All questions
Question 1
A neuron releases neurotransmitter N into a synaptic cleft. On the postsynaptic cell, N binds a receptor that spans the membrane and opens an ion channel within milliseconds. When an enzyme that degrades N is added to the cleft, the postsynaptic ion flow stops quickly even though the presynaptic neuron continues releasing N. The postsynaptic cell has no direct cytoplasmic connection to the presynaptic neuron. Which outcome is most likely if the degrading enzyme concentration is increased?
Which outcome is most likely if the degrading enzyme concentration is increased?
- Postsynaptic ion flow will persist longer because receptors bind N more tightly
- Postsynaptic ion flow will stop sooner because N is removed from the cleft faster (correct answer)
- Postsynaptic ion flow will increase because more N is released to compensate
- Postsynaptic ion flow will be unchanged because N acts only inside presynaptic cells
- Postsynaptic ion flow will stop only after N alters transcription in the postsynaptic nucleus
Explanation: This question assesses understanding of cell communication via signal transduction pathways, particularly in synaptic signaling where neurotransmitters trigger rapid responses. The enzyme degrading N causes postsynaptic ion flow to stop quickly, and increasing its concentration would accelerate N removal from the cleft, shortening the duration of receptor activation and ion channel opening. The millisecond-scale response and lack of cytoplasmic connections indicate N acts extracellularly on postsynaptic receptors, with signal termination relying on degradation rather than internalization or compensation. Even as the presynaptic neuron continues releasing N, faster degradation prevents sustained binding, leading to quicker cessation of the response. A tempting distractor is choice A, stemming from the misconception that tighter binding prolongs signaling, but here degradation rate controls duration, not binding affinity. When evaluating neurotransmitter effects, consider mechanisms of signal termination like degradation to predict changes in response timing.
Question 2
A diploid cell (2n=6) undergoes meiosis. During prophase I, synapsis occurs normally, but crossing over is experimentally prevented. Homologous chromosomes still align as tetrads at metaphase I and segregate to opposite poles in anaphase I. Sister chromatids separate in meiosis II. Which outcome is most likely regarding the chromatid composition of the resulting gametes?
- Gametes contain recombinant chromatids because synapsis alone causes segment exchange.
- Gametes contain chromatids that match the original parental chromatids for each chromosome. (correct answer)
- Gametes contain diploid sets because crossing over is required for homolog separation.
- Gametes contain identical chromatids because meiosis II is equivalent to mitosis.
- Gametes contain extra chromosomes because preventing crossing over causes nondisjunction.
Explanation: This question examines meiosis when crossing over is prevented but synapsis still occurs. Without crossing over, homologous chromosomes pair normally during prophase I but do not exchange DNA segments between non-sister chromatids. Each chromatid remains identical to its original parental form throughout meiosis. When homologs separate in meiosis I and sister chromatids separate in meiosis II, each gamete receives chromatids that exactly match the original parental chromosomes without any recombination. Students choosing A incorrectly believe synapsis alone causes DNA exchange, not recognizing that crossing over is a separate process requiring specific enzymatic machinery. The key insight is that crossing over and synapsis are distinct processes: chromosomes can pair without exchanging segments, producing gametes with purely parental chromosome combinations.
Question 3
A clonal population of yeast cells was split into two flasks: one at 20°C and one at 37°C. After 6 hours, cells at 37°C produced higher levels of a heat-shock protein, measured by antibody staining, than cells at 20°C. DNA sequencing of the heat-shock gene was identical in both flasks. When 37°C cells were returned to 20°C, heat-shock protein levels decreased within hours. Which explanation best accounts for the temperature-dependent protein levels?
- Higher temperature increased transcription and translation of the heat-shock gene, changing protein abundance without DNA change. (correct answer)
- Higher temperature caused a permanent mutation that duplicated the heat-shock gene in the 37°C flask.
- Cells at 37°C underwent meiosis, producing new genotypes that expressed more heat-shock protein.
- Cells at 20°C lost chromosomes, reducing heat-shock protein production compared with the 37°C cells.
- The 37°C flask formed a new population with altered allele frequencies for heat-shock protein expression.
Explanation: This question examines environmental effects on phenotype through temperature-induced protein expression. The correct answer is A because higher temperature increases the transcription and translation of heat-shock genes, producing more heat-shock protein without any changes to the DNA sequence. The rapid increase in protein levels at 37°C and equally rapid decrease when returned to 20°C, along with identical DNA sequences, clearly demonstrate this is gene expression regulation. Answer B is incorrect because it suggests a permanent gene duplication mutation, which would be detectable by DNA sequencing and wouldn't allow protein levels to decrease quickly when temperature dropped. When analyzing environmental responses, rapid and reversible changes in protein levels with unchanged DNA indicate transcriptional and translational regulation, not genetic mutations.
