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AP Biology

AP Biology Practice Test: Practice Test 99

Practice Test 99 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A cell is monitored for chromosome structure. In G1, each chromosome consists of a single DNA molecule. During S phase, each chromosome becomes two sister chromatids joined at the centromere. In metaphase, duplicated chromosomes align at the equator. In anaphase, sister chromatids separate and are considered individual chromosomes. The cell then completes telophase and cytokinesis. Which description best matches chromosome composition in a daughter cell immediately after cytokinesis?

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Question 1

A cell is monitored for chromosome structure. In G1, each chromosome consists of a single DNA molecule. During S phase, each chromosome becomes two sister chromatids joined at the centromere. In metaphase, duplicated chromosomes align at the equator. In anaphase, sister chromatids separate and are considered individual chromosomes. The cell then completes telophase and cytokinesis. Which description best matches chromosome composition in a daughter cell immediately after cytokinesis?

  1. Each chromosome has two sister chromatids because replication occurred during telophase.
  2. Each chromosome has one chromatid because sister chromatids separated during anaphase. (correct answer)
  3. Each chromosome is paired with its homolog because mitosis includes synapsis.
  4. Each chromosome has four chromatids because S phase duplicates chromatids twice.
  5. Each chromosome is absent because chromosomes remain at the metaphase plate after cytokinesis.

Explanation: This question assesses the skill of analyzing the cell cycle, detailing chromosome structure changes. From G1 single chromatids to S phase duplication, metaphase alignment, and anaphase separation, post-cytokinesis daughters have chromosomes with one chromatid each, matching choice B and AP Biology's mitotic process where sisters become independent chromosomes. The monitoring confirms separation as the key event. Telophase and cytokinesis finalize this state. A tempting distractor is choice C, suggesting homolog pairing in mitosis, but this arises from a phase confusion misconception, conflating mitotic with meiotic events. For structure-tracking questions, recall transformations at each phase and verify post-division composition.

Question 2

A bacterial cell imports lactose using a membrane symporter that requires a preexisting H+ gradient: extracellular pH is 6.0 and cytosolic pH is 7.5. When a chemical collapses the H+ gradient without affecting ATP levels, lactose uptake stops. Lactose is more concentrated inside than outside during uptake. Which mechanism best explains lactose entry under normal conditions?

  1. Simple diffusion of lactose through the lipid bilayer down its gradient
  2. Facilitated diffusion of lactose through a carrier down its gradient
  3. Secondary active transport coupling lactose import to H+ moving down gradient (correct answer)
  4. Primary active transport of lactose using ATP binding and hydrolysis
  5. Exocytosis exporting H+ to allow lactose to diffuse inward freely

Explanation: This question tests understanding of secondary active transport mechanisms. The lactose symporter uses the pre-existing H+ gradient (lower pH outside means higher H+ concentration outside) to power lactose uptake against its concentration gradient (lactose is more concentrated inside). When the H+ gradient is collapsed, transport stops even though ATP is available, proving the transporter doesn't use ATP directly but rather harnesses the H+ gradient energy. This coupling of lactose uphill movement to H+ downhill movement defines secondary active transport via symport. Students might choose primary active transport (D) because lactose moves uphill, but the dependence on H+ gradient rather than ATP directly distinguishes secondary from primary transport. To identify secondary transport, look for uphill movement of one substance coupled to downhill movement of another.

Question 3

A phospholipid bilayer separates two solutions. Two solutes are tested for simple diffusion through the lipid portion only: NH3 (ammonia; small, uncharged) and NH4+ (ammonium; small, positively charged). The solutions have the same total nitrogen concentration, but the membrane is impermeable to charged species unless they can enter the hydrophobic core. Charged solutes remain strongly associated with water molecules, making entry into the nonpolar interior unfavorable. The uncharged form has fewer electrostatic interactions with water and can partition into the bilayer more readily.​

  1. NH4+, because its positive charge is attracted to the negative phosphate groups
  2. NH3, because the uncharged form more readily enters the hydrophobic interior (correct answer)
  3. NH4+, because ions diffuse faster than neutral molecules in membranes
  4. NH3, because it is polar and therefore dissolves best in the lipid core
  5. Both equally, because they are the same size and differ only by one proton

Explanation: This question examines membrane permeability differences between charged and uncharged forms of the same molecule. The phospholipid bilayer's hydrophobic core strongly excludes charged species because they remain tightly associated with water molecules through electrostatic interactions. NH3 is uncharged and can partition into the nonpolar interior, while NH4+ carries a positive charge that keeps it strongly hydrated and prevents membrane entry. Choice A incorrectly assumes that positive charges are attracted to negative phosphate groups, but phosphate groups face outward toward water, not into the hydrophobic core where diffusion occurs. For weak acids and bases, remember that only the uncharged form can cross membranes by simple diffusion—charge creates an insurmountable barrier.

