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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

Mitosis and meiosis: ap® biology crash course review.

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

As a eukaryotic organism grows, its cells are constantly dividing and creating new cells according to the “genetic blueprint” of its DNA. The processes by which these new cells are developed are known as mitosis and meiosis. Mitosis is the method by which somatic (or non-reproductive) are created, while meiosis   is the method that creates gametes (reproductive cells like sperm and eggs).

Keep in mind: prokaryotic cells do not have membrane-bound organelles like nuclei, and therefore do not undergo mitosis and meiosis as eukaryotic cells do (instead, they undergo binary fission). Throughout our discussion of mitosis and meiosis, we will be talking only about eukaryotes.

Before we get into the specifics of each process, let’s go over some AP® Biology background information that will help us understand the differences between them.

Chromatin, Chromatids and Chromosomes

These are essentially the three forms of a cell’s genetic material. Chromatin is its loosest, least-organized form, which usually floats freely around inside the defined envelope of the nucleus. Chromatids are formed from condensed chromatin and serve as one-half of each chromosome. In its most complete form, two identical “sister chromatids” are joined together by a centromere to form a full chromosome .

Diploid vs. Haploid Cells

Cells come in essentially two “flavors”: diploid and haploid . As the names imply, a diploid cell contains two sets of genetic information in homologous chromosome pairs, while a haploid cell contains only one set of genetic information in single copies of each chromosome.

Non-reproductive somatic cells are diploid cells, containing two sets of chromosomes. Human cells, for example, have 23 chromosome pairs (46 total chromosomes), with one set of genetic information inherited from each of that human’s parents.

Reproductive gametes, on the other hand, are haploid cells, containing only one set of chromosomes. In humans, egg and sperm cells contain only 23 chromosomes. When gametes combine during sexual reproduction, the sets of chromosomes from both parents provide the chromosome pairs for future diploid cells.

Now that we’ve reviewed the necessary AP® Bio background, let’s get to the meat of this section: the actual processes of mitosis and meiosis.

The process of cellular mitosis occurs in four primary phases: prophase, metaphase, anaphase and telophase . A fifth “phase,” known as interphase , is the state in which a somatic cell spends most of its lifespan.

Note: you will not need to know the names of these phases for the AP® Biology exam, but you will still be required to describe the steps.

Take a look at how each of these phases breaks down.

Not necessarily a true “phase” of mitosis, interphase is the normal, non-division state of somatic cells. If you throw a prepared slide of cells under a microscope, chances are the majority of them will be sitting in interphase, looking relatively inactive and uninteresting. If a cell is not in interphase, it is undergoing mitosis (which is sometimes referred to as “M phase”).

Interphase itself is split into three stages, as follows:

•  G1: cell simply grows

•  S phase: cell continues growing, starts duplicating DNA

•  G2: growth continues while cell prepares for mitotic division

This is where the action begins. As a cell prepares to divide, it enters prophase, in which the nucleoli—spherical structures inside the nucleus that contain RNA and protein—disappear and the chromatin of the nucleus condenses into tightly-packaged chromosomes. Note that because the DNA was duplicated in S-Interphase, each chromosome now contains two copies of the cell’s DNA.

The membrane that surrounds the genetic material of the cell (known as the nuclear envelope) then disappears, and a mitotic spindle is created as the microtubule organization centers (MTOCs) move toward opposite ends of the nucleus. These MTOCs are specialized structures that control the arrangement of a protein called tubulin into long microtubules that can manipulate the positioning of the cell’s genetic material. The mitotic spindle is simply the term for the overall structure of microtubules that guide this material.

As the MTOCs move apart, the microtubules they’ve built increase in length and connect to the centromeres of the chromosomes via a region called the kinetochore . The MTOCs are then capable of moving the chromosomes toward or away from the poles of the cell by shortening or lengthening the microtubules.

