An original Thinka practice paper modelled on the structure and difficulty of the May 2025 AP AP Biology paper. Not affiliated with or reproduced from AP.
部分 II: Free-Response
Section II contains 6 free-response questions. Questions 1 and 2 are long free-response questions (9 points each; recommended 25 minutes each). Questions 3 through 6 are short free-response questions (4 points each; recommended 10 minutes each). Answers must be written in complete paragraph form.
6 題目 · 34 分
題目 1 · Long Free-Response
9 分
Proteins destined for the peroxisomal matrix are synthesized on free cytosolic ribosomes and must be translocated across the peroxisomal membrane. Most peroxisomal matrix proteins possess specific peroxisomal targeting signals (PTS). Type 1 targeting signals (PTS1) are recognized by the receptor protein PEX5, whereas type 2 targeting signals (PTS2) are recognized by the receptor protein PEX7. Upon binding their respective cargo proteins, these receptor complexes facilitate docking and translocation into the peroxisomal lumen.
A. Describe the role of a targeting signal peptide sequence within a polypeptide in determining its final cellular destination.
To investigate the distinct import pathways of peroxisomal matrix proteins, researchers utilized short hairpin RNAs (shRNAs) to selectively reduce the expression of either the PEX5 gene or the PEX7 gene in cultured human cells. Cells were treated with a non-targeting control shRNA, PEX5 shRNA, or PEX7 shRNA. The researchers quantified the relative protein levels of PEX5 and PEX7 in each group compared with the control (Figure 1).
Figure 1. Relative cellular levels of PEX5 and PEX7 proteins in cells treated with control shRNA, PEX5 shRNA, or PEX7 shRNA. Error bars represent \(\pm 2\text{SE}_{\bar{x}}\).
B. i. Identify the dependent variable in the experiment shown in Figure 1. ii. Justify why the researchers measured the relative levels of both PEX5 and PEX7 proteins in cells treated only with PEX5 shRNA. iii. Based on the data in Figure 1, describe the effect of PEX5 shRNA on the expression of PEX7 protein.
The researchers then evaluated the translocation efficiency of three candidate peroxisomal enzymes (Enzyme 1, Enzyme 2, and Enzyme 3). They isolated peroxisomes from each treatment group and measured the amount of each enzyme imported into the peroxisomal matrix relative to the control group (Figure 2).
Figure 2. Relative peroxisomal import of Enzyme 1, Enzyme 2, and Enzyme 3 in cells treated with control shRNA, PEX5 shRNA, or PEX7 shRNA. Error bars represent \(\pm 2\text{SE}_{\bar{x}}\).
C. i. Identify the independent variable manipulated in the second experiment (data shown in Figure 2). ii. Based on Figure 2, identify the enzyme(s) that showed a statistically significant decrease in peroxisomal import in cells treated with PEX5 shRNA compared to the control. iii. The coding region of the mRNA encoding Enzyme 1 contains 816 nucleotides, while the coding region of the mRNA encoding Enzyme 2 contains 1,275 nucleotides. Assuming all nucleotides in both coding regions specify amino acids, calculate the difference in the number of amino acids between Enzyme 1 and Enzyme 2.
D. i. Researchers claim that Enzyme 2 is transported into peroxisomes exclusively via the PTS2 pathway. Using the data in Figure 2, support the researchers' claim. ii. A mutation in the gene encoding Enzyme 1 alters the amino acid sequence of its PTS1 recognition domain, changing a key polar amino acid to a hydrophobic amino acid and disrupting its interaction with PEX5. Justify the researchers' claim that this mutation will lead to an accumulation of Enzyme 1 in the cytosol.
查看答案詳解收起答案詳解
解題
Part A (1 point) - Targeting signal sequences (signal peptides) are specific amino acid motifs that bind to specific cytosolic receptor proteins or translocation machinery, directing the nascent or folded polypeptide to its intended cellular destination (such as the endoplasmic reticulum, peroxisome, or mitochondrion).
