AP® Biology

AP® Biology Free-Response Questions Past Paper 2025 Solution

AP® Biology 2025 Free-Response Questions: Detailed Solutions

A Note from Your AP Bio Educator: Welcome, biologists! This guide provides a detailed walkthrough of the 2025 AP Biology Free-Response Questions. Success on this exam requires a deep understanding of biological concepts, the ability to analyze data, and the skill to construct clear, logical, and evidence-based explanations. For each question, we'll break down the prompt, outline a strategic approach, and present a model answer that demonstrates the level of thinking required.

Question 1: Protein Transport and siRNA

A. Describe the function of ribosomes.

Thinking Process: This is a foundational knowledge question. What do ribosomes do? They are the site of protein synthesis. I need to describe this process, mentioning mRNA and translation.

Ribosomes are cellular organelles responsible for protein synthesis, a process called translation. They read the sequence of codons on a messenger RNA (mRNA) molecule and, with the help of transfer RNA (tRNA), link amino acids together in the specific order dictated by the mRNA sequence to form a polypeptide chain.

B. i. Identify the dependent variable in the experiments shown in Figure 1.

Thinking Process: The dependent variable is what is being measured. Look at the y-axis of Figure 1. It is labeled "Relative Protein Content (% of control)." This is the variable that is expected to change in response to the treatment (the independent variable).

The dependent variable in the experiments shown in Figure 1 is the relative amount of Sec62 and SR protein, measured as a percentage of the control.

B. ii. Justify why the researchers included the control of measuring the relative amounts of both Sec62 and SR proteins in cells that were treated with Sec62 siRNA only (data shown in Figure 1).

Thinking Process: The question asks why, when treating with Sec62 siRNA, they also measured the SR protein level. The purpose of siRNA is to be specific. Sec62 siRNA should only reduce Sec62 protein, not other proteins. Measuring SR protein is a control to check for this specificity. If SR protein levels were also reduced, it would mean the siRNA was having non-specific, off-target effects, which would invalidate the results.

The researchers included this control to ensure the specificity of the siRNA treatment. By measuring the amount of SR protein in cells treated with Sec62 siRNA, they could verify that the siRNA was only reducing the expression of its target protein (Sec62) and not having unintended, off-target effects on the expression of other proteins like SR. The data show that the Sec62 siRNA did not significantly affect the level of SR protein, confirming the specificity of the treatment.

B. iii. Based on Figure 1, describe the effect on the production of SR protein when cells are treated with Sec62 siRNA.

Thinking Process: I need to look at the data in Figure 1. The relevant bars are the second group, labeled "Sec62 siRNA." In this group, I need to look at the bar for "SR Protein." This bar is at approximately 100% of the control level. This means there was no significant change.

Based on Figure 1, treating cells with Sec62 siRNA has little to no effect on the production of SR protein. The relative amount of SR protein in these cells remains at approximately 100% of the control level, and the error bars for the control and the Sec62 siRNA-treated SR protein levels overlap.

C. i. Identify the independent variable in the researchers’ second experiment (data shown in Figure 2).

Thinking Process: The independent variable is what the researchers are manipulating. In Figure 2, they are comparing three groups of cells: those treated with control siRNA, Sec62 siRNA, or SR siRNA. Therefore, the type of siRNA added to the cells is the independent variable.

The independent variable in the second experiment is the type of siRNA (control, Sec62, or SR) added to the cells.

C. ii. Based on Figure 2, identify the protein(s) that when treated with Sec62 siRNA showed an increase in percent transport to the ER compared with the control.

Thinking Process: I need to look at Figure 2 and find the bars for "Sec62 siRNA" treatment. For each protein (1, 2, and 3), I need to compare the "Sec62" bar to the "Control" bar. - Protein 1: Sec62 bar is lower than Control. - Protein 2: Sec62 bar is higher than Control. - Protein 3: Sec62 bar is lower than Control. So, only Protein 2 showed an increase.

Based on Figure 2, Protein 2 is the only protein that showed an increase in percent transport to the ER when treated with Sec62 siRNA compared with the control.

C. iii. Protein 1 is encoded by 234 nucleotides, while protein 2 is encoded by 495 nucleotides. Assuming all nucleotides for both proteins encode amino acids, calculate the difference in the number of amino acids between the two proteins.