Question 4

A cell’s plasma membrane contains ion channels that allow passive diffusion of K+\text{K}^+K+. Two cells have the same channel density (channels per μm2\mu m^2μm2) and similar cytoplasmic composition. One cell doubles its linear dimensions, increasing its volume eightfold. If channel density stays constant, the cell’s ability to maintain the same cytoplasmic K+\text{K}^+K+ concentration becomes less stable over time. Which explanation best accounts for this change?

  1. Total membrane surface area decreases when linear dimensions increase, reducing total ion flux.
  2. The surface area–to–volume ratio decreases, reducing ion exchange capacity per unit cytoplasm. (correct answer)
  3. The diffusion coefficient of K+\text{K}^+K+ decreases in larger cells due to increased cytoplasmic pressure.
  4. Larger cells always have more channels per area, increasing passive loss of K+\text{K}^+K+.
  5. Volume increases faster than surface area, so intracellular K+\text{K}^+K+ must increase automatically.

Explanation: This question tests understanding of how surface area-to-volume ratio affects cellular processes like ion maintenance. The correct answer is B because doubling linear dimensions decreases the surface area-to-volume ratio, reducing the membrane area for ion exchange relative to the cytoplasmic volume that requires stable K+ levels. This makes it harder to counteract passive loss through channels, destabilizing concentration. This transport efficiency logic illustrates the challenges larger cells face in ion homeostasis via membrane channels. A tempting distractor is A, which wrongly states total surface area decreases, based on the misconception that scaling reduces absolute area when it actually increases it. To approach similar problems, compute how changes in linear dimensions affect surface area-to-volume ratios and transport capacity.

Question 5

A synthetic membrane is made of phospholipids and has no embedded proteins. Two solutes are placed on one side at equal concentrations: ethanol (2 carbons, one hydroxyl; uncharged) and ethylene glycol (2 carbons, two hydroxyls; uncharged). The bilayer interior is nonpolar, so solutes must leave their hydration shell to enter it. Molecules with more hydroxyl groups form more hydrogen bonds with water and are more strongly hydrated. Because ethanol and ethylene glycol are similar in size, the primary determinant of permeability is polarity rather than molecular mass.​

  1. Ethylene glycol, because additional hydroxyl groups increase attraction to membrane lipids
  2. Ethanol, because fewer hydrogen bonds with water increase partitioning into the bilayer (correct answer)
  3. Ethylene glycol, because it is uncharged and therefore crosses without resistance
  4. Ethanol, because its hydroxyl group becomes positively charged in the membrane core
  5. Both equally, because they have the same number of carbon atoms

Explanation: This problem tests membrane permeability analysis based on the number of polar groups. The synthetic membrane's nonpolar interior requires solutes to leave their hydration shell to enter, which is energetically costly for polar molecules. Ethanol has one hydroxyl group while ethylene glycol has two, making ethylene glycol more polar and more strongly hydrogen-bonded to water molecules. Therefore, ethanol can more readily partition into the bilayer and cross faster. Choice A incorrectly suggests that additional hydroxyl groups increase attraction to membrane lipids, when they actually increase attraction to water and decrease membrane permeability. When comparing molecules of similar size, count hydroxyl groups—each additional OH group decreases membrane permeability by increasing water interactions.

Question 6

A student stains two cell types from the same individual for a transcription factor (TF) that binds the enhancer of Gene Y. The TF is present in the nucleus of cell type 1 but absent from the nucleus of cell type 2. Only cell type 1 contains detectable Gene Y mRNA. Both cell types contain the same Gene Y DNA sequence. Which explanation best accounts for Gene Y expression in cell type 1 but not cell type 2?

  1. Gene Y is expressed only in cell type 1 because its DNA is replicated more times in that cell type.
  2. Cell type 2 has a different genome that lacks the enhancer sequence for Gene Y.
  3. Nuclear localization of the TF in cell type 1 enables transcription of Gene Y, producing mRNA. (correct answer)
  4. Cell type 1 produces Gene Y mRNA because its ribosomes bind DNA directly during translation.
  5. Cell type 2 cannot produce Gene Y mRNA because mRNA synthesis occurs only in mitochondria.

Explanation: This question examines gene expression and cell specialization through transcription factor localization. The correct answer C explains that nuclear localization of the transcription factor in cell type 1 enables it to bind Gene Y's enhancer and activate transcription, producing mRNA, while absence from the nucleus in cell type 2 prevents Gene Y expression. Transcription factors must be in the nucleus to access DNA and regulate genes - their cellular localization is a key control point in cell specialization. Answer B incorrectly suggests cell type 2 has a different genome lacking the enhancer, failing to recognize that regulatory differences, not DNA differences, drive cell specialization. When analyzing gene expression patterns, consider not just which regulatory proteins are present, but whether they can access their target DNA sequences in the nucleus.

