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AP Biology Free Response Questions (FRQ) – Past Prompts

34 min read • december 29, 2020

Dylan Black

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.

👉 AP Bio 2019 FRQs

Long FRQ #1

Gene expression and regulation, ecology (gene expression and symbiosis).

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 .

Circle ONE arrow that represents transcription on the template pathway. Identify the molecule that would be absent if enzyme YUC is nonfunctional.

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.

Explain one feedback mechanism by which a cell could prevent production of too much IAA without limiting I3PA production

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.

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.

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).

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.

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.

Predict TWO factors that will most likely limit the population growth of species A in treatment group I.

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.

Short FRQ #1

Cellular energetics, heredity (cellular respiration and sex linked inheritance).

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.

Identify the cellular location where PDC is most active.

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.

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.

Short FRQ #2

Cell communication and cell cycle (cell signaling).

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.

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.

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.

Short FRQ #3

Natural selection (cladograms and evolutionary relationships).

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).

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.

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.

On the cladogram, draw a circle around all of the species that are descended from the species indicated by the node within the square.

Short FRQ #4

Cell communication and cell cycle (gene expression).

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.

Identify the role of treatment I in the experiment.

Provide reasoning to explain how Mutant 1 can grow on treatment I medium but cannot grow on treatment III medium.

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.

Short FRQ #5

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.

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.

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.

Short FRQ #6

Cell structure and function (cellular transport).

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.

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.

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.

👉 AP Bio 2018 FRQs

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.

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.

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.

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.

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.

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.

Cell Communication and Cell Cycle (Gene Regulation)

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.

Describe the effect of inhibiting step 3 on the formation of pores AND on the release of interleukin from the cell.

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.

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.

Describe the most likely effect of gasdermin pore formation on water balance in the cell in a hypotonic environment.

Explain how gasdermin pore formation AND interleukin release contribute to an organism’s defense against a bacterial pathogen.

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.

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.

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.

Cell Structure and Function (Cell Transport and Experimental Design)

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.

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.

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.

Ecology (Symbiotic Relationships)

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).

Describe the symbiotic relationship that exists between the cuckoo and warbler in an environment without predators.

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:

the cuckoos were added to the nest containing only warblers (group 1)

the cuckoos were removed from the nest containing warblers and cuckoos (group 2)

Identify the symbiotic relationship that exists between the cuckoo and the warbler in the presence of predators.

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.

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.

Identify the most likely cellular location of the ribosomes that synthesize CFTR protein.

Identify the most likely cellular location of a mutant CFTR protein that has an amino acid substitution in the ATP-binding site.

Heredity (Sex Linked Heredity)

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*). Z*W 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 Z*Z male is shown in the Punnett square below.

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℃.

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.

Cell Communication and Cell Cycle

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.

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

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.

👉 AP Bio 2017 FRQs

Experimental Design and Cellular Energetics

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.

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).

Based on the results, describe the effect of caffeine on each of the following:

Short-term (10 minute) memory of a nectar source

Long-term (24 hour) memory of a nectar source

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.

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.

Ecology (Ecological Succession)

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.

The researchers made the following claims about the effect of KAR and the effect of TMB on seed germination relative to the control treatment.

KAR alone affects the timing of seed germination.

KAR alone affects the percentage of seeds that germinate.

TMB alone affects the timing of seed germination.

TMB alone affects the percentage of seeds that germinate.

Provide support using data from Figure 1 for each of the researchers’ claims.

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.

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.

Cell Communication and Cell Cycle (Gene Regulation and Expression)

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.

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.

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.

Describe how individuals with one (heterozygous) or two (homozygous) copies of the wild-type GA3H allele can have the same phenotype.

Ecology (Energy Diagrams)

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.

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.

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.

Cellular Energetics

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.

Cell Communication and Cell Cycle (DNA Structure and Function)

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).

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.

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.

