AP Biology
Practice Test 29 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.
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Researchers culture myoblasts and differentiate them into muscle fibers. DNA sequencing shows no change in the actin gene sequence during differentiation. Yet actin mRNA increases greatly in muscle fibers compared with myoblasts. A DNA-binding protein is detected in muscle fiber nuclei that binds a regulatory region upstream of the actin gene, but it is absent in myoblast nuclei. Which explanation best accounts for the increased actin mRNA in muscle fibers?
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Researchers culture myoblasts and differentiate them into muscle fibers. DNA sequencing shows no change in the actin gene sequence during differentiation. Yet actin mRNA increases greatly in muscle fibers compared with myoblasts. A DNA-binding protein is detected in muscle fiber nuclei that binds a regulatory region upstream of the actin gene, but it is absent in myoblast nuclei. Which explanation best accounts for the increased actin mRNA in muscle fibers?
Explanation: This question tests gene expression and cell specialization by examining how muscle-specific transcription factors drive differentiation. The correct answer B explains that a muscle-specific transcription factor appears during differentiation and binds to regulatory sequences upstream of the actin gene, recruiting RNA polymerase and increasing transcription. This demonstrates how cell-type-specific proteins regulate gene expression without changing DNA sequence. The presence of this DNA-binding protein only in differentiated muscle fibers, combined with unchanged DNA sequences, clearly indicates transcriptional regulation as the mechanism. Answer E incorrectly claims myoblasts lack ribosomes, which would prevent all protein synthesis and cell survival, not just affect transcription. When analyzing gene expression changes, look for regulatory proteins that appear or disappear during differentiation to control transcription of specific genes.
In a cell line, ligand U triggers a rapid increase in IP3 and a subsequent rise in cytosolic Ca2+. When IP3 receptors on the endoplasmic reticulum (ER) are blocked, ligand binding and IP3 production are unchanged, but cytosolic Ca2+ does not rise. Which of the following best explains the earliest role of the ER IP3 receptor in transduction?
Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on receptor-ligand interactions and early transduction events. The correct answer is A because blocking ER IP3 receptors prevents Ca2+ rise despite IP3 production, indicating the receptor is a channel releasing Ca2+ from ER stores upon IP3 binding. This matches principles where IP3 gates ER channels for second messenger action. Unchanged upstream steps confirm the role. A tempting distractor is B, which incorrectly makes the ER receptor synthesize IP3, stemming from the misconception that receptors produce messengers, though PLC does. When analyzing signal transduction questions, block intracellular components to map second messenger targets.
Cells are labeled with a dye that binds DNA, and fluorescence intensity reflects DNA content. A cell in G1 shows baseline intensity (2C). As it progresses through S phase, intensity rises until it reaches double baseline (4C) in G2. The cell then enters mitosis and completes anaphase and telophase, followed by cytokinesis. If fluorescence is measured in each daughter cell immediately after cytokinesis, which intensity is most likely relative to the original G1 cell?
Explanation: This question assesses the skill of analyzing the cell cycle, tracking DNA content via fluorescence post-mitosis. The stimulus follows a cell from baseline 2C in G1 to 4C in G2, then through division restoring 2C per daughter. In AP Biology, equal chromatid segregation and cytokinesis return each daughter to baseline DNA content. Therefore, intensity is baseline as DNA returns to 2C. A tempting distractor is C, claiming double baseline, but this is incorrect due to a teleological misconception that mitosis preserves rather than halves per-cell content. To approach similar questions, calculate content changes phase-by-phase, accounting for replication and division.
In immune cells, a soluble ligand binds a receptor tyrosine kinase (RTK) on the plasma membrane. Within 1 minute, the receptor becomes phosphorylated on cytosolic tyrosines, and a cytosolic adaptor protein binds only to the phosphorylated receptor. A mutation prevents receptor dimerization but does not affect ligand binding. Which outcome is most consistent with early transduction in the mutant cells?
Explanation: This question tests your ability to analyze signal transduction in receptor tyrosine kinases, specifically the role of dimerization. The correct answer B explains that while ligand binding will still occur in the mutant, the lack of dimerization prevents trans-autophosphorylation between receptor monomers, greatly reducing tyrosine phosphorylation and subsequent adaptor protein recruitment. RTK signaling requires two receptors to come together so each can phosphorylate the other's tyrosines, creating docking sites for adaptors. Answer A incorrectly assumes phosphorylation can occur without dimerization, reflecting the misconception that individual RTKs can fully activate themselves. When analyzing RTK signaling questions, remember that dimerization is essential for cross-phosphorylation, which creates the phosphotyrosine sites needed for downstream protein recruitment.
