AP Biology

AP Biology Practice Test: Practice Test 5

Practice Test 5 for AP Biology: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

Two membranes have the same phospholipid bilayer composition but differ in cholesterol content. Membrane A has high cholesterol; Membrane B has low cholesterol. At a low temperature, small nonpolar molecules diffuse more readily through Membrane A than through Membrane B. Cholesterol is embedded among phospholipid tails. Which membrane property change best explains the observation?

Which change would most likely explain higher permeability in the high-cholesterol membrane at low temperature?

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Question 1

Which change would most likely explain higher permeability in the high-cholesterol membrane at low temperature?

Cholesterol reduces tight packing of phospholipid tails at low temperature, maintaining fluidity (correct answer)
Cholesterol increases the polarity of the bilayer interior, dissolving nonpolar molecules faster
Cholesterol forms open channels that allow nonpolar molecules to pass through the membrane
Cholesterol binds to nonpolar molecules and carries them across by active transport
Cholesterol increases phospholipid head charge, pulling nonpolar molecules across the surface

Explanation: This question assesses the skill of analyzing plasma membrane structure and transport. At low temperatures, Membrane A with high cholesterol allows better diffusion of small nonpolar molecules by reducing tight packing of phospholipid tails, thus maintaining membrane fluidity. Cholesterol embeds among the tails and prevents crystallization-like solidification that would occur in low-cholesterol Membrane B, increasing permeability. Both membranes have the same phospholipids, so cholesterol's moderating effect on fluidity explains the difference. A tempting distractor is choice C, which falsely claims cholesterol forms open channels, confusing its structural role with that of transport proteins. To assess permeability at varying temperatures, analyze how cholesterol influences bilayer fluidity and packing.

Question 2

A human red blood cell lacks a nucleus and most internal membrane-bound organelles. Despite this, it maintains a flexible biconcave shape while squeezing through narrow capillaries. Which cellular feature best explains this flexibility at the cellular level?

A cytoskeletal network beneath the membrane that supports shape while allowing deformation (correct answer)
A thick peptidoglycan cell wall that provides both rigidity and flexibility
Large central vacuoles that press outward to keep the cell biconcave
Chloroplast membranes that fold to permit passage through capillaries
Mitochondrial cristae that act as springs to restore cell shape

Explanation: This question assesses the skill of analyzing cell structure-function relationships. The cytoskeletal network, including spectrin and actin, beneath the red blood cell's plasma membrane provides mechanical support, allowing flexibility to deform through capillaries while maintaining the biconcave shape. This is evident despite the lack of nucleus and organelles, as noted in the stimulus. In AP Biology, the erythrocyte cytoskeleton exemplifies adaptation for circulation without rigid walls. A tempting distractor is B, suggesting a peptidoglycan wall, but this is incorrect due to level-of-organization error, as animal cells lack bacterial cell walls. For shape-related questions, evaluate internal support structures and their mechanical properties.

Question 3

In immune cells, ligand L binds a receptor tyrosine kinase (RTK) on the plasma membrane. Within 30 seconds of adding L, the receptor becomes phosphorylated on cytosolic tyrosines, and a cytosolic adaptor protein binds the phosphorylated receptor. If a mutation removes the receptor’s cytosolic kinase domain, L still binds extracellularly but receptor phosphorylation is not detected. Which of the following best explains the earliest disrupted step?

Ligand binding to the extracellular domain is prevented because the kinase domain determines ligand specificity.
The receptor cannot catalyze phosphorylation needed to create docking sites for adaptor proteins. (correct answer)
The mutation forces L to bind intracellular receptors instead of membrane receptors.
The receptor becomes constitutively phosphorylated because kinase domains inhibit phosphorylation.
Adaptor proteins bind directly to L, so receptor phosphorylation is not relevant to early signaling.

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on receptor-ligand interactions and early transduction events. The correct answer is B because removing the kinase domain prevents receptor phosphorylation despite normal L binding, indicating the domain is essential for catalyzing phosphorylation that creates docking sites for adaptor proteins. This aligns with RTK signaling principles where ligand binding induces autophosphorylation, enabling adaptor binding within 30 seconds. The rapid timeframe and specific loss of phosphorylation confirm the kinase domain's role in early transduction. A tempting distractor is A, which incorrectly ties the kinase to ligand specificity, stemming from the misconception that intracellular domains control extracellular binding, though binding is unchanged. When analyzing signal transduction questions, use mutation effects on phosphorylation and binding to pinpoint roles in reception versus transduction.