Cellular Energetics (Anaerobic Respiration)

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.

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.

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.

Cell Structure and Function Cell Transport

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.

Describe ONE characteristic of the plasma membrane that allows estrogens to passively cross the membrane.

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.

👉 AP Bio 2016 FRQs

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

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).

On the axes provided, construct an appropriately labeled bar graph to illustrate the observed frequencies of the lap 94 allele in the study populations.

Based on the data, describe the most likely effect of salinity on the frequency of the lap 94 allele in the marine mussel populations in Long Island Sound. Predict the likely lap 94 allele frequency at a sampling site between site 1 and site 2 in Long Island Sound.

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.

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.

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

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.

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.

Using the data, calculate the growth rate in cells/mL×hour of the bacterial population between hours 2 and 4.

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.

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.

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).

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.

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.

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.

The figure represents the process of expression of gene X in a eukaryotic cell.

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.

Predict the length of the mature gene X mRNA if the full-length gene is introduced and expressed in prokaryotic cells. Justify your prediction.

Ecology (Ecological Relationships)

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.

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).

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.

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.

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.

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.

Predict the possible phenotypes and their ratios in the offspring of a testcross between an F1 individual and a ggdd individual.

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.

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.

Justify the use of the without-exercise treatment as the control in the study design.

Using evidence from the specific treatments, determine whether prolactin release changes after exercise. Justify your answer.

👉 AP Bio 2015 FRQs

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

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.

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.

A photoreceptor in the retina of the eye

A motor neuron

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.

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.

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.

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.

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

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.

Using the information above, describe ONE contribution of each of the following in ATP synthesis.

Catabolism of glucose in glycolysis and pyruvate oxidation

Oxidation of intermediates in the Krebs cycle

Formation of a proton gradient by the electron transport chain

Use each of the following observations to justify the claim that glycolysis first occurred in a common ancestor of all living organisms.

Nearly all existing organisms perform glycolysis.

Glycolysis occurs under anaerobic conditions.

Glycolysis occurs only in the cytosol.

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.

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.

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

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.

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.

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.

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.

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

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.

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.

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.

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.

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.

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

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

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

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 transmission of information to the brain. Mammalian genomes typically have approximately 1,000 functional odorant-receptor genes, each encoding a unique odorant receptor.

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.

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

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.

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.

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.

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

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AP® Biology

Photosynthesis and cellular respiration: ap® biology crash course.

• The Albert Team
• Last Updated On: March 1, 2022

The AP® Biology course has so many different concepts and ideas within it that it may seems like there are way too many to understand.  We get that, and want to make sure that you know the most important concepts as well as possible.  We’ve picked out a few of the most important AP® Bio concepts for you which fall under the umbrella of two critical processes: photosynthesis and cellular respiration.

Photosynthesis

The process of photosynthesis is an essential AP® Biology concept to understand.  Photosynthesis is the means by which most of the plant life on Earth gets its energy.  All animals receive nutrients either from plants or from animals that get their energy from plants.  This means that plants are the basis for all of the ecosystems on Earth, making photosynthesis a very important process.  So, in order to understand how photosynthesis happens, biologists have to break it down into two parts, the light and dark reactions.

1. Light Reaction (or the Light-Dependent Reactions)

This is the first step in the process of photosynthesis.  The light reaction is the process by which the leaf of a plant will absorb energy from the sun (in the form of photons) and will transfer it into electron carriers and energy (in the form of A denosine T ri p hosphate, or ATP).

First, light will enter the leaf and will be absorbed by pigments within photosystem II, which is the first half of the light reaction.  After the light is absorbed by the pigments, the photon is bounced around inside the photosystem until it reaches the pigment that is correct for that wave of light.  Once it reaches that pigment, it will excite an electron and will move the electron up to the primary electron acceptor.

