AP Biology 2023 FRQ Content Summary

Apr 3, 2026

All files

summarize_document

This document contains a series of AP Biology 2025 Free-Response Questions, covering various biological concepts. The questions are designed to assess students' understanding of ecological principles, evolutionary mechanisms, molecular biology, and cellular processes.

Question 3: Ecology and Invasive Species

This question focuses on the ecological impact of invasive species and experimental design.

  • Scenario: Buffelgrass, an invasive species in southwestern deserts, threatens the saguaro cactus (a keystone species). Buffelgrass is drought-tolerant and fire-resistant but increases wildfire severity due to its dry biomass. Young saguaros are particularly vulnerable to these severe fires.
  • Experiment: Scientists investigated controlling buffelgrass by growing it with native grass species under varying water conditions (nondrought vs. drought) to measure buffelgrass height and dry weight.

A. Effect of Removing a Keystone Species

  • Removing a keystone species can have a disproportionately large impact on an ecosystem, leading to significant changes in community structure and function. This can result in a decrease in biodiversity and the collapse of other populations that depend on the keystone species.

B. Control Group for the Experiment

  • A suitable control group would be buffelgrass grown alone under both nondrought and drought conditions, without any native grass species. This allows for comparison to see the effect of native grasses specifically.

C. Null Hypothesis

  • The null hypothesis would state that the presence of native grass species has no significant effect on the abundance (measured by height and dry weight) of buffelgrass, regardless of watering conditions.

D. Justification for Scientists' Claim on Wildfire Impact

  • The scientists claim wildfires increase buffelgrass abundance because native grasses have slower population growth rates after a wildfire compared to buffelgrass. This means buffelgrass can recover and proliferate more quickly in the post-fire environment, outcompeting the slower-recovering native species.

Question 4: Evolution and Biogeography

This question explores evolutionary processes driven by geographical changes.

  • Scenario: The formation of the Isthmus of Panama millions of years ago separated the Caribbean Sea and the Pacific Ocean, which were once connected and shared marine species. This land formation altered ocean currents and temperatures.

A. Genetic Evidence for Evolution

  • Genetic evidence for evolution includes changes in allele frequencies within a population over time, the presence of homologous genes in different species, and patterns of DNA similarity that reflect evolutionary relationships.

B. Divergent Evolution Due to Isolation

  • The isolation of marine species by the Isthmus of Panama created separate gene pools. Over time, different environmental pressures (e.g., cooler, nutrient-rich Pacific water vs. warmer Caribbean water) acted on these isolated populations. This led to the accumulation of different genetic mutations and adaptations, resulting in divergent evolution, where the isolated populations evolved into distinct species.

C. Predicted Effect on Resource Availability for South American Species

  • The formation of the Isthmus allowed North American land animals to migrate to South America. This would likely decrease resource availability for native South American species as they would face new competition from the migrating species for similar ecological niches.

D. Justification for Prediction in Part C

  • When species with similar ecological requirements (occupying similar niches) are introduced into the same environment, they compete for limited resources such as food, water, and space. This increased competition would reduce the resources available for the original South American species.

Question 5: Metabolic Pathways and Enzymes

This question examines enzyme function and metabolic regulation.

  • Scenario: Figure 1 illustrates a metabolic pathway where amino acid B is synthesized from amino acid A through a series of enzymatic reactions.

A. Enzyme Active Site Characteristic

  • A characteristic of an enzyme's active site that allows it to catalyze a specific reaction is its unique three-dimensional shape, which is complementary to the shape of its specific substrate(s). This precise fit ensures specificity.

B. Regulation of Enzyme 1 by Amino Acid B

  • Based on Figure 1, amino acid B likely acts as an allosteric inhibitor of enzyme 1. When amino acid B (the end product) is present in sufficient amounts, it binds to enzyme 1 at a site other than the active site, causing a conformational change that reduces or prevents enzyme 1's activity. This is a form of feedback inhibition.

C. Product of Enzyme 2

  • Based on Figure 1, enzyme 2 catalyzes the conversion of intermediate X to intermediate Y. Therefore, the product of the reaction catalyzed by enzyme 2 is intermediate Y.

D. Effect of pH Change on Enzyme 3

  • A change in pH can alter the ionization state of amino acid residues within an enzyme's active site and overall structure. If the pH deviates significantly from the enzyme's optimal range, the tertiary structure of enzyme 3 could be disrupted, altering the shape of its active site. This would prevent the substrate from binding effectively, thus inhibiting its catalytic activity and preventing the production of amino acid B.

Question 6: Genetics and Meiosis in Fruit Flies

This question investigates the role of a specific gene in meiosis using fruit fly genetics.

  • Scenario: The ald gene in fruit flies encodes the ALD protein, crucial for chromosome alignment during meiosis I. Mutations in ald can lead to aneuploid gametes. Scientists studied various ald genotypes, measuring ALD-associated filaments (Figure 1A) and ALD protein amount (Figure 1B).

A. Fly Genotype with ~12% ALD-Associated Filaments

  • Based on Figure 1A, the fly genotype where the average percent of metaphase cells with ALD-associated filaments is close to 12% is ald23/del.

B. Difference in ALD Protein Production

  • Flies with the genotype ald3/ald23 produce a significantly greater amount of ALD protein compared to flies with the genotype ald23/del, as indicated by the thicker band for ald3/ald23 in Figure 1B.

C. Support for Hypothesis on ALD Protein and Filaments

  • The hypothesis is that cells can produce normal filaments with about half the ALD protein of wild-type.
    • Wild-type (WT/WT) cells produce a high amount of ALD protein and have a high percentage of cells with filaments (~90%).
    • ald1/del cells produce a similar amount of ALD protein as wild-type (~100% relative) but have a lower percentage of cells with filaments (~60%).
    • ald3/ald23 cells produce about 75% of the ALD protein compared to wild-type and have ~70% of cells with filaments.
    • ald23/del cells produce about 50% of the ALD protein compared to wild-type and have ~12% of cells with filaments.
    • WT/del cells produce about 50% of the ALD protein compared to wild-type and have ~50% of cells with filaments.
  • The data from WT/del and ald23/del genotypes support the hypothesis. These genotypes produce approximately 50% of the ALD protein compared to wild-type, and the WT/del genotype shows a relatively high percentage of cells with filaments (around 50%), suggesting that even with reduced protein, filaments can form.

D. Phenotype Difference in WT/del and ald1/del Flies

  • For WT/del and ald1/del flies, Figure 1B shows similar amounts of ALD protein produced (around 50% of wild-type for WT/del, and slightly less than wild-type for ald1/del, but comparable to WT/del). However, Figure 1A shows a significant difference in the percentage of cells with ALD-associated filaments: WT/del has about 50%, while ald1/del has about 60%. This difference in filament formation, despite similar protein levels, suggests that the quality or functionality of the ALD protein produced by these different genotypes might differ, or that other genetic interactions influence filament assembly. The ald1 allele might produce a protein that is less functional in filament formation even if the amount produced is sufficient.

Question 2: Cell Signaling and Behavior

This question examines a signaling pathway regulating insect behavior.

  • Scenario: Male moths use pheromones to find mates, regulated by the extracellular signaling molecule 20E. Scientists investigate how the receptor DopEcR and its signaling pathway affect male moth behavior.

A. Polarity of Receptor Portion Inside Membrane

  • The portion of a plasma membrane receptor that is inside the membrane is typically hydrophobic (nonpolar) to interact with the lipid tails of the phospholipid bilayer.

B. Graphing and Analyzing Activity Data

  • i. Graph Construction: An appropriate graph would be a bar graph comparing "General Activity" and "Oriented Activity" for the two treatment groups (saline vs. siRNA). Each treatment group would have two bars: one for general activity and one for oriented activity.
  • ii. Activity Affected by Inhibiting DopEcR: Inhibiting the expression of the DopEcR receptor significantly affected oriented activity. While general activity remained high in both groups, the percentage of general activity that was oriented decreased dramatically in the siRNA group.
  • iii. Treatment Group with >50% Oriented Activity: The treatment group in which oriented activity was greater than 50% of the general activity is Male moths injected with saline (control solution).
  • iv. Effect of G Protein Mutation: If GTP cannot displace GDP on the G protein, the G protein signaling pathway will not be activated. Based on Figure 1, this means the downstream signaling cascade leading to gene transcription for oriented activity will be blocked. Therefore, the mutation would decrease or eliminate oriented activity in male moths exposed to the pheromone.

C. Support for Scientists' Claim on Gene Expression and Mating

  • i. Evidence: The scientists claim increased gene expression of DopEcR increases the likelihood of males finding females. This is supported by the information that DopEcR gene expression is low in sexually immature males and rapidly increases as they reach sexual maturity. This timing correlates with the period when successful mating becomes more likely.
  • ii. Inhibitor as Crop Protection: An inhibitor of the DopEcR pathway could protect crops from moth damage by disrupting the moths' ability to locate mates. If the pathway is inhibited, male moths would be less likely to engage in oriented activity towards pheromones released by females, thus reducing mating success and subsequent reproduction, which controls the moth population.

Question 1: Protein Transport and Cellular Processes

This question delves into protein synthesis, transport, and experimental analysis of these processes.

  • Scenario: Proteins secreted from cells are transported to the endoplasmic reticulum (ER) either during or after translation. ER membrane proteins SR and Sec62 are involved in these two different transport mechanisms. Researchers used siRNAs to reduce the expression of SR or Sec62 and measured protein levels and ER transport.

A. Ribosome Function

  • Function of Ribosomes: Ribosomes are cellular machinery responsible for protein synthesis (translation). They read messenger RNA (mRNA) sequences and assemble amino acids into polypeptide chains.

B. Analysis of siRNA Experiments (Figure 1)

  • i. Dependent Variable: The dependent variables in the experiments shown in Figure 1 are the relative amounts of Sec62 protein and SR protein.
  • ii. Justification for Measuring Both Proteins: Researchers included the control of measuring both Sec62 and SR proteins in cells treated with Sec62 siRNA to ensure that the siRNA specifically targeted and reduced the expression of Sec62 without affecting the production of SR protein. This confirms the specificity of the siRNA treatment.
  • iii. Effect of Sec62 siRNA on SR Protein: Based on Figure 1, treating cells with Sec62 siRNA has no significant effect on the production of SR protein. The relative amount of SR protein remains close to 100% of the control.

C. Analysis of ER Transport Experiment (Figure 2)

  • i. Independent Variable: The independent variable in the researchers' second experiment (Figure 2) is the type of siRNA treatment (Control siRNA, Sec62 siRNA, or SR siRNA).
  • ii. Protein(s) Showing Increased Transport with Sec62 siRNA: When treated with Sec62 siRNA, Protein 1 showed an increase in percent transport to the ER compared with the control.
  • iii. Difference in Amino Acids:
    • Protein 1 length: 234 nucleotides / 3 nucleotides/amino acid = 78 amino acids.
    • Protein 2 length: 495 nucleotides / 3 nucleotides/amino acid = 165 amino acids.
    • Difference: 165 - 78 = 87 amino acids.

D. Interpretation of Transport Mechanisms

  • i. Support for Protein 1 Transport Mechanism: Researchers claim Protein 1 is the only tested protein transported after complete cytosolic translation. Figure 2 shows that when SR siRNA (involved in co-translational transport) is added, Protein 1 transport is significantly reduced (around 20% of control). However, when Sec62 siRNA (involved in post-translational transport) is added, Protein 1 transport increases significantly (around 120% of control). This pattern supports the claim that Protein 1 is transported post-translationally, meaning it is fully translated in the cytosol before entering the ER.
  • ii. Justification for Amino Terminus Role: Amino acids near the protein's amino terminus often contain signal sequences or specific hydrophobic/hydrophilic regions. These sequences are recognized by cellular machinery (like signal recognition particles or translocon channels) that mediate the passage of proteins through membranes. The sequence and properties of these initial amino acids dictate how the protein interacts with the protein channel in the ER membrane, thus determining its likelihood of transport.