During metaphase, the fully-formed chromosomes are aligned by the microtubules at the center of the cell in a plane known as the metaphase plate . Then, the attached microtubules retract, splitting each chromosome into its individual sister chromatids. These resulting chromatids still have a centromere each, however, and therefore are referred to as individual chromosomes from this point forward.

Metaphase ends as soon as the original chromosomes are split.

Top tip : to determine the number of chromosomes at any time during the process, simply count the number of centromeres.

After the initial separation of the chromosomes, the new chromosomes (the split chromatids) are pulled to the poles of the cell via the shortening of the microtubules. At the end of this phase, each pole contains a complete set of identical chromosomes.

Since the DNA copies made during the S phase of interphase have now split, the chromosomes at the poles consist of single chromatids with only a single copy of the parent cell’s DNA.

To wrap-up the division process, normal cell organelles start to re-build and the newly-formed daughter cells begin to take shape for their own interphase. Nuclear envelopes develop around the genetic material at each pole, the chromosomes unwind and return to loosely-floating chromatin, and the nucleoli appear once more.

While the nucleus reforms, the dividing cell undergoes cytokinesis , which refers to the splitting of the unit and the division of cytoplasm across the two new cells. A cleavage furrow develops at the center of the dividing unit and cinches closed like a drawstring, leaving two separate cells with enclosed cell membranes .

Final result: two diploid daughter cells containing identical genetic material to the parent cell.

Because meiosis has the special task of creating new sex cells for reproduction, its process is unique, though similar to mitosis in many ways. Meiosis essentially goes through the stages of mitosis twice, with some key variations.

Perhaps the most important thing about meiosis is that it enables the independent assortment of genetic material. The determination of which chromosomes end up in which gametes is random, allowing for natural variation in the gene pool. It is this variation and biological diversity that keeps species naturally resilient.

Now that you’re inspired by the beauty of natural genetic diversity, let’s discuss how it happens.

This phase begins similarly to prophase in mitosis, with the nuclear envelope breaking down and the chromatin condensing into chromosomes. In meiosis, however, homologous chromosomes pair up into groups of four chromatids (known as tetrads or bivalents ) in a process called synapsis .

During synapsis, genetic material may cross over between non-sister homologous chromatids (chromatids that are not connected by a centromere and are therefore not part of the same chromosome).

Metaphase I

Next, homologous chromosome pairs are arranged at the metaphase plate. Instead of a line of single chromosomes, as in mitosis, meiosis sees a line of pairs. Microtubules from each pole then attach to the kinetochore of one chromosome from each pair.

This next phase starts as soon as the tetrads begin to separate. Like in mitosis, the separate chromosomes are pulled by the microtubules to opposite ends of the cell. Unlike mitosis, however, these chromosomes still comprise two sister chromatids.

Telophase I

The first half of the process completes with the formation of nuclear membranes around the chromosomes at the poles. Unlike in mitosis, the cleavage furrow does not yet develop. Note that once this process repeats to form the final four daughter cells, the resulting cells will be haploid.

Prophase II

The fun starts again with prophase II, in which the two newly-formed nuclear envelopes break down again and the mitotic spindle forms. This time, there is no crossing over.

Metaphase II

Metaphase II is nearly identical to metaphase in mitosis, with single chromosomes aligning at the metaphase plate. In this case, however, there is half the number of chromosomes present as in mitosis.

Anaphase II

Just like metaphase II, anaphase II mirrors the happenings of anaphase in mitosis, but with half as many chromosomes. Each single chromosome is pulled apart by microtubules and the new chromosomes (formerly sister chromatids) are pulled to opposite poles.

Telophase II

The entire process wraps up in telophase II. Four new nuclei form and cytokinesis occurs to form the four final cells. Note that the resulting cells’ chromosomes comprise only one chromatid each, and even when these are replicated during the S phase of interphase the haploid cell will still only contain half the number of chromosomes of the parent cell.

Final result: four haploid daughter cells, each containing copies of half the genetic material of the parent cell.

