Answer Key & Rubric

AP Biology — Full Practice Exam Answer Key & Scoring Rubric

SECTION I: MULTIPLE CHOICE ANSWER KEY

1. Correct Answer: C

Explanation: A polypeptide (protein) is composed of amino acids, which each contain nitrogen, carbon, oxygen, and hydrogen. The presence of nitrogen without phosphorus or sulfur is characteristic of proteins. Phospholipids and nucleic acids contain phosphorus, and triglycerides contain no nitrogen.

• Distractor A (A): Triglycerides are composed of glycerol and fatty acids, which contain carbon, hydrogen, and oxygen but not nitrogen.
• Distractor B (B): Phospholipids contain phosphorus, which contradicts the observation of no phosphorus.
• Distractor D (D): Nucleic acids contain phosphorus (in the sugar-phosphate backbone), which contradicts the data.

  2. Correct Answer: B

  Explanation: Buffers resist pH changes because they contain a weak acid and its conjugate base. When acid is added, the conjugate base accepts excess H⁺ ions; when base is added, the weak acid donates H⁺ ions to neutralize it.

• Distractor A (A): Buffers do not prevent dissociation of ions; they work by accepting or donating H⁺ ions.
• Distractor C (C): Buffers do not affect solubility of acids and bases; they moderate pH changes through the equilibrium between a weak acid and its conjugate base.
• Distractor D (D): Buffers maintain a relatively stable pH but not necessarily 7.0; different buffer systems maintain different pH ranges.

  3. Correct Answer: C

  Explanation: Dividing the protein's molar mass (44,000 g/mol) by the average amino acid molar mass (110 g/mol) gives approximately 400 amino acids (44,000 ÷ 110 ≈ 400).

• Distractor A (A): 44 would be the result if the amino acid molar mass were 1,000, which is incorrect.
• Distractor B (B): 110 is simply the molar mass of one amino acid, not a calculation of the number of residues.
• Distractor D (D): 4,400 would result from multiplying instead of dividing, which is the wrong operation.

  4. Correct Answer: B

  Explanation: The Biuret test turns violet in the presence of peptide bonds (proteins), indicating proteins are present. The Benedict's test turns blue to orange/red in the presence of reducing sugars; since it remained blue, no reducing sugars are present.

• Distractor A (A): The Benedict's test was negative, so reducing sugars are not present.
• Distractor C (C): There is no indication of lipids in these test results; neither test detects lipids.
• Distractor D (D): The Biuret test was positive, so proteins are present, not absent.

  5. Correct Answer: C

  Explanation: Water's polarity (II) allows it to interact with and dissolve ionic and polar substances, and its ability to form hydrogen bonds (III) helps it surround and stabilize dissolved ions. Cohesion (I) describes water-water attraction, not its solvent properties.

• Distractor A (A): Cohesion refers to water molecules sticking to each other, not to their ability to dissolve substances.
• Distractor B (B): While polarity alone partially explains solvent ability, the ability to form hydrogen bonds with solutes is also critical; listing only II is incomplete.
• Distractor D (D): Cohesion (I) does not directly contribute to water's solvent properties.

  6. Correct Answer: B

  Explanation: The data show the highest reaction rate at pH 7, with much lower rates at pH 4 and pH 10. This indicates the enzyme's three-dimensional active site structure is optimal near neutral pH, and it becomes less effective at extreme pH values.

• Distractor A (A): The enzyme is NOT denatured at pH 7 — it has its highest activity there, meaning it is properly folded and functional.
• Distractor C (C): The experiment measures reaction rate at fixed pH values; differences in rate are due to enzyme conformation, not substrate concentration changes.
• Distractor D (D): The enzyme functions poorly at pH 4 (rate of 12), so it does not require an acidic environment.

  7. Correct Answer: B

  Explanation: An extensive rough ER (for protein synthesis), many Golgi apparatuses (for protein modification and packaging), and numerous lysosomes (for degradation of materials) together indicate a cell specialized for producing and secreting proteins, such as digestive enzymes.

• Distractor A (A): Steroid hormone synthesis occurs primarily in the smooth ER, not the rough ER.
• Distractor C (C): ATP generation through oxidative phosphorylation is associated with many mitochondria, not lysosomes or rough ER.
• Distractor D (D): Lipid storage is associated with lipid droplets and smooth ER, not the combination of organelles described.

  8. Correct Answer: B

  Explanation: The mitochondrial inner membrane houses the electron transport chain complexes and ATP synthase. The proton gradient built across this membrane drives chemiosmosis and ATP production.

• Distractor A (A): Pyruvate oxidation to acetyl-CoA occurs in the mitochondrial matrix, not on the inner membrane.
• Distractor C (C): The inner membrane is NOT permeable to H⁺; its impermeability is essential for maintaining the proton gradient.
• Distractor D (D): The citric acid cycle occurs in the mitochondrial matrix.

  9. Correct Answer: B

  Explanation: The presence of 70S ribosomes and a cell wall containing peptidoglycan are defining characteristics of Bacteria. Archaea also lack a true nucleus and have 70S ribosomes, but their cell walls do not contain peptidoglycan.

• Distractor A (A): Archaea have cell walls without peptidoglycan (they contain pseudopeptidoglycan or other polymers).
• Distractor C (C): Eukaryotic cells have a visible nucleus, 80S ribosomes, and lack peptidoglycan in their cell walls.
• Distractor D (D): The combination of 70S ribosomes and peptidoglycan is diagnostic for Bacteria, so the domain can be determined.

  10. Correct Answer: B

  Explanation: The CFTR protein is translocated into the rough ER after translation. Proper folding and quality control in the rough ER are required for the protein to be transported to the Golgi and then to the plasma membrane. Misfolding in the rough ER would prevent proper trafficking.

• Distractor A (A): The nucleus is the site of transcription, not protein processing or transport.
• Distractor C (C): The smooth ER is involved in lipid synthesis and detoxification, not protein quality control.
• Distractor D (D): Lysosomes are involved in degradation, not the initial processing of secretory proteins.

  11. Correct Answer: C

  Explanation: Osmosis is the diffusion of water across a selectively permeable membrane and does not require membrane proteins. Since water is a small uncharged molecule allowed by this membrane, osmosis would still occur.

• Distractor A (A): Active transport requires membrane proteins (pumps) and energy, and ions are blocked by this membrane.
• Distractor B (B): Facilitated diffusion requires transport proteins, which are not present in this artificial membrane.
• Distractor D (D): Cotransport involves membrane proteins and ions, both of which are blocked by this membrane.

  12. Correct Answer: B

  Explanation: Mitochondria and chloroplasts contain their own circular DNA (similar to bacterial chromosomes) and 70S ribosomes (the same size as bacterial ribosomes, not the 80S ribosomes found in the eukaryotic cytoplasm). This strongly supports their prokaryotic origin.

• Distractor A (A): While both organelles have double membranes, this alone could be explained by other mechanisms and is not as strong as genetic evidence.
• Distractor C (C): Chloroplasts are NOT found in all eukaryotic cells (only in photosynthetic organisms), so this statement is false.
• Distractor D (D): While both organelles produce ATP, chloroplasts produce it during photosynthesis rather than through the same mechanism as mitochondria; independence of ATP production does not directly support endosymbiosis.

  13. Correct Answer: C

  Explanation: A flaccid cell indicates water has left the cell, causing the central vacuole to shrink. The 0.3 M sucrose solution is hypertonic relative to the cell interior, causing water to move out by osmosis.