Thinking Process: The genetic code is read in triplets (codons). Three nucleotides code for one amino acid. 1. Calculate amino acids for Protein 1: $234 \text{ nucleotides} \div 3 \text{ nucleotides/amino acid} = 78 \text{ amino acids}$. 2. Calculate amino acids for Protein 2: $495 \text{ nucleotides} \div 3 \text{ nucleotides/amino acid} = 165 \text{ amino acids}$. 3. Calculate the difference: $165 \text{ amino acids} - 78 \text{ amino acids} = 87 \text{ amino acids}$.
Number of amino acids in Protein 1 = 234 nucleotides / 3 = 78 amino acids
Number of amino acids in Protein 2 = 495 nucleotides / 3 = 165 amino acids
Difference = 165 - 78 = 87 amino acids

The difference in the number of amino acids between the two proteins is 87.

D. i. Researchers claim that protein 1 is the only tested protein that is transported to the ER following its complete translation in the cytosol. Using data from Figure 2, support the researchers’ claim.

Thinking Process: 1. Recall the background info: The prompt states that Sec62 is necessary for transport *after* translation, and SR is necessary for transport *during* translation. 2. Connect to the claim: "Transported...following its complete translation" means it uses the post-translational pathway, which requires Sec62. So, the claim is that Protein 1 is the only one that uses the Sec62 pathway. 3. Find supporting data in Figure 2: If Protein 1 uses the Sec62 pathway, its transport should be inhibited when Sec62 is knocked down by siRNA. Looking at Figure 2, when cells are treated with Sec62 siRNA, the transport of Protein 1 drops to about 50% of the control. The transport of Proteins 2 and 3 is not inhibited by Sec62 siRNA (in fact, Protein 2's transport increases). This supports the claim that Protein 1's transport is dependent on Sec62, and thus it uses the post-translational pathway.

The data in Figure 2 support the researchers' claim. The background information states that the Sec62 protein is required for the transport of proteins to the ER *after* their complete translation. The experiment shows that when cells are treated with Sec62 siRNA to reduce the amount of Sec62 protein, the transport of Protein 1 to the ER is significantly inhibited, dropping to approximately 50% of the control level. In contrast, the transport of Proteins 2 and 3 is not inhibited under these conditions. This indicates that Protein 1's transport is dependent on Sec62, and therefore it is the only one of the three proteins tested that uses the post-translational transport pathway.

D. ii. For any protein that enters the ER, researchers claim that amino acids close to the protein’s amino terminus determine how likely the protein is to pass through the protein channel within the ER membrane. Justify the researchers’ claim based on your understanding of factors that affect the transport of proteins across membranes.

Thinking Process: 1. Recall protein structure: The amino terminus is the beginning of the polypeptide chain. 2. Recall protein transport into the ER: This process is initiated by a "signal peptide" or "signal sequence." This is a short stretch of amino acids, usually at the N-terminus, that acts as a zip code, directing the ribosome-protein complex to the ER. 3. Connect to the claim: The properties of this signal sequence—specifically its chemical properties like hydrophobicity—determine how it interacts with the protein channel (translocon) in the ER membrane. The claim is essentially about the role of this signal peptide. 4. Justify the claim: The sequence of amino acids at the amino terminus forms the signal peptide. The chemical properties of these amino acids (e.g., their charge, polarity, size) determine the overall three-dimensional shape and hydrophobicity of this signal sequence. This structure must be complementary to the shape and chemical properties of the binding site within the ER protein channel (the translocon) for the protein to be recognized and threaded through the membrane. Therefore, the specific amino acids at the N-terminus are critical determinants of successful protein transport across the ER membrane.

The researchers' claim is justified because the sequence of amino acids at the amino terminus of a protein often acts as a signal peptide. The specific R-groups of these amino acids determine their chemical properties, such as hydrophobicity and charge. These properties dictate the three-dimensional shape of the signal peptide, which must be chemically and sterically compatible with the binding site of the protein channel (translocon) in the ER membrane. This interaction, based on molecular complementarity, is what allows the channel to recognize the protein and initiate its transport across the membrane. Therefore, the specific amino acids at the N-terminus are a critical factor in determining the efficiency of protein transport into the ER.

Question 2: Cell Signaling in Moths

A. Many receptors are embedded in the plasma membrane. Describe the polarity of the portion of the receptor that is inside the membrane.

Thinking Process: The plasma membrane is a lipid bilayer. The interior of the bilayer is made of hydrophobic fatty acid tails. For a protein (the receptor) to be embedded within this region, the amino acids in that part of the protein must be compatible with the hydrophobic environment. Therefore, the R-groups of those amino acids must be nonpolar.

The portion of a receptor that is embedded inside the plasma membrane is composed of amino acids with nonpolar R-groups. This is because the interior of the plasma membrane consists of the hydrophobic, nonpolar fatty acid tails of phospholipids. For the receptor to be stable within the membrane, its transmembrane domain must also be hydrophobic and therefore nonpolar, allowing it to interact favorably with the surrounding lipid tails.