Part B (3 points) - B.i (1 point): The dependent variable in Figure 1 is the relative cellular amount (or content/concentration) of PEX5 and PEX7 proteins (measured as a percentage of control). - B.ii (1 point): Justification: Measuring both PEX5 and PEX7 confirms the knockdown specificity of the PEX5 shRNA, verifying that the PEX5 shRNA specifically reduces PEX5 expression without causing off-target downregulation of the alternative import receptor PEX7. - B.iii (1 point): Based on Figure 1, treatment with PEX5 shRNA has no significant effect on the relative level of PEX7 protein (the level remains at approximately 102%, and error bars overlap with the 100% control).
Part C (3 points) - C.i (1 point): The independent variable is the type of shRNA added to the cells (PEX5 shRNA vs. PEX7 shRNA vs. control shRNA), or the specific peroxisomal enzyme being assayed (Enzyme 1, 2, or 3). - C.ii (1 point): Enzyme 1 and Enzyme 3 (both show non-overlapping error bars with the control, with import dropping from 100% down to ~22% and ~20% respectively). - C.iii (1 point): - Number of amino acids in Enzyme 1: \(816 / 3 = 272\) amino acids. - Number of amino acids in Enzyme 2: \(1,275 / 3 = 425\) amino acids. - Difference: \(425 - 272 = 153\) amino acids (or \((1275 - 816) / 3 = 459 / 3 = 153\)).
Part D (2 points) - D.i (1 point): In Figure 2, Enzyme 2 exhibits a significant decrease in peroxisomal import (down to \(16\%\)) specifically in cells treated with PEX7 shRNA, while treatment with PEX5 shRNA results in no significant difference in import (\(99\%\)) compared to the control (\(100\%\)), demonstrating that Enzyme 2 relies exclusively on PEX7 (the PTS2 receptor) for peroxisomal entry. - D.ii (1 point): Matrix proteins are synthesized on ribosomes in the cytosol; if the PTS1 signal is mutated and cannot interact with the cytosolic receptor PEX5, the protein cannot be targeted to or translocated across the peroxisomal membrane and thus remains in the cytosol.
評分準則
Point A1 (1 point): - Accepts: Description that the signal sequence serves as a molecular 'tag' / address recognized by receptors/translocons that direct the protein to a specific organelle/membrane. - Rejects: Vague statements like 'it tells the cell what to do' without mentioning protein targeting or destination.
Point B1 (1 point): - Accepts: Relative amount / level / content of PEX5 and/or PEX7 protein.
Point B2 (1 point): - Accepts: To determine whether the PEX5 shRNA was specific to PEX5 / to confirm that PEX5 shRNA did not alter PEX7 levels / to rule out off-target effects on the alternative receptor.
Point B3 (1 point): - Accepts: No effect / no significant change / PEX7 levels remain at control levels (~100%).
Point C1 (1 point): - Accepts: The type of shRNA treatment / knockdown condition OR the type of enzyme tested (Enzyme 1, 2, or 3).
Point C2 (1 point): - Accepts: Enzyme 1 and Enzyme 3 (must identify both; partial credit not awarded for single point).
Point C3 (1 point): - Accepts: 153 (amino acids). Must show correct calculation: \((1275 - 816) / 3 = 153\) or \(425 - 272 = 153\).
Point D1 (1 point): - Accepts: Support stating that only PEX7 shRNA reduced the import of Enzyme 2 (to ~16%), whereas PEX5 shRNA had no effect on Enzyme 2 import (~99% / overlapping with control).
Point D2 (1 point): - Accepts: Justification connecting cytosolic translation of Enzyme 1 to the requirement of PEX5 binding for transport through the peroxisomal membrane channel; without binding, the protein cannot cross the membrane and remains in the cytosol.
