[
 {
  "instruction":"Osteomyelitis is a bone infection often caused by the bacterium Staphylococcus aureus, which spreads through the blood to the bones. The infection is quite common after surgeries when soft tissues become infected and the infection moves to the adjacent bones. The long bones of children are particularly susceptible as they have excellent circulation to them during the growth period. In adults, the best circulation is to the vertebrae, and these are the bones that are most likely to develop osteomyelitis.\nOsteomyelitis is treated with various antibiotics, but some patients have chronic infections that are not eradicated by antibiotic therapy due to a unique feature of S. aureus. S. aureus can be internalized by osteoblasts where the bacteria can continue to survive in the intracellular environment. Inside the osteoblasts, bacteria can induce apoptosis, killing their host cell. Surgical debridement is often required to remove the damaged and dead bone tissue created by the infection.\nA major research question deals with how an S. aureus infection induces apoptosis in the host osteoblasts. In order for apoptosis to initiate in osteoblasts, the enzyme capsase-8 must be activated. This enzyme serves as a protease that cleaves critical proteins in the cell, causing cell death. Capsase-8 is only expressed under certain circumstances. In the case of osteomyelitis, the S. aureus bacteria cause the induction of tumor necrosis factor-related apoptosis-inducing ligand (TRAIL) in the osteoblasts. Then TRAIL activates capsase-8, which triggers a series of events leading to apoptosis and death of osteoblasts.\nWhich of the following potential drugs might be helpful in preventing the death of osteoblasts caused by S. aureus infection?\n\nOptions:\nA. one that increases capsase-8 activity\nB. one that increases TRAIL activity\nC. one that is an antagonist to TRAIL\nD. all of the above",
  "response":"Death of osteoblasts due to S. aureus infection is the result of induction of apoptosis. Apoptosis is triggered by capsase-8, so increasing the activity of capsase-8 would increase the rate of apoptosis and osteoblast death. Further, capsase-8 is activated by TRAIL made by S. aureus, so increasing the TRAIL activity would also increase osteoblast death rates. A drug that is antagonistic to TRAIL would decrease capsase-8 activity and decrease the rate of apoptosis in osteoblasts. The correct answer is C."
 },
 {
  "instruction":"Osteomyelitis is a bone infection often caused by the bacterium Staphylococcus aureus, which spreads through the blood to the bones. The infection is quite common after surgeries when soft tissues become infected and the infection moves to the adjacent bones. The long bones of children are particularly susceptible as they have excellent circulation to them during the growth period. In adults, the best circulation is to the vertebrae, and these are the bones that are most likely to develop osteomyelitis.\nOsteomyelitis is treated with various antibiotics, but some patients have chronic infections that are not eradicated by antibiotic therapy due to a unique feature of S. aureus. S. aureus can be internalized by osteoblasts where the bacteria can continue to survive in the intracellular environment. Inside the osteoblasts, bacteria can induce apoptosis, killing their host cell. Surgical debridement is often required to remove the damaged and dead bone tissue created by the infection.\nA major research question deals with how an S. aureus infection induces apoptosis in the host osteoblasts. In order for apoptosis to initiate in osteoblasts, the enzyme capsase-8 must be activated. This enzyme serves as a protease that cleaves critical proteins in the cell, causing cell death. Capsase-8 is only expressed under certain circumstances. In the case of osteomyelitis, the S. aureus bacteria cause the induction of tumor necrosis factor-related apoptosis-inducing ligand (TRAIL) in the osteoblasts. Then TRAIL activates capsase-8, which triggers a series of events leading to apoptosis and death of osteoblasts.\nMany strains of $S$. aureus that cause osteomyelitis are termed methicillin-resistant Staphylococcus aureus (MRSA). These strains have developed resistance to the antibiotic methicillin as well as to other antibiotics. The mechanism by which this antibiotic resistance occurs is by:\n\nOptions:\nA. the overuse of antibiotics, which causes patients to become resistant to the antibiotics\nB. the overuse of antibiotics, which selects for bacteria that have mutations that allow them to survive in the presence of the antibiotics\nC. mutations, which occur as the bacteria are exposed to antibiotics that can serve as mutagens\nD. all of the above",
  "response":"Bacteria spontaneously mutate in order to produce diversity in the population. Some bacteria have spontaneous mutations, usually on plasmids, that allow them to survive in the presence of certain antibiotics. If the bacterial population is repeatedly exposed to that antibiotic due to overuse, it exerts selective pressure where only the resistant bacteria survive. These resistant bacteria multiply, and in the next generation a higher frequency of resistance is observed. Antibiotic resistance has nothing to do with people becoming resistant to antibiotics-it is all about the bacterial mutations. The correct answer is B."
 },
 {
  "instruction":"Human immunodeficiency virus (HIV) is known to infect T cells and macrophages possessing the CD4 receptor. In addition to the CD4 receptor, there are a variety of coreceptors that are also needed for certain strains of HIV to infect cells. One of these coreceptors is called CCR5 and is needed for the most common strains of HIV (called R5) to infect their host cells. The more CCR5 receptors a cell has, the greater the rate of infection for the cell. Some individuals have the delta-32 allele of CCR5 that leads to a decreased risk of HIV infection.\nWomen are at a much higher risk for contracting HIV from a male partner than a male is for contracting HIV from a female partner. Some of the increased risk for women is dependent on sex hormones. It has been hypothesized that women have different risks of contracting HIV at different points in their reproductive cycles. For example, women may be at a greater risk of contracting HIV following ovulation during the last 2 weeks of their cycle as opposed to the first 2 weeks of their cycle prior to ovulation. The difference between risk of infection before and after ovulation relates to sex hormones produced at various points in the reproductive cycle. It seems that the estrogen that is produced prior to ovulation provides a somewhat protective, although certainly not absolute, role against HIV infection. For women who do become infected with HIV, studies have shown that women in the first 3 to 5 years after HIV infection carry lower HIV viral loads than men, perhaps due to the influence of estrogen. This suggests a major role for sex hormones in the infection and progression of HIV.\nSome populations with the delta-32 CCR5 mutation seem to have close to $100 \\%$ protection against HIV, while other populations with the delta-32 CCR5 mutation seem to have only partial protection. What could account for this difference?\n\nOptions:\nA. There is an environmental influence on expression of the delta-32 CCR5 phenotype.\nB. The delta-32 CCR5 allele exhibits incomplete dominance. Homozygotes have complete protection while heterozygotes have only partial protection.\nC. The expression of the delta-32 CCR5 allele can be masked by epistasis.\nD. Those that have only partial protection against HIV infection have been exposed to larger doses of HIV.",
  "response":"To answer this question, you need to rely on your knowledge of genetics. The passage states that the delta-32 CCR5 mutation is protective against HIV infection. The question indicated that there are two different phenotypes noted within the delta-32 CCR5 mutation-one that offers near 100 percent protection against HIV infection and one that offers only partial protection. The best explanation for why two phenotypes (complete protection and partial protection) are observed would be that the delta-32 allele exhibits incomplete dominance. An individual that is homozygous for the allele (meaning he or she inherited two copies of the delta-32 allele) would have full protection, while an individual who is heterozygous (meaning he or she has one normal copy of the allele and one copy of the delta-32 allele) would have an intermediate phenotype (being partial protection). This is indicated by choice B. Since the question indicates that the differences in phenotype (protection) are related to the delta32 mutation, choices A and D can be eliminated as they suppose that something other than the allele is providing protection. Choice C suggests that epistasis is involved in masking the delta-32 allele. While this might sound remotely plausible, choice B is a much more likely explanation. The correct answer is B."