Question 7

In endothelial cells, ligand L binds receptor tyrosine kinase RTK, causing receptor dimerization and autophosphorylation on cytosolic tyrosines. An adaptor protein (Ad) binds the phosphotyrosines and recruits kinase K1, which phosphorylates kinase K2. Active K2 phosphorylates ion channel C, increasing Ca2+^{2+}2+ influx. A cytosolic phosphatase P removes phosphate groups from RTK and from K2. In cells lacking P, a brief 5 s pulse of L produces Ca2+^{2+}2+ influx that persists for 2 min; in wild-type cells, influx returns to baseline within 20 s. Which explanation is best supported by the results?

  1. P normally increases ligand binding affinity, reducing receptor activation time
  2. P normally dephosphorylates pathway components, limiting signal duration after L removal (correct answer)
  3. P normally phosphorylates ion channel C, preventing sustained Ca2+^{2+}2+ entry
  4. Loss of P increases ATP synthesis, amplifying phosphorylation of all kinases
  5. Loss of P causes RTK to enter the nucleus and directly open Ca2+^{2+}2+ channels

Explanation: This question assesses the skill of analyzing signal transduction pathways by interpreting experimental outcomes to explain the role of a regulatory enzyme in signal duration. The pathway entails ligand L inducing RTK dimerization and autophosphorylation, recruiting adaptor Ad to activate kinase K1, which phosphorylates K2 to activate ion channel C and increase Ca²⁺ influx, with phosphatase P dephosphorylating RTK and K2 to terminate signaling. In cells lacking P, a brief L pulse results in prolonged Ca²⁺ influx because phosphates on RTK and K2 persist, sustaining downstream activation beyond the normal 20 seconds observed in wild-type cells. This supports that P normally dephosphorylates pathway components to limit signal duration after ligand removal, as its absence allows unchecked persistence. A tempting distractor is choice C, which claims P phosphorylates channel C to prevent sustained Ca²⁺ entry, but this misidentifies P as a kinase rather than a phosphatase, reflecting a confusion in enzyme function. A transferable strategy is to compare wild-type and mutant responses in pathways to deduce the roles of regulators like phosphatases in controlling signal amplitude and timing.

Question 8

In a neuron, Neurotransmitter Y binds a receptor embedded in the postsynaptic membrane. Immediately after binding, the membrane potential rapidly depolarizes, and the effect ends when Y is removed from the synaptic cleft. No cytosolic second messenger increase is detected during the first second. Which prediction is most consistent with the receptor’s role in early signal transduction?

  1. The receptor functions as a ligand-gated ion channel that changes ion permeability upon Y binding. (correct answer)
  2. The receptor must enter the nucleus to bind DNA and initiate the depolarization response.
  3. The receptor catalyzes transcription of ion channel genes, causing rapid depolarization within one second.
  4. The receptor is a soluble enzyme that diffuses out of the cell to bind Y in the synaptic cleft.
  5. Y binds phospholipids directly, and the membrane depolarizes because lipids become phosphorylated.

Explanation: This question requires analyzing signal transduction to determine how a neurotransmitter causes rapid membrane depolarization. The stimulus indicates that Neurotransmitter Y causes immediate depolarization that ends when Y is removed, with no cytosolic second messenger detected in the first second. This rapid, reversible response is characteristic of ligand-gated ion channels, where neurotransmitter binding directly opens the channel to allow ion flow and change membrane potential. Choice B incorrectly suggests the receptor enters the nucleus, which would take far longer than the immediate response observed and wouldn't directly cause depolarization. For rapid signaling events measured in milliseconds to seconds, look for direct mechanisms like ion channels rather than slower pathways involving gene expression or second messengers.

Question 9

Two macromolecules are compared. Molecule X is a long chain of repeating units, each containing a central carbon bonded to an amino group, a carboxyl group, a hydrogen, and a variable R group; adjacent units are covalently linked with water released. Molecule Y is composed of a sugar-phosphate backbone with nitrogenous bases attached; adjacent units are covalently linked with water released. Which statement best predicts how the monomer differences relate to polymer classification?

  1. X is a nucleic acid and Y is a protein because both contain nitrogen in their monomers.
  2. X is a protein and Y is a nucleic acid because their monomers form peptide versus phosphodiester bonds. (correct answer)
  3. X is a carbohydrate and Y is a lipid because both are formed by dehydration reactions.
  4. X is a lipid and Y is a carbohydrate because neither forms true polymers of monomers.
  5. X and Y are both carbohydrates because their backbones contain repeating carbon atoms.

Explanation: This question assesses the skill of analyzing macromolecule categories and structure-function relationships. Molecule X's repeating units with a central carbon, amino group, carboxyl group, hydrogen, and variable R group, linked covalently with water release, identify it as a protein formed by peptide bonds, while Y's sugar-phosphate backbone with nitrogenous bases matches nucleic acids formed by phosphodiester bonds, aligning with AP Biology macromolecule classifications. The monomer differences—amino acids versus nucleotides—directly relate to polymer types, where peptide bonds enable diverse protein folding via R groups, and phosphodiester bonds support informational sequences in nucleic acids. This distinction underscores how bonding types and monomer structures determine overall function, such as catalysis in proteins versus genetic coding in nucleic acids. A tempting distractor is choice A, which is incorrect due to a level-of-organization error, as nitrogen presence alone does not classify polymers, ignoring specific bonding and backbone differences. To approach similar questions, contrast monomer functional groups and linkage types to predict and differentiate macromolecule classes accurately.

