Home

Tutoring

Subjects

Live Classes

Study Coach

Essay Review

On-Demand Courses

Colleges

Games

Opening subject page...

Loading your content

AP Biology

AP Biology Practice Test: Practice Test 17

Practice Test 17 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.

0%

0 / 25 answered

Question 1 of 25

A kinase cascade activated by extracellular growth factor signaling increases CDK activity by promoting removal of an inhibitory phosphate from a G1 CDK. In the absence of growth factor, the inhibitory phosphate remains on the CDK, and cells do not pass the G1 checkpoint. Researchers introduce a small molecule that mimics the growth factor signal downstream by directly activating the phosphatase that removes the inhibitory phosphate. Nutrients and ATP are sufficient, and DNA is undamaged. Which outcome is most likely after adding the small molecule?

Question Navigator

All questions

Question 1

A kinase cascade activated by extracellular growth factor signaling increases CDK activity by promoting removal of an inhibitory phosphate from a G1 CDK. In the absence of growth factor, the inhibitory phosphate remains on the CDK, and cells do not pass the G1 checkpoint. Researchers introduce a small molecule that mimics the growth factor signal downstream by directly activating the phosphatase that removes the inhibitory phosphate. Nutrients and ATP are sufficient, and DNA is undamaged. Which outcome is most likely after adding the small molecule?

  1. Cells bypass G1 arrest and enter S phase because inhibitory CDK phosphorylation is removed (correct answer)
  2. Cells arrest in metaphase because APC/C cannot target securin for degradation
  3. Cells stop DNA replication because cohesin cleavage requires phosphatase activation
  4. Cells delay cytokinesis because microtubule attachment increases checkpoint signaling
  5. Cells remain in G0 because cyclins are permanently degraded during nutrient abundance

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, focusing on growth factor signaling and CDK activation at the G1 checkpoint. The small molecule mimics growth factor signaling by activating the phosphatase that removes the inhibitory phosphate from G1 CDK, even without actual growth factor. This enables CDK activation, allowing cells to pass the G1 checkpoint and enter S phase under sufficient nutrient conditions. Cells bypass G1 arrest because the inhibitory phosphorylation is removed, promoting progression. A tempting distractor is B, which suggests metaphase arrest due to APC/C issues, but APC/C is mitotic, stemming from the misconception that G1 signals directly affect later checkpoints. To approach similar problems, map signaling pathways to specific checkpoints and predict outcomes based on which regulatory step is targeted.

Question 2

A cell expresses Receptor R, a transmembrane protein with an extracellular ligand-binding domain and an intracellular kinase domain. Ligand L binding leads to rapid dimerization of R and phosphorylation of specific tyrosines on the intracellular domains. A mutant R that still binds L but cannot dimerize shows no detectable tyrosine phosphorylation. Which of the following best explains why phosphorylation fails in the mutant?

  1. Dimerization positions kinase domains to phosphorylate each other, enabling early signal transduction. (correct answer)
  2. Dimerization is required for L to cross the membrane and directly phosphorylate cytosolic proteins.
  3. Dimerization prevents phosphatases from entering the nucleus, which otherwise blocks phosphorylation.
  4. Dimerization allows receptor mRNA to be translated, and translation produces phosphorylated receptors.
  5. Dimerization increases extracellular L concentration, and high L directly adds phosphate groups to R.

Explanation: This question requires analyzing signal transduction to understand why receptor dimerization is essential for kinase activation. The stimulus describes Receptor R with a kinase domain that undergoes tyrosine phosphorylation after ligand-induced dimerization, but a dimerization-deficient mutant shows no phosphorylation despite normal ligand binding. This indicates dimerization brings two receptor kinase domains into proximity, allowing them to trans-phosphorylate each other's tyrosine residues - a classic receptor tyrosine kinase mechanism. Choice B incorrectly suggests ligands cross membranes to phosphorylate proteins directly, misunderstanding that phosphorylation is catalyzed by the receptor's own kinase domains. For receptor tyrosine kinases, remember that dimerization enables the kinase domains to phosphorylate each other, initiating the signaling cascade.

Question 3

A biochemist notes that ATP hydrolysis releases heat into the surrounding aqueous cytosol, but the cytosol temperature changes only slightly. Water molecules are polar, with partial negative charge on oxygen and partial positive charges on hydrogen. Neighboring water molecules form hydrogen bonds, creating an intermolecular network. When heat is added, some of the energy increases molecular motion, and some is used to disrupt hydrogen bonds before the average kinetic energy rises substantially. This buffering reduces rapid temperature shifts in the solution. Which feature best explains water’s ability to limit temperature change after heat release?

  1. Hydrogen bonds absorb added energy when disrupted, so more heat is required to raise water’s temperature. (correct answer)
  2. Hydrogen bonds cause water to become nonpolar, preventing it from absorbing energy released by reactions.
  3. Hydrogen bonds store energy by forming new covalent bonds, keeping kinetic energy constant.
  4. Hydrogen bonds reduce water’s mass, so the same heat input produces a smaller temperature increase.
  5. Hydrogen bonds prevent any molecular collisions, so heat cannot spread through the cytosol.