題目 2 · FRQ
9 分
Plants initiate defense responses against herbivorous insects through cell-surface receptor signaling pathways. In Arabidopsis thaliana, physical wounding by aphid feeding triggers the release of an endogenous peptide signal, PEP1, into the extracellular space. PEP1 binds to the extracellular domain of the transmembrane receptor kinase PEPR1. Activation of PEPR1 stimulates a phosphorylation cascade involving mitogen-activated protein kinases (MAPKs), which ultimately promotes the transcription of genes responsible for synthesizing defensive glucosinolate compounds that deter aphids.
To investigate the role of the PEP1-PEPR1 signaling pathway in plant defense, researchers treated wild-type (WT) plants and mutant plants lacking functional PEPR1 (pepr1) with either a mock control solution or a solution containing PEP1. They measured the relative expression of glucosinolate defense genes (as a percentage relative to untreated wild-type plants) and the aphid avoidance rate (defined as the percentage of aphids choosing to feed on an untreated control leaf rather than the test leaf) (Table 1).
**Table 1. Relative glucosinolate gene expression and aphid avoidance in wild-type and pepr1 mutant plants.**
A. Transmembrane receptor kinases, such as PEPR1, span the phospholipid bilayer of the plasma membrane. Describe the characteristic chemical properties (polarity/hydrophobicity) of the amino acids found in the transmembrane domain of PEPR1 compared to those in the extracellular ligand-binding domain.
B. (i) Using the data in Table 1, construct an appropriately labeled bar graph on the template provided to represent both glucosinolate gene expression and aphid avoidance across all three experimental groups. Include appropriate scaling and error bars representing ±2SEx̄. (ii) Based on the data in Table 1, determine the effect of the loss of functional PEPR1 receptors on aphid avoidance behavior in plants exposed to PEP1.
C. (i) Based on Table 1, identify the treatment group in which aphid avoidance was greater than 50%. (ii) Scientists identified a gain-of-function mutation in the PEPR1 gene that makes the intracellular kinase domain constitutively (permanently) active, even in the absence of PEP1. Predict the effect of this mutation on glucosinolate gene expression in mutant plants treated with the mock solution.
D. (i) Researchers claim that PEP1-mediated defense activation is dependent on the PEPR1 receptor. Using data from Table 1, support the researchers' claim. (ii) Explain how applying a synthetic chemical that mimics PEP1 (an agonist) to agricultural crops could reduce crop damage caused by herbivorous insects.
查看答案詳解收起答案詳解
解題
### Part A (1 point) - Point A1: Transmembrane domains interact with the hydrophobic interior of the phospholipid fatty acid tails, requiring nonpolar (hydrophobic) amino acid residues. In contrast, the extracellular domain interacts with the aqueous extracellular fluid and polar peptide ligands, requiring hydrophilic (polar or charged) amino acids.
### Part B (4 points) - Point B1 (Graph Type & Axes): Bar graph format (or clustered/paired bar graph) with clearly labeled axes (e.g., Plant Genotype and Treatment on the x-axis, and appropriate % units on the y-axis/axes) and appropriate scaling. - Point B2 (Data Plotting): All bars accurately plotted based on the means from Table 1. - Point B3 (Error Bars): Error bars representing ±2SE accurately plotted on each corresponding bar. - Point B4 (Data Interpretation): Loss of PEPR1 prevents the increase in aphid avoidance in response to PEP1 (avoidance remains low at 22%, compared to 76% in wild-type).
### Part C (2 points) - Point C1 (Identification): The Wild-type + PEP1 treatment group. - Point C2 (Prediction): Glucosinolate gene expression will be constitutively high/increased (similar to wild-type treated with PEP1) because the kinase cascade will remain active independently of ligand binding.