Question 4
A phospholipid bilayer’s hydrophobic core is a barrier to solutes that are highly polar or charged. Permeability generally decreases with increasing size and increasing polarity. Compare two carbohydrates: glucose (a monosaccharide) and maltose (a disaccharide). Both are uncharged but contain many hydroxyl groups, making them strongly polar. No transport proteins are present.
Which statement best predicts their relative permeability through the bilayer?
- Maltose is more permeable because it has more bonds that can rotate
- Glucose is more permeable because it is smaller while both are polar (correct answer)
- Maltose is more permeable because larger molecules diffuse faster
- They are equally permeable because both are uncharged
- Neither can diffuse because all carbohydrates require ATP to cross membranes
Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Glucose is more permeable than maltose because it is smaller while both are polar, reducing the energy barrier for entering the hydrophobic core compared to the larger disaccharide. Both have many hydroxyl groups making them strongly polar, but size further slows maltose in simple diffusion. Without transport proteins, permeability decreases with increasing size for polar molecules. A tempting distractor is maltose because larger molecules diffuse faster (choice C), but this stems from the misconception that size increases diffusion rate, whereas larger size actually decreases it. To analyze similar problems, for polar molecules of the same type, choose the smaller one for higher permeability across bilayers.
Question 5
In a meiosis simulation, a student tracks alleles on homologous chromosomes without considering gene expression. A heterozygous cell has alleles M/m on one homologous pair and N/n on a different pair. No crossing over occurs. The student observes four gamete types (MN, Mn, mN, mn) produced in roughly equal proportions. The chromatids remain intact, and the only variability arises from how whole homologs are distributed to daughter cells. Which meiotic event most directly produces these four gamete combinations?
- Synapsis of homologous chromosomes forming tetrads during prophase I
- Random orientation of homologous pairs at metaphase I (correct answer)
- Segregation of sister chromatids at anaphase I
- Cytokinesis after meiosis II producing four cells
- Nondisjunction of homologous chromosomes producing aneuploid gametes
Explanation: This question tests understanding of how meiosis generates genetic diversity through independent assortment of unlinked genes. The correct answer is B because the scenario describes two gene pairs on different chromosomes producing four gamete types (MN, Mn, mN, mn) in equal proportions, which results from random orientation of homologous pairs at metaphase I. Since M/m and N/n are on different chromosome pairs, each pair orients independently, creating 2² = 4 possible combinations when homologs separate. Answer C (segregation of sister chromatids at anaphase I) is incorrect because it confuses the timing of chromatid separation—sister chromatids remain together during anaphase I and only separate during anaphase II, making this a common misconception about meiotic divisions. To identify independent assortment, look for equal proportions of all possible gamete combinations when genes are on different chromosomes.
Question 6
A researcher compares two glucose polymers in animal tissues. Polymer X is highly branched, with α-1,6 glycosidic bonds at frequent branch points in addition to α-1,4 bonds along each chain. Polymer Y is unbranched and consists only of α-1,4 bonds, forming long helices. When enzymes bind to polymer ends to release glucose monomers, the rate depends on the number of nonreducing ends available for enzyme attachment. Which statement best predicts how branching affects the rate of monomer release from polymer X compared with polymer Y?
- Polymer X is slower because α-1,6 bonds cannot be hydrolyzed by enzymes at chain ends.
- Polymer X is faster because branching increases the number of nonreducing ends for enzymes. (correct answer)
- Polymer X is slower because branching decreases solubility and prevents enzyme binding.
- Polymer X is faster because β-1,4 bonds form straight chains that expose more ends.
- Polymer X and Y are equal because glycosidic bond type does not affect enzyme access.
Explanation: This question requires analysis of carbohydrate structure-function relationships to understand how branching affects enzyme accessibility. The correct answer B recognizes that polymer X's frequent α-1,6 branch points create many more nonreducing ends compared to unbranched polymer Y, and since enzymes work by binding to these ends, more ends means faster glucose release. The branched structure of polymer X provides multiple simultaneous sites for enzyme attachment, while polymer Y has only two ends per molecule regardless of its length. Option A incorrectly claims α-1,6 bonds cannot be hydrolyzed at chain ends (a bond-specificity misconception), when the question states enzymes bind to polymer ends, not specifically to α-1,6 bonds. The fundamental principle is that branching increases surface area and access points for enzymatic action. When evaluating polymer degradation rates, count the number of accessible ends rather than focusing on bond types or overall polymer size.