 },
 {
  "instruction":"A scientist suspects that the food in an ecosystem may have been contaminated with radioactive phosphates over a period of months. Which of the following substances could be examined for radioactive phosphate to test the hypothesis?\n\nOptions:\nA. the carbohydrates produced by plants in the area\nB. the DNA of the organisms in the area\nC. the lipids produced by organisms living in the area\nD. all of the above",
  "response":"This question relies on a knowledge of the basic chemical composition of the biologically important molecules. Carbohydrates and lipids are composed of C, O, and H. However, DNA is a nucleic acid and contains phosphate groups. The correct answer is B."
 },
 {
  "instruction":"At the end of the electron transport chain in aerobic cellular respiration, the final acceptor of the electrons is which will then produce a molecule\n\nOptions:\nA. $\\mathrm{CO}_{2} ; \\mathrm{O}_{2}$\nB. $\\mathrm{NAD}^{+} ; \\mathrm{NADH}$\nC. $\\mathrm{O}_{2} ; \\mathrm{H}_{2} \\mathrm{O}$\nD. ADP; ATP",
  "response":"During the electron transport chain, oxygen serves as the terminal electron acceptor during aerobic respiration. In addition to serving as an electron acceptor, the oxygen also gathers protons from the electron transport chain, producing water in the process. The correct answer is C."
 },
 {
  "instruction":"Endothelin is a vasoactive peptide involved in a variety of processes, including the control of blood flow. Endothelin is capable of binding to two different receptors. There are receptors for endothelin located on smooth muscle (ETA) and receptors for endothelin located on vascular endothelium (ETB). When endothelin binds to ETA, it initiates vasoconstriction by narrowing blood vessels. However, when endothelin binds to ETB, vasodilation occurs due to the widening of blood vessels. The activity of the ETB receptor seems to counterregulate the ETA effects and prevents excessive vasoconstriction.\nDrug $X$ is a selective receptor antagonist for endothelin. It is designed to bind preferentially to certain endothelin receptors. Drug $\\mathrm{Y}$ is another selective receptor antagonist for endothelin. It is also designed to preferentially bind to one of the endothelin receptors. Researchers are hopeful that drugs $\\mathrm{X}$ and $\\mathrm{Y}$ will be of value in the treatment of conditions such as pulmonary arterial hypertension (PAH). In PAH, excess endothelin is produced, which causes an increase in blood pressure within the pulmonary arteries. Normal pressure in the pulmonary arteries is expected to be about $14 \\mathrm{~mm} \\mathrm{Hg}$, and pressure at or above $25 \\mathrm{~mm}$ Hg constitutes PAH. Both drugs have been tested in patients with PAH. There were five patients in each group. At the end of the study, the percent decrease in mean pulmonary artery pressure was measured for each patient and reported in the table.\n\\begin{tabular}{cccc}\n\\hline Patient & Drug X & DrugY & Placebo \\\\\n\\hline 1 & $+10 \\%$ & $-10 \\%$ & $+5 \\%$ \\\\\n2 & $+3 \\%$ & $-5 \\%$ & $0 \\%$ \\\\\n3 & $-5 \\%$ & $+2 \\%$ & $+13 \\%$ \\\\\n4 & $+2 \\%$ & $0 \\%$ & $-4 \\%$ \\\\\n5 & $0 \\%$ & $-7 \\%$ & $-2 \\%$ \\\\\nAverage decrease & $+2 \\%$ & $-4 \\%$ & $+2.4 \\%$ \\\\\n\\hline\n\\end{tabular}\nWhich endothelin receptor is drug X selectively binding to?\n\nOptions:\nA. ETA only\nB. ETB only\nC. both ETA and ETB\nD. neither ETA nor ETB",
  "response":"The results from the drug $\\mathrm{X}$ group indicate that the treatment caused an increase in mean pulmonary artery pressure. This helps us to eliminate choice $\\mathrm{A}$, which suggests that drug $\\mathrm{X}$ is binding to ETA. If this were the case, the drug would be bound to ETA so that endothelin could only bind to ETB. When endothelin binds to ETB, vasodilation occurs, implying a decrease in pressure. This was not seen in the drug $\\mathrm{X}$ group. Choice $\\mathrm{B}$ suggests that drug $\\mathrm{X}$ is selectively binding to the ETB receptor. This would mean that endothelin could only bind to ETA, which does in fact cause an increase in pressure. This would be the correct choice. Answers and Explanations Passage V MCAT The correct answer is B."
 },
 {
  "instruction":"Sperm count in human males is influenced by a variety of factors. One of the most important variables affecting sperm count is the number of Sertoli cells located in seminiferous tubules of the testes. The Sertoli cells' primary function is to nurture cells through the process of spermatogenesis by providing them with nutrients and an environment conducive to spermatogenesis. Each Sertoli cell has a set number of spermatozoa that it can support. Sertoli cells have an additional role in forming the blood-testis barrier that provides immune privilege for the seminiferous tubules. This forms a physical barricade that keeps the developing sperm from coming into contact with cells of the immune system.\nWhile it is not clear as to all the factors involved in determining Sertoli cell number, it is apparent that their development is most critical during the first nine months of life and prior to puberty. Events that affect the development of Sertoli cells in young males can influence fertility later in life. There may also be genetic factors involved in determining the number of Sertoli cells formed in a particular male.\nThe following data were collected from two groups of men. The men in group 1 had a normal sperm count while the men in group 2 were seeking infertility treatment due to a low sperm count. Levels of follicle-stimulating hormone (FSH) and luteinizing hormone (LH) from the pituitary were measured as were the testicular hormones inhibin-B and testosterone in an attempt to determine which hormones seem to be linked to reduced sperm count.\n\\begin{tabular}{ccccc}\n\\hline Sample Group & LH & FSH & Inhibin-B & Testosterone \\\\\n\\hline 1 & $5.3 \\mathrm{IU} \/ \\mathrm{L}$ & $5.0 \\mathrm{IU} \/ \\mathrm{L}$ & $190 \\mathrm{pg} \/ \\mathrm{mL}$ & $18.0 \\mathrm{nmol} \/ \\mathrm{L}$ \\\\\n2 & $3.1 \\mathrm{IU} \/ \\mathrm{L}$ & $17.6 \\mathrm{IU} \/ \\mathrm{L}$ & $51 \\mathrm{pg} \/ \\mathrm{mL}$ & $17.8 \\mathrm{nmol} \/ \\mathrm{L}$ \\\\\n\\hline\n\\end{tabular}\nWhy would it be important for the Sertoli cells to establish a blood-testis barrier and immune privilege?\n\nOptions:\nA. to prevent the immune cells from attacking the sperm cells, which display different antigens from the other cells in the body\nB. to prevent the developing sperm cells from leaving the reproductive system\nC. to prevent blood from flowing into the testes\nD. to keep the developing sperm cells from entering into the epididymis before they are mature",
  "response":"The blood-testis barrier is described in the passage as a barricade to establish immune privilege for the testes. To answer this question, you need to think about how the sperm cells in the testes would compare to other cells in the body. Sperm cells are haploid, and all are genetically different from each other and from other cells in the body. Because they are genetically different, they should express unique surface antigens. If the cells of the immune system were to contact the sperm, they would be perceived as foreign and could be subject to immune system attack, which would destroy the sperm. For this reason, the blood-testis barrier is required to keep the sperm away from the immune cells. The remaining choices do not make good sense. Sperm cells will eventually leave the reproductive system during ejaculation. It also would not make sense to indicate that the blood flow to the testes needs to be cut off. This would prevent oxygen and nutrient flow to the area. Preventing the sperm from moving to the epididymis before they are ready does not provide an explanation for why the sperm cells need to be kept separate from the cells of the immune system. The correct answer is A."