B. i. Using the template in the space provided for your response, construct an appropriate type of graph that represents the data in Table 1. Your graph should be appropriately plotted and labeled.

Thinking Process: 1. Choose the graph type: The independent variable is the treatment (Saline vs. siRNA), which is a discrete category. The dependent variables are the percentages of activity. A bar graph is the most appropriate choice for comparing discrete categories. 2. Set up the axes: The x-axis will be the Treatment (Saline, siRNA). The y-axis will be the Mean Percent of Activity. The y-axis should go up to 100 to accommodate the General Activity values. 3. Plot the data: I need to plot two sets of bars for each treatment: one for General Activity and one for Oriented Activity. I should use a legend to distinguish them. - **Saline:** General Activity at 95%, Oriented Activity at 60%. - **siRNA:** General Activity at 90%, Oriented Activity at 25%. 4. Add labels and title: The graph needs a descriptive title (e.g., "Effect of DopEcR siRNA on Moth Activity"). The axes must be labeled with units ("Treatment" and "Mean Percent of Activity (%)"). Error bars (±2SE) should be included as requested by the exam format, though drawing them perfectly isn't the main point—plotting the means correctly is.

(Student response would be a hand-drawn or digitally created bar graph. A description is provided here.)

Graph Title: Effect of DopEcR Inhibition on Moth Activity

A bar graph showing two groups on the x-axis: 'Saline Control' and 'siRNA Treatment'. The y-axis is labeled 'Mean Percent of Activity (%)' and goes from 0 to 100. For each group, there are two bars. For the Saline group, the 'General Activity' bar is at 95% and the 'Oriented Activity' bar is at 60%. For the siRNA group, the 'General Activity' bar is at 90% and the 'Oriented Activity' bar is at 25%. A legend indicates which bars represent General vs. Oriented activity. Error bars are drawn on top of each bar.

Description of the Constructed Graph: The graph is a bar chart with the "Treatment" (Saline Control and siRNA) on the x-axis and "Mean Percent Activity" on the y-axis. For each treatment, two bars are plotted: one for "General Activity" and one for "Oriented Activity," distinguished by a legend. The bars for the saline control are at 95% (general) and 60% (oriented). The bars for the siRNA treatment are at 90% (general) and 25% (oriented). Error bars representing ±2SE are included on each bar.

B. ii. Based on the data in Table 1, determine the type of activity that was affected by inhibiting the expression of the DopEcR receptor.

Thinking Process: Compare the saline (control) and siRNA (treatment) groups for both types of activity. - **General Activity:** 95% (saline) vs. 90% (siRNA). The error bars are ±5 and ±8, so the ranges (90-100 and 82-98) overlap significantly. This is likely not a significant difference. - **Oriented Activity:** 60% (saline) vs. 25% (siRNA). The error bars are ±4 and ±6, so the ranges (56-64 and 19-31) do not overlap at all. This is a very large and significant difference. Therefore, oriented activity was the one primarily affected.

Oriented activity was the type of activity significantly affected by inhibiting the expression of the DopEcR receptor. While general activity showed only a small decrease from 95% to 90% with overlapping error bars, oriented activity dropped dramatically from 60% in the control group to 25% in the siRNA-treated group, and their respective error bar ranges do not overlap.

C. i. Based on Table 1, identify the treatment group in which the oriented activity was greater than 50% of the general activity.

Thinking Process: - **Saline group:** Oriented activity is 60%. Is this > 50% of its general activity (95%)? $0.50 \times 95 = 47.5$. Since $60 > 47.5$, yes. - **siRNA group:** Oriented activity is 25%. Is this > 50% of its general activity (90%)? $0.50 \times 90 = 45$. Since $25 < 45$, no. The answer is the male moths injected with saline.

The treatment group in which the oriented activity was greater than 50% of the general activity was the male moths injected with saline (control solution).

C. ii. Based on Figure 1, predict the effect of this mutation on the oriented activity in male moths exposed to the pheromone.

Thinking Process: 1. Understand the mutation: It "prevents GTP from displacing the GDP bound to the G protein." 2. Look at the pathway in Figure 1: The G protein is "Inactive" when bound to GDP. It becomes "Active" only when GTP displaces GDP. This activation is required for the "Increased Expression of Genes" that leads to oriented activity. 3. Predict the effect: If GTP cannot displace GDP, the G protein will remain stuck in its inactive state. If the G protein cannot be activated, the entire signaling pathway will be blocked. Therefore, there will be no increased expression of the necessary genes, and the oriented activity will be significantly reduced or absent, similar to the effect seen in the siRNA-treated moths.