### Part D (2 points) - Point D1 (Supporting Claim): In wild-type plants, PEP1 treatment induces a substantial increase in both glucosinolate gene expression (from 100% to 340%) and aphid avoidance (from 18% to 76%). In contrast, in pepr1 mutants treated with PEP1, gene expression (115%) and avoidance (22%) remain close to mock levels (error bars overlap or remain near baseline), demonstrating that PEPR1 is required for the response. - Point D2 (Application/Explanation): The PEP1 mimic binds to PEPR1 receptors, activating the intracellular MAPK signaling pathway and inducing systemic production of glucosinolates (or defense chemicals) throughout the crop plants, which deters pests and prevents insect herbivory without using synthetic neurotoxic pesticides.
評分準則
Part A (1 point max): - Point A1: Description that the transmembrane domain consists of nonpolar/hydrophobic amino acids while the extracellular domain consists of polar/hydrophilic/charged amino acids (1 pt).
Part B (4 points max): - Point B1: Construction of a bar graph with appropriate labels and scaling (1 pt). - Point B2: Accurate plotting of all mean values across categories (1 pt). - Point B3: Accurate plotting of error bars representing ±2SEx̄ (1 pt). - Point B4: Determination that inhibiting/lacking the PEPR1 receptor results in significantly lower/reduced aphid avoidance in the presence of PEP1 (1 pt).
Part C (2 points max): - Point C1: Identification of the "Wild-type + PEP1" group (1 pt). - Point C2: Prediction that glucosinolate gene expression will increase/remain high even without PEP1 treatment (1 pt).
Part D (2 points max): - Point D1: Support using data: stating that PEP1 elevates defense gene expression and avoidance in WT plants, but in pepr1 mutants both metrics show no significant increase relative to the mock control (1 pt). - Point D2: Explanation connecting receptor agonist binding to pathway activation, leading to pre-emptive defense gene transcription/deterrent production, thereby reducing pest herbivory (1 pt).
題目 3 · free-response
4 分
Giant salvinia (Salvinia molesta) is an invasive floating aquatic fern in freshwater ecosystems that forms dense mats across the water surface, severely reducing light penetration into the water column. Native submerged macrophytes, such as wild celery (Vallisneria americana), serve as essential habitat and produce oxygen necessary to support diverse aquatic communities. When dense mats of S. molesta shade the water column, native submerged plants fail to thrive.
Ecologists conducted an experiment to test whether a host-specific biological control insect, the salvinia weevil (Cyrtobagous salviniae), could control the abundance of S. molesta and facilitate the growth of V. americana. The researchers set up identical outdoor aquatic tanks (mesocosms) containing equal initial amounts of S. molesta and V. americana. The tanks were assigned to four treatment groups receiving different stocking densities of C. salviniae (0, 10, 20, or 40 weevils per \(\text{m}^2\)). All tanks were maintained under identical temperature, nutrient, and photoperiod conditions for 16 weeks. At the end of 16 weeks, the researchers measured the percent surface area covered by S. molesta and the dry biomass of V. americana in each tank.
A. Describe how primary producers such as Vallisneria americana contribute to energy flow in an aquatic ecosystem.
B. Identify the control group the ecologists included in their experiment.
C. State the null hypothesis for the experiment regarding the effect of Cyrtobagous salviniae density on the percent surface area covered by Salvinia molesta.
D. The ecologists observed that in ponds where Salvinia molesta completely blocks light, massive fish die-offs occur. Justify the claim that the loss of submerged aquatic plants leads to fish mortality in these ecosystems.
查看答案詳解收起答案詳解
解題
A. Primary producers occupy the first trophic level in an ecosystem. They absorb radiant/solar energy and convert it via photosynthesis into chemical potential energy stored in organic macromolecules (biomass). This chemical energy is subsequently transferred to herbivores/primary consumers and higher trophic levels throughout the food web.
B. The control group is the set of mesocosms with a stocking density of 0 weevils per \(\text{m}^2\) (i.e., the tanks in which no biological control agent C. salviniae was introduced), serving as a baseline comparison for the weevil treatments.