 },
 {
  "instruction":"The one principal effect not attributed to the hormone insulin is:\n\nOptions:\nA. glycogen formation and storage\nB. conversion of glycogen into glucose\nC. carbohydrate oxidation\nD. inhibition of gluconeogenesis",
  "response":"The pancreas is the source for two hormones: insulin and glucagon. Among its principal effects, insulin stimulates glycogen formation and storage, stimulates carbohydrate oxidation, and inhibits gluconeogenesis. All of these effects are described in choices A, C, and D. The other hormone, glucagon, stimulates conversion of glycogen into glucose. So choice B is the correct answer. The correct answer is B."
 },
 {
  "instruction":"Human immunodeficiency virus (HIV) is known to infect T cells and macrophages possessing the CD4 receptor. In addition to the CD4 receptor, there are a variety of coreceptors that are also needed for certain strains of HIV to infect cells. One of these coreceptors is called CCR5 and is needed for the most common strains of HIV (called R5) to infect their host cells. The more CCR5 receptors a cell has, the greater the rate of infection for the cell. Some individuals have the delta-32 allele of CCR5 that leads to a decreased risk of HIV infection.\nWomen are at a much higher risk for contracting HIV from a male partner than a male is for contracting HIV from a female partner. Some of the increased risk for women is dependent on sex hormones. It has been hypothesized that women have different risks of contracting HIV at different points in their reproductive cycles. For example, women may be at a greater risk of contracting HIV following ovulation during the last 2 weeks of their cycle as opposed to the first 2 weeks of their cycle prior to ovulation. The difference between risk of infection before and after ovulation relates to sex hormones produced at various points in the reproductive cycle. It seems that the estrogen that is produced prior to ovulation provides a somewhat protective, although certainly not absolute, role against HIV infection. For women who do become infected with HIV, studies have shown that women in the first 3 to 5 years after HIV infection carry lower HIV viral loads than men, perhaps due to the influence of estrogen. This suggests a major role for sex hormones in the infection and progression of HIV.\nDrugs such as AZT and ddI are nucleoside analogs used in HIV-infected patients. Both are chemically modified versions of the nucleotides that make up DNA and RNA. These modified nucleosides interfere with normal replication and transcription. The MOST important goal of these drugs would be:\n\nOptions:\nA. to prevent the replication of host cell DNA so that the host cell cannot function\nB. to prevent the transcription of host cell RNA so that no proteins can be expressed\nC. to prevent the viral nucleic acid from entering the host cells\nD. to prevent replication of the viral genome",
  "response":"The question explains that AZT and ddI interfere with replication and transcription. During a viral infection, the ideal situation would be to interfere with viral processes while leaving host cell processes unaffected. Choice A suggests that host cell DNA replication should be prevented in order to keep the host cell from functioning and mentions nothing about how this would affect the virus. Choice B is similar in that it does not mention how the virus would be affected. Choice $\\mathrm{C}$ is not appropriate because it discusses preventing infection. If the patient is already confirmed as infected, then this choice wouldn't make sense. If a person is infected with HIV, then the viral genetic material has already entered the host cells. If replication of the viral genome can be prevented as suggested by choice $\\mathrm{D}$, then the progression of the infection can be slowed. The correct answer is D."
 },
 {
  "instruction":"The discovery and development of various types of anesthesia have revolutionized modern medicine, allowing many medical procedures to be performed with minimal discomfort to the patient. Local anesthesia is used to induce loss of sensation in a particular area, while general anesthesia is used to induce loss of sensation along with loss of consciousness. Both procaine (also known as novocaine) and lidocaine are drugs used for local anesthesia, and they have similar mechanisms of action. The drugs affect neurons in the local areas to which they are applied by preventing the opening of $\\mathrm{Na}^{+}$gated channels in the neurons.\nProcaine and lidocaine have their own advantages and disadvantages. For example, procaine has the ability to cause vasoconstriction in addition to its analgesic properties. However, lidocaine induces anesthesia faster and for a longer period of time than procaine. In addition to being used as an anesthetic, lidocaine is often used in the treatment of ventricular fibrillation, a condition where the cardiac muscle displays rapid, uncoordinated, and weak rhythms. The drug causes decreased excitability of the ventricles by blocking $\\mathrm{Na}^{+}$gated channels, while the sinoatrial node in the right atrium is unaffected by the drug.\nProcaine has a half-life of up to $0.5$ hours in patients. When procaine is metabolized in the body, it is converted to para-aminobenzoic acid (PABA) in the plasma by the enzyme pseudocholinesterase, and PABA is then excreted by the kidneys. Lidocaine is twice as potent as procaine, but lidocaine is also twice as toxic. The half-life of lidocaine in patients is typically between $1.5$ and 2 hours. Lidocaine is metabolized in the liver by two enzymes that produce the metabolites monoethylglycinexylidide and glycinexylidide. The passage states that when lidocaine is given for ventricular fibrillation the sinoatrial node is unaffected. The sinoatrial node normally functions as:\n\nOptions:\nA. a valve in the heart\nB. the pacemaker of the heart\nC. the barricade between the right and left side of the heart that keeps oxygenated and deoxygenated blood from mixing\nD. a regulator of blood pressure in the heart",
  "response":"This question relies on your knowledge of the anatomy of the heart and the conducting system of the heart. The sinoatrial node (SA node) is a group of conducting cells located in the right atrium. Recall that the valves of the heart are atrioventricular (AV) and semilunar valves. The barricade that divides the right and left side of the heart is the septum. The SA node generates electrical signals that control cardiac muscle contractions and serves as the pacemaker for the heart. The correct answer is B."