Explanation: This question assesses the analysis of water structure and hydrogen bonding in temperature buffering in cellular environments. The correct answer is choice A because the stimulus indicates that added heat from ATP hydrolysis disrupts hydrogen bonds, absorbing energy and requiring more heat to increase molecular kinetic energy, thus limiting the cytosol's temperature rise. This connects to AP Biology principles of water's high specific heat capacity, which stabilizes cellular conditions during metabolic heat release. Moreover, the network of hydrogen bonds due to water's polarity enhances this buffering effect. A tempting distractor is choice C, which is incorrect due to a teleology misconception by suggesting hydrogen bonds purposefully store energy in new bonds to maintain constancy. To approach similar questions, consider how hydrogen bonding affects energy distribution in aqueous solutions during biochemical reactions.

Question 4

In a cell, a ligand activates a receptor that opens a Ca2+^{2+}2+ channel in the plasma membrane. Incoming Ca2+^{2+}2+ activates a Ca2+^{2+}2+-dependent kinase that phosphorylates the same Ca2+^{2+}2+ channel, increasing its open probability. A kinase inhibitor reduces the steep rise in Ca2+^{2+}2+ during ligand exposure but does not affect ligand binding. Which outcome best illustrates how feedback changes pathway behavior?

  1. Reduced amplification because Ca2+^{2+}2+-dependent positive feedback on channel opening is blocked (correct answer)
  2. Increased amplification because Ca2+^{2+}2+-dependent negative feedback on channel opening is blocked
  3. No change because Ca2+^{2+}2+ influx cannot regulate ion channel gating in cells
  4. Longer signaling because kinase inhibition prevents Ca2+^{2+}2+ export pump activation
  5. Lower signaling because kinase inhibition decreases the extracellular Ca2+^{2+}2+ concentration

Explanation: This question assesses the skill of understanding feedback regulation in signal transduction pathways. The correct answer is reduced amplification because Ca²⁺-dependent positive feedback on channel opening is blocked, as the stimulus shows Ca²⁺ activates a kinase that phosphorylates the channel to increase open probability. This positive feedback steepens Ca²⁺ rise. The kinase inhibitor reduces the rise without affecting binding, confirming blocked feedback diminishes amplification. A tempting distractor is increased amplification because Ca²⁺-dependent negative feedback on channel opening is blocked, which confuses positive with negative, a common misconception. For similar questions, evaluate amplification by comparing response curves with feedback intact or disrupted.

Question 5

A meiotic cell from a diploid organism (2n = 4) is observed during metaphase I. Each homologous pair is synapsed as a tetrad, and the two tetrads align at the metaphase plate with random orientation relative to the poles. Homologs separate in anaphase I, and sister chromatids separate in anaphase II. Which number of distinct chromosome combinations is most likely possible in gametes from independent assortment alone?​​​

  1. 2
  2. 4 (correct answer)
  3. 6
  4. 8
  5. 16

Explanation: This question tests calculation of possible gamete combinations from independent assortment. With 2n = 4 (two homologous pairs), each pair can orient independently at metaphase I, with either homolog facing either pole. The number of possible combinations equals 2^n where n is the haploid number, so with n = 2, there are 2^2 = 4 possible chromosome combinations in the gametes. Choice D incorrectly calculates based on the diploid number rather than the number of homologous pairs. To solve independent assortment problems, use the formula 2^n where n represents the number of homologous pairs (or the haploid chromosome number), as each pair has two possible orientations at metaphase I.

Question 6

A eukaryotic gene is transcribed into pre-mRNA that is complementary to the template DNA strand. In a mutant cell, RNA polymerase II has an altered active site that occasionally incorporates U opposite template-strand G, but otherwise transcription proceeds. Sequencing of newly synthesized pre-mRNA shows some positions where A is expected (from pairing with template T) are unchanged, but some positions where C is expected (from pairing with template G) are instead U. Which explanation best accounts for the base changes observed in the mutant pre-mRNA?

  1. The mutant polymerase mispairs U with template G, producing U where C would normally be added. (correct answer)
  2. The mutant polymerase replaces thymine with uracil in the DNA template during transcription.
  3. Spliceosomes substitute uracil for cytosine during intron removal at exon junctions.
  4. Poly(A) polymerase adds uracil residues throughout the transcript instead of adenines.
  5. 5′ capping enzymes deaminate cytosines to uracils at random sites in the pre-mRNA.

Explanation: This question tests understanding of transcription and RNA processing, specifically the fidelity of RNA synthesis. The mutant RNA polymerase II has an altered active site that occasionally incorporates U opposite template G instead of the correct C, explaining why some expected C positions are replaced with U. This is a transcription error where the polymerase mispairs U with template G during RNA synthesis. Positions where A is expected (from pairing with template T) remain unchanged because the mutation specifically affects G-pairing. Choice B incorrectly suggests the polymerase changes DNA, but RNA polymerase only synthesizes RNA and cannot modify the DNA template. To understand transcription errors, remember that RNA polymerase reads the template strand and adds complementary ribonucleotides, and mutations in the polymerase can cause specific mispairing patterns.