Here's a summary of the provided content, structured according to your requirements:

Summary of AP Biology Exam Materials

This document contains excerpts from the 2016 AP Biology Exam, including free-response questions, scoring guidelines, and multiple-choice questions. The content covers a range of biological topics, including ecology, genetics, molecular biology, evolution, and physiology.

Section II: Free-Response Questions

This section includes long and short free-response questions designed to assess students' ability to apply biological concepts and analyze data.

Question 1: Insecticide Effect on Plant Growth

  • Experiment: Investigated the effect of an insecticide on Haplopappus squarrosus shrubs.
  • Treatments:
    • Experimental: Sprayed with insecticide.
    • Control: Sprayed with water only.
  • Reasoning for Control: Using water only, rather than unsprayed plants, ensures that any observed effects are due to the insecticide and not the act of spraying or the water itself.
  • Student Graph Error: A plotting error was identified in the student-constructed graph (Figure 1), specifically with the data point for seed formation/immature seeds from the water-only treatment.
  • Logarithmic Scale Justification: A logarithmic scale on the Y-axis is appropriate because the data spans several orders of magnitude, allowing for the visualization of both large and small values on a single graph.
  • Life Stage Impact Calculation: Students were asked to calculate the percentage of plants/parts remaining between life stages and identify the stage most impacted by insects.
    • Greatest Impact Stage: Between immature seeds and mature seeds.
    • Justification: This stage shows the largest difference in remaining plant parts between the insecticide and water-only treatments, indicating significant insect impact.
    • Interaction: Insects likely eat or damage developing seeds.
  • Second Experiment (Different Insecticide): A different insecticide targeting a different insect group significantly reduced seed formation.
    • Likely Interaction: The affected insects are likely pollinators. Their absence or reduced activity prevents seed development.

Question 2: Mosquito Host Preference

  • Organism: Yellow-fever mosquito (Aedes aegypti), with forest and domestic forms.
  • Experiment: Used a choice chamber (Figure 1) to test host preference (human vs. guinea pig).
  • Data: Figure 2 shows the host-preference index.
  • Key Gene: OR4 gene encodes an olfactory receptor for sulcatone, a volatile odorant produced more by humans than guinea pigs.
  • Host Preferences:
    • Forest form: Prefers guinea pig (negative index).
    • Domestic form: Prefers human (positive index).
  • Experimental Refinements:
    • Rule out Visual Cues: Use a dark/no-light box, cover hosts, use models, or stop host movement.
    • Test Sulcatone Attractant: Use a sulcatone-soaked cotton ball vs. a control, use a gloved hand to prevent odor release, or genetically alter the OR4 gene.
  • OR4 Gene Mutations and Sensitivity to Sulcatone:
    • Removal of Intracellular Domain: Decreased sensitivity; the receptor cannot initiate an intracellular signal.
    • Substitution of Similar Hydrophobic Amino Acid in Ligand-Binding Site: No change in sensitivity; the ligand can still bind, and the binding site's structure is largely unaffected.
  • Speciation Hypothesis: If the two forms are evolving into different species, a potential postzygotic isolating mechanism is reduced hybrid fitness (hybrid breakdown). This maintains reproductive isolation because hybrids are less viable or fertile, preventing gene flow between the populations.

Question 3: Turtle Embryo Response to Thermal Cues

  • Experiment: Investigated turtle embryos' ability to shift position in response to directional heat.
  • Treatments:
    • Dorsal Heating: Used as the control.
    • Lateral Heating: Heat applied from the side, with the source switched on day 15.
  • Data: Figure 1 shows the mean angle of shift in embryo position.
  • Claim Support: The researcher's claim is supported.
    • Justification: The lateral heating treatment showed a statistically significant shift in embryo position after the heat source was switched, while the dorsal heating (control) did not show a significant shift. The mean angles before and after the shift in lateral heat source did not overlap, unlike in the control.
  • Alternative Control Treatment: Uniform heat source, ventral heat source, or a lateral heat source that does not shift.
  • Advantage of Shorter Embryonic Period: A shorter embryonic period can decrease vulnerability to predation, reduce exposure to parasitic infections, or improve the chances of obtaining resources compared to nest mates or other species.

Question 4: Plasmodium falciparum and Apicoplasts

  • Organism: Plasmodium falciparum, the parasite causing malaria.
  • Organelle: Apicoplast, which synthesizes precursors for biomolecules.
  • Evidence for Prokaryotic Origin of Apicoplasts:
    1. Double Membrane: Possesses a double membrane, characteristic of endosymbiotic organelles.
    2. Prokaryotic Ribosomes: Contains 70S ribosomes (similar to prokaryotes).
    3. Circular DNA: Contains its own circular DNA molecule (similar to prokaryotes).
  • Doxycycline Treatment: Doxycycline targets the prokaryotic 30S ribosomal subunit.
    • Effectiveness: Doxycycline is effective because the apicoplast contains prokaryotic 30S ribosomes. By inhibiting these ribosomes, doxycycline prevents the synthesis of essential apicoplast proteins, thus inhibiting parasite growth and reproduction.

Question 5: Bacterial Transformation and Plasmid Insertion

  • Plasmid: pTET-KAN, containing ori, tet resistance gene, and kan resistance gene.
  • Transformation: E. coli bacteria were transformed with the plasmid.
  • Culture in Tetracycline: Transformed E. coli cultured in the presence of tetracycline will survive and reproduce.
    • Justification: The bacteria express the tet-resistance gene, producing a protein that confers resistance to tetracycline.
  • Modified Plasmid: Human DNA fragment inserted into pTET-KAN.
  • Survival Results: Transformed cells survived in tetracycline but not kanamycin.
  • Insertion Site: The human DNA fragment was most likely inserted within the kanamycin resistance gene (kan) or its promoter.
    • Justification: The inability to survive in kanamycin indicates that the kan resistance gene was disrupted. The human DNA insertion likely inactivated the kan gene or its promoter, preventing the production of kanamycin resistance.

Question 6: Neuromodulin Evolution and Cladogram

  • Protein: Neuromodulin, highly conserved among vertebrates.
  • Data: Table shows the number of differences in the first 60 amino acids of neuromodulin among five species.
  • Cladogram Construction: A cladogram should be constructed to show evolutionary relatedness based on the number of amino acid differences.
    • Placement of Zebrafish: Zebrafish should be placed furthest from the other species (or at the base of the tree) because it has the greatest number of differences in the neuromodulin sequence compared to the other four species.
    • Most Recent Common Ancestor: The node (position) representing the most recent common ancestor of the two most closely related species (those with the fewest differences between them) should be circled.

Question 7: Starch Biosynthesis and Temperature

  • Pathway: Starch biosynthesis involves sucrose hydrolysis and glucose polymerization (Figure 1).
  • Enzymes & Optimal Temperatures:
    • Starch Synthase: 27°C-30°C
    • Sucrose Synthase: 30°C-35°C
  • Effect of Temperatures > 35°C on Starch Synthase:
    • Structure: The high temperature will likely cause the starch synthase enzyme to change its conformation (shape) and become partially denatured.
    • Function: The enzyme's activity will decrease.
  • Consequences of Growing Rice at 33°C:
    • The average temperature (33°C) is above the optimal range for starch synthase (27-30°C) but within the optimal range for sucrose synthase (30-35°C).
    • Predicted Starch Content: Starch content will likely be less than if grown at optimal temperatures for starch synthase. While sucrose synthase activity might be high, the subsequent step of starch synthesis will be less efficient due to suboptimal temperature for starch synthase.

Question 8: Intestinal Glucose and Sodium Transport

  • Process: Glucose and sodium transport from the intestinal lumen to the blood via epithelial cells (Figure 1).
  • Direct Energy Source for Glucose Uptake (Lumen to Epithelial Cell): The energy comes from the sodium gradient. The Na+/Glucose symporter uses the electrochemical gradient of Na+ (maintained by the Na+/K+ pump) to drive glucose uptake.
  • Maximizing Glucose Absorption:
    • Lumen to Epithelial Cell: The Na+/K+ pump (not explicitly shown but implied by "High Na+ Medium") actively maintains a low intracellular Na+ concentration. This creates a strong electrochemical gradient for Na+ to flow into the cell. The Na+/Glucose symporter couples the downhill movement of Na+ with the uphill movement of glucose, effectively using the energy of the Na+ gradient to absorb glucose.
    • Epithelial Cell to Blood: Glucose moves from the epithelial cell to the blood via facilitated diffusion through glucose transporters (GLUTs) on the basolateral membrane. This movement is driven by a glucose concentration gradient, which is maintained because glucose is continuously being transported into the epithelial cell from the lumen and is also being consumed or transported further into the body.
    • Surface Area: Microvilli on the intestinal lining (implied by the structure) significantly increase the surface area available for absorption.

Section I: Multiple-Choice and Grid-In Questions

This section contains multiple-choice and grid-in questions covering various biological topics.

Question 1 (Squirrel Populations and Disease):

  • Scenario: Two isolated squirrel populations (one larger, one smaller) face a virulent bacterial disease.
  • Prediction: The smaller population will be more affected.
  • Justification: Smaller populations generally have less genetic variation, making them more vulnerable to diseases that exploit specific genetic susceptibilities.

Question 2 (Mitochondria and Endosymbiosis):

  • Observation: Mitochondria have their own DNA and ribosomes similar to prokaryotes.
  • Justified Statement: An ancestral cell most likely engulfed an aerobic prokaryote in a mutually beneficial relationship (Endosymbiotic Theory).

Question 3 (Sickle-Cell Anemia Mutation):

  • Condition: Sickle-cell anemia results from a single amino acid change (glutamic acid to valine at position 6) in the beta-hemoglobin subunit.
  • Most Likely Mutation: A single base-pair substitution in the gene encoding the beta subunit. This is the simplest mutation that can change a single codon, leading to a single amino acid change.

Question 4 (Invasive Species - Caulerpa taxifolia):

  • Scenario: Caulerpa taxifolia, an aggressive invasive alga, is introduced to the California coast.
  • Predicted Consequence: Without natural herbivores or competitors, C. taxifolia will grow rapidly and crowd out native species of producers.

Question 5 (Predator-Prey Dynamics - Wolves and Moose):

  • Graph: Shows population sizes of wolves and moose over time.
  • Supported Statement: The wolves were predators of the moose, which were otherwise reproductively successful. The cyclical pattern suggests a predator-prey relationship where moose population increases are followed by wolf population increases, and vice versa.

Question 6 (Keystone Species - Salmon and Bears):

  • Data: Figure 1 shows nitrogen influx based on the presence of salmon, bears, or both.
  • Best Supported Statement: The presence of black bears and salmon correlates with a significant increase in nitrogen influx. The highest nitrogen influx occurs when both species are present (SB).

Question 7 (Bears, Salmon, and Nitrogen Dynamics):

  • Interaction: Bears consume salmon.
  • Mechanism: When bears consume salmon, they leave parts of the carcasses on the ground, which decompose and release nitrogen into the environment.

Question 8 (Investigating Nitrogen Influx):

  • Additional Data Needed: Set up a net to catch salmon before they enter the area and then measure nitrogen influx. This would help isolate the effect of salmon migration on nitrogen influx, independent of bear presence.

Question 10 (Honeybee Communication - Waggle Dance):

  • Model: Waggle dance communicates direction and distance.
  • Consistent Statement: The farther the target flowers from the hive, the longer the waggle phase.

Question 13 (Speciation Model):

  • Model: Shows stages of speciation with varying hybrid viability/fertility.
  • Most Likely Consequence: Hybrid individuals are less likely to pass their genetic information on to subsequent generations (reduced hybrid fitness/viability). This is a key driver of reproductive isolation and divergence into separate species.

Question 15 (Hybrid Infertility in Mice):

  • Experiment: Hybrid mouse offspring from two isolated populations were mated with original populations. Only female hybrids were fertile.
  • Consistent Stage: This scenario aligns with the Late Stage of speciation (Figure 3), where one sex of the hybrid is fertile, and the other is sterile, indicating significant divergence.