Question 4

A digestive protease shows maximal activity at pH 2.0. When placed at pH 8.0, activity decreases to near zero, but returns to near maximal when moved back to pH 2.0. Which condition would most likely reduce activity at pH 8.0?

Deprotonation of key active-site residues disrupts charge-based interactions needed for substrate binding and catalysis. (correct answer)
Increased pH increases peptide bond formation in the enzyme, locking the active site in an open conformation.
Increased pH increases enzyme gene expression, producing many immature enzymes with incorrect amino acids.
Increased pH causes substrate molecules to disappear from solution by converting them into gas.
Increased pH permanently denatures the enzyme by cleaving disulfide bonds into separate amino acids.

Explanation: This question examines environmental impacts on enzyme function, specifically how pH affects a protease adapted to acidic conditions. The correct answer is A because at pH 8.0, amino acid side chains that are normally protonated at pH 2.0 become deprotonated, disrupting the charge-based interactions essential for substrate binding and catalytic mechanism in the active site. The reversibility when returned to pH 2.0 confirms this involves ionization changes rather than permanent damage. For acid-adapted enzymes, key catalytic residues like aspartate or glutamate need to be protonated to function properly. Answer E is incorrect because it suggests permanent denaturation through disulfide bond cleavage, which would prevent activity recovery—this reflects the misconception that pH changes break covalent bonds. To analyze pH effects, consider how changing protonation states of ionizable groups affects both structure and catalytic mechanism.

Question 5

A group of 30 small birds is observed on a cold morning ( $2^{\circ}\mathrm{C}$ ) and a mild afternoon ( $18^{\circ}\mathrm{C}$ ). In the morning, birds fluff feathers and huddle; in the afternoon, birds spread out and feathers lie flat. Which response best explains the short-term mechanism underlying these behavioral changes?

Fluffing feathers and huddling reduce heat loss by trapping air and decreasing exposed surface area. (correct answer)
Birds increase external temperature by releasing heat into the air, warming the environment for hours.
Birds alter feather genes in the morning, producing permanently thicker plumage by the afternoon.
Birds flatten feathers in the cold to increase conduction, which raises body temperature quickly.
Birds huddle in the cold to ensure more food appears later, so they can forage efficiently.

Explanation: This question assesses the skill of analyzing how organisms respond to changes in their external environment. In the cold morning, birds fluff feathers to trap insulating air and huddle to reduce exposed surface area, minimizing heat loss through convection and radiation. In the milder afternoon, they spread out and flatten feathers as less insulation is needed, allowing normal activity. These behaviors represent short-term thermoregulatory responses that conserve body heat without generating external warmth or altering plumage genetically. A tempting distractor is choice C, which suggests altering feather genes for thicker plumage, but this misconceptions mixes behavioral adjustments with long-term genetic adaptations. A transferable strategy is to identify insulation and grouping behaviors as reversible short-term responses to temperature, distinguishing them from metabolic or evolutionary strategies.

Question 6

At a replication fork, single-strand DNA-binding proteins (SSBs) bind to exposed parental DNA strands after helicase separates them. DNA polymerase then uses each parental strand as a template, extending new DNA by complementary base pairing. In a mutant extract lacking functional SSBs, the parental strands frequently re-form hydrogen bonds with each other soon after unwinding. Which outcome is most likely in the mutant extract compared with a normal extract?

Replication fork progression will slow because reannealing reduces the availability of single-stranded templates. (correct answer)
Replication accuracy will increase because reannealing forces correct base pairing in the new strands.
Okazaki fragments will become longer because polymerase can add nucleotides in both directions.
The leading strand will be synthesized as RNA because SSBs are required for DNA nucleotide selection.
The parental strands will be degraded because SSBs normally catalyze phosphodiester bond formation.

Explanation: This question assesses the skill of analyzing DNA replication, addressing the function of single-strand binding proteins in maintaining template availability. Without SSBs, parental strands reanneal after unwinding, reducing single-stranded template exposure and slowing replication fork progression as polymerase has less access to templates. This reannealing disrupts the normal process where SSBs stabilize separated strands, allowing continuous complementary synthesis. Consequently, the mutant extract exhibits delayed replication compared to normal, where SSBs prevent such interference. A tempting distractor is choice B, suggesting accuracy increases due to forced base pairing, but this misconceives reannealing as aiding new strand fidelity when it actually hinders synthesis by reforming parental duplexes. When evaluating replication mutants, consider how each component stabilizes the fork and predict effects on speed versus accuracy.