Question 10

In a species with 2n = 10, a meiotic cell completes prophase I and enters metaphase I. If one homologous pair fails to synapse, which outcome is most likely during anaphase I for that pair?

  1. The homologs may not segregate reliably to opposite poles, increasing nondisjunction risk. (correct answer)
  2. The sister chromatids will separate early, producing haploid cells after meiosis I.
  3. The pair will replicate again, restoring synapsis before anaphase I begins.
  4. The homologs will separate normally because synapsis occurs only in meiosis II.
  5. The homologs will be excluded from the spindle and degraded before cytokinesis.

Explanation: This question analyzes the consequences of failed synapsis during prophase I and its impact on chromosome segregation. When homologous chromosomes fail to synapse (pair up), they cannot form the normal bivalent structure that ensures proper attachment to opposite spindle poles. Without synapsis, the homologs may both attach to the same pole or attach randomly, significantly increasing the risk of nondisjunction where both homologs move to the same daughter cell instead of segregating properly. This creates aneuploid gametes with incorrect chromosome numbers. Choice D incorrectly claims synapsis occurs in meiosis II, when homologous pairing actually occurs exclusively in prophase I. When analyzing meiosis problems involving synapsis failure, recognize that proper pairing is essential for accurate chromosome segregation during anaphase I.

Question 11

A peptide ligand binds a receptor on immune cells. When extracellular pH is lowered, ligand binding decreases sharply, but receptor abundance on the membrane is unchanged. Which of the following best explains the reduced binding at low pH?

  1. Low pH alters charge interactions and protein conformation, reducing ligand–receptor affinity (correct answer)
  2. Low pH increases receptor gene transcription, temporarily diluting receptors across the membrane
  3. Low pH causes the ligand to become nonpolar and diffuse through the membrane, avoiding receptors
  4. Low pH blocks blood flow, preventing long-distance endocrine delivery of the peptide to cells
  5. Low pH prevents vesicles from forming, so receptors cannot be inserted into the membrane

Explanation: This question assesses understanding of cell communication via signal transduction pathways. The correct answer is A because low extracellular pH alters charge interactions and protein conformation in the ligand or receptor, reducing their binding affinity without changing receptor abundance on the membrane. Evidence from the stimulus indicates that binding decreases sharply at low pH, but receptor levels remain unchanged, pointing to a direct effect on interaction dynamics. This is common in immune signaling where pH influences receptor-ligand pairs. A tempting distractor is C, which claims low pH makes the ligand nonpolar, but this is incorrect due to the misconception that pH changes molecular polarity, ignoring that peptides remain polar. A transferable strategy is to consider environmental factors like pH when binding affinity changes without alterations in receptor number.

Question 12

Some hypotheses propose hydrothermal vents as sites for early chemistry. In a study, microporous iron-sulfide “chimney” structures were placed in a solution containing CO2, H2, and dissolved minerals. After several days, researchers detected increased concentrations of simple organic acids inside pores compared with the surrounding solution, even without added enzymes or cells. Which conclusion is best supported by this evidence about the origin of life?

  1. Organic acids inside pores indicate that early Earth required ultraviolet light to produce any carbon-based molecules.
  2. The pore enrichment supports the idea that natural mineral compartments could concentrate and promote abiotic organic synthesis. (correct answer)
  3. Because organics formed, the first organisms must have been photosynthetic and oxygen-producing.
  4. The results show that only ATP-dependent enzymes can catalyze carbon fixation reactions.
  5. The experiment confirms that eukaryotic cells evolved before any prebiotic chemistry occurred.

Explanation: This question assesses the skill of evaluating evidence about the origins of life on Earth. The microporous iron-sulfide structures, mimicking hydrothermal vents, showed higher concentrations of organic acids inside pores than outside, even without enzymes, supporting that mineral compartments could naturally concentrate and promote abiotic synthesis. This aligns with AP Biology concepts of prebiotic chemistry where environmental gradients at vents provide energy and surfaces for organic reactions. The absence of cells or added catalysts in the experiment underscores the plausibility of non-biological origins for life's precursors. A tempting distractor, choice C, is wrong by assuming photosynthetic organisms formed first, which is a structure–function confusion linking organic formation directly to oxygen production instead of abiotic processes. When tackling such questions, evaluate how the data support specific hypotheses about early chemical environments and avoid assuming modern metabolic pathways.

