AP Biology Free Response Questions (FRQ) – Past Prompts

Overview

We’ve compiled a list of a bunch of the AP Biology past prompts! The AP Bio FRQs are 60% of the exam including 2 long questions and 4 short questions. It’s important that you understand the rubrics and question styles going into the exam. Use this list to practice! 

By practicing with previously released free-response questions (FRQs), you’ll build critical-thinking and analytical skills that will prepare you for the exam. These past prompts have been designed to help you connect concepts and ideas to each other while applying your knowledge to real-life scenarios. You’ll also learn how to tackle the exam in a better format, and you won’t be surprised come test day with certain questions.

All recredit to College Board.

2019 FRQs

👉AP Bio 2019 FRQs

Long FRQ #1

Gene Expression and Regulation, Ecology (Gene Expression and Symbiosis)

1. Auxins are plant hormones that coordinate several aspects of root growth and development. Indole-3-acetic acid (IAA) is an auxin that is usually synthesized from the amino acid tryptophan (Figure 1). Gene Trp-T encodes an enzyme that converts tryptophan to indole-3-pyruvic acid (I3PA), which is then converted to IAA by an enzyme encoded by the gene YUC.1. Circle ONE arrow that represents transcription on the template pathway. Identify the molecule that would be absent if enzyme YUC is nonfunctional.1. Predict how the deletion of one base pair in the fourth codon of the coding region of gene Trp-T would most likely affect the production of IAA. Justify your prediction.1. Explain one feedback mechanism by which a cell could prevent production of too much IAA without limiting I3PA production1. Rhizobacteria are a group of bacteria that live in nodules on plant roots. Rhizobacteria can produce IAA and convert atmospheric nitrogen into forms that can be used by plants. Plants release carbon-containing molecules into the nodules. Based on this information, identify the most likely ecological relationship between plants and rhizobacteria. Describe ONE advantage to the bacteria of producing IAA.1. A researcher removed a plant nodule and identified several “cheater” rhizobacteria that do not produce IAA or fix nitrogen. Describe the evolutionary advantage of being a bacterial cheater in a population composed predominantly of noncheater bacteria. Plants can adjust the amount of carbon-containing molecules released into nodules in response to the amount of nitrogen fixed in the nodule. Predict the change in the bacterial population that would cause the plant to reduce the amount of carbon-containing molecules provided to the nodule.

Long FRQ #2

Units 2 and 8 (Competition and Osmoregulation)

A student studying two different aquatic, plant-eating, unicellular protist species (species A and B) designed an experiment to investigate the ecological relationship between the two species (Table 1).

In treatment group I, the student placed 10 individuals of species A into a container with liquid growth medium and 10 individuals of species B into a separate container with an equal amount of the same liquid growth medium. In treatment group II, the student placed 5 individuals of each species into a single container with the liquid growth medium. The student then maintained the containers under the same environmental conditions and recorded the number of individuals in each population at various time points. The results are shown in Table 2. 

1. The growth curves for species B in group I and for species A in group II (shaded columns) have been plotted on the template. Use the template to complete an appropriately labeled line graph to illustrate the growth of species A in treatment group I and species B in treatment group II (unshaded columns).
2. As shown in the table, the student established treatment group II with 5 individuals of each species. Provide reasoning for the reduced initial population sizes.
3. The student claims that species A and B compete for the same food source. Provide TWO pieces of evidence from the data that support the student’s claim.
4. Predict TWO factors that will most likely limit the population growth of species A in treatment group I.
5. Many protists contain an organelle called a contractile vacuole that pumps water out of the cell. The student repeated the experiment using a growth medium with a lower solute concentration. Predict how the activity of the contractile vacuole will change under the new experimental conditions. Justify your prediction.

Mean Score: 4.53/10

Short FRQ #1

Cellular Energetics, Heredity (Cellular Respiration and Sex Linked Inheritance)

1. The pyruvate dehydrogenase complex (PDC) catalyzes the conversion of pyruvate to acetyl-CoA, a substrate for the Krebs (citric acid) cycle. The rate of pyruvate conversion is greatly reduced in individuals with PDC deficiency, a rare disorder. 1. Identify the cellular location where PDC is most active.1. Make a claim about how PDC deficiency affects the amount of NADH produced by glycolysis AND the amount of NADH produced by the Krebs (citric acid) cycle in a cell. Provide reasoning to support your claims based on the position of the PDC-catalyzed reaction in the sequence of the cellular respiration pathway.1. PDC deficiency is caused by mutations in the PDHA1 gene, which is located on the X chromosome. A male with PDC deficiency and a homozygous female with no family history of PDC deficiency have a male offspring. Calculate the probability that the male offspring will have PDC deficiency.