During a sudden loud sound, a rabbit freezes for several seconds and its breathing rate increases. When the environment becomes quiet again, the rabbit resumes movement and breathing slows. Which response best explains the short-term mechanism causing the rabbit’s immediate changes?
Explanation: This question assesses the skill of analyzing how organisms respond to changes in their external environment. The loud sound activates the sympathetic nervous system, leading to freezing behavior and increased breathing rate to prepare for potential threats by enhancing oxygen delivery. When the sound stops, the parasympathetic system restores normal movement and breathing, showing the response is tied to the stimulus duration. This short-term fight-or-flight mechanism enhances survival without altering cellular or organ structures permanently. A tempting distractor is choice C, which describes developing larger lungs, but this misconceptions blends immediate neural responses with long-term anatomical changes. A transferable strategy is to link autonomic nervous system activation to short-term behavioral and physiological shifts in stress responses, distinguishing them from developmental or metabolic alterations.
A protein-free phospholipid bilayer is exposed to equal concentrations of sucrose (342 Da, polar uncharged) and glycerol (92 Da, polar uncharged). Which molecule would most likely diffuse across the bilayer faster?
Explanation: This question tests the skill of analyzing membrane permeability based on solute properties in a phospholipid bilayer. Glycerol diffuses faster because it is smaller than sucrose, allowing better passage despite both being polar uncharged. Sucrose's larger size increases the barrier to entering the hydrophobic core. The stimulus notes both are polar uncharged with no proteins, focusing on size's role. A tempting distractor is choice A, suggesting larger size aids bonding, but this reflects the misconception that mass enhances polar diffusion. For transferable strategy, always compare sizes for polar uncharged solutes, as smaller ones permeate faster in pure bilayers.
A biologist monitors a rabbit population in two fenced grassland plots of equal size. In Plot A, rabbit density increases from 10 to 40 rabbits per hectare over two months, and the percentage of rabbits with visible skin lesions rises from 2% to 28%. In Plot B, density remains near 10 rabbits per hectare and lesions stay below 3%. No changes in weather or food availability are recorded during the study. Which outcome is most consistent with density-dependent regulation in Plot A?
Explanation: This question assesses understanding of how population density affects growth and regulation in biological systems, focusing on density-dependent mechanisms. Higher disease prevalence, indicated by rising skin lesions, acts as a density-dependent factor because transmission increases with closer rabbit proximity in Plot A as density rises. This regulates the population by reducing survival or reproduction more intensely at higher densities, consistent with the observed increase in lesions from 2% to 28%. No weather or food changes support that density itself drives the regulation through disease. A tempting distractor is choice B, a drought reducing grass growth equally, which is wrong due to the misconception of density independence, as it affects both plots uniformly regardless of rabbit numbers. For transferable strategy, identify density-dependent regulation by looking for effects that intensify with increasing population size.
In eukaryotic cells, many enzymes that add phosphate groups to proteins are located on the cytosolic side of specific membranes, while enzymes that remove phosphate groups may be enriched in different regions of the cytosol. If a mutation causes these enzymes to become uniformly dispersed throughout the cytosol, the same total amount of each enzyme remains present, but the rate of phosphorylation of membrane-associated target proteins decreases. Which feature best explains how localized environments within cells influence reaction efficiency?
Explanation: This question requires analysis of cell compartmentalization to understand how enzyme localization affects reaction efficiency. The correct answer (A) explains that localizing kinases near membrane-associated substrates increases the effective concentration of both enzyme and substrate in a confined space, increasing collision frequency and reaction rates compared to uniform dispersion throughout the larger cytosolic volume. When mutations cause uniform distribution, the same total enzyme amount is diluted across the entire cytosol, reducing encounters with membrane-bound substrates and decreasing phosphorylation rates, demonstrating that spatial organization within compartments optimizes biochemical reactions. Option E commits a teleological error by suggesting enzymes localize "so cells can increase phosphorylation when conditions change," attributing purposeful intent to an evolutionary outcome. The strategy for analyzing intracellular organization is to consider how proximity between enzymes and substrates affects reaction kinetics through local concentration effects.
A bacterial cell is placed in pure water. The cell does not burst, while a similar cell that has had its cell wall enzymatically removed swells and lyses. Which feature best explains the difference?
