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AP Biology_Practice Exam #3 and Notes. Spring 2020 Exam. Contains AP BIOLOGY EQUATIONS AND FORMULAS. 60 MCQ with Skill, Learning Objective, and the Topic answers at the end.

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Practice Exam #3 and Notes AP® Biology For the AP® BIOLOGY EQUATIONS AND FORMULAS population AP Biology Practice Exam 9 10 AP Biology Practice Exam BIOLOGY SECTION I Time—1 hou... r and 30 minutes 60 Questions Directions: Each of the questions or incomplete statements below is followed by four suggested answers or completions. Select the one that is best in each case and then enter the letter in the corresponding space on the answer sheet. 1. Many species of corals are threatened by the increasing temperatures and decreasing pH of ocean waters. One species, Stylophora pistillata, has been found to thrive in water that is warmer and has a lower pH than the water that corals typically thrive in. Additionally, researchers have found that the tolerance for the new water conditions is heritable. Which of the following statements best explains the changes seen in S. pistillata in response to the changing water conditions? (A) The corals’ adaptation is an example of natural selection because the tolerance is in response to a changing environment and has a genetic basis. (B) The corals’ adaptation is an example of the founder effect because the majority of corals do not have a tolerance for warmer water. (C) The corals’ adaptation is an example of genetic drift because the change was a chance event and not the result of selection in response to environmental change. (D) The corals’ adaptation is an example of adaptive radiation because it has resulted in a wide range of species adapting to the new ocean conditions. 2. A researcher measured the temperature at which two different samples of double-stranded DNA denature (separate into single strands). Sample 1 denatured at a significantly lower temperature than sample 2 did. Based on the data, the researcher claims that the DNA in sample 2 is composed of a higher percentage of guanine and cytosine than the DNA in sample 1 is. Which of the following best supports the researcher’s claim? (A) The bonds between guanine and cytosine are covalent bonds, which require more energy to disrupt than those between adenine and thymine. (B) Guanine-cytosine pairs denature at a higher temperature because they have more hydrogen bonds between them than adenine-thymine pairs do. (C) Adenine-thymine pairs require less energy to separate because adenine and thymine are both single-ring bases. (D) Guanine-cytosine pairs require more energy to separate because one is a purine and one is a pyrimidine. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 11 3. Researchers studying the bacterium Escherichia coli split a population of the bacteria into two samples. Sample 1 was transformed with a plasmid containing a gene that makes the bacteria resistant to the antibiotic kanamycin. Sample 2 was transformed with a plasmid lacking the antibiotic resistance gene. A portion of each sample was then added to growth plates containing just nutrients or growth plates containing nutrients and kanamycin. After being allowed to grow for 24 hours at 37˜C, the number of colonies on each growth plate was counted (Table 1). TABLE 1. BACTERIAL GROWTH FOLLOWING TRANSFORMATION WITH OR WITHOUT THE KANAMYCIN RESISTANCE GENE Plate # Transforming Plasmid Contains Kanamycin Resistance Gene Growth Media Results 1 No Nutrient only Too many colonies to count (lawn) 2 Yes Nutrient only Too many colonies to count (lawn) 3 No Nutrient with antibiotic No colonies observed 4 Yes Nutrient with antibiotic 7 colonies observed Which of the following claims is best supported by the data in Table 1 ? (A) The transformation procedure killed all the bacteria that were added to plate 3. (B) More bacteria on plates 1 and 2 were successfully transformed than on any other plate. (C) None of the bacteria on plate 2 were successfully transformed with the kanamycin resistance gene. (D) Only the bacteria that were successfully transformed with the kanamycin resistance gene grew on plate 4. GO ON TO THE NEXT PAGE. 12 AP Biology Practice Exam 4. The fertilization of ovules from plant Q by pollen from plant R results in the production of seeds. What percent of the genes in each offspring’s chloroplasts will have been inherited from plant R ? (A) 0% (B) 25% (C) 50% (D) 100% GO ON TO THE NEXT PAGE. AP Biology Practice Exam 13 Questions 5 - 8 High levels of certain plant nutrients in runoff can lead to rapid growth of algae (an algal bloom) in aquatic ecosystems. These algal blooms are generally followed by algal death and decomposition, which consumes large amounts of dissolved oxygen in the water and results in oxygen levels insufficient to support aerobic respiration. This process is known as eutrophication. The amount of algae present in a body of water can be estimated from the amount of chlorophyll a in a sample of the water. A researcher studying eutrophication collected samples at different times of the year in a freshwater ecosystem. The samples were analyzed for total nitrogen and chlorophyll a concentration (Figure 1) as well as total phosphorus and chlorophyll a concentration (Figure 2). Figure 1. Amount of chlorophyll a in relation to the amount of total nitrogen Figure 2. Amount of chlorophyll a in relation to the amount of total phosphorous 5. Which of the following best explains how higher concentrations of nitrogen and phosphorus contribute to eutrophication? (A) An increase in the population of algae results in more nitrogen and phosphorus in the water, causing severe eutrophication. (B) Both bacteria and algae require nitrogen and phosphorus, so the algae must grow faster to compete with bacteria. (C) Nitrogen and phosphorus stimulate oxidative phosphorylation, which consumes the available oxygen in the water. (D) Algae require nitrogen and phosphorus to build macromolecules, so higher concentrations of these nutrients can result in algal blooms. 6. Which of the following was the dependent variable in the researcher’s study? (A) The concentration of chlorophyll a (B) The concentration of total nitrogen and phosphorus (C) The slope of the trend line showing the rate of change (D) The variance of the data points from the trend line GO ON TO THE NEXT PAGE. 14 AP Biology Practice Exam 7. Which of the following describes the relationship between the amount of chlorophyll a in a water sample and the concentration of nitrogen in that sample? (A) As the concentration of chlorophyll a increases, the concentration of nitrogen decreases. (B) The concentrations of nitrogen and chlorophyll a are directly correlated. (C) The concentrations of chlorophyll a and nitrogen increase throughout the year. (D) There is no relationship between the concentrations of chlorophyll a and nitrogen. 8. Which of the following investigations would enable researchers to test the claim that an increased concentration of algae has a negative effect on the number of aquatic invertebrates in the ecosystem? (A) Examining the contents of the digestive tracts of aquatic invertebrates and looking for the presence of algae (B) Examining the growth rate of algae in the absence of aquatic invertebrates (C) Counting the number of aquatic invertebrates at different concentrations of chlorophyll a in the water (D) Counting the number of aquatic invertebrates at different concentrations of nitrogen and phosphorus in the water GO ON TO THE NEXT PAGE. AP Biology Practice Exam 15 9. Amylase is an enzyme that converts carbohydrate polymers into monomers. Glycogen synthase is one of the enzymes involved in converting carbohydrate monomers into polymers. Which of the following best explains the reactions of these enzymes? (A) Amylase aids in the removal of a water molecule to break covalent bonds whereas glycogen synthase aids in the addition of a water molecule to form covalent bonds. (B) Amylase aids in the addition of a water molecule to break covalent bonds whereas glycogen synthase aids in the removal of a water molecule to form covalent bonds. (C) Amylase aids in the addition of a water molecule to form covalent bonds whereas glycogen synthase aids in the removal of a water molecule to break covalent bonds. (D) Amylase aids in the removal of a water molecule to form covalent bonds whereas glycogen synthase aids in the addition of a water molecule to break covalent bonds. 