Mean Score: 1.68/4

Short FRQ #2

Cell Communication and Cell Cycle (Cell Signaling)

1. Acetylcholine is a neurotransmitter that can activate an action potential in a postsynaptic neuron (Figures 1 and 2). A researcher is investigating the effect of a particular neurotoxin that causes the amount of acetylcholine released from presynaptic neurons to increase. 1. Describe the immediate effect of the neurotoxin on the number of action potentials in a postsynaptic neuron. Predict whether the maximum membrane potential of the postsynaptic neuron will increase, decrease, or stay the same.1. The researcher proposes two models, A and B, for using acetylcholinesterase (AChE), an enzyme that degrades acetylcholine, to prevent the effect of the neurotoxin. In model A, AChE is added to the synapse. In model B, AChE is added to the cytoplasm of the postsynaptic cell. Predict the effectiveness of EACH proposed model. Provide reasoning to support your predictions.

Mean Score: 1.46/4

Short FRQ #3

Natural Selection (Cladograms and evolutionary relationships)

1. A researcher studying the evolutionary relationship among five primate species obtained data from a sequence of mitochondrial DNA (mtDNA) from a representative individual of each species. The researcher then calculated the percent divergence in the sequences between each pair of primate species (Table 1). 1. Based on fossil data, the researcher estimates that humans and their most closely related species in the data set diverged approximately seven million years ago. Using these data, calculate the rate of mtDNA percent divergence per million years between humans and their most closely related species in the data set. Round your answer to two decimal places.1. Using the data in the table, construct a cladogram on the template provided. Provide reasoning for the placement of gibbons as the outgroup on the cladogram.1. On the cladogram, draw a circle around all of the species that are descended from the species indicated by the node within the square.

Mean Score: 1.88/4

Short FRQ #4

Cell Communication and Cell Cycle (Gene Expression)

1. The yeast Saccharomyces cerevisiae is a single-celled organism. Amino acid synthesis in yeast cells occurs through metabolic pathways, and enzymes in the synthesis pathways are encoded by different genes. The synthesis of a particular amino acid can be prevented by mutation of a gene encoding an enzyme in the required pathway. A researcher conducted an experiment to determine the ability of yeast to grow on media that differ in amino acid content. Yeast can grow as both haploid and diploid cells. The researcher tested two different haploid yeast strains (Mutant 1 and Mutant 2), each of which has a single recessive mutation, and a haploid wild-type strain. The resulting data are shown in Table 1. 1. Identify the role of treatment I in the experiment.1. Provide reasoning to explain how Mutant 1 can grow on treatment I medium but cannot grow on treatment III medium.1. Yeast mate by fusing two haploid cells to make a diploid cell. In a second experiment, the researcher mates the Mutant 1 and Mutant 2 haploid strains to produce diploid cells. Using the table provided, predict whether the diploid cells will grow on each of the four media. Use a plus sign (+) to indicate growth and a minus sign (−) to indicate no growth.

Mean Score: 0.88/3

Short FRQ #5

Cell Communication and Cell Cycle (Gene Expression)

1. A researcher is studying patterns of gene expression in mice. The researcher collected samples from six different tissues in a healthy mouse and measured the amount of mRNA from six genes. The data are shown in Figure 1. 1. Based on the data provided, identify the gene that is most likely to encode a protein that is an essential component of glycolysis. Provide reasoning to support your identification.1. The researcher observed that tissues with a high level of gene H mRNA did not always have gene H protein. Provide reasoning to explain how tissues with high gene H mRNA levels can have no gene H protein.

Mean Score: 0.69/3

Short FRQ #6

Cell Structure and Function (Cellular Transport)

1. The petal color of the Mexican morning glory (Ipomoea tricolor) changes from red to blue, and the petal cells swell during flower opening. The pigment heavenly blue anthocyanin is found in the vacuole of petal cells. Petal color is determined by the pH of the vacuole. A model of a morning glory petal cell before and after flower opening is shown in Table 1. 1. Identify the cellular component in the model that is responsible for the increase in the pH of the vacuole during flower opening AND describe the component’s role in changing the pH of the vacuole.1. A researcher claims that the activation of the K+/H+ transport protein causes the vacuole to swell with water. Provide reasoning to support the researcher’s claim.

Mean Score: 1.04/3

2018 FRQs

👉AP Bio 2018 FRQs

Long FRQ #1

Natural Selection (Cladograms and evolutionary relationships)