Question 7
Two phospholipid bilayers are assembled at the same temperature. Bilayer 1 contains phospholipids with shorter fatty acid tails, while bilayer 2 contains phospholipids with longer tails; both have the same head groups and the same degree of saturation. Longer hydrocarbon chains have greater surface area and therefore more van der Waals interactions with neighboring tails, increasing packing strength. Shorter chains have fewer such interactions. Which statement best predicts the relative permeability of the two bilayers to small nonpolar molecules?
- Bilayer 1 is more permeable because shorter tails reduce van der Waals interactions and loosen packing. (correct answer)
- Bilayer 2 is more permeable because longer tails create more space between phospholipids.
- Bilayer 1 is less permeable because shorter tails increase hydrogen bonding within the bilayer interior.
- Bilayer 2 is less permeable because longer tails make phosphate heads more polar and attract solutes.
- Both bilayers have identical permeability because tail length does not affect intermolecular forces.
Explanation: This question analyzes lipid structure-function relationships affecting membrane permeability. The correct answer is A because shorter fatty acid tails have less surface area for van der Waals interactions with neighboring tails, resulting in weaker intermolecular forces, looser packing, and more transient gaps through which small nonpolar molecules can pass. Option C incorrectly invokes hydrogen bonding in the bilayer interior (a chemistry error—hydrocarbon tails cannot form hydrogen bonds as they lack the necessary functional groups). The transferable principle is that membrane permeability inversely correlates with tail length: shorter tails mean weaker interactions, looser packing, and higher permeability.
Question 8
A phospholipid bilayer separates two aqueous solutions. The membrane interior is hydrophobic, so permeability is highest for small nonpolar molecules, lower for small polar molecules, and lowest for ions. Compare hydrogen cyanide (HCN), which is small and weakly polar but uncharged, and cyanide ion (CN−), which is charged. Assume conditions keep one species uncharged and the other charged. No proteins are present.
Which species would most likely diffuse across the bilayer more readily?
- Cyanide ion (CN−), because it is smaller than HCN
- Hydrogen cyanide (HCN), because it is uncharged and more membrane-soluble (correct answer)
- Cyanide ion (CN−), because charge increases diffusion through lipids
- Both, because they have the same atoms and similar size
- Neither, because polar molecules cannot diffuse across bilayers
Explanation: This question assesses the skill of analyzing membrane permeability based on molecular properties in a phospholipid bilayer. Hydrogen cyanide (HCN) diffuses more readily than cyanide ion (CN−) because it is uncharged and more membrane-soluble, crossing the hydrophobic interior easily, while CN− is charged and faces a high barrier. Conditions maintaining charge states emphasize the role of charge in permeability. No proteins mean ions are poorly permeable via simple diffusion. A tempting distractor is CN− because it is smaller (choice A), but this ignores the misconception that size trumps charge, whereas charge greatly reduces solubility. To analyze similar problems, favor uncharged weak acids or bases over their ionized forms for faster bilayer crossing.
Question 9
A gene encodes a nuclear protein. In the wild type, codon 200 in the coding strand is 5'-CGA-3' (mRNA 5'-CGA-3'), encoding arginine. A mutant changes codon 200 to 5'-TGA-3' (mRNA 5'-UGA-3'). The nuclear localization signal is located near the C-terminus, encoded after codon 240. No other sequence changes occur.
Which outcome is most likely from this mutation?
- A silent mutation occurs because UGA can be read as arginine by most tRNAs in eukaryotic cells.
- A premature stop codon produces a shortened protein lacking the C-terminal localization signal. (correct answer)
- A missense mutation substitutes arginine with lysine, preserving protein length and localization.
- A frameshift begins at codon 200 because single-base substitutions shift the reading frame.
- Transcription stops at codon 200 because RNA polymerase recognizes stop codons in DNA.
Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The mutation alters codon 200 from CGA (arginine) to UGA, a stop codon that terminates translation prematurely, producing a truncated protein. This shortened polypeptide lacks the C-terminal nuclear localization signal encoded after codon 240, preventing proper nuclear import. Molecularly, nonsense mutations introduce early stops, recruiting release factors and halting elongation before the full sequence is translated. A tempting distractor is choice C, proposing a missense substitution of arginine with lysine, stemming from the misconception that UGA codes for an amino acid instead of a termination signal. For suspected nonsense mutations, verify the mutated codon against the genetic code and assess the position relative to key functional domains.
Question 10
A student compares the chemical structures of thymine and uracil. Thymine has a methyl group at the 5-carbon position that uracil lacks. In a duplex, both thymine (in DNA) and uracil (in RNA) can hydrogen-bond with adenine using the same donor-acceptor pattern. However, the methyl group increases hydrophobic surface area and can influence base stacking interactions with neighboring bases. Considering only these molecular interactions within a duplex, which statement best predicts a consequence of replacing thymine with uracil in a DNA-like double helix?
