AP Biology Free Response Questions (FRQ) – Past Prompts

Score Higher on AP Biology 2024: Past FRQ Prompts & Practice 33 min read • July 11, 2024

Dylan Black & Dalia Savy

Overview

We’ve compiled a comprehensive list of past AP Biology Free Response Questions (FRQs) to help you maximize your exam prep! The AP Bio FRQs make up 60% of the exam, which includes 2 long questions and 4 short questions. It’s important to understand the question styles and rubrics as you practice, so you’re fully prepared for whatever comes your way. Use this guide to navigate and practice with real FRQ prompts, understand how they are graded, and learn how to structure your answers effectively.

By practicing with previously released free-response questions, you’ll build critical-thinking and analytical skills that will prepare you for the real exam. These prompts are designed to help you connect different biological concepts and apply them to real-life scenarios. Practicing will also familiarize you with the exam structure and reduce surprises on test day.

All credit goes to College Board for these released questions.

2019 FRQs

Long FRQ #1: Gene Expression and Regulation, Ecology (Gene Expression and Symbiosis)

Auxins are plant hormones that coordinate aspects of root growth and development. One specific auxin, Indole-3-acetic acid (IAA), is synthesized from tryptophan through an enzyme encoded by the gene Trp-T. A follow-up step is catalyzed by another enzyme encoded by gene YUC. Here’s a glimpse at some parts of the questions:

1. Transcription & Enzyme Malfunction: You were asked to circle an arrow representing transcription and identify a molecule that would be absent if enzyme YUC were nonfunctional.

2. Base Pair Deletion Impact: Predict how the deletion of a single base pair in the coding region of gene Trp-T would affect IAA production.

3. Feedback Mechanisms: Explain a feedback mechanism that prevents excess IAA production.

4. Rhizobacteria Relationship: Identify the ecological relationship between rhizobacteria and plants and describe one advantage of IAA production to bacteria.

Mean Score: 4.53/10

Long FRQ #2: Competition and Osmoregulation

This question looked at the growth of two different aquatic, plant-eating protists and analyzed their competitive relationships in controlled environments:

1. Population Dynamics: Graph the growth of species A and B under different treatment groups.

2. Competition Analysis: Provide reasoning for reduced initial population sizes and use the data to support the claim that species A and B compete for the same food source.

3. Environmental Impact: Predict how a reduced solute concentration would impact the activity of contractile vacuoles in these protists.

Mean Score: 4.53/10

Short FRQs

Short FRQ #1: Cellular Energetics, Heredity (Cellular Respiration and Sex-Linked Inheritance)

This question focused on the pyruvate dehydrogenase complex (PDC), a critical enzyme in cellular respiration.

1. Cellular Location: Identify where PDC is most active in the cell.

2. Impact of PDC Deficiency: Make a claim about how PDC deficiency would affect NADH production in both glycolysis and the Krebs cycle.

3. Genetic Inheritance: Given that the gene is on the X chromosome, calculate the probability that offspring of specific parental combinations will have PDC deficiency.

Mean Score: 1.68/4

Short FRQ #2: Cell Signaling (Neurotransmitters)

Acetylcholine and its effect on postsynaptic neurons were explored in this question:

1. Neurotoxin Effect: Describe the immediate effect of a neurotoxin on action potential frequency in the postsynaptic neuron.

2. Effectiveness of Proposed Models: Predict whether two proposed models using acetylcholinesterase would mitigate the effects of the neurotoxin, and explain why.

Mean Score: 1.46/4

2018 FRQs

Long FRQ #1: Natural Selection (Cladograms and Evolutionary Relationships)

This prompt asked about polar bears, brown bears, and other related bear species. You were provided data on mitochondrial DNA and asked to:

1. Phylogenetic Tree: Estimate the age of the common ancestor for brown bears and construct a phylogenetic tree based on provided data.

2. Evolutionary Argument: Justify which type of data (mitochondrial DNA or protein sequences) is better for phylogenetic studies.

Mean Score: 4.73/10

Short FRQ #1: Ecology (Seagrass Pollination)

Seagrasses are aquatic plants that reproduce by using sticky pollen to achieve fertilization.

1. Experimental Setup: Use symbols to illustrate the experimental design to test if invertebrates can act as pollinators in the absence of water flow.

2. Dependent Variable: Identify the dependent variable and predict the results that support the claim.

Mean Score: 2.38/4

Short FRQ #2: Cell Structure and Function (Cell Transport and Experimental Design)

This question explored insecticide resistance in bedbugs, specifically looking at gene deletions and how they affect resistance:

1. Gene Function & Resistance: Identify a control strain and explain the effect of deleting certain genes related to insecticide resistance.

Mean Score: 1.41/4

Short FRQ #2 – Cell Structure and Function (Cell Transport and Experimental Design)

The Common Bedbug’s Insecticide Resistance
The common bedbug (Cimex lectularius) is increasingly resistant to insecticides. Bedbugs possess several genes suspected of contributing to this resistance, including P450, Abc8, and Cps. Researchers deleted one or more of these genes in different strains of bedbugs (as indicated in Figure 1) and treated them with beta-cyfluthrin. All strains were genetically identical, except for the deleted gene(s), and had equal fitness in the absence of beta-cyfluthrin. The percent survival of each strain following beta-cyfluthrin treatment is shown in Figure 1.

1. Identify the control strain in the experiment. Use the means and confidence intervals in Figure 1 to justify the claim that Abc8 is effective at providing resistance to beta-cyfluthrin.