•  Mitosis creates two diploid somatic daughter cells that are clones of the parent cell.

•  A somatic cell spends most of its time in interphase, growing and replicating DNA in preparation for mitosis.

•  The four phases of actual mitotic division are prophase, metaphase, anaphase and telophase. These names will not need to be memorized for the AP® Biology exam.

•   Meiosis creates four haploid gamete daughter cells, each containing half of the original cell’s genetic material.

•  The phases of meiosis vary in a few key ways from those of mitosis, but follow the same general phase order twice. Again, the names of the phases will not need to be memorized.

That’s all there is to it! Can you describe each of the stages and key structures of mitosis and meiosis for the AP® Bio exam?

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AP®︎/College Biology

Course: ap®︎/college biology   >   unit 9.

• 1a-c, Responses to the environment
• 1d-e, Responses to the environment & natural selection
• 2a-b, Cellular respiration & common ancestry
• 2c-d, Cellular respiration & cell compartmentalization and its origins
• 3a-b, Phylogeny

4a-b, Meiosis and genetic diversity

• 5a-b, Responses to the environment
• 6a-c, Population ecology

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AP Biology : Understanding Meiosis

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

A scientist is interested in studying the event of crossing over (recombination) in mouse gametes. She wants to label some of the mouse DNA with fluorescent dyes to better visualize it. During what phase of the cell cycle should she look at these fluorescent DNA strands in order to most effectively study recombination?

Telophase I

Metaphase II

Anaphase II

Recombination or crossing over happens primarily in Prophase I. In anaphase and telophase, the DNA strands are separated and cannot recombine. In metaphase, homologous chromosomes line up with each other, but do not recombine.

Which of these statements is FALSE concerning mitosis and meiosis?

Both processes require DNA synthesis beforehand.

Meiosis results in a halving of chromosomes in the daughter cells, while mitosis reserves the same amount of genetic material in the daughter cells.

Both use the same order of events: prophase, metaphase, anaphase, and telophase.

Both of these processes allow genetic variation.

Mitosis and meiosis share many procedural similarities, however, it is important to remember that mitosis makes identical cells while meiosis allows genetic variability between cells.

Meiosis takes place in sex cells, which are capable of creating millions of different variations of offspring. Mitosis takes place in normal cells involved in growth and regeneration within a single, uniform organism.

Example Question #3 : Understanding Meiosis

Which of the following highlights a key difference between meiosis and mitosis?

Meiosis involves two divisions, while mitosis involves only one

All answer choices are correct

Genetic mutations can only occur in meiosis; mitosis always results in identical daughter and parent cells

There is no metaphase stage in meiosis

Meiosis includes two divisions, resulting in two pairs of haploid cells, while mitosis only involves a single division.

Both meiotic and mitotic divisions share the same phases, though there are some differences in the activity of these phases. Mutation can, and does, occur in both mitosis and meiosis. Crossing over, however, is unique to meiosis.

Example Question #4 : Understanding Meiosis

Crossing over during prophase occurs during which cycle of division?

Crossing over ensures genetic variability as it results in daughter cells with different genetic material than their parent cells. This occurs during meiosis I, but is not seen in mitosis or meiosis II.

Which of the following statements regarding mitosis and meiosis is correct?

None of the other answer choices are correct

In the human body, mitosis occurs in somatic cells while meiosis occurs in sex cells

Mitosis only occurs in prokaryotes, including all bacteria, while meiosis occurs in higher life forms

Mitosis results in higher genetic variability than meiosis

Both mitosis and meiosis occur in humans. Somatic cells (body cells) divide via mitosis, while gametes (sex cells) divide via meiosis. Because of actions such as crossing over, meiosis results in a higher genetic variability than mitosis.

Prokaryotes, such as bacteria, reproduce asexually, and are incapable of meiosis.

Which of the following chromosomal abnormalities is an example of monosomy?