Question 7

A coastal plant population includes individuals with either high leaf salt-gland density or low salt-gland density. A storm surge increases soil salinity for several growing seasons. Plants with high gland density have higher survival and set more seeds than plants with low gland density under these conditions. In the next generation, the proportion of seedlings with high gland density increases. Genetic analyses show both phenotypes were present before the storm surge.

Which statement best explains the role of variation in this evolutionary change?

The salt surge induced high gland density in all plants, and the new phenotype was inherited.
Existing variation in gland density led to differential reproductive success, changing trait frequencies in the population. (correct answer)
Low gland density plants increased gland density by using more salt, so their offspring had higher density.
The population increased gland density because individuals needed to remove salt to survive.
Random mating alone increased high gland density without any differences in survival or seed set.

Explanation: This question requires analyzing how existing variation in populations enables evolutionary responses to environmental change. The correct answer (B) recognizes that genetic variation in salt-gland density was already present before the storm surge, and this variation led to differential reproductive success when salinity increased—plants with high gland density survived better and produced more seeds. This differential reproduction changed the trait frequencies in the next generation, with more seedlings having high gland density. Answer A incorrectly suggests the salt induced the trait in all plants (acquired characteristic), but the question explicitly states both phenotypes existed before the storm. To solve variation problems, always check whether the variation was preexisting and heritable, then trace how differential survival or reproduction changes population frequencies.

Question 8

A cell is heterozygous for two genes (A/a and B/b) located on the same homologous chromosome pair. In prophase I, homologs synapse and a single crossover occurs between the A and B loci on non-sister chromatids. After meiosis, some gametes carry Ab and aB combinations not present on either original homolog. Which process best explains the appearance of these recombinant gametes?

Crossing over between non-sister chromatids during prophase I producing recombinant chromatids (correct answer)
Independent assortment of sister chromatids during metaphase II producing new allele combinations
Replication errors in S phase creating additional homologous chromosomes with new alleles
Random fusion of gametes after meiosis producing Ab and aB zygotes
Segregation of homologous chromosomes in anaphase I creating alleles absent from chromatids

Explanation: This question tests understanding of how meiosis generates genetic diversity through crossing over between homologous chromosomes. The scenario describes genes A/a and B/b on the same chromosome producing recombinant gametes (Ab and aB) that weren't present on the original homologs, which can only occur through crossing over during prophase I. During crossing over, non-sister chromatids (one from each homolog) exchange corresponding DNA segments at the crossover point, physically recombining alleles that were originally linked on the same chromosome. This creates chromatids with new allele combinations - if the original homologs were AB and ab, crossing over between the loci produces Ab and aB recombinants. Students often confuse this with independent assortment (answer B), but independent assortment of sister chromatids in metaphase II cannot create new combinations since sister chromatids are identical. When genes are on the same chromosome, only crossing over can produce recombinant gametes with allele combinations not found on the original parental chromosomes.

Question 9

In a fish population, allele $T$ has frequency $p=0.4$ and allele $t$ has frequency $q=0.6$ . The population meets Hardy-Weinberg conditions. Which expected genotype frequency is correct?

$TT=0.16$ (correct answer)
$Tt=0.24$
$tt=0.36$
$TT=0.40$
$Tt=0.60$

Explanation: This question tests Hardy-Weinberg equilibrium analysis by calculating expected genotype frequencies from given allele frequencies. With p(T) = 0.4 and q(t) = 0.6, we apply Hardy-Weinberg formulas: TT = p² = (0.4)² = 0.16, Tt = 2pq = 2(0.4)(0.6) = 0.48, and tt = q² = (0.6)² = 0.36. Therefore, the expected frequency of TT is 0.16. Choice E incorrectly states Tt = 0.60, which would be the frequency of allele t, not the heterozygote genotype frequency. To calculate Hardy-Weinberg genotype frequencies, always use p² for dominant homozygotes, 2pq for heterozygotes, and q² for recessive homozygotes.

Question 10

In a diploid cell, homologous chromosomes pair during prophase I, forming tetrads. A crossover occurs between two non-sister chromatids, producing chromatids that contain segments derived from both homologs. The cell then proceeds through meiosis I and II, separating homologs first and sister chromatids second. The resulting gametes contain chromatids that are not identical to either original parental chromatid due to exchanged segments. Which stage and process directly created these recombinant chromatids?