Question 17 (Aquaculture Impact - Dependent Variable):

  • Experiment: Investigating the effect of fish farming (ocean pens) on nearby ecosystems.
  • Appropriate Dependent Variable: The amount of metabolic wastes in the water where the fish are being raised. This directly measures the impact of the farmed fish on their immediate environment.

Question 19 (Membrane Transport Model):

  • Model: Depicts movement of molecules across a membrane.
  • Most Relevant Scientific Question: Which molecular substance is actively transported across the plasma membrane? This question directly relates to understanding the mechanisms of transport shown in the model, which often involve active processes requiring energy.

Question 20 (Compost Effect on Plant Growth - Experimental Validity):

  • Conclusion: Pea plants grew better in compost than melon plants.
  • Validity: The conclusion is invalid.
  • Reason: Other variables (biotic: fungal growth on melon plants; abiotic: rainy weather) affected the results, confounding the effect of compost.

Question 21 (Enzyme Variants and Temperature Adaptation):

  • Observation: Butterflies in the same population have glycolytic enzyme variants optimal at different temperatures (29°C, 40°C, or both).
  • Consistent Claim: Butterflies that express two variants of the enzyme are active over a greater range of temperatures. This suggests heterozygosity or expression of multiple alleles provides a broader thermal tolerance.

Grid-In Questions (Examples)

  • Question 121 (Phylogenetic Characters): Requires identifying a derived character shared by alligators and manatees but not salamanders, based on a provided cladogram.
  • Question 122 (Drosophila Genetics): Calculates the expected proportion of offspring with specific traits (male, wild-type wings, white eyes) from a given cross, considering autosomal and sex-linked genes.
  • Question 123 (Hardy-Weinberg Equilibrium - ABO Blood Type): Calculates the frequency of the 'i' allele in a population using Hardy-Weinberg equations for a three-allele system.
  • Question 124 (Simpson's Index of Diversity): Calculates the Simpson's index of diversity for a tree community based on species abundance data.
  • Question 125 (Water Potential Calculation): Calculates the water potential of zucchini squash tissue using data from a mass change experiment in different sucrose solutions, based on a provided graph and formula.

This summary provides a structured overview of the key concepts, experimental designs, data interpretations, and biological principles assessed in the provided AP Biology exam materials.

Summary of AP Biology Exam Materials (2018 Administration)

This document contains materials from the 2018 AP Biology Exam, including administrative instructions, exam sections, equations and formulas, and scoring guidelines. It is presented as a modified version for practice purposes.

Section I: Multiple-Choice and Grid-In Questions

  • Total Time: 1 hour and 30 minutes
  • Number of Questions: 58 (53 multiple-choice, 5 grid-in)
  • Percent of Total Score: 50%
  • Writing Instrument: Pencil required
  • Electronic Device: Calculator allowed

Key Instructions for Students:

  • Answers for multiple-choice questions (1-53) must be marked on the answer sheet using a No. 2 pencil. Only options A-D should be used.
  • Grid-in questions (121-125) require writing the numeric answer in the boxes at the top of the grid and filling in the corresponding circles.
  • No credit is given for anything written in the exam booklet.
  • Margins and blank spaces in the booklet can be used for scratch work.
  • The score is based solely on the number of questions answered correctly; no points are deducted for incorrect answers.
  • Students must use a pen for Section II.

Section II: Free-Response Questions

  • Total Time: 1 hour and 30 minutes (10-minute reading period, 1 hour and 20-minute writing period)
  • Number of Questions: 8
  • Percent of Total Score: 50%
  • Writing Instrument: Pen with black or dark blue ink
  • Calculator: Allowed

Question Breakdown:

  • Questions 1 & 2: Long free-response questions (approx. 22 minutes each, 10 points each).
  • Questions 3-5: Short free-response questions (approx. 6 minutes each, 4 points each).
  • Questions 6-8: Short free-response questions (approx. 6 minutes each, 3 points each).

Key Instructions for Students:

  • Answers must be written in paragraph form; outlines or bulleted lists alone are not acceptable.
  • Labeled diagrams may supplement discussion but are not sufficient for credit unless specifically requested.
  • Write clearly and legibly.
  • Begin each answer on a new page.
  • Cross out any errors; crossed-out work will not be scored.
  • Manage time carefully and proceed freely between questions.
  • Use unlined pages for notes and planning during the 10-minute reading period.

AP Biology Equations and Formulas

The document includes a comprehensive list of formulas and equations relevant to AP Biology, categorized as:

  • Statistical Analysis and Probability: Standard Error of the Mean, probability rules (mutually exclusive and independent events).
  • Hardy-Weinberg Equations: Allele and genotype frequencies.
  • Descriptive Statistics: Mode, Median, Mean, Range.
  • Population Growth: Exponential Growth, Logistic Growth.
  • Physiology and Ecology: Temperature Coefficient (Q10), Primary Productivity Calculation, Water Potential (Y = Yp + Ys), Solute Potential (Ys = -iCRT).
  • Biochemistry and Cell Biology: Gibbs Free Energy (ΔG), Entropy (ΔS), Enthalpy (ΔH).
  • Geometry: Volume and Surface Area of Rectangular Solids, Cubes, Right Cylinders, and Spheres.
  • Solutions: Dilution (CiVi = CfVf).

Sample Questions and Scoring Guidelines

The document includes examples of free-response questions and their corresponding scoring guidelines, demonstrating how student responses are evaluated. These examples cover topics such as:

  • Bacterial Communication and Biofilms (Question 1): Investigating the role of signaling proteins (S and R) in bacterial survival and biofilm formation.
  • Animal Behavior and Speciation (Question 2): Analyzing the influence of visual and auditory cues on mate competition in flycatchers and identifying prezygotic barriers.
  • Ecology and Bee-Flower Interactions (Question 3): Examining the relationship between bee tongue length, flower depth, and pollination success.
  • Molecular Biology and Gene Expression (Question 4): Exploring alternative splicing and the effect of mutations on protein structure.
  • Endocrinology and Signal Transduction (Question 5): Understanding the role of leptin in appetite suppression and the effectiveness of different treatments.
  • Genetics and Evolution (Question 6): Constructing cladograms based on SNP data and describing positive selection in the EPAS1 gene.
  • Malaria Parasite Biology (Question 7): Analyzing the effect of atovaquone on Plasmodium falciparum growth and identifying cellular locations.
  • Cell Biology and Membrane Structure (Question 8): Representing plasma membrane models in different environments.

Administrative Information

  • Exam Dates: May 14, 2018 (regular), May 24, 2018 (late-testing).
  • Proctor Instructions: Detailed guidance for administering both Section I and Section II, including managing student materials, time, and security protocols.
  • Answer Sheet: A sample of the student answer sheet is provided, detailing areas for personal information, exam labels, multiple-choice answers, and grid-in responses.
  • Scoring: Information on scoring worksheets and conversion charts is included, indicating how raw scores are converted to AP scores.
  • Copyright: Materials are copyrighted by the College Board and are for educational use only. Redistribution is prohibited.

AP® Biology 2024 Free-Response Questions Summary

This document outlines the structure and content of the AP Biology Free-Response Questions for 2024. It includes instructions for students and presents the text of six questions, covering various biological concepts.

General Instructions for Students

  • Time Allotment:
    • Questions 1 and 2 (Long FRQ): Approximately 25 minutes each.
    • Questions 3-6 (Short FRQ): Approximately 10 minutes each.
  • Answering Format:
    • Answers must be written in paragraph form.
    • Outlines, bulleted lists, or diagrams alone are not acceptable.
  • Response Booklet:
    • Planning can be done in the provided booklet, but only responses written in the separate Free Response booklet will be graded.

Question 1: Meiosis, Crossing Over, and Chromosome Structure

This question focuses on the process of meiosis, specifically crossing over, and its relationship with chromosome structure and gene expression.

  • Core Concept: Crossing over in meiosis I is crucial for proper chromosome alignment and segregation.
  • (i) Function of S phase:
    • This part asks to describe the function of the S phase of interphase. (The provided text does not contain the answer to this specific sub-question but introduces the topic).
  • Hotspots and Centromere Suppression:
    • Hotspots: Regions on chromosomes with a higher frequency of crossing over.
    • Centromere Suppression: Crossing over is suppressed near centromeres.
    • Hypothesis: Kinetochore proteins near the centromere are responsible for suppressing crossing over.
  • Experimental Setup:
    • Yeast chromosome 8 was modified: one homologous chromosome carried the gene for Red Fluorescent Protein (RFP), and the other carried the gene for Green Fluorescent Protein (GFP).
    • RFP-expressing cells emit red light; GFP-expressing cells emit green light.
    • Figure 1 illustrates chromosome 8 before and after crossing over.
  • (ii) Haploid cells with one fluorescent marker:
    • This part asks to explain why some haploid cells formed after meiosis will have only one fluorescent marker. (The provided text sets up the experiment but doesn't detail the explanation).
  • Investigating Kinetochore Proteins:
    • Scientists tested if attaching kinetochore proteins (CTF and IML) to a crossing-over hotspot affected crossing over frequency.
    • Experimental Groups:
      • Group 1: No kinetochore proteins attached.
      • Group 2: Kinetochore protein CTF attached.
      • Group 3: Kinetochore protein IML attached.
    • Frequency Calculation: (Cells with both red and green light + cells with no light) / Total cells.
    • Figure 2 shows the frequency of crossing over for each group.
  • (i) Control Group:
    • Identify the control group for the first experiment shown in Figure 2.
  • (ii) Follow-up Experiment Control:
    • A modified CTF protein lacking its DNA-binding portion was used.
    • Justify why this modified CTF was used as a control in a follow-up experiment.
  • (iii) Independent Variable:
    • Identify the independent variable in the follow-up experiment.
  • (c) Effect of CTF vs. IML:
    • Based on Figure 2, describe the effect on crossing over frequency when CTF is attached compared to when IML is attached.
  • (i) Prediction of Chromosome Copies:
    • Predict the effect on the number of chromosome 8 copies in daughter cells when CTF is attached to the hotspot.
  • (ii) Reasoning for Prediction:
    • Provide reasoning to justify the prediction in (i).
  • (iii) Hotspots and Survival:
    • Explain how the presence of hotspots could increase population survival under selective pressures.

AP® Biology Equations and Formulas

This section provides a reference sheet for statistical analysis, probability, Hardy-Weinberg equations, water potential, surface area and volume calculations, and the Simpson's Diversity Index.

  • Statistical Analysis and Probability:
    • Standard Error of the Mean (SEM) formula.
    • Degrees of freedom definition.
    • Probability rules for mutually exclusive events (P(A or B)) and independent events (P(A and B)).
  • Hardy-Weinberg Equations:
    • Definitions of allele frequencies (p, q).
  • Descriptive Statistics:
    • Mode, Median, Mean, Range definitions.
  • Water Potential:
    • Formula: Ψ = Ψp + Ψs
    • Ψs (solute potential) formula: Ψs = -iCRT
      • i = ionization constant
      • C = molar concentration
      • R = pressure constant (0.0831 liter bars/mole K)
      • T = temperature in Kelvin
    • Note: Pressure potential (Ψp) is zero in an open container.
  • Surface Area and Volume:
    • Formulas for Sphere, Rectangular Solid, and Cylinder.
    • Formula for Cube.
  • Population Growth:
    • Logistic growth equation: dN/dt = rmaxN(1 - N/K)
      • K = carrying capacity
      • rmax = maximum per capita growth rate
  • Simpson's Diversity Index:
    • Formula: Diversity Index = 1 - Σ(ni/N)²
      • ni = total number of organisms of a particular species
      • N = total number of organisms of all species

Question 2: Metabolism and Temperature Effects

This question investigates the relationship between environmental temperature and metabolic activity in toad liver cells.