Question 7

In a city park, a population of pigeons includes two alleles at a gene affecting beak depth: D (deeper) and d (shallower). After a shift in available food toward larger, harder seeds, researchers sample allele frequencies each breeding season for 5 seasons. Allele D increases from 0.33 to 0.58. Which observation best demonstrates continuing evolution in the pigeon population?

  1. Individual pigeons developed stronger jaw muscles after eating harder seeds for several weeks.
  2. Allele D increased from 0.33 to 0.58 across five breeding seasons in the park population. (correct answer)
  3. Pigeons spent more time foraging near benches where visitors dropped food scraps.
  4. The average seed size in the park increased when a new tree species was planted.
  5. Some pigeons switched from seeds to insects during summer when insects were abundant.

Explanation: Continuing evolution refers to genetic shifts in populations adapting to new conditions, such as these pigeons facing harder seeds. The correct answer, choice B, demonstrates this as allele D for deeper beaks rose from 0.33 to 0.58 over five breeding seasons, implying better food handling led to higher survival and reproduction. This population trend reflects natural selection favoring deeper beaks in the changed food environment. Consistent sampling across seasons highlights the heritable nature of the change. Choice A tempts with muscle development from eating harder seeds, a misconception of acquired traits or plasticity rather than genetic evolution. A transferable approach is to examine allele frequency data over generations to confirm evolution, avoiding confusion with non-heritable changes.

Question 8

In a diploid organism, homologous chromosome 7 carries alleles A and B on one homolog (AB) and a and b on the other (ab). During prophase I, the homologs synapse and form a tetrad. A single crossover occurs between the loci, and chromatids separate normally through meiosis I and II. As a result, the four gametes produced from this meiosis include two parental chromatid types and two recombinant chromatid types. Which meiotic process best explains the appearance of recombinant gametes containing Ab and aB allele combinations?

  1. Random fertilization combining two haploid gametes into a diploid zygote
  2. Crossing over between nonsister chromatids of homologous chromosomes in prophase I (correct answer)
  3. DNA replication during S phase creating sister chromatids with new allele combinations
  4. Separation of sister chromatids during anaphase I producing Ab and aB chromatids
  5. Point mutation in one chromatid during metaphase I changing B to b

Explanation: This question tests understanding of how meiosis generates genetic diversity through recombination. Crossing over between nonsister chromatids of homologous chromosomes during prophase I (B) is the correct answer because it physically exchanges DNA segments between maternal and paternal chromosomes, creating new allele combinations (Ab and aB) from the original parental types (AB and ab). The question specifically describes a crossover between the A/a and B/b loci, which would produce exactly these recombinant types. DNA replication during S phase (C) is incorrect because it creates identical sister chromatids, not new allele combinations—this represents a common misconception that DNA replication itself generates diversity. When analyzing recombination problems, always identify whether genes are on the same chromosome (linked) and whether crossing over can produce the observed recombinants.

Question 9

An enzyme-catalyzed reaction is started with fixed enzyme and substrate concentrations. After 2 minutes, an additional amount of substrate is added, but the measured reaction rate immediately after the addition does not increase. Prior measurements showed the rate had already plateaued at the original substrate concentration. Temperature and pH are constant and no inhibitor is added. Which explanation best accounts for the unchanged rate after adding more substrate?

  1. Enzyme active sites were already saturated, so additional substrate does not increase ES formation rate. (correct answer)
  2. Adding substrate decreases enzyme concentration by chemical reaction, reducing the number of active sites.
  3. Adding substrate increases activation energy by stabilizing the transition state less effectively.
  4. Adding substrate causes irreversible denaturation of the enzyme due to increased solution viscosity.
  5. Adding substrate forces product to bind the active site, preventing substrate binding through feedback.

Explanation: This question assesses the skill of analyzing enzyme function by evaluating rate responses to additional substrate after initial conditions. The rate does not increase after adding more substrate because the enzyme active sites were already saturated at the original concentration, as prior measurements showed a plateau, meaning further substrate cannot enhance ES complex formation. At saturation, the reaction rate is at Vmax, limited by the enzyme's turnover rate rather than substrate availability, with constant temperature, pH, and no inhibitor confirming this. The unchanged rate immediately after addition aligns with Michaelis-Menten kinetics, where beyond saturation, rate is independent of substrate concentration. A tempting distractor is choice C, claiming added substrate increases activation energy, but this reflects a teleology misconception, assuming substrate actively hinders catalysis, whereas activation energy is enzyme-determined and not altered by concentration alone. For enzyme questions involving rate changes over time or additions, check if conditions indicate saturation to predict whether rate will respond to more substrate.

Question 10

A membrane-associated molecule contains a glycerol backbone attached to two fatty acid tails and a phosphate-containing head group. Formation of the ester bonds between glycerol and fatty acids involves dehydration reactions, but the molecule is not an extended chain of repeating monomers. The phosphate-containing region is polar, while the hydrocarbon tails are nonpolar, influencing how the molecule interacts with water. Which statement best predicts the molecule’s behavior in an aqueous environment?