 },
 {
  "instruction":"Leukemia is a term that describes a wide variety of blood cell cancers. Most leukemia cases involve elevated leukocyte counts. In acute cases of leukemia, immature leukocytes proliferate quickly, which leads to many abnormal immature cells in the bone marrow. This inhibits the ability of the bone marrow to produce healthy new leukocytes. This form of the cancer progresses rapidly, and the cells metastasize easily as they move into the bloodstream from the bone marrow. It can be fatal within weeks or months if not treated promptly. In contrast, chronic cases of leukemia typically involve the proliferation of abnormal mature leukocytes. This form of the disease progresses more slowly, and treatment need not always be immediate.\nOnce classified as acute or chronic, leukemia can be further subdivided into lymphocytic or myelogenous. This distinction depends on which type of leukocyte is abnormal in the patient. In lymphocytic leukemia, the lymphoid cells, such as the lymphocytes, are abnormal. In the myelogenous form, the myeloid cells, such as basophils, neutrophils, and eosinophils, are affected.\nChemotherapy drugs are often used in the management of leukemia cases. There are numerous categories of these drugs, and they have varied mechanisms of action. In treatment, several drugs are used in combination because some of the drugs are cell-cycle specific while others are not. Alkylating agents are capable of disrupting the function of DNA. Tubulin-binding agents are designed to interfere with microtubule formation. Other drugs are designed to interfere with DNA polymerase, DNA repair mechanisms, or protein synthesis, or to act as purine analogs. Many of these drugs cannot enter into the brain when injected or ingested, so if the cancer has spread to\nthis area, special methods must be used to introduce the drugs to the brain.\nWhat is the most logical explanation for why certain drugs may NOT be able to reach the brain if they have been ingested or injected?\n\nOptions:\nA. These drugs have a chemical nature that does not allow them to cross the blood-brain barrier.\nB. These drugs are so toxic that they are broken down before they reach the brain.\nC. Most of these drugs target only cancerous cells that normally would not be in the brain.\nD. The drugs that have been ingested have likely been denatured by the acids in the stomach.",
  "response":"The blood-brain barrier is a change in the permeability of blood vessels leading to the brain. Depending on the chemical composition, some drugs can cross and some cannot. Of the choices listed, it seems that if the drugs cannot reach the brain, then they have not crossed the blood-brain barrier. While choice D might seem plausible if a drug was ingested, it would not make sense for a drug that was injected. The correct answer is A."
 },
 {
  "instruction":"Osteomyelitis is a bone infection often caused by the bacterium Staphylococcus aureus, which spreads through the blood to the bones. The infection is quite common after surgeries when soft tissues become infected and the infection moves to the adjacent bones. The long bones of children are particularly susceptible as they have excellent circulation to them during the growth period. In adults, the best circulation is to the vertebrae, and these are the bones that are most likely to develop osteomyelitis.\nOsteomyelitis is treated with various antibiotics, but some patients have chronic infections that are not eradicated by antibiotic therapy due to a unique feature of S. aureus. S. aureus can be internalized by osteoblasts where the bacteria can continue to survive in the intracellular environment. Inside the osteoblasts, bacteria can induce apoptosis, killing their host cell. Surgical debridement is often required to remove the damaged and dead bone tissue created by the infection.\nA major research question deals with how an S. aureus infection induces apoptosis in the host osteoblasts. In order for apoptosis to initiate in osteoblasts, the enzyme capsase-8 must be activated. This enzyme serves as a protease that cleaves critical proteins in the cell, causing cell death. Capsase-8 is only expressed under certain circumstances. In the case of osteomyelitis, the S. aureus bacteria cause the induction of tumor necrosis factor-related apoptosis-inducing ligand (TRAIL) in the osteoblasts. Then TRAIL activates capsase-8, which triggers a series of events leading to apoptosis and death of osteoblasts.\nOsteoblasts are normally involved in building new bone matrix. Their activity is stimulated by the hormone made by the\n\nOptions:\nA. calcitonin; thyroid\nB. parathyroid hormone; parathyroid glands\nC. osteotonin; bone marrow\nD. none of the above",
  "response":"Knowledge of the endocrine structures and hormones is needed to answer this question. The hormone calcitonin stimulates the storage of calcium in bone matrix. Its target is the osteoblasts that build new bone matrix. The calcitonin hormone is made by the thyroid gland. Parathyroid hormone is antagonistic to calcitonin and stimulates osteoclasts. Osteotonin is not a real hormone. The correct answer is A."
 },
 {
  "instruction":"Neurons contain a high concentration of potassium ions relative to the fluids surrounding them. How could a neuron acquire even more potassium?\n\nOptions:\nA. active transport\nB. osmosis\nC. endocytosis\nD. diffusion",
  "response":"This question is asking how to move a potassium ion (which is small) from an area that has a low concentration to an area with a relatively higher concentration. In this case we are moving against the concentration gradient. The transport type that could do this is active transport, using energy to move the ions against their concentration gradient. The correct answer is A."
 },
 {
  "instruction":"Passage $\\mathrm{V}$\nFor many years, researchers have been interested in developing edible vaccines that would provide protection against various bacterial and viral diseases. An edible vaccine is a genetically modified organism (GMO) that is created when genes from the bacteria or virus being vaccinated against are inserted into plant cells. As the plants grow, they produce proteins from the bacteria or virus that the researchers are attempting to vaccinate against. When a person eats the plant, the viral or bacteria proteins serve as antigens, which initiates a primary immune response.\nSome of the plants that are of interest in edible vaccine development are bananas, potatoes, tomatoes, corn, rice, tobacco, and soybeans. While early research focused on growing crops that produced edible products such as bananas, the most recent research is focused on extracting material from the leaves and seeds of the plants, dehydrating it, and putting it into capsules that can be swallowed. This allows for a more predictable dosage of the vaccine and eliminates concerns that the crops containing the vaccine would be accidentally eaten by individuals who are not in need of the vaccine. The term edible vaccine has now been replaced with the term plantderived oral vaccine.\nThe development of plant-derived oral vaccines solves many problems associated with injectable vaccines. Because plant-derived oral vaccines are grown in crops, there is no problem with storage or refrigeration that is required of most traditional vaccines. These plant-derived vaccines also require no special training to administer. Additionally, when a vaccine is taken orally, it provides mucosal immunity due to the production of IgA antibodies by the immune system-in addition to systemic immunity. Since 90 percent of diseases enter the body through a mucus membrane, a vaccine that coerces mucosal immunity is of great benefit in the prevention of many diseases.\nPlant-derived oral vaccines produce bacterial or viral proteins that serve as antigens in the body. Which of these events would occur first when a new antigen enters the body?\n\nOptions:\nA. production of antibodies from plasma cells\nB. presentation of the antigen by a macrophage\nC. secretion of interleukin-2 from helper T cells\nD. attack of the antigen by cytotoxic $T$ cells",
  "response":"This question requires an understanding of the specific defenses of the immune system. The ultimate goal of the specific defenses is to allow for the production of antibodies from plasma cells and the production of cytotoxic $\\mathrm{T}$ cells. However, neither of these things can occur without the activation of a helper $\\mathrm{T}$ cell. In order to activate the helper $\\mathrm{T}$ cell, macrophages must phagocytize foreign antigens. The fragments of these antigens will be displayed on the surface of the macrophage. This antigen presenting cell will then activate the helper $\\mathrm{T}$ cell. The correct answer is B."