The mutation will significantly decrease or eliminate oriented activity. According to Figure 1, the G protein must be activated by GTP displacing GDP to initiate the signaling pathway that leads to oriented activity. If the mutation prevents this displacement, the G protein will remain in its inactive state, the signaling cascade will be blocked, and the genes responsible for oriented activity will not be expressed, even when the moth is exposed to the pheromone.

D. i. Use evidence from the information provided to support the scientists’ claim.

Thinking Process: 1. Identify the claim: "increase in [DopEcR] gene expression increases the likelihood of males finding females." 2. Connect the pieces of information: - The background states that males find mating partners by responding to female pheromones. - Table 1 shows that oriented activity (movement toward pheromones) is dependent on the DopEcR receptor. When DopEcR is inhibited (siRNA group), oriented activity drops from 60% to 25%. - The prompt states DopEcR expression increases as moths reach sexual maturity. 3. Synthesize the support: The data show that the DopEcR receptor is essential for oriented activity toward pheromones. Since female moths release pheromones to attract mates, successful oriented activity is required for a male to find a female. Therefore, the increase in DopEcR expression at sexual maturity would lead to greater oriented activity, directly increasing the likelihood of the male successfully locating a female to mate with.

The scientists' claim is supported by the experimental data in Table 1. The data show that inhibiting the DopEcR receptor via siRNA drastically reduces oriented activity (from 60% to 25%), which is defined as movement toward a high concentration of female pheromones. Since finding a female requires moving toward her pheromones, and this behavior is dependent on the DopEcR receptor, the described increase in DopEcR gene expression as moths reach sexual maturity would enhance this critical mate-finding behavior, thus increasing the likelihood of successful mating.

D. ii. Based on Figure 1, explain how an inhibitor of the DopEcR pathway might serve as an effective chemical to protect crops from moth damage.

Thinking Process: 1. Understand the goal: Protect crops from moth damage. Moth damage is often caused by larvae (caterpillars), which result from moth reproduction. 2. Understand the pathway's function: The DopEcR pathway is essential for male moths to find female moths to reproduce. 3. Explain the mechanism of protection: If a chemical inhibitor blocks any part of this pathway (e.g., blocks the DopEcR receptor itself, or prevents G protein activation), it will prevent male moths from finding females. This disruption of mating would lead to a decrease in moth reproduction. A lower reproductive rate would mean fewer eggs are laid on the crops, resulting in fewer larvae to eat the plants, thus protecting the crops from damage.

An inhibitor of the DopEcR pathway could protect crops by disrupting the moths' reproductive cycle. Moth larvae are a common cause of crop damage. The signaling pathway shown in Figure 1 is essential for male moths to locate females for mating. By applying a chemical that inhibits this pathway—for example, by blocking the DopEcR receptor so 20E cannot bind—male moths would be unable to exhibit the oriented activity needed to find mates. This would lead to a significant reduction in successful mating events, a lower reproductive rate for the moth population, and consequently, fewer larvae to damage the crops.

Questions 3-6: Short Free-Response

Solutions for the short free-response questions follow the same principles of breaking down the prompt, applying biological knowledge, and forming a clear, concise answer.

Question 3: Invasive Species

A. Describe the effect that removing a keystone species will have on an ecosystem.

Removing a keystone species will have a disproportionately large and often negative effect on its ecosystem relative to its abundance. The removal can lead to a trophic cascade, causing a drastic change in the populations of other species and altering the overall structure and biodiversity of the ecosystem. For example, removing a keystone predator can lead to an explosion in the herbivore population, which can then overgraze and decimate the plant community.

B. Identify a control group the scientists should include in their experiment.

The scientists should include a control group where buffelgrass is grown by itself, without the presence of any native grass species, under both the nondrought and drought watering conditions.

C. State the null hypothesis of the experiment in which buffelgrass is grown in the presence of native grass species.

The null hypothesis is that the presence of native grass species will have no significant effect on the height or dry weight of buffelgrass.

D. Based on the information given, justify the scientists’ claim.

The scientists' claim is justified because buffelgrass has two key advantages in a fire-prone environment. First, it is drought-tolerant and can survive wildfires, while many other plants, including young saguaro cacti, cannot. Second, its population growth rate is much faster than that of native grasses after a fire. This means that after a wildfire clears an area of competition, the surviving buffelgrass will be able to repopulate the area and outcompete the slow-growing native species for resources like water and space, thereby increasing its abundance and dominance in the ecosystem.