Question 13

In a coastal fish population, a locus with alleles M and m is monitored. At time 1, allele frequencies are p(M)=0.55p(M)=0.55p(M)=0.55 and q(m)=0.45q(m)=0.45q(m)=0.45. Over the next year, a nearby river floods and connects the lagoon to an upstream population in which p(M)=0.20p(M)=0.20p(M)=0.20 and q(m)=0.80q(m)=0.80q(m)=0.80. After the connection, researchers measure the lagoon population and find p(M)=0.42p(M)=0.42p(M)=0.42 and q(m)=0.58q(m)=0.58q(m)=0.58. No evidence indicates differential survival among genotypes in the lagoon during this period.

Which Hardy-Weinberg condition is most clearly violated in the lagoon population?

  1. No gene flow, because movement of individuals between populations changed allele frequencies. (correct answer)
  2. Random mating, because flooding forces fish to choose mates nonrandomly.
  3. No mutation, because floods increase mutation rates at the M locus.
  4. No selection, because allele-frequency change always indicates selection.
  5. Large population size, because p(M)=0.42p(M)=0.42p(M)=0.42 can occur only in small populations.

Explanation: This question assesses the skill of analyzing Hardy-Weinberg equilibrium in populations. No gene flow is most clearly violated because the flood introduced individuals from an upstream population with different allele frequencies, shifting the lagoon's p(M) from 0.55 to 0.42 and q(m) from 0.45 to 0.58. This immigration mixes alleles, directly altering frequencies without evidence of selection or other forces. The connection via flooding confirms gene flow as the mechanism disrupting equilibrium. A tempting distractor is choice D, which claims no selection is violated because allele-frequency changes always indicate selection, based on the misconception that any shift implies fitness differences, but gene flow can cause changes independently. When allele frequencies change abruptly, investigate migration or gene flow before assuming internal forces like selection.

Question 14

A cell enters mitosis with duplicated chromosomes. During metaphase, each sister chromatid pair is attached to microtubules from opposite poles. The cell then undergoes anaphase and telophase normally. However, cytokinesis is experimentally inhibited, so the cleavage furrow does not form and the cytoplasm does not divide. The cell reforms nuclear envelopes around the separated chromosome sets. Which outcome is most likely immediately after telophase in this cell?

  1. A single nucleus with 4C DNA because chromatids rejoin at centromeres after anaphase.
  2. Two nuclei in one cell, each with 2C DNA. (correct answer)
  3. Two nuclei in one cell, each with 4C DNA.
  4. Two cells, each with 1C DNA because cytokinesis drives reduction division.
  5. A single nucleus with 2C DNA because DNA is degraded when cytokinesis fails.

Explanation: This question assesses the skill of analyzing the cell cycle, investigating failed cytokinesis outcomes. After normal anaphase and telophase but inhibited cytokinesis, the cell reforms envelopes around separated 2C chromosome sets, yielding one cell with two 2C nuclei, per choice B and AP Biology's distinction between nuclear and cytoplasmic division. The stimulus confirms no cytoplasmic split despite segregation. This produces a binucleate cell. A tempting distractor is choice C, suggesting two 4C nuclei, but this stems from a structure-function confusion, assuming segregation does not halve DNA until cytoplasm divides. For division perturbation questions, separate nuclear and cytoplasmic events to predict cellular configuration.

Question 15

Two monomers are mixed in a test tube. Monomer 1 has a carboxyl group (–COOH) and monomer 2 has an amino group (–NH2). After incubation, the solution contains a covalently linked dimer and a measurable increase in water molecules. Repeating this reaction can produce a longer chain whose properties depend on the sequence and interactions among side groups. Which statement best describes the bond formed and the macromolecule produced?

  1. A glycosidic bond forms, producing a polysaccharide that stores energy in glucose linkages.
  2. A phosphodiester bond forms, producing a nucleic acid with a sugar-phosphate backbone.
  3. A peptide bond forms, producing a polypeptide from amino acid monomers via dehydration synthesis. (correct answer)
  4. An ester bond forms, producing a triglyceride from glycerol and three fatty acids.
  5. A hydrogen bond forms, producing a lipid bilayer held together by base pairing.

Explanation: This question requires identifying bond formation and macromolecule type from functional group interactions. The carboxyl group (–COOH) of monomer 1 reacts with the amino group (–NH2) of monomer 2 through dehydration synthesis, forming a peptide bond and releasing water—the hallmark of protein formation from amino acid monomers. The mention of sequence-dependent properties and side group interactions further confirms polypeptide/protein identity. Option A incorrectly identifies this as glycosidic bond formation, representing a functional group confusion since glycosidic bonds form between hydroxyl groups of sugars, not between carboxyl and amino groups. To identify polymerization reactions, match the reacting functional groups (carboxyl + amino = peptide bond) to the correct bond type and resulting macromolecule.