 },
 {
  "instruction":"Genetically modified (GM) foods contain DNA from more than one source. The goal of creating GM foods is to modify the organisms to express new and useful traits. A few examples involve engineering crops to be resistant to bacterial or viral infections, to be resistant to herbicides, to ripen more slowly, to survive extreme temperature changes, to contain vaccines, or to create crops that are more nutritious.\nOne GM food that has received media attention over the years is golden rice. This GM product was first introduced in 2000 and was engineered to provide carotene that is a precursor to vitamin $\\mathrm{A}$. The carotene in the rice provides the golden color. Vitamin A deficiency is a major problem in underdeveloped parts of the world, killing more than one million people a year, most of whom are children. This vitamin is critical for the development and maintenance of epithelial tissues, for vision, and for growth.\nRice plants normally produce carotene in the inedible parts of the plant. To be able to access and use the carotene, it was necessary to engineer the rice to produce it in the edible grains. In order to achieve this goal, two genes from daffodil plants, psy and $l y c$, and a bacterial gene called $c r t 1$ were added to the rice genome. These genes engineered a pathway that produces lycopene, which is eventually converted to carotene. Each of these genes was placed under the control of a grain-specific promoter so that the products were produced in the rice grains.\nWhile golden rice was produced to provide carotene, which the human body can convert into the essential vitamin A, the rice provided much less carotene than was anticipated. Golden rice 2, a modified version released in 2005, contains a much larger amount of carotene and, as such, is a better source of the vitamin. Field tests were conducted in 2009, and human consumption trials were completed in 2012. However, neither version has been grown in large-scale quantities at this time.\nGenetically modified organisms (GMOs) can be produced in a variety of ways. In the case of golden rice, genes from the daffodil plant and bacteria were added to the rice genome. Which of the following would be necessary to remove the genes from these other organisms so that they could be added to rice?\n\nOptions:\nA. reverse transcriptase\nB. a viral vector\nC. restriction enzymes\nD. polymerase chain reaction",
  "response":"Restriction enzymes are known to cut DNA at specific recognition sequences and are valuable tools for excising genes to be transferred from one source to another. Reverse transcriptase is used to copy RNA to DNA. Viral vectors can carry DNA of interest into other cells, but they are not used to remove genes from an organism. Polymerase chain reaction (PCR) is used to produce multiple copies of a target DNA sequence. The correct answer is C."
 },
 {
  "instruction":"The discovery and development of various types of anesthesia have revolutionized modern medicine, allowing many medical procedures to be performed with minimal discomfort to the patient. Local anesthesia is used to induce loss of sensation in a particular area, while general anesthesia is used to induce loss of sensation along with loss of consciousness. Both procaine (also known as novocaine) and lidocaine are drugs used for local anesthesia, and they have similar mechanisms of action. The drugs affect neurons in the local areas to which they are applied by preventing the opening of $\\mathrm{Na}^{+}$gated channels in the neurons.\nProcaine and lidocaine have their own advantages and disadvantages. For example, procaine has the ability to cause vasoconstriction in addition to its analgesic properties. However, lidocaine induces anesthesia faster and for a longer period of time than procaine. In addition to being used as an anesthetic, lidocaine is often used in the treatment of ventricular fibrillation, a condition where the cardiac muscle displays rapid, uncoordinated, and weak rhythms. The drug causes decreased excitability of the ventricles by blocking $\\mathrm{Na}^{+}$gated channels, while the sinoatrial node in the right atrium is unaffected by the drug.\nProcaine has a half-life of up to $0.5$ hours in patients. When procaine is metabolized in the body, it is converted to para-aminobenzoic acid (PABA) in the plasma by the enzyme pseudocholinesterase, and PABA is then excreted by the kidneys. Lidocaine is twice as potent as procaine, but lidocaine is also twice as toxic. The half-life of lidocaine in patients is typically between $1.5$ and 2 hours. Lidocaine is metabolized in the liver by two enzymes that produce the metabolites monoethylglycinexylidide and glycinexylidide. What benefits would come from using an anesthetic such as novocaine, which causes vasoconstriction in addition to analgesic effects, during a surgical procedure?\n\nOptions:\nA. to keep blood pressure low\nB. to prevent infection\nC. to minimize bleeding during the procedure\nD. to prevent inflammation following the surgery",
  "response":"Vasoconstriction means that vessel diameter narrows, restricting blood flow to the area. The question introduces the idea of using novocaine during surgery. Since the drug causes vasoconstriction, this would actually increase blood pressure, eliminating choice A. The best choice is C. Bleeding is likely to occur during surgery, so vasoconstriction should slow the blood flow, which results in less blood loss. The correct answer is C."
 },
 {
  "instruction":"An endocrinologist has been studying an enzyme that she suspects is the rate-limiting step in the conversion of cholesterol to estrogen. It is reported in the literature that female mice with a mutation in the gene coding for this enzyme reach sexual maturity earlier than mice without the mutation. Controlled experiments are completed using breeding pairs of wild-type mice and the mutant strain of mice. The plasma estrogen concentration is measured in female offspring at 3 weeks of age, prior to sexual maturity, which typically occurs by 6 weeks of age. The results, in estrogen $\\mathrm{pg} \/ \\mathrm{mL}$ of plasma, are seen in the following table.   \\begin{tabular}{ccc} line Trial & Wild Type* & Mutant Strain* \\\\ line 1 & $2.3$ & $8.5$ \\\\ 2 & $8.1$ & $4.5$ \\\\ 3 & $4.4$ & $16.2$ \\\\ 4 & $7.6$ & $5.5$ \\\\ 5 & $3.2$ & $3.8$ \\\\ 6 & $8.2$ & $12.5$ \\\\ 7 & $5.5$ & $10.7$ \\\\ 8 & $6.3$ & $8.6$ \\\\ \\hline \\end{tabular}. The rate-limiting step in a metabolic pathway: \n\nOptions:\nA. A. requires the lowest activation energy\nB. B. is the fastest step in a metabolic pathway\nC. C. is always the last reaction in the pathway\nD. D. is the slowest step in a metabolic pathway",
  "response":"Rate-limiting steps generally require the highest activation energies and are the slowest steps in a metabolic pathway. The correct answer is D."