Prophase I crossing over between non-sister chromatids within a tetrad (correct answer)
Metaphase I alignment of tetrads at the equator without DNA exchange
Anaphase I separation of sister chromatids producing new allele sequences
Telophase II re-formation of nuclei causing chromosomal segment swapping
S phase DNA replication generating chromatids with mixed homolog segments

Explanation: This question tests understanding of how meiosis generates genetic diversity through crossing over during prophase I. The correct answer is A because the scenario explicitly describes crossover occurring between non-sister chromatids within a tetrad during prophase I, producing recombinant chromatids containing segments from both homologs. The question traces the complete process: tetrad formation, crossover between non-sisters, and subsequent separation producing gametes with exchanged segments. Answer C (anaphase I separation of sister chromatids) is incorrect because it contains a critical error about meiotic mechanics—sister chromatids remain attached at anaphase I and only homologs separate, making this a common misconception about when different chromosomal elements divide. To identify crossing over, look for DNA exchange between non-sister chromatids specifically during prophase I when homologs are paired as tetrads.

Question 11

In a eukaryotic cell, the cytosol is near neutral pH, while lysosomes maintain an acidic lumen using proton pumps. Many lysosomal hydrolases show maximal activity only at low pH and have reduced activity at neutral pH. If a lysosomal membrane becomes leaky, protons diffuse into the cytosol and the lysosomal pH rises toward neutral. Shortly after, the rate of macromolecule breakdown inside lysosomes decreases even though the hydrolase proteins remain present. Which feature best explains how compartmentalization influences this change in cellular function?

Acidic lysosomal pH creates a microenvironment that increases hydrolase activity and confines digestion from the cytosol. (correct answer)
Lysosomes contain different DNA than the nucleus, producing enzymes that function only inside lysosomes.
Ribosomes on lysosomes synthesize hydrolases directly into the lumen to increase reaction speed.
Mitochondria supply ATP to lysosomes so hydrolases can catalyze reactions at neutral pH.
Lysosomal membranes exist so the cell can break down molecules faster when it needs additional nutrients.

Explanation: This question requires analysis of cell compartmentalization to understand how pH-dependent enzyme function relies on membrane-bounded organelles. The correct answer (A) accurately describes how lysosomes maintain an acidic pH through proton pumps, creating a microenvironment where hydrolases achieve maximal activity while protecting the neutral cytosol from potentially damaging digestive enzymes. When the membrane becomes leaky, protons diffuse out and the lysosomal pH rises toward neutral, causing hydrolases to lose activity even though the proteins remain intact, demonstrating that compartmentalization creates specialized chemical environments essential for enzyme function. Option E incorrectly suggests a teleological explanation where membranes exist "so the cell can break down molecules faster when needed," confusing evolutionary adaptation with purposeful design. The key strategy for compartmentalization questions is to identify how membrane boundaries create and maintain distinct chemical environments (pH, ion concentrations, redox states) that optimize specific biochemical processes.

Question 12

In a meiosis with no crossing over, a cell contains three homologous chromosome pairs. For each pair, either the maternal or paternal homolog can move to a given pole during anaphase I. The student focuses only on how whole homologs segregate, not on alleles within a chromosome. Which statement best explains how this process generates different combinations of chromosomes in gametes from different meioses?

Independent assortment produces multiple possible sets of maternal and paternal homologs in gametes (correct answer)
Crossing over is required to separate homologous chromosomes into different gametes
Meiosis II determines which homologs enter gametes by separating homologous pairs
DNA replication during interphase changes chromosome identity, creating new homolog combinations
Fertilization determines which chromosomes were assorted during meiosis in the parent cell

Explanation: This question tests understanding of how meiosis generates genetic diversity through independent assortment at the whole chromosome level. Independent assortment produces multiple possible sets of maternal and paternal homologs (A) is correct because during metaphase I, each homologous pair orients randomly, creating 2^n possible combinations of maternal and paternal chromosomes in gametes (where n = number of chromosome pairs). With three pairs, this gives 8 possible combinations. Meiosis II determines which homologs enter gametes (C) is incorrect because homolog segregation occurs in meiosis I, not meiosis II—meiosis II separates sister chromatids, which doesn't affect which homologs are present. To understand chromosomal diversity, focus on how random orientation of multiple chromosome pairs multiplies the possible combinations exponentially.