Klinefelter syndrome

Down syndrome

Turner syndrome

Edwards syndrome

Turner syndrome occurs when a person is missing one sex chromosome, and only has one X-chromosome. As a result, they will be female, and may suffer a variety of symptoms. This is an example of monosomy, in which a person only has one chromosome, when they should have two.

The other three choices are examples of trisomy. Klinefelter is an instance of sex-linked trisomy, with a karyotype of XXY. Down syndrome is cause by trisomy 21, and Edwards syndrome is caused by trisomy 18.

Example Question #7 : Understanding Meiosis

Non-disjunction can result in which of the following?

Neither monosomy, nor trisomy can be caused by non-disjunction

Both monosomy and trisomy

Non-disjunction occurs when sister chromatids fail to separate during meiosis, leading subsequent daughter cells to have an unequal number of chromosomes. This can result in a cell having one extra chromosome (trisomy), or missing one chromosome (monosomy).

The following shows non-disjuction occuring in metaphase II of meiosis. "X" represents a chromosome, "|" represents a cell membrane, and "\" and "/" represent chromatids. When a sperm cell fuses with the egg formed to the left of the membrane (|), it will result in trisomy. When a sperm fuses with the egg to the right of the membrane (|), it will result in monosomy.

Metaphase I: X  X

Telophase I: X | X

Metaphase II: / \

Telophase II: / \ |

Example Question #331 : Cellular Biology

Which of the following would result in a cell with an abnormal number of chromosomes after meiosis?

Nondisjunction

Recombination

Separation of sister chromatids

Cleavage of the securin protein

Crossing over, or recombination, is a process that takes place in the earlier stages of meiosis and promotes genetic diversity. During recombination, genetic material is exchanged between two homologous chromosomes. The chromosomes ultimately contain the same amount of genetic material after recombination, and are properly separated during subsequent divisions.

The cleavage of the protein securin is actually what allows the sister chromatids to separate, a process that is essential to maintaining the correct number of chromosomes in each daughter cell.

Nondisjunction is the name given to the phenomenon in which separation of genetic material fails to occur. Either homologous chromosomes fail to separate properly in meiosis I, or sister chromatids fail to separate properly during meiosis II. The result of these nondisjunction events is one cell with an abnormally high number of chromosomes (for example trisomy) and one cell with an abnormally low number of chromosomes (for example monosomy).

Example Question #332 : Cellular Biology

Crossing over is an event that contributes to the non-identical nature of gametes. Which of the following is true regarding crossing over?

I. It occurs during prophase I

II. It involves exchange of genetic material between sister chromatids

III. It involves exchange of genetic material between homologous chromosomes

Crossing over occurs during prophase of meiosis I (prophase I). This process requires tetrad formation, where the homologous chromosomes (with their sister chromatids) pair with each other. Following tetrad formation, the genetic material from one homologous chromosome can be exchanged with that of the other. This exchange of genetic material leads to genetic recombination and results in production of non-identical gametes. Crossing over occurs only between homologous chromosomes. Sister chromatids are situated to form a single chromosome; crossing over does not include recombination of genetic material within a single chromosome.

Remember that crossing over is not a mutation and is a completely natural process for every sexually reproducing organism.

Example Question #333 : Cellular Biology

Independent assortment of traits on different chromosomes is due to the random alignment of different pairs of homologues. This alignment occurs during which of the given phases?

Prophase II

Metaphase I

Remember that the law of independent assortment states that genes on different chromosomes are passed independently of one another to offspring. This phenomenon results from the random alignment of the chromosomes along the metaphase plate. This random alignment allows genes to be segregated independently, and occurs during metaphase I.

Metaphase II involves the alignment of single chromosomes along the metaphase plate for segregation of identical sister chromatids. Remember that independent assortment is only valid for genes on different chromosomes. Genes on the same chromosomes are not passed independently of one another from parent to offspring.

Independent assortment can, thus, only occur during metaphase I, since this phase involves alignment of independent, non-identical chromosomes.