1. Polar bears are highly adapted for life in cold climates around the North Pole. Brown bears, black bears, and pandas are found in warmer environments. Researchers collected complete mitochondrial DNA sequences from several populations of bears and constructed a phylogenetic tree to represent their evolutionary relatedness (Figure 1). A researcher studying adaptation in bears sequenced the nuclear gene encoding a lysosomal trafficking protein (LYST) in polar bears, brown bears, black bears, and panda bears. There are seven inferred amino acid substitutions that are found only in polar bears. Mutations that cause similar substitutions in the human LYST protein are associated with Chediak-Higashi syndrome, an autosomal recessive condition in which pigment is absent from the hair and eyes. The researcher used the inferred amino acid sequences to build the distance matrix shown in Table 1. 1. Use the phylogenetic tree in Figure 1 to estimate the age in hundreds of thousands of years of the most recent common ancestor of all brown bears. Identify the population of brown bears to which polar bears are most closely related based on the mitochondrial DNA sequence comparison. Identify two populations whose positions could be switched without affecting the relationships illustrated in the phylogenetic tree.1. Construct a cladogram on the template to represent a model of the evolutionary relatedness among the bear species based on the differences in LYST protein sequences (Table 1). Circle the position on the cladogram that represents the out-group.1. A student claims that mitochondrial DNA sequence comparisons provide a more accurate phylogeny of bear species than do LYST protein sequence comparisons. Provide ONE piece of reasoning to support the student’s claim.1. A researcher genetically engineers a mouse strain by deleting the mouse lyst gene and replacing it with the polar bear lyst gene. Predict the most likely difference in phenotype of the transgenic mouse strain compared to the wild-type mouse strain. Justify your prediction.1. Describe how the mutation in the lyst gene became common in the polar bear population. If the lyst gene were the only determinant of fur color, predict the percent of white offspring produced by a mating between a polar bear and a brown bear.

**Mean Score: 4.73/10 **

Long FRQ #2

Cell Communication and Cell Cycle (Gene Regulation)

1. Some pathogenic bacteria enter cells, replicate, and spread to other cells, causing illness in the host organism. Host cells respond to these infections in a number of ways, one of which involves activating particular enzymatic pathways (Figure 1). Cells normally produce a steady supply of inactive caspase-1 protein. In response to intracellular pathogens, the inactive caspase-1 is cleaved and forms an active caspase-1 (step 1). Active caspase-1 can cleave two other proteins. When caspase-1 cleaves an inactive interleukin (step 2), the active portion of the interleukin is released from the cell. An interleukin is a signaling molecule that can activate the immune response. When caspase-1 cleaves gasdermin (step 3), the N-terminal portions of several gasdermin proteins associate in the cell membrane to form large, nonspecific pores. Researchers created the model in Figure 1 using data from cell fractionation studies. In the experiments, various parts of the cell were separated into fractions by mechanical and chemical methods. Specific proteins known to be located in different parts of the cell were used as markers to determine the location of other proteins. The table below shows the presence of known proteins in specific cellular fractions. 1. Describe the effect of inhibiting step 3 on the formation of pores AND on the release of interleukin from the cell.1. Make a claim about how cleaving inactive caspase-1 results in activation of caspase-1. A student claims that preinfection production of inactive precursors shortens the response time of a cell to a bacterial infection. Provide ONE reason to support the student’s claim.1. A student claims that the NF-kB protein is located in the cytoplasm until the protein is needed for transcription. Justify the student’s claim with evidence. Identify TWO fractions where N-terminal gasdermin would be found in cells infected with pathogenic bacteria.1. Describe the most likely effect of gasdermin pore formation on water balance in the cell in a hypotonic environment.1. Explain how gasdermin pore formation AND interleukin release contribute to an organism’s defense against a bacterial pathogen.

Mean Score: 2.68/10

Short FRQ #1

Ecology

1. Seagrasses are aquatic plants that reproduce sexually. Male seagrass flowers produce sticky pollen that is carried by circulating water to female flowers, resulting in fertilization. A researcher claims that mobile aquatic invertebrates can also transfer pollen from male to female flowers in the absence of circulating water. To investigate this claim, the researcher set up aquariums to model the possible interactions between the invertebrates and seagrasses. 1. Use the symbols below and the template aquariums to demonstrate the experimental design for testing the researcher’s claim that mobile aquatic invertebrates can pollinate seagrass in the absence of circulating water. Draw the appropriate symbols in the negative control aquarium AND the experimental aquarium. Do not use any symbol more than once in the same aquarium.1. Identify the dependent variable in the experiment. Predict the experimental results that would support the researcher’s claim that mobile aquatic invertebrates can also transfer pollen from male to female flowers in the absence of circulating water.

Mean Score: 2.38/4

Short FRQ #2

Cell Structure and Function (Cell Transport and Experimental Design)

1. The common bedbug (Cimex lectularius) is a species of insect that is becoming increasingly resistant to insecticides. Bedbugs possess several genes suspected of contributing to the resistance, including P450, Abc8, and Cps. To investigate the role of these genes in insecticide resistance, researchers deleted one or more of these genes in different strains of bedbugs, as indicated in Figure 1, and treated the strains with the insecticide beta-cyfluthrin. Each strain was genetically identical except for the deleted gene(s) and was equally fit 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.1. P450 encodes an enzyme that detoxifies insecticides. Abc8 encodes a transporter protein that pumps insecticides out of cells. Cps encodes an external structural protein located in the exoskeleton that greatly reduces the absorption of insecticides. Based on this information and the data in Figure 1, explain how a deletion of both P450 and Abc8 results in lower survival in bedbugs compared with a deletion of Cps only.