After reaching the primary electron acceptor, the electron will be moved to photosystem I.  It will be transported across what is known as the electron transport chain.  All you need to know about the ETC is that the electron will move down it and as it moves down it will transfer energy, which is used by the plant to combine ADP and a phosphate group into ATP.

Once the electron reaches photosystem I (just remember that photosystems go in reverse, II is first and I is second), it is excited by another photon and the electron again is bounced to a primary electron acceptor.  Only this time when the electron falls, it is used to combine two hydrogen ions and one NADP+ ion to create NADPH.  This will be used later on in the dark reactions to further the process.  Once the NADPH has been made, the ATP and the NADPH move into the dark reactions, the second part of the cycle.

2. Dark Reaction (or the Light Independent Reaction/Calvin Cycle)

The next and final process of photosynthesis is the Calvin cycle.  This is the process that takes the NADPH, ATP and CO 2 and and converts them into G3P (which can be turned into glucose, the basic unit of energy).  This process is a cycle, so part of the product, G3P, will be used to start and end the cycle.

The cycle starts with three five-carbon chains that have a phosphorus attached to each of them using some of the ATP, creating three RuBPs.  After this, carbon dioxide is fixed (carbon will be added) to the RuBP and will create three unstable six-carbon compounds.  Then the compound will fall apart and will make six three-carbon chains.  Next, an electron will be donated to the three carbon chain and it will create the end product, 6 G3Ps.  One of these G3Ps will be removed from the cycle and the other five will be used to repeat the cycle and create the five-carbon chain.  Finally, after enough G3P is made, it will be turned into glucose and thus photosynthesis is complete.

The Different Parts of Cellular Respiration

Cellular respiration is what cells will do in order create energy from glucose.  The currency of energy that is used by cells is called ATP (which pops up often on the AP® Biology exam), which is the product of this reaction.  Cell respiration can be broken down into five different steps: glycolysis, formation of acetyl COA, Krebs cycle, oxidative phosphorylation and chemiosmosis.

1. Glycolysis and Formation of Acetyl COA

The first step in the essential AP® Biology concept of cellular respiration is glycolysis, which is a simple reaction that moves in a linear chain.  All you need to know for the AP® Biology exam is that one glucose will be turned into two pyruvate, two ATP and two NADH (similar to NADPH, it is an electron carrier).  The only other thing you will need to know about glycolysis is that it takes place in the cytoplasm of the cell.

The next step is another simple process that takes place in the cytoplasm of the cell.  This step takes one pyruvate and will turn it into one acetyl COA using coenzyme A.  This process will also create two NADH during the process.  Now, onto the Krebs cycle.

2. The Krebs Cycle

This is again another cycle, which is common among AP® Biology concepts, but this takes place in the matrix of the mitochondria.  It will take the acetyl COA and will use it to create electron carriers and energy for the next part of the reaction.  This process will create one ATP, three NADHs and one FADH 2 .  In this process acetyl COA will be combined with oxaloacetate, which will then turn into citric acid.  After this, this compound will go around the cycle, releasing two carbons along the cycle and creating two carbon dioxides in the process.  This process will then repeat itself and allow the electron carriers to move onto the next cycle.

3. Oxidative Phosphorylation and Chemiosmosis

These two steps will occur at the same time.  All of those electron acceptors that were made in the previous steps will be used to release the electrons that will go down the electron transport chain.  As these electrons move down the chain, protons will be pumped out of the matrix of the mitochondria.  This will cause a higher concentration of protons to exist, increasing the proton concentration  Because of this, the protons will want to re-enter the matrix, and they will do so using a channel known as ATP Synthase.  This will cause this protein to create more ATP as the protons travel across it.  Finally after the ATP is made, the electrons will be accepted by the final electron acceptor, oxygen.  This will combine with the protons (also called hydrogen ions) to make water.

How to Get Better?

If you have read this and you understand these AP® Biology concepts, then you are that much closer to succeeding on your AP® Biology exam!  If you want to test your knowledge of these topics, practice questions are a great way to do this.