2. Explain how P450 and Abc8 deletion impacts survival. P450 encodes an enzyme that detoxifies insecticides, Abc8 encodes a transporter protein that pumps insecticides out of cells, and Cps encodes a structural protein reducing absorption. Explain why the deletion of P450 and Abc8 results in lower survival compared to the deletion of Cps only.

Improvised Explanation:
The control strain in the experiment is the bedbug strain without any genes deleted. Using Figure 1, if we compare the survival rates, we observe that the deletion of Abc8 significantly reduces survival compared to the control, suggesting that Abc8 provides resistance by effectively removing beta-cyfluthrin. When both P450 and Abc8 are deleted, survival is even lower, indicating a loss of detoxification and transport functions, leading to increased sensitivity to insecticides.

Short FRQ #3 – Ecology (Symbiotic Relationships)

Cuckoo and Warbler Relationship
Some birds, like the great spotted cuckoo, lay eggs in the nests of other birds, such as reed warblers. The warbler parents raise the unrelated chicks. Researchers investigated the relationship between warblers and cuckoos in predator-free environments. It was found that nests containing only warblers had higher success rates compared to nests with both warblers and cuckoos.

1. Describe the symbiotic relationship without predators. Explain the relationship between cuckoos and warblers in an environment without predators.

2. Predict relative nest success with or without cuckoos. Draw bars to predict nest success probabilities when cuckoos are added or removed.

3. Identify symbiotic relationship with predators. When predators are present, explain the relationship between cuckoos and warblers.

Improvised Explanation:
Without predators, the relationship between cuckoos and warblers is parasitic, as the warblers expend energy raising cuckoo chicks instead of their own. When predators are present, however, the relationship becomes mutualistic, as cuckoo chicks produce a deterrent to protect the nest from predators, increasing the survival of both cuckoo and warbler offspring.

Short FRQ #4 – Cell Structure and Function (Cellular Transport)

Cystic Fibrosis and CFTR Protein
Cystic fibrosis is linked to defects in the CFTR protein, a gated ion channel that requires ATP to allow chloride ions (Cl-) to diffuse across the membrane.

1. Draw the pathway of CFTR production. On the provided model of a cell, draw arrows to describe the pathway for the production of a normal CFTR protein, from gene expression to its final location.

2. Identify ribosome location. Identify where in the cell ribosomes synthesize the CFTR protein.

3. Identify cellular location of mutant CFTR. Determine where a CFTR protein with a defective ATP-binding site would likely be found.

Improvised Explanation:
The CFTR protein is synthesized on ribosomes bound to the rough endoplasmic reticulum (ER). After synthesis, it is transported to the Golgi apparatus for processing and then moved to the cell membrane. A CFTR protein with a defective ATP-binding site would most likely fail to reach the cell membrane, remaining stuck in the ER or Golgi, unable to function properly.

Short FRQ #5 – Heredity (Sex-Linked Heredity)

Sex Determination in Tongue Sole Fish
In tongue sole fish, sex determination is influenced by genetics and environmental temperature. Genetically male fish have two Z chromosomes (ZZ), and genetically female fish have one Z and one W chromosome (ZW). At 28°C, ZW fish develop as phenotypic males and can pass on the modified Z chromosome (denoted as Z*).

1. Predict phenotypic males in offspring. Predict the percentage of phenotypic males among F1 offspring from a ZW female and ZZ male at 22°C.

2. Identify genotype and fitness cost. Based on observed ratios, identify the genotype of a male parent and describe one fitness cost to the female of mating with this male.

Improvised Explanation:
At 22°C, ZW offspring develop as females, while ZZ offspring develop as males, resulting in 50% phenotypic males. If a male produces a 2:1 male-to-female ratio, it suggests a ZZ genotype, with some female offspring not surviving. A fitness cost to the female is that mating with such a male reduces the number of viable female offspring, impacting population growth.

Short FRQ #6 – Cell Communication and Cell Cycle

Acetylcholine Receptor (AChR) in Muscle Contraction
Acetylcholine (ACh) is a neurotransmitter that binds to receptor proteins (AChRs) at synapses between neurons and skeletal muscle cells, causing sodium ions to enter the muscle cells and initiate contraction. Nicotine can also bind to certain AChR types.

1. Describe differences between AChR types. Describe the difference in the structure and function between AChR type 1 and type 2.

2. Effect of acetylcholinesterase inhibition. Acetylcholinesterase breaks down acetylcholine. Describe what happens if acetylcholinesterase is inhibited in muscle cells with AChR type 2.

Improvised Explanation:
AChR type 1 may have a higher affinity for acetylcholine, whereas type 2 may also respond to other molecules, like nicotine. Inhibiting acetylcholinesterase prevents the breakdown of acetylcholine, leading to continuous stimulation of AChR type 2, which can cause prolonged muscle contraction and potentially lead to muscle fatigue or paralysis.

Short FRQ #6 (2016)
Cell Communication and Cell Cycle (Cell Communication, Immune System)

An individual has lost the ability to activate B cells and mount a humoral immune response.

1. Propose ONE direct consequence of the loss of B-cell activity on the individual’s humoral immune response to the initial exposure to a bacterial pathogen.

   • Answer: One direct consequence is that the individual will be unable to produce specific antibodies against the bacterial pathogen. Without B cells to differentiate into plasma cells, no antibodies are synthesized to target and neutralize the bacteria during initial exposure.

2. Propose ONE direct consequence of the loss of B-cell activity on the speed of the individual’s humoral immune response to a second exposure to the bacterial pathogen.