  • Experimental Setup:
    • Toad liver cells were incubated at different temperatures.
    • Two markers of metabolic activity were measured:
      • Rate of oxygen consumption.
      • Rate of ATP synthesis.
    • Table 1 presents the data collected at 15°C and 25°C.
  • (a) Role of Water in ATP Hydrolysis:
    • Describe the role of water in the hydrolysis of ATP. (The provided text does not contain the answer but introduces the topic).
  • (i) Bar Graph Construction:
    • Construct a bar graph representing the data from Table 1. The graph must be appropriately plotted and labeled.
  • (ii) Oxygen Consumption Difference:
    • Based on the data, determine the temperature (°C) at which the rate of oxygen consumption differs from the rate at 25°C.
  • (i) Effect of Temperature on ATP Synthesis:
    • Based on Table 1, describe the effect of temperature on the rate of ATP synthesis.
  • (ii) Calculation of Oxygen Consumption:
    • Calculate the average amount of oxygen consumed (in nmol) for 10 mg of mitochondrial protein after 10 minutes at 25°C.
  • (i) Effect of Oligomycin:
    • Oligomycin blocks the channel protein function of ATP synthase. Predict the effect of oligomycin on the proton gradient across the inner mitochondrial membrane.
  • (ii) Justification of Prediction:
    • Justify the prediction made in (i).

Question 3: Glucose Uptake and Aging in Red Blood Cells

This question examines how the ability of red blood cells to take up glucose changes with age.

  • Experimental Setup:
    • Red blood cells from guinea pigs of varying ages (1 day to 7 months) were collected.
    • Equal numbers of red blood cells were incubated in separate dishes with radioactively labeled glucose (300 nM).
    • The amount of labeled glucose inside the cells was measured over time.
  • (a) Passive vs. Active Transport:
    • Describe a difference between passive transport and active transport.
  • (b) Control Justification:
    • Justify why the scientists used an equal number of red blood cells in each culture dish as a control.
  • (c) Prediction of Glucose Uptake:
    • Scientists claim that the gene expression for glucose transporters decreases with age. If this claim is supported, predict the effect of increased age on the amount of radioactively labeled glucose inside the cells.
  • (d) Justification of Prediction:
    • Justify the prediction made in part (c).

Question 4: Invasive Species and Ecosystem Dynamics

This question explores the ecological impact of an invasive species, the common wild oat, on native bunchgrass ecosystems.

  • Scenario:
    • Common wild oat (invasive) has replaced native bunchgrasses in California.
    • Aphids (virus carriers) have higher reproductive rates in areas with common wild oat.
    • Viruses affect only native bunchgrasses, increasing their death rates.
  • (a) Ecosystem Resilience:
    • Describe the change in ecosystem resilience when the number of species decreases.
  • (b) Effect of Invasive Species:
    • Explain how the addition of the common wild oat affects the number of native bunchgrass plants that can be supported by the ecosystem.
  • (c) Predicted Effect of Ladybugs:
    • Researchers suggest adding ladybugs (aphid predators). Predict the effect of adding ladybugs on the abundance of the native bunchgrass population.
  • (d) Justification of Prediction:
    • Justify the prediction made in part (c).

Question 5: Speciation and Gene Evolution

This question delves into the mechanisms of speciation and the evolution of new genes.

  • Speciation: Researchers study mechanisms that enable or prevent speciation.
  • (a) Post-Zygotic Mechanism:
    • Describe a post-zygotic mechanism that prevents gene flow and enables speciation.
  • Gene Evolution:
    • New genes can evolve from noncoding DNA with the presence of regulatory elements.
    • Functional Gene Structure (Figure 1): Promoter, 5' UTR, start codon, coding region, stop codon, 3' UTR.
  • Antifreeze Glycoprotein (AG) Genes in Cods (Figure 2):
    • AG glycoproteins reduce the freezing temperature of fish.
    • Researchers studied AG genes in nine cod species and one non-cod species (B. brosme).
    • They noted the presence/absence of functional AG genes and AG-like sequences.
    • Figure 2 shows a phylogenetic tree illustrating the evolution of AG genes in relation to periods of freezing temperatures and habitat water temperatures.
  • (b) Survival After Freezing:
    • Based on Figure 2, explain how genome changes enabled cods to survive and reproduce after a period of freezing temperatures (10-15 million years ago).
  • (c) Origin of Functional AG Gene:
    • Place an "X" on the phylogenetic tree to represent the origin of the functional AG gene.
  • (d) Genetic Differences and Habitats:
    • Based on Figure 2, explain how genetic differences among Gadidae species determine their survival habitats.

Question 6: Translation Rates and Protein Production

This question focuses on the process of translation, specifically the variation in translation rates for different codons and its implications.

  • Ribosome Profiling:
    • A technique to measure how long ribosomes pause at each codon.
    • Average translation rate: 5.2 amino acids/second.
    • Variation in translation rates for specific codons can occur.
    • Hypothesis: Variations facilitate correct protein folding.
  • Experimental Data (Figure 1):
    • The distribution of translation rates (time per codon) was measured for three codons (GAC, AUU, UGG) across 100 mRNAs.
    • Figure 1A shows data for GAC.
    • Figure 1B shows data for AUU.
    • Figure 1C shows data for UGG.
  • (a) Most Frequent Rate for GAC:
    • Using Figure 1A, identify the translation rate (ms/codon) recorded most frequently for the GAC codon.
  • (b) Variation Comparison:
    • Using Figures 1B and 1C, describe the variation in translation rate for the AUU codon compared to the UGG codon.
  • (c) tRNA Abundance Hypothesis:
    • Scientists hypothesize that tRNAs for UGG codons are less abundant than tRNAs for AUU codons.
    • Support this hypothesis using the data in Figure 1.
  • (d) Codon Choice and Protein Production:
    • Amino acids can be encoded by multiple codons.
    • Explain why using one codon over another for the same amino acid might result in increased protein production from an mRNA, based on the provided data.

AP Biology Practice Exam #1 and Notes (Spring 2020)

This document serves as a practice exam and accompanying notes for the AP Biology exam, specifically for the Spring 2020 administration. It is provided by the College Board for AP Exam preparation and has strict distribution policies.

I. Exam Content and Format

The AP Biology Exam is approximately 3 hours in length and is divided into two sections:

  • Section I: Multiple Choice

    • Duration: 1 hour and 30 minutes
    • Number of Questions: 60
    • Weight: 50% of the final score
    • Allowed Electronic Device: Calculator (four-function, scientific, or graphing)
    • Scoring: Based solely on the number of questions answered correctly; no penalty for incorrect answers or unanswered questions.

  • Section II: Free Response

    • Duration: 1 hour and 30 minutes
    • Number of Questions: 2 long free-response questions and 4 short free-response questions
    • Weight: 50% of the final score
    • Writing Instrument: Pen (black or dark blue ink)
    • Suggested Time: Approximately 25 minutes per long question, 10 minutes per short question.
    • Scoring: Answers must be in paragraph form. Outlines, bulleted lists, or diagrams alone are not acceptable.

II. Administering the Practice Exam

These instructions are designed to simulate an actual AP Exam administration.

  • Before Testing: Ensure all exam materials (test booklets, answer sheets) are ready.
  • Section I: Multiple Choice
    • Students are instructed to mark all responses on their answer sheet.
    • A calculator is permitted.
    • Time allotted: 1 hour and 30 minutes.
    • Announcements: "10 minutes remaining" and "Stop working."
    • Collection: Section I booklet and answer sheet are collected.
  • Break: A 10-minute break is scheduled between Section I and Section II.
  • Section II: Free Response
    • Students are responsible for pacing themselves.
    • Answers must be written on the lined pages provided for each question.
    • Extra paper can be requested and must be clearly labeled with the student's name and question number.
    • Time allotted: 1 hour and 30 minutes.
    • Announcements: "10 minutes remaining" and "Stop working and close your exam booklet."
    • Collection: Section II booklet is collected. Extra paper must be stapled to the corresponding question page.

III. AP® Biology Equations and Formulas

The exam provides a reference sheet with formulas and equations for:

  • Statistical Analysis and Probability:
    • Standard Error of the Mean
    • Probability rules (mutually exclusive and independent events)
  • Hardy-Weinberg Equations:
    • Allele frequencies (p, q)
    • Genotype frequencies (p², 2pq, q²)
  • Descriptive Statistics: Mode, Median, Mean, Range
  • Population Growth:
    • Rate of change (dY/dt)
    • Population growth (dN/dt = B - D)
    • Logistic Growth (dN/dt = rmax N(K-N)/K)
  • Water Potential:
    • Total water potential (Ψ = Ψp + Ψs)
    • Solute potential (Ψs = -iCRT)
  • Surface Area and Volume: Formulas for spheres and rectangular solids.
  • Simpson's Diversity Index: Formula for calculating diversity.

IV. Sample Multiple-Choice Questions and Concepts Covered

The practice exam includes questions covering a wide range of AP Biology topics, including:

  • Properties of Water: High specific heat capacity, high heat of vaporization, surface tension, melting temperature.
  • Cellular Processes: Osmosis, photosynthesis, cellular respiration (Krebs cycle, pyruvate dehydrogenase), gene expression, signal transduction.
  • Evolution: Compartmentalization in eukaryotes, homologous hormones, convergent evolution, phylogenetic trees, Hardy-Weinberg equilibrium.
  • Ecology: DMSP production in corals, symbiotic relationships, environmental stress (temperature), feedback mechanisms (positive and negative), population dynamics, keystone species, ecosystem disruptions, nutrient pollution, climate change impacts.
  • Genetics: Gene mutations, DNA damage (UV radiation), inheritance patterns (pedigrees, X-linked, Y-linked, mitochondrial), genetic disorders, PCR, DNA gel electrophoresis, gene expression regulation (transcription factors, oncogenes).
  • Molecular Biology: DNA structure, RNA polymerase function, protein synthesis, retroviruses, phospholipid bilayers, transport vesicles.
  • Cellular Respiration and Photosynthesis: Electron transport chains, role of water, role of DCMU, rubisco activity, carbon fixation.
  • Cell Biology: Cell cycle regulation, mitosis, chromosome number, surface area-to-volume ratio, mitochondrial structure.
  • Experimental Design and Data Analysis: Identifying variables, controls, interpreting graphs and tables, statistical analysis (chi-square).

V. Free-Response Section Examples

The free-response section includes questions that require students to:

  • Interpret and Evaluate Experimental Results: Analyze data from experiments, describe components of biological molecules, explain the function of enzymes (e.g., RNA polymerase), identify variables and controls, and justify claims with evidence.
  • Graphing and Data Representation: Construct and label graphs based on provided data, interpret statistical differences, and make predictions based on trends.
  • Scientific Investigation: Justify experimental design choices (e.g., controlling variables), predict the effects of environmental factors on populations, and provide reasoning for predictions.
  • Conceptual Analysis: Describe factors influencing phenotypes, explain gene expression mechanisms, and predict outcomes based on biological models.
  • Analyze Models or Visual Representations: Describe the role of cellular processes (e.g., mitosis), explain differences in cell populations, refine models, and explain how cellular structures facilitate functions.
  • Analyze Data (Surface Area-to-Volume): Identify relationships between dimensions and SA:V ratios, evaluate hypotheses based on data, and explain the functional significance of cellular structures.

VI. Important Notes on Exam Security

  • This exam is for AP Exam preparation only.
  • It may not be posted on school or personal websites or electronically redistributed.
  • Teachers may download and copy materials for classroom use only.
  • All exam materials must be collected and kept secure after administration.
  • Unauthorized distribution violates College Board copyright policies and can lead to termination of access to Practice Exams and other online services.

This summary is based on the provided text and aims to capture the essential information regarding the AP Biology Practice Exam structure, content, and administration guidelines.

Here's a summary of the provided content, organized by the questions and topics presented:

AP Biology Exam Content Summary

This document appears to be a collection of questions and scoring guidelines from an AP Biology exam, covering various biological concepts. The content is structured around specific questions, each addressing a different area of biology.