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AP Biology Test: Mitosis And Meiosis- ChAPters 12-13

This test covers the essentials of Mitosis and Meiosis cell divisions and is helpful for all the students preparing for the Advanced Placement Biology exam. It features forty basic to advanced questions to test out your knowledge on the same.

The centromere is a region in which:

Chromatids are attached to one another.

Metaphase chromosomes become aligned.

Chromosomes are grouped during telophase.

The nucleus is located prior to mitosis.

New spindle microtubules form.

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What is a chromatid?

A chromosome in G1 of the cell cycle

A replicated chromosome

A chromosome found outside the nucleus

A special region that holds two centromeres together

Another name for the chromosomes found in genetics

What is the name for the special region on a duplicated chromosome that holds the sister chromatids together?

Kinetochore

Microtubule organizer region

How many cells will be present after a zygote undergoes 5 mitotic cell divisions?

If there are 20 chromatids in a cell, how many centromeres are there, which of the following statements is not true.

Mitosis produces new nuclei with exactly the same chromosomal endowment as the parent nucleus.

Mitosis may occur without cytokinesis.

Mitosis and cytokinesis are required for asexual reproduction.

All cells come from a preexisting cell.

The mitotic spindles in prokaryotic cells are composed of microtubules.

The lettered circle in the figure above shows a diploid nucleus with four chromosomes. There are two pairs of homologous chromosomes, one long and the other short. One haploid set is symbolized as black and the other haploid set is gray. The chromosomes in the unlettered circle have not yet replicated. Choose the correct chromosomal conditions for the following stage. *At prometaphase of mitosis

The lettered circle in the figure above shows a diploid nucleus with four chromosomes. There are two pairs of homologous chromosomes, one long and the other short. One haploid set is symbolized as black and the other haploid set is gray. The chromosomes in the unlettered circle have not yet replicated. Choose the correct chromosomal conditions for the following stage. one daughter nucleus at telophase of mitosis

                a.     telophase                 b.     anaphase                 c.     prometaphase                 d.     metaphase                 e.     prophase choose the answer from the above that best fits the description: two centrosomes are arranged at opposite poles of the cell.,                 a.     telophase                 b.     anaphase                 c.     prometaphase                 d.     metaphase                 e.     prophase   choose the answer from above that best fits this description: centrioles begin to move apart in animal cells.,                 a.     telophase                 b.     anaphase                 c.     prometaphase                 d.     metaphase                 e.     prophase   choose the answer from above that best fits this description: this is the longest of the mitotic stages. ,                 a.     telophase                 b.     anaphase                 c.     prometaphase                 d.     metaphase                 e.     prophase   choose the answer from above that best fits this description: centromeres uncouple, sister chromatids are separated, and the two new chromosomes move to opposite poles of the cell., if cells in the process of dividing are subjected to colchicine, a drug that interferes with the functioning of the spindle apparatus, at which stage will mitosis be arrested, a cell containing 92 chromatids at metaphase of mitosis would, at its completion, produce two nuclei containing how many chromosomes.

If the cell whose nuclear material is shown above continues toward completion of mitosis, which of the following events would occur next?

Cell membrane synthesis

Spindle fiber formation

Nuclear envelope breakdown

Formation of telophase nuclei

Synthesis of chromatids

All of the following occur during prophase of mitosis in animal cells except

The centrioles move toward opposite poles.

The nucleolus can no longer be seen.

The nuclear envelope disappears.

Chromosomes are duplicated.

The spindle is organized.

If there are 20 centromeres in a cell at anaphase, how many chromosomes are there in each daughter cell following cytokinesis?

If there are 20 chromatids in a cell at metaphase, how many chromosomes are there in each daughter cell following cytokinesis, where do the microtubules of the spindle originate during mitosis in both plant and animal cells, all of the following occur during mitosis except the.

Condensing of chromosomes.

Uncoupling of chromatids at the centromere.