   • Answer: The loss of B-cell activity means that no memory B cells are formed during the first exposure. As a result, during a second exposure, the immune response will be much slower, resembling a primary response rather than the rapid and amplified secondary immune response typically seen when memory cells are present.

3. Describe ONE characteristic of the individual’s immune response to the bacterial pathogen that is not affected by the loss of B cells.

   • Answer: The individual’s innate immune response is not affected by the loss of B cells. This response includes the activation of macrophages, neutrophils, and other cells that phagocytose pathogens and release inflammatory signals to help control the infection.

Short FRQ #5 (2015)
Cell Communication and Cell Cycle (Cell Signaling, Transmissions)

Smell perception in mammals involves the interactions of airborne odorant molecules from the environment with receptor proteins on the olfactory neurons in the nasal cavity. The binding of odorant molecules to the receptor proteins triggers action potentials in the olfactory neurons and results in the transmission of information to the brain. Mammalian genomes typically have approximately 1,000 functional odorant-receptor genes, each encoding a unique odorant receptor.

1. Describe how the signal is transmitted across the synapse from an activated olfactory sensory neuron to the interneuron that transmits the information to the brain.

   • Answer: The signal is transmitted across the synapse when the activated olfactory sensory neuron releases neurotransmitters. These chemical signals diffuse across the synaptic cleft and bind to receptors on the postsynaptic membrane of the interneuron, initiating an action potential that propagates the signal to the brain.

2. Explain how the expression of a limited number of odorant receptor genes can lead to the perception of thousands of odors. Use the evidence about the number of odorant receptor genes to support your answer.

   • Answer: The perception of thousands of odors is possible due to the combinatorial coding of olfactory receptors. Each odorant receptor can bind multiple odorant molecules with different affinities, and each odorant molecule can activate multiple receptors. This overlap and combination allow the brain to interpret a vast range of odors by analyzing the specific pattern of activated receptors.

Short FRQ #4 (2015)
Ecology (Population Dynamics)

In an attempt to rescue a small isolated population of snakes from decline, a few male snakes from several larger populations of the same species were introduced into the population in 1992. The snakes reproduce sexually, and there are abundant resources in the environment. The figure below shows the results of a study of the snake population both before and after the introduction of the outside males. In the study, the numbers of captured snakes indicate the overall population size.

1. Describe ONE characteristic of the original population that may have led to the population’s decline in size between 1989 and 1993.

   • Answer: One characteristic that may have led to the population’s decline is inbreeding, which is more likely in a small, isolated population. Inbreeding can increase the prevalence of deleterious alleles, reducing genetic diversity and making the population more susceptible to disease or environmental changes.

2. Propose ONE reason that the introduction of the outside males rescued the snake population from decline.

   • Answer: The introduction of outside males increased genetic diversity through gene flow, which can improve fitness by reducing the likelihood of inbreeding depression and increasing the population’s ability to adapt to changing conditions.

3. Describe how the data support the statement that there are abundant resources in the environment.

   • Answer: The population size increased following the introduction of new males, indicating that food and other necessary resources were sufficient to support the growth and expansion of the population.

Short FRQ #3 (2015)
Cellular Energetics (Photosynthesis)

Phototropism in plants is a response in which a plant shoot grows toward a light source. The results of five different experimental treatments from classic investigations of phototropism are shown above.

1. Give support for the claim that the cells located in the tip of the plant shoot detect the light by comparing the results from treatment group I with the results from treatment group II and treatment group III.

   • Answer: In treatment group I, the plant grew toward the light, indicating that the tip detected the light. In treatment group II, where the tip was covered, no phototropic response occurred, showing that the tip is necessary for light detection. In treatment group III, where the lower part of the plant was covered, the plant still grew toward the light, indicating that the tip, not the lower part, is responsible for detecting light.

2. In treatment groups IV and V, the tips of the plants are removed and placed back onto the shoot on either a permeable or impermeable barrier. Using the results from treatment groups IV and V, describe TWO additional characteristics of the phototropism response.

   • Answer: The response in treatment group IV, where the tip was placed on a permeable barrier and phototropism occurred, indicates that a chemical signal from the tip is responsible for the response. In treatment group V, where no response occurred due to the impermeable barrier, it suggests that the chemical signal must move down the shoot to initiate the phototropic response.

Short FRQ #2 (2015)
Cell Communication and Cell Cycle (Mitosis, Meiosis)

Both mitosis and meiosis are forms of cell division that produce daughter cells containing genetic information from the parent cell.

1. Describe TWO events that are common to both mitosis and meiosis that ensure the resulting daughter cells inherit the appropriate number of chromosomes.

   • Answer: Both mitosis and meiosis involve DNA replication during the S phase of the cell cycle to ensure that each daughter cell inherits a complete set of chromosomes. Additionally, both processes include the alignment of chromosomes along the metaphase plate to ensure accurate segregation during cell division.

2. The genetic composition of daughter cells produced by mitosis differs from that of the daughter cells produced by meiosis. Describe TWO features of the cell division processes that lead to these differences.

   • Answer: In mitosis, daughter cells are genetically identical to the parent cell because there is no crossing over or independent assortment. In contrast, meiosis includes crossing over during prophase I and independent assortment during metaphase I, resulting in daughter cells with unique combinations of alleles and half the chromosome number of the parent cell.