Question 1: Cell Cycle and Cancer

  • Cell Division: Human cells divide for growth, replacement, and repair, progressing through three phases of the cell cycle.
  • Cell Signaling in Tissue Repair:
    • Damaged skin cells release epidermal growth factor (EGF).
    • Skin cells have epidermal growth factor receptors (EGFR) on their membranes.
    • EGF binding to EGFR initiates a signaling pathway that promotes cell cycle progression and division, leading to tissue repair.
  • Cancer and EGFR:
    • Mutations increasing EGFR production are linked to head and neck skin cancer.
    • Mutations outside the coding region (e.g., in promoter or enhancer regions) can lead to overproduction of EGFR protein by increasing transcription rates, gene duplication, or inactivating repressors.
  • Antibody Therapy for Cancer:
    • Antibodies targeting the extracellular portion of EGFR can be effective.
    • Model: Antibodies likely bind to EGFR, preventing EGF from binding and activating the receptor, thus inhibiting the cell signaling pathway and blocking cell division.
    • Effectiveness: This therapy is effective because it targets the overactive signaling pathway driving cancer cell proliferation.

Question 2: Plant Growth and Respiration

  • Experiment Setup: Thirty corn seedlings were divided into three groups:
    • Group I: Initial dry mass measured.
    • Group II: Maintained in light for one week.
    • Group III: Maintained in the dark for one week.
  • Light-Grown Plants (Group II):
    • Increase in mass: Explained by photosynthesis.
    • Inorganic source of mass: Carbon dioxide (CO2) incorporated into carbohydrates, or water (H2O) incorporated into carbohydrates.
  • Dark-Grown Plants (Group III):
    • Decrease in mass: Explained by cellular respiration.
    • Overall chemical reaction: C6H12O6 + 6O2 → 6CO2 + 6H2O (sugar + oxygen → carbon dioxide + water).
    • Process: Plants consume stored sugars and oxygen, releasing CO2 and water, leading to a net loss of dry mass.

Question 3: Action Potentials in Neurons

  • Experiment: Studying the effect of a stimulus on a motor neuron's action potential, observing changes in membrane permeability to Na+ and K+.
  • Figure 1 (Permeability): Shows changes in relative permeability to Na+ and K+ over time.
  • Figure 2 (Membrane Potential): Shows changes in membrane potential (millivolts) during an action potential.
  • Key Observations:
    • Difference in Permeability Curves: Peak permeability to Na+ occurs before K+; changes in Na+ permeability are more rapid and of greater magnitude than K+.
    • Structural Feature: Voltage-gated ion channels (for Na+ and K+) regulate ion movement. The phospholipid bilayer also restricts unregulated ion diffusion. Na+/K+ pumps maintain resting potential.
    • Connection to Membrane Potential:
      • Increased Na+ permeability leads to depolarization (membrane potential becomes more positive) during stage II.
      • Increased K+ permeability leads to repolarization (membrane potential becomes more negative) during stage III. (Decreased Na+ permeability also contributes to repolarization).

Question 4: Aquatic Animal Dives and Physiology

  • Observation: Aquatic animals build up lactic acid during long dives.
  • Data: Graph shows blood lactate levels after dives of varying durations for three species (Weddell seals, Baikal seals, Emperor penguins).
  • Hypothesis (Weddell Seals > 20 min): Increased lactate levels are due to fermentation or anaerobic metabolism occurring during prolonged dives when oxygen is depleted.
  • Blood Oxygen Levels (Weddell Seals): The curve would likely show oxygen levels starting high and declining steadily during the dive. The decline might slow down or plateau after 20 minutes as anaerobic metabolism becomes more dominant.
  • Evolutionary Hypothesis (Baikal Seals vs. Emperor Penguins): Baikal seals likely evolved a greater capacity for sustained dives due to genetic adaptations that provide a selective advantage (e.g., access to food, predator avoidance), which may not have been as strong a selective pressure for penguins.

Question 5: Protein Secretion Pathway

  • Experiment: Tracking radioactively labeled polypeptides in pancreatic cells using electron microscopy.
  • Results (Figure 1): Dark dots (labeled polypeptides) move sequentially from the Rough Endoplasmic Reticulum (RER) to vesicles, then to the Golgi apparatus, and finally to vesicles for release outside the cell.
  • Pathway of Secretory Proteins: Synthesis on ribosomes attached to the RER → modification and folding within the RER → transport via vesicles → further processing in the Golgi apparatus → packaging into secretory vesicles → exocytosis (release from the cell).
  • Prediction (Labeled mRNA): If mRNA were labeled, the dots would initially appear in the nucleus (where transcription occurs) and then move to the cytoplasm/RER (where translation occurs).

Question 6: Plant Distribution and Climate Change

  • Observation: Plant distribution shifts to higher elevations/latitudes with warming climates (Humboldt's observation).
  • Contradictory Finding: In the western US, 72% of species shifted to lower elevations despite increasing temperatures.
  • Explanation for Lower Elevation Shift:
    • Hypothesis (Environmental Factor): Factors other than temperature, such as increased soil moisture/rainfall at lower elevations, decreased UV light at lower elevations, increased CO2 at lower elevations, increased nutrient availability in richer soil at lower elevations, or decreased herbivory at lower elevations.
    • Evidence: Researchers could collect data on soil moisture, UV light intensity, CO2 levels, soil nutrient content, or herbivore populations at different elevations.
    • Reasoning: Connect the evidence to plant growth; e.g., higher soil moisture at lower elevations supports plant growth, allowing species to migrate downwards even as temperatures rise.

Question 7: Enzyme Kinetics and Inhibition

  • Graph: Shows the initial reaction rate of an enzyme at different substrate concentrations.
  • Enzyme Structure:
    • Primary structure (amino acid sequence) determines the enzyme's overall 3D shape through interactions between amino acid side chains (R-groups).
  • Noncompetitive Inhibitor:
    • Prediction: A noncompetitive inhibitor will decrease the reaction rate, even at high substrate concentrations.
    • Mechanism: The inhibitor binds to an allosteric site (not the active site), changing the enzyme's shape and reducing its ability to bind the substrate or catalyze the reaction effectively.

Question 8: Antibiotic Resistance

  • Study Data: Table shows the incidence of antibiotic resistance in bacteria from ICUs (1994-2000) and the overall change in resistance.
  • Most/Least Effective Antibiotics (Dec 2000):
    • Most effective: The antibiotic with the lowest percentage of resistance in 2000 (e.g., Imipenem).
    • Least effective: The antibiotic with the highest percentage of resistance in 2000 (e.g., Ampicillin).
  • Justification for Resistance Arising: The overall increase in percent resistance from 1994 to 2000 indicates that resistance is developing or spreading within the bacterial population.
  • Processes of Resistance Development:
    1. Random Mutations: Changes in DNA sequences can confer resistance.
    2. Transformation: Uptake of DNA from the environment.
    3. Conjugation: Transfer of DNA between bacteria.
    4. Transduction: Transfer of DNA via viruses.
    5. Horizontal Gene Transfer: General term for acquiring new genes.
  • Natural Selection and Spread of Resistance:
    • Features: Variants with resistance have higher survival/reproduction rates; variants without resistance have lower survival rates; resistant individuals produce more offspring; phenotypes/allele frequencies change over generations.
    • Evolution Example: The development of antibiotic resistance is evolution because there is a change in the heritable characteristics (allele frequencies) of the bacterial population over time due to differential survival and reproduction.
  • Ethical Question (Antibiotics in Cattle): Should low doses of antibiotics be used in beef cattle if it poses a risk to human health (e.g., by promoting resistance) or harms the environment/ecosystem?

Question 9: Circadian Rhythms in Mice

  • Experiment: Investigating mouse activity patterns using running wheels and actograms under different light conditions (L12:D12 and continuous darkness - DD).
  • Figure 1: Method for recording activity.
  • Figure 2 (L12:D12): Mice are active during the dark periods and inactive during the light periods.
  • Figure 3 (DD): Mice maintain a rhythm, but the cycle is slightly less than 24 hours (e.g., activity starts slightly earlier each day), indicating a genetically controlled circadian rhythm independent of the light-dark cycle.
  • Roles in Light-Dark Response:
    • Photoreceptor (Retina): Detects light/dark stimulus and initiates nerve signals.
    • Nervous System (e.g., Brain): Integrates/processes signals and transmits commands to effectors.
    • Effector (e.g., Muscles): Transmits signals from the brain to carry out actions (e.g., movement).
  • Researchers' Claim (Circadian Rhythm ≠ 24 hours): The daily activity pattern in DD is slightly shorter than 24 hours, causing the pattern to shift earlier each day. If it were a strict 24-hour rhythm, the pattern would remain consistent daily, even in DD.
  • Mutant Mice (No Circadian Gene):
    • Under L12:D12: Activity would likely be random throughout the 24-hour period, or potentially suppressed during light and slightly more active during dark, but without a clear rhythm.
    • Under DD: Activity would be completely random, showing no cyclical pattern.
    • Support for Light Overriding: This pattern would support the claim that light can impose a rhythm (even if imperfect) in the absence of the internal clock, or that the internal clock is essential for a coherent rhythm.
  • Evolution of Activity Pattern (Predator-Prey):
    • Features: Selection favors individuals active at night (avoiding daytime predators like birds); selection against individuals active during the day; night-active variants have higher survival and reproductive success.

Question 10: Cellular Respiration and ATP Synthesis

  • Pathways: Glycolysis, Krebs cycle, Electron Transport Chain (ETC).
  • Contributions to ATP Synthesis:
    • Glycolysis/Pyruvate Oxidation: Produces NADH (for ETC), Acetyl-CoA (for Krebs cycle), and some ATP via substrate-level phosphorylation.
    • Krebs Cycle: Produces NADH and FADH2 (for ETC), releases high-energy electrons, and produces some ATP (GTP) via substrate-level phosphorylation.
    • ETC: Creates a proton gradient across the inner mitochondrial membrane, which drives ATP synthesis via oxidative phosphorylation through ATP synthase.
  • Evidence for Glycolysis in Common Ancestor:
    • Ubiquity: Performed by nearly all organisms (highly conserved).
    • Anaerobic Conditions: Occurs without oxygen, suggesting it predates the oxygen-rich atmosphere.
    • Cytosolic Location: Occurs in the cytosol, suggesting it predates membrane-bound organelles like mitochondria.
  • ATP Production Efficiency:
    • Given: 30 ATP per glucose, 686 kcal/mol for glucose oxidation, 7.3 kcal/mol for ATP hydrolysis.
    • Energy in ATP: 30 moles ATP * 7.3 kcal/mol = 219 kcal/mol.
    • Efficiency: (219 kcal / 686 kcal) * 100% ≈ 31.9%.
    • Excess Energy: Released as heat, increasing entropy.
  • Krebs Cycle Location and Eukaryote Evolution: Scientific question: "Since the Krebs cycle enzymes are in the mitochondrial matrix (cytoplasm of mitochondria), does this support the endosymbiotic theory that mitochondria originated from engulfed prokaryotes?"

Question 11: Molecular Evolution (Cytochrome c)

  • Data: Table shows amino acid differences in cytochrome c among five vertebrate species.
  • Phylogenetic Tree: Constructed based on the number of differences. Species with the fewest differences are most closely related. The species with the most differences from all others (e.g., D. polylepis) is placed furthest from the others.
  • Morphological vs. Amino Acid Data:
    • Amino acid sequence data is generally considered more accurate for representing evolutionary relationships.
    • Reasoning: Morphological similarities can arise from convergent evolution (analogous structures), while molecular data directly reflects genetic relatedness. Similar protein sequences indicate shared ancestry.

Question 12: Mitosis vs. Meiosis

  • Common Events: DNA replication (before division), chromatin condensation, spindle formation, chromosome alignment, chromatid separation (in mitosis and meiosis II), checkpoints.
  • Differences Leading to Genetic Variation:
    • Mitosis: One division → two genetically identical diploid (2n) daughter cells.
    • Meiosis: Two divisions → four genetically unique haploid (n) daughter cells.
    • Key Meiotic Events: Crossing over (recombination) between homologous chromosomes in Prophase I, and independent assortment of homologous chromosomes in Metaphase I lead to genetic variation.