10. Which of the following describes the most likely location of cholesterol in an animal cell? (A) Embedded in the plasma membrane (B) Dissolved in the cytosol (C) Suspended in the stroma of the chloroplast (D) Bound to free ribosomes 11. Lactase is the enzyme needed to digest lactose, the sugar found in milk. Most mammals produce lactase when they are young but stop once nursing ends. In humans however, many people continue to produce lactase into adulthood and are referred to as lactase-persistent. Which of the following mutations is most likely to cause lactase persistence in humans? (A) A nucleotide substitution in the coding region of the lactase gene that interferes with the interaction between lactase and lactose (B) A mutation that turns off the expression of transcription factors that activate the expression of lactase (C) A mutation that increases the binding of transcription factors to the promoter of the lactase gene (D) The insertion of a single nucleotide into the lactase gene that results in the formation of a stop codon GO ON TO THE NEXT PAGE. 16 AP Biology Practice Exam Figure 1. The relative concentrations of both the cyclin and CDK components of MPF 12. Maturation promoting factor, MPF, is a cyclin-CDK complex that catalyzes the phosphorylation of other proteins to start mitosis. The activity level of MPF is dependent on the relative concentrations of the cyclin and CDK components of MPF (Figure 1). Based on Figure 1, which of the following describes the role of cyclin in the regulation of the cell cycle? (A) During G1 phase, the cyclin level decreases to signal the start of the resting phase of the cell cycle. (B) During M phase, the cyclin level peaks, resulting in an increased binding frequency with CDK. (C) During S phase, the cyclin level remains the same because DNA replication is occurring. (D) During G2 phase, the cyclin level remains low, causing MPF activity to decrease, which leads cells to initiate mitosis. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 17 13. Modern bananas originated from a cross between a tetraploid banana species and a diploid banana species. The product of this cross was the triploid Cavendish banana strain, a sterile hybrid that is only grown asexually. Recently, the Panama fungus has been observed more frequently parasitizing the Cavendish banana, and scientists claim that this type of banana is on the brink of extinction. Which of the following provides the best explanation of the scientists’ claim regarding the threat to the Cavendish banana? (A) The triploid genome of the Cavendish banana makes it susceptible to the negative impact of recessive alleles. (B) Having three of each chromosome interferes with normal function of the cells and increases the likelihood of contracting a disease. (C) The lack of genetic diversity of the Cavendish banana decreases the chance that a variation exists in the population that is immune to the fungus. (D) Asexual reproduction increases the mutation rate during replication, resulting in a greater chance that the offspring have a dysfunctional immune system. 14. Belding’s ground squirrels (Spermophilus beldingi) live in closely related groups. When they feed in the open, certain individuals (guard squirrels) watch for predators instead of feeding. The guard squirrels give an alarm call when a predator is sighted, allowing the rest of the group to run to safety. Researchers have noted that, because the alarm call draws attention to the guard, the guard is more likely to be caught by the predator and therefore has a lower survival rate in comparison with the other squirrels. Which of the following best explains the behavior of the guard squirrels? (A) The behavior of the guard squirrels increases the survival of close relatives that share the genes of the guard squirrels. (B) The guard squirrels confuse the predator, lowering the predator’s success rate because the predator cannot tell which squirrel is producing the sound. (C) Guard squirrels typically have recessive alleles, and by sacrificing themselves, they lessen the chance that recessive alleles will get passed on. (D) Guard squirrels are typically females who have already reproduced, so they are no longer needed by the group. 15. If 30% of the nucleotides in a single-stranded RNA molecule are adenine, then what percent are expected to be thymine? (A) 0% (B) 20% (C) 30% (D) 70% GO ON TO THE NEXT PAGE. 18 AP Biology Practice Exam Questions 16 - 19 Melanocytes are skin cells that can become cancerous and develop into a cancer known as melanoma. Some cancerous melanocytes have developed resistance to the drugs currently used to treat melanoma. As a result, researchers are investigating the effects of a new compound (drug X) on four different melanoma cell lines. Researchers analyzed cell survival in two cell lines (Figure 1) and oxygen consumption in the presence of drug X in all four cell lines (Figure 2). Figure 3 shows the proposed mechanism by which drug X affects cells. Figure 1. Percent survival of normal melanocytes and cancerous melanocyte (melanoma) lines 1 and 2 after treatment with different concentrations of drug X Figure 2. Oxygen consumption per cell in four melanoma lines after treatment with either solvent alone or solvent containing drug X. Error bars represent ±2SEx. Figure 3. Pathway leading to cell survival, growth, and proliferation and the likely effect of drug X GO ON TO THE NEXT PAGE. AP Biology Practice Exam 19 16. Which of the following best describes the data in Figure 1 ? (A) As the concentration of drug X increases, there is an increase in melanoma cell survival. (B) At a concentration above 10 mM, drug X reduces melanoma cell survival. (C) At a concentration below 25 mM, drug X increases survival in all melanoma cell lines. (D) At a concentration of 25 mM, drug X has a greater effect on melanoma line 1 than on melanoma line 2. 17. Based on Figure 2, which of the following best supports the claim that drug X inhibits oxygen consumption? (A) In the absence of drug X, melanoma lines 1 and 4 consume similar amounts of oxygen. (B) In the presence of drug X, melanoma line 2 consumes statistically more oxygen per cell than does melanoma line 3. (C) Melanoma line 3 consumes statistically less oxygen per cell in the presence of drug X than it does in the presence of the solvent alone. (D) Melanoma line 2 in the presence of drug X consumes statistically less oxygen than does melanoma line 4 in the absence of drug X. 18. Based on the information presented, which of the following best explains why the researchers measured oxygen consumption as an indicator of the effectiveness of drug X ? (A) Oxygen provides the source of electrons for cellular respiration and is necessary for energy production. (B) Oxygen consumption increases the mutation rate and causes cells to become cancerous. (C) Oxygen activates apoptosis, which results in the death of melanoma cells. (D) Oxygen accepts electrons in oxidative phosphorylation, a process necessary for melanoma cell survival. GO ON TO THE NEXT PAGE. 20 AP Biology Practice