Short FRQ #6 – 2017: Cell Communication and Cell Cycle (Cell Communication, Immune System)

1. Problem Description: An individual has lost the ability to activate B cells and mount a humoral immune response.

   • Propose ONE direct consequence of the loss of B-cell activity on the individual’s humoral immune response to the initial exposure to a bacterial pathogen: Without the activation of B cells, the individual will be unable to produce antibodies specific to the bacterial pathogen upon initial exposure. This will result in a significantly reduced capacity to neutralize or mark the pathogen for destruction by other immune cells, impairing the body’s ability to fight the infection.

   • Propose ONE direct consequence of the loss of B-cell activity on the speed of the individual’s humoral immune response to a second exposure to the bacterial pathogen: The loss of B-cell activity also means there is no production of memory B cells during the first exposure, which would normally facilitate a faster and more effective response to a subsequent exposure. Therefore, the response during a second exposure will be slow, as if the individual is encountering the pathogen for the first time.

   • Describe ONE characteristic of the individual’s immune response to the bacterial pathogen that is not affected by the loss of B cells: The cell-mediated immune response, involving T cells, would not be directly affected by the loss of B-cell activity. Cytotoxic T cells can still identify and kill infected host cells, and helper T cells can still activate other immune components, allowing for some level of pathogen control.

Short FRQ #1 – 2016: Cellular Energetics

1. Problem Description: The graph illustrates the percent dry weight of different parts of a particular annual plant from early May to late August. The percent dry weight can be used to estimate the amount of energy a plant uses to produce its leaves, vegetative buds, stems, roots, and reproductive parts.

   • Identify the direct source of the energy used for plant growth during the first week of May, and identify the part of the plant that grew the most during the same period: The direct source of energy used for plant growth during the first week of May is photosynthesis, which uses light energy to synthesize carbohydrates. During this period, the leaves of the plant exhibited the most growth, as they are essential for capturing sunlight for photosynthesis.

   • Estimate the percent of the total energy that the plant has allocated to the growth of leaves on the first day of July based on the graph: On the first day of July, the plant allocated approximately 40% of its energy to the growth of leaves, as indicated by the graph showing the percentage of dry weight dedicated to this organ.

   • Propose ONE evolutionary advantage of the energy allocation strategy in annual plants compared with that in perennial plants: Annual plants allocate a greater percentage of their energy to the growth of reproductive parts because they must complete their life cycle within a single growing season. This rapid reproductive strategy increases their chances of producing offspring before adverse conditions (e.g., frost) occur, which is particularly advantageous in environments with short growing seasons.

Short FRQ #2 – 2016: Cell Communication and Cell Cycle (Gene Regulation)

1. Problem Description: The figure represents the process of expression of gene X in a eukaryotic cell.

   • Describe the modification that most likely resulted in the 8 kb difference in length of the mature mRNA molecule. Identify in your response the location in the cell where the change occurs: The modification responsible for the 8 kb difference is RNA splicing, where introns (non-coding regions) are removed from the primary transcript. This occurs in the nucleus of the cell, resulting in a shorter mature mRNA molecule composed only of exons (coding regions).

   • Predict the length of the mature gene X mRNA if the full-length gene is introduced and expressed in prokaryotic cells. Justify your prediction: If the full-length gene X is introduced and expressed in prokaryotic cells, the mature mRNA will be 15 kb in length. This is because prokaryotic cells do not have a nucleus and do not undergo RNA splicing; therefore, the entire primary transcript, including both introns and exons, will be translated.

Short FRQ #3 – 2016: Ecology (Ecological Relationships)

1. Problem Description: The graph shows the mass of plants from two different species over time. The plants were grown while attached to each other and were later separated at the time indicated by the vertical line in the graph.

   • Graph the predicted shape of the plant-mass lines after separation of the two plants if they were in an obligate mutualistic relationship: If the plants were in an obligate mutualistic relationship, both plant masses would likely decline after separation, as each plant depends on the other for a critical resource or service. The decline in mass will be evident as neither plant can thrive without the other.

   • Graph the predicted shape of the plant-mass lines after separation if the species 2 plant was a parasite of the species 1 plant: If species 2 was a parasite of species 1, the mass of species 1 would increase after separation, as it is no longer being negatively affected by the parasitic relationship. Conversely, species 2 would show a significant decline in mass, as it can no longer exploit species 1 for nutrients or resources.

Short FRQ #4 – 2016: Cell Communication and Cell Cycle (Gene Regulation)

1. Problem Description: Living and dead organisms continuously shed DNA fragments, known as environmental DNA (eDNA), into the environment. To detect eDNA fragments in the environment, the polymerase chain reaction (PCR) can be used to amplify specific eDNA fragments.

   • Justify the use of eDNA sampling as an appropriate technique for detecting the presence of silver carp in an environment where many different species of fish are found: eDNA sampling is appropriate because it is non-invasive and allows for the detection of specific species without the need to physically capture or observe them. This is particularly useful in environments with diverse fish populations, as eDNA can provide species-specific identification based on genetic sequences.

   • Propose ONE advantage of identifying long eDNA fragments as opposed to short fragments for detecting silver carp: Identifying long eDNA fragments is advantageous because they are more likely to contain unique sequences that can distinguish silver carp from other closely related species, reducing the chances of false positives due to genetic similarity.

   • Provide reasoning other than human error to support the researchers’ claim that the single positive sample was a false positive: The single positive sample could be a false positive due to contamination of the water sample, either from the equipment used or from the presence of DNA from dead fish or fish parts that were transported from other areas by birds or human activities.