- Duplex stability may decrease slightly because loss of thymine’s methyl group can weaken base stacking interactions. (correct answer)
- Duplex stability increases because uracil forms three hydrogen bonds with adenine instead of two.
- Duplex stability is unchanged because methyl groups participate directly in hydrogen bonding between bases.
- Duplex stability increases because uracil adds an extra phosphate group to the backbone.
- Duplex stability is unchanged because base stacking depends only on the sugar, not the base.
Explanation: This question assesses the analysis of nucleic acids as macromolecules. Duplex stability may decrease slightly when replacing thymine with uracil, as in choice A, because the loss of the methyl group reduces hydrophobic surface area, potentially weakening base stacking interactions. Both bases pair with adenine via the same two hydrogen bonds, but the methyl influences stacking with adjacent bases. In AP Biology, base structure affects duplex stability through stacking and hydrogen bonding, with thymine's methyl contributing to stronger pi-pi interactions in DNA. A tempting distractor is choice B, which is incorrect due to a structure-function confusion by claiming uracil forms three hydrogen bonds with adenine, misrepresenting standard base pairing. To handle these questions, compare structural differences in bases and assess their impact on non-covalent interactions in duplexes.
Question 11
A student digests a 2,000 bp plasmid with restriction enzyme EcoRI, which cuts once in the plasmid, and runs the digest on an agarose gel next to undigested plasmid. The undigested sample shows multiple bands due to different plasmid conformations. The EcoRI-digested sample shows a single band at ~2,000 bp. Which result best supports that EcoRI cut the plasmid at one site?
- The digested sample produces one linear DNA fragment the same length as the plasmid. (correct answer)
- The digested sample produces two fragments whose sizes add to more than 2,000 bp.
- The undigested sample produces one band because supercoiled DNA is always linear.
- The digested sample produces no bands because restriction enzymes remove phosphate groups.
- The digested sample produces many bands because EcoRI amplifies DNA at its cut site.
Explanation: This question assesses the skill of analyzing biotechnology techniques, specifically restriction enzyme digestion and gel electrophoresis of plasmids. EcoRI cuts the 2,000 bp plasmid once, converting the circular DNA into a single linear fragment of the same length, which migrates as one band at ~2,000 bp. The undigested plasmid shows multiple bands due to supercoiled, relaxed circular, and linear conformations that migrate differently in the gel. This single band in the digested sample confirms a single cut site, as multiple cuts would produce smaller fragments. A tempting distractor is choice B, which wrongly claims the digested sample produces fragments adding to more than 2,000 bp, based on the misconception that digestion increases total DNA length rather than just linearizing it. When evaluating restriction digests on gels, compare digested and undigested samples to confirm the number of cut sites and resulting fragment sizes.
Question 12
A single strain of bacteria is split into two flasks. Flask 1 contains antibiotic; Flask 2 does not. After 2 hours, cells in Flask 1 show increased mRNA for an efflux pump and increased pump protein in the membrane, but sequencing shows no DNA sequence differences between flasks. Which explanation best accounts for the increased efflux pump protein in Flask 1?
- Antibiotic exposure increased transcription of the efflux pump gene through regulatory responses. (correct answer)
- Antibiotic exposure caused mutations that created a new efflux pump gene in the genome.
- Antibiotic exposure changed the bacterial population genotype by eliminating susceptible alleles.
- Antibiotic exposure increased efflux by doubling the entire bacterial chromosome in each cell.
- Antibiotic exposure increased efflux by converting membrane proteins into DNA that encodes pumps.
Explanation: This question assesses understanding of environmental effects on phenotype, specifically how antibiotics can influence protein levels without altering the genetic code. The correct answer, A, is right because antibiotic exposure activates stress-response regulators that increase transcription of the efflux pump gene, leading to more mRNA and protein to expel the drug, without DNA sequence changes. This inducible resistance mechanism allows rapid adaptation in bacteria. The short timeframe and identical sequencing confirm expression-level changes. A tempting distractor is C, which wrongly attributes changes to population-level selection, stemming from the misconception that short-term exposure evolves the genotype rather than regulates existing genes. To approach similar questions, distinguish between immediate regulatory responses and evolutionary changes when phenotypes shift quickly in identical strains.
Question 13
A gene encodes a membrane channel protein. A missense mutation changes one codon in the coding region so that the mRNA codon changes from 5′-GCU-3′ (alanine) to 5′-GAA-3′ (glutamic acid). The substituted amino acid lies within a transmembrane helix composed mostly of nonpolar residues. No other changes occur to transcription or translation. Which outcome is most likely at the protein level?