Question 13

A bird population on an island was counted annually. For four years, the population stayed near 1,200 birds. In year 5, a hurricane reduced the population to 700 birds. In years 6–8, the population increased to 900, then 1,050, then 1,170 birds. Habitat area and food availability returned to pre-hurricane levels by year 6, and no new predators arrived. Which factor most likely caused the sharp decline in year 5?

Density-dependent competition intensified gradually as the population approached carrying capacity
A density-independent disturbance abruptly increased mortality regardless of population density (correct answer)
Improved resource availability reduced births, lowering the population size suddenly
Genetic changes reduced fitness, causing an immediate population decline in one year
Long-term emigration steadily reduced the population, producing a single-year crash

Explanation: This question tests understanding of population ecology by distinguishing density-dependent from density-independent factors. The hurricane represents a density-independent disturbance that reduced the population from 1,200 to 700 birds regardless of population density - hurricanes affect mortality through physical destruction, not through population-mediated mechanisms. After the disturbance, the population recovered following a logistic pattern back toward the original carrying capacity as habitat recovered. Choice A incorrectly suggests gradual density-dependent competition, but the data shows an abrupt one-year decline rather than gradual change, which is characteristic of catastrophic events. When analyzing population crashes, examine whether the decline is sudden (suggesting density-independent factors) or gradual (suggesting density-dependent factors).

Question 14

A diploid cell begins meiosis with 12 chromosomes, each consisting of two sister chromatids. After meiosis I completes cytokinesis, which description best matches each resulting cell’s chromosome state?

Each cell has 12 chromosomes, each still composed of two sister chromatids.
Each cell has 6 chromosomes, each still composed of two sister chromatids. (correct answer)
Each cell has 6 chromosomes, each composed of a single chromatid.
Each cell has 3 chromosomes because DNA replication halves chromosome number.
Each cell has 24 chromosomes because homologs replicate during meiosis I.

Explanation: This question requires tracking chromosome structure through meiosis I to understand the difference between chromosome number and chromatid number. Starting with 12 chromosomes (each consisting of two sister chromatids), meiosis I separates homologous chromosomes but not sister chromatids. Since the cell is diploid with 12 chromosomes, it has 6 homologous pairs. After meiosis I, each daughter cell receives one member from each homologous pair, resulting in 6 chromosomes per cell. Crucially, each chromosome still consists of two sister chromatids joined at the centromere because sister chromatids don't separate until meiosis II. Choice C incorrectly suggests sister chromatids have already separated, which would make each chromosome a single chromatid. When analyzing meiosis problems, distinguish between chromosome number (counting centromeres) and chromatid number (counting DNA molecules).

Question 15

In fruit flies, gray body (G) is completely dominant to black body (g), and normal wings (N) are completely dominant to vestigial wings (n). Two flies heterozygous for both traits (GgNn) are crossed, assuming independent assortment. Offspring with black bodies and vestigial wings must be homozygous recessive at both genes. Which proportion of offspring is expected to show both recessive phenotypes?

1/16 (correct answer)
1/8
3/16
1/4
9/16

Explanation: This question examines dihybrid crosses between heterozygotes using Mendelian inheritance with independent assortment. In GgNn × GgNn, each parent produces four gamete types (GN, Gn, gN, gn) at 1/4 frequency each. The double recessive phenotype (black body, vestigial wings) requires the ggnn genotype, which only occurs when both parents contribute gn gametes. The probability is 1/4 (gn from parent 1) × 1/4 (gn from parent 2) = 1/16. Students often confuse this with single-gene crosses where recessive phenotypes appear at 1/4 frequency, but with two genes, you must consider both simultaneously. For any specific two-gene phenotype in a dihybrid cross, multiply the individual gene probabilities: for recessive at both genes, it's 1/4 × 1/4 = 1/16.

Question 16

A researcher designs a short nucleic acid probe to bind a target RNA sequence in solution. The probe is made of DNA rather than RNA but still binds the RNA target strongly. Which feature best explains why a DNA probe can base-pair with an RNA target?

DNA and RNA share the same sugar, enabling identical backbone geometry for pairing.
Complementary base pairing depends on hydrogen bonding patterns of bases that occur in both DNA and RNA. (correct answer)
DNA contains uracil, allowing perfect matching with adenine in RNA targets.
DNA forms peptide cross-links that stabilize binding to RNA through covalent attachment.
DNA is positively charged, so it electrostatically binds the negatively charged RNA backbone.