Question 4: Evolution and Biogeography

A. Describe the genetic evidence that evolution is occurring in a population.

Genetic evidence that evolution is occurring is a change in the allele frequencies within a population's gene pool over successive generations. This can be measured directly through genetic sequencing of individuals over time.

B. Explain how the isolation of marine species by the formation of a land barrier can lead to divergent evolution of those species.

The formation of a land barrier causes allopatric speciation. The barrier splits an ancestral population into two isolated populations, cutting off gene flow between them. The two populations are then subjected to different selective pressures in their new, distinct environments (e.g., cooler, nutrient-rich Pacific vs. warmer Caribbean). Over time, different mutations arise and are selected for in each population, leading to the accumulation of genetic differences. Eventually, these differences become so significant that the two populations can no longer interbreed, even if the barrier were removed, meaning they have diverged into two separate species.

C. Predict the effect the formation of the isthmus had on resource availability for South American species.

The formation of the isthmus likely led to a decrease in resource availability for many native South American species.

D. Justify your prediction in part C.

My prediction is justified because the formation of the isthmus created a land bridge that allowed North American land animals to migrate south. These migrating species would have increased the overall number of animals in South America, leading to increased interspecific competition for resources such as food, water, and territory. Since the North American species occupied similar niches as the native South American species, this new competition would have reduced the amount of resources available for the original inhabitants.


Question 5: Enzymes and Metabolism

A. Describe a characteristic of an enzyme’s active site that allows it to catalyze a specific chemical reaction.

An enzyme's active site has a unique three-dimensional shape and chemical environment that is highly specific and complementary to its substrate. This specificity, often described by the induced-fit model, ensures that only the correct substrate molecule can bind to the active site, allowing the enzyme to catalyze a specific reaction.

B. Based on Figure 1, explain how the binding of amino acid A to enzyme 1 is regulated by amino acid B.

The pathway shows that amino acid B, the end product, regulates enzyme 1 through feedback inhibition. Amino acid B binds to an allosteric site on enzyme 1, changing the enzyme's conformation. This change alters the shape of the active site, making it less effective at binding to its substrate, amino acid A. This prevents the pathway from producing more of amino acid B when it is already abundant.

C. Using the information in Figure 1, identify the product of the reaction catalyzed by enzyme 2.

The product of the reaction catalyzed by enzyme 2 is Intermediate Y.

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

A significant change in pH outside of the optimal range for enzyme 3 could cause the enzyme to denature. This denaturation would alter the three-dimensional structure of the enzyme, specifically changing the shape of its active site. If the active site shape is changed, it will no longer be complementary to its substrate, intermediate Y, and will be unable to catalyze the conversion of intermediate Y to amino acid B, thus halting the production of the final product.


Question 6: Meiosis and Genetics

A. Based on Figure 1A, identify the fly genotype in which the average percent of metaphase cells with ALD-associated filaments is close to 12%.

The fly genotype in which the average percent of metaphase cells with ALD-associated filaments is close to 12% is ald1/del.

B. Based on Figure 1B, describe the difference in ALD protein production between gamete-forming metaphase cells of flies with the genotype ald3/ald23 and flies with the genotype ald23/del.

Based on Figure 1B, flies with the genotype ald3/ald23 produce a visibly greater amount of ALD protein than flies with the genotype ald23/del, as indicated by the thicker protein band for the ald3/ald23 genotype.

C. Use the data in Figures 1A and 1B to support the scientists’ hypothesis.

The data from the WT/del genotype support the hypothesis. In Figure 1B, the WT/del flies produce about half as much ALD protein as the wild-type (WT/WT) flies. Despite this 50% reduction in protein level, Figure 1A shows that the percent of metaphase cells with ALD-associated filaments in WT/del flies (around 85%) is nearly identical to that of the wild-type flies (around 90%), and the error bars overlap. This indicates that a normal amount of filaments can be produced even with half the amount of ALD protein.

D. For gamete-forming metaphase cells of the WT/del and ald1/del flies, explain why the phenotypes observed in Figure 1A differ even though the amount of ALD protein produced (Figure 1B) does not.

Even though both genotypes produce a similar, reduced amount of ALD protein, the phenotypes differ because the structure and function of the protein produced by the ald1 allele are different from the wild-type (WT) allele. The ald1 allele is a mutation, not a deletion, meaning it produces a full-length but likely altered or non-functional protein. In the WT/del flies, all the protein present is functional wild-type protein, which is sufficient for near-normal filament formation. In the ald1/del flies, the protein produced by the ald1 allele may interfere with the function of the remaining cellular machinery or may be simply non-functional, leading to a much lower percentage of filament formation, even with a similar amount of total protein present.

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