Short FRQ #6 – Cell Communication and Cell Cycle (Cell Transport)

Topic: Estrogen Transport and Cancer Treatment
Question Overview:
Estrogens are small hydrophobic lipid hormones that promote cell division and the development of reproductive structures in mammals. Estrogens passively diffuse across the plasma membrane and bind to their receptor proteins in the cytoplasm of target cells.

1. Describe ONE characteristic of the plasma membrane that allows estrogens to passively cross the membrane.
2. In a laboratory experiment, a researcher generates antibodies that bind to purified estrogen receptors extracted from cells. The researcher uses the antibodies in an attempt to treat estrogen-dependent cancers but finds that the treatment is ineffective. Explain the ineffectiveness of the antibodies for treating estrogen-dependent cancers.

Improvised Responses:

1. The plasma membrane has a phospholipid bilayer structure with hydrophobic tails, which allows nonpolar, hydrophobic molecules like estrogen to pass through by simple diffusion.
2. The antibodies used in this experiment are unable to enter the cell to bind to the intracellular estrogen receptors since antibodies are large, hydrophilic proteins. As a result, they cannot effectively prevent the estrogens from activating the receptors within the target cells.

Long FRQ #1 – Gene Regulation and Experimental Design (Marine Mussels and LAPs)

Topic: Leucine Aminopeptidase (LAP) Evolution in Marine Mussels
Question Overview:
Leucine aminopeptidases (LAPs) are enzymes found in all living organisms and are associated with salinity response in the marine mussel, Mytilus edulis. The researchers sampled adult mussels from different locations along the northeast coast of the US, determining the percentage of individuals with a particular allele, lap94, in each population.

1. Construct an appropriately labeled bar graph to illustrate the observed frequencies of the lap94 allele in the study populations.
2. Based on the data, describe the most likely effect of salinity on the frequency of the lap94 allele in the marine mussel populations. Predict the likely lap94 allele frequency at a sampling site between Site 1 and Site 2 in Long Island Sound.
3. Describe how LAP94 activity affects osmolarity and how it helps maintain water balance in mussels.
4. Explain differences in lap94 allele frequency among adult mussel populations at different sites, given that larvae are evenly dispersed throughout the area by water movement.

Improvised Responses:

1. (Graph omitted, but the graph should include proper labeling of allele frequencies for each sample site with a clear legend and title).
2. Higher salinity correlates with higher frequencies of lap94, suggesting positive selection for lap94 in high salinity environments. The likely allele frequency at a site between Site 1 and Site 2 would be intermediate.
3. LAP94 likely helps regulate osmolarity by breaking down proteins and releasing amino acids that help balance cellular osmolarity in high salinity environments, thus maintaining water balance.
4. Despite larvae dispersal, adult survival and reproductive success at each site depend on salinity, selecting for the lap94 allele in higher salinity sites. Local adaptation leads to differences in allele frequency among populations.

Long FRQ #2 – Bacterial Population Growth and Nutrient Regulation

Topic: Bacterial Population Growth in Media with Limiting Nutrients
Question Overview:
Bacteria were grown in a medium with limiting amounts of two nutrients, I and II. A graph shows the population growth over time.

1. Estimate the maximum population density for the culture. Describe what prevents further growth of the bacterial population.
2. Calculate the growth rate between hours 2 and 4.
3. Identify the preferred nutrient source and justify your response. Propose ONE advantage of the nutrient preference.
4. Describe how nutrient I likely regulates the genes for metabolism of both nutrients. Provide TWO reasons the population stops growing between hours 5 and 6.

Improvised Responses:

1. The maximum population density is around 8 × 10⁸ cells/mL. Further growth is prevented by depletion of nutrients I and II, leading to nutrient limitation.
2. (Calculation omitted, but it should show the slope of the growth curve between hours 2 and 4.)
3. Nutrient I is the preferred nutrient as it is utilized first, shown by rapid growth at the beginning of the experiment. The advantage is efficient energy acquisition due to a higher yield of ATP per unit of nutrient I.
4. Nutrient I represses the metabolism of nutrient II by acting as a repressor or feedback inhibitor. Population stops growing due to depletion of both nutrients and accumulation of toxic by-products.

Short FRQ #1 – Cellular Energetics (Annual Plant Growth)

Topic: Energy Allocation in Annual Plants
Question Overview:
A graph shows the percent dry weight of different parts of an annual plant from early May to late August.

1. Identify the direct source of energy for growth in the first week of May and the part of the plant that grew most during this period.
2. Estimate the percent of total energy allocated to leaf growth on July 1st.
3. Propose ONE evolutionary advantage of the energy allocation strategy in annual plants compared to perennial plants.

Improvised Responses:

1. The direct energy source is stored carbohydrates, and the part of the plant that grew the most during the first week of May is the leaves, as they capture light for photosynthesis.
2. (Estimation omitted, but the answer should reference the graph values for leaf weight as a percentage of total plant weight.)
3. An evolutionary advantage is rapid reproduction, allowing annual plants to complete their life cycle in one season and take advantage of temporary favorable conditions.

Short FRQ #2 – Gene Regulation (Modification of mRNA)

Topic: Gene Expression in Eukaryotic Cells
Question Overview:
The figure represents the process of expression of gene X in a eukaryotic cell.

1. The primary transcript is 15 kb, but the mature mRNA is 7 kb. Describe the modification that resulted in the 8 kb difference and identify where this change occurs.
2. Predict the length of the mature gene X mRNA if expressed in prokaryotic cells. Justify your prediction.