Explanation: This question requires analyzing nucleic acids as macromolecules to understand cross-type base pairing. DNA probes can bind RNA targets because complementary base pairing depends on the hydrogen bonding patterns of the nitrogenous bases (A, U/T, G, C), which are present in both nucleic acid types regardless of sugar differences. The bases adenine, guanine, and cytosine are identical in DNA and RNA, while thymine in DNA can hydrogen bond with adenine just as uracil does in RNA, allowing DNA-RNA hybrid formation through standard Watson-Crick pairing. Choice C incorrectly states that DNA contains uracil, representing a structure-function confusion where students mix up the characteristic bases of each nucleic acid type. The key strategy for nucleic acid hybridization questions is to focus on base complementarity rather than sugar identity, as hydrogen bonding between bases drives specificity.

Question 17

A population of dividing cells is exposed to a chemical that prevents completion of DNA replication. Cells enter S phase and begin copying chromosomes, but replication forks stall and many regions remain unreplicated. When these cells reach the G2 checkpoint, the checkpoint detects incomplete DNA replication and blocks entry into M phase. Which result is most likely observed in the treated cells after several hours?

Cells accumulate in M phase with sister chromatids separating normally
Cells accumulate in G2 with partially replicated DNA and no mitotic spindle (correct answer)
Cells skip S phase and enter mitosis with unreplicated chromosomes
Cells complete cytokinesis, producing gametes with half the chromosome number
Cells enter G1 and immediately replicate DNA again without mitosis

Explanation: This question requires analysis of the cell cycle, specifically how the G2 checkpoint prevents cells with incomplete DNA replication from entering mitosis. The G2 checkpoint monitors DNA integrity and replication completion, and when it detects unreplicated DNA regions, it blocks progression into M phase, causing cells to accumulate in G2. Since the chemical prevents completion of DNA replication but cells can still enter and remain in S phase, and the G2 checkpoint blocks their progression, cells will accumulate in G2 with partially replicated DNA and no mitotic spindle formation (answer B). Answer C incorrectly suggests cells skip S phase, but the question states cells do enter S phase and begin replication—this represents a sequence error where students misunderstand that checkpoints prevent progression forward, not cause cells to skip phases. The strategy is to trace the cell cycle sequence and identify where checkpoint activation causes accumulation.

Question 18

The fossil record provides strong evidence for evolution. Which aspect of the fossil record best supports the concept that evolution is a continuous, ongoing process rather than an event that occurred only in the distant past?

The existence of fossils of extinct organisms, which proves that life forms have changed over Earth's history.
The discovery of living fossils, organisms that have remained morphologically unchanged for millions of years.
The consistent ordering of fossils in rock strata, with simpler organisms in older layers and more complex ones in younger layers.
The presence of transitional fossils that show intermediate anatomical features between ancestral and more recently evolved groups of organisms. (correct answer)

Explanation: Transitional fossils, such as Archaeopteryx (linking reptiles and birds) or Tiktaalik (linking fish and amphibians), demonstrate the incremental changes that occur over time as one lineage evolves into another. This provides a snapshot of evolution in progress, supporting its continuous nature. Choice A shows that change occurs but not necessarily that it's continuous. Choice B demonstrates evolutionary stasis in some lineages, not ongoing change. Choice C shows a general pattern of change over time but doesn't illustrate the continuous, step-by-step process as well as transitional forms do.

Question 19

A person enters a cold room; within minutes, skin blood vessels constrict and shivering begins. Which response mechanism best explains these changes?

The person’s genome changes to code for more heat-producing proteins immediately.
Vasoconstriction and muscle contractions increase heat conservation and production. (correct answer)
The body stops cellular respiration to prevent heat loss to the environment.
Shivering occurs to keep the population’s average temperature stable over time.
The person grows a thicker layer of skin during the cold exposure.

Explanation: This question tests understanding of human thermoregulatory responses to cold exposure. When body temperature sensors detect cold, the hypothalamus triggers two key responses: vasoconstriction reduces blood flow to the skin (conserving heat in the core), while shivering produces heat through rapid muscle contractions that increase metabolic rate. These are homeostatic mechanisms that maintain core body temperature through the autonomic nervous system and do not involve genetic changes or structural modifications. Option A incorrectly suggests immediate genomic changes, which cannot occur in minutes. For temperature regulation questions, identify whether responses involve behavioral changes, physiological adjustments, or long-term adaptations.