Explanation: This question assesses the skill of analyzing cell structure-function relationships. The rigid cell wall in the intact bacterial cell counteracts turgor pressure from water influx in pure water (hypotonic), preventing lysis, unlike the wall-removed cell that swells without this support. This highlights the prokaryotic cell wall's role in AP Biology for structural integrity and protection against osmotic rupture. Composed of peptidoglycan, the wall limits expansion despite water entry. A tempting distractor is choice C, which is incorrect due to structure-function confusion, as the cell wall does not replace the plasma membrane but surrounds it, allowing water diffusion while providing mechanical resistance. To approach similar questions, compare cells with and without extracellular structures to infer their protective functions in osmotic challenges.
A deer population in a fenced reserve was estimated each year: Year 1 = 120, Year 2 = 180, Year 3 = 250, Year 4 = 290, Year 5 = 300, Year 6 = 298. Vegetation surveys showed increasing browse damage after Year 3. Which pattern best illustrates the deer population growth shown by the data?
Explanation: This question assesses the skill of analyzing population ecology trends by identifying growth models from sequential population estimates. The deer population increased from 120 to nearly 300 over six years, with the rate of increase slowing as it approached 300, indicating density-dependent limitations from vegetation browse damage after Year 3. This slowdown reflects logistic growth, where factors like food scarcity reduce net growth as the population nears the carrying capacity of about 300 individuals. The stable counts in Years 5 and 6 further support that the population equilibrated at this level. A tempting distractor is choice C, which claims exponential growth with a constant per capita rate, but this overlooks the decelerating growth rate, a common misconception ignoring density dependence. For future problems, plot the data on a graph to visualize if the curve is J-shaped (exponential) or S-shaped (logistic).
A researcher observes a cell at the G1 checkpoint. The cell has a DNA content of 2C and unreplicated chromosomes. The researcher then damages the DNA and notes that the cell does not enter S phase during the experiment. Later, a different cell with no DNA damage passes the same checkpoint and enters S phase, increasing DNA content toward 4C. No other phases are manipulated. Which outcome is most likely for the first cell during the observation period?
Explanation: This question assesses the skill of analyzing the cell cycle, examining checkpoint responses to DNA damage in G1. The damaged cell at the G1 checkpoint with 2C DNA does not enter S phase, remaining in G1 to allow repair, as per choice A and AP Biology's G1 checkpoint function in blocking replication of damaged DNA. In contrast, the undamaged cell proceeds normally, highlighting the checkpoint's role. No manipulation of other phases supports this arrest. A tempting distractor is choice B, suggesting direct entry into mitosis triggered by damage, but this reflects a teleology misconception, assuming damage accelerates progression rather than halting it for repair. For damage-related questions, recall which checkpoint detects issues and the typical cellular response of arrest.
Two cell types from the same plant—leaf mesophyll cells and root hair cells—are grown in identical conditions. DNA sequencing shows the cells have the same genome. When exposed to light, mesophyll cells show high levels of mRNA for a photosystem protein, while root hair cells show little to none; root hair cells instead show high levels of mRNA for a membrane transporter. Which explanation best accounts for these differences in mRNA abundance?
Explanation: This question examines gene expression and cell specialization in plant cells with identical genomes. The correct answer (B) identifies that different regulatory proteins control transcription in mesophyll versus root hair cells, causing mesophyll cells to express photosystem genes while root hair cells express transporter genes. Option A incorrectly claims root cells permanently lose photosystem genes, contradicting the stated identical genomes. Option C impossibly suggests light converts one mRNA type into another post-translation, which cannot occur. The fundamental principle is that cell-type-specific transcription factors, not DNA differences or environmental cues directly changing mRNA, determine which genes are expressed.
A lab tests two polysaccharides made of glucose: Polymer M is unbranched and forms a tight helix; Polymer N is highly branched with many short chains. When iodine solution is added, M produces a deep blue color while N produces a reddish-brown color. The color change occurs because iodine molecules fit into helical cavities of certain polymers. Which structural feature best explains the deep blue result for M? Which structural feature best explains Polymer Mbcs iodine color change?
Explanation: This question assesses the analysis of carbohydrate structure–function. Polymer M's unbranched α(1→4) chains, as per the stimulus, form a helical structure with internal cavities that can accommodate iodine molecules, leading to the deep blue color in the classic starch-iodine test from AP Biology. This helix arises from the α linkage's geometry, allowing iodine to bind noncovalently and alter light absorption without permanent attachment. In contrast, the branching in Polymer N disrupts helix formation, resulting in weaker color changes. A tempting distractor like choice C is incorrect because it suggests covalent binding via branches, which misrepresents the reversible, noncovalent interaction, indicating a level-of-organization error. When facing such questions, link polymer conformation to specific molecular interactions observed in diagnostic tests.