Improvised Responses:

1. The modification is splicing, where introns are removed from the primary transcript. This occurs in the nucleus.
2. The mature mRNA in prokaryotes would be 15 kb, as prokaryotes lack the machinery for splicing introns out of the transcript.
   
   

   Short FRQ #3 – Ecology (Ecological Relationships)

   Topic: Mutualism and Parasitism in Plants
   Question Overview:
   A graph shows the mass of plants from two different species over time while they are attached to each other. The plants were separated at the indicated time.

   1. On Template 1, graph the predicted shape of plant-mass lines if the plants were in an obligate mutualistic relationship.
   2. On Template 2, graph the predicted shape of plant-mass lines if the species 2 plant was a parasite of the species 1 plant.
   3. Justify each of your predictions.

   Improvised Responses:

   1. In an obligate mutualistic relationship, both plant species will decline in mass after separation as they rely on each other for essential resources. The lines should both show a decrease after the separation.
   2. In a parasitic relationship, species 1 (the host) will increase in mass after separation, while species 2 (the parasite) will decrease in mass after separation. Species 1 benefits from no longer having its resources taken.
   3. In mutualism, both species benefit from the association, leading to a decline in mass when separated. In parasitism, the host’s mass increases due to the absence of resource drainage by the parasite.

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   Short FRQ #4 – Cell Communication and Cell Cycle (Gene Regulation)

   Topic: Environmental DNA (eDNA) and Silver Carp Detection
   Question Overview:
   Environmental DNA (eDNA) sampling was used to detect the presence of invasive silver carp in Lake Michigan.

   1. Justify the use of eDNA sampling to detect the presence of silver carp in an environment with other fish species.
   2. Propose ONE advantage of identifying long eDNA fragments over short fragments.
   3. Researchers detected eDNA specific to silver carp in one water sample but concluded it was a false positive. Provide reasoning other than human error to support the researchers’ claim.

   Improvised Responses:

   1. eDNA sampling is effective because it can detect trace DNA left behind by fish without needing to physically observe or capture the organisms, which is especially useful in a mixed species environment.
   2. Longer eDNA fragments provide greater specificity, reducing the risk of false positives due to similarity with DNA from closely related species.
   3. Silver carp DNA could have been transported into the lake via water currents, boats, or other vectors, leading to the detection of DNA without the presence of live fish.

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   Short FRQ #5 – Heredity (Genetic Crosses and Linkage)

   Topic: Genetic Crosses in Diploid Plants
   Question Overview:
   In a plant species, the diploid number is 4 (2n = 4). Flower color is controlled by a single gene with green (G) dominant over purple (g). Height is controlled by a separate gene with dwarf (D) dominant over tall (d).

   1. Construct a diagram of the four possible normal products of meiosis produced by the F1 progeny. Show chromosomes and allele(s).
   2. Predict the phenotypes and ratios in the offspring of a testcross between an F1 individual and a ggdd individual.
   3. If the two genes were genetically linked, describe how the proportions of phenotypes in the offspring of the testcross would differ from those predicted.

   Improvised Responses:

   1. (Diagram omitted, but it should show each gamete with one copy of chromosome 1 and one of chromosome 2, each carrying one allele of each gene).
   2. The offspring would have a phenotypic ratio of 1 green/dwarf : 1 green/tall : 1 purple/dwarf : 1 purple/tall.
   3. If the genes are linked, the phenotypes will not follow a typical Mendelian ratio. Instead, offspring with parental combinations (green/dwarf and purple/tall) will occur more frequently than recombinants (green/tall and purple/dwarf).

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   Short FRQ #6 – Cellular Energetics (Prolactin Release and Exercise)

   Topic: Exercise and Prolactin Release
   Question Overview:
   Researchers measured blood prolactin in eight adult males before and after one hour of vigorous exercise. As a control, prolactin was measured again without exercise.

   1. Justify the use of the without-exercise treatment as a control.
   2. Using evidence, determine if prolactin release changes after exercise. Justify your answer.

   Improvised Responses:

   1. The without-exercise treatment serves as a baseline to compare changes in prolactin levels, ensuring that observed effects are due to exercise and not normal daily fluctuations.
   2. The data show an increase in prolactin after exercise compared to the control treatment, suggesting that exercise stimulates prolactin release. This conclusion is supported by a statistically significant increase in prolactin levels only in the exercise condition.

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   Long FRQ #1 – Circadian Rhythms in Mice (Gene Regulation and Ecology)

   Topic: Circadian Rhythms in Mice and Light-Dark Exposure
   Question Overview:
   Mice were exposed to 12 hours of light and 12 hours of dark (L12

    

   ) for 14 days, then switched to continuous darkness (DD). Their activity was recorded.

   1. Describe one role of each of the following in responding to light-dark stimuli:
      • A photoreceptor in the retina
      • The brain
      • A motor neuron
   2. Describe the activity pattern of mice under L12
       
      .
   3. Compare the activity pattern of mice in L12
       
      to DD to support the researchers’ claim about circadian rhythm.
   4. Predict the activity pattern of mutant mice lacking a circadian gene under L12
       
      and DD.
   5. Describe two features of a predator-prey relationship model that explain the evolution of this activity pattern.

   Improvised Responses:
   1.