 },
 {
  "instruction":"The discovery and development of various types of anesthesia have revolutionized modern medicine, allowing many medical procedures to be performed with minimal discomfort to the patient. Local anesthesia is used to induce loss of sensation in a particular area, while general anesthesia is used to induce loss of sensation along with loss of consciousness. Both procaine (also known as novocaine) and lidocaine are drugs used for local anesthesia, and they have similar mechanisms of action. The drugs affect neurons in the local areas to which they are applied by preventing the opening of $\\mathrm{Na}^{+}$gated channels in the neurons.\nProcaine and lidocaine have their own advantages and disadvantages. For example, procaine has the ability to cause vasoconstriction in addition to its analgesic properties. However, lidocaine induces anesthesia faster and for a longer period of time than procaine. In addition to being used as an anesthetic, lidocaine is often used in the treatment of ventricular fibrillation, a condition where the cardiac muscle displays rapid, uncoordinated, and weak rhythms. The drug causes decreased excitability of the ventricles by blocking $\\mathrm{Na}^{+}$gated channels, while the sinoatrial node in the right atrium is unaffected by the drug.\nProcaine has a half-life of up to $0.5$ hours in patients. When procaine is metabolized in the body, it is converted to para-aminobenzoic acid (PABA) in the plasma by the enzyme pseudocholinesterase, and PABA is then excreted by the kidneys. Lidocaine is twice as potent as procaine, but lidocaine is also twice as toxic. The half-life of lidocaine in patients is typically between $1.5$ and 2 hours. Lidocaine is metabolized in the liver by two enzymes that produce the metabolites monoethylglycinexylidide and glycinexylidide. Which type of neuron would be affected by lidocaine?\n\nOptions:\nA. motor neurons\nB. interneurons\nC. sensory neurons\nD. glial cells",
  "response":"The passage indicates that lidocaine is an analgesic (prevents pain) by blocking $\\mathrm{Na}^{+}$gates channels in neurons. Since glial cells are not neurons, choice D can immediately be eliminated. In order to prevent pain from being processed, the messages must be prevented from entering the central nervous system where interneurons are located. Therefore, sensory neurons must be blocked. If these are blocked, the interneurons do not receive the pain messages and motor neurons will not act on them. The correct answer is C."
 },
 {
  "instruction":"Osteomyelitis is a bone infection often caused by the bacterium Staphylococcus aureus, which spreads through the blood to the bones. The infection is quite common after surgeries when soft tissues become infected and the infection moves to the adjacent bones. The long bones of children are particularly susceptible as they have excellent circulation to them during the growth period. In adults, the best circulation is to the vertebrae, and these are the bones that are most likely to develop osteomyelitis.\nOsteomyelitis is treated with various antibiotics, but some patients have chronic infections that are not eradicated by antibiotic therapy due to a unique feature of S. aureus. S. aureus can be internalized by osteoblasts where the bacteria can continue to survive in the intracellular environment. Inside the osteoblasts, bacteria can induce apoptosis, killing their host cell. Surgical debridement is often required to remove the damaged and dead bone tissue created by the infection.\nA major research question deals with how an S. aureus infection induces apoptosis in the host osteoblasts. In order for apoptosis to initiate in osteoblasts, the enzyme capsase-8 must be activated. This enzyme serves as a protease that cleaves critical proteins in the cell, causing cell death. Capsase-8 is only expressed under certain circumstances. In the case of osteomyelitis, the S. aureus bacteria cause the induction of tumor necrosis factor-related apoptosis-inducing ligand (TRAIL) in the osteoblasts. Then TRAIL activates capsase-8, which triggers a series of events leading to apoptosis and death of osteoblasts.\nThe strains of $S$. aureus associated with osteomyelitis have a variety of cell adhesion molecules called adhesins that help them to attach to and be internalized by osteoblasts. These molecules are likely found in which of the following bacterial structures?\n\nOptions:\nA. the nucleoid\nB. the ribosomes\nC. the capsule\nD. spores",
  "response":"This question relies on your knowledge of bacterial cell structures and functions. The nucleoid is the region of the cell where the one circular loop of chromosomal DNA is located. The ribosomes are responsible for protein synthesis. Bacteria can convert themselves to spores when environmental conditions are poor, allowing them to survive for extended periods of time in harsh conditions. The capsule is a sticky layer surrounding the cell that helps in the attachment of the cell to a surface. It is likely that the adhesins are located here. The correct answer is C."
 },
 {
  "instruction":"The process of quorum sensing is observed in a variety of species. It involves a system of communication between organisms that is dependent on population density. Bacteria are notorious for using quorum sensing (QS) via the secretion of secreted signaling molecules, called autoinducers (AI), to coordinate their gene expression once a threshold density of population is reached. Not only can bacteria communicate via QS within their species, but they can also cross-talk through QS between different species. In order to participate in quorum sensing, bacteria require a way to produce an autoinducer and a way to respond to autoinducers via binding to specific receptors.\nOne example of how bacteria use quorum sensing to their advantage is to regulate the gene expression of virulence factors. During infection, if a small number of bacteria were to express their virulence factors, this would largely have no impact on the host. Instead, bacteria wait until a quorum is reached in terms of population size, indicating that the population is large enough to express virulence factors. These virulence factors relate to such abilities as escaping the immune response, invasion, and spreading within the host.\nWe know that we are facing a current healthcare crisis related to the loss of effectiveness of many traditional types of antibiotics based on bacterial resistance. This provides a major impetus for identifying novel ways to target bacterial pathogens that avoid the use of antibiotics. Several chemical compounds and molecules have been identified as facilitators of quorum sensing inhibition. Typically, these factors can serve as signal-targeting enzymes or small-molecule inhibitors of signal synthases and receptors.\nQuorum sensing is used frequently within populations of bacteria such as Pseudomonas aeruginosa that form biofilms within their host. Once a quorum is reached, expression of the genes required for biofilm formation occurs. From the bacterial perspective, what direct advantage does the formation of biofilms provide?\n\nOptions:\nA. It provides a strong attachment to surfaces of the host.\nB. It enhances virulence expression of the organisms involved in the biofilm.\nC. It provides increased resistance to antibiotics.\nD. It enhances the secretions of toxins.",
  "response":"The advantage of biofilm formation is that it provides a direct and strong attachment to its host. Biofilms are notoriously difficult to eradicate, and the bacteria within them are provided protection against many variables, sometimes including antibiotics. The correct answer is A."
