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Practice Test 9 - AP Biology

20 questions. Answer them in any order, then submit to see your score and every explanation.

  1. 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?
  2. 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?
  3. 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?
  4. 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?
  5. 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?
  6. 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?
  7. 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?
  8. 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?
  9. 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?
  10. 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?
  11. 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?
  12. 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?
  13. 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?
  14. 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?
  15. 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?
  16. 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?
  17. 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?
  18. 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?
  19. A plant population is split between two soil types that occur in adjacent patches with no physical barrier. Plants in patch X grow on serpentine soil; patch Y is normal soil. Pollinator observations show that pollinators move frequently between patches. However, pollen from patch X plants germinates poorly on patch Y stigmas, and the reverse cross also shows low pollen tube growth. Seeds are rarely produced from cross-patch hand pollinations, even though flowering times overlap. Which reproductive barrier most directly contributes to reduced gene flow between the patches?
  20. A membrane receptor binds ligand K and rapidly activates a cytosolic kinase. In vitro, the kinase becomes active only when it binds a phosphorylated tyrosine motif on the receptor's cytosolic side. A point mutation changes the tyrosine to phenylalanine; ligand binding remains normal but kinase activation is absent. Which of the following best explains the missing early transduction step?

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