- The protein may misfold or insert into the membrane less effectively due to a polar residue in the helix. (correct answer)
- The mutation prevents mRNA from being synthesized because alanine codons are required for transcription.
- The mutation is silent because alanine and glutamic acid have identical side-chain properties.
- A frameshift occurs because changing one codon changes the grouping of all downstream codons.
- Translation stops at the mutated codon because GAA is a stop codon in the genetic code.
Explanation: This question assesses the skill of analyzing mutations and their effects on protein synthesis, particularly missense mutations in functional protein domains. The missense mutation changes the codon from GCU (alanine, nonpolar) to GAA (glutamic acid, polar and charged), introducing a charged residue into a transmembrane helix that typically requires nonpolar amino acids for proper membrane insertion and stability. This substitution can disrupt the hydrophobic environment of the helix, leading to misfolding or inefficient membrane integration of the channel protein. Transcription and translation otherwise proceed normally, producing a full-length protein with this single change. A tempting distractor is choice C, which claims the mutation is silent due to identical side-chain properties, but this misconceives the distinct chemical properties of alanine (nonpolar) versus glutamic acid (charged). To analyze missense mutations, evaluate the chemical differences between the original and new amino acids and consider the structural context, such as membrane domains.
Question 14
In a cave-dwelling fish species, some populations have reduced eyes and lack pigmentation, while nearby surface populations have functional eyes and pigmentation. Genetic mapping identifies the same loss-of-function mutation in a pigmentation gene in multiple cave populations, but different mutations affecting eye development in different caves. Surface populations do not carry these cave-associated alleles. Which conclusion is best supported by these genetic patterns across populations?
- Pigmentation loss likely evolved once and spread among cave populations, while eye reduction evolved independently in different caves. (correct answer)
- Eye reduction likely evolved once and spread among caves, while pigmentation loss evolved independently in each cave population.
- Both traits must have evolved at the same time in every cave because cave environments cause identical mutations.
- Cave fish lost eyes because individuals stopped using them, and the unused organs disappeared across generations.
- Surface fish do not carry cave alleles because genes cannot change in populations without intentional selection.
Explanation: This question assesses the skill of analyzing evidence of evolution, specifically using genetic patterns to infer convergent or shared evolutionary histories. The identical loss-of-function mutation in the pigmentation gene across multiple cave populations suggests this trait evolved once and was shared, possibly through common ancestry or gene flow. In contrast, different mutations for eye reduction in each cave indicate independent evolution of this trait under similar dark environments. The absence of these alleles in surface populations supports cave-specific adaptations via natural selection. A tempting distractor is choice D, which attributes eye loss to disuse, reflecting the misconception of Lamarckian evolution through lack of use. When comparing populations, examine mutation patterns to distinguish single-origin traits from those evolving convergently in parallel environments.
Question 15
In plant guard cells, peptide P binds a membrane receptor that activates a kinase cascade leading to opening of anion channel A1. Opening of A1 depolarizes the membrane and triggers opening of an outward K+ channel, causing K+ efflux and reduced turgor. A phosphatase PP2 normally dephosphorylates the receptor-associated kinase, limiting the duration of A1 opening. In a PP2 loss-of-function mutant, P causes prolonged A1 opening and extended K+ efflux compared with wild type. Which explanation is best supported by these observations?
- PP2 acts upstream by preventing peptide P from binding the receptor at the membrane
- PP2 limits signal duration by reversing phosphorylation in the transduction pathway (correct answer)
- PP2 increases K+ channel conductance by directly phosphorylating the channel protein
- PP2 is required to generate anion gradients that drive peptide P diffusion to receptors
- PP2 prevents depolarization by moving anion channel A1 from cytosol into the nucleus
Explanation: This question assesses the skill of analyzing signal transduction pathways by evaluating phosphatase roles in plant guard cell signaling. In the PP2 mutant, peptide P causes prolonged A1 opening and K+ efflux, indicating PP2 normally limits signal duration by dephosphorylating the receptor-associated kinase, reversing activation. This negative regulation shortens the cascade's effect on channels, controlling turgor. The observations support PP2 acting downstream to terminate signaling, as its loss extends the response. A tempting distractor is choice A, suggesting PP2 prevents P binding, but this is incorrect because the mutant shows a response, just prolonged, confusing initiation with duration control. For interpreting mutants, assess how altered components affect timing and amplitude to determine regulatory roles.
Question 16
A botanist studies algae growth in identical lab flasks with the same light and nutrients. Flask A begins with 1,000 cells/mL and reaches 8,000 cells/mL after four days. Flask B begins with 10,000 cells/mL and reaches 12,000 cells/mL after four days. The botanist notes that waste products accumulate in the flasks over time and can inhibit cell division when concentrations are high. Which factor best accounts for the slower per-capita growth in Flask B?