• Distractor A (A): If the solution were hypotonic, water would enter the cell, causing it to become turgid, not flaccid.
• Distractor B (B): If the solution were isotonic, there would be no net water movement, and the cell would remain in its original state, not become flaccid.
• Distractor D (D): Sucrose is too large to readily cross the plant cell membrane; the flaccidity is due to water loss, not sucrose entry.

  14. Correct Answer: B

  Explanation: Vesicles originating from the trans face of the Golgi that contain fully modified glycoproteins are secretory vesicles. They travel to the plasma membrane and release their contents outside the cell via exocytosis.

• Distractor A (A): Vesicles from the ER to the Golgi are transport vesicles and would carry unmodified or partially modified proteins, not fully modified ones.
• Distractor C (C): Vesicles to lysosomes carry hydrolytic enzymes tagged with mannose-6-phosphate, not general secretory glycoproteins.
• Distractor D (D): Peroxisomal vesicles originate from the ER, not the Golgi, and serve entirely different functions.

  15. Correct Answer: B

  Explanation: When ADP is abundant, ATP synthase can readily use the proton gradient to phosphorylate ADP to ATP. This allows protons to flow back through ATP synthase, dissipating the gradient and enabling the electron transport chain to continue pumping protons. Without ADP, the proton gradient builds up and eventually inhibits further electron transport (respiratory control).

• Distractor A (A): ADP is not a substrate of the ETC; it is a substrate of ATP synthase.
• Distractor C (C): ADP does not directly reduce oxygen; oxygen reduction is carried out by cytochrome c oxidase (Complex IV).
• Distractor D (D): Mitochondria do not switch to fermentation; fermentation occurs in the cytoplasm and is an anaerobic process.

  16. Correct Answer: A

  Explanation: Glycolysis produces a net gain of 2 ATP (4 produced, 2 consumed in the investment phase) and 2 NADH (from the oxidation of glyceraldehyde-3-phosphate) per molecule of glucose.

• Distractor B (B): 4 ATP represents the gross ATP production, not the net after subtracting the 2 ATP used in the investment phase.
• Distractor C (C): 4 NADH would be incorrect; only 2 NADH are produced during glycolysis.
• Distractor D (D): Both values are incorrect; this would represent the gross ATP plus a wrong NADH count.

  17. Correct Answer: B

  Explanation: The data show high oxygen production at 420 nm (blue-violet) and 680 nm (red), but very low production at 550 nm (green). This is consistent with chlorophyll pigments absorbing red and blue-violet light most effectively for photosynthesis.

• Distractor A (A): Green light (550 nm) produces the LEAST oxygen (5 μmol/hr), making it the least effective wavelength.
• Distractor C (C): Oxygen production clearly varies with wavelength, so it is not independent.
• Distractor D (D): The plant absorbs the LEAST light at 550 nm (green light is mostly reflected, which is why plants appear green).

  18. Correct Answer: B

  Explanation: DNP dissipates the proton gradient by making the membrane leaky to protons. Without a proton gradient, the proton-motive force that drives ATP synthase is lost, so ATP production decreases sharply.

• Distractor A (A): The ETC would actually speed up because the lack of a proton gradient removes the resistance to proton pumping; it does not stop.
• Distractor C (C): The citric acid cycle might slow down, not speed up, because NAD⁺ regeneration would be impaired if the ETC is uncoupled from ATP synthesis.
• Distractor D (D): Oxygen consumption would actually INCREASE because the ETC runs faster without the back-pressure of the proton gradient.

  19. Correct Answer: A

  Explanation: In the Calvin cycle, CO₂ is fixed by RuBisCO when it reacts with RuBP. The carbon atoms from CO₂ become incorporated into organic molecules and ultimately into G3P.

• Distractor B (B): RuBP is the carbon acceptor molecule, not the source of new carbon; it is regenerated each cycle.
• Distractor C (C): O₂ is not incorporated into G3P; in fact, O₂ competes with CO₂ at the RuBisCO active site during photorespiration.
• Distractor D (D): Glucose is a product formed from G3P molecules after the Calvin cycle, not the source of carbon.

  20. Correct Answer: C

  Explanation: Under anaerobic conditions, yeast undergo alcoholic fermentation, producing both ethanol and CO₂. Under aerobic conditions, yeast perform cellular respiration, which produces CO₂ but not ethanol.

• Distractor A (A): Ethanol production is higher under anaerobic conditions, not aerobic conditions.
• Distractor B (B): CO₂ is produced under both conditions, but ethanol is produced only anaerobically, so they are not both higher anaerobically.
• Distractor D (D): CO₂ is produced under BOTH conditions, not only under aerobic conditions.

  21. Correct Answer: A

  Explanation: At 100 meters depth, primarily blue-green light penetrates. Phycoerythrin absorbs blue-green light (around 500–570 nm), making it the most advantageous pigment for capturing available light at this depth.

• Distractor B (B): Beta-carotene absorbs blue light (around 450 nm), which does not penetrate to 100 meters effectively.
• Distractor C (C): Chlorophyll a absorbs red (~680 nm) and blue-violet (~420 nm) light, neither of which is abundant at 100 meters.
• Distractor D (D): Xanthophyll absorbs primarily blue light, which is not the dominant wavelength at 100 meters.

  22. Correct Answer: A

  Explanation: At 40°C, the oxygenase activity of RuBisCO increases in C₃ plants, leading to photorespiration, which wastes carbon and reduces net carbon fixation. C₄ plants concentrate CO₂ near RuBisCO, effectively suppressing photorespiration even at high temperatures.

• Distractor B (B): C₄ plants do not use CAM metabolism; CAM is a separate photosynthetic adaptation found in different plants (e.g., cacti).
• Distractor C (C): C₃ plants DO have PEP carboxylase (used in some metabolic pathways), but they lack the spatial separation of carbon fixation that C₄ plants have.
• Distractor D (D): C₄ plants often have FEWER stomata, not more, because their CO₂ concentration mechanism allows them to fix carbon more efficiently per unit of gas exchange.

  23. Correct Answer: B

  Explanation: When a ligand binds to an RTK, the receptor dimerizes (two receptors come together) and autophosphorylates tyrosine residues on its intracellular domain. These phosphorylated tyrosines serve as docking sites for intracellular signaling proteins.

• Distractor A (A): RTKs do not enter the nucleus; they relay signals through intracellular signaling cascades.
• Distractor C (C): G-protein activation and cAMP production are associated with GPCRs, not receptor tyrosine kinases.
• Distractor D (D): Direct ion channel opening is associated with ligand-gated ion channels, not RTKs.

  24. Correct Answer: A

  Explanation: CDK2, when bound to cyclin E, is required for the cell to pass the G₁/S checkpoint (the restriction point). Inhibiting CDK2 would prevent the cell from entering S phase.

• Distractor B (B): The G₂/M checkpoint is regulated primarily by CDK1 (cdc2), not CDK2.
• Distractor C (C): Inhibiting CDK2 would arrest the cell cycle, not trigger apoptosis.
• Distractor D (D): Inhibiting CDK2 would slow or stop division, not accelerate it.

  25. Correct Answer: B

  Explanation: Epinephrine binding activates a GPCR, which activates a G protein that stimulates adenylyl cyclase to produce cAMP. cAMP then activates protein kinase A (PKA), which phosphorylates enzymes that promote glycogen breakdown (e.g., phosphorylase kinase).

• Distractor A (A): cAMP does not directly break down glycogen; it works through an enzyme cascade involving PKA.
• Distractor C (C): cAMP does not inhibit adenylyl cyclase; rather, it is a product of adenylyl cyclase.
• Distractor D (D): cAMP does not bind to the GPCR; it acts as a second messenger in the cytoplasm.