  1. It will orient with polar heads toward water and nonpolar tails away, due to differences in polarity. (correct answer)
  2. It will form a linear polymer because phosphate groups link repeating monomers into long chains.
  3. It will dissolve completely as individual monomers because dehydration synthesis prevents aggregation.
  4. It will fold like a protein because R-group interactions determine its tertiary structure.
  5. It will store genetic information because nitrogenous bases pair by hydrogen bonding.

Explanation: This question examines phospholipid behavior based on amphipathic structure and molecular polarity. The molecule (glycerol + two fatty acids + phosphate head group) is a phospholipid with distinct polar (phosphate) and nonpolar (hydrocarbon tail) regions, causing it to orient with polar heads toward water and nonpolar tails away from water. This amphipathic nature drives membrane bilayer formation, not polymer formation since phospholipids don't link into chains. Option B incorrectly predicts polymer formation, representing a structural misconception since phospholipids exist as individual molecules that aggregate through hydrophobic interactions rather than covalent polymerization. To predict lipid behavior, identify polar and nonpolar regions—amphipathic molecules self-organize to minimize unfavorable water-hydrocarbon contacts.

Question 11

A coastal snail population shows variation in shell thickness controlled largely by a heritable allele T (thick) versus t (thin). A crab species preys on these snails and is more successful at crushing thin-shelled individuals, reducing their survival to reproduction. Thick- and thin-shelled snails occur in the same interbreeding population, and offspring shell thickness resembles that of their parents. Over multiple years, crab abundance remains high and other conditions are similar. Which outcome is most likely in the snail population over time?

  1. The t allele becomes more common because predation increases the need for thin shells to grow quickly.
  2. The frequency of allele T increases because thick-shelled snails survive and reproduce more often. (correct answer)
  3. Allele frequencies do not change because shell thickness is acquired from diet, not inherited.
  4. Both alleles disappear because crabs will eventually consume all snails regardless of shell type.
  5. Thin-shelled snails develop thicker shells during their lifetime and transmit that change to offspring.

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. The presence of crabs that preferentially crush thin-shelled snails with allele t creates a selective pressure favoring thick-shelled snails with allele T, as they survive predation better and reproduce more. Over time, this leads to an increase in the frequency of allele T in the population, as thick-shelled individuals contribute a larger proportion of offspring that inherit the trait. Population-level change occurs because the heritable variation in shell thickness directly affects reproductive success under consistent predation. A tempting distractor is choice E, which wrongly implies that snails acquire thicker shells during life and pass them on, embodying the misconception of inheritance of acquired characteristics. 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 12

A population of freshwater snails was counted monthly in a pond after a fish predator was removed. Counts were: Month 0 = 40, Month 1 = 60, Month 2 = 90, Month 3 = 135, Month 4 = 200, Month 5 = 205, Month 6 = 198. No new predators were added, and food algae became visibly scarce by Month 4. Which explanation best accounts for the population trend over time?

  1. Logistic growth as density-dependent limits reduced net growth near carrying capacity (correct answer)
  2. Exponential growth because resources remained unlimited throughout the entire period
  3. A constant death rate caused a linear decline beginning immediately after Month 0
  4. Directional change in traits increased survival, producing a stable population by Month 5
  5. Seasonal immigration alone produced the rise, while local reproduction stayed unchanged

Explanation: This question assesses the skill of analyzing population ecology trends by interpreting growth patterns in response to environmental factors. The snail population initially grew rapidly from 40 to 200 individuals, reflecting exponential growth when resources were abundant after predator removal. However, as algae became scarce by Month 4, density-dependent factors like resource limitation reduced the net growth rate, causing the population to level off near a carrying capacity around 200 by Months 5 and 6. This pattern exemplifies logistic growth, where population size stabilizes as it approaches the environment's carrying capacity. A tempting distractor is choice B, which suggests exponential growth due to unlimited resources, but this misinterprets the data as the growth rate slowed and stabilized rather than continuing to accelerate. To analyze similar trends, always calculate per capita growth rates between intervals to distinguish between exponential and logistic patterns.

Question 13

During translation, the ribosome moves along mRNA in the 5′ to 3′ direction, decoding each codon in order. A researcher introduces an mRNA in which the start codon AUG is unchanged, but the stop codon is mutated to a sense codon that is recognized by a tRNA. The mRNA sequence downstream contains additional codons before a later stop codon. Assume initiation occurs normally and the ribosome remains bound to the mRNA. Which outcome is most likely for the translated product compared with the original mRNA?

  1. A longer polypeptide is produced because translation continues to the next in-frame stop codon. (correct answer)
  2. A shorter polypeptide is produced because mutating a stop codon prevents initiation at AUG.
  3. No polypeptide is produced because a sense codon cannot be read by the ribosome.
  4. An identical polypeptide is produced because stop codons do not affect translation length.
  5. A longer mRNA is produced because translation adds nucleotides after the mutated stop codon.