Exam 19. A researcher has identified a compound that reverses the effect of drug X. Based on Figure 3, which of the following best explains how the compound acts in the pathway to reverse the effects of drug X ? (A) (B) (C) (D) GO ON TO THE NEXT PAGE. AP Biology Practice Exam 21 20. Which of the following best explains how molecules such as O2 and CO2 can move across the membrane of a cell? (A) The majority of the cell membrane contains protein channels that allow this type of molecule into the cell. (B) The majority of the cell membrane is nonpolar, which allows small, nonpolar molecules to freely cross. (C) The phospholipids of the membrane are tightly packed, so only small molecules and ions can fit between phospholipids. (D) ATP is hydrolyzed to provide energy to help O2 and CO2 move against their concentration gradient and across the membrane. GO ON TO THE NEXT PAGE. 22 AP Biology Practice Exam 21. Scientists investigated the effect of oxygen levels on the net rate of carbon fixation in two types of plants. The plants were grown in either well-watered soil (control) or dry soil and then exposed to either 21% or 1% O2. The net rate of CO2 fixation for both types of plants was measured. Data are shown in Figure 1 and Figure 2. Figure 1. Net rate of CO2 fixation in two types of plants grown in wet (control) or dry soil at 21% O2 Figure 2. Net rate of CO2 fixation in two types of plants grown in wet (control) or dry soil at 1% O2 Which of the following statements about the rate of CO2 fixation in the two types of plants is supported by the data shown in the figures? (A) At 21% O2, plant type 2 has a lower rate of CO2 fixation than plant type 1 does in both types of soil. (B) At 1% O2, plant type 2 has a higher rate of CO2 fixation than plant type 1 does in the dry soil but not in the control soil. (C) Plant types 1 and 2 have a statistically different rate of CO2 fixation in both soil types at both oxygen levels. (D) The rate of CO2 fixation is the same in both types of plants in the control soil at both oxygen levels. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 23 22. A researcher is conducting an experiment in which cells in different phases of the cell cycle are fused together. The researcher then records what happens to the nuclei of the resulting cell (Table 1). TABLE 1. COMBINATIONS OF CELLS THAT WERE FUSED AND THE PHASE OF NUCLEI IN THE RESULTING CELL Phase of Cell 1 Phase of Cell 2 Phase of Nuclei in Resulting Cell S phase G1 phase Two S-phase nuclei S phase G2 phase One S-phase nucleus and one G2-phase nucleus G1 phase G2 phase One G1-phase nucleus and one G2-phase nucleus Interphase M phase Two M-phase nuclei Which of the following research questions is best addressed by the experiment? (A) How do chemical messengers affect a cell’s transition between the phases of the cell cycle? (B) How does the number of chromosomes affect when a cell transitions to the next phase of the cell cycle? (C) How does the amount of genetic information change throughout the cell cycle? (D) How does the checkpoint at G2 serve to prevent the transmission of mutations? GO ON TO THE NEXT PAGE. 24 AP Biology Practice Exam Figure 1. Protein synthesis in a prokaryotic cell 23. Which of the following best describes a characteristic of the process shown in Figure 1 that is unique to prokaryotes? (A) The mRNA is synthesized in a 5¢ to 3¢ direction. (B) A single strand of the DNA is being used as a template for the transcription of the mRNA. (C) The translation of the mRNA is occurring while the mRNA is still being transcribed. (D) The enzyme that is transcribing the mRNA is RNA polymerase. 24. A scientist is investigating the possibility that two traits in a particular plant are determined by genes that are on the same chromosome. The scientist crossed a plant that is homozygous dominant for both traits with a plant that is homozygous recessive for both traits. The heterozygous offspring in the F1 generation were then crossed with a plant that is homozygous recessive for both traits. The results expected if the genes independently assort and the observed results are presented in the table. Phenotype Expected Number in F2 Observed Number in F2 Long stems, white flowers 25 17 Short stems, red flowers 25 19 Long stems, red flowers 25 31 Short stems, white flowers 25 33 Total number of plants 100 100 Which of the following critical values should the scientist use for the chi-square analysis of the data? (A) 3.00 (B) 3.84 (C) 7.81 (D) 8.00 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 25 25. Sugar gliders and Northern flying squirrels are mammals that have many morphological similarities despite being genetically different. Both are nocturnal and have large eyes to help them see in the dark. Both live in trees and have flaps of skin connecting their front and back legs, and when their front and back legs are extended, the flaps allow the animals to glide from one tree branch to another. Sugar gliders are native to Australia, and Northern flying squirrels are native to North America. Which of the following best explains the morphological similarities between these two species? (A) The similarities in traits indicate the species are the result of divergent evolution from a common ancestor that had the same traits that these two species share. (B) The traits evolved through sympatric speciation, which often results in species being highly similar because they evolve in the same area. (C) The two species evolved as the result of two separate bottleneck events that reduced the existing populations to a few individuals who happened to have the same traits. (D) The similarities between the species evolved independently as a result of similar selective pressures in each species’ environment. GO ON TO THE NEXT PAGE. 26 AP Biology Practice Exam Figure 1. Change in the population size of sea lions over time. Error bars represent ±2SE x. 26. Which of the following best estimates the population size of the sea lions in 2000 based on the data shown in Figure 1 ? (A) 100,000 (B) 125,000 (C) 175,000 (D) 285,000 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 27 27. The figure shows a phylogenetic tree of various members of the order Proboscidea, which includes modern elephants. Which of the following claims is best supported by the information in the figure ? (A) The Asian and African elephants are the most closely related species shown on the tree. (B) The mammoth diverged from its most recent common ancestor with African elephants before the mastodon diverged from its most recent common ancestor with Stegodons. (C) The mastodon and the Stegodon diverged from their common ancestor 2 million years ago. (D) The common ancestor of the African elephant and the mastodon is the Palaeomastodon. 28. Which of the following describes the most direct effect of a mutation in the DNA that encodes a cell’s rRNA ? (A) The cell’s ability to transport the amino acids needed for translation will be reduced. (B) The cell’s ability to transcribe RNA transcripts that will be translated will be reduced. (C) The cell’s ability to properly assemble ribosomes and initiate translation will be reduced. (D) The cell’s ability to modify proteins after they have been assembled will be reduced. GO ON TO THE NEXT PAGE. 28 AP Biology Practice Exam Questions 29 - 32 Figure 1. Incidence of protoporphyria in a particular family Protoporphyria is a genetic disorder characterized by an extreme sensitivity to sunlight. One form of protoporphyria is caused by a mutation in the ALAS2 gene that results in the accumulation of protoporphyrin, an organic compound, in the blood, liver, and skin. The pedigree in Figure 1 shows the incidence of protoporphyria in a particular family. 