  26. Correct Answer: A

  Explanation: Internal signals such as irreparable DNA damage can activate intrinsic apoptotic pathways that release cytochrome c from mitochondria, which activates caspases — proteases that dismantle the cell in an orderly fashion.

• Distractor B (B): Apoptosis is a highly regulated process, not a random event; it requires specific internal or external signals.
• Distractor C (C): Apoptosis can be triggered by both internal (intrinsic) and external (extrinsic) signals, not only external ones.
• Distractor D (D): Apoptosis can occur at any point in the cell cycle and is not tied to entering S phase.

  27. Correct Answer: B

  Explanation: p53 is a tumor suppressor that normally halts the cell cycle when DNA damage is detected, allowing time for repair or triggering apoptosis. When p53 is nonfunctional, cells with damaged DNA can continue dividing, accumulating mutations that promote cancer.

• Distractor A (A): p53 does NOT promote cell division; it inhibits it when DNA is damaged.
• Distractor C (C): p53 triggers apoptosis in DAMAGED cells, not healthy ones; its loss allows damaged cells to survive.
• Distractor D (D): p53 is not primarily responsible for promoting angiogenesis; it is a cell cycle regulator and tumor suppressor.

  28. Correct Answer: B

  Explanation: Quorum sensing is a process in which bacteria produce and release signaling molecules (autoinducers). As the bacterial population density increases, the concentration of these molecules rises. When a threshold concentration is reached, bacteria detect the signal and coordinately change gene expression, enabling group behaviors like biofilm formation.

• Distractor A (A): Conjugation is a separate process of direct genetic exchange and is not quorum sensing.
• Distractor C (C): Flagella are used for motility, not chemical communication between cells.
• Distractor D (D): Quorum sensing coordinates behavior at HIGH population density, not low density.

  29. Correct Answer: B

  Explanation: The spindle checkpoint (also called the mitotic checkpoint) prevents anaphase from beginning until all kinetochores are properly attached to spindle fibers. If a chromosome is unattached, the checkpoint delays anaphase, preventing unequal chromosome distribution (nondisjunction).

• Distractor A (A): The cell does not immediately undergo apoptosis; the checkpoint delays anaphase to allow correction.
• Distractor C (C): Unattached chromosomes are not degraded; the checkpoint delays the cell cycle until proper attachment occurs.
• Distractor D (D): The cell does not skip mitosis; it pauses at the checkpoint within mitosis.

  30. Correct Answer: B

  Explanation: The offspring include approximately equal numbers of parental types (tall/round and short/wrinkled, ~42 each) and recombinant types (tall/wrinkled and short/round, ~8 each). This 1:1:1:1 ratio (with approximately equal parental and recombinant groups) is consistent with a dihybrid cross (TtRr × ttrr). The tall/round parent must therefore be TtRr.

• Distractor A (A): If the parent were TTRR, all offspring would be tall and round (or tall/wrinkled if the other gene segregated differently), but we see short offspring, so the parent must carry the t allele.
• Distractor C (C): If the parent were TTRr, all offspring would be tall (since TT would always be passed), and no short offspring would appear, but short offspring are observed.
• Distractor D (D): If the parent were TtRR, all offspring would have round seeds (since RR would always be passed), but wrinkled offspring are observed.

  31. Correct Answer: C

  Explanation: The mother is a carrier (XᴺXⁿ) and the father is normal (XᴺY). For a son, the mother passes one of her X chromosomes (Xᴺ or Xⁿ) and the father passes his Y chromosome. There is a 50% chance the son receives Xⁿ (and thus has the disorder XⁿY) and a 50% chance he receives Xᴺ (and is unaffected XᴺY).

• Distractor A (A): There is definitely a chance of the son having the disorder because the mother carries the recessive allele.
• Distractor B (B): 25% would be the probability for a daughter to have the disorder (she would need to receive Xⁿ from both parents, requiring the father to have it).
• Distractor D (D): Not all sons will be affected; only those who inherit the recessive X chromosome from the mother.

  32. Correct Answer: A

  Explanation: With two genes showing recessive epistasis (aa masks B), the dihybrid cross AaBb × AaBb produces a 9:3:4 ratio. Of the 9 A_B_ offspring, those with B_ are purple (9 purple); of the 3 A_bb offspring, they are red (3 red); and the 4 aa__ offspring are white (4 white).

• Distractor B (B): This ratio is incorrect; the 4 white should correspond to the aa genotype, not the red.
• Distractor C (C): A 12:3:1 ratio occurs with dominant epistasis, not recessive epistasis.
• Distractor D (D): A 9:7 ratio occurs with complementary gene interaction, not the described scenario.

  33. Correct Answer: B

  Explanation: During Anaphase I of meiosis, homologous chromosomes (each still consisting of two sister chromatids) separate and move to opposite poles. The fact that sister chromatids remain attached distinguishes this from Anaphase II.

• Distractor A (A): In Metaphase I, homologous chromosome pairs are aligned at the metaphase plate but have not yet separated.
• Distractor C (C): In Anaphase II, sister chromatids (not homologous chromosomes) separate.
• Distractor D (D): Telophase II is the final stage where chromosomes arrive at the poles and nuclear envelopes reform.

  34. Correct Answer: C

  Explanation: The recombinant offspring are Mn (8) and mN (10), totaling 18 out of 100. The recombination frequency is (8 + 10) / (42 + 8 + 10 + 40) × 100 = 18/100 × 100 = 18%.

• Distractor A (A): 9% would be 9 recombinants out of 100, but there are 18 recombinants.
• Distractor B (B): 10% would only account for one recombinant type, not both.
• Distractor D (D): 20% does not match the actual calculation of 18%.

  35. Correct Answer: B

  Explanation: In incomplete dominance, heterozygotes show an intermediate phenotype. Crossing two pink (Rr) individuals gives the genotypic ratio 1 RR (red) : 2 Rr (pink) : 1 rr (white), which translates to the phenotypic ratio 1 red : 2 pink : 1 white.

• Distractor A (A): This ratio (1:1:1) does not result from any standard monohybrid cross.
• Distractor C (C): A 3:1 ratio is expected for complete dominance, not incomplete dominance.
• Distractor D (D): This ratio does not result from crossing two heterozygous individuals with incomplete dominance.

  36. Correct Answer: B

  Explanation: cDNA is synthesized from mature mRNA using reverse transcriptase. Since introns have been removed from mature mRNA during RNA processing, cDNA contains only the expressed exons (coding sequences). Genomic DNA includes both exons and introns.

• Distractor A (A): This is the opposite of the correct relationship; cDNA lacks introns.
• Distractor C (C): cDNA is NOT identical to genomic DNA because it lacks introns, regulatory regions, and other non-coding sequences.
• Distractor D (D): This is the opposite of the truth; cDNA contains exons but not introns.

  37. Correct Answer: A

  Explanation: When tryptophan levels are high, tryptophan acts as a corepressor by binding to the trp repressor protein. The activated repressor then binds to the operator, blocking RNA polymerase and preventing transcription of the trp operon genes.

• Distractor B (B): Tryptophan acts as a corepressor (activating the repressor), not an inducer (inactivating the repressor).
• Distractor C (C): High tryptophan REDUCES, not increases, transcription of the trp operon through repressor activation.
• Distractor D (D): The operon is repressed, not deleted; repression is reversible.

  38. Correct Answer: B

  Explanation: The promoter region is where RNA polymerase and transcription factors bind to initiate transcription. A mutation in the promoter could alter the binding affinity of these proteins, affecting the rate of transcription without necessarily changing the protein's amino acid sequence.