Question 16

A lab uses flow cytometry to measure DNA content. A sample shows most cells at 2C, with a smaller population at 4C. Cells at 2C have unreplicated chromosomes and intact nuclei, consistent with G1. Cells at 4C have completed DNA replication and are either in G2 or early M before chromatid separation. The lab then adds a compound that prevents entry into S phase by blocking initiation of DNA replication, while leaving mitosis unaffected. After one full cycle time, which change in the DNA-content distribution is most likely?

  1. The 4C population decreases as cells divide, and few new 4C cells appear (correct answer)
  2. The 4C population increases as cells replicate DNA repeatedly without mitosis
  3. A new 1C peak appears as homologous chromosomes separate during meiosis
  4. The 2C population disappears because cells cannot survive without spindle attachment
  5. Both 2C and 4C peaks shift upward because DNA is synthesized during cytokinesis

Explanation: This question assesses the skill of analyzing the cell cycle, interpreting flow cytometry changes under replication blockade. The stimulus shows initial 2C (G1) and 4C (G2/M) populations, then a compound blocking S phase entry while allowing mitosis. In AP Biology, blocking S prevents new 4C cells, while existing 4C cells divide to 2C, reducing the 4C population. Thus, the 4C population decreases as cells divide, with few new 4C cells forming. A tempting distractor is B, suggesting 4C increase via repeated replication, but this is incorrect due to a structure-function confusion, mistaking the block's target for enabling endocycles. To approach similar questions, simulate population shifts by considering blocked and permitted transitions in the cycle.

Question 17

Cells treated with a microtubule-depolymerizing drug show many unattached kinetochores. The spindle checkpoint signal inhibits APC/C, stabilizing securin and cyclin B. A second compound is added that forces APC/C to remain active even when checkpoint signaling is present. Which outcome is most likely with both drugs present?

  1. Sister chromatids separate despite unattached kinetochores, increasing segregation errors (correct answer)
  2. Cells arrest in G1 because cyclin D–CDK4/6 requires microtubules to function
  3. Cells remain in metaphase because APC/C activation strengthens checkpoint inhibition
  4. DNA replication increases because APC/C activation directly activates DNA polymerase
  5. Cyclin B levels rise because APC/C activity promotes cyclin accumulation

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, specifically the spindle checkpoint inhibiting APC/C in response to unattached kinetochores. The microtubule drug creates unattached kinetochores, but the second compound forces APC/C activity, overriding inhibition. This causes chromatid separation with errors, as in choice A, due to missegregation. The outcome stems from checkpoint bypass in the stimulus. A tempting distractor is choice C, which predicts metaphase arrest, but this misconceives that forced APC/C sustains rather than overcomes inhibition. A transferable strategy is to combine perturbations to evaluate checkpoint robustness and failure modes.

Question 18

In a reptile species, eggs from the same mother are incubated at either 26°C or 32°C. Hatchlings from 26°C develop testes, while hatchlings from 32°C develop ovaries. DNA sequencing of hatchlings shows no differences in sex-determining genes between temperature groups. Which explanation best accounts for the different gonad phenotypes?

  1. Incubation temperature altered expression of sex-determining genes during development without changing DNA. (correct answer)
  2. Higher temperature caused point mutations that converted testes genes into ovary genes.
  3. Temperature changed the species’ allele frequencies by favoring ovary genotypes in warm incubators.
  4. Warm incubation doubled chromosome sets in embryos, producing ovaries through polyploidy.
  5. Temperature replaced nuclear DNA with mitochondrial DNA, changing gonad development pathways.

Explanation: This question assesses understanding of environmental effects on phenotype, specifically how temperature can influence sex determination without altering the genetic code. The correct answer, A, is right because incubation temperature affects the expression of sex-determining genes through temperature-sensitive regulatory mechanisms, such as hormone signaling or epigenetic modifications, leading to testes or ovaries without DNA changes. This demonstrates phenotypic plasticity where the same genotype produces different phenotypes based on environmental cues during critical developmental windows. DNA sequencing confirming no differences reinforces that the variation is due to gene regulation rather than sequence alterations. A tempting distractor is B, which incorrectly claims temperature causes point mutations, reflecting the misconception that environmental factors directly edit the genome instead of modulating expression. To approach similar questions, consider how environmental signals can activate or repress gene expression pathways to produce variable phenotypes in identical genotypes.

Question 19

An enzyme catalyzes an endergonic reaction A→BA \rightarrow BA→B in the cytosol only when ATP is present. Measurements show that during catalysis, ATP is converted to ADP and a phosphorylated intermediate forms transiently on molecule A. When ATP is absent, no intermediate forms and little B is produced. The enzyme does not change the overall free energy of ATP hydrolysis, but it increases reaction rate. This scenario emphasizes energetic coupling by using ATP hydrolysis to create a higher-energy intermediate that can proceed to product formation.