Cellulose and starch are both polymers of glucose. In cellulose, adjacent monomers are connected by b2(1\to4) glycosidic bonds, producing straight chains that align closely. In starch (amylose), monomers are connected by b1(1\to4) bonds, producing a helical chain. A plant cell wall sample resists stretching when pulled, even when hydrated. Which molecular feature best explains the wallbcs tensile strength at the polymer level? Which feature best explains high tensile strength in the hydrated cell wall?
Explanation: This question assesses the analysis of carbohydrate structure–function. The β(1→4) glycosidic bonds in cellulose, as per the stimulus, produce straight, unbranched chains that align parallel and form extensive interchain hydrogen bonds, contributing to the tensile strength observed in plant cell walls. This alignment allows for the creation of microfibrils, where hydrogen bonding between hydroxyl groups on adjacent chains provides resistance to stretching, a key concept in AP Biology for structural polysaccharides. In hydrated conditions, these noncovalent interactions maintain integrity without dissolving, unlike the helical starch chains that coil and interact less rigidly. A tempting distractor like choice D is incorrect because it attributes branching to cellulose, which actually reduces strength by introducing flexibility, representing a structure–function confusion. When tackling such questions, evaluate how linkage stereochemistry dictates chain shape and intermolecular forces for mechanical properties.
Genetically identical hydrangea plants were grown in pots with different soil pH. Plants in acidic soil produced blue flowers, while plants in basic soil produced pink flowers. DNA sequencing of a pigment gene was identical in both groups. Which explanation best accounts for the different flower colors under different soil pH conditions?
Explanation: This question assesses understanding of environmental effects on phenotype, where external factors influence traits without altering the underlying DNA sequence. The correct answer, A, is right because soil pH can affect the expression of pigment genes or the chemical form of pigments like anthocyanins, leading to blue in acidic conditions and pink in basic ones through pH-dependent mechanisms. In hydrangeas, aluminum availability in acidic soil alters pigment chemistry post-translationally, changing color without DNA modifications. Since the plants are genetically identical and sequencing is the same, the color variation is an environmental effect on phenotype. A tempting distractor is B, which is wrong because it suggests pH causes mutations, a misconception that confuses chemical influences on phenotype with genetic alterations. To approach similar questions, always check if phenotypic differences in identical genotypes under varying conditions point to gene expression changes rather than genetic mutations or evolution.
A weed population in soybean fields is exposed to the same herbicide each spring. Seeds are collected and genotyped yearly at a locus with alleles H (herbicide resistance) and h. Over 9 years, allele H increases from 0.08 to 0.61, while planting density and field size stay similar. Which observation best demonstrates evolution occurring in this weed population?
Explanation: Continuing evolution involves populations developing genetic resistance to pressures like herbicides, as seen in these weeds. Choice C correctly shows evolution with allele H for resistance increasing from 0.08 to 0.61 over nine years, indicating weeds carrying H survived applications better and seeded more, changing the population genetically. Stable planting density and field size suggest consistent selection pressure drove this trend. The yearly sampling reveals a clear heritable shift toward resistance. Choice A is tempting, describing temporary growth slowdowns from herbicide, but this misconceptions mixes short-term physiological recovery with evolutionary genetic changes. To identify evolution, prioritize data on allele frequency trends over time, distinguishing them from individual responses.
DNA replication is semiconservative: each daughter DNA molecule contains one parental strand and one newly synthesized strand. Replication depends on strand separation and complementary base pairing (A–T, G–C) as polymerase builds new DNA 5′→3′. A researcher grows cells for one generation in medium containing a heavy isotope of nitrogen, then transfers them to light nitrogen for one generation and isolates DNA. Which DNA composition is most likely after the first generation in light nitrogen?
Explanation: This question tests understanding of semiconservative DNA replication using the classic Meselson-Stahl experiment design. In semiconservative replication, each daughter DNA molecule retains one original parental strand and contains one newly synthesized strand. After one generation in light nitrogen, cells that previously incorporated heavy nitrogen will produce hybrid DNA molecules with one heavy strand (original) and one light strand (newly synthesized). Choice A incorrectly suggests conservative replication where parental DNA remains intact, which was disproven by this exact experimental approach. The key is recognizing that semiconservative replication produces hybrid molecules after one replication cycle.