Formation of a spindle.

Synthesis of DNA.

Disappearance of the nucleolus.

If a cell has 8 chromosomes at metaphase of mitosis, how many chromosomes will it have during anaphase?

Cytokinesis usually, but not always, follows mitosis. if a cell completed mitosis but not cytokinesis, the result would be a cell with.

A single large nucleus.

High concentrations of actin and myosin.

Two abnormally small nuclei.

Two nuclei.

Two nuclei but with half the amount of DNA.

Regarding mitosis and cytokinesis, one difference between higher plants and animals is that in plants

The spindles contain microfibrils in addition to microtubules, whereas animal spindles do not contain microfibrils.

Sister chromatids are identical, but they differ from one another in animals.

A cell plate begins to form at telophase, whereas animals a cleavage furrow is initiated at that stage.

Chromosomes become attached to the spindle at prophase, whereas in animals chromosomes do not become attached until anaphase.

Spindle poles contain centrioles, whereas spindle poles in animals do not.

How do the daughter cells at the end of mitosis and cytokinesis compare with their parent cell when it was in G1 of the cell cycle?

The daughter cells have half the amount of cytoplasm and half the amount of DNA.

The daughter cells have half the number of chromosomes and half the amount of DNA.

The daughter cells have the same number of chromosomes and half the amount of DNA.

The daughter cells have the same number of chromosomes and the same amount of DNA.

The daughter cells have the same number of chromosomes and twice the amount of DNA.

The formation of a cell plate is beginning across the middle of a cell and nuclei are re-forming at opposite ends of the cell. What kind of cell is this?

An animal cell in metaphase

An animal cell in telophase

An animal cell undergoing cytokinesis

A plant cell in metaphase

A plant cell undergoing cytokinesis

Taxol is an anticancer drug extracted from the Pacific yew tree. In animal cells, taxol disrupts microtubule formation by binding to microtubules and accelerating their assembly from the protein precursor, tubulin. Surprisingly, this stops mitosis. Specifically, taxol must affect

The fibers of the mitotic spindle.

Formation of the centrioles.

Chromatid assembly.

The S phase of the cell cycle.

Which of the following are primarily responsible for cytokinesis in plant cells?

Kinetochores

Golgi-derived vesicles

Actin and myosin

Centrioles and basal bodies

Cyclin-dependent kinases

Which of the following organisms does not reproduce cells by mitosis and cytokinesis?

Banana tree

Chromosomes first become visible during ________ of mitosis.

Prometaphase

The correct sequence of steps in the M phase of the cell cycle is

Prophase, prometaphase, metaphase, anaphase, telophase.

Prophase, metaphase, prometaphase, anaphase, telophase.

Prophase, prometaphase, metaphase, anaphase, telophase, cytokinesis.

Prophase, metaphase, anaphase, telophase, cytokinesis.

Cytokinesis, telophase, prophase, prometaphase, metaphase, anaphase.

During which phases of mitosis are chromosomes composed of two chromatids?

From interphase through anaphase

From G1 of interphase through metaphase

From metaphase through telophase

From anaphase through telophase

From G2 of interphase through metaphase

Which of the following is false regarding the bacterial chromosome?

It consists of a single, circular DNA molecule.

DNA replication begins at the origin of replication.

Its centromeres uncouple during metaphase of mitosis.

It is highly folded within the cell.

It has genes that control binary fission.

In which group of eukaryotic organisms does the nuclear envelope remain intact during mitosis?

Seedless plants

Dinoflagellates

B and C only

A, B, and C

Movement of the chromosomes during anaphase would be most affected by a drug that

Reduces cyclin concentrations.

Increases cyclin concentrations.

Prevents elongation of microtubules.

Prevents shortening of microtubules.

Prevents attachment of the microtubules to the kinetochore.

Measurements of the amount of DNA per nucleus were taken on a large number of cells from a growing fungus. The measured DNA levels ranged from 3 to 6 picograms per nucleus. In which stage of the cell cycle was the nucleus with 6 picograms of DNA?