   • A photoreceptor in the retina detects light and converts it into neural signals.
   • The brain processes these signals and regulates sleep-wake cycles.
   • A motor neuron carries signals from the brain to muscles to initiate activity.
   2. Mice are primarily active during the dark period, exhibiting a nocturnal behavior pattern.
   3. Under DD conditions, mice show a free-running rhythm slightly longer or shorter than 24 hours, supporting the claim that their circadian rhythm does not follow an exact 24-hour cycle.
   4. Mutant mice would lose their rhythmic activity under DD, but maintain activity under L12
       
      , showing that light can override the circadian rhythm.
   5. The predator-prey model includes nocturnal activity to avoid diurnal predators and reduced movement during the day, minimizing the risk of detection by visual predators.
      
      
      

      Long FRQ #2 – Cellular Energetics (Cellular Respiration and ATP)

      Topic: Cellular Respiration Processes and ATP Production
      Question Overview:
      Cellular respiration involves glycolysis, the Krebs cycle, and the electron transport chain (ETC). Carbohydrates and metabolites are oxidized to produce ATP.

      1. Describe one contribution of each process to ATP synthesis:
         • Glycolysis and pyruvate oxidation
         • Krebs cycle
         • Formation of a proton gradient by the ETC
      2. Use the following observations to justify the claim that glycolysis is an ancient metabolic pathway:
         • Nearly all organisms perform glycolysis.
         • Glycolysis occurs under anaerobic conditions.
         • Glycolysis occurs in the cytosol.
      3. Calculate the amount of energy from hydrolysis of 30 moles of ATP. Calculate the efficiency of ATP production from glucose metabolism if 30 ATP are produced. Describe the fate of excess energy.

      Improvised Responses:
      1.

      • Glycolysis and pyruvate oxidation produce ATP directly through substrate-level phosphorylation and generate NADH for the ETC.
      • The Krebs cycle generates NADH and FADH₂, which are used to produce ATP via oxidative phosphorylation in the ETC.
      • The ETC creates a proton gradient across the mitochondrial membrane, which drives ATP synthesis via chemiosmosis.
      2. • Nearly all organisms performing glycolysis suggest it evolved early and was conserved.
         • Glycolysis occurring anaerobically indicates it evolved before significant atmospheric oxygen was present.
         • Glycolysis occurring in the cytosol shows that it predates the development of specialized organelles like mitochondria.
      3. The energy from hydrolysis of 30 moles of ATP is $30 \times 7.3 \text{ kcal} = 219 \text{ kcal}$. The efficiency is calculated as $\frac{219}{686} \times 100 = 31.9\%$. Excess energy is released as heat, which helps maintain body temperature.

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      Short FRQ #1 – Natural Selection (Phylogenetic Trees)

      Topic: Constructing Phylogenetic Trees with Amino Acid Data
      Question Overview:
      The amino acid sequence of cytochrome c was determined for five vertebrate species. Differences in sequences are shown in a table.

      1. Create a phylogenetic tree using the data, and explain the placement of the species least related to others.
      2. State whether morphological data or amino acid sequences better represent true evolutionary relationships and justify your answer.

      Improvised Responses:

      1. The species with the most differences in its amino acid sequence compared to others should be placed farthest from the root, indicating it is least closely related.
      2. Amino acid sequence data better represents true evolutionary relationships because genetic differences accumulate at a consistent rate and provide direct information about shared ancestry, whereas morphological data can be influenced by convergent evolution.

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      Short FRQ #2 – Cell Communication and Cell Cycle (Mitosis and Meiosis)

      Topic: Comparing Mitosis and Meiosis
      Question Overview:
      Mitosis and meiosis produce daughter cells containing genetic information from parent cells.

      1. Describe two events common to both mitosis and meiosis that ensure the resulting daughter cells inherit the correct number of chromosomes.
      2. Describe two features of mitosis and meiosis that lead to different genetic compositions in the daughter cells.

      Improvised Responses:

      1. Both processes involve chromosome duplication during interphase, ensuring each chromosome is replicated. Additionally, both mitosis and meiosis feature chromosome alignment and separation, ensuring chromosomes are properly distributed to daughter cells.
      2. Meiosis involves crossing over during prophase I, which creates genetic variation. Also, meiosis results in two rounds of division, reducing chromosome number by half, while mitosis maintains the diploid number.

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      Short FRQ #3 – Cellular Energetics (Photosynthesis and Phototropism)

      Topic: Plant Growth Toward Light
      Question Overview:
      Phototropism in plants leads to shoots growing towards light.

      1. Support the claim that cells in the tip of the plant shoot detect light by comparing treatment groups I, II, and III.
      2. Describe two additional characteristics of the phototropism response based on treatments IV and V.

      Improvised Responses:

      1. Treatment group I (control) shows normal growth towards light, while treatment group II (with tip removed) shows no response. Treatment group III (tip covered) also shows no growth towards light, suggesting cells in the tip detect light.
      2. In treatment IV, the tip was replaced on a permeable barrier, and the plant grew towards light, indicating that a signal from the tip travels downward. In treatment V, the tip was replaced on an impermeable barrier, and no growth occurred, suggesting the signal is a diffusible chemical.

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      Short FRQ #4 – Ecology (Population Dynamics)

      Topic: Population Rescue of Isolated Snake Population
      Question Overview:
      An isolated snake population was rescued by introducing males from larger populations.

      1. Describe one characteristic of the original population that may have led to its decline.
      2. Propose one reason the introduction of outside males helped rescue the population.
      3. Explain how the data support the statement that resources were abundant.

      Improvised Responses:

      1. The original population may have experienced inbreeding depression, which reduces genetic diversity and increases the frequency of harmful alleles.
      2. The introduction of males increased genetic diversity, leading to healthier offspring with higher fitness.
      3. The data show that after introducing outside males, the population size increased significantly, which indicates that resources were not a limiting factor and allowed the population to grow.