Question 13: Phototropism in Plants

  • Experiment: Investigating the role of the plant shoot tip in response to light.
  • Claim: Shoot tip cells detect light.
  • Support:
    • Treatment II (Tip Removed): No bending occurs, indicating the tip is necessary.
    • Treatment III (Opaque Cap on Tip): No bending occurs, indicating the tip itself detects the light.
  • Additional Characteristics (Treatments IV & V):
    • Treatment IV (Permeable Barrier): Bending occurs, suggesting the tip produces a signal (auxin) that diffuses through the barrier.
    • Treatment V (Impermeable Barrier): No bending occurs, indicating the signal must move down from the tip through the shoot tissue.
    • Conclusion: The tip produces a mobile signal (auxin) that travels downwards to cause bending.

Question 14: Population Genetics (Snake Introduction)

  • Scenario: Small, declining snake population receives males from larger populations.
  • Initial Decline (1989-1993): Possible causes include low genetic diversity, an unfavorable sex ratio (too few males), an aged population, or a harmful mutation/disease.
  • Rescue by Introduced Males: Increased genetic diversity and/or corrected the sex ratio, leading to increased reproductive success.
  • Abundant Resources: The population size grows after the introduction, indicating that resources are not limiting growth.

Question 15: Sensory Perception (Olfactory Neurons)

  • System: Smell involves odorant molecules binding to receptors on olfactory neurons, triggering action potentials transmitted to the brain. Mammals have ~1000 odorant receptor genes.
  • Synaptic Transmission: Activated olfactory neuron releases neurotransmitters into the synapse, which bind to receptors on the postsynaptic interneuron, transmitting the signal.
  • Perception of Thousands of Odors:
    • Combinatorial Coding: A single odorant molecule can bind to multiple receptor types, and a single receptor type can bind multiple odorant molecules. The brain interprets the combination of activated receptors to perceive a specific odor.
    • Alternative processing/splicing of mRNA can also generate multiple receptor variants from a single gene.

Question 16: Immune Response (B Cells)

  • Condition: Individual lacks functional B cells and cannot mount a humoral immune response.
  • Initial Exposure:
    • Consequence: No antibody production and no memory B cell formation.
  • Second Exposure:
    • Consequence: The response will not be faster or stronger than the initial exposure; it will be slow and similar to the primary response because memory B cells are absent.
  • Unaffected Immune Response: Cell-mediated immunity (involving T cells, NK cells) and non-specific immunity (phagocytes like macrophages) remain functional.

Question 17: Cell Membrane Permeability and Pollutants

  • Experiment: Testing the effect of isopropanol and acetone (water-soluble pollutants) on beet root cell membrane permeability. Betacyanin release (measured by absorbance) indicates increased permeability.
  • Dependent Variable: Absorbance of 460 nm light by the surrounding solution.
  • Results Interpretation: Higher absorbance indicates more betacyanin released, meaning higher membrane permeability. Isopropanol and acetone increase membrane permeability.
  • Scientific Question: Will organisms in polluted water experience detrimental effects due to increased cellular membrane permeability?

Question 18: Meiosis and Genetic Diversity

  • Concept: Meiosis generates genetic diversity in daughter cells.
  • Stages Contributing Most to Diversity:
    • Prophase I: Crossing over (exchange of genetic material between homologous chromosomes).
    • Metaphase I: Independent assortment of homologous chromosome pairs.

Question 19: Population Genetics and Fur Color

  • Scenario: Mouse population changes fur color distribution over 20 years.
  • Possible Explanation: The field environment (light soil, sparse vegetation) provided camouflage (protection from predators) for gray mice, leading to their increased survival and reproduction (natural selection).

Question 20: Cell Structure and Function

  • Observation: Liver cells (synthesize glycoproteins) likely have more prominent rough endoplasmic reticulum (RER) and Golgi apparatus than adipose cells (store fat).

Question 21: Osmosis and IV Fluids

  • Scenario: Distilled water (hypotonic) IV administered.
  • Consequence: Water will move osmotically from the hypotonic IV fluid into the body's cells (which are hypertonic relative to distilled water), causing the cells to expand and potentially burst.

Question 22: Prokaryotic Cell Structures

  • Diagram: Rod-shaped bacterium.
  • Feature Unique to Archaea and Bacteria: Lack of a nuclear membrane surrounding the genetic material (nucleoid region).

Question 23: Common Ancestry Evidence

  • Claim: Organisms from different domains share a common ancestor.
  • Supporting Evidence: Glycolysis occurs in both prokaryotic and eukaryotic cells. This fundamental metabolic pathway's presence across diverse life forms suggests it originated in a common ancestor.

Question 24: Predator-Prey Population Dynamics

  • Model: Graph showing cyclical changes in predator and prey populations.
  • Disruptions: The presence of additional predator species would increase competition for prey, potentially disrupting the established predator-prey cycle.

Question 25: Gene Regulation and Phenotypic Variation (Sticklebacks)

  • Scenario: Sticklebacks have identical copies of the Pitx1 gene but show phenotypic variation in pelvic spines.
  • Explanation: Expression of the Pitx1 gene is affected by regulatory elements (like enhancers) and potentially other genetic loci. Differences in enhancer activity or other regulatory factors can lead to different levels or patterns of Pitx1 gene expression, resulting in phenotypic variation even with identical gene sequences.
  • Mutation Affecting All Tissues: A mutation in a region common to all enhancers or in the gene's core promoter would likely affect Pitx1 expression in all tissue types.
  • Jaw Enhancer Disabled: If the jaw enhancer is disabled, Pitx1 expression in the developing jaw would be affected, but expression in the pelvis (regulated by the hindlimb enhancer) might remain normal, leading to a normal jaw but reduced pelvic spine.
  • Selective Mechanism (Freshwater Sticklebacks): Reduced pelvic spines might offer an advantage in freshwater environments, possibly by reducing predation risk (e.g., fewer spines make them less visible or easier to hide) or by improving maneuverability.

Question 26: Plant Physiology (Flower Color)

  • Scenario: Hydrangea flower color depends on soil conditions, not inheritance.
  • Evidence for Soil Conditions: Growing cuttings from the same plant under controlled conditions with varying soil pH is the most direct way to show that soil, not genetics, determines flower color.

Question 27: Phylogenetics and Molecular Data

  • Scenario: Morphological phylogenetic tree for four species; DNA sequencing data is also available.
  • Question Answered by DNA Sequencing: DNA sequencing can help determine if species III and IV are the most closely related, providing molecular evidence to refine or confirm the morphological tree.

Question 28: Neuron Communication

  • Process: Transmission of information from sensory to interneurons.
  • Mechanism: Involves the release of chemical messengers (neurotransmitters) from the axon terminal of the sensory neuron into the synaptic cleft, which then bind to receptors on the interneuron.

Question 29: Respiratory System Regulation

  • Scenario: Increased cellular respiration leads to increased blood CO2 and decreased pH.
  • Physiological Response: pH sensors signal the brain → brain signals the diaphragm → increased breathing rate to eliminate excess CO2.

Question 30: Endocrine System Regulation (Thyroxin)

  • Hormonal Loop: Thyroxin secretion is regulated by a negative feedback loop involving TSH. Low thyroxin → increased TSH → increased thyroxin. High thyroxin → decreased TSH.
  • Conclusion: The negative feedback mechanism maintains relatively constant levels of thyroxin.

Question 31: Iron Metabolism and Ferritin

  • Experiment: Measuring ferritin protein levels in rats with and without iron, and with inhibitors of transcription (actinomycin D) and translation (cycloheximide).
  • Figure 1 (Actinomycin D): Iron increases ferritin levels. Actinomycin D blocks this increase, suggesting iron's effect involves transcription.
  • Figure 2 (Cycloheximide): Iron increases ferritin levels. Cycloheximide blocks this increase, suggesting iron's effect involves translation.
  • Biotechnology Approach (Identify Ferritin mRNA): Use Northern blotting – separate RNA by size (gel electrophoresis) and probe with labeled DNA complementary to ferritin mRNA.
  • Regulation Mechanism: Iron stimulates the translation of ferritin mRNA.
  • Mechanism Support: Iron likely increases ribosome binding to ferritin mRNA (or stabilizes the mRNA).
  • IREs (Iron Response Elements):
    • Observation: Ferritin IREs are conserved across multicellular organisms; others are specific to vertebrates.
    • Explanation: Ferritin IREs likely evolved early in multicellular organisms; non-ferritin IREs evolved later in specific lineages (vertebrates).
  • Hereditary Hemochromatosis (HHC): Reduced iron uptake into cells.
    • Effect on Ferritin: Reduced iron in the cytosol would likely lead to reduced levels of ferritin protein because the stimulus for its synthesis (high intracellular iron) is diminished.

Question 32: Protein Synthesis and Secretion (Ovalbumin)

  • Observation: Ovalbumin production in chick oviduct cells in response to estrogen, visualized with fluorescent dye.
  • Conclusion: Ovalbumin is synthesized on ribosomes bound to the rough endoplasmic reticulum (RER), moves through the ER and Golgi apparatus, and is secreted via vesicles. Option (C) best describes this pathway.

Question 33: Osmoregulation in Fish

  • Scenario: Freshwater fish secrete dilute urine; marine fish secrete concentrated urine.
  • Explanation for Differences: These differences arose during divergence from a common ancestor, as fish adapted to different environmental pressures (osmolarity) in marine vs. freshwater habitats.

Question 34: Evolution of Drug Resistance (Malaria)

  • Scenario: Plasmodium resistance to antimalarials is increasing.
  • Hypothesis Support (Preexisting Mutations): Some mutations conferring resistance are already present in Plasmodium populations before drug treatment begins. This is supported if these mutations are detected in samples collected prior to drug exposure.

Question 35: Type 2 Diabetes and Insulin Signaling

  • Scenario: Insulin signaling pathway defects in type 2 diabetes. Insulin mobilizes GLUT4 transporters to the cell surface for glucose uptake.
  • Figures 2, 3, 4: Show reduced glucose uptake, normal insulin receptor activation, but reduced IRS-1 activation in diabetic muscle samples.
  • Valid Interpretation: Reduced glucose uptake in type 2 diabetes indicates reduced mobilization of GLUT4 to the cell surface.
  • Most Likely Defect: Reduced IRS-1 activation suggests a defect downstream of the insulin receptor but upstream of GLUT4 mobilization, possibly involving IRS-1 itself or subsequent signaling molecules.
  • Genetic Change Resulting in Disrupted Signaling: A deletion in the cytoplasmic domain of the insulin receptor would prevent downstream signaling (like IRS-1 activation), mimicking the diabetic phenotype.

Question 36: Biological Amplification (Trypsinogen Activation)

  • Scenario: Trypsinogen activated to trypsin by enterokinase.
  • Positive Feedback Confirmation: Trypsin activates additional trypsinogen molecules. This self-amplification is a hallmark of positive feedback.

Question 37: Cancer Initiation

  • Scenario: Cancer results from mutations affecting cell cycle control.
  • Least Likely to Initiate Cancer: A defect in a cell-cycle checkpoint that prevents a cell from entering S phase (e.g., G1 checkpoint failure). This would halt the cell cycle, preventing uncontrolled division, unlike mutations causing continuous signaling or loss of checkpoint control.

Question 38: Cell Membrane Transport

  • Statement: Ions cannot easily cross the phospholipid bilayer due to the hydrophobic tails.
  • Correct Statement: Ions are unable to move through the phospholipid bilayer because the nonpolar tail regions are hydrophobic. (Options B, C, D contain inaccuracies about water/ion movement and phospholipid properties).

Question 39: Hardy-Weinberg Equilibrium

  • Scenario: Graph showing genotype frequencies vs. allele frequencies.
  • Calculation: If frequency of homozygous recessive (q²) = 0.75, then frequency of recessive allele (q) = √0.75 ≈ 0.87. Frequency of dominant allele (p) = 1 - q = 1 - 0.87 = 0.13.
  • Result: Frequency of dominant allele ≈ 0.13.

Question 40: Population Genetics (PKU Probability)

  • Scenario: Autosomal recessive PKU. Two carrier parents (genotype Pp).
  • Punnett Square:
    • Offspring genotypes: PP, Pp, Pp, pp.
    • Probability of not having PKU (PP or Pp) = 3/4.
  • Result: Probability = 0.75.