29. Which of the following best describes the genotype of the individual identified with an asterisk in the pedigree in Figure 1 ? (A) Two dominant ALAS2 alleles (B) Two recessive ALAS2 alleles (C) One dominant ALAS2 allele and one recessive ALAS2 allele (D) One recessive ALAS2 allele and no second allele for the ALAS2 gene 30. Which of the following best describes the inheritance pattern illustrated in Figure 1 ? (A) Protoporphyria has an autosomal recessive inheritance pattern. (B) Protoporphyria has an X-linked dominant inheritance pattern. (C) Protoporphyria has an X-linked recessive inheritance pattern. (D) Protoporphyria has a mitochondrial inheritance pattern. 31. One mutation in ALAS2 that is associated with protoporphyria is a four-nucleotide deletion. The protein expressed from the mutant allele is 20 amino acids shorter than the wild-type protein. Which of the following best explains why a shortened protein is produced? (A) The mutation disrupts the start codon, preventing the ribosome from beginning translation. (B) The mutation introduces a premature stop codon, causing translation to end early. (C) The mutation changes the gene’s regulatory region, causing unregulated gene expression. (D) The mutation affects posttranscriptional modifications by preventing the removal of introns. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 29 32. A researcher claims that an individual has protoporphyria, based on a physical exam. Which of the following techniques would most likely confirm the researcher’s claim? (A) Transforming bacteria with the mutant variation of ALAS2 to measure gene expression (B) Culturing cells from the individual in the lab and measuring the cells’ growth rate (C) Using light microscopy to examine the individual’s chromosomes during metaphase (D) Determining the nucleotide sequence of the individual’s ALAS2 alleles 33. Which of the following is evidence that eukaryotes and prokaryotes share a common ancestor? (A) All eukaryotes and prokaryotes contain linear DNA. (B) All eukaryotes and prokaryotes contain ribosomes. (C) All eukaryotes and prokaryotes use organic molecules as an energy source. (D) All eukaryotes and prokaryotes are capable of mitosis. 34. Wolves, once native to Yellowstone National Park, were hunted to the point of complete extinction in the park. As a result, the elk population in the park flourished, putting extra demands on the carrying capacity of the park. Many other species, such as certain trees and beavers, were negatively affected by the increase in elk. Years later, wolves were reintroduced into the park. While the elk population decreased after the wolves returned, the beaver and songbird populations began to increase as did the populations of various plant species. Which of the following best explains how wolves are a keystone species in this ecosystem? (A) Wolves help balance the population sizes of other species, allowing more species to thrive in the wolves’ presence than in their absence. (B) Wolves prey on certain species, increasing competition among other species, which results in a decrease in biodiversity in the ecosystem. (C) Wolves are not affected by factors that typically limit the population size; therefore their population can grow exponentially, increasing the demand on the parks’ resources. (D) Wolves prey on other species without having any natural predators themselves; therefore their presence significantly decreases many populations within the community. GO ON TO THE NEXT PAGE. 30 AP Biology Practice Exam Figure 1. A model of epinephrine signaling 35. Two types of cells, alpha and beta cells, produce signaling molecules that affect blood sugar levels in opposite ways (Figure 1). Epinephrine is a chemical, often released during periods of exercise, that ultimately causes an increase in blood sugar levels in the body. Based on Figure 1, which of the following best explains how exercise causes blood glucose levels to rise? (A) Epinephrine inhibits alpha cells, causing the release of glucagon, and activates beta cells, blocking the release of insulin. (B) Epinephrine activates alpha cells, blocking the release of glucagon, and inhibits beta cells, causing the release of insulin. (C) Epinephrine activates alpha cells, causing the release of glucagon, and inhibits beta cells, blocking the release of insulin. (D) Epinephrine inhibits alpha cells, blocking the release of glucagon, and activates beta cells, causing the release of insulin. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 31 36. Pesticides are chemicals that are often sprayed on crops to kill plant-eating insects, preventing damage to the crops. While pesticides are effective initially, many researchers claim that any single pesticide will see reduced effectiveness in as little as ten to fifteen years. Which of the following best supports the claim by scientists that the pesticides will eventually lose their effectiveness? (A) Insects will evolve to avoid any plant sprayed with pesticides to increase their chances of survival. (B) Insects that are naturally resistant to the pesticide will survive and reproduce more than the insects that are sensitive to the pesticide. (C) The insects will build up a tolerance to the pesticides, and eventually the pesticide will not affect them. (D) The pesticides will increase the mutation rate in the insects, resulting in higher genetic diversity and higher survival rate. GO ON TO THE NEXT PAGE. 32 AP Biology Practice Exam 37. Newborn babies and hibernating animals contain a large amount of brown adipose (fat) tissue (BAT). Certain proteins in the BAT cells increase the permeability of the inner mitochondrial membrane to protons, disrupting the proton gradient. Which of the following best predicts the effect of disrupting the proton gradient in BAT ? (A) The pH of the matrix will increase, allowing the production of more ATP per gram of substrate. (B) The pH of the intermembrane space will decrease, allowing a steeper proton gradient to form. (C) Electron transport and oxidative phosphorylation will be decoupled, generating more heat but less ATP. (D) The number of protons available to pass through ATP synthase will increase, resulting in more ATP. 38. In pea plants, flower color and the length of the flower’s pollen grains are genetically determined. Researchers studying pea plants crossed homozygous dominant pea plants with homozygous recessive pea plants. The F1 plants were then crossed, and the number of offspring with each phenotype was recorded. The researchers’ observed data, however, differed from the expected data. The researchers did a chi-square analysis and calculated the chi-square value to be 5.5. Based on their calculation, the researchers would most likely conclude which of the following? (A) The genes that determine these two traits are likely on the same chromosome. (B) The differences between observed data and expected data are due to chance. (C) The genes mutated since the researchers began the experiment and now have a different inheritance pattern. (D) The allele frequencies of the offspring have changed, suggesting the population is evolving. 