  1. ATP hydrolysis is coupled by phosphorylating A, creating an intermediate that can proceed to B (correct answer)
  2. ATP provides activation energy by being stored unchanged in the enzyme active site
  3. ATP increases enzyme concentration, which raises the equilibrium amount of B
  4. ATP is required because A must be reduced by ATP before it can become B
  5. ATP is required because ADP has higher free energy than ATP and drives A→BA \rightarrow BA→B

Explanation: This question assesses the skill of analyzing cellular energy transformations, specifically how ATP couples to endergonic reactions via phosphorylated intermediates. The correct answer is A because ATP hydrolysis phosphorylates A, forming a high-energy intermediate that drives the endergonic conversion to B, coupling the exergonic hydrolysis to the reaction. Evidence reveals a transient phosphorylated A during catalysis with ATP, absent without it, and the enzyme accelerates the rate without altering ATP's free energy. This adheres to energy principles where ATP provides energy through group transfer, making unfavorable reactions feasible via intermediates. A tempting distractor is E, which is incorrect by stating ADP has higher energy than ATP, based on the misconception of reversed ATP/ADP energetics. A transferable strategy for cellular energy questions is to identify intermediate forms that link ATP hydrolysis to endergonic processes in metabolic pathways.

Question 20

DNA replication relies on complementary base pairing and synthesis of new strands in the 5′→3′ direction. DNA polymerase selects incoming nucleotides by matching them to the template base, and many polymerases also remove a mispaired nucleotide immediately after insertion before continuing synthesis. A researcher compares two cell populations: one with normal polymerase and one with a polymerase variant that cannot remove mispaired nucleotides but still synthesizes DNA at the same rate. After one round of replication, the researcher sequences the new DNA. Which result is most likely in the variant population?

  1. Fewer mutations because replication proceeds faster and reduces exposure to damage.
  2. More base-substitution mutations because mispaired nucleotides remain in the new strand. (correct answer)
  3. More silent mutations because polymerase inserts only purines on the leading strand.
  4. No change in mutations because complementary pairing alone fully prevents errors.
  5. More insertions because polymerase begins copying RNA templates instead of DNA.

Explanation: This question examines DNA replication fidelity and the proofreading function of DNA polymerase. Normal DNA polymerase has 3'→5' exonuclease activity that allows it to remove incorrectly paired nucleotides immediately after insertion, maintaining high replication accuracy. Without this proofreading ability, mismatched base pairs that occur during nucleotide selection remain in the newly synthesized strand, leading to more base-substitution mutations after replication is complete. Choice D incorrectly assumes that complementary base pairing alone prevents all errors, ignoring that polymerase occasionally makes mistakes during nucleotide selection. To solve such problems, focus on what specific enzyme functions are lost and their direct consequences for replication accuracy.

Question 21

A polar solute is placed outside cells at high concentration. Net solute entry occurs only when a specific channel protein is open. Entry rate increases with the solute gradient but does not show saturation over the tested range. No ATP is required, and net movement stops when the inside and outside concentrations become equal. Which mechanism best explains the solute’s movement?

  1. Primary active transport by an ATPase moving solute against its gradient
  2. Facilitated diffusion through a channel down the solute’s electrochemical gradient (correct answer)
  3. Simple diffusion directly through the phospholipid bilayer
  4. Secondary active transport requiring a Na+^++ gradient and cotransport
  5. Exocytosis of solute from intracellular vesicles to the outside

Explanation: This question assesses the skill of analyzing membrane transport mechanisms. Facilitated diffusion is correct because the polar solute moves down its gradient through an open channel protein, without ATP, stopping at equilibrium. The rate increases with gradient but lacks saturation, typical of channels rather than carriers. Protein dependence distinguishes it from simple diffusion. A tempting distractor is simple diffusion, incorrect due to the misconception that polar solutes easily cross bilayers, overlooking the need for protein facilitation. To analyze transport, always evaluate if movement is down a gradient (passive) or against it (active) and check for energy requirements.

Question 22

In a cultured mammalian cell line, entry into mitosis depends on a checkpoint where cyclin B binds Cdk1 to form active MPF. MPF activity rises sharply only when inhibitory phosphates on Cdk1 are removed by a phosphatase, and MPF activity falls when cyclin B is rapidly degraded by a ubiquitin-ligase complex activated later in mitosis. A researcher adds a small molecule that prevents removal of the inhibitory phosphates on Cdk1 without changing cyclin B abundance. Which change would most likely occur in cell cycle progression after treatment?

  1. Cells accumulate in G2 because Cdk1 remains inhibited despite cyclin B binding. (correct answer)
  2. Cells bypass the G2 checkpoint and enter mitosis early due to increased MPF activity.
  3. Cells accelerate S phase because cyclin B directly activates DNA polymerase at replication forks.
  4. Cells arrest in G1 because cyclin B degradation prevents activation of G1/S cyclin-Cdk complexes.
  5. Cells complete mitosis faster because inhibitory phosphorylation stabilizes spindle microtubule attachments.