Mean Score: 1.41/4

Short FRQ #3

Ecology (Symbiotic Relationships)

1. Some birds, including great spotted cuckoos, lay their eggs in the nests of other birds, such as reed warblers. The warbler parents raise the unrelated chicks and provide them with food that would otherwise be given to their biological offspring. A researcher conducted an investigation to determine the type of relationship between warblers and cuckoos in an environment without predators. The researcher found that nests containing only warblers were more likely to be successful than nests containing warblers and cuckoos (data not shown). A successful nest is defined as a nest where at least one chick becomes an adult warbler. In some geographic areas, several species of nest predators are present. Researchers have found that cuckoo chicks, while in the nest, produce a smelly substance that deters nest predators. The substance does not remain in the nest if cuckoo chicks are removed. Figure 1 shows the probability that nests containing only warblers or containing both warblers and cuckoos will be successful in an environment with predators. In a follow-up experiment, the researchers added cuckoos to a nest that contained only warblers (group 1) and removed cuckoos from a nest containing warblers and cuckoos (group 2). 1. Describe the symbiotic relationship that exists between the cuckoo and warbler in an environment without predators.1. On the template provided, draw bars in the appropriate locations to predict the relative probability of success for the nest in the presence of predators where: 1. the cuckoos were added to the nest containing only warblers (group 1)1. the cuckoos were removed from the nest containing warblers and cuckoos (group 2)1. Identify the symbiotic relationship that exists between the cuckoo and the warbler in the presence of predators.

Mean Score: 2.66/4

Short FRQ #4

Cell Structure and Function (Cellular Transport)

1. Cystic fibrosis is a genetic condition that is associated with defects in the CFTR protein. The CFTR protein is a gated ion channel that requires ATP binding in order to allow chloride ions (Cl−) to diffuse across the membrane.1. In the provided model of a cell, draw arrows to describe the pathway for production of a normal CFTR protein from gene expression to final cellular location.1. Identify the most likely cellular location of the ribosomes that synthesize CFTR protein.1. Identify the most likely cellular location of a mutant CFTR protein that has an amino acid substitution in the ATP-binding site.

Mean Score: 0.89/3

Short FRQ #5

Heredity (Sex Linked Heredity)

1. In the tongue sole fish (Cynoglossus semilaevis), sex is determined by a combination of genetics and environmental temperature. Genetically male fish have two Z chromosomes (ZZ), and genetically female fish have one Z chromosome and one W chromosome (ZW). When fish are raised at 22℃, ZZ fish develop into phenotypic males and ZW fish develop into phenotypic females. However, when fish are raised at 28℃, the Z chromosome is modified (denoted as Z*). ZW individuals develop as phenotypic males that are fertile and can pass on the Z chromosome to their offspring even when the offspring are raised at 22℃. A cross between a ZW female and a ZZ male is shown in the Punnett square below. 1. Predict the percent of phenotypic males among the F1 offspring of the cross shown in the Punnett square if the offspring are raised at 22℃.1. At least one Z or Z chromosome is necessary for survival of the fish. A researcher crossed two fish and observed a 2:1 ratio of males to females among the offspring. Based on the information, identify the genotype of the male parent in the cross. Describe ONE fitness cost to the female of mating with this particular male.

Mean Score: 1.10/3

Short FRQ #6

Cell Communication and Cell Cycle

1. Acetylcholine receptor (AChR) proteins are found at the synapse between neurons and skeletal muscle cells. Acetylcholine released from neurons binds to a specific site on the receptor proteins, which causes an ion channel in the receptors to open and allow sodium ions (Na+) to enter muscle cells. The resulting depolarization of muscle cells initiates muscle contractions. Another molecule, nicotine, can also bind to certain types of AChR proteins and activate the receptors.A researcher is investigating two different types of AChR proteins: type 1 and type 2. To determine which stimuli activate the receptors, the researcher exposes muscle cells expressing the different types of receptor proteins to stimuli and observes the results indicated in Table 1. 1. Describe the difference in the structure AND function between AChR type 1 and AChR type 2.1. Acetylcholinesterase is an enzyme that breaks down acetylcholine in the synapse. Describe the effect of inhibiting acetylcholinesterase on the muscle cells with AChR type 2.