A group of cells is assayed for dna content immediately following mitosis and is found to have an average of 8 picograms of dna per nucleus. those cells would have ________ picograms at the end of the s phase and ________ picograms at the end of g2., the somatic cells derived from a single-celled zygote divide by which process.

Replication

Cytokinesis alone

Binary fission

Cytoskeletal elements play important roles in cell division.  The mitotic spindle apparatus is made of ________ and pulls sister chromatids apart, whereas the contractile ring is made of ________ and required for the separation of daughter cells at the end of the mitotic phase of the cell cycle.

Intermediate filaments; actin microfilaments

Microtubules; actin microfilaments

Microtubules; contractile filaments

Intermediate filaments; contractile filaments

Actin microfilaments; myosin

Imagine looking through a microscope at a squashed onion root tip. The chromosomes of many of the cells are plainly visible. In some cells, replicated chromosomes are aligned along the center (equator) of the cell. These particular cells are in which stage of mitosis?

If mammalian cells receive a go-ahead signal at the g1 checkpoint, they will.

Move directly into telophase.

Complete the cycle and divide.

Exit the cycle and switch to a nondividing state.

Show a drop in MPF concentration.

Complete cytokinesis and form new cell walls.

Cells that are in a nondividing state are in which phase?

What causes the decrease in the amount of cyclin at a specific point in the cell cycle.

An increase in production once the restriction point is passed

The cascade of increased production once its protein is phosphorylated by Cdk

The changing ratio of cytoplasm to genome

Its destruction by a process initiated by the activity of MPF complexes

The binding of PDGF to receptors on the cell surface

Select the term below that is most closely related to this phrase: released by platelets in the vicinity of an injury     A.    PDGF     B.    MPF     C.    protein kinase     D.    cyclin     E.    Cdk

Select the term below that is most closely related to this phrase: a general term for enzymes that activate or inactivate other proteins by phosphorylating them     a.    pdgf     b.    mpf     c.    protein kinase     d.    cyclin     e.    cdk, select the term below that is most closely related to this phrase: fibroblasts have receptors for this substance on their plasma membranes.     a.    pdgf     b.    mpf     c.    protein kinase     d.    cyclin     e.    cdk, select the term below that is most closely related to this phrase: a protein synthesized at specific times during the cell cycle that associates with a kinase to form a catalytically active complex     a.    pdgf     b.    mpf     c.    protein kinase     d.    cyclin     e.    cdk, select the term below that is most closely related to this phrase: a protein maintained at constant levels throughout the cell cycle that requires cyclin to become catalytically active     a.    pdgf     b.    mpf     c.    protein kinase     d.    cyclin     e.    cdk, select the term below that is most closely related to this phrase: triggers the cell's passage past the g2 checkpoint into mitosis     a.    pdgf     b.    mpf     c.    protein kinase     d.    cyclin     e.    cdk, select the term below that is most closely related to this phrase: the "restriction point" occurs here.     a.    g0     b.    g1     c.    s     d.    g2     e.    m, select the term below that is most closely related to this phrase: nerve and muscle cells are in this phase.     a.    g0     b.    g1     c.    s     d.    g2     e.    m.