• Distractor A (A): The promoter is not translated into protein, so a promoter mutation does not alter the amino acid sequence.
• Distractor C (C): A shorter mRNA would result from a mutation in the coding region (such as a premature stop codon), not the promoter.
• Distractor D (D): A frameshift requires insertion or deletion of nucleotides in the coding region, not a substitution in the promoter.

  39. Correct Answer: B

  Explanation: An oncogene is a gene that, when overexpressed or constitutively active, promotes uncontrolled cell division and cancer. A 20-fold increase in expression in cancer cells suggests this gene may be acting as an oncogene.

• Distractor A (A): A tumor suppressor gene would be UNDEREXPRESSED (not overexpressed) in cancer cells.
• Distractor C (A): A 20-fold expression difference is significant and suggests a potential role in cancer, so it should not be dismissed.
• Distractor D (D): An apoptosis-promoting gene would likely show DECREASED expression in cancer cells, not increased expression.

  40. Correct Answer: B

  Explanation: The 5' end of pre-mRNA receives a modified guanine nucleotide cap (7-methylguanosine cap), which protects the mRNA from degradation and assists in ribosome binding during translation initiation.

• Distractor A (A): The poly-A tail is added to the 3' end, not the 5' end.
• Distractor C (C): Intron removal by spliceosomes occurs in the middle of the pre-mRNA, not at the 5' end specifically.
• Distractor D (D): The start codon (AUG) is part of the coding sequence, not a modification added during processing.

  41. Correct Answer: B

  Explanation: Histone acetylation adds acetyl groups to lysine residues on histone tails, which reduces the positive charge and weakens the interaction between histones and negatively charged DNA. This loosens chromatin structure (euchromatin), making DNA more accessible to transcription factors and RNA polymerase.

• Distractor A (A): Acetylation LOOSENS, not tightens, chromatin packing, leading to increased, not decreased, expression.
• Distractor C (C): Histone modifications are a major mechanism of gene regulation and definitely affect transcription.
• Distractor D (D): Histone acetylation promotes transcription; RNA interference is a separate mechanism involving small RNAs.

  42. Correct Answer: B

  Explanation: Frameshift mutations result from insertions or deletions of nucleotides that are not in multiples of three, altering the reading frame of the gene from the point of mutation onward.

• Distractor A (A): Substitutions (point mutations) are caused by different mutagens and do not cause frameshifts.
• Distractor C (C): Chromosomal inversions involve large segments of DNA and are not caused by typical frameshift-inducing chemicals.
• Distractor D (D): Trinucleotide repeat expansions involve repeating three-nucleotide sequences and are a specific type of mutation, not a frameshift.

  43. Correct Answer: C

  Explanation: When both glucose and lactose are present, glucose is metabolized preferentially (catabolite repression). High glucose leads to low cAMP levels, which means CAP (catabolite activator protein) cannot bind to the promoter. Even though lactose inactivates the repressor, without CAP binding, the lac operon is only transcribed at very low levels.

• Distractor A (A): Lactose can still enter the cell when glucose is present; the issue is catabolite repression, not lactose uptake.
• Distractor B (B): Glucose does not affect the repressor directly; it affects cAMP/CAP levels.
• Distractor D (D): The promoter is not mutated; the reduced transcription is a normal regulatory response.

  44. Correct Answer: B

  Explanation: miRNAs are small non-coding RNA molecules that bind to complementary sequences on target mRNAs through base pairing. This binding can lead to degradation of the mRNA or block its translation, effectively reducing or silencing gene expression.

• Distractor A (A): miRNAs bind to mRNA, not DNA, and they repress rather than promote gene expression.
• Distractor C (C): miRNAs are not second messengers; they are regulatory RNA molecules.
• Distractor D (D): tRNA, not miRNA, carries amino acids to the ribosome during translation.

  45. Correct Answer: B

  Explanation: The green allele frequency is increasing consistently over generations (0.40 → 0.54), indicating directional selection favoring the green phenotype. In the forested environment, green beetles have a selective advantage due to better camouflage.

• Distractor A (A): Genetic drift causes random, unpredictable changes, but the data show a consistent directional trend.
• Distractor C (C): Disruptive selection would favor both extremes and reduce intermediate phenotypes, which is not what the data show.
• Distractor D (D): Stabilizing selection would favor intermediate phenotypes and reduce both extremes, which is not observed.

  46. Correct Answer: B

  Explanation: To find the frequency of the recessive allele (b), first determine the total number of b alleles. Each bb individual has 2 b alleles, and each Bb individual has 1 b allele. Total b alleles = (2 × 60) + (1 × 120) = 240. Total alleles = 2 × 500 = 1,000. Frequency of b = 240/1,000 = 0.24.

• Distractor A (A): 0.12 is the frequency of the bb genotype (60/500), not the allele frequency.
• Distractor C (C): 0.36 would be the result of an incorrect calculation using only the bb individuals.
• Distractor D (D): 0.48 would result from doubling the bb frequency incorrectly.

  47. Correct Answer: C

  Explanation: Hardy-Weinberg equilibrium requires five conditions: no mutations, random mating, no natural selection, extremely large population size, and no gene flow. Random mating is one of these required conditions.

• Distractor A (A): Hardy-Weinberg equilibrium requires a VERY LARGE population; small populations experience genetic drift.
• Distractor B (B): Gene flow (migration) must be ABSENT, not present, for Hardy-Weinberg equilibrium.
• Distractor D (D): Natural selection must NOT be occurring; selection would cause allele frequencies to change.

  48. Correct Answer: B

  Explanation: The loss of 30 out of 50 individuals (60% of the population) represents a severe bottleneck event. In small populations, genetic drift has a much larger effect, and this bottleneck would cause a significant, random change in allele frequencies.

• Distractor A (A): Even though the deaths were random, the small population size means random events cause large changes in allele frequencies.
• Distractor C (C): The deaths were random, not selective, so this is not an example of natural selection.
• Distractor D (D): The populations are described as isolated, so gene flow from Population A is unlikely.

  49. Correct Answer: B

  Explanation: Antibiotic resistance arises through pre-existing random mutations in bacterial populations. When antibiotics are applied, susceptible bacteria die while resistant ones survive and reproduce, passing resistance alleles to their offspring. This is natural selection acting on existing genetic variation.

• Distractor A (A): Lamarckian inheritance (acquired traits passed to offspring) has been discredited; resistance is genetic, not acquired during exposure.
• Distractor C (C): Mutations are random, not directed toward the antibiotic; the antibiotic selects for pre-existing resistant individuals.
• Distractor D (D): While genetic drift can play a role, the primary mechanism is natural selection, not random drift.

  50. Correct Answer: A

  Explanation: Allopatric speciation occurs when a geographic barrier (the mountain range) physically separates a population into two groups, preventing gene flow and allowing each population to evolve independently until they become reproductively isolated.

• Distractor B (B): Sympatric speciation occurs without geographic separation; here, the mountain range provides clear geographic isolation.
• Distractor C (C): Parapatric speciation occurs in adjacent populations with a narrow hybrid zone, not populations separated by a mountain range.
• Distractor D (D): Polyploidy is a mechanism of sympatric speciation, not geographic separation.

  51. Correct Answer: A

  Explanation: The aa genotype has the highest survival rate (0.90), while the AA genotype has the lowest (0.20). This indicates directional selection favoring the a allele, as homozygous recessive individuals survive best.

• Distractor B (B): The A allele is being selected AGAINST (low survival of AA), not favored.
• Distractor C (C): Disruptive selection would favor both extremes (AA and aa) and disadvantage heterozygotes, but here Aa survives better than AA.
• Distractor D (D): Heterozygote advantage would mean Aa has the highest survival rate, but aa has the highest (0.90 vs. 0.70).