Question 41: Enzyme Kinetics and Temperature

  • Scenario: Respiration rate of bacteria at different water temperatures.
  • Calculation: Find the temperature change associated with the observed change in respiration rate. (Requires specific data points from Figures 1 and 2, which are not fully detailed here but imply a temperature increase led to increased respiration).

Question 42: DNA Replication and Mismatches

  • Scenario: DNA mismatch repair followed by replication.
  • Daughter Cells: After replication, one daughter DNA molecule will contain the original sequence, and the other will contain the sequence with the mutation (mismatched base pair). The illustration shows a T-G mismatch. After replication, one strand will have the original sequence, and the complementary strand will have the mismatch. The daughter cells will inherit one original strand and one replicated strand.

Question 43: Enzyme Structure and Function

  • Primary Structure: The sequence of amino acids determines the protein's 3D shape.
  • Noncompetitive Inhibitor: Decreases reaction rate by binding to an allosteric site, altering enzyme shape and function.

Question 44: Molecular Biology Techniques

  • Goal: Identify ferritin mRNA.
  • Method: Northern blotting (RNA gel electrophoresis followed by hybridization with a labeled complementary DNA probe).

Question 45: Gene Regulation (Iron Response)

  • Scenario: Iron regulation of ferritin protein levels.
  • Explanation: Iron stimulates translation of ferritin mRNA.

Question 46: Gene Regulation (Iron Response)

  • Data Interpretation: Iron increases ferritin protein levels. This effect is blocked by cycloheximide (inhibits translation) but not actinomycin D (inhibits transcription).
  • Conclusion: Iron increases ribosome binding to ferritin mRNA (or affects translation efficiency/mRNA stability).

Question 47: Evolutionary Conservation (IREs)

  • Observation: Ferritin IREs are conserved in all multicellular organisms; non-ferritin IREs are specific to vertebrates.
  • Explanation: Ferritin IREs likely originated in the common ancestor of multicellular organisms, while non-ferritin IREs evolved later in specific lineages (vertebrates).

Question 48: Genetic Diseases and Iron Metabolism

  • Scenario: Hereditary hemochromatosis (HHC) reduces iron uptake.
  • Effect on Ferritin: Reduced iron uptake means less iron in the cytosol, leading to reduced ferritin protein levels because the stimulus for ferritin synthesis is decreased.

Question 49: Protein Synthesis and Secretion (Ovalbumin)

  • Observation: Ovalbumin production visualized over time.
  • Conclusion: Ovalbumin is synthesized on ribosomes attached to the RER, processed through the Golgi, and secreted via vesicles. Option (C) accurately describes this pathway.

Question 50: Multiple Choice Questions (Various Topics)

  • Questions 1-28: Cover topics including genetics (Huntington's disease, PKU), evolution (antibiotic resistance, tetrapod evolution, prairie dog behavior, population growth, sticklebacks, common ancestry), physiology (muscle metabolism, respiration, endocrine system, osmosis), ecology (predator-prey dynamics), molecular biology (DNA, gene regulation), and cell biology (membrane permeability, cell structures).
  • Specific Examples:
    • Q2: Antibiotic resistance graph supports natural selection.
    • Q3: Comparing bone structure in fins/limbs tests tetrapod evolution hypothesis.
    • Q4: Shortage of oxygen leads to lactic acid buildup via fermentation.
    • Q5: Prairie dog barking explained by kin selection.
    • Q6: Highest population growth rate is typically at intermediate population sizes (K/2).
    • Q7-11: Beet root experiment on membrane permeability.
    • Q12: Meiosis (crossing over, independent assortment) contributes most to diversity.
    • Q13: Fur color change explained by predation and camouflage.
    • Q14: Liver cells have prominent RER/Golgi for glycoprotein synthesis.
    • Q15: Distilled water IV causes cells to swell/burst (osmosis).
    • Q16: Lack of nuclear membrane is unique to bacteria/archaea.
    • Q17: Glycolysis in both prokaryotes and eukaryotes supports common ancestry.
    • Q18: Additional predator species can disrupt predator-prey cycles.
    • Q19-22: Stickleback Pitx1 gene regulation and evolution.
    • Q23: Flower color determined by soil pH shown by growing cuttings in different soils.
    • Q24: DNA sequencing can refine phylogenetic trees.
    • Q25: Neuron communication involves neurotransmitter release.
    • Q26: Increased cellular respiration leads to increased breathing rate to eliminate CO2.
    • Q27: Thyroxin regulation is a negative feedback loop.
    • Q28-53: Continue covering diverse biological topics.

Question 121-125: Grid-In Questions (Numerical Answers)

  • These questions require calculation or determination of numerical values based on provided data (tables, graphs). Examples include predicting internal temperature, calculating allele frequencies, determining temperature changes, calculating probabilities, and finding temperature changes associated with physiological rates.

Section II: Free-Response Questions

  • Structure: Includes long (10 points) and short (4 or 3 points) free-response questions.
  • Content: Covers topics similar to multiple-choice, requiring detailed explanations, hypotheses, data interpretation, and connections between concepts.
    • Q1: Antibiotic resistance, natural selection, ethics.
    • Q2: Circadian rhythms, nervous system function, evolution.
    • Q3: Cellular respiration, ATP synthesis, evolution of metabolic pathways.
    • Q4: Molecular evolution, phylogenetic trees, data interpretation.
    • Q5: Cell division (mitosis vs. meiosis), genetic variation.
    • Q6: Plant physiology (phototropism), signal transduction.
    • Q7: Population genetics, conservation biology.
    • Q8: Sensory systems (olfaction), signal transmission, evolution of perception.
    • Q9: Immunology (B cells, humoral immunity).

Here's a structured summary of the provided AP® Biology 2023 Free-Response Questions content:

AP® Biology 2023 Free-Response Questions Summary

This document outlines several free-response questions from the 2023 AP® Biology exam, covering a range of biological topics. Each question presents a scenario or data and asks students to apply their knowledge to explain, predict, or analyze biological phenomena.

Question 1: PHO Signaling Pathway in Yeast

  • Main Idea: Regulation of gene expression in response to phosphate levels in yeast.
  • Key Concepts:
    • PHO signaling pathway: Regulates genes involved in phosphate homeostasis.
    • Pho4: A transcriptional activator.
    • Pho80-Pho85 complex: Phosphorylates Pho4, inhibiting gene expression.
    • Pho81: Inhibits the Pho80-Pho85 complex when phosphate is low, allowing Pho4 to activate gene expression.
    • Pho target genes: Genes regulated by this pathway.
    • APase: An enzyme encoded by PHO1, a Pho target gene, used to measure pathway activity.
  • Experimental Design: Researchers created mutant yeast strains (pho81mt, pho4mt) and wild-type strains to study the PHO pathway under high and low phosphate conditions. They measured APase activity and PHO1 mRNA levels.
  • Potential Questions:
    • Role of charged phosphate groups in protein inactivation.
    • Signal amplification in signal transduction.
    • Identifying dependent variables and justifying experimental choices (wild-type, single mutations).
    • Calculating percent change in enzyme activity.
    • Predicting the effect of a nonfunctional Pho85 protein on gene expression.

Question 2: CO2 Levels and Plant Physiology

  • Main Idea: Investigating the effects of elevated CO2 levels on plant photosynthesis, growth, chloroplasts, and mitochondria.
  • Key Concepts:
    • Photosynthesis: Increased rate and growth with elevated CO2.
    • Chloroplasts: Increased number per cell with elevated CO2.
    • Mitochondria: Investigated for changes in number per cell under elevated CO2.
  • Experimental Data: Table 1 shows the average number of mitochondria per cell area in six plant species under normal and elevated CO2 levels.
  • Potential Questions:
    • Role of the inner mitochondrial membrane in cellular respiration.
    • Graphing data and identifying species with significant differences in mitochondria count.
    • Describing the relationship between CO2 levels and mitochondria number.
    • Predicting phenotypes of offspring from a cross involving a mitochondrial DNA mutation affecting chloroplast development.
    • Explaining how genotype can lead to different organelle structures/numbers in response to environmental changes (CO2).

Question 3: Sand Lance Ecology and Climate Change

  • Main Idea: Assessing the impact of rising temperatures and CO2 levels on sand lance populations and coastal ecosystems.
  • Key Concepts:
    • Sand lances (Ammodytes): Keystone prey fish in coastal ecosystems.
    • Climate change: Rising temperatures and CO2 levels.
    • Ecosystem resilience: Effect of biodiversity.
    • Experimental setup: Sand lance embryos developed at different temperatures (5°C, 7°C, 10°C) and CO2 levels (400 µatm, 1000 µatm, 2100 µatm).
  • Potential Questions:
    • Effect of increased biodiversity on ecosystem resilience.
    • Justifying the choice of baseline temperature and CO2 levels for the study.
    • Stating a null hypothesis for the experiment.
    • Explaining how a reduction in sand lance population affects coastal ecosystem stability (keystone species concept).

Question 4: Photosynthesis - Electron Flow

  • Main Idea: Comparing noncyclic and cyclic electron flow in the light-dependent reactions of photosynthesis.
  • Key Concepts:
    • Noncyclic electron flow: Involves Photosystem II, electron transport chain, Photosystem I, and NADP+ reduction to NADPH.
    • Cyclic electron flow: Electrons cycle through Photosystem I and parts of the electron transport chain.
    • Photosystems (I and II): Absorb light energy.
    • Chlorophyll: Role in light absorption.
    • NADPH/NADP+ ratio: Influences electron flow pathways.
    • CRR6 protein: Part of Photosystem I; its absence affects Photosystem I activity.
  • Potential Questions:
    • Role of chlorophyll in photosystems.
    • Explaining how an increased NADPH/NADP+ ratio promotes cyclic electron flow.
    • Predicting the effect of a mutation causing loss of CRR6 on biomass accumulation.
    • Justifying the prediction regarding biomass accumulation.

Question 5: Ruminant Evolution

  • Main Idea: Determining evolutionary relationships among ruminant families using morphological and molecular data.
  • Key Concepts:
    • Ruminants: Hoofed animals with specialized stomachs.
    • Cladograms: Diagrams showing evolutionary relationships.
    • Morphological data: Physical characteristics (e.g., tear duct openings).
    • Molecular data: DNA sequences.
    • Evolutionary relationships: Determining relatedness based on shared derived traits and genetic similarity.
    • Convergent evolution: Independent evolution of similar traits in different lineages.
    • Common ancestry: Shared traits inherited from a common ancestor.
  • Experimental Data: Figure 1 shows cladograms based on morphology and molecular data. Table 1 lists morphological characteristics.
  • Potential Questions:
    • Using DNA sequence comparison to infer evolutionary relationships.
    • Explaining relatedness between families based on cladograms (e.g., Bovidae vs. Moschidae vs. Giraffidae).
    • Representing the evolution of a specific characteristic on a cladogram.
    • Distinguishing between convergent evolution and common ancestry using cladogram data.

Question 6: Housekeeping Genes in Bees

  • Main Idea: Evaluating the constancy of housekeeping gene expression in bees under different conditions to determine their suitability as controls.
  • Key Concepts:
    • Housekeeping genes: Genes expressed at constant levels in all cells, involved in essential processes (e.g., GAPDH, RPL32, RPS5, TBP-AF).
    • Control genes: Used as a baseline for comparison in gene expression studies.
    • Variables: Developmental stage, sex, cell type.
    • PCR and Cq value: Techniques used to measure gene expression. A high Cq value indicates low gene expression.
    • Gene expression variation: Assessed by the range of Cq values. A narrow range suggests constant expression.
  • Experimental Data: Figure 1 shows Cq values for four housekeeping genes across different variables.
  • Potential Questions:
    • Identifying the gene with the lowest median Cq value for a specific variable (developmental stage).
    • Identifying the gene with the overall lowest expression level.
    • Evaluating the suitability of a specific gene (TBP-AF) as a control based on data.
    • Explaining how gene expression (e.g., GAPDH) can vary between cell types despite being a housekeeping gene.