39. A massive increase in the growth of a specific species of algae resulted in record-breaking levels of a potentially dangerous toxin being released into the water. A researcher hypothesizes that the unusual growth of this algal species was caused by an increase in water temperature. The researcher designs an experiment to test the hypothesis. Which of the following is the dependent variable in the researcher’s experiment? (A) The growth of the algae (B) The temperature of the water (C) The concentration of toxin in the water (D) The different species of algae growing in the water GO ON TO THE NEXT PAGE. AP Biology Practice Exam 33 40. Which of the following is the most likely effect of a mutation in the gene coding for a DNA repair enzyme? (A) The cell containing the mutation will divide more frequently because the cell cycle checkpoints will not function properly. (B) Mutations will accumulate more quickly because the cell will not be able to fix errors in replication. (C) The mutated gene will not be transcribed because RNA polymerase cannot transcribe mutated DNA. (D) The cell will immediately undergo apoptosis so that mutated DNA is not replicated in future rounds of cell division. GO ON TO THE NEXT PAGE. 34 AP Biology Practice Exam Questions 41 - 45 Certain chemicals, including sodium fluoride (NaF), are capable of inhibiting specific steps of glycolysis. Figure 1 shows the steps of the glycolysis pathway, indicating where various macromolecules enter the pathway as well as the specific reaction inhibited by NaF. Figure 1. Key steps in the metabolic pathway of glucose GO ON TO THE NEXT PAGE. AP Biology Practice Exam 35 41. If NaF is added to cells undergoing cellular respiration, which of the following will most likely accumulate in the cells? (A) Glucose Transporter (B) Pyruvate (C) Phosphoenolpyruvate (PEP) (D) 2-phosphoglycerate 42. Based on Figure 1, the net number of ATP molecules produced during glycolysis from the metabolism of a single glucose molecule is closest to which of the following? (A) 0 (B) 2 (C) 4 (D) 8 43. An increase in the concentration of protons in the cytosol will most likely have which of the following effects on glycolysis? (A) Glycolytic enzymes will denature as a result of the increased H+ concentration. (B) Reaction rate will increase as a result of the increased H+ concentration. (C) H+ will replace phosphorous and inhibit ATP formation from ADP. (D) The water potential will increase, resulting in a decrease in the rate of glycolysis. 44. Which of the following describes why a glucose transporter is needed to move glucose into the cell? (A) Glucose is nonpolar and requires ATP to move across the membrane. (B) Glucose molecules are polar and need to move from low concentration to high concentration. (C) Glucose molecules are charged, and charged molecules are only ever actively transported. (D) Glucose is large and polar and cannot pass through the phospholipid bilayer. 45. Tarui disease is an inherited disorder that is caused by mutations in PFKM, the gene that encodes a subunit of phosphofructokinase, an enzyme in the glycolysis pathway. Individuals with Tarui disease produce little or no functional phosphofructokinase in skeletal muscle cells. Based on Figure 1, which of the following best explains why a low carbohydrate diet is recommended for those with Tarui disease? (A) Carbohydrates are capable of undergoing lactic acid fermentation, and amino acids and fatty acids are not. (B) Carbohydrate metabolism requires all the reactions of glycolysis, and amino acids and fatty acids do not. (C) Carbohydrates cannot be used to synthesize important metabolic enzymes like amino acids and fatty acids can be. (D) Carbohydrates cannot be stored, while amino acids and fatty acids can be. GO ON TO THE NEXT PAGE. 36 AP Biology Practice Exam 46. A mutation in the upland cotton plant causes the development of chloroplasts with a single outer membrane and no internal membranes. Which of the following would most likely be observed in chloroplasts of cotton plants with this mutation? (A) They would be unable to remove waste products, because internal transport proteins would not be present. (B) They would be unable to generate the ATP and NADPH needed to make sugars, because these processes occur on membranes within the chloroplast. (C) They would be unable to take up carbon dioxide, because CO2 is transported into the chloroplast in membrane-bound vesicles. (D) They would be unable to move within the cell, because only organelles with double membranes are mobile. 47. A researcher is crossing two organisms that are heterozygous for three Mendelian, unlinked traits (XxYyZz). Which of the following is the fraction of offspring that are predicted to have the genotype xxyyzz ? (A) 1 / 64 (B) 1 / 32 (C) 1 / 16 (D) 1 / 8 48. Mosquitoes are frequently a target of insect control strategies because of their ability to spread disease. One strategy is to introduce guppies, a type of freshwater fish, into areas where mosquitoes are a problem. Guppies can consume large numbers of mosquito larvae, which cuts down on the number of mosquitoes living to adulthood. Researchers assessing the effectiveness of the guppy solution are concerned that the introduction of guppies might cause more problems than it is fixing. Guppies are hardy, tolerant of a wide range of environmental conditions, and fast-reproducing. Which of the following predicts the most likely ecological problem that would occur if guppies are introduced into new areas to control the mosquito population? (A) The guppies might have no natural predators, which will result in a dramatic increase in the guppy population and an increase in competition for other native species. (B) The guppy population will likely evolve to consume another food source instead of the mosquito larvae. (C) The larvae will evolve a mechanism that will enable them to avoid predation from the guppies. (D) Predators in the area will consume the guppies, requiring the addition of more guppies on a regular basis. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 37 Figure 1. Diagram of the cell cycle with key checkpoints 49. Which of the following describes a mutation that would lead to an increase in the frequency of nondisjunction? (A) A mutation affecting checkpoint 1 proteins that forces cells to enter G0 (B) A mutation affecting checkpoint 2 proteins that allows cells to divide with DNA damage (C) A mutation affecting checkpoint 3 proteins that prevents attachment of spindle fibers (D) A mutation affecting checkpoint 2 proteins that prevents duplication of the chromosomes 50. Scientists studying a wild population of mantled howler monkeys found the average birth rate to be 0.22 and the average death rate to be 0.12. At the start of the study, the population consisted of 13 monkeys. Assuming no immigration or emigration, which of the following best describes the change in population size that will occur over the next year? (A) The population size will decrease because a population of 13 is too small to sustain itself. (B) The population will decrease because more monkeys are dying each year than are being born. (C) The population will increase exponentially because it is so small that there are likely few competitors and abundant resources. (D) The population will increase because more monkeys are being born each year than are dying. 51. Which of the following best explains how the prokaryotic expression of a metabolic protein can be regulated when the protein is already present at a high concentration? (A) Repressor proteins can be activated and bind to regulatory sequences to block transcription. (B) Transcription factors can bind to regulatory sequences to increase RNA polymerase binding. (C) Regulatory proteins can be inactivated to increase gene expression. (D) Histone modification can prevent transcription of the gene. GO ON TO THE NEXT PAGE. 38 AP Biology Practice Exam Figure 1. Activity levels of two digestive enzymes over a range of pH 52. Trypsin and pepsin are enzymes that function in different areas of the digestive tract. One functions in the stomach, where the pH is between 1.5 and 3.5, while the other functions in the small intestines, where the pH is between 6 and 8. Based on Figure 1, which of the following best describes where each enzyme functions? (A) Pepsin works in the intestines because the optimal pH for pepsin is basic. (B) Trypsin works in the stomach because the optimal pH for trypsin is basic. (C) Pepsin works in the stomach because the optimal pH for pepsin is acidic. (D) Trypsin works in the stomach because the optimal pH for trypsin is acidic. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 39 Questions 53 - 57 Figure 1. Experimental steps taken to determine whether the nutrient source in a growth chamber affects mating preference A population of fruit flies (Drosophila pseudoobscura) grown on a typical nutrient source was separated into several growth chambers. Each of the new chambers was assigned a different source of carbohydrates for nutrition, either starch or maltose, and maintained for many generations (Figure 1). After a year, male and female flies were paired up in different combinations. The instances of mating between the males and females grown on the same nutrient source and males and females grown on different nutrient sources were recorded (Table 1). TABLE 1. MATING INSTANCES BETWEEN MALES AND FEMALES GROWN ON THE SAME OR DIFFERENT NUTRIENT SOURCES Females Grown on Starch Females Grown on Maltose Males Grown on Starch 22 9 Males Grown on Maltose 8 20 GO ON TO THE NEXT PAGE. 40 AP Biology Practice Exam To test whether the mating preferences were simply the result of being isolated in separate growth chambers, the researchers tested the mating preferences of flies that were both grown on starch, but in either the same or different chambers (Table 2). TABLE 2. MATING INSTANCES BETWEEN MALES AND FEMALES GROWN ON THE SAME NUTRIENT SOURCE IN THE SAME OR DIFFERENT CHAMBERS Females Grown on Starch in Chamber 1 Females Grown on Starch in Chamber 2 Males Grown on Starch in Chamber 1 18 15 Males Grown on Starch in Chamber 2 12 15 53. Which of the following best represents the null hypothesis for the experiment detailed in Figure 1 ? (A) The mating preferences of flies are not dependent on the nutrient source on which they are grown. (B) Prolonged exposure to a different diet can change behavior in fruit flies. (C) When crossed, flies grown on the same medium will not produce viable offspring. (D) Growing flies in a confined chamber will not create a barrier to mating in fruit flies. 54. For a population to be in Hardy-Weinberg equilibrium, certain conditions must be met. Which of the following best explains a characteristic of the fly populations that meets a condition of Hardy-Weinberg equilibrium? (A) The population size of the flies was very small, which increases the likelihood of random mutations having a greater impact. (B) Since the flies were confined in chambers, they did not experience any migration, so no new alleles were introduced through migration. (C) Flies have a relatively short generation time and therefore are more likely to pass mutations to the next generation. (D) The different food sources may have exerted selective pressure on the fly populations, giving an advantage to certain flies. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 41 55. Which of the following claims is best supported by the data in Tables 1 and 2 ? (A) The population of flies has undergone sympatric speciation. (B) The flies have developed a form of behavioral preference that could lead to divergence. (C) A postzygotic barrier has evolved between the populations of flies that prevents the birth of viable offspring. (D) The flies have undergone allopatric speciation, since only flies from the same chamber mated. 56. Starch is a complex carbohydrate that is digested by the enzyme amylase. A mutation in the gene that encodes amylase prevents the digestion of starch. The inability to digest starch is an autosomal recessive trait. In a population of 1,000 flies, 410 were unable to break down starch. Which of the following is closest to the number of flies expected to be heterozygous for the amylase mutation assuming all conditions of Hardy-Weinberg equilibrium are met? (A) 0.13 (B) 0.36 (C) 0.46 (D) 0.64 57. Which of the following combinations of flies showed the highest number of mating instances? (A) Males grown on maltose paired with females grown on starch (B) Males grown on starch paired with females grown on starch (C) Males grown on maltose paired with females grown on maltose (D) Males grown on starch paired with females grown on maltose 58. The primary function of the kidney is to exchange molecules across a membrane between the blood and the urine. One type of kidney cell has a basic rectangular shape, except for a single surface, which is lined with tiny, finger-like projections that extend into the surrounding extracellular space. Which of the following best explains the advantage these projections provide the cell? (A) The projections increase the volume of the cell without affecting the surface area, which increases the metabolic needs of the cell. (B) The projections increase the surface area–to-volume ratio of the cell, which allows for more efficient nutrient exchange with the environment. (C) The projections increase the speed at which an individual molecule can move, resulting in faster nutrient exchange with the environment. (D) The projections increase the selectivity of the membrane because the small size of the projections limits the number of transport proteins that can be embedded in the membrane. GO ON TO THE NEXT PAGE. 42 AP Biology Practice Exam Figure 1. Diagram of the electron transport chain and ATP synthase in the membrane of mitochondria 59. On average, more ATP can be produced from an NADH molecule than can be produced from a molecule of FADH2. Based on Figure 1, which of the following best explains the difference in ATP production between these two molecules? (A) NADH contributes more electrons to the electron transport chain than FADH2 does and therefore provides more energy to pump protons. (B) The electrons of FADH2 are transferred through three complexes of the electron transport chain whereas those of NADH are transferred through all four complexes. (C) FADH2 contributes more protons to the mitochondrial matrix, which decreases the proton gradient. (D) The protons contributed by FADH2 are combined with O2 to make water and are not pumped across the membrane. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 43 60. Which of the following best explains why ligase is required for DNA replication? (A) The lagging strand cannot be replicated continuously, and ligase is needed to join the fragments. (B) Ligase forms the hydrogen bonds between complementary bases in the two strands of DNA. (C) Ligase facilitates the binding of RNA polymerase to the promoter region. (D) Ligase enables the newly synthesized DNA to twist into a double helix. GO ON TO THE NEXT PAGE. 44 AP Biology Practice Exam ____________________________________________________________________________ END OF SECTION I IF YOU FINISH BEFORE TIME IS CALLED, YOU MAY CHECK YOUR WORK ON THIS SECTION. DO NOT GO ON TO SECTION II UNTIL YOU ARE TOLD TO DO SO. MAKE SURE YOU HAVE DONE THE FOLLOWING: • PLACED YOUR AP ID LABEL ON YOUR ANSWER SHEET • WRITTEN AND GRIDDED YOUR AP ID CORRECTLY ON YOUR ANSWER SHEET • TAKEN THE AP EXAM LABEL FROM THE FRONT OF THIS BOOKLET AND PLACED IT ON YOUR ANSWER SHEET AP Biology Practice Exam 45 Number of Questions Percent of Total Score Writing Instrument Electronic Device Suggested Time AP® Biology Exam SECTION II: Free Response DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO. At a Glance Total Time 1 hour and 30 minutes 6 50% Pen with black or dark blue ink Calculator allowed Weight Approximately 25 minutes per long question, and 10 minutes per short question. Approximate weights: Questions 1 and 2: 26% each Questions 3 6: 12% each Instructions The questions for Section II are printed in this booklet. You may use the unlined pages to organize your answers and for scratch work, but you must write your answers on the labeled pages provided for each question. Each