Explanation: This question assesses the skill of analyzing translation by predicting the effect of mutating a stop codon to a sense codon on polypeptide length. Mutating the stop codon to a sense codon allows a tRNA to bind and add an amino acid at that position, preventing normal termination. The ribosome continues translating downstream codons until it reaches the next in-frame stop codon further along the mRNA. As a result, the translated product is a longer polypeptide incorporating additional amino acids beyond the original termination site. A tempting distractor is B, which claims a shorter polypeptide because the mutation prevents initiation at AUG, but this is incorrect due to the misconception that downstream mutations affect upstream initiation rather than only elongation and termination. To evaluate similar mutations, trace the translation path from the unchanged start codon and note how changes in stop signals extend or shorten the coding sequence.

Question 14

A researcher compares DNA replication in two cell extracts. In both, helicase separates strands and DNA polymerase extends new DNA by adding nucleotides to the 3′ end, guided by base complementarity. In extract 1, the polymerase has normal proofreading ability that removes mismatched nucleotides soon after insertion. In extract 2, the polymerase lacks this mismatch-removal activity but still polymerizes at the same rate. After one round of replication, both samples are sequenced. Which explanation best accounts for a higher frequency of point mutations in extract 2?

  1. Without mismatch removal, incorrect bases remain paired and become fixed after the next replication cycle. (correct answer)
  2. Without mismatch removal, helicase cannot unwind DNA efficiently, causing strand breaks that appear as mutations.
  3. Without mismatch removal, DNA polymerase will synthesize RNA instead of DNA, increasing base substitutions.
  4. Without mismatch removal, complementary base pairing rules change so A pairs with C more often than T.
  5. Without mismatch removal, replication becomes conservative so the original strands accumulate copying errors.

Explanation: This question assesses the skill of analyzing DNA replication, emphasizing the importance of proofreading in maintaining fidelity. Without mismatch removal, incorrect bases inserted during synthesis are not excised, allowing them to persist in the new strand and become permanently incorporated as mutations after the next replication cycle when the mismatched base serves as a template. This increases point mutations because the polymerase, while still adding nucleotides via base complementarity, lacks the exonuclease activity to correct errors immediately after insertion. In extract 1, normal proofreading reduces errors, explaining the difference in mutation frequency between the extracts. A tempting distractor is choice B, which claims helicase cannot unwind DNA efficiently without mismatch removal, but this arises from the misconception that proofreading is linked to unwinding, whereas they are separate processes with proofreading occurring post-insertion. For similar questions, focus on distinguishing the roles of replication enzymes and how defects in one specifically impact fidelity versus progression.

Question 15

A bacterial population uses quorum sensing by releasing small molecule A into the environment. At low cell density, A remains below a threshold and cells show no change in membrane activity. At high density, A accumulates and binds a cytosolic receptor only after A enters cells by diffusion. If the medium is continuously replaced to prevent A buildup, cells remain unresponsive even at high density. Which outcome is most likely when medium replacement stops? Which outcome is most likely when medium replacement stops?

  1. Cells will remain unresponsive because diffusion prevents A from ever entering the cytosol
  2. Cells will become responsive as A accumulates extracellularly and then diffuses into cells to bind receptor (correct answer)
  3. Cells will become responsive only if A is transported through plasmodesmata between cells
  4. Cells will become responsive because A directly phosphorylates membrane lipids outside the cell
  5. Cells will become responsive only after A is translated into a protein inside the cell

Explanation: This question assesses understanding of cell communication via signal transduction pathways in quorum sensing, where density-dependent signals accumulate to trigger responses. Stopping medium replacement allows A to accumulate extracellularly at high density, then diffuse into cells to bind cytosolic receptors and activate membrane activity, making cells responsive. The threshold requirement and diffusion-based entry explain the unresponsiveness at low density or with continuous replacement, without needing plasmodesmata or phosphorylation outside cells. The bacterial context and small molecule nature support passive diffusion rather than translation or active transport for signaling initiation. A tempting distractor is choice A, based on the misconception that diffusion cannot enable intracellular binding, but small molecules readily cross membranes in quorum sensing. For density-dependent signaling, evaluate accumulation and entry mechanisms to predict response changes upon environmental alterations.

Question 16

A checkpoint monitors completion of DNA replication before mitosis by regulating CDK activation. When replication forks are stalled, signaling increases an inhibitory phosphate on the mitotic CDK, preventing activation even if cyclin is bound. A replication-stalling agent is applied, and cells arrest in G2. A second drug is added that blocks the kinase responsible for adding the inhibitory phosphate, while leaving cyclin levels unchanged. Which outcome is most likely after adding the second drug?