  52. Correct Answer: B

  Explanation: Homologous structures (similar structures derived from a common ancestor) and similar embryonic development patterns provide strong anatomical and developmental evidence for common ancestry. These shared features indicate descent from a common ancestor.

• Distractor A (A): Living in the same geographic region could result from convergent evolution or coincidence, not necessarily common ancestry.
• Distractor C (C): Having the same chromosome number is not strong evidence of common ancestry; chromosome numbers can change through evolutionary mechanisms.
• Distractor D (D): Body size is influenced by environmental factors and natural selection and does not reliably indicate common ancestry.

  53. Correct Answer: B

  Explanation: Using Hardy-Weinberg: frequency of C (p) = 0.7, frequency of c (q) = 1 - 0.7 = 0.3. The frequency of heterozygotes (Cc) = 2pq = 2(0.7)(0.3) = 0.42, or 42%.

• Distractor A (A): 21% is q² = (0.3)² = 0.09 (9%), not 21%; this is incorrect.
• Distractor C (C): 49% is p² = (0.7)², the frequency of the homozygous dominant genotype, not heterozygous.
• Distractor D (D): 70% is the frequency of the C allele (p), not the heterozygous genotype.

  54. Correct Answer: B

  Explanation: This statement reflects the correct understanding of natural selection: genetic variation (including resistance alleles) exists randomly through mutation prior to exposure. The pesticide then selectively kills susceptible individuals, leaving resistant individuals to reproduce and increase the frequency of resistance alleles.

• Distractor A (A): This contradicts the claim; resistance alleles arise through random mutation, not in response to the pesticide.
• Distractor C (C): Individuals do not develop new traits during their lifetime in response to the pesticide; this is a Lamarckian idea.
• Distractor D (D): Genetic variation exists in populations before exposure; not all individuals have the same genetic makeup.

  55. Correct Answer: B

  Explanation: Convergent evolution occurs when unrelated species independently evolve similar traits due to similar environmental pressures. Bat wings (mammalian forelimbs modified for flight) and insect wings (outgrowths of the exoskeleton) evolved independently but serve the same function.

• Distractor A (A): Bat and human forelimbs are homologous structures (shared common ancestor), not convergent evolution.
• Distractor C (C): Similar DNA sequences in closely related fish indicate common ancestry, not convergent evolution.
• Distractor D (D): Whale pelvic bones are vestigial structures homologous to land mammal pelvic bones, not convergent evolution.

  56. Correct Answer: B

  Explanation: The population shows rapid initial growth (years 1–6), then the growth rate slows progressively (years 7–10) as the population approaches and levels off near approximately 1,350 individuals. This is characteristic of logistic growth approaching a carrying capacity.

• Distractor A (A): Exponential growth would show a continuously accelerating increase without leveling off.
• Distractor C (C): Oscillating populations show repeated increases and decreases; this population steadily levels off.
• Distractor D (D): The population is increasing, not declining, throughout the study period.

  57. Correct Answer: B

  Explanation: Energy transfer between trophic levels follows the 10% rule. If primary producers store 20,000 kcal, primary consumers would store approximately 2,000 kcal (10%), and secondary consumers would store approximately 200 kcal (10% of 2,000).

• Distractor A (A): 2,000 kcal represents one trophic level transfer, corresponding to primary consumers, not secondary consumers.
• Distractor C (C): 20 kcal would represent three levels of transfer (tertiary consumers), not secondary consumers.
• Distractor D (D): 2 kcal would represent four levels of transfer, which is not what was asked.

  58. Correct Answer: A

  Explanation: Primary succession occurs on newly formed land or surfaces where no soil or organisms previously existed, such as land exposed after a volcanic eruption. The progressive colonization from lichens to mature communities is the classic pattern of primary succession.

• Distractor B (B): Secondary succession occurs when an existing ecosystem is disturbed but soil remains; here, lava destroyed everything including soil.
• Distractor C (C): Ecological climax refers to the stable, final community, not the process of succession itself.
• Distractor D (D): Biomagnification is the increasing concentration of toxins at higher trophic levels, not a succession process.

  59. Correct Answer: B

  Explanation: Excess nitrogen and phosphorus cause eutrophication. Algal populations grow rapidly (algal bloom), then die and decompose. Decomposition by bacteria consumes dissolved oxygen, leading to hypoxic conditions that can cause fish die-offs.

• Distractor A (A): Oligotrophic lakes have LOW nutrient levels and productivity; adding nutrients makes a lake eutrophic, not oligotrophic.
• Distractor C (C): High nutrient levels have profound effects on lake ecosystems, particularly through eutrophication.
• Distractor D (D): Fish populations would initially increase but then DECREASE as dissolved oxygen is depleted.

  60. Correct Answer: C

  Explanation: When Species Q is present, Species P is restricted to the upper zone. When Species Q is removed, Species P expands into the lower zone. This indicates that Species Q was competitively excluding Species P from the lower zone, which is an example of competitive exclusion (or at least interspecific competition limiting a species' fundamental niche).

• Distractor A (A): Mutualism involves both species benefiting; here, Species Q limits Species P's range.
• Distractor B (B): Commensalism involves one species benefiting and the other being unaffected; here, the interaction is clearly competitive.
• Distractor D (D): Parasitism involves one organism harming another while benefiting; the barnacles are competing, not parasitizing each other.

SECTION II: FREE RESPONSE — SCORING RUBRICS

LONG FRQ 1: Natural Selection / Population Genetics

Total Points: 10

Question text repeated for reference:

A researcher is studying a population of wild mice that live in a field consisting of light-colored sandy soil and patches of dark volcanic rock. The fur color of the mice is controlled by a single gene with two alleles: the D allele (dark fur, dominant) and the d allele (light fur, recessive). Homozygous dominant (DD) and heterozygous (Dd) mice have dark fur, while homozygous recessive (dd) mice have light fur.

The researcher conducts a study over several years and observes the following:

• In a sandy area, light-colored mice have a higher survival rate because they are better camouflaged from hawks (visual predators).
• In a rocky area, dark-colored mice have a higher survival rate.
• Mice occasionally migrate between the sandy and rocky areas.

  The researcher estimates the following data at the start of the study in the sandy area:

  | Genotype | Number of Individuals | |----------|-----------------------| | DD | 40 | | Dd | 160 | | dd | 300 |

  (a) Calculate the observed frequency of the D allele and the d allele in the sandy area population. Show your work. (2 points)

  (b) The researcher claims that the mouse population in the sandy area is NOT in Hardy-Weinberg equilibrium. Using the data provided, perform a chi-square test to evaluate this claim. Use the expected genotype frequencies under Hardy-Weinberg equilibrium based on your calculated allele frequencies from part (a). (Chi-square critical value at p = 0.05 with df = 1 is 3.84.) (3 points)

  (c) Predict how the frequency of the d allele in the sandy area would change over the next ten generations if the environment remains stable and migration between areas ceases. Justify your prediction using the principles of natural selection. (2 points)

  (d) Describe how the migration of mice between the sandy and rocky areas would affect genetic diversity in each subpopulation. Use the term gene flow in your response. (3 points)

SCORING RUBRIC

Part (a) — 2 points total

Point 1 (1 point): Correct calculation of total alleles.

• Earning the point: The response must show that total alleles = 2 × (40 + 160 + 300) = 1,000.

  Point 2 (1 point): Correct allele frequency calculations.

• Earning the point: The response must calculate D = (2×40 + 160)/1,000 = 240/1,000 = 0.24, and d = (2×300 + 160)/1,000 = 760/1,000 = 0.76.

  Part (b) — 3 points total

  Point 3 (1 point): Correct calculation of expected genotype numbers under HWE.