Explanation: This question assesses understanding of signaling-based regulation of the cell cycle, emphasizing the activation of mitosis-promoting factor (MPF) for G2/M transition. Normally, MPF forms when cyclin B binds Cdk1, but full activity requires removal of inhibitory phosphates by a phosphatase to trigger mitosis entry. The small molecule prevents this dephosphorylation, keeping Cdk1 inhibited even with cyclin B present, causing cells to accumulate in G2 without progressing to mitosis, as in choice A. This blocks the sharp rise in MPF activity needed for mitotic entry. A tempting distractor is choice B, suggesting early mitosis due to increased MPF, but this ignores the necessity of dephosphorylation for activation, confusing binding with full enzymatic function. A transferable strategy is to identify key activation steps in cyclin-CDK complexes and evaluate how inhibitors affect checkpoint passage.

Question 23

In a temperate lake, an introduced planktivorous fish increases and reduces large zooplankton density by 70%. Phytoplankton biomass then doubles. Which consequence is most likely for the lake ecosystem?

  1. Water clarity decreases and deeper aquatic plants decline because increased phytoplankton reduces light penetration. (correct answer)
  2. Water clarity increases because fewer zooplankton means less suspended organic matter in the water column.
  3. Aquatic plant growth increases because phytoplankton release nutrients directly to plants through photosynthesis.
  4. Phytoplankton biomass decreases because fewer zooplankton reduces carbon dioxide availability in surface waters.
  5. Lake stability increases because the food web becomes shorter, eliminating indirect effects among trophic levels.

Explanation: This question assesses the skill of analyzing disruptions in ecosystems by analyzing a trophic cascade in a temperate lake following fish introduction. The reduction in large zooplankton allows phytoplankton biomass to double, decreasing water clarity and light penetration, which harms deeper aquatic plants. System-level impacts include reduced primary production from submerged vegetation and altered energy flow through the aquatic food web. This cascade demonstrates how top-down control shifts can amplify productivity at lower levels while suppressing others. A tempting distractor is choice B, suggesting clarity increases with fewer zooplankton, but this reflects the misconception that zooplankton are the primary source of suspended matter rather than controllers of phytoplankton. When evaluating aquatic invasions, trace predator effects through trophic levels to predict changes in light availability and producer dynamics.

Question 24

A cell’s size increases but the number of membrane transport proteins remains constant. Each protein transports a fixed number of molecules per second, and cellular demand scales with cytoplasmic volume. Which explanation best accounts for the change in transport capacity per unit cytoplasm as the cell enlarges?

  1. Transport capacity per unit cytoplasm decreases because total transport stays constant while volume increases. (correct answer)
  2. Transport capacity per unit cytoplasm increases because volume increase raises protein activity.
  3. Transport capacity per unit cytoplasm stays constant because proteins set a fixed concentration.
  4. Transport capacity per unit cytoplasm increases because surface area increases faster than volume.
  5. Transport capacity per unit cytoplasm decreases because larger cells have higher surface area–to–volume ratios.

Explanation: This question examines how fixed transport capacity interacts with changing cell volume. When a cell enlarges but maintains the same number of transport proteins, the total transport capacity remains constant (each protein moves a fixed number of molecules per second). However, cellular demand increases proportionally with cytoplasmic volume. This means the transport capacity per unit cytoplasm decreases as the cell grows larger, since the same number of proteins must now serve a larger volume of cytoplasm. Choice B incorrectly suggests that volume increase somehow raises protein activity, confusing total capacity with capacity per unit volume. The key insight is that without proportionally increasing transport proteins, larger cells face transport limitations even with active transport mechanisms.

Question 25

In a grassland, flowering plants receive visits from native bees that collect nectar and pollen. In cages that exclude bees but allow wind, plants produce fewer seeds per flower than in uncaged controls, while bee abundance outside cages remains unchanged. Herbivore damage and soil moisture are similar between treatments. Which interaction best explains the higher seed set in the uncaged controls?

  1. Parasitism by bees decreases plant reproduction in controls
  2. Competition between bees and plants for pollen limits seed production
  3. Mutualism between bees and flowering plants increases pollination success (correct answer)
  4. Predation by bees on herbivores increases seed set directly
  5. Amensalism by plants reduces bee foraging and raises seed output

Explanation: This question assesses the skill of analyzing community ecology by identifying species interactions based on experimental outcomes. Mutualism between bees and flowering plants enhances pollination, as bees transfer pollen while collecting nectar, leading to higher seed production in uncaged controls. Exclusion of bees reduces seed set, showing that the interaction benefits both the plants (via reproduction) and bees (via food), with no changes in herbivores or moisture. This explains the difference in seed output, as wind alone is insufficient for effective pollination. A tempting distractor is choice B, proposing competition for pollen, but this is wrong due to the misconception that mutual benefits are confused with resource rivalry, ignoring the facilitative nature of pollination. To evaluate mutualistic interactions, assess how exclusion affects reproductive success and resource acquisition in both species.