- Higher waste-product inhibition at higher initial algal density (correct answer)
- A change in light intensity affecting only the high-density flask
- Random contamination that reduces growth independent of density
- A seasonal temperature shift decreasing division in all flasks
- A long‑term trait change that increases carrying capacity immediately
Explanation: This question assesses understanding of how population density affects growth and regulation in biological systems, via waste accumulation. Higher waste-product inhibition at higher initial algal density slows per-capita growth in Flask B, as toxins build up faster and limit division more than in Flask A. This density-dependent factor explains reaching only 12,000 cells from 10,000, versus 8,000 from 1,000, despite identical conditions. Waste concentration scales with cell numbers, inhibiting growth proportionally. A tempting distractor is choice C, random contamination reducing growth independent of density, which is wrong due to the density-independent misconception, not accounting for the initial density's role in growth rates. For similar lab scenarios, calculate per-capita changes to detect density-dependent self-limitation.
Question 17
In secretory cells, newly synthesized proteins enter the lumen of the rough endoplasmic reticulum (ER), where they begin folding. A drug that prevents proteins from entering the ER lumen causes many of those proteins to aggregate in the cytosol. Total protein synthesis remains similar. Which feature best explains how compartmentalization influences protein processing in this experiment?
- The ER lumen provides a localized environment that separates nascent proteins from cytosolic conditions that promote aggregation. (correct answer)
- The ER prevents aggregation by changing the DNA sequence of each protein after translation.
- Protein aggregation increases because ER membranes normally generate ATP for ribosomes in the cytosol.
- Blocking ER entry improves folding because proteins remain closer to the nucleus for instructions.
- Secretory proteins enter the ER so the cell can fold them faster for the purpose of rapid growth.
Explanation: This question assesses the skill of analyzing cell compartmentalization in eukaryotic cells. The correct answer is A because the ER lumen offers a protected space for folding secretory proteins, preventing aggregation in the cytosol, as evidenced by aggregation when entry is blocked despite unchanged synthesis rates. This ties to AP Biology's description of the ER as a folding chaperone-rich compartment separate from cytosolic stressors. Compartmentalization thus supports proper protein maturation by isolating nascent chains. A tempting distractor is D, which is incorrect due to a structure-function confusion by claiming proximity to the nucleus aids folding, ignoring that folding occurs post-translationally in the ER. To approach similar questions, examine how blocking access to compartments affects processing pathways and outcomes.
Question 18
In a meiosis lab, a cell is heterozygous at two loci (C/c and D/d) on different chromosomes. No crossing over is observed. The instructor notes that the orientation of the C/c homologs is independent of the orientation of the D/d homologs at metaphase I. Students then observe gametes with Cd, CD, cd, and cD combinations. Which meiotic process best explains these observed gamete combinations?
- Crossing over between sister chromatids during prophase II generating new allele combinations
- Independent assortment of homologous chromosome pairs during metaphase I producing varied chromosome combinations (correct answer)
- Random fertilization mixing alleles from different individuals to create Cd and cD
- Nondisjunction in anaphase II producing gametes with different allele combinations but extra chromosomes
- DNA replication after meiosis I producing chromatids with different alleles in each cell
Explanation: This question tests understanding of how meiosis generates genetic diversity through independent assortment when genes are on different chromosomes. The scenario describes C/c and D/d on different chromosomes producing four gamete types (Cd, CD, cd, cD) without crossing over, which results from independent chromosome alignment at metaphase I. During metaphase I, the C/c homologous pair and D/d homologous pair align independently at the metaphase plate - each pair's orientation is random and doesn't influence the other pair's orientation. This creates four equally likely gamete combinations depending on which homolog from each pair migrates to which pole during anaphase I. Students often incorrectly choose answer C about random fertilization, confusing the mixing of alleles between individuals with the segregation of alleles within a single meiotic cell - the question asks specifically about gamete formation, not zygote formation. To recognize independent assortment, look for multiple gene pairs on different chromosomes producing all possible allele combinations in equal frequencies.
Question 19
A cell contains an enzyme that generates reactive oxygen species (ROS) as a byproduct during substrate oxidation. The enzyme is localized inside a membrane-bound organelle, and ROS levels in the cytosol remain low. When the organelle membrane is disrupted, cytosolic ROS increases and nearby cytosolic proteins become damaged. Which feature best explains how compartmentalization limits the impact of this reaction on the rest of the cell?
- A membrane-bound compartment confines ROS production, reducing exposure of cytosolic proteins to reactive byproducts. (correct answer)
- The compartment prevents ROS formation by changing the DNA sequence of the oxidation enzyme inside the organelle.