Question 20

In an experiment, cells are grown for one generation in medium containing a heavy nitrogen isotope so that newly synthesized DNA strands incorporate the heavy isotope. The cells are then shifted to light nitrogen medium and allowed to complete exactly one round of DNA replication. DNA molecules are separated by density. The results show a single band at an intermediate density. The student recalls that replication uses each parental strand as a template and that complementary base pairing determines the sequence of the new strand. Which explanation best accounts for the intermediate-density band after one round in light medium?

Replication is semiconservative, producing DNA molecules with one heavy parental strand and one light new strand. (correct answer)
Replication is conservative, producing one heavy molecule and one light molecule that overlap into one band.
Replication is dispersive, producing DNA molecules with alternating heavy and light bases within each strand.
Transcription occurs during S phase, producing heavy RNA that shifts DNA density to an intermediate value.
Complementary base pairing causes heavy nitrogen to pair only with light nitrogen, yielding intermediate density.

Explanation: This question assesses the skill of analyzing DNA replication, particularly the semiconservative model demonstrated by density labeling. The intermediate-density band after one round in light medium results from semiconservative replication, where each daughter molecule consists of one heavy parental strand and one light new strand, yielding a hybrid density. This aligns with the mechanism where each parental strand templates a new complementary strand via base pairing, dispersing the heavy isotope into hybrid molecules. No heavy-heavy or light-light molecules form after one round, confirming the single intermediate band. A tempting distractor is choice C, suggesting dispersive replication with alternating heavy and light bases, but this misconceives the model by implying random isotope distribution within strands rather than whole-strand conservation. To tackle such experiments, remember to track isotope distribution across generations based on the replication model being tested.

Question 21

A scientist isolates mitochondria and measures the pH in the intermembrane space and matrix. The intermembrane space is more acidic than the matrix. Which structure best explains how this pH difference is maintained during respiration?

The inner mitochondrial membrane contains electron transport proteins that pump protons into the intermembrane space (correct answer)
The outer mitochondrial membrane contains chlorophyll that releases protons into the intermembrane space
The mitochondrial matrix contains ribosomes that secrete protons to lower the intermembrane pH
The mitochondrial DNA forms channels that move protons against their gradient into the matrix
The cytoskeleton surrounding mitochondria actively transports protons into the intermembrane space

Explanation: This question assesses the skill of analyzing cell structure-function relationships. The inner mitochondrial membrane houses electron transport chain proteins that pump protons into the intermembrane space during respiration, creating the acidic pH difference observed. This ties to chemiosmosis in AP Biology, where the proton gradient across the cristae drives ATP synthesis, maintaining lower pH in the intermembrane space. The membrane's impermeability to protons sustains this gradient for energy production. A tempting distractor is choice B, which is incorrect due to structure-function confusion, as mitochondria lack chlorophyll; that's in chloroplasts, and the outer membrane is permeable but not directly involved in proton pumping. To approach similar questions, trace energy transfer processes through organelle compartments and identify key membrane functions.

Question 22

A membrane receptor activates JAK kinase, which phosphorylates a cytosolic protein STAT. Phosphorylated STAT (STAT-P) forms dimers in the cytosol within 1 minute. A phosphatase dephosphorylates STAT-P, returning STAT to monomers. When the phosphatase is inhibited, STAT remains phosphorylated longer and dimer levels stay high. Which change would most likely decrease STAT dimer levels without altering receptor activation?

Prevent STAT phosphorylation by inhibiting JAK catalytic activity (correct answer)
Increase ligand concentration so STAT dimers dissociate due to receptor saturation
Block phosphatase activity to accelerate STAT dephosphorylation and reduce dimers
Add more receptor molecules so STAT dimers are diluted in the cytosol
Chelate extracellular Na $^+$ so STAT cannot dimerize in low-ionic conditions

Explanation: This question assesses the skill of analyzing signal transduction pathways by decreasing dimer levels without affecting receptor activation in JAK-STAT signaling. Preventing STAT phosphorylation by inhibiting JAK stops STAT-P formation, reducing dimers, as phosphatase inhibition shows phosphorylation drives dimerization. This targets the modification required for dimer assembly. Receptor activation remains intact, isolating the effect. A tempting distractor is choice C, blocking phosphatase to accelerate dephosphorylation, but blocking slows it, increasing STAT-P, confusing inhibition with activation. To modulate assemblies, target the enzymes controlling the necessary post-translational modifications.