Mean Score: 0.95/3

2017 FRQs

👉AP Bio 2017 FRQs

Long FRQ #1

Experimental Design and Cellular Energetics

1. In flowering plants, pollination is a process that leads to the fertilization of an egg and the production of seeds. Some flowers attract pollinators, such as bees, using visual and chemical cues. When a bee visits a flower, in addition to transferring pollen, the bee can take nectar from the flower and use it to make honey for the colony. Nectar contains sugar, but certain plants also produce caffeine in the nectar. Caffeine is a bitter-tasting compound that can be toxic to insects at high concentrations. To investigate the role of caffeine in nectar, a group of researchers studied the effect of 0.1 mM caffeine on bee behavior. The results of an experiment to test the effect of caffeine on bees’ memory of a nectar source are shown in Table 1. 1. On the axes provided, construct an appropriately labeled graph to illustrate the effect of caffeine on the probability of bees revisiting a nectar source (memory).1. Based on the results, describe the effect of caffeine on each of the following: 1. Short-term (10 minute) memory of a nectar source1. Long-term (24 hour) memory of a nectar source1. Design an experiment using artificial flowers to investigate potential negative effects of increasing caffeine concentrations in nectar on the number of floral visits by bees. Identify the null hypothesis, an appropriate control treatment, and the predicted results that could be used to reject the null hypothesis.1. Researchers found that nectar with caffeine tends to have a lower sugar content than nectar without caffeine. Plants use less energy to produce the caffeine in nectar than they do to produce the sugar in nectar. Propose ONE benefit to plants that produce nectar with caffeine and a lower sugar content. Propose ONE cost to bees that visit the flowers of plants that produce nectar with caffeine and a lower sugar content.

Mean Score: 3.83/10

Long FRQ #2

Ecology (Ecological Succession)

1. Fires frequently occur in some ecosystems and can destroy all above-ground vegetation. Many species of plants in these ecosystems respond to compounds in smoke that regulate seed germination after a major fire. Karrikins (KAR) and trimethylbutenolides (TMB) are water-soluble compounds found in smoke that are deposited in the soil as a result of a fire. KAR and TMB bind to receptor proteins in a seed. In a study on the effects of smoke on seeds, researchers recorded the timing and percent of seed germination in the presence of various combinations of KAR and TMB. The results are shown in Figure 1. In a second investigation into the effect of available water on seed germination after a fire, researchers treated seeds with KAR or TMB. The treated seeds were then divided into two treatment groups. One group received a water rinse and the other group received no water rinse. The seeds were then incubated along with a group of control seeds that were not treated. The results are shown in the table.1. The researchers made the following claims about the effect of KAR and the effect of TMB on seed germination relative to the control treatment. 1. KAR alone affects the timing of seed germination.1. KAR alone affects the percentage of seeds that germinate.1. TMB alone affects the timing of seed germination.1. TMB alone affects the percentage of seeds that germinate.1. Provide support using data from Figure 1 for each of the researchers’ claims.1. Make a claim about the effect of rinsing on the binding of KAR to the receptor in the seed and about the effect of rinsing on the binding of TMB to the receptor in the seed. Identify the appropriate treatment groups and results from the table that, when compared with the controls, provide support for each claim.1. There is intense competition by plants to successfully colonize areas that have been recently cleared by a fire. Describe ONE advantage of KAR regulation and ONE advantage of TMB regulation to plants that live in an ecosystem with regular fires.

Mean Score: 4.04/10

Short FRQ #1

Cell Communication and Cell Cycle (Gene Regulation and Expression)

1. Gibberellin is the primary plant hormone that promotes stem elongation. GA 3-beta-hydroxylase (GA3H) is the enzyme that catalyzes the reaction that converts a precursor of gibberellin to the active form of gibberellin. A mutation in the GA3H gene results in a short plant phenotype. When a pure-breeding tall plant is crossed with a pure-breeding short plant, all offspring in the F1 generation are tall. When the F1 plants are crossed with each other, 75 percent of the plants in the F2 generation are tall and 25 percent of the plants are short. 1. The wild-type allele encodes a GA3H enzyme with alanine (Ala), a nonpolar amino acid, at position 229. The mutant allele encodes a GA3H enzyme with threonine (Thr), a polar amino acid, at position 229. Describe the effect of the mutation on the enzyme and provide reasoning to support how this mutation results in a short plant phenotype in homozygous recessive plants.1. Using the codon chart provided, predict the change in the codon sequence that resulted in the substitution of alanine for threonine at amino acid position 229.1. Describe how individuals with one (heterozygous) or two (homozygous) copies of the wild-type GA3H allele can have the same phenotype.

Mean Score: 1.10/4

Short FRQ #2

Ecology (Energy Diagrams)

1. The table above shows how much each organism in an aquatic ecosystem relies on various food sources. The rows represent the organisms in the ecosystem, and the columns represent the food source. The percentages indicate the proportional dietary composition of each organism. High percentages indicate strong dependence of an organism on a food source. 1. Based on the food sources indicated in the data table, construct a food web in the template below. Write the organism names on the appropriate lines AND draw the arrows necessary to indicate the energy flow between organisms in the ecosystem.1. In an effort to control the number of midges, an area within the ecosystem was sprayed with the fungus Metarhizium anisopliae, which significantly decreased the midge population. Based on the data in the table, predict whether the spraying of the fungus will have the greatest short-term impact on the population of the stoneflies, the caddisflies, or the hellgrammites. Justify your prediction.