  1. Cells remain arrested because cyclin cannot bind CDK when replication is incomplete
  2. Cells enter mitosis more readily because inhibitory phosphorylation signaling is reduced (correct answer)
  3. Cells trigger anaphase immediately because cohesin is degraded by the replication checkpoint
  4. Cells skip S phase because APC/C activation requires stalled replication forks
  5. Cells complete cytokinesis because spindle checkpoint signaling increases inhibitory phosphorylation

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, focusing on the replication checkpoint and CDK inhibitory phosphorylation. The replication-stalling agent activates signaling that increases inhibitory phosphate on mitotic CDK, causing G2 arrest despite bound cyclin. The second drug blocks the kinase adding the inhibitory phosphate, reducing this negative regulation and allowing CDK activation. Cells enter mitosis more readily because the checkpoint-enforced inhibition is alleviated. A tempting distractor is A, which suggests cells remain arrested due to cyclin binding issues, but cyclin binds normally, stemming from the misconception that replication status directly affects cyclin-CDK association. To approach similar problems, analyze sequential perturbations to checkpoints and predict how they alter progression barriers.

Question 17

Two populations of a grass species occur in adjacent fields with no physical barrier. Population M grows on soil contaminated with heavy metals; population N grows on nearby uncontaminated soil. Pollen moves freely between fields, and flowering times overlap. Crosses produce viable seeds, but in reciprocal transplant experiments, seedlings with mixed ancestry have much lower survival on metal-contaminated soil than seedlings from population M, while survival on uncontaminated soil is similar across types. Genetic data show reduced gene flow near loci associated with metal tolerance. Which process most directly explains the increasing reproductive isolation?

  1. Reinforcement driven by selection against hybrids in the contaminated habitat, reducing effective gene flow (correct answer)
  2. Mechanical isolation because flower parts prevent pollen transfer between the two fields
  3. Allopatric speciation because a canyon blocks movement of pollen between populations
  4. Hybrid vigor because hybrids outperform both parent populations in both soil types
  5. Inheritance of acquired metal tolerance by individual plants exposed to contamination

Explanation: This question examines reinforcement as a process strengthening reproductive isolation through selection against hybrids. Hybrids have "much lower survival on metal-contaminated soil" creating strong selection against gene flow at metal-tolerance loci, while pollen moves freely between adjacent fields. This reduced hybrid fitness drives the evolution of stronger reproductive barriers through reinforcement - selection favors individuals that avoid producing unfit hybrid offspring. The pattern of "reduced gene flow near loci associated with metal tolerance" indicates genome-wide isolation is evolving from initial ecological divergence. Choice E incorrectly invokes Lamarckian inheritance of acquired traits, misunderstanding that metal tolerance evolves through natural selection on genetic variation. When analyzing reinforcement, look for reduced hybrid fitness driving the evolution of additional reproductive barriers to minimize maladaptive hybridization.

Question 18

A cultured animal cell is exposed to signaling molecule X, which cannot cross the plasma membrane. Within seconds, intracellular cAMP levels increase. When researchers add a competitive antagonist that binds the extracellular side of receptor R, the cAMP increase is eliminated. However, when cells are treated with a membrane-permeable cAMP analog, the intracellular response occurs even in the presence of the antagonist. Receptor R spans the membrane and is present at similar levels in all treatments. Which of the following best explains how molecule X initiates the intracellular response?

  1. Molecule X diffuses through the membrane and directly activates adenylyl cyclase in the cytosol.
  2. Molecule X binds receptor R, triggering a conformational change that activates a G protein leading to cAMP production. (correct answer)
  3. Molecule X binds receptor R and is transported into the cell by receptor-mediated endocytosis to raise cAMP.
  4. Molecule X increases transcription of the adenylyl cyclase gene, causing higher cAMP within seconds.
  5. Molecule X binds receptor R to supply ATP for cAMP synthesis in the extracellular fluid.

Explanation: This question assesses the skill of analyzing signal transduction pathways in cells. The correct answer is B because molecule X cannot cross the plasma membrane, indicating it must bind to an extracellular receptor site, and the rapid increase in cAMP within seconds suggests activation of a second messenger pathway rather than slower processes like transcription. The competitive antagonist blocking the extracellular side of receptor R eliminates the cAMP increase, confirming that X binds R to initiate signaling, while the membrane-permeable cAMP analog bypassing this block shows the issue is upstream of cAMP production. This aligns with basic signaling principles where ligand binding to a G protein-coupled receptor (GPCR) causes a conformational change that activates an associated G protein, which then stimulates adenylyl cyclase to produce cAMP. A tempting distractor is D, which is wrong due to the misconception that transcriptional changes can occur within seconds, whereas gene expression typically takes minutes to hours. For signal transduction questions, always consider the ligand's ability to cross the membrane and the response timeline to differentiate between direct intracellular actions and receptor-mediated pathways.

Question 19

In a cell line, APC/C activation requires phosphorylation by an upstream mitotic kinase. Once active, APC/C ubiquitinates securin and M-cyclin. A drug inhibits the upstream mitotic kinase but does not affect APC/C abundance or substrate binding. Cells form spindles and align chromosomes but do not initiate chromatid separation.

Which change would most likely trigger chromatid separation despite the drug?