Need help preparing for your AP® Biology exam?

Albert has hundreds of AP® Biology practice questions, free response, and full-length practice tests to try out.

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Excel at Science

• Feb 16, 2021

AP Biology Past FRQs by Topic

Updated: Jan 31

**Updated on 1/31/24 to include the 2022-23 FRQ exams!**

If you are looking for past AP Biology free-response questions (FRQs) that are organized by topic, then you have come to the right place. In this post, we have linked every freely available past FRQ there is from College Board and organized it into the following major topics of AP Biology .

(Please note that we are not associated with College Board and are simply sharing the resources they have made available to students.)

Biochemistry

Metabolism & energetics.

Physiology (note that this topic will not be tested on the official AP Biology exam this year in 2021, although many questions about physiology could also cover concepts that will be tested)

Experiment design & data analysis

Need more AP-style practice problems?

Intensively doing and reviewing practice questions is proven to be much more effective than spending hours studying. Check out our AP Bio Practice Portal , which is an easy-to-use database of 300+ AP-style MCQ and FRQ practice questions. Students love the Practice Portal because it includes answers and explanations for every problem, tracks progress, and saves time from Googling practice problems.

Try the Practice Portal >

How to make the most of past frqs from college board.

As noted above, the diversity of organisms, plants, and physiology will not be on the 2021 AP Biology exam. However, the exam could include questions about topics or hypothetical situations that are related to those topics. One great example is cell communication, which is used in multiple systems inside our bodies. Let’s say an FRQ was to appear about the immune system and how the immune cells communicate. That would be fair game as long as the question focuses on the cell signaling part, not the details of the immune system. If the question requires some background knowledge about the immune system, it will be provided.

If you want to do a whole practice FRQ set just like the ones on the real exam (which we highly recommend), all the freely available past FRQs by year are available here on the College Board website. Tip: time yourself and take the practice FRQ set in an environment that mimics how you imagine your actual testing environment to be.

If you would like to focus on a particular topic, then the section coming up is for you. Some FRQs will show up under multiple topics because they truly do test students’ understanding of multiple different topics.

Tip : Whether you are doing individual free-response questions or doing a full problem set in one go, it is extremely important and effective to do test corrections! Don’t only consult the scoring guidelines and model responses when you have no clue how to answer a question. You should be checking them for all the FRQs you do. When you find a difference between your answer and the scoring guidelines, it is important that you pause and analyze why your response is incorrect. Take the time to understand your mistakes and see how your answer could have been better. This will help you boost your scores the most efficiently.

AP BIOLOGY FRQs BY TOPIC

Below are the linked FRQs organized by topic. The header for each topic will also lead you to the corresponding study guide that will help you review the unit in detail!

Basic and organic chemistry concepts do not come up often on the FRQs (but of course, it’s better to be prepared). The properties of water and macromolecules come up occasionally.

2017 #7 and 8

Includes cell structure and function, cell transport and the proteins involved.

2019 #3 and 8

2018 #2, 6, and 8

2006 #1, 3, and 4

2001 #1 and 4

(study guide coming soon!)

This unit includes enzymes, cellular respiration, and photosynthesis.

2023 #2 (cell respiration & photosynthesis)

2023 #4 (photosynthesis)

2022 #3 (enzymes)

2021 #3 (cell respiration)

2019 #3 (cell respiration)

2018 #2 (cell respiration)

2017 #7 (cell respiration)

2017 #5 (photosynthesis)

2015 #2 (cell respiration)

2013 #2 (photosynthesis) and 4 (cell respiration & photosynthesis)

2012 #2 (cell respiration) and 4 (cell respiration & photosynthesis)

2010 #2 (enzymes)

2007 #3 (photosynthesis)

2006 #4 (photosynthesis)

2005 #1 (cell respiration & photosynthesis)

2004 #3 (photosynthesis)

Cell cycle & cell signaling

This topic has shown up more frequently and in more difficult FRQs in recent years, especially cell communication. The trend will most likely continue so definitely prioritize reviewing and practicing this topic!

