Practice Test 11 - AP Biology

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

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1. In yeast, gene Y has an enhancer located 1 kb upstream. An activator binds the enhancer and recruits a mediator complex that helps RNA polymerase II assemble at the promoter. In nutrient-rich conditions, the activator is phosphorylated and localized in the nucleus; under starvation, the activator remains unphosphorylated in the cytosol. Starved cells show reduced gene Y pre-mRNA levels compared with nutrient-rich cells. Which change would most likely restore high transcription of gene Y during starvation?
2. A cell is experimentally synchronized so that it exits mitosis and begins G1 at time 0. DNA content is measured every 2 hours and reported as relative units: at 0 h, 2C; at 2 h, 2C; at 4 h, 3C; at 6 h, 4C; at 8 h, 4C. Microscopy at 8 h shows duplicated chromatids but no spindle apparatus. The cell has not divided. Based on the pattern, the cell has completed S phase by 6 h and is in G2 at 8 h. Which observation would best indicate entry into M phase next?
3. In cultured mammalian cells, entry into mitosis correlates with a sharp rise in M-cyclin bound to Cdk1. The complex is inactive when Cdk1 carries an inhibitory phosphate. A checkpoint phosphatase removes this phosphate, rapidly increasing kinase activity and promoting phosphorylation of mitotic target proteins. In an experiment, cells are treated with a small molecule that specifically inhibits the checkpoint phosphatase but does not affect cyclin levels or Cdk1 abundance. After treatment, cells accumulate with duplicated DNA and intact nuclear envelopes. Which change would most likely restore progression into mitosis under these conditions?
4. A researcher adds a chemical uncoupler to respiring cells. After uncoupler addition, oxygen consumption increases, but ATP production decreases. The uncoupler allows protons to cross the inner mitochondrial membrane without passing through ATP synthase. Electron transport continues transferring electrons to oxygen, and proton pumping continues, but the proton gradient is dissipated as heat rather than used to phosphorylate ADP. Because ATP synthase receives less proton-motive force, less ATP is made per oxygen consumed. Which explanation best accounts for increased oxygen consumption despite decreased ATP production?
5. In an experiment, nuclei from a frog skin cell and a frog intestinal cell are analyzed and found to contain the same DNA sequences for several tested genes. Yet, the skin cell has high keratin mRNA and low digestive enzyme mRNA, while the intestinal cell shows the opposite pattern. Both samples are taken from healthy adult tissue. Which explanation best accounts for the different mRNA profiles?
6. A scientist increases the size of a spherical cell without changing its shape. The density of membrane channels (channels per µm²) remains constant, and channel flux per channel remains constant. The cell requires a fixed number of nutrient molecules per unit volume each minute. Under the same external nutrient concentration, the larger cell shows a nutrient deficit. Which explanation best accounts for the deficit?
7. In a tropical rainforest, a species of ant nests in hollow thorns of an acacia tree and patrols the tree's leaves. When ants are experimentally removed from some trees, leaf-chewing insects increase on those trees and leaf area decreases compared with control trees that retain ants. Soil nutrients and rainfall are similar for all trees. Which interaction best explains the higher leaf area in control trees?
8. A DNA segment is transcribed into RNA. In the resulting RNA, every position that corresponds to an adenine (A) on the DNA template strand contains uracil (U). The RNA nucleotides pair transiently with the DNA template during synthesis according to base-pairing rules. Which statement best explains the presence of U in the RNA at those positions?
9. A student compares two spherical cells made of the same membrane and placed in identical solution. Cell 1 has radius $5\,\mu m$; Cell 2 has radius $10\,\mu m$. Both cells rely on diffusion across the plasma membrane for oxygen entry and carbon dioxide exit, and both have similar cytoplasm composition. When respiration rate per unit volume is similar in both cells, measurements show that oxygen concentration is lower at the center of Cell 2 than at the center of Cell 1. Which explanation best accounts for the difference in internal oxygen concentration?
10. Red blood cells (internal solute concentration 0.30 M) are placed in Solution Z containing 0.30 M nonpenetrating solute. The membrane is permeable to water but not to the solute. The cells begin at normal volume and are observed for 10 minutes while the external concentration remains constant. Water movement depends on relative solute concentrations across the membrane. Which outcome is most likely?
11. In a cell lineage with an extensive endomembrane system, the nuclear envelope is continuous with the ER, and both have a similar lipid composition to the plasma membrane. Genes encoding nuclear pore proteins are homologous to genes encoding proteins involved in vesicle coating and budding. No organelle-like DNA is associated with the nucleus or ER. Which evidence best supports a model for the origin of the nuclear envelope?
12. A lab tests two molecules of similar size. Molecule X contains many C–H bonds and few oxygen atoms; molecule Y contains several hydroxyl (–OH) groups with O and H attached to carbon. When placed in water, Y dissolves readily, while X separates into a distinct layer. The –OH groups can form hydrogen bonds with water because oxygen is electronegative, creating partial charges that align with water's polarity. In contrast, C–H bonds are largely nonpolar and interact weakly with water. Which feature best explains molecule Y's higher solubility in water?
13. A bacterial culture is shifted from 37°C to 50°C; within minutes, heat-shock proteins increase. Which response best explains this change?
14. A researcher adds a drug that prevents acidification of lysosomes by inhibiting proton pumping across the lysosomal membrane. After treatment, cells accumulate undigested material inside membrane-bound vesicles. Lysosomes normally maintain a low internal pH, and many hydrolytic enzymes inside lysosomes function best under acidic conditions. Which outcome is most likely caused directly by the disrupted lysosomal pH?
15. A signaling molecule V is released from a gland and travels in blood bound to a carrier protein. Only free (unbound) V can bind receptors on target cells. If carrier protein concentration increases, total V in blood stays the same. Which of the following outcomes is most likely in target cells?
16. Cells are exposed to Ligand S, which binds a GPCR. In untreated cells, a rapid decrease in cytosolic cAMP is observed. When cells are treated with pertussis toxin, which prevents activation of certain G proteins, Ligand S still binds the receptor but cAMP no longer decreases. Which of the following best explains the early transduction pathway affected by the toxin?
17. After replication, each daughter DNA molecule contains one parental strand and one newly synthesized complementary strand. A researcher observes a point mutation present in only one of the two sister chromatids immediately after S phase. Which explanation best accounts for this observation?
18. A researcher compares RNA and DNA stability in alkaline solution. RNA breaks more rapidly due to cleavage of its backbone. Which structural feature best explains this difference?
19. A single population of wild wheat includes occasional individuals with a chromosome-doubling error during meiosis, producing $4n$ individuals. The $4n$ plants flower at the same time and in the same fields as the $2n$ plants. Crosses between $4n$ and $2n$ plants produce seeds that rarely develop, but $4n \times 4n$ crosses produce viable, fertile offspring. Over several generations, a stable $4n$ population persists. Which process most directly caused reproductive isolation and speciation?
20. A meiosis lab compares two cells with the same genotype RrSs, where R/r and S/s are on different chromosomes. Cell 1 undergoes meiosis with crossing over suppressed; cell 2 undergoes meiosis with normal crossing over. The lab records only allele combinations in gametes at the chromosomal level, not DNA sequence changes. Both cells produce RS, Rs, rS, and rs gametes. Which conclusion best explains why cell 1 can still produce all four gamete types?

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