  1. Provide a constitutively active APC/C that does not require phosphorylation for activity (correct answer)
  2. Increase inhibitory phosphorylation of Cdk1 to reduce mitotic substrate phosphorylation
  3. Stabilize securin so separase remains inhibited until kinetochores detach
  4. Block replication origin firing so cells cannot enter metaphase with duplicated DNA
  5. Increase cyclin synthesis to raise Cdk activity and promote APC/C inhibition

Explanation: Signaling-based regulation of the cell cycle is a key skill in understanding how upstream modifications enable APC/C function in anaphase. A constitutively active APC/C bypasses the need for phosphorylation by the inhibited mitotic kinase, allowing ubiquitination of securin and triggering separation despite the drug. This directly activates APC/C without the upstream input. Spindle formation and alignment occur, but separation is blocked by inactive APC/C. A tempting distractor is choice B, increasing Cdk1 inhibition, but this would hinder mitosis further, misconstruing reduced activity as promoting exit. To approach similar problems, seek interventions that render the regulator independent of the blocked step.

Question 20

A membrane contains many identical transport proteins for solute S. When extracellular S is increased from 1 mM to 10 mM, the net influx rate increases. Increasing extracellular S further from 10 mM to 100 mM produces little additional increase in influx. ATP inhibitors do not reduce influx at any concentration. Which explanation best accounts for the observed rate pattern?

  1. S influx is facilitated diffusion, and transport rate plateaus when transport proteins become saturated. (correct answer)
  2. S influx is simple diffusion, and diffusion plateaus when concentration gradients are steep.
  3. S influx is active transport, and inhibitors increase ATP availability for pumping.
  4. S influx is osmosis, and water movement limits solute entry at high concentrations.
  5. S influx is vesicular transport, and vesicle number is fixed by extracellular concentration.

Explanation: This question tests understanding of facilitated diffusion. The correct answer is A, as S influx via facilitated diffusion increases with concentration but plateaus due to protein saturation, unaffected by ATP inhibitors. Limited transporters cause the rate limit at high gradients. Passive process driven by gradients. A tempting distractor is C, which claims active transport and ATP effects, confusing saturation causes. For rate patterns, look for saturation without energy involvement to spot facilitated diffusion.

Question 21

A researcher isolates neurons and skin fibroblasts from the same mouse. Genomic analysis indicates both cell types contain the same set of genes. When the researcher measures RNA levels, neurons have high mRNA for a neurotransmitter receptor gene, while fibroblasts have high mRNA for a collagen gene. The cells are kept at identical temperature and nutrient conditions. Which explanation best accounts for the observed differences in mRNA abundance between the two cell types?

  1. Neurons and fibroblasts contain identical DNA but activate different promoters using cell-specific regulatory proteins. (correct answer)
  2. Fibroblasts lack the neurotransmitter receptor gene because it is deleted during early embryonic divisions.
  3. Neurons have more ribosomes, so they produce more receptor mRNA than fibroblasts can produce.
  4. The collagen mRNA is higher in fibroblasts because collagen genes are located only on fibroblast chromosomes.
  5. The difference occurs because neurons intentionally avoid expressing collagen to maintain their specialized form.

Explanation: This question evaluates knowledge of gene expression and cell specialization, emphasizing differences in mRNA levels despite identical genomes. The best explanation is that neurons and fibroblasts contain identical DNA but activate different promoters using cell-specific regulatory proteins, leading to high neurotransmitter receptor mRNA in neurons and high collagen mRNA in fibroblasts. Regulatory proteins, such as transcription factors, interact with promoters to initiate transcription of specific genes suited to each cell's role. This mechanism allows for tissue-specific gene expression without altering the underlying DNA. A common distractor suggests fibroblasts lack the receptor gene due to deletion, which is incorrect as it confuses gene regulation with permanent DNA loss during development. A useful strategy is to recall that all cells in an organism share the same DNA, with differences arising from regulatory controls on transcription.

Question 22

A tundra ecosystem has producers (mosses), primary consumers (lemmings), and secondary consumers (foxes). Researchers found that mosses store 4,000 kJ/m2^22/yr as new biomass, lemmings store 200 kJ/m2^22/yr, and foxes store 8 kJ/m2^22/yr. Which conclusion is best supported about energy transfer efficiencies between these trophic levels?

  1. Moss-to-lemming efficiency is ~5%, and lemming-to-fox efficiency is ~4%, both within typical ranges. (correct answer)
  2. Moss-to-lemming efficiency is ~20%, because 200 is one-fifth of 4,000.
  3. Lemming-to-fox efficiency is ~0.2%, because secondary consumers receive little energy from producers.
  4. Efficiencies must sum to 100% across the chain, so moss-to-lemming is 96% and lemming-to-fox is 4%.
  5. Efficiencies cannot be compared because energy is recycled and therefore does not decrease with trophic level.

Explanation: This question assesses the skill of analyzing energy flow through ecosystems by calculating efficiencies in a tundra chain. Moss-to-lemming efficiency is ~5% (200/4,000), and lemming-to-fox is ~4% (8/200), both within typical low ranges due to harsh conditions limiting energy capture and conversion. These values highlight dissipative processes like respiration dominating over biomass accumulation. Choice A is supported for recognizing these efficiencies as ecologically reasonable. A tempting distractor is choice B, which miscalculates moss-to-lemming as 20% by erroneous fraction interpretation, a common math misconception in ratios. To assess efficiencies across levels, calculate each pairwise percentage and contextualize against known ecological benchmarks for validation.