 },
 {
  "instruction":"The discovery and development of various types of anesthesia have revolutionized modern medicine, allowing many medical procedures to be performed with minimal discomfort to the patient. Local anesthesia is used to induce loss of sensation in a particular area, while general anesthesia is used to induce loss of sensation along with loss of consciousness. Both procaine (also known as novocaine) and lidocaine are drugs used for local anesthesia, and they have similar mechanisms of action. The drugs affect neurons in the local areas to which they are applied by preventing the opening of $\\mathrm{Na}^{+}$gated channels in the neurons.\nProcaine and lidocaine have their own advantages and disadvantages. For example, procaine has the ability to cause vasoconstriction in addition to its analgesic properties. However, lidocaine induces anesthesia faster and for a longer period of time than procaine. In addition to being used as an anesthetic, lidocaine is often used in the treatment of ventricular fibrillation, a condition where the cardiac muscle displays rapid, uncoordinated, and weak rhythms. The drug causes decreased excitability of the ventricles by blocking $\\mathrm{Na}^{+}$gated channels, while the sinoatrial node in the right atrium is unaffected by the drug.\nProcaine has a half-life of up to $0.5$ hours in patients. When procaine is metabolized in the body, it is converted to para-aminobenzoic acid (PABA) in the plasma by the enzyme pseudocholinesterase, and PABA is then excreted by the kidneys. Lidocaine is twice as potent as procaine, but lidocaine is also twice as toxic. The half-life of lidocaine in patients is typically between $1.5$ and 2 hours. Lidocaine is metabolized in the liver by two enzymes that produce the metabolites monoethylglycinexylidide and glycinexylidide. A neuron under the influence of an anesthetic such as lidocaine would MOST likely:\n\nOptions:\nA. display decreased neurotransmitter release at the synapse\nB. show increased enzymatic degradation of neurotransmitters in the synapse\nC. show increased reuptake of neurotransmitters in the synapse\nD. display a more negative charge outside the neuron as compared to inside the neuron",
  "response":"Since lidocaine blocks $\\mathrm{Na}^{+}$gated channels, we have already established that it prevents the initiation of action potentials in the affected neurons. Normally, as action potentials propagate, they eventually cause the release of neurotransmitters into the synapse. Since action potentials are impaired in lidocaine-treated neurons, we should expect that there will be fewer neurotransmitters released into the synapse. The correct answer is A."
 },
 {
  "instruction":"During pulmonary gas exchange, oxygen and carbon dioxide always move:\n\nOptions:\nA. into the alveoli\nB. into the blood\nC. from high to low concentration\nD. out of the blood",
  "response":"Pulmonary gas exchange is always based on simple diffusion. Diffusion allows for the movement of a substance from an area of high concentration of the substance to an area of low concentration of the substance. Depending on the concentrations, oxygen and carbon dioxide will move in variable directions. Choices $\\mathrm{A}, \\mathrm{B}$, and $\\mathrm{D}$ indicate that the movement of gases always occurs in a fixed direction, which is incorrect. The correct answer is C."
 },
 {
  "instruction":"An endocrinologist has been studying an enzyme that she suspects is the rate-limiting step in the conversion of cholesterol to estrogen. It is reported in the literature that female mice with a mutation in the gene coding for this enzyme reach sexual maturity earlier than mice without the mutation. Controlled experiments are completed using breeding pairs of wild-type mice and the mutant strain of mice. The plasma estrogen concentration is measured in female offspring at 3 weeks of age, prior to sexual maturity, which typically occurs by 6 weeks of age. The results, in estrogen $\\mathrm{pg} \/ \\mathrm{mL}$ of plasma, are seen in the following table.   \\begin{tabular}{ccc} line Trial & Wild Type* & Mutant Strain* \\\\ line 1 & $2.3$ & $8.5$ \\\\ 2 & $8.1$ & $4.5$ \\\\ 3 & $4.4$ & $16.2$ \\\\ 4 & $7.6$ & $5.5$ \\\\ 5 & $3.2$ & $3.8$ \\\\ 6 & $8.2$ & $12.5$ \\\\ 7 & $5.5$ & $10.7$ \\\\ 8 & $6.3$ & $8.6$ \\\\ \\hline \\end{tabular}. What conclusion can be made based on the data?\n\nOptions:\nA. A. The presence of the wild-type enzyme increases estrogen concentration.\nB. B. The presence of the wild-type enzyme increases cholesterol concentration.\nC. C. The presence of the mutant enzyme increases estrogen concentration.\nD. D. The presence of the mutant enzyme increases cholesterol concentration.",
  "response":"The data report estrogen concentration. From reviewing the data, it can be seen that the mutant strain generally has a higher estrogen concentration than the wild-type strain. The correct answer is C."
 },
 {
  "instruction":"Passage $\\mathrm{V}$\nFor many years, researchers have been interested in developing edible vaccines that would provide protection against various bacterial and viral diseases. An edible vaccine is a genetically modified organism (GMO) that is created when genes from the bacteria or virus being vaccinated against are inserted into plant cells. As the plants grow, they produce proteins from the bacteria or virus that the researchers are attempting to vaccinate against. When a person eats the plant, the viral or bacteria proteins serve as antigens, which initiates a primary immune response.\nSome of the plants that are of interest in edible vaccine development are bananas, potatoes, tomatoes, corn, rice, tobacco, and soybeans. While early research focused on growing crops that produced edible products such as bananas, the most recent research is focused on extracting material from the leaves and seeds of the plants, dehydrating it, and putting it into capsules that can be swallowed. This allows for a more predictable dosage of the vaccine and eliminates concerns that the crops containing the vaccine would be accidentally eaten by individuals who are not in need of the vaccine. The term edible vaccine has now been replaced with the term plantderived oral vaccine.\nThe development of plant-derived oral vaccines solves many problems associated with injectable vaccines. Because plant-derived oral vaccines are grown in crops, there is no problem with storage or refrigeration that is required of most traditional vaccines. These plant-derived vaccines also require no special training to administer. Additionally, when a vaccine is taken orally, it provides mucosal immunity due to the production of IgA antibodies by the immune system-in addition to systemic immunity. Since 90 percent of diseases enter the body through a mucus membrane, a vaccine that coerces mucosal immunity is of great benefit in the prevention of many diseases.\nWhen a vaccine is given orally, it must pass through the digestive system before being absorbed into the bloodstream, whereas injected vaccines enter the bloodstream directly. What problems might vaccine developers be concerned about with a plant-derived oral vaccine that has to pass through the digestive tract?\n\nOptions:\nA. There would be concern that the vaccine would not work quickly enough if it had to travel through the digestive tract.\nB. There would be concern that stomach acidity might damage the bacterial or viral proteins that were expressed by the plant cells of the vaccine.\nC. There would be concern that the vaccine might elicit an inflammatory response in the digestive tract.\nD. all of the above",
  "response":"All vaccines take time to work (often weeks before full immunity occurs), so waiting for an oral vaccine to pass through the digestive tract (which takes no more than a few hours) is not of major concern. This eliminates choice A. Choice $\\mathrm{C}$ suggests that inflammation in the digestive tract might be a problem. Since the inflammatory response is expected as one of the nonspecific immune defenses, it is not of major concern. Since two choices have been eliminated, choice D is no longer an option. This leaves choice B as the correct response. Since the plant has been engineered to express proteins unique to the virus or bacterium being vaccinated against, it is possible that these proteins might be damaged in the stomach. Recall that the function of the stomach is to digest proteins by using the enzyme pepsin. The correct answer is B."