• Earning the point: The response calculates p² (DD) = (0.24)² = 0.0576, so expected DD = 57.6; 2pq (Dd) = 2(0.24)(0.76) = 0.3648, so expected Dd = 364.8; q² (dd) = (0.76)² = 0.5776, so expected dd = 577.6.

  Point 4 (1 point): Correct chi-square calculation.

• Earning the point: χ² = Σ (observed − expected)²/expected = (40−57.6)²/57.6 + (160−364.8)²/364.8 + (300−577.6)²/577.6. The student needs to show the setup and obtain a chi-square value much greater than 3.84 (the actual χ² ≈ 181.6). Accept any reasonable calculation attempt that shows the formula and a value > 3.84.

  Point 5 (1 point): Correct conclusion with justification.

• Earning the point: The response must state that since the calculated χ² value (≈181.6) far exceeds the critical value (3.84), the null hypothesis is rejected, and the population is NOT in Hardy-Weinberg equilibrium.

  Common mistakes that lose points: Using the wrong total for allele calculations; forgetting to multiply DD and dd counts by 2; failing to show work; making an arithmetic error that leads to an incorrect chi-square value below 3.84.

  Part (c) — 2 points total

  Point 6 (1 point): Correct prediction.

• Earning the point: The response predicts that the frequency of the d allele will INCREASE over the next ten generations.

  Point 7 (1 point): Correct justification using natural selection.

• Earning the point: The response explains that in the sandy area, light-colored mice (dd) have higher survival due to camouflage from predators. Therefore, the d allele confers a selective advantage, and natural selection will increase its frequency in the population over time.

  Common mistakes that lose points: Predicting the D allele will increase; failing to connect the camouflage advantage to differential survival and reproduction.

  Part (d) — 3 points total

  Point 8 (1 point): Definition/application of gene flow.

• Earning the point: The response must use the term gene flow and describe it as the movement of alleles between populations through migration.

  Point 9 (1 point): Effect on genetic diversity.

• Earning the point: The response must state that gene flow INCREASES genetic diversity in each subpopulation by introducing alleles that may have been rare or absent.

  Point 10 (1 point): Specific connection to the scenario.

• Earning the point: The response explains that migration from the rocky area (where D is favored) to the sandy area introduces more D alleles, and migration from the sandy area (where d is favored) to the rocky area introduces more d alleles, counteracting the effects of natural selection in each area and maintaining higher genetic variation.

  Common mistakes that lose points: Not using the term "gene flow"; stating that gene flow decreases diversity; failing to connect the specific alleles (D and d) to the movement between areas.

MODEL RESPONSE (Full Credit — 10/10 points)

(a) Total number of individuals = 40 + 160 + 300 = 500. Total alleles = 500 × 2 = 1,000.

D allele count = (2 × 40) + 160 = 240. Frequency of D (p) = 240/1,000 = 0.24. d allele count = (2 × 300) + 160 = 760. Frequency of d (q) = 760/1,000 = 0.76.

(b) Expected frequencies under HWE: p² = (0.24)² = 0.0576 (DD); 2pq = 2(0.24)(0.76) = 0.3648 (Dd); q² = (0.76)² = 0.5776 (dd).

Expected numbers: DD = 0.0576 × 500 = 28.8; Dd = 0.3648 × 500 = 182.4; dd = 0.5776 × 500 = 288.8.

Chi-square: χ² = (40 − 28.8)²/28.8 + (160 − 182.4)²/182.4 + (300 − 288.8)²/288.8 = 4.36 + 2.75 + 0.43 = 7.54.

Since χ² = 7.54 > 3.84 (critical value), we reject the null hypothesis. The population is NOT in Hardy-Weinberg equilibrium.

(c) The frequency of the d allele will increase over the next ten generations. In the sandy area, light-colored (dd) mice have better camouflage from hawks, giving them a survival advantage. Because the d allele confers higher fitness in this environment, natural selection will favor dd individuals, leading them to survive and reproduce more successfully than DD or Dd mice. Over generations, the proportion of the d allele in the gene pool will increase.

(d) Migration between the sandy and rocky areas results in gene flow, which is the transfer of alleles between populations. Gene flow would increase genetic diversity in each subpopulation by introducing alleles that are rare or absent. For example, mice migrating from the rocky area (where D is favored and more common) would bring D alleles into the sandy area, and mice migrating from the sandy area (where d is favored) would bring d alleles into the rocky area. This gene flow counteracts natural selection in each area and maintains greater genetic variation in both subpopulations than would exist without migration.

LONG FRQ 2: Cellular Energetics / Photosynthesis & Respiration

Total Points: 10

Question text repeated for reference:

A student designs an experiment to investigate the effect of light intensity on the rate of photosynthesis in an aquatic plant (Elodea). The student places several identical Elodea specimens in separate test tubes containing a bicarbonate solution (which provides dissolved CO₂). Each test tube is placed at a different distance from a fixed light source. The student measures the number of oxygen bubbles produced per minute as an indicator of photosynthesis rate.

The following data were collected:

(a) Based on the data, describe the relationship between light intensity and the rate of photosynthesis. Identify the dependent and independent variables in this experiment. (2 points)

(b) The light-dependent reactions of photosynthesis produce both ATP and NADPH. Describe the specific role of each of these molecules in the Calvin cycle. (3 points)

(c) The student hypothesizes that at very high light intensities, the rate of photosynthesis will plateau. Identify a factor, other than light intensity, that could limit the rate of photosynthesis at this point, and explain how it would limit the rate. (2 points)

(d) In a separate experiment, the student adds a chemical inhibitor that blocks electron transport between Photosystem II and Photosystem I. Predict the effect of this inhibitor on: oxygen production, NADPH production, and the Calvin cycle. Justify each prediction. (3 points)

SCORING RUBRIC

Part (a) — 2 points total

Point 1 (1 point): Description of the relationship.

• Earning the point: The response must describe a positive/direct relationship — as light intensity increases, the rate of photosynthesis (oxygen bubble production) increases.

  Point 2 (1 point): Correct identification of variables.

• Earning the point: The response must identify the independent variable as light intensity (or distance from the light source) and the dependent variable as the rate of photosynthesis (measured as O₂ bubbles per minute).

  Common mistakes that lose points: Reversing dependent and independent variables; describing an inverse relationship; failing to identify both variables.

  Part (b) — 3 points total

  Point 3 (1.5 points): Role of ATP in the Calvin cycle.

• Earning the point: The response must state that ATP provides the energy needed to power the reduction of 3-PGA to G3P by phosphorylating 3-PGA and other intermediates.

  Point 4 (1.5 points): Role of NADPH in the Calvin cycle.

• Earning the point: The response must state that NADPH provides the reducing power (electrons/hydrogen) to reduce 3-PGA (or 1,3-BPG) to G3P.

  Common mistakes that lose points: Saying ATP and NADPH "make" G3P without specifying their distinct roles; confusing energy provision (ATP) with electron donation (NADPH); saying they are used in the light-dependent reactions.

  Part (c) — 2 points total

  Point 5 (1 point): Identification of a limiting factor.

• Earning the point: The response identifies a valid factor such as CO₂ concentration (availability), temperature, or water availability.

  Point 6 (1 point): Explanation of how the factor limits the rate.

• Earning the point: For CO₂: the response explains that CO₂ is a reactant (substrate) in the Calvin cycle, and if CO₂ levels are low, RuBisCO cannot fix carbon, limiting G3P production. For temperature: the response explains that enzyme activity (e.g., RuBisCO) has an optimal temperature, and at non-optimal temperatures, enzyme efficiency decreases. For water: water is a reactant in the light-dependent reactions, and insufficient water limits photolysis.