2023 #1 (cell communication)

2022 #1 (cell communication)

2022 #2 (cell cycle, meiosis)

2021 #1 (cell communication)

2019 #4 (cell communication)

2018 #8 (cell communication)

2017 #8 (cell communication)

2016 # 7 (cell division)

2015 # 4 (cell division)

2015 #5 and 7 (cell communication)

2013 #8 (cell communication)

2011 #1B (cell division)

2010 #1 (cell communication)

2006 #1B (cell division)

2004 #1 (cell division)

Genetics, Gene Expression and Regulation

Genetics Pt 1 and Genetics Pt 2 Study Guides

This section includes the classic Mendelian genetics, with Punnett squares, crosses, and Mendel’s laws. It also includes DNA replication, protein synthesis, and gene expression regulation for both eukaryotes and prokaryotes.

2023 #6 (gene expression)

2022 #6 (protein synthesis, gene expression)

2021 #6 (gene expression)

2021 #2 (heredity + pedigrees)

2020 #1 parts a-b

2019 #1 and 3

2018 #1, 4, and 7

2016 #4 and 7

2023 #5 (Cladistics)

2022 #4 (speciation)

2020 #1 parts f-j

2015 #3 and 6

2014 #2 and 4

2015 #2 (nervous system)

2014 #2 (immune system) and 6 (musculoskeletal system) and 7

2017 #2, 4, and 7b

2016 #3 and 5

2014 #3 and 4

Experimental design & analysis

This is an additional section that isn’t focused on any particular topic or has significant data analysis involved. While most FRQs do pertain to a specific topic(s), some are simply there to test your knowledge of experimental design and understanding of statistical concepts such as performing Chi-Square tests and interpreting error bars on graphs. These types of questions have become more and more common on the AP exam, so it is important to feel comfortable and confident with them.

2023 #6 (data analysis)

2022 #3 (experiment design)

2020 #1 parts c-e

2016 #2 , 6 and 8

2014 #1 and 5

2013 #1 and 7

Hope these organized FRQs saved you some time so you can focus more on actually doing them and practicing! You can easily share this post with friends who may find it helpful as well.

How to Improve AP Biology FRQ Scores, Fast

Do a lot of FRQ practice problems and review the answers! Practice is key, especially for a subject as dense as AP Bio. Check out the AP Bio Practice Portal , which is our popular vault of 300+ AP-style MCQ and FRQ problem sets with answers and explanations for every question. Don't waste any more time Googling practice problems or answers - try it out now!

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AP Biology : Cellular Respiration

Study concepts, example questions & explanations for ap biology, all ap biology resources, example questions, example question #1 : understanding glycolysis.

Where does glycolysis take place in the cell? 

Mitochondria

Endoplasmic reticulum 

Glycolysis happens in the cytosol (the fluid containing the organelles) of the cell. The next step in cellular respiration, the citric acid cycle, occurs in the mitochondria.

Example Question #1 : Cellular Respiration

Which of the following reflects a function of fermentation?

It leads to the production of lactic acid in yeast cells

It produes two molecules of ATP

It oxidizes NADH to NAD +

It leads to the production of ethanol in muscle cells

Fermentation oxidizes molecules of NADH to NAD + so the cell can have oxidizing agents for any subsequent glycolysis reactions. It does not, however, produce any usable energy in the process.

Fermentation leads to the production of ethanol in yeast cells and lactic acid in muscle cells.

Example Question #108 : Cellular Biology

Where in the cell does glycolysis take place?

Endoplasmic reticulum

Glycolysis occurs in the cytosol of cells. Once finished, the two pyruvate products are transported into the mitochondria to go through the citric acid cycle, at a cost of 1 ATP per pyruvate. Neither the nucleus, nor the endoplasmic reticulum have any function in glycolysis or the citric acid cycle.