Question 23

In an epithelial cell, intracellular glucose is 20 mM while extracellular glucose is 2 mM. A cotransporter moves glucose into the cell only when Na+ is present outside at high concentration (140 mM outside, 10 mM inside). When ATP is depleted, the Na+ gradient dissipates and glucose uptake stops, even though the cotransporter protein remains intact. Which mechanism best explains glucose uptake when ATP is available?

Which mechanism best explains glucose uptake into the cell?

Facilitated diffusion of glucose down its concentration gradient
Secondary active transport of glucose coupled to Na+ influx (correct answer)
Simple diffusion of glucose through the lipid bilayer
Primary active transport of glucose by direct ATP hydrolysis
Osmosis-driven glucose movement through aquaporin channels

Explanation: This question requires analyzing secondary active transport mechanisms. Glucose is moving against its concentration gradient (2 mM outside to 20 mM inside), which requires energy. The cotransporter uses the Na+ gradient (140 mM outside, 10 mM inside) as its energy source—Na+ moves down its gradient while glucose moves up its gradient. When ATP is depleted, the Na+ gradient dissipates, stopping glucose uptake even though the transporter remains functional. Facilitated diffusion (A) is incorrect because glucose is moving against its gradient, which passive transport cannot accomplish. Remember that secondary active transport couples uphill movement of one substance to downhill movement of another.

Question 24

Cells are exposed to ligand V, and within 12 seconds a fluorescent reporter shows increased diacylglycerol (DAG) at the inner leaflet of the plasma membrane. When a drug blocks cleavage of PIP $_2$ , ligand binding still occurs but DAG does not increase. Which of the following best explains the earliest biochemical event producing DAG?

DAG is generated when an activated enzyme cleaves PIP $_2$ in the membrane during early transduction. (correct answer)
DAG diffuses into the cell from the extracellular fluid after V binds the receptor.
DAG is synthesized by ribosomes directly on the receptor’s extracellular domain.
DAG increases because V is phosphorylated into DAG by the receptor’s ligand-binding site.
DAG increases only after new membrane phospholipids are transcribed and inserted into the membrane.

Explanation: This question assesses the skill of analyzing signal transduction pathways, focusing on receptor-ligand interactions and early transduction events. The correct answer is A because blocking PIP2 cleavage prevents DAG increase despite V binding, indicating an activated enzyme like PLC cleaves PIP2 to generate DAG at the membrane within 12 seconds. Fluorescent data show DAG at the inner leaflet, aligning with principles of phospholipid-derived messengers. The rapid timing supports enzymatic production. A tempting distractor is B, which incorrectly suggests DAG diffusion from outside, arising from the misconception that messengers are imported, though they are generated internally. When analyzing signal transduction questions, track second messenger location and inhibitors to trace their biochemical origins.

Question 25

A chemist synthesizes a nucleic acid analog in which the phosphate group is replaced with an uncharged linkage, while the bases and sugars remain unchanged. Compared with normal DNA, the analog shows reduced attraction to positively charged proteins. Which feature best explains this change?

Removing phosphate eliminates negative charges on the backbone, decreasing electrostatic interactions with positive charges. (correct answer)
Removing phosphate prevents hydrogen bonding between bases, so proteins cannot bind to the major groove.
Removing phosphate converts nucleotides into amino acids, changing the polymer into a polypeptide.
Removing phosphate increases the number of purines, which reduces binding by disrupting base pairing.
Removing phosphate adds methyl groups to thymine, which blocks all protein interactions with nucleic acids.

Explanation: This question requires analyzing nucleic acids as macromolecules to understand charge-based interactions. Replacing phosphate groups with uncharged linkages eliminates the negative charges normally present on the nucleic acid backbone, reducing electrostatic attraction to positively charged amino acids in proteins. Normal DNA-protein interactions often involve ionic bonds between negatively charged phosphates and positively charged lysine or arginine residues, so removing this charge prevents these stabilizing interactions. Choice C incorrectly claims that removing phosphate converts nucleotides to amino acids, representing a level-of-organization error where students confuse modifications within a molecule type with conversion between different macromolecule classes. The strategy for understanding nucleic acid-protein interactions is to recognize that backbone phosphates provide negative charges essential for ionic interactions with positive protein residues.

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