   Short FRQ #5 – Cell Communication and Cell Cycle (Signal Transmission)

   Topic: Smell Perception in Mammals
   Question Overview:
   Odorant molecules interact with receptor proteins on olfactory neurons, triggering action potentials and transmitting information to the brain.

   1. Describe how the signal is transmitted across the synapse from an activated olfactory sensory neuron to the interneuron transmitting information to the brain.
   2. Explain how the limited number of odorant receptor genes can lead to the perception of thousands of odors.

   Improvised Responses:

   1. When an action potential reaches the axon terminal of the olfactory sensory neuron, synaptic vesicles release neurotransmitters into the synaptic cleft. These neurotransmitters bind to receptors on the postsynaptic membrane of the interneuron, generating an action potential that continues the signal transmission to the brain.
   2. Combinations of activated receptor proteins generate unique patterns of action potentials, allowing the brain to distinguish among thousands of odors. The limited number of odorant receptor genes can form many combinations, similar to how letters form different words.

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   Short FRQ #6 – Cell Communication and Immune System (B-Cell Deficiency)

   Topic: Impact of B-Cell Deficiency on Immune Response
   Question Overview:
   An individual has lost the ability to activate B cells and mount a humoral immune response.

   1. Propose one direct consequence of losing B-cell activity for the initial exposure to a bacterial pathogen.
   2. Propose one direct consequence for the speed of the response to a second exposure to the same pathogen.
   3. Describe one aspect of the immune response to the pathogen that is not affected by the loss of B cells.

   Improvised Responses:

   1. The loss of B-cell activity would prevent the individual from producing specific antibodies, making it harder to neutralize or eliminate the bacterial pathogen during the initial exposure.
   2. The memory B cells would not be generated, leading to a delayed or absent antibody response to a second exposure to the same pathogen.
   3. The innate immune response, such as phagocytosis by macrophages and activation of T cells, would still occur as it does not depend on B cells.

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   Long FRQ #1 – Experimental Design and Cellular Energetics (Pollination and Bee Behavior)

   Topic: Impact of Caffeine in Nectar on Bee Behavior
   Question Overview:
   Researchers studied the effect of caffeine in nectar on bee memory and behavior.

   1. Construct a graph to illustrate the effect of caffeine on bees’ probability of revisiting a nectar source.
   2. Describe the effect of caffeine on short-term and long-term memory of a nectar source.
   3. Design an experiment using artificial flowers to investigate potential negative effects of increasing caffeine concentrations on bee visits. Identify the null hypothesis, a control, and predicted results to reject the null hypothesis.
   4. Propose one benefit for plants that produce nectar with caffeine and lower sugar content, and one cost for bees.

   Improvised Responses:

   1. Graph should have time on the x-axis and the probability of revisit on the y-axis, with lines for caffeine and no-caffeine groups.
   2. Caffeine improved short-term memory retention, increasing the probability of revisiting a nectar source within 10 minutes. It also enhanced long-term memory, as evidenced by increased revisits after 24 hours.
   3. Null hypothesis: Increasing caffeine concentrations do not affect the number of floral visits by bees. Control: Artificial flowers without caffeine. Predicted results: Bees visit fewer flowers with high caffeine, allowing rejection of the null hypothesis if significant.
   4. Benefit for plants: Caffeine attracts bees and improves their memory, increasing pollination rates. Cost for bees: They may become less efficient in finding the best nectar sources due to reduced sugar intake.

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   Long FRQ #2 – Ecology (Ecological Succession and Seed Germination)

   Topic: Seed Germination Regulation by Smoke Compounds
   Question Overview:
   Researchers studied how smoke compounds regulate seed germination after a fire.

   1. Provide support using data for claims about the effect of KAR and TMB on seed germination timing and percentage.
   2. Make claims about the effect of rinsing on KAR and TMB binding to receptors. Identify appropriate treatment groups.
   3. Describe an advantage of KAR and TMB regulation in ecosystems with regular fires.

   Improvised Responses:

   1. KAR alone led to earlier seed germination compared to controls, indicating it affects timing. TMB alone increased the percentage of seeds germinated compared to the control.
   2. Rinsing after KAR treatment reduced germination, suggesting that rinsing removes KAR from its receptor. The appropriate treatment groups are the rinsed vs. non-rinsed KAR and TMB groups.
   3. KAR regulation ensures rapid germination following a fire, giving plants a competitive advantage in colonizing new spaces. TMB regulation enhances the percentage of seeds germinating, increasing the chances of plant establishment.

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   Short FRQ #1 – Cellular Energetics (Plant Growth Energy Allocation)

   Topic: Energy Use in Annual Plant Growth
   Question Overview:
   The graph shows percent dry weight of different plant parts from early May to late August.

   1. Identify the direct energy source for plant growth in early May and the plant part with the most growth during this time.
   2. Estimate the percent of energy allocated to leaves on the first day of July.
   3. Propose an evolutionary advantage of the energy allocation strategy in annual plants compared to perennials.

   Improvised Responses:

   1. The direct source of energy for plant growth in early May is sunlight, which powers photosynthesis. The part of the plant that grew the most during this time is the leaves.
   2. The estimated percent of energy allocated to leaf growth on July 1 is approximately 25%.
   3. Annual plants allocate more energy to reproductive structures to ensure successful seed production within their short life cycle, maximizing reproduction in a limited time compared to perennials.