How many direct ATP are made if fructose-1,6-bisphosphate is put through glycolysis?

The conversion of glucose to two pyruvate molecules in glycolysis produces a net total of two direct ATP. When fructose-1,6-bisphosphate enters glycolysis, it bypasses the two steps involved that normally cost one ATP each, therefore, there is no required input and the net total is four produced ATP.

We have to remember that each step beyond the conversion of fructose-1,6-bisphosphate to glyceraldehyde-3-phosphate and dihydroxyacetone phosphate happens twice, or we would come up with two ATP created.

Which of the following products is not created during glycolysis?

Glycolysis is the first step in cellular respiration, and is seen in both aerobic and anaerobic respiration. The products of glycolysis are pyruvate, NADH, ATP, and water. Oxygen is only a product of the light reactions of photosynthesis; it is consumed as a reactant in the electron transport chain.

Example Question #2 : Cellular Respiration

Which of the following processes does not take place during glycolysis?

Four ATP molecules are produced for every molecule of glucose, but the net production is only two molecules of ATP

One molecule of glucose eventually yields two molecules of pyruvate

An end result is the production of 3-carbon molecules that are later fed into the citric acid cycle

Unlike the next two steps (the citric acid cycle and oxidative phosphorylation), glycolysis can occur in the absence of oxygen. The only step given that is  not part of glycolysis is the transfer of electrons from carrier molecules to oxygen via a series of steps. This happens during oxidative phosphorylation and, unlike glycolysis, is an aerobic process.

In comparison to fermentation, the aerobic pathways of glucose metabolism yield more __________ .

acetaldehyde 

adenosine triphosphate 

Adenosine triphosphate (ATP) is the main product of cellular respiration, and the molecular energy of the cell. Aerobic metabolism results in a much higher yield of these energy carrying molecules due to the fact that it can use oxygen as a final electron acceptor in the electron transport chain.

Where does glycolysis take place in the cell?

Glycolysis takes place in the cell cytosol, and can take place under anaerobic conditions. After the completion of glycolysis, the product pyruvate is transported to the mitochondria for the citric acid cycle and electron transport chain.

The nucleus houses the cell's DNA, and the endoplasmic reticulum is involved with protein modification.

Which of the following stages of cellular respiration generates ATP, regardless of the presence of oxygen?

Citric acid cycle

Electron transport chain

Oxidative phosphorylation

Krebs cycle

Glycolysis is the process that converts glucose to pyruvate. It produces a total of four ATP, but consumes two ATP, for a net yield of two ATP. Glycolysis is not dependent on the presence of oxygen and can occur in either aerobic or anaerobic environments.

The citric acid cycle, or Krebs cycle, is used to generate NADH from pyruvate. The NADH is then used in the electron transport chain to generate a proton gradient, which fuels oxidative phosphorylation. Since oxidative phosphorylation requires an oxygen molecule, the citric acid cycle and electron transport chain cannot continue in anaerobic environments.

What is the net production of ATP from one glucose molecule in glycolysis?

Glycolysis produces four molecules of ATP, but two molecules are used to complete reactions during the initial steps of the process. With four molecules produced and two molecules consumed in the process, there is a net yield of two ATP from each glucose molecule in glycolysis

During ATP synthesis in the electron transport chain, approximately 32 additional ATP are generated.

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7.7 Regulation of Cellular Respiration

Learning objectives.

In this section, you will explore the following question:

• What mechanisms control cellular respiration?

Connection for AP ® Courses

Cellular respiration is controlled by a variety of means. For example, the entry of glucose into a cell is controlled by the transport proteins that aid glucose passage through the cell membrane. However, most of the control of the respiration processes is accomplished through negative feedback inhibition of specific enzymes that respond to the intracellular concentrations of ATP, ADP, NAD + , and FAD, etc.