- The compartment increases cytosolic pH, which permanently neutralizes all ROS produced anywhere in the cell.
- The compartment blocks substrate entry, so oxidation occurs only when substrates are absent from the cell.
- The compartment exists so the cell can make ROS only when it needs to remove damaged proteins.
Explanation: This question assesses the skill of analyzing cell compartmentalization by examining how organelles limit reactive oxygen species damage. The membrane-bound compartment confines ROS production, minimizing cytosolic exposure, as per the stimulus, which connects to the AP Biology principle that compartmentalization isolates harmful byproducts to protect other cellular components. Disrupting the membrane increases cytosolic ROS, damaging proteins, confirming the barrier's protective function. This enables necessary oxidation without widespread harm. A tempting distractor is choice C, which involves a level-of-organization error by claiming compartmentalization affects whole-cell pH, whereas it specifically sequesters ROS. To approach similar questions, evaluate how enclosures contain toxic intermediates to safeguard broader cellular integrity.
Question 20
A phylogeny is inferred from shared derived characters. In a group of four vertebrates (G, H, I, J), the derived character “amniotic egg” is present in H, I, and J but absent in G. The derived character “hair” is present only in I and J. The derived character “opposable thumb” is present only in J. Assume each character evolved once and was not lost. Which pair of species shares the most recent common ancestor?
- G and H
- H and I
- I and J (correct answer)
- G and J
- H and J
Explanation: This question assesses the skill of inferring phylogenetic relatedness using shared derived characters in vertebrates. The traits nest as amniotic egg grouping H-I-J (excluding G), hair grouping I-J, and opposable thumb only in J, so I and J share the most recent common ancestor via the hair synapomorphy, with J diverging last. This parsimony-based approach in AP Biology assumes one evolution per trait without loss, defining clades progressively. G is the outgroup lacking amniotic egg. A tempting distractor is choice B (H and I), but H lacks hair shared by I and J, reflecting a level-of-organization error confusing broader clade membership with sister taxa. Map traits to a cladogram by nesting groups to identify the innermost shared ancestors for relatedness questions.
Question 21
In a species with 2n = 6, a meiotic cell is treated with a chemical that prevents separation of homologous chromosomes during anaphase I, but sister chromatids can still separate during anaphase II. The cell proceeds through cytokinesis after each division. Which outcome is most likely for chromosome number in the final meiotic products from this cell?
- Four cells with n = 3 because sister chromatids separate in meiosis II
- Two cells with 2n = 6 because meiosis I fails but meiosis II occurs
- Four cells with 2n = 6 because homologs never segregate into different cells (correct answer)
- Four cells with n = 6 because chromosome number doubles after S phase
- Two cells with n = 3 because meiosis stops before meiosis II begins
Explanation: This question tests understanding of how preventing homolog separation affects meiosis outcomes. In a cell with 2n = 6, if homologous chromosomes cannot separate during anaphase I, both homologs of each pair remain together, so after meiosis I, each "daughter cell" still has all 6 chromosomes (remaining diploid). During meiosis II, sister chromatids separate normally, but since homologs never separated, each of the four final cells still contains both members of each homologous pair, maintaining the diploid number of 2n = 6. Choice A incorrectly assumes that preventing homolog separation would somehow still result in haploid cells. To analyze meiosis modifications, trace what happens when specific steps are blocked: preventing homolog separation in meiosis I means cells never become haploid, regardless of normal sister chromatid separation in meiosis II.
Question 22
In a diploid organism, a meiotic cell is arrested at metaphase I. Each homologous pair is attached to spindle fibers from opposite poles. Which explanation best accounts for how meiosis I reduces chromosome number in the resulting daughter cells?
- Sister chromatids separate, producing cells with half the DNA content.
- Homologous chromosomes separate, placing one from each pair into each cell. (correct answer)
- Chromosomes replicate again, producing cells with fewer chromosomes.
- Centromeres split in meiosis I, yielding haploid daughter cells.
- Spindle fibers detach, causing random loss of chromosomes from the nucleus.
Explanation: This question tests understanding of how meiosis I reduces chromosome number from diploid to haploid. During metaphase I, homologous pairs align at the cell equator with each homolog attached to spindle fibers from opposite poles. When anaphase I begins, homologous chromosomes separate and move to opposite poles, with one homolog from each pair going to each pole. This separation of homologs (not sister chromatids) reduces the chromosome number by half, creating two haploid cells each containing one member of every homologous pair. Choice A incorrectly describes meiosis II events where sister chromatids separate, confusing the two divisions. The key strategy for meiosis questions is remembering that homologs separate in meiosis I (reducing chromosome number) while sister chromatids separate in meiosis II (reducing DNA content).