A researcher amplifies a 220-bp region of a gene using PCR with primers flanking the region. After 30 cycles, the products are run on an agarose gel stained to visualize DNA. The negative control (no template DNA) shows no bands. Patient 1 shows a single band at ~220 bp. Patient 2 shows no visible band at any size. The thermocycler settings and reagents were identical for both patients. Which explanation best accounts for Patient 2’s gel result?
Explanation: This question assesses biotechnology analysis skills by interpreting PCR gel electrophoresis results to identify why no band appears for Patient 2. The correct answer is that Patient 2’s DNA lacks primer-binding sites due to sequence changes, which prevents the primers from annealing to the target DNA during the PCR annealing step. Without primer binding, DNA polymerase cannot initiate extension, resulting in no amplification of the 220-bp fragment after 30 cycles. Since the negative control shows no bands and conditions were identical, this points to a template-specific issue like mutations in the primer sites rather than reagent or thermocycler errors. A tempting distractor is choice B, which is wrong because PCR typically produces fragments of a specific size rather than many variable sizes that could migrate off the gel, reflecting the misconception that PCR is inherently nonspecific without optimized conditions. To troubleshoot absent PCR bands, always verify primer-template complementarity using sequence alignment tools before assuming other failures.
A fish population has genetic variation at a locus affecting toxin resistance: allele R (resistant) and r (nonresistant). A factory begins releasing a toxin into the river, and within 8 generations the frequency of R increases from 0.10 to 0.46. Researchers find no evidence of increased mutation rate at this locus. Which statement best explains how genetic variation enabled the population-level change?
Explanation: This question examines how genetic variation present in a population before an environmental change enables evolutionary adaptation. The fish population already contained both R (resistant) and r (non-resistant) alleles at low frequency before the factory began releasing toxins. When toxins entered the river, fish carrying the R allele survived and reproduced at much higher rates than those with the r allele, leading to differential reproductive success. This natural selection process caused the R allele frequency to increase from 0.10 to 0.46 in just 8 generations, demonstrating rapid evolution. Choice A incorrectly suggests that environmental exposure can cause directed mutations, a common misconception that confuses the random nature of mutation with the non-random process of selection. The strategy is to recognize that selection acts on existing variation—the R allele was already present at low frequency and increased because R-carriers had higher fitness, not because new mutations arose in response to the toxin.
In a species of mouse, black fur (B) is completely dominant to brown fur (b). A black-furred mouse with genotype Bb is crossed with a brown-furred mouse with genotype bb. Each offspring receives one allele from each parent. Assume this trait is controlled by one gene and that dominance is complete. Which proportion of offspring is expected to have brown fur?
Explanation: This question tests the skill of analyzing Mendelian inheritance patterns. In the cross between a heterozygous black-furred mouse (Bb) and a homozygous brown-furred mouse (bb), the Bb parent produces gametes with 1/2 B and 1/2 b alleles, while the bb parent produces only b alleles. The offspring genotypes are Bb and bb in equal proportions, with brown fur appearing in the bb genotype due to the recessive allele. Therefore, half of the offspring are expected to have brown fur, reflecting the segregation of alleles during gamete formation. A tempting distractor is 1/4, which could stem from the misconception of treating this as a dihybrid cross instead of a monohybrid one. To solve similar inheritance questions, always identify the parental genotypes and use a Punnett square to determine phenotypic ratios based on dominance and segregation.
Researchers create two liposomes. Liposome M contains phospholipids with short fatty acid tails; Liposome N contains phospholipids with longer tails. Both lack transport proteins. The hydrophobic tails form the interior barrier, and small nonpolar molecules cross by simple diffusion. The liposomes are placed in a solution containing a small nonpolar dye, and dye entry is measured over time. Which feature best explains why dye enters Liposome M faster than Liposome N?
Explanation: This question assesses the skill of analyzing plasma membrane structure and transport. Shorter fatty acid tails in Liposome M result in a thinner bilayer, shortening the diffusion path through the hydrophobic core for the small nonpolar dye to cross by simple diffusion. Longer tails in Liposome N create a thicker barrier, increasing resistance and slowing dye entry. Both lack transport proteins, so diffusion depends solely on bilayer properties like thickness. A tempting distractor is choice B, which claims longer tails increase fluidity, reflecting the misconception that tail length enhances rather than potentially restricts diffusion by increasing thickness. In experiments comparing lipid structures, evaluate how physical dimensions of the bilayer influence the rate of passive diffusion for nonpolar solutes.