Information presented and the examples highlighted in the section support concepts and learning objectives outlined in Big Idea 2 of the AP ® Biology Curriculum Framework, as shown in the table. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP ® Biology course, an inquiry-based laboratory experience, instructional activities, and AP ® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices.

Cellular respiration must be regulated in order to provide balanced amounts of energy in the form of ATP. The cell also must generate a number of intermediate compounds that are used in the anabolism and catabolism of macromolecules. Without controls, metabolic reactions would quickly come to a stand-still as the forward and backward reactions reached a state of equilibrium. Resources would be used inappropriately. A cell does not need the maximum amount of ATP that it can make all the time: At times, the cell needs to shunt some of the intermediates to pathways for amino acid, protein, glycogen, lipid, and nucleic acid production. In short, the cell needs to control its metabolism.

Regulatory Mechanisms

A variety of mechanisms is used to control cellular respiration. Some type of control exists at each stage of glucose metabolism. Access of glucose to the cell can be regulated using the GLUT proteins that transport glucose ( Figure 7.19 ). Different forms of the GLUT protein control passage of glucose into the cells of specific tissues.

Some reactions are controlled by having two different enzymes—one each for the two directions of a reversible reaction. Reactions that are catalyzed by only one enzyme can go to equilibrium, stalling the reaction. In contrast, if two different enzymes (each specific for a given direction) are necessary for a reversible reaction, the opportunity to control the rate of the reaction increases, and equilibrium is not reached.

A number of enzymes involved in each of the pathways—in particular, the enzyme catalyzing the first committed reaction of the pathway—are controlled by attachment of a molecule to an allosteric site on the protein. The molecules most commonly used in this capacity are the nucleotides ATP, ADP, AMP, NAD + , and NADH. These regulators, allosteric effectors, may increase or decrease enzyme activity, depending on the prevailing conditions. The allosteric effector alters the steric structure of the enzyme, usually affecting the configuration of the active site. This alteration of the protein’s (the enzyme’s) structure either increases or decreases its affinity for its substrate, with the effect of increasing or decreasing the rate of the reaction. The attachment signals to the enzyme. This binding can increase or decrease the enzyme’s activity, providing feedback. This feedback type of control is effective as long as the chemical affecting it is attached to the enzyme. Once the overall concentration of the chemical decreases, it will diffuse away from the protein, and the control is relaxed.

Control of Catabolic Pathways

Enzymes, proteins, electron carriers, and pumps that play roles in glycolysis, the citric acid cycle, and the electron transport chain tend to catalyze non-reversible reactions. In other words, if the initial reaction takes place, the pathway is committed to proceeding with the remaining reactions. Whether a particular enzyme activity is released depends upon the energy needs of the cell (as reflected by the levels of ATP, ADP, and AMP).

The control of glycolysis begins with the first enzyme in the pathway, hexokinase ( Figure 7.20 ). This enzyme catalyzes the phosphorylation of glucose, which helps to prepare the compound for cleavage in a later step. The presence of the negatively charged phosphate in the molecule also prevents the sugar from leaving the cell. When hexokinase is inhibited, glucose diffuses out of the cell and does not become a substrate for the respiration pathways in that tissue. The product of the hexokinase reaction is glucose-6-phosphate, which accumulates when a later enzyme, phosphofructokinase, is inhibited.

Phosphofructokinase is the main enzyme controlled in glycolysis. High levels of ATP, citrate, or a lower, more acidic pH decrease the enzyme’s activity. An increase in citrate concentration can occur because of a blockage in the citric acid cycle. Fermentation, with its production of organic acids like lactic acid, frequently accounts for the increased acidity in a cell; however, the products of fermentation do not typically accumulate in cells.