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Critical Thinking Questions

• Prophase I of meiosis forms the tetrads. They line up at the midway point between the two poles of the cell to form the metaphase plate. There is equal chance of a microtubule fiber to encounter a maternally or a paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads.
• Prophase II of meiosis forms the tetrads. They line up at the midway point between the two poles of the cell to form the metaphase plate. There is equal chance of microtubule fiber to encounter maternally or paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads.
• Prophase I of mitosis forms the tetrads. They line up at the midway between the two poles of the cell to form the metaphase plate. There is a low chance of a microtubule fiber to encounter both a maternally and a paternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads.
• Prophase I of meiosis forms the tetrads. They line up at the midway between the two poles of the cell to form the metaphase plate. There is a chance of microtubule fiber to encounter maternally inherited chromosome. Orientation of each tetrad is independent of the orientation of other tetrads.
• Metaphase I occurs when homologous chromosome pairs align on the metaphase plate. Metaphase II has sister chromatids of chromosomes aligned at the metaphase plate.
• Metaphase I occurs when chromosomes appear in homologous pairs at the metaphase plate. Metaphase II has single sister chromatids of chromosomes on the spindle.
• Metaphase I occurs when chromosomes separate the homologous pairs on the spindle. Metaphase II has a single line of chromosomes on the plate.
• Metaphase I occurs when chromosomes appear in homologous pairs on the spindle. During metaphase II, the chromosomes line up in a double line across the spindle.
• Meiosis differs from mitosis in that the number of chromosomes is halved and genetic variation is reduced in meiosis, but not in mitosis.
• Meiosis differs from mitosis in that the number of chromosomes is halved and genetic variation is introduced in meiosis, but not in mitosis.
• The metaphase and telophase portions of meiosis and mitosis are the same, but anaphase and prophase portions are different. Meiosis and mitosis differ overall in the number of chromosomes involved.
• The prophase and telophase portions of meiosis and mitosis are the same, but anaphase I and anaphase are different. Meiosis II and mitosis are also the same and have the same number of chromosomes.
• The random alignment of homologous chromosomes at the metaphase plate ensures the random destination of the chromosomes in the daughter cells.
• Because homologous chromosomes dissociate from the spindle fibers during metaphase I, they move randomly to the daughter cells.
• The homologous chromosomes are paired tightly during metaphase I and undergo crossover as the synaptonemal complex forms a lattice around them.
• Recombination of maternal and paternal chromosomes occurs in metaphase I because the homologous chromosomes are not connected at their centromeres.
• When a sexually reproducing species and an asexually reproducing species compete for the same resources, they both “run [evolve] in the same place” because the increased genetic variation in the sexually reproducing species balances the loss in energy that species uses to find and attract mates.
• When one species gains an advantage with a favorable variation, selection pressure increases on another species with which it competes. This species must also develop an advantage or it will be outcompeted. The two species “run [evolve] to stay in the same place.”
• When one species develops a mutation that decreases its ability to survive, a competing species will become better able to survive even though it has not changed in any way. In effect, this second species “runs [evolves] to stay in the same place.”
• When two asexually reproducing species encounter rapid environmental change, the species that is also able to reproduce sexually will outcompete the other. This way it can “run [evolve] to stay in the same place.”
• genetic recombination, fertilization, meiosis
• crossing over, random chromosome assortment, genetic recombination
• meiosis, crossing over, genetic recombination
• fertilization, crossing over, random chromosome assortment
• In a diploid-dominant life cycle, the diploid multicellular stage is present, as in humans. Haploid-dominant life cycles have a multicellular haploid stage, as in fungi. In alternation of generations, both haploid-dominant and diploid-dominant stages are multicellular and the stages alternate, as in plants.
• In a diploid-dominant life cycle, the unicellular stage is present, as in humans. Haploid-dominant life cycles have a multicellular haploid stage, as in fungi. In alternation of generations, both haploid-dominant and diploid-dominant stages are multicellular and the stages alternate, as in plants.
• In a diploid-dominant life cycle, a haploid multicellular stage is present, as in humans. Haploid-dominant life cycles have a multicellular haploid stage, as in fungi. In alternation of generations, both haploid-dominant and diploid-dominant stages are multicellular and the stages alternate, as in plants.
• In a diploid-dominant life cycle, a multicellular diploid stage is present, as in algae. In a haploid-dominant life cycle, a multicellular haploid stage is present, as in plants. In alternation of generations, both haploid-dominant and diploid-dominant stages are multicellular and the stages alternate, as in humans.

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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/11-critical-thinking-questions

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