Question 23
A student compares two mRNAs translated in the same cytosol. mRNA 1 begins with AUG and has many codons before a stop codon. mRNA 2 lacks an AUG near its 5′ end but contains the same internal codons as mRNA 1. Translation requires an initiator tRNA to pair with a start codon to establish the reading frame, after which the ribosome matches tRNAs to codons sequentially. Which outcome is most likely for mRNA 2?
- A polypeptide identical to mRNA 1 forms because internal codons determine initiation.
- No polypeptide forms because initiation is unlikely without a start codon to set the frame. (correct answer)
- A shorter polypeptide forms because ribosomes always begin at the second codon present.
- A longer polypeptide forms because lacking AUG prevents stop codon recognition.
- The mRNA is converted into DNA first because AUG is required for reverse transcription.
Explanation: This question assesses the skill of analyzing translation processes in protein synthesis. The correct answer is B because mRNA 2 lacks an AUG start codon near its 5′ end, making initiation unlikely as the initiator tRNA requires it to set the reading frame. Without proper initiation, the ribosome does not reliably begin translation, even if internal codons match those in mRNA 1. Translation on mRNA 1 proceeds normally from AUG to the stop codon, but mRNA 2's absence of a start prevents polypeptide synthesis. A tempting distractor is A, which falsely claims an identical polypeptide forms, arising from the misconception that internal codons can independently trigger initiation. A transferable strategy is to check for the presence and position of the start codon when comparing mRNAs, as it is crucial for establishing the translation reading frame.
Question 24
A bacterial gene encodes a 300–amino acid enzyme. In the wild type, codon 10 in the coding strand is 5'-GGC-3' (mRNA 5'-GGC-3'), which encodes glycine. A mutant has codon 10 changed to 5'-GGT-3' (mRNA 5'-GGU-3'). No other DNA changes are present, and transcription and translation initiation occur normally.
Which outcome is most likely from this mutation?
- A premature stop codon forms at codon 10, preventing translation of most of the enzyme.
- The amino acid sequence remains unchanged because both codons specify glycine, producing the same protein. (correct answer)
- A frameshift begins at codon 10, altering the reading frame and changing all downstream amino acids.
- The mutation prevents transcription because RNA polymerase cannot bind to genes with synonymous codons.
- The enzyme becomes longer because silent mutations cause ribosomes to read through the normal stop codon.
Explanation: This question assesses the skill of analyzing mutations in DNA sequences and their effects on protein synthesis. The mutation changes codon 10 from GGC to GGT, both of which encode glycine due to the redundancy of the genetic code, resulting in no alteration to the amino acid sequence. Consequently, the 300-amino acid enzyme is produced unchanged, maintaining its normal structure and function. Molecularly, this is a synonymous or silent mutation where the nucleotide change does not affect the translated protein because multiple codons specify the same amino acid. A tempting distractor is choice A, suggesting a premature stop codon at codon 10, arising from the misconception that any codon change involving G to T creates a stop signal like TGA. For point mutations in coding regions, compare the original and mutated codons using the genetic code to classify them as silent, missense, or nonsense.
Question 25
A lab uses a fluorescent reporter that increases signal when Cdk1 phosphorylates a specific peptide. In control cells, reporter fluorescence rises sharply at mitotic entry and falls at mitotic exit. In experimental cells, fluorescence rises normally but fails to fall, and cells do not reform nuclear envelopes. APC/C substrate tagging is normal, but proteasome function is impaired.
Which molecular signaling state best matches the experimental cells?
- Sustained Cdk1 signaling because M-cyclin is tagged but not degraded, preventing activity decline (correct answer)
- Reduced Cdk1 signaling because proteasome inhibition removes cyclins from cyclin–Cdk complexes
- Normal exit because nuclear envelope reformation depends only on kinetochore attachment signals
- Early G1 arrest because proteasome inhibition activates origin firing kinases at centrosomes
- Mitotic exit occurs because APC/C tagging alone is sufficient to inactivate Cdk1
Explanation: Signaling-based regulation of the cell cycle is a key skill in understanding how degradation is essential for activity shifts, like mitotic exit. Impaired proteasome function allows APC/C tagging but prevents M-cyclin degradation, sustaining Cdk1 signaling and high fluorescence, blocking envelope reformation. This matches the persistent phosphorylation observed. Tagging alone isn't sufficient; removal is needed. A tempting distractor is choice B, claiming reduced signaling from cyclin removal, but inhibition stabilizes cyclins, misconstruing blockage as enhancement of degradation. To approach similar problems, differentiate between preparatory tagging and the degradative step in regulation.