The last step in glycolysis is catalyzed by pyruvate kinase. The pyruvate produced can proceed to be catabolized or converted into the amino acid alanine. If no more energy is needed and alanine is in adequate supply, the enzyme is inhibited. The enzyme’s activity is increased when fructose-1,6-bisphosphate levels increase. (Recall that fructose-1,6-bisphosphate is an intermediate in the first half of glycolysis.) The regulation of pyruvate kinase involves phosphorylation by a kinase (pyruvate kinase kinase), resulting in a less-active enzyme. Dephosphorylation by a phosphatase reactivates it. Pyruvate kinase is also regulated by ATP (a negative allosteric effect).

If more energy is needed, more pyruvate will be converted into acetyl CoA through the action of pyruvate dehydrogenase. If either acetyl groups or NADH accumulate, there is less need for the reaction and the rate decreases. Pyruvate dehydrogenase is also regulated by phosphorylation: A kinase phosphorylates it to form an inactive enzyme, and a phosphatase reactivates it. The kinase and the phosphatase are also regulated.

Citric Acid Cycle

The citric acid cycle is controlled through the enzymes that catalyze the reactions that make the first two molecules of NADH ( Figure 7.10 ). These enzymes are isocitrate dehydrogenase and α ‑ketoglutarate dehydrogenase. When adequate ATP and NADH levels are available, the rates of these reactions decrease. When more ATP is needed, as reflected in rising ADP levels, the rate increases. α -ketoglutarate dehydrogenase will also be affected by the levels of succinyl CoA—a subsequent intermediate in the cycle—causing a decrease in activity. A decrease in the rate of operation of the pathway at this point is not necessarily negative, as the increased levels of the α -ketoglutarate not used by the citric acid cycle can be used by the cell for amino acid (glutamate) synthesis.

Electron Transport Chain

Specific enzymes of the electron transport chain are unaffected by feedback inhibition, but the rate of electron transport through the pathway is affected by the levels of ADP and ATP. Greater ATP consumption by a cell is indicated by a buildup of ADP. As ATP usage decreases, the concentration of ADP decreases, and now, ATP begins to build up in the cell. This change in the relative concentration of ADP to ATP triggers the cell to slow down the electron transport chain.

Link to Learning

Visit this site to see an animation of the electron transport chain and ATP synthesis.

• A hydrogen ion gradient across the membrane establishes a concentration gradient and not an electrical gradient, thus assisting during the electron transport chain.
• A hydrogen ion gradient is established by pumping hydrogen ions across the membrane from the matrix in the intermembrane space. Its uneven distribution across the membrane establishes both concentration and electrical gradients.
• A hydrogen ion gradient is established by pumping hydrogen ions across the membrane from the intermembrane space into the matrix and its uneven distribution across the membrane establishes concentration and electrical gradients.
• Hydrogen ions are present in the intermembrane space from the beginning and results in the formation of gradients necessary for the function of ATP synthase.

For a summary of feedback controls in cellular respiration, see Table 7.1 .

Science Practice Connection for AP® Courses

Think about it.

Phosphofructokinase is a key enzyme in glycolysis. High levels of ATP or citrate or low pH can decrease the enzyme’s activity. Explain why this is beneficial to the cell.

Teacher Support

This question is an application of Learning Objective 2.16 and Science Practice 7.1 and Learning Objective 2.17 and Science Practice 5.3 because students are connecting changes in concentrations of key molecules (ATP and citrate) and a change in an environmental variable (pH) to the regulation of glycolysis via negative feedback.

Possible answer:

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• Authors: Julianne Zedalis, John Eggebrecht
• Publisher/website: OpenStax
• Book title: Biology for AP® Courses
• Publication date: Mar 8, 2018
• Location: Houston, Texas
• Book URL: https://openstax.org/books/biology-ap-courses/pages/1-introduction
• Section URL: https://openstax.org/books/biology-ap-courses/pages/7-7-regulation-of-cellular-respiration

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Unit 3: Cellular energetics

About this unit, enzyme structure and catalysis.

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