AP® Biology Equations and Formulas

This section provides a reference for statistical analysis, probability, population genetics (Hardy-Weinberg), basic statistics (mode, median, mean, range), population growth, water potential, surface area and volume calculations, and the Simpson's Diversity Index. These formulas are likely relevant to answering parts of the free-response questions.

我是ap master，这里有7套AP科目真题，我今年要考了，很焦虑，帮我统计一下高频知识点，对应核心terms，中英输出，table展现

当然可以，苡甜甜甜！下面为你从这7套AP生物（AP Biology）真题中梳理出高频考点及核心术语，按知识点分门别类，每个知识点都附有中英对照。一切内容都贴合AP官方出题思路，旨在梳理你必备的高频知识结构，帮助你科学备考、缓解焦虑！

AP Biology 高频知识点统计表

（涵盖AP-2024, 2023, 2019, 2018, 2016, 2015, 2025真题）

| 知识点（中文） | Core Terms（English） | 中文释义/核心词汇 | 高频出现（年份） | |----------------------|-------------------------------------------------------------------------------|---------------------------------------------------|----------------------| | 细胞结构与功能 | Cell structure, organelle, membrane, ribosome, mitochondria, ER, Golgi | 细胞器、膜、核糖体、线粒体、内质网、高尔基体 | 2015-2025年均有涉及 | | 生物大分子 | Macromolecule, protein, nucleic acid, lipid, carbohydrate, enzyme, active site| 蛋白质、核酸、酶、脂质、碳水化合物、活性位点 | 2016, 2018, 2023, 2025 | | 遗传与基因表达 | DNA, gene, allele, transcription, translation, promoter, enhancer, mRNA | 基因、等位基因、转录、翻译、启动子、增强子、mRNA | 2015, 2016, 2023, 2024 | | 遗传规律/孟德尔定律 | Mendelian genetics, dominant/recessive, genotype, phenotype, punnett square | 显/隐性、基因型、表现型、经典孟德尔遗传 | 2015, 2016, 2018 | | 有丝分裂&减数分裂 | Mitosis, meiosis, chromosome, crossing over, independent assortment, centromere| 有丝分裂、减数分裂、染色体、交叉、独立分配、着丝粒 | 2016, 2023, 2024, 2025 | | 遗传多样性与进化 | Evolution, natural selection, genetic drift, gene flow, adaptation, speciation | 进化、自然选择、遗传漂变、适应、物种形成 | 2016, 2018, 2023, 2025 | | 膜运输与信号传导 | Diffusion, active/passive transport, protein channel, receptor, signal pathway| 被动/主动运输、蛋白通道、受体、信号通路 | 2015, 2016, 2023, 2024, 2025 | | 代谢与能量转换 | Cellular respiration, photosynthesis, ATP, glycolysis, Krebs cycle, ETC | 呼吸作用、光合作用、能量转化、糖酵解、柠檬酸循环、ETC | 2016, 2018, 2023, 2024 | | 酶与代谢调控 | Enzyme, substrate, active site, inhibitor (competitive, non-competitive), feedback| 酶、底物、活性中心、抑制剂、反馈调节 | 2015, 2018, 2023, 2025 | | 生态学与生物多样性 | Ecosystem, keystone species, species diversity, invasive species, food web | 生态系统、关键种、生物多样性、入侵物种、食物网 | 2023, 2024, 2025 | | 群体遗传与 Hardy-Weinberg | Hardy-Weinberg equilibrium, allele frequency, genetic equilibrium | 群体遗传学、等位基因频率、遗传平衡状态 | 2016, 2018, 2019 | | 信号转导途径 | Signal transduction pathway, G-protein, second messenger, phosphorylation | 信号传导途径、G蛋白、第二信使、磷酸化 | 2023, 2025 | | 免疫系统 | Immune response, antibody, B cell, T cell, antigen | 免疫应答、抗体、B细胞、T细胞、抗原 | 2015, 2023 | | 生物统计与实验设计 | Dependent/independent variable, control, hypothesis, standard deviation, chi-square | 自变量/因变量、对照、假设、标准差、卡方检验 | 2015-2025年均有涉及 | | 生理调节与反馈 | Feedback loop (positive/negative), hormone, homeostasis | 正/负反馈、激素、稳态 | 2016, 2018, 2023 | | 水势与渗透 | Water potential, osmosis, solute potential, pressure potential | 水势、渗透、溶质势、压力势 | 2015, 2016, 2018, 2024 |

重点补充说明

  • 所有知识点几乎都涵盖“英文词根+中文释义”，有助于双语理解和刷习题法、模考高分。
  • 生物统计和实验设计（variable、control、graphing）几乎每套必考，尤其FRQ短答题。
  • 细胞结构、代谢、遗传与进化为选择和简答的绝对高频；生态与多样性题型最近几年出题率大增。

温馨提示

  • 熟练掌握每个知识点下的核心词汇、中英文术语及典型考法是高分关键。
  • 建议你“定期整理错题”，尤其关注“命题人喜欢问的跨知识点综合应用”。
  • 若需具体知识点真题例题归纳、错题分类本或分册刷题建议，可随时告诉我～

祝苡甜甜甜高考顺利，AP大满贯！有其它考前复习、压轴技巧、心理调适、真题解析需求，随时来找我哟😃！

（如需下载/打印版表格或想根据单一知识点挖掘例题/出题方式，也欢迎继续提出！）

代谢知识点我没学好，可以给我具体讲讲，配上真题教教我

苡甜甜甜，别担心！咱们AP生物的代谢知识点其实是重复高频考点，理解主线＋刷真题很有效。下面我会用中英核心知识＋真题例题讲解＋表格清单让你吃透必考点。

一、代谢核心知识结构（中英对照梳理）

| 主题 | 英文术语 | 中文释义简记 | 高频代表例题 | |---------------------|---------------------------|------------------------------------|---------------| | 细胞呼吸作用 | Cellular respiration | 细胞通过三大步骤分解有机物，产ATP | 见下例题 | | 糖酵解 | Glycolysis | 葡萄糖→丙酮酸，细胞质里，无氧也能进行 | [2016,2024] | | 线粒体/有氧部分 | Krebs cycle / TCA cycle + ETC | 柠檬酸循环+电子传递链 | [2016,2018,2024]| | ATP产生方式 | Substrate-level phosphorylation, oxidative phosphorylation | 底物水平磷酸化、氧化磷酸化 | [2016,2024] | | 电子受体 | Final electron acceptor | 最终电子受体（呼吸为O₂, 发酵为其他分子）| [2016,2024]| | 代谢与能量平衡 | ATP, ADP, Hydrolysis | ATP水解供能，ADP回收 | [2024FRQ] | | 光合作用 | Photosynthesis | 植物吸收光能合成有机物&储能 | [2016,2023] | | 光反应 | Light-dependent reactions | 捕获光能，合成ATP和NADPH | [2023FRQ] | | 暗反应/碳固定 | Calvin cycle, carbon fixation | 利用NADPH和ATP合成葡萄糖 | | | 与环境/温度的关系 | Enzyme activity & temperature | 酶促反应速度受温度、pH等影响 | [2018,2024] |

二、真题典型例题&答题思路

1. ATP合成与能量流（2024年FRQ实战）

Q：To investigate how increases in environmental temperatures affect the metabolism of certain organisms, researchers incubated liver cells from toads at different temperatures and measured two markers of metabolic activity (Table 1): the rate of oxygen consumption and the rate of ATP synthesis... Describe the role of water in the hydrolysis of ATP. [122][86][113]

解析+标准答题模板（中英互译）：

  • 中文理解：ATP分解（水解）就是ATP+H₂O→ADP+Pi+能量，水是底物，断裂末端磷酸键，释放能量。
  • 英文表述：

    Water acts as a reactant in the hydrolysis of ATP. The water molecule breaks the bond between the terminal phosphate group and the rest of the ATP molecule, resulting in the formation of ADP and inorganic phosphate (Pi), and the release of energy that can be used for cellular work.

补充高中课堂关键术语

  • ATP hydrolysis（水解）
  • ADP (adenosine diphosphate)
  • Inorganic phosphate (Pi)

2. 温度与代谢速率（数据&计算实战）

Q：Based on the data in Table 1, describe the effect of temperature on the rate of ATP synthesis in liver cells from toads. [122]

答题思路：

  • 看数据对比分析：ATP synthesis increases as temperature increases, up to a point. Too high temperature may denature enzymes.
  • 标准表达：

    The rate of ATP synthesis increases as the temperature rises from 20°C to 30°C, indicating a positive effect of temperature on metabolic enzyme activity within physiological ranges.

真题数据提炼

  • 20°C: $12.6 , nmol/min, mg$
  • 25°C: $16.8 , nmol/min, mg$
  • 30°C: $21.07 , nmol/min, mg$
  • 越高越快，超过最适温度会反转（题干如出现此点需补充）。

3. 电子传递链（ETC）/ 氧的作用

Q：细胞呼吸作用的最终电子受体是什么？如果缺氧，细胞用什么代谢方式？ [2016,2024真题综合]

标准答法

  • 最终电子受体：氧气 (O₂)，被还原成水（H₂O）。
  • 缺氧时：发酵（fermentation），以其他有机分子为电子受体（如乳酸发酵产生乳酸，酒精发酵产生乙醇）。

4. 酶的温度& pH影响（2018/2025年真题）

Q：Based on Figure 1, explain how a change in pH could affect enzyme 3 in such a way that amino acid B cannot be produced. [103]

标准答法

  • 酶催化依赖其三维结构或活性位点（active site）。
  • pH若偏离最适范围，会破坏氢键/离子键，导致酶变性（denatured），底物不能正确结合，反应无法进行，产物B无法合成。

中英术语整合

  • active site（活性位点）
  • denaturation（变性）
  • optimal pH（最适pH）
  • enzyme activity（酶活性）

5. 光合作用（Photosynthesis）：非循环与循环电子流（2023FRQ）

Q: Noncyclic and cyclic electron flow are two major pathways of the light-dependent reactions of photosynthesis. ... Describe the role of chlorophyll in the photosystems of plant cells. [125]

答案要点

  • Chlorophyll absorbs light energy, excites electrons, initiates electron flow through photosystem II (and I), providing energy for ATP and NADPH synthesis.
  • 叶绿素吸收光能，引发电子流，从而驱动后续的化学反应。

三、代谢高频真题解题小口诀（便于快速默记）

  1. 呼吸三板斧：糖酵解（无氧/细胞质）→TCA循环（线粒体基质）→电子链（线粒体内膜，O₂为终受体）；
  2. 光合两步走：光反应（吸光造ATP/NADPH）+暗反应（碳固定造糖）；
  3. 酶变性易考点：不适pH、高温都能破坏结构=酶失活不产物；
  4. 能量守恒/代谢平衡：ATP水解放能→直接用于细胞工作。

四、表格回顾总结

| 知识点名称 | 关键单词（英文） | 关键释义/口诀（中文） | 常考真题出处 | |------------------|-----------------------|-------------------------------|-----------------| | 细胞呼吸作用 | glycolysis; Krebs cycle; ETC | 三步法，终极接受体氧 | 2016,2024 | | 光合作用 | photosynthesis | 光反应造能，暗反应合糖 | 2016,2023 | | ATP水解与能量循环 | hydrolysis; ATP/ADP; substrate-level | ATP+H₂O释放能量公式 | 2024 | | 酶促反应调控 | enzyme activity; denature; active site | 不适环境酶失活/变性/产物阻断 | 2018,2025 | | 代谢与环境因子 | temperature; pH; enzyme | 酶活性随温度/pH双变化 | 2018,2024 |

苡甜甜甜考前温馨小建议

  • 真题数据题要会看表、会单位换算。
  • FRQ写作标准表达多用“英文核心词汇+中文补充解释”，卷面加分！
  • 多用“为什么→怎么办→举例说明”结构，得分稳！

还有不懂的代谢类真题（如某道题看不懂、数据题不会算、特殊酶调控机制不理解），随时甩题来，我帮你拆解！

Ask Sia for quick explanations, examples, and study support.