answer should be written in paragraph form; an outline or bulleted list alone is not acceptable. Do not spend time restating the questions or providing more than the number of examples called for. For instance, if a question calls for two examples, you can earn credit only for the first two examples that you provide. Labeled diagrams may be used to supplement discussion, but unless specifically called for by the question, a diagram alone will not receive credit. Write clearly and legibly. Begin each answer on a new page. Do not skip lines. Cross out any errors you make; crossed-out work will not be scored. Manage your time carefully. You may proceed freely from one question to the next. You may review your responses if you finish before the end of the exam is announced. Mean Standard Deviation Standard Error of the Mean Chi-Square n 1 x = Âxi n i=1 Â(xi - x)2 s = n - 1 s SE = x n 2 o e     e  2 Chi-Square Table Laws of Probability Hardy-Weinberg Equations Statistical Analysis and Probability If A and B are mutually exclusive, then: P(A or B) = P(A) + P(B) If A and B are independent, then: P(A and B) = P(A)  P(B) p + q = 1 q = frequency of allele 2 in a population x = sample mean n = sample size s = sample standard deviation (i.e., the sample-based estimate of the standard deviation of the population) o = observed results e = expected results  = sum of all Degrees of freedom are equal to the number of distinct possible outcomes minus one. Metric Prefixes Mode = value that occurs most frequently in a data set Median = middle value that separates the greater and lesser halves of a data set Mean = sum of all data points divided by number of data points Range = value obtained by subtracting the smallest observation (sample minimum) from the greatest (sample maximum) AP® BIOLOGY EQUATIONS AND FORMULAS p value Degrees of Freedom 1 2 3 4 5 6 7 8 0.05 3.84 5.99 7.81 9.49 11.07 12.59 14.07 15.51 0.01 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 p2 + 2pq + q2 = 1 p = frequency of allele 1 in a population Factor Prefix Symbol 109 giga G 106 mega M 103 kilo k 10– 1 deci d 10– 2 centi c 10– 3 milli m 10– 6 micro μ 10 – 9 nano n 10– 12 pico p 46 AP Biology Practice Exam AP Biology Practice Exam 47 Population Growth Exponential Growth dY dt dN = B - D dt dN = r N dt max dN K - N = r N dt max ( K ) Logistic Growth Rate Rate and Growth dY = amount of change dt = change in time B = birth rate D = death rate N = population size K = carrying capacity r max = maximum per capita growth rate of population The Solute Potential of a Solution Water Potential ( Y ) Y = Y P + YS Y = P pressure potential YS = solute potential The water potential will be equal to the solute potential of a solution in an open container because the pressure potential of the solution in an open container is zero. YS = -iCRT i = ionization constant (1.0 for sucrose because sucrose does not ionize in water) C = molar concentration R = pressure constant ( R = 0.0831 liter bars/mole K) T = temperature in Kelvin (ºC + 273) Surface Area of a Rectangular Solid Surface Area of a Cube 4 3 V  r 3 Surface Area of a Sphere Surface Area of a Cylinder Volume of a Sphere Volume of a Rectangular Solid Volume of a Cylinder Volume of a Cube Surface Area and Volume r = radius 2 SA  4r l = length h = height V  lwh w = width SA  2lh  2lw  2wh s = length of one side of a cube 2 V   r h 2 SA  2rh  2r SA = surface area 3 V  s SA  6s2 V = volume 2 n = 1 - Â( ) N Simpson’s Diversity Index Diversity Index ݊ ൌ total number of organisms of a particular species ܰ ൌ total number of organisms of all species pH = – log[H+] 48 AP Biology Practice Exam BIOLOGY SECTION II Time—1 hour and 30 minutes 6 Questions Directions: Questions 1 and 2 are long free-response questions that require about 25 minutes each to answer. Questions 3 through 6 are short free-response questions that require about 10 minutes each to answer. Read each question carefully and completely. Write your response in the space provided for each question. Only material written in the space provided will be scored. Answers must be written out in paragraph form. Outlines, bulleted lists, or diagrams alone are not acceptable. Question 1 is on the following page. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 49 1. Cryptosporidium parvum (C. parvum) is a single-celled, eukaryotic parasite that infects human cells in the digestive system and causes illness. Although it is a eukaryote, C. parvum does not have functional mitochondria and generates ATP only through glycolysis. C. parvum uses the enzyme lactate dehydrogenase to perform fermentation after glycolysis. Two chemicals, gossypol and FX11, are noncompetitive inhibitors of lactate dehydrogenase. Researchers investigated the effectiveness of gossypol and FX11 as drugs to kill C. parvum. In the experiment, human cells were treated with different concentrations of either gossypol or FX11 after infection with C. parvum, and the relative growth of C. parvum compared with that of control cells was measured (Figures 1 and 2). Figure 1. Effect of different concentrations of gossypol on C. parvum growth Figure 2. Effect of different concentrations of FX11 on C. parvum growth (a) Describe how C. parvum obtains the glucose it needs for glycolysis after it has infected another cell. Explain the role of lactate dehydrogenase in enabling C. parvum to continue producing ATP by glycolysis. (b) Identify the independent variable used in the experiment. Identify the difference between the control cells and the experimental cells used in the experiment. Justify the researchers using a different range of concentrations for FX11 than was used for gossypol. (c) Based on the data in Figure 1, identify the concentration of gossypol that reduced C. parvum growth to 50% of that in control cells. (d) Researchers discovered a strain of C. parvum that expresses a functional variation of the lactate dehydrogenase gene. A DNA sequence comparison showed that the variant differs from the normal sequence in the region that codes for the enzyme’s allosteric site. Predict the effect of FX11 treatment on C. parvum cells that express this variant of lactase dehydrogenase. Provide reasoning to support your prediction. Explain how gossypol and FX11 might be used as drugs to treat C. parvum infections in humans without negatively affecting human cells. GO ON TO THE NEXT PAGE. 50 AP Biology Practice Exam THIS PAGE MAY BE USED FOR TAKING NOTES AND PLANNING YOUR ANSWERS. NOTES WRITTEN ON THIS PAGE WILL NOT BE SCORED. WRITE ALL YOUR RESPONSES ON THE LINED PAGES. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 51 PAGE FOR ANSWERING QUESTION 1 GO ON TO THE NEXT PAGE. 52 AP Biology Practice Exam ADDITIONAL PAGE FOR ANSWERING QUESTION 1 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 53 ADDITIONAL PAGE FOR ANSWERING QUESTION 1 GO ON TO THE NEXT PAGE. 54 AP Biology Practice Exam ADDITIONAL PAGE FOR ANSWERING QUESTION 1 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 55 2. Many types of cancer are treated with a combination of therapies. In lung cancer, some tumors respond well to the drug paclitaxel followed by radiation treatment. Paclitaxel is a chemical that disrupts mitosis. Instead of spindle fibers originating from the two sides (poles) of the cell, paclitaxel-treated cells develop three poles and then divide into three cells (tripolar division). Radiation therapy is more effective on tumor cells that have undergone tripolar division than on cells that have undergone normal mitosis. Researchers treated cancer cells in the lab with different concentrations of paclitaxel for 15 hours. The researchers then determined the average percent of mitotic cells that were tripolar. The results are shown in Table 1. TABLE 1. EFFECT OF PACLITAXEL CONCENTRATION ON PERCENT