Mean Score: 2.05/4

Short FRQ #3

Cellular Energetics

1. Microcystis aeruginosis is a freshwater photosynthetic cyanobacterium. When temperatures increase and nutrients are readily available in its pond habitat, M. aeruginosis undergoes rapid cell division and forms an extremely large, visible mass of cells called an algal bloom. M. aeruginosis has a short life span and is decomposed by aerobic bacteria and fungi. Identify the metabolic pathway and the organism that is primarily responsible for the change in oxygen level in the pond between times I and II AND between times III and IV.

Mean Score: 1.81/4

Short FRQ #4

Cell Communication and Cell Cycle (DNA Structure and Function)

1. A comet assay is a technique used to determine the amount of double-strand breaks in DNA (DNA damage) in cells. The nucleus of an individual cell is placed on a microscope slide coated with an agarose gel. An electric current is applied to the gel that causes DNA to move (electrophoresis), and the DNA is stained with a fluorescent dye. When viewed using a microscope, undamaged DNA from the nucleus appears as a round shape (the head), and the fragments of damaged DNA extend out from the head (the tail). The length of the tail corresponds to the amount of the damage in the DNA (see Figure 1). 1. To explain the movement of DNA fragments in the comet assay, identify one property of DNA and provide reasoning to support how the property contributes to the movement during the comet assay technique.1. In a different experiment, cells are treated with a chemical mutagen that causes only nucleotide substitutions in DNA. Predict the likely results of a comet assay for this treatment.

Mean Score: 0.75/3

Short FRQ #5

Cellular Energetics (Anaerobic Respiration)

1. Many species of bacteria grow in the mouths of animals and can form biofilms on teeth (plaque). Within plaque, the outer layers contain high levels of oxygen and the layers closest to the tooth contain low levels of oxygen. The surface of the tooth is covered in a hard layer of enamel, which can be dissolved under acidic conditions. When the enamel breaks down, the bacteria in plaque can extract nutrients from the tooth and cause cavities.Certain types of bacteria (e.g., Streptococcus mutans) thrive in the innermost anaerobic layers of the plaque and are associated with cavities. Other types of bacteria (Streptococcus sanguinis) compete with S. mutans but are unable to thrive in acidic environments. 1. Identify the biochemical pathway S. mutans uses for metabolizing sugar and describe how the pathway contributes to the low pH in the inner layers of plaque.1. Normal tooth brushing effectively removes much of the plaque from the flat surfaces of teeth but cannot reach the surfaces between teeth. Many commercial toothpastes contain alkaline components, which raise the pH of the mouth. Predict how the population sizes of S. mutans AND S. sanguinis in the bacterial community in the plaque between the teeth are likely to change when these toothpastes are used.

Mean Score: 0.76/3

Short FRQ #6

Cell Structure and Function Cell Transport

1. 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.1. 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.

Mean Score: 0.60/3

2016 FRQs

👉AP Bio 2016 FRQs

Long FRQ #1

Cell Communication and Cell Cycle (Gene Regulation and Experimental Design)

1. Leucine aminopeptidases (LAPs) are found in all living organisms and have been associated with the response of the marine mussel, Mytilus edulis, to changes in salinity. LAPs are enzymes that remove N-terminal amino acids from proteins and release the free amino acids into the cytosol. To investigate the evolution of LAPs in wild populations of M. edulis, researchers sampled adult mussels from several different locations along a part of the northeast coast of the United States, as shown in Figure 1. The researchers then determined the percent of individuals possessing a particular lap allele, lap94, in mussels from each sample site (table 1). 1. On the axes provided, construct an appropriately labeled bar graph to illustrate the observed frequencies of the lap94 allele in the study populations.1. Based on the data, describe the most likely effect of salinity on the frequency of the lap94 allele in the marine mussel populations in Long Island Sound. Predict the likely lap94 allele frequency at a sampling site between site 1 and site 2 in Long Island Sound.1. Describe the most likely effect of LAP94 activity on the osmolarity of the cytosol. Describe the function of LAP94 in maintaining water balance in the mussels living in the Atlantic Ocean.1. Marine mussel larvae are evenly dispersed throughout the study area by water movement. As larvae mature, they attach to the rocks in the water. Explain the differences in lap94 allele frequency among adult mussel populations at the sample sites despite the dispersal of larvae throughout the entire study area. Predict the likely effect on distribution of mussels in Long Island Sound if the lap94 allele was found in all of the mussels in the population. Justify your prediction.

Mean Score: 5.20/10

Long FRQ #2

Cell Communication and Cell Cycle and Ecology (Gene Regulation, Ecology, and Experimental Design)

1. Bacteria can be cultured in media with a carefully controlled nutrient composition. The graph above shows the growth of a bacterial population in a medium with limiting amounts of two nutrients, I and II. 1. Estimate the maximum population density in cells/mL for the culture. Using the data, describe what prevents further growth of the bacterial population in the culture.1. Using the data, calculate the growth rate in cells/mL×hour of the bacterial population between hours 2 and 4.1. Identify the preferred nutrient source of the bacteria in the culture over the course of the experiment. Use the graph to justify your response. Propose ONE advantage of the nutrient preference for an individual bacterium.1. Describe how nutrient I most likely regulates the genes for metabolism of nutrient I and the genes for metabolism of nutrient II. Provide TWO reasons that the population does not grow between hours 5 and 6.