Question 23

A lake food chain shows 12,000 kJ stored in phytoplankton biomass, 1,200 kJ in zooplankton biomass, and 120 kJ in small fish biomass over the same interval. Which outcome is most likely if the phytoplankton energy storage drops to 6,000 kJ, with similar transfer efficiencies?

  1. Zooplankton biomass energy would most likely decrease to about 600 kJ over the interval. (correct answer)
  2. Zooplankton biomass energy would most likely increase to about 2,400 kJ over the interval.
  3. Small fish biomass energy would most likely remain near 120 kJ because consumers regulate energy flow.
  4. Small fish biomass energy would most likely increase because less producer energy reduces respiration losses.
  5. Zooplankton biomass energy would most likely stay constant because energy cycles within ecosystems.

Explanation: This question requires analyzing energy flow to predict ecosystem changes when producer energy decreases. The original transfer efficiency from phytoplankton to zooplankton is 10% (1,200/12,000 = 0.10). If phytoplankton energy drops to 6,000 kJ and the same 10% efficiency applies, zooplankton would receive 600 kJ (6,000 × 0.10 = 600). This matches answer A, showing that energy at higher trophic levels decreases proportionally when producer energy decreases. Answer E incorrectly claims energy cycles within ecosystems, confusing energy flow (one-way) with nutrient cycling (circular). When producer energy changes, multiply the new producer value by the original transfer efficiency to predict consumer energy.

Question 24

Amylase activity is measured by the time required to reduce a starch-iodine color to half intensity. At 37°C the time is 20 s; at 5°C the time is 120 s. Amylase is a protein enzyme whose catalysis depends on frequent collisions between enzyme and substrate and proper alignment at the active site. Lower temperature decreases kinetic energy and molecular movement but does not directly break covalent bonds. Substrate concentration and pH are constant. Which explanation best accounts for the slower reaction at 5°C?

  1. Lower temperature reduces molecular motion, decreasing collision frequency and slowing formation of enzyme-substrate complexes. (correct answer)
  2. Lower temperature denatures amylase by disrupting all hydrogen bonds, permanently eliminating the active site.
  3. Lower temperature increases kinetic energy, causing more effective collisions and faster starch hydrolysis.
  4. Lower temperature decreases amylase concentration by stopping translation of the amylase gene in the reaction mixture.
  5. Lower temperature increases reaction rate because enzymes function best when their structures are most rigid.

Explanation: This question assesses the skill of analyzing environmental impacts on enzyme function, specifically how temperature affects enzyme activity. The correct answer is choice A because the stimulus highlights that lower temperature decreases kinetic energy and molecular motion, reducing collision frequency between amylase and starch, which slows enzyme-substrate complex formation. This is supported by the longer time to halve starch-iodine color intensity from 20 s at 37°C to 120 s at 5°C, indicating a slower hydrolysis rate. Protein structure logic confirms that while cold temperatures do not break bonds, they hinder the dynamic movements necessary for effective catalysis without causing denaturation. A tempting distractor is choice B, which is wrong because it claims low temperature denatures the enzyme by disrupting all hydrogen bonds, reflecting the misconception that cooling causes structural loss like heating does, rather than just kinetic slowdown. A transferable strategy for interpreting enzyme-environment questions is to assess temperature's dual role in kinetics and stability, evaluating if the change promotes collisions or risks denaturation based on the enzyme's optimal range.

Question 25

In yeast cells, a pheromone binds a specific membrane receptor and triggers rapid activation of a cytosolic MAP kinase within 2 minutes. A mutant receptor binds pheromone normally but lacks most of its cytosolic tail. In the mutant, MAP kinase activation is greatly reduced. Which of the following best explains why the cytosolic tail is important for early transduction?​

  1. The cytosolic tail provides binding sites for intracellular signaling proteins needed to relay the signal. (correct answer)
  2. The cytosolic tail prevents pheromone degradation by secreting protease inhibitors into the medium.
  3. The cytosolic tail transports pheromone into the nucleus where MAP kinases are stored.
  4. The cytosolic tail converts pheromone into a lipid so it can cross the membrane and activate MAP kinase.
  5. The cytosolic tail increases pheromone concentration outside the cell by pumping it across the membrane.

Explanation: This question tests your ability to analyze signal transduction by examining structure-function relationships in receptors. The correct answer A explains that the cytosolic tail provides essential binding sites for intracellular signaling proteins that relay the pheromone signal to activate the MAP kinase cascade. The evidence that tail deletion reduces MAP kinase activation despite normal pheromone binding demonstrates the tail's role in coupling receptor activation to downstream signaling. Answer C incorrectly suggests the tail transports pheromone to the nucleus, reflecting the misconception that extracellular ligands must physically reach their ultimate targets rather than triggering cascades. When analyzing receptor mutations, distinguish between effects on ligand binding (extracellular) versus signal transmission (intracellular domains).