OF MITOTIC CELLS THAT WERE TRIPOLAR Concentration of Paclitaxel (nM) Average Percent of Mitotic Cells that were Tripolar (±2SE x) 0 0.0 ± 0.0 2 17.0 ± 3.0 4 48.0 ± 3.5 6 65.0 ± 5.0 8 70.0 ± 4.0 10 50.0 ± 2.0 The AURKA gene encodes an enzyme that helps assemble the spindle fibers, which signals the cells to continue through mitosis. When researchers analyzed the levels of AURKA protein in different types of cancer cells, they found that cancer cells expressing high levels of AURKA protein had more tripolar divisions when treated with paclitaxel, than did cancer cells expressing low levels of AURKA protein. (a) Describe the situations in which a normal human cell would enter the cell cycle and undergo mitotic cell division. Explain how spindle fibers help ensure the products of mitosis are two identical cells with a full set of chromosomes. (b) Using the template in the space provided for your response, construct an appropriately labeled graph that represents the data shown in Table 1. Based on the data, determine the concentration(s) of paclitaxel that is (are) most effective in causing tripolar cell division. (c) Based on the data, identify the lowest level of paclitaxel that will allow for at least 50% of the cells to be tripolar. From the start codon through the stop codon, the length of the fully processed AURKA mRNA is 1,212 nucleotides. Calculate the number of amino acids in the polypeptide chain coded for by the mRNA. (d) Predict the effect of a mutation that prevents the expression of AURKA on a normal (noncancerous) cell. GO ON TO THE NEXT PAGE. 56 AP Biology Practice Exam THIS PAGE MAY BE USED FOR TAKING NOTES AND PLANNING YOUR ANSWERS. NOTES WRITTEN ON THIS PAGE WILL NOT BE SCORED. WRITE ALL YOUR RESPONSES ON THE LINED PAGES. GO ON TO THE NEXT PAGE. AP Biology Practice Exam 57 PAGE FOR ANSWERING QUESTION 2 (data table reprinted for reference) TABLE 1. EFFECT OF PACLITAXEL CONCENTRATION ON PERCENT OF MITOTIC CELLS THAT WERE TRIPOLAR Concentration of Paclitaxel (nM) Average Percent of Mitotic Cells that were Tripolar (±2SE x) 0 0.0 ± 0.0 2 17.0 ± 3.0 4 48.0 ± 3.5 6 65.0 ± 5.0 8 70.0 ± 4.0 10 50.0 ± 2.0 GO ON TO THE NEXT PAGE. 58 AP Biology Practice Exam ADDITIONAL PAGE FOR ANSWERING QUESTION 2 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 59 ADDITIONAL PAGE FOR ANSWERING QUESTION 2 GO ON TO THE NEXT PAGE. 60 AP Biology Practice Exam ADDITIONAL PAGE FOR ANSWERING QUESTION 2 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 61 3. In mice, fur color is a genetically determined trait. To observe the effects of natural selection on fur color in mice, scientists set up six enclosures with either light- or dark-colored sand on the ground. The enclosures were isolated from all ground predators and wild mice but accessible to predatory birds. The scientists placed equal numbers of light- and dark-colored mice into each enclosure. A total of 500 mice were used in the experiment. After several generations, the scientists sampled the mice and found that populations in the light sand enclosures were, on average, lighter in color than the original population, while those in the dark sand enclosures were, on average, darker in color than the mice in the original population. (a) Describe the way the scientists will determine the evolutionary fitness of the mice in the experiment. (b) Identify the independent variable in the scientists’ experiment. (c) State the null hypothesis of this experiment. (d) The scientists claim that the changes in the frequency of fur color were the result of natural selection. Justify the researchers’ claim. PAGE FOR ANSWERING QUESTION 3 GO ON TO THE NEXT PAGE. 62 AP Biology Practice Exam ADDITIONAL PAGE FOR ANSWERING QUESTION 3 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 63 4. The pinewood nematode is a eukaryote that infects certain species of pine trees, feeds on the cells surrounding the trees’ transport system, and ultimately kills the trees. Trees are infected when nematode-carrying beetles feed off the trees and inject the nematode into the trees when they bite through the bark. Once infected, pine trees increase the production of chemicals that serve as a defense mechanism for the trees by negatively affecting the nematodes. Researchers have found that pinewood nematodes contain symbiotic bacteria that can degrade the pine trees’ defensive chemicals. To investigate the role these bacteria play in nematode survival in the presence of these defensive chemicals, researchers pretreated nematodes with antibiotics and then exposed them to α–pinene, one of the defensive chemicals produced by the pine trees. (a) Describe the relationship between a parasite and its host. (b) Explain how producing the enzymes that digest α-pinene is beneficial to the bacterial species living within the nematodes. (c) Predict the effect of the antibiotic treatments on the mortality rate of the nematodes when exposed to α-pinene. (d) Provide reasoning to justify your prediction in part (c). PAGE FOR ANSWERING QUESTION 4 GO ON TO THE NEXT PAGE. 64 AP Biology Practice Exam ADDITIONAL PAGE FOR ANSWERING QUESTION 4 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 65 5. A cross section of a chloroplast showing membranes and the spaces between membranes is shown in Figure 1. Figure 1. A diagram of the cross section of a chloroplast (a) Describe the major process that takes place in this eukaryotic organelle. (b) Explain the function of the structure labeled with an X in Figure 1. (c) On the template in the space provided for your response, represent the location where carbon fixation takes place by writing � CF � and the location of the electron transport proteins by writing � ETP �. (d) Explain how the shape and stacking of the thylakoids contributes to the rate of carbon fixation by the chloroplast. GO ON TO THE NEXT PAGE TO BEGIN YOUR ANSWER TO QUESTION 5. GO ON TO THE NEXT PAGE. 66 AP Biology Practice Exam PAGE FOR ANSWERING QUESTION 5 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 67 ADDITIONAL PAGE FOR ANSWERING QUESTION 5 GO ON TO THE NEXT PAGE. 68 AP Biology Practice Exam NO TEST MATERIAL ON THIS PAGE GO ON TO THE NEXT PAGE. AP Biology Practice Exam 69 6. Drosophila melanogaster (D. melanogaster) is a species of fruit fly frequently used by researchers in genetic studies. Members of this species have two of each of four different chromosomes: the sex chromosome (flies have X and Y) and three autosomes (chromosomes 2, 3, and 4). Researchers studying D. melanogaster conducted genetic crosses to investigate a particular X-linked recessive trait encoded by a single gene (Table 1). Affected flies have the trait. TABLE 1. CROSSES PERFORMED AND THE PHENOTYPES OF THE RESULTING OFFSPRING Cross Number Phenotype of Females Used in the Cross Phenotype of Males Used in the Cross Percent of Total Population That Are Affected Percent of Male Offspring That Are Affected Percent of Female Offspring That Are Affected 1 Unaffected Affected 0 0 0 2 Affected Unaffected 50 100 0 3 Unaffected Unaffected 25 50 0 4 Affected Affected 100 100 100 (a) Identify the genotypes of the male and female flies used in cross 2. (b) Identify the cross in which the female parent was most likely heterozygous. (c) The researchers hypothesize that crossing any unaffected female and an affected male will result in a 0% chance of producing an affected male offspring. Evaluate the validity of the hypothesis. (d) Explain how the results exclude the possibility that the trait is encoded by a mitochondrial gene. PAGE FOR ANSWERING QUESTION 6 GO ON TO THE NEXT PAGE. 70 AP Biology Practice Exam ADDITIONAL PAGE FOR ANSWERING QUESTION 6 GO ON TO THE NEXT PAGE. AP Biology Practice Exam 71 ___________________________________________ [Show More]

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