Mean Score: 3.09/10

Short FRQ #1

Cellular Energetics

1. The graph above illustrates the percent dry weight of different parts of a particular annual plant (plants that live less than one year) 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 (seeds, receptacles, and flowers). 1. 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.1. Based on the data on the graph, estimate the percent of the total energy that the plant has allocated to the growth of leaves on the first day of July.1. Compared with perennials (plants that live more than two years), annual plants often allocate a much greater percentage of their total energy to growth of their reproductive parts in any given year. Propose ONE evolutionary advantage of the energy allocation strategy in annual plants compared with that in perennial plants.

Mean Score: 1.36/4

Short FRQ #2

Cell Communication and Cell Cycle (Gene Regulation)

1. The figure represents the process of expression of gene X in a eukaryotic cell.1. The primary transcript in the figure is 15 kilobases (kb) long, but the mature mRNA is 7 kb in length. 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.1. Predict the length of the mature gene X mRNA if the full-length gene is introduced and expressed in prokaryotic cells. Justify your prediction.

Mean Score: 1.43/4

Short FRQ #3

Ecology (Ecological Relationships)

1. The graph above shows the mass of plants from two different species over time. The plants grew while attached to each other. The plants were separated at the time indicated by the vertical line in the graph. Using template 1, graph the predicted shape of the plant-mass lines after separation of the two plants if the plants were in an obligate mutualistic relationship. On template 2, graph the predicted shape of the plant-mass lines if the species 2 plant was a parasite of the species 1 plant. Justify each of your predictions.

Mean Score: 1.92/4

Short FRQ #4

Cell Communication and Cell Cycle (Gene Regulation)

1. Living and dead organisms continuously shed DNA fragments, known as eDNA, into the environment. To detect eDNA fragments in the environment, the polymerase chain reaction (PCR) can be used to amplify specific eDNA fragments. eDNA fragments of different lengths persist in the environment for varying amounts of time before becoming undetectable (Figure 1). To investigate whether silver carp, an invasive fish, have moved from a nearby river system into Lake Michigan, researchers tested water samples for the presence of eDNA specific to silver carp (Figure 2). 1. 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. Propose ONE advantage of identifying long eDNA fragments as opposed to short fragments for detecting silver carp.1. The researchers tested a large number of water samples from Lake Michigan and found eDNA specific to silver carp in a single sample in the lake, as indicated in Figure 2. The researchers concluded that the single positive sample was a false positive and that no silver carp had entered Lake Michigan. Provide reasoning other than human error to support the researchers’ claim.

Mean Score: 0.79/3

Short FRQ #5

Heredity

1. In a certain species of plant, the diploid number of chromosomes is 4 (2n = 4). Flower color is controlled by a single gene in which the green allele (G) is dominant to the purple allele (g). Plant height is controlled by a different gene in which the dwarf allele (D) is dominant to the tall allele (d). Individuals of the parental (P) generation with the genotypes GGDD and ggdd were crossed to produce F1 progeny. 1. Construct a diagram below to depict the four possible normal products of meiosis that would be produced by the F1 progeny. Show the chromosomes and the allele(s) they carry. Assume the genes are located on different chromosomes and the gene for flower color is on chromosome 1.1. Predict the possible phenotypes and their ratios in the offspring of a testcross between an F1 individual and a ggdd individual.1. If the two genes were genetically linked, describe how the proportions of phenotypes of the resulting offspring would most likely differ from those of the testcross between an F1 individual and a ggdd individual.

Mean Score: 0.51/3

Short FRQ #6

Cellular Energetics

1. Researchers conducted a study to investigate the effect of exercise on the release of prolactin into the blood. The researchers measured the concentration of prolactin in the blood of eight adult males before (T = 0 hour) and after one hour (T = 1 hour) of vigorous exercise. As a control, the researchers measured the concentration of blood prolactin in the same group of individuals at the same times of day one week later, but without having them exercise. The results are shown in Figure 1. 1. Justify the use of the without-exercise treatment as the control in the study design.1. Using evidence from the specific treatments, determine whether prolactin release changes after exercise. Justify your answer.