C. An appropriate null hypothesis states that the independent variable will have no effect on the dependent variable: 'There is no significant difference in the percent surface area covered by Salvinia molesta among tanks with different stocking densities of Cyrtobagous salviniae (or between weevil-treated tanks and the untreated control).'
D. Without sunlight penetration, submerged plants cannot carry out photosynthesis to generate dissolved oxygen (\(\text{O}_2\)). Furthermore, as plants die, populations of aerobic microbial decomposers rapidly increase and consume large amounts of dissolved oxygen during cellular respiration. This causes severe hypoxia/anoxia in the water column, depriving fish of the oxygen necessary for aerobic ATP synthesis and leading to mortality.
評分準則
Part A (1 point) * 1 point for describing that primary producers fix/convert light/solar energy into chemical energy/biomass/organic molecules accessible to primary consumers/higher trophic levels.
Part B (1 point) * 1 point for identifying the control group as the treatment group with 0 weevils/\(\text{m}^2\) (or tanks without Cyrtobagous salviniae / no weevils added).
Part C (1 point) * 1 point for stating a valid null hypothesis indicating that C. salviniae density has no effect on the surface area coverage / abundance of S. molesta (e.g., 'There will be no difference in the percent surface area covered by S. molesta regardless of C. salviniae density').
Part D (1 point) * 1 point for justifying that the decrease in photosynthesis and/or increased aerobic cellular respiration by decomposers depletes dissolved oxygen (causing hypoxia), preventing fish from carrying out aerobic respiration.
題目 4 · frq
4 分
Approximately five million years ago, tectonic activity caused the uplift of a high mountain range across a large landmass, dividing a continuous temperate forest into two distinct geographic zones. The windward western slopes receive high annual rainfall, sustaining a dense, humid temperate rainforest. The leeward eastern slopes lie in a severe rain shadow, creating an arid, open grassland. Prior to mountain formation, a single ancestral species of flightless ground beetle inhabited the entire region. Today, two morphologically distinct, reproductively isolated beetle species occupy the two opposing sides of the mountain range.
A. Describe the condition required for two populations to be considered separate species under the biological species concept.
B. Explain how the formation of the mountain barrier and the resulting environmental differences led to allopatric speciation in the two beetle populations.
A nonnative, generalist insectivorous rodent species was recently introduced to the eastern grassland. This introduced rodent feeds aggressively on native ground beetles as well as on native grasshoppers, which are the primary herbivores consuming the foliage of the dominant native bunchgrass species.
C. Predict the effect that the introduction of the nonnative rodent will have on the biomass of the dominant native bunchgrass species in the eastern grassland.
D. Justify your prediction in part C.
查看答案詳解收起答案詳解
解題
A. According to the biological species concept, organisms belong to different species if they are reproductively isolated—meaning they are incapable of interbreeding under natural conditions to produce viable, fertile offspring.
B. The geographic barrier (mountain range) stopped gene flow between the separated populations. In the humid forest and arid grassland, natural selection acted on distinct variations, favoring traits suited to the local environment (e.g., moisture retention, camouflage, dietary adaptations). Over time, random mutations, genetic drift, and natural selection led to accumulated genetic differences that prevented members of the two populations from successfully interbreeding even if they were reunited.
C. Bunchgrass biomass will increase.
D. The introduced rodent consumes grasshoppers, leading to a decline in the grasshopper population. Because grasshoppers are the primary herbivores feeding on the bunchgrass, a reduction in the grasshopper population decreases herbivory on the bunchgrass, resulting in an increase in bunchgrass growth and total biomass.
評分準則
Part A (1 point): - Acceptable descriptions include: • Members of the two populations cannot interbreed to produce viable, fertile offspring (reproductive isolation).
Part B (1 point): - Acceptable explanations include: • The physical/geographic barrier prevents gene flow, allowing different selective pressures (or mutations/genetic drift) in the two environments to accumulate genetic differences over time until reproductive isolation occurs.