  Common mistakes that lose points: Naming a factor without explaining how it limits the rate; naming light intensity (which was excluded in the question prompt).

  Part (d) — 3 points total

  Point 7 (1 point): Prediction for oxygen production.

• Earning the point: The response predicts that oxygen production will decrease or stop. Justification: oxygen is produced by the splitting of water at Photosystem II, and blocking electron transport will cause a backup that inhibits further water splitting.

  Point 8 (1 point): Prediction for NADPH production.

• Earning the point: The response predicts that NADPH production will decrease or stop. Justification: electrons cannot reach Photosystem I and then NADP⁺ reductase, so NADP⁺ cannot be reduced to NADPH.

  Point 9 (1 point): Prediction for the Calvin cycle.

• Earning the point: The response predicts that the Calvin cycle will slow down or stop. Justification: the Calvin cycle requires ATP and NADPH from the light-dependent reactions, and the inhibitor reduces or eliminates their production.

  Common mistakes that lose points: Correctly predicting the outcome without justification; stating oxygen production would increase; failing to connect the light-dependent and light-independent reactions.

MODEL RESPONSE (Full Credit — 10/10 points)

(a) As light intensity increases, the rate of photosynthesis increases, as shown by the increasing number of oxygen bubbles produced per minute. The independent variable is light intensity (measured in arbitrary units, manipulated by changing distance from the light source), and the dependent variable is the rate of photosynthesis (measured as the number of O₂ bubbles produced per minute).

(b) In the Calvin cycle, ATP provides the energy required to phosphorylate 3-PGA to 1,3-bisphosphoglycerate, which is an energy-requiring step. NADPH provides the reducing power (high-energy electrons and hydrogen ions) needed to reduce 1,3-bisphosphoglycerate to G3P. Together, ATP and NADPH convert 3-PGA into the energy-rich sugar G3P.

(c) One factor that could limit the rate of photosynthesis at high light intensity is CO₂ concentration. CO₂ is a substrate for RuBisCO in the Calvin cycle; it is the source of carbon atoms incorporated into G3P. If CO₂ availability is limited, RuBisCO cannot fix carbon fast enough to keep up with the energy (ATP and NADPH) supplied by the light-dependent reactions, and the overall rate of photosynthesis will plateau.

(d)

• Oxygen production: Oxygen production will decrease or stop. Oxygen is produced when water is split (photolysis) at Photosystem II. If electron transport between PSII and PSI is blocked, electrons from water cannot be passed along the chain, causing a backup that prevents further water splitting.
• NADPH production: NADPH production will decrease or stop. Electrons from PSII cannot reach PSI and then NADP⁺ reductase, so NADP⁺ cannot be reduced to NADPH.
• Calvin cycle: The Calvin cycle will slow or stop. The Calvin cycle depends on ATP and NADPH produced by the light-dependent reactions. Without these products, the cycle cannot reduce 3-PGA to G3P.

SHORT FRQ 3: Gene Expression

Total Points: 4

Question text repeated for reference:

A mutation in the DNA of a gene changes the codon GAA to GUA. The original codon (GAA) codes for the amino acid glutamic acid, and the mutant codon (GUA) codes for valine.

(a) Identify the type of mutation that occurred. (1 point)

(b) Explain how this mutation could affect the structure and function of the resulting protein. (2 points)

(c) Describe a scenario in which this type of mutation would have no effect on the phenotype of the organism. (1 point)

SCORING RUBRIC

Part (a) — 1 point total

Point 1 (1 point): Correct identification.

• Earning the point: The response identifies this as a missense mutation (a substitution mutation that changes one amino acid to another). Accepting "point mutation" or "substitution mutation" is also acceptable if the student specifies it changes the amino acid.

  Common mistakes that lose points: Calling it a frameshift, nonsense, or silent mutation.

  Part (b) — 2 points total

  Point 2 (1 point): Effect on protein structure.

• Earning the point: The response explains that substituting glutamic acid (a polar, negatively charged amino acid) for valine (a nonpolar, hydrophobic amino acid) could alter the protein's secondary, tertiary, or quaternary structure by disrupting hydrogen bonding, ionic interactions, or hydrophobic interactions at that position.

  Point 3 (1 point): Effect on protein function.

• Earning the point: The response explains that the change in structure could affect the protein's function by altering the active site shape, reducing substrate binding, or preventing proper folding.

  Common mistakes that lose points: Discussing only structure or only function without connecting them; stating the protein will definitely be nonfunctional (it might still work); not mentioning that the chemical properties of the amino acids differ.

  Part (c) — 1 point total

  Point 4 (1 point): Scenario with no phenotypic effect.

• Earning the point: The response provides a valid scenario such as: (1) the mutation occurs in a non-coding region of the DNA; (2) the mutation occurs in a gene that is not expressed in the organism's tissues; (3) the amino acid substitution does not significantly alter the protein's structure or function (e.g., if it occurs in a non-critical region of the protein); or (4) the organism has a backup copy of the gene that compensates.

  Common mistakes that lose points: Stating that the mutation never affects phenotype; providing an implausible scenario.

MODEL RESPONSE (Full Credit — 4/4 points)

(a) This is a missense mutation, which is a type of point mutation (substitution) in which a single nucleotide change results in a codon that codes for a different amino acid.

(b) The substitution replaces glutamic acid (a polar, negatively charged amino acid) with valine (a nonpolar, hydrophobic amino acid). This change in chemical properties could disrupt the protein's three-dimensional structure by interfering with hydrogen bonds, ionic interactions, or proper folding at that location. If the protein's structure is altered, it may no longer function correctly — for example, the active site of an enzyme might change shape and lose the ability to bind its substrate.

(c) This mutation would have no effect on the phenotype if the amino acid change occurs in a region of the protein that is not critical for its function, such as a flexible region far from the active site. In this case, the protein could still fold correctly and perform its normal function despite the single amino acid substitution.

SHORT FRQ 4: Cell Communication

Total Points: 4

Question text repeated for reference:

A researcher discovers a new hormone, Hormone X, that is secreted by the pancreas in response to elevated blood glucose levels. Hormone X binds to a G-protein-coupled receptor (GPCR) on the surface of muscle cells. Upon binding, a signaling cascade is initiated that results in the translocation of glucose transporter proteins (GLUT4) to the plasma membrane of the muscle cells.

(a) Describe the role of the G-protein in the signal transduction pathway initiated by Hormone X. (1 point)

(b) Explain how the translocation of GLUT4 transporters to the plasma membrane affects blood glucose levels. (2 points)

(c) If a mutation caused the GPCR on muscle cells to be nonfunctional, predict the effect on blood glucose homeostasis and justify your prediction. (1 point)

SCORING RUBRIC

Part (a) — 1 point total

Point 1 (1 point): Description of G-protein role.

• Earning the point: The response must describe that when Hormone X binds the GPCR, the G-protein is activated (GTP replaces GDP on the alpha subunit), and the activated G-protein then activates an effector protein (such as adenylyl cyclase or phospholipase C), which generates a second messenger to propagate the signal.

  Common mistakes that lose points: Saying the G-protein directly transports glucose; describing the G-protein as a receptor; failing to mention activation of a downstream effector.

  Part (b) — 2 points total

  Point 2 (1 point): Mechanism of GLUT4 action.

• Earning the point: The response explains that when GLUT4 transporters are inserted into the plasma membrane, they facilitate the transport of glucose from the blood into the muscle cells by facilitated diffusion.

  Point 3 (1 point): Effect on blood glucose levels.

• Earning the point: The response states that as more glucose enters the muscle cells, blood glucose levels decrease (are lowered), helping to restore blood glucose to normal homeostatic levels.

