNA and the presence of only free cytosolic ribosomes further indicate that these compartments did not arise from endosymbiotic events, as they lack independent genetic material typical of organelles like mitochondria. This mechanism reflects how early eukaryotic cells could have developed internal compartments to separate biochemical reactions without engulfing prokaryotes, consistent with AP Biology concepts of eukaryotic evolution. A tempting distractor, choice B, is incorrect due to a structure-function confusion, as it assumes endosymbiosis despite the lack of internal DNA or separate ribosomes, which would be expected in an engulfed bacterium. To approach similar questions, compare the presence of independent genomes and membrane continuity to distinguish between infolding and endosymbiotic origins.

Question 24

A scientist isolates cells that have just completed cytokinesis. In these cells, chromosomes have decondensed, nuclear envelopes are intact, and each cell contains a single copy of each chromosome (one chromatid per chromosome). The cells have not begun DNA replication, and total DNA content is 2C. The scientist then provides nutrients that allow cells to continue cycling. Assuming normal progression, which change will occur during the next S phase in each cell?

DNA content will increase from 2C to 4C as sister chromatids are synthesized (correct answer)
DNA content will decrease from 2C to 1C as homologs separate into gametes
Chromosomes will condense and align at the metaphase plate before replication begins
Cohesin will be removed and chromatids will segregate to poles during S phase
The spindle checkpoint will prevent DNA synthesis until kinetochores attach to microtubules

Explanation: This question assesses the skill of analyzing the cell cycle, tracking changes during S phase after cytokinesis. The stimulus notes post-cytokinesis cells with 2C DNA and unreplicated chromosomes, ready for nutrient-driven cycling. In AP Biology, S phase duplicates DNA, increasing content from 2C to 4C by synthesizing sister chromatids for each chromosome. Thus, during S phase, DNA content rises to 4C with new sister chromatids formed. A tempting distractor is B, suggesting a decrease to 1C via homolog separation, but this is incorrect due to a teleological misconception that mitosis reduces ploidy like meiosis. To approach similar questions, map DNA content and chromosome structure across phases, predicting changes based on replication and division events.

Question 25

A kinase cascade activated by cell stress increases activity of a Cdk inhibitor (CKI) by phosphorylating it, which strengthens CKI binding to cyclin–Cdk complexes. In treated cells, cyclin levels are unchanged, but overall Cdk activity decreases. The stress signal is then removed, and a phosphatase normally dephosphorylates CKI to weaken binding.

Which change would most likely prolong cell-cycle arrest after stress removal?

Inhibit the CKI phosphatase so CKI remains phosphorylated and tightly bound to cyclin–Cdk (correct answer)
Activate APC/C to destroy securin and trigger anaphase in G1 phase
Increase kinetochore inhibitor production to delay APC/C only during metaphase
Block origin licensing factor phosphorylation to prevent relicensing after S phase
Increase cyclin gene expression so cyclin–Cdk outcompetes CKI binding

Explanation: Signaling-based regulation of the cell cycle is a key skill in understanding how stress signals modulate inhibitors to enforce arrest. Inhibiting the CKI phosphatase keeps CKI phosphorylated and bound to cyclin–Cdk, prolonging low Cdk activity and arrest after stress removal. This prevents dephosphorylation needed to weaken binding. Stress is removed, but phosphatase block sustains inhibition. A tempting distractor is choice E, increasing cyclin expression to outcompete CKI, but this doesn't address persistent binding, misconstruing quantity over affinity regulation. To approach similar problems, identify modifications that lock in the inhibitory state post-stimulus.

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