A large population of wildflowers has two alleles at a pigment locus, R and r. Before a drought, p(R)=0.60. After three years of drought, surveys show p(R)=0.82. Field data indicate plants with genotype RR and Rr produce more seeds than rr during drought years, and the population size remains large. Which factor most directly explains the observed allele frequency change?
Explanation: This question examines population genetics mechanisms driving allele frequency change over multiple generations. The wildflower population shows an increase in allele R frequency from 0.60 to 0.82 over three drought years, with field data explicitly showing that RR and Rr genotypes produce more seeds than rr during drought. This differential reproductive success based on genotype is the definition of natural selection - specifically directional selection favoring the R allele. The large population size rules out genetic drift as a major factor, and there's no mention of immigration or unusually high mutation rates. Students might be tempted by genetic drift due to the drought (B), confusing environmental stress with population bottlenecks, but the problem states the population remains large. When genotypes show consistent fitness differences and allele frequencies change predictably in response, natural selection is the driving force.
In a large moth population, allele D at a wing-color locus has frequency 0.50. Over ten generations, p(D) remains near 0.50, but genotype surveys show heterozygotes (Dd) consistently have higher survival than either homozygote. Population size is large, and there is no immigration. Which evolutionary process most likely explains the stable intermediate allele frequency?
Explanation: This question illustrates balancing selection maintaining genetic variation in populations. Despite the large population size and absence of immigration, allele D remains stable at frequency 0.50 over ten generations, with heterozygotes (Dd) consistently showing higher survival than either homozygote. This heterozygote advantage is a form of balancing selection that maintains both alleles in the population because neither can go to fixation - when either allele becomes rare, it appears mostly in high-fitness heterozygotes. Students might choose disruptive selection (A), but that favors both homozygotes over heterozygotes, which is the opposite of what's described. When heterozygotes have the highest fitness, balancing selection maintains stable intermediate allele frequencies regardless of population size.
In a desert ecosystem, a shrub species produces abundant nectar that attracts ants. Ants patrol the shrub and are observed attacking caterpillars feeding on shrub leaves. Experimental shrubs with ants excluded by sticky barriers show higher caterpillar abundance and increased leaf damage compared with control shrubs with ants present. Nectar production and water availability are similar between treatments. Which interaction best explains the relationship between the shrub and the ants in this system?
Explanation: This question assesses the skill of analyzing community ecology by evaluating protective interactions via exclusion experiments. The shrub provides nectar to ants, which in turn attack caterpillars, reducing herbivory on the shrub, exemplifying mutualism where both species benefit: the shrub gains protection, and ants receive food. Experimental exclusion of ants leads to higher caterpillar abundance and leaf damage, confirming that ants actively defend the shrub, enhancing its fitness in exchange for nectar. This interaction logic underscores mutualism as reciprocal benefits, with the shrub's nectar production facilitating ant presence and defense. A tempting distractor is A, parasitism, which is incorrect because both species benefit rather than one harming the other, stemming from the misconception that resource exchange always involves exploitation. When studying symbioses, use exclusion methods to quantify benefits like reduced damage or increased growth for each partner.
A eukaryotic cell line is engineered so the 5′ splice site consensus sequence at the exon 1–intron 1 boundary is mutated, while all other splice signals remain unchanged. After transcription, nuclear RNA is analyzed. Compared with wild type, the predominant RNA species retains intron 1 sequences but does not retain intron 2 sequences, and exon 2 and exon 3 are still joined. Transcription initiation and elongation rates are unchanged. Which explanation best accounts for the selective retention of intron 1?
Explanation: This question tests understanding of transcription and RNA processing, specifically how splice site mutations affect intron removal. The correct answer is A because the spliceosome requires specific consensus sequences at both the 5' splice site (beginning of intron) and 3' splice site (end of intron) to recognize and remove introns—mutation of the 5' splice site at the exon 1-intron 1 boundary prevents the spliceosome from recognizing intron 1 as an intron. Without proper recognition, intron 1 cannot be excised even though intron 2, with intact splice sites, is removed normally. This selective retention demonstrates that each intron is independently recognized and processed based on its own splice signals. Answer E is incorrect because it suggests introns are removed in the cytosol, but splicing occurs exclusively in the nucleus—this represents a misconception about the cellular location of RNA processing. The strategy is to connect specific splice site mutations with retention of the corresponding intron.