  Common mistakes that lose points: Saying GLUT4 uses active transport (it uses facilitated diffusion); stating blood glucose increases; failing to connect GLUT4 to glucose uptake.

  Part (c) — 1 point total

  Point 4 (1 point): Prediction with justification.

• Earning the point: The response predicts that blood glucose levels would remain elevated (fail to return to normal) because the nonfunctional GPCR cannot initiate the signaling cascade that leads to GLUT4 translocation, so muscle cells cannot take up glucose efficiently.

  Common mistakes that lose points: Predicting lower blood glucose; failing to justify the prediction.

MODEL RESPONSE (Full Credit — 4/4 points)

(a) When Hormone X binds to the GPCR on the muscle cell surface, the receptor changes shape and activates the associated G-protein by causing GTP to replace GDP on the G-protein's alpha subunit. The activated G-protein then activates a downstream effector protein, which produces second messengers that continue the signal transduction cascade.

(b) The translocation of GLUT4 transporters to the plasma membrane allows glucose to move from the blood into the muscle cells through facilitated diffusion. As more glucose is transported into the muscle cells for storage (as glycogen) or use (in cellular respiration), the concentration of glucose in the blood decreases, helping to restore blood glucose to normal homeostatic levels.

(c) If the GPCR were nonfunctional, blood glucose homeostasis would be disrupted. Blood glucose levels would remain abnormally high after a meal because the signal from Hormone X could not be received by the muscle cells. Without the signal cascade, GLUT4 transporters would not translocate to the plasma membrane, and glucose uptake by muscle cells would be impaired.

SHORT FRQ 5: Ecology / Population Dynamics

Total Points: 4

Question text repeated for reference:

A population of deer is introduced to an island that has never previously had deer. The island has abundant vegetation and no natural predators of deer. The initial population size is 20 individuals.

(a) Describe the expected growth pattern of the deer population during the first several years. (2 points)

(b) Identify two density-dependent factors that could eventually limit the growth of the deer population on the island, and explain how each factor would limit population growth. (2 points)

SCORING RUBRIC

Part (a) — 2 points total

Point 1 (1 point): Identification of the growth pattern.

• Earning the point: The response identifies the growth pattern as exponential (or J-shaped) growth during the initial years.

  Point 2 (1 point): Justification of the growth pattern.

• Earning the point: The response explains that because resources (vegetation) are abundant and there are no predators, the per capita growth rate remains high and the population increases at an accelerating rate.

  Common mistakes that lose points: Describing logistic growth without explaining the initial exponential phase; saying growth will be linear; not justifying the answer.

  Part (b) — 2 points total

  Point 3 (1 point): First density-dependent factor with explanation.

• Earning the point: The response identifies one valid factor (e.g., food/vegetation availability, disease, competition for resources, waste accumulation) AND explains how it limits growth (e.g., as the population grows, vegetation is depleted faster than it can regrow, leading to starvation and reduced reproduction).

  Point 4 (1 point): Second density-dependent factor with explanation.

• Earning the point: The response identifies a second valid factor AND explains its mechanism (e.g., at higher population density, disease spreads more easily through contact between individuals, increasing mortality).

  Common mistakes that lose points: Naming density-independent factors (e.g., weather, natural disasters); naming a factor without explaining how it limits growth; naming two factors that are essentially the same (e.g., food shortage and starvation).

MODEL RESPONSE (Full Credit — 4/4 points)

(a) During the first several years, the deer population will exhibit exponential (J-shaped) growth. Because the island has abundant vegetation and no predators, there are no significant limiting factors initially. The population will grow at an accelerating rate as each generation produces more offspring than the last, with birth rates far exceeding death rates.

(b)

• Food availability (vegetation): As the deer population increases, more vegetation will be consumed. Eventually, the rate of consumption will exceed the rate of plant regrowth, leading to food shortages. Deer that cannot find enough food will have lower survival and reproductive rates, limiting population growth.
• Disease transmission: At higher population densities, deer are in closer contact with one another, making it easier for pathogens and parasites to spread. Increased disease incidence would raise mortality rates and reduce population growth.

SHORT FRQ 6: Heredity / Genetics

Total Points: 4

Question text repeated for reference:

In fruit flies (Drosophila melanogaster), the gene for body color is located on the X chromosome. The allele for gray body (Xᴮ) is dominant over the allele for ebony body (Xᴙ). A female fruit fly with a gray body is crossed with a male fruit fly with an ebony body. All of the male offspring have gray bodies, and approximately half of the female offspring have gray bodies while half have ebony bodies.

(a) Determine the genotypes of both parents. Show your reasoning. (2 points)

(b) If one of the gray-bodied female offspring from this cross is mated with an ebony-bodied male, what is the expected phenotypic ratio among their offspring? Show your work. (2 points)

SCORING RUBRIC

Part (a) — 2 points total

Point 1 (1 point): Correct identification of the female parent's genotype.

• Earning the point: The response must determine that the female parent is XᴮXᴙ (heterozygous). Reasoning: she has a gray body (so she must have at least one Xᴮ), and she produces both gray and ebony daughters (so she must carry Xᴙ and pass it to some daughters).

  Point 2 (1 point): Correct identification of the male parent's genotype.

• Earning the point: The response must determine that the male parent is XᴙY. Reasoning: he has an ebony body, and since the gene is X-linked, an ebony male must be XᴙY. This is confirmed because all male offspring receive Xᴮ from the mother and Y from the father, giving them all gray bodies.

  Common mistakes that lose points: Writing the genotypes incorrectly (e.g., using autosomal notation); not showing reasoning; confusing which parent contributes which chromosome to male vs. female offspring.

  Part (b) — 2 points total

  Point 3 (1 point): Correct setup of the cross.

• Earning the point: The response must set up the cross as XᴮXᴙ (gray female offspring) × XᴙY (ebony male) and show a Punnett square or list the expected offspring genotypes: XᴮXᴙ, XᴙXᴙ, XᴮY, XᴙY.

  Point 4 (1 point): Correct phenotypic ratio.

• Earning the point: The response must state the expected phenotypic ratio as 1 gray female : 1 ebony female : 1 gray male : 1 ebony male (or equivalently, 1:1:1:1 among all offspring, or 1 gray : 1 ebony within each sex).

  Common mistakes that lose points: Using the wrong female genotype (e.g., XᴮXᴮ instead of XᴮXᴙ); incorrect phenotypic ratio; not showing work.

MODEL RESPONSE (Full Credit — 4/4 points)

(a)

• Female parent: XᴮXᴙ. The female has a gray body, so she must carry at least one Xᴮ allele. Since half of her female offspring have ebony bodies, she must also carry the Xᴙ allele (if she were XᴮXᴮ, all daughters would receive Xᴮ and have gray bodies).
• Male parent: XᴙY. The male has an ebony body, and since the gene is X-linked, he must be XᴙY. All male offspring receive their X chromosome from the mother (Xᴮ) and their Y from the father, giving them all XᴮY (gray body), which is consistent with the data.

  (b) One of the gray-bodied female offspring has the genotype XᴮXᴙ (she received Xᴮ from the mother and Xᴙ from the father). She is crossed with an ebony-bodied male (XᴙY).

  Punnett square:

  | | Xᴮ (from female) | Xᴙ (from female) | |-------|-------------------|-------------------| | Xᴙ (from male) | XᴮXᴙ (gray female) | XᴙXᴙ (ebony female) | | Y (from male) | XᴮY (gray male) | XᴙY (ebony male) |

  The expected phenotypic ratio among the offspring is: 1 gray female : 1 ebony female : 1 gray male : 1 ebony male (1:1:1:1).

  End of Answer Key