Mean Score: 0.56/3

2015 FRQs

👉AP Bio 2015 FRQs

Long FRQ #1

Gene Expression and Regulation, Ecology (Symbiosis, Genetic Expression)

1. Many species have circadian rhythms that exhibit an approximately 24-hour cycle. Circadian rhythms are controlled by both genetics and environmental conditions, including light. Researchers investigated the effect of light on mouse behavior by using a running wheel with a motion sensor to record activity on actograms, as shown in Figure 1. For the investigation, adult male mice were individually housed in cages in a soundproof room at 25°C. Each mouse was provided with adequate food, water, bedding material, and a running wheel. The mice were exposed to daily periods of 12 hours of light (L) and 12 hours of dark (D) (L12:D12) for 14 days, and their activity was continuously monitored. The activity data are shown in Figure 2. After 14 days in L12:D12, the mice were placed in continuous darkness (DD), and their activity on the running wheel was recorded as before. The activity data under DD conditions are shown in Figure 3.1. The nervous system plays a role in coordinating the observed activity pattern of the mice in response to light-dark stimuli. Describe ONE role of each of the following anatomical structures in responding to lightdark stimuli. 1. A photoreceptor in the retina of the eye1. The brain1. A motor neuron1. Based on an analysis of the data in Figure 2, describe the activity pattern of the mice during the light and dark periods of the L12:D12 cycle.1. The researchers claim that the genetically controlled circadian rhythm in the mice does not follow a 24-hour cycle. Describe ONE difference between the daily pattern of activity under L12:D12 conditions (Figure 2) and under DD conditions (Figure 3), and use the data to support the researchers’ claim.1. To investigate the claim that exposure to light overrides the genetically controlled circadian rhythm, the researchers plan to repeat the experiment with mutant mice lacking a gene that controls the circadian rhythm. Predict the observed activity pattern of the mutant mice under L12:D12 conditions and under DD conditions that would support the claim that light overrides the genetically controlled circadian rhythm.1. In nature, mice are potential prey for some predatory birds that hunt during the day. Describe TWO features of a model that represents how the predator-prey relationship between the birds and the mice may have resulted in the evolution of the observed activity pattern of the mice.

Mean Score: 3.81/10

Long FRQ #2

Cellular Energetics (Cellular Respiration, ATP, Krebs Cycle, Glycolysis)

1. Cellular respiration includes the metabolic pathways of glycolysis, the Krebs cycle, and the electron transport chain, as represented in the figures. In cellular respiration, carbohydrates and other metabolites are oxidized, and the resulting energy-transfer reactions support the synthesis of ATP. 1. Using the information above, describe ONE contribution of each of the following in ATP synthesis.1. Catabolism of glucose in glycolysis and pyruvate oxidation1. Oxidation of intermediates in the Krebs cycle1. Formation of a proton gradient by the electron transport chain1. Use each of the following observations to justify the claim that glycolysis first occurred in a common ancestor of all living organisms. 1. Nearly all existing organisms perform glycolysis.1. Glycolysis occurs under anaerobic conditions.1. Glycolysis occurs only in the cytosol.1. A researcher estimates that, in a certain organism, the complete metabolism of glucose produces 30 molecules of ATP for each molecule of glucose. The energy released from the total oxidation of glucose under standard conditions is 686 kcal/mol. The energy released from the hydrolysis of ATP to ADP and inorganic phosphate under standard conditions is 7.3 kcal/mol. Calculate the amount of energy available from the hydrolysis of 30 moles of ATP. Calculate the efficiency of total ATP production from 1 mole of glucose in the organism. Describe what happens to the excess energy that is released from the metabolism of glucose.1. The enzymes of the Krebs cycle function in the cytosol of bacteria, but among eukaryotes the enzymes function mostly in the mitochondria. Pose a scientific question that connects the subcellular location of the enzymes in the Krebs cycle to the evolution of eukaryotes.

Mean Score: 2.89/10

Short FRQ #1

Natural Selection, Chemical Structure of Life (Amino Acids, Phylogenic Tree)

1. The amino acid sequence of cytochrome c was determined for five different species of vertebrates. The table below shows the number of differences in the sequences between each pair of species. 1. Using the data in the table, create a phylogenetic tree on the template provided to reflect the evolutionary relationships of the organisms. Provide reasoning for the placement on the tree of the species that is least related to the others.1. Identify whether morphological data or amino acid sequence data are more likely to accurately represent the true evolutionary relationships among the species, and provide reasoning for your answer.

Mean Score: 1.90/4

Short FRQ #2

Cell Communication and Cell Cycle (Mitosis, Meiosis)

1. 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.1. 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.

Mean Score: 1.47/4

Short FRQ #3

Cellular Energetics (Photosynthesis)

1. 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.1. 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.

Mean Score: 1.95/4

Short FRQ #4

Ecology (Population Dynamics)

1. 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 1993.1. Propose ONE reason that the introduction of the outside males rescued the snake population from decline.1. Describe how the data support the statement that there are abundant resources in the environment.

Mean Score: 1.95/3

Short FRQ #5

Cell Communication and Cell Cycle (Cell signaling, transmissions)

1. 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 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.1. 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

Mean Score: 0.43/3

Short FRQ #6

Cell Communication and Cell Cycle (Cell Communication, Immune System)

1. 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.1. 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.1. Describe ONE characteristic of the individual’s immune response to the bacterial pathogen that is not affected by the loss of B cells.

Mean Score: 0.82/3