 },
 {
  "instruction":"With which theoretical paradigm is the German political philosopher Karl Marx MOST associated?\n\nOptions:\nA. conflict theory\nB. functionalism\nC. symbolic interactionism\nD. avoidance theory",
  "response":"Conflict theory states that society is fundamentally in conflict when there is an uneven distribution of resources, and that power struggles between groups of people create divisiveness across gender, race, and socioeconomic status. The correct answer is A."
 },
 {
  "instruction":"Passage $\\mathrm{V}$\nFor many years, researchers have been interested in developing edible vaccines that would provide protection against various bacterial and viral diseases. An edible vaccine is a genetically modified organism (GMO) that is created when genes from the bacteria or virus being vaccinated against are inserted into plant cells. As the plants grow, they produce proteins from the bacteria or virus that the researchers are attempting to vaccinate against. When a person eats the plant, the viral or bacteria proteins serve as antigens, which initiates a primary immune response.\nSome of the plants that are of interest in edible vaccine development are bananas, potatoes, tomatoes, corn, rice, tobacco, and soybeans. While early research focused on growing crops that produced edible products such as bananas, the most recent research is focused on extracting material from the leaves and seeds of the plants, dehydrating it, and putting it into capsules that can be swallowed. This allows for a more predictable dosage of the vaccine and eliminates concerns that the crops containing the vaccine would be accidentally eaten by individuals who are not in need of the vaccine. The term edible vaccine has now been replaced with the term plantderived oral vaccine.\nThe development of plant-derived oral vaccines solves many problems associated with injectable vaccines. Because plant-derived oral vaccines are grown in crops, there is no problem with storage or refrigeration that is required of most traditional vaccines. These plant-derived vaccines also require no special training to administer. Additionally, when a vaccine is taken orally, it provides mucosal immunity due to the production of IgA antibodies by the immune system-in addition to systemic immunity. Since 90 percent of diseases enter the body through a mucus membrane, a vaccine that coerces mucosal immunity is of great benefit in the prevention of many diseases.\nSuppose that an individual has been given a plant-derived oral vaccine against a specific virus. How could you determine whether the vaccine was effective?\n\nOptions:\nA. Test for the presence of interleukins in the blood of the patient.\nB. Test for the presence of interferons in the blood of the patient.\nC. Test to determine whether the complement system in the blood had been activated.\nD. Test for the presence of antibodies to the viral antigen.",
  "response":"This question relies on knowledge of the immune response. Interleukins are cytokines that are secreted during the process of clonal selection in primary immune responses. Since they are not specific toward any one particular antigen, choice A can be eliminated. Interferons are produced as nonspecific defenses against viral infections. Virally infected cells secrete interferons that are detected by neighboring cells. Secretion of interferons gives neighboring cells the chance to increase their immune defenses so that they do not become infected by the virus. Since interferons are nonspecific, their presence would not indicate immunity against a specific virus, which eliminates choice B. The components of the complement system circulate in the plasma and are activated by bacteria that may enter the blood. Activation of the complement system ultimately lyses and kills the bacterial cell. Since the question is asking about a virus, choice C can be eliminated because it deals with bacteria. This leaves choice $\\mathrm{D}$ as the correct response. The detection of antibodies against the viral antigen would indicate that specific defenses were initiated and that plasma cells were activated to secrete antibodies against the viral antigen that was presented by the vaccine. The correct answer is D."
 },
 {
  "instruction":"The major constituent of blood plasma is:\n\nOptions:\nA. protein\nB. $\\mathrm{NaCl}$\nC. water\nD. cholesterol",
  "response":"Blood is a fluid made up of liquid plasma (55\\%), the primary component of which is water, and floating cells (45\\%). The plasma is richly endowed with dissolved proteins, lipids, and carbohydrates. The correct answer is C."
 },
 {
  "instruction":"Passage $\\mathrm{V}$\nFor many years, researchers have been interested in developing edible vaccines that would provide protection against various bacterial and viral diseases. An edible vaccine is a genetically modified organism (GMO) that is created when genes from the bacteria or virus being vaccinated against are inserted into plant cells. As the plants grow, they produce proteins from the bacteria or virus that the researchers are attempting to vaccinate against. When a person eats the plant, the viral or bacteria proteins serve as antigens, which initiates a primary immune response.\nSome of the plants that are of interest in edible vaccine development are bananas, potatoes, tomatoes, corn, rice, tobacco, and soybeans. While early research focused on growing crops that produced edible products such as bananas, the most recent research is focused on extracting material from the leaves and seeds of the plants, dehydrating it, and putting it into capsules that can be swallowed. This allows for a more predictable dosage of the vaccine and eliminates concerns that the crops containing the vaccine would be accidentally eaten by individuals who are not in need of the vaccine. The term edible vaccine has now been replaced with the term plantderived oral vaccine.\nThe development of plant-derived oral vaccines solves many problems associated with injectable vaccines. Because plant-derived oral vaccines are grown in crops, there is no problem with storage or refrigeration that is required of most traditional vaccines. These plant-derived vaccines also require no special training to administer. Additionally, when a vaccine is taken orally, it provides mucosal immunity due to the production of IgA antibodies by the immune system-in addition to systemic immunity. Since 90 percent of diseases enter the body through a mucus membrane, a vaccine that coerces mucosal immunity is of great benefit in the prevention of many diseases.\nIn order to engineer a plant-derived oral vaccine, genes from the microbe to be vaccinated against must be placed into the plant genome. Which of the following choices would be the most reasonable way to achieve this goal?\n\nOptions:\nA. Perform polymerase chain reaction (PCR) to amplify copies of the bacterial or viral genes and then insert the copies into the plant cells by transduction.\nB. Insert the bacterial or viral genes of interest into a plasmid, which can serve as a vector to transfer the gene to the plant cell.\nC. Clone the genes into a bacterial cell and allow it to conjugate with a plant cell.\nD. Use DNA hybridization to allow the genes of interest from the bacterial or viral genome to hybridize with the plant DNA.",
  "response":"While polymerase chain reaction (PCR) is used to amplify copies of target DNA, it would not help to get the viral or bacterial genes of interest into plant cells. This eliminates choice A. Choice $C$ suggests that the genes could be cloned into a bacterial cell and transferred to a plant cell by conjugation. Conjugation is unique to certain species of bacteria and occasionally is seen in fungi, but it does not happen in plants. So choice $C$ is eliminated. Choice D suggests that the genes of interest should be hybridized to the DNA of the plant. Since hybridization occurs between complementary base pairs, this would suggest that the DNA sequence of the gene of interest and the plant would have to be the same, which is not the case. This leaves choice B as the best option. Bacterial or viral genes can be spliced into plasmids. The plasmids serve as vectors to deliver the gene of interest to a variety of cell types, including plants. The correct answer is B."
 },
 {
  "instruction":"Sperm count in human males is influenced by a variety of factors. One of the most important variables affecting sperm count is the number of Sertoli cells located in seminiferous tubules of the testes. The Sertoli cells' primary function is to nurture cells through the process