Part C (1 point): - Acceptable predictions include: • The biomass of the dominant native bunchgrass will increase.
Part D (1 point): - Acceptable justifications include: • The rodent preys on grasshoppers, reducing the herbivore population and decreasing herbivory/consumption of the bunchgrass, thereby allowing plant biomass to increase (top-down regulation / trophic cascade).
題目 5 · free-response
4 分
Figure 1 shows a model of a four-step metabolic pathway in bacterial cells used to synthesize the essential amino acid leucine (Leu) from the precursor compound alpha-ketoisovalerate (KIV).
Figure 1. Model of the enzymatic pathway for leucine synthesis from KIV
Note: Enzyme 1 possesses an active site for KIV and a separate allosteric site that binds leucine.
A. Describe how the shape and chemical properties of an enzyme's active site enable it to bind specifically to its substrate.
B. Based on Figure 1, explain how the accumulation of leucine regulates its own synthesis via enzyme 1.
C. Using the information in Figure 1, identify the substrate of the reaction catalyzed by enzyme 3: precursor KIV, intermediate IPM, intermediate KIC, or leucine.
D. Based on Figure 1, explain how an increase in environmental temperature well beyond the optimum temperature for enzyme 2 would affect the cellular concentration of intermediate IPM.
查看答案詳解收起答案詳解
解題
A. The specificity of an enzyme is determined by its active site, which has a distinct three-dimensional tertiary shape formed by folded polypeptide chains. The functional groups (R-groups) lining the active site possess compatible chemical characteristics (such as polar, nonpolar, positively charged, or negatively charged properties) that interact specifically through noncovalent bonds with the substrate.
B. When leucine accumulates in the cell, it acts as an allosteric inhibitor by binding to a non-active allosteric regulatory site on enzyme 1. This binding induces a conformational change in enzyme 1 that alters the shape of its active site, reducing its affinity for precursor KIV or preventing KIV from binding. This negative feedback loop downregulates further flux through the pathway, preventing overproduction of leucine.
C. In the sequential linear pathway shown, enzyme 3 catalyzes the final step where intermediate KIC is converted into the end-product leucine. Therefore, intermediate KIC is the substrate of enzyme 3.
D. High temperatures above the thermal optimum increase molecular kinetic energy enough to disrupt the weak interactions (hydrogen bonds, ionic bonds, hydrophobic interactions) that stabilize the tertiary/quaternary structure of enzyme 2. Denaturation alters the conformation of the active site so that enzyme 2 can no longer bind or catalyze the reaction involving intermediate IPM. As enzyme 1 continues producing IPM while enzyme 2 cannot consume it, the concentration of intermediate IPM will accumulate/increase.
評分準則
Part A (1 point): - Acceptable responses include: • The active site has a specific shape/conformation that is complementary to the substrate. • The R-groups/chemical properties (charges, polarity, hydrophobicity) within the active site form compatible/complementary interactions with the substrate.
Part B (1 point): - Acceptable responses include: • Leucine binds to the allosteric site on enzyme 1, altering the active site conformation and preventing precursor KIV from binding/inhibiting enzyme 1 activity. • Leucine acts as a noncompetitive/allosteric inhibitor to shut down or reduce the activity of enzyme 1 via negative feedback.
Part C (1 point): - Acceptable responses include: • Intermediate KIC (or KIC).
Part D (1 point): - Acceptable responses include: • The concentration of IPM will increase/accumulate because enzyme 2 denatures/loses function and cannot convert IPM into KIC.
題目 6 · free-response
4 分
In the model plant Arabidopsis thaliana, exposure to low nonfreezing temperatures induces the expression of cold-regulated (COR) genes, which protect plant tissues from subsequent freezing damage. The CBF3 gene encodes a transcription factor that binds to specific regulatory DNA sequences in the promoters of COR genes to activate their transcription. Activation of CBF3 requires phosphorylation by the protein kinase CIPK1.
Scientists investigated how different mutations in the CIPK1 gene affect freezing tolerance. They generated three plant lines: cipk1-null (a deletion of the entire CIPK1 gene), cipk1-K40A (a single missense mutation that substitutes an alanine for a critical lysine in the catalytic kinase domain, producing a full-length but kinase-inactive protein), and CIPK1-OE (a line engineered to overexpress wild-type CIPK1 protein). Following cold acclimation at 4 °C for 24 hours, the scientists measured the relative expression level of COR15A mRNA (one of the major COR genes) and the percentage of seedlings surviving subsequent freezing stress at -6 °C (Table 1).
**Table 1. Relative COR15A mRNA Expression and Seedling Survival Rate in Different Arabidopsis Lines Following Freezing Stress**
A. Based on Table 1, identify the plant line that exhibited the lowest relative expression of COR15A mRNA.
B. Based on Table 1, describe the difference in COR15A mRNA expression between the WT plants and the CIPK1-OE plants.
C. Researchers hypothesize that the catalytic kinase activity of CIPK1, rather than merely the physical presence of the protein, is required to activate COR15A expression during cold acclimation. Support the researchers' hypothesis using the data from Table 1.
D. A loss-of-function mutation occurs in the promoter region of the COR15A gene that prevents the CBF3 transcription factor from binding. Explain how this promoter mutation would affect the survival of CIPK1-OE seedlings exposed to freezing stress.
查看答案詳解收起答案詳解
解題
Part A (1 point) * Task: Identify the plant line showing the lowest COR15A mRNA level. * Analysis: Looking at Table 1 under 'Relative COR15A mRNA Expression', the lowest value listed is 11 ± 3, which corresponds to cipk1-null.
Part B (1 point) * Task: Describe the difference in mRNA expression between WT and CIPK1-OE. * Analysis:WT has a mean expression of 100 (± 9), while CIPK1-OE has a mean expression of 245 (± 18). Describing that CIPK1-OE produces significantly higher/greater amounts of COR15A mRNA (or an increase of ~145% / 2.45-fold) correctly characterizes the difference.
Part C (1 point) * Task: Support the hypothesis that catalytic kinase activity is essential using data from Table 1. * Analysis: The cipk1-K40A line expresses the full-length protein but lacks catalytic activity due to a point mutation. Its COR15A mRNA expression (14 ± 4) is severely reduced relative to WT (100 ± 9) and is statistically indistinguishable from the complete absence of the protein in cipk1-null (11 ± 3, as the error bars overlap). This indicates that the presence of the inactive protein alone is insufficient to drive gene expression.
Part D (1 point) * Task: Explain the effect of a promoter mutation preventing CBF3 binding in CIPK1-OE plants. * Analysis: Even though CIPK1-OE produces high levels of active CIPK1 and phosphorylated CBF3, CBF3 functions as a sequence-specific transcription factor. If it cannot bind the COR15A promoter, RNA polymerase will not be recruited/transcription will not occur, leading to a lack of protective COR15A protein and thus a reduction in freezing survival.
評分準則
Part A (1 point max): * 1 point for identifying cipk1-null.
Part B (1 point max): * 1 point for describing that COR15A mRNA expression is higher/increased in CIPK1-OE plants compared to WT plants (or increased by ~145% / 2.45-fold).
Part C (1 point max): * 1 point for supporting the hypothesis by stating that plants with the kinase-inactive protein (cipk1-K40A) show drastically reduced COR15A mRNA expression compared to WT / have expression levels comparable to/overlapping with the deletion mutant (cipk1-null).
Part D (1 point max): * 1 point for explaining that survival will decrease because CBF3 cannot bind the promoter to induce/activate transcription of COR15A (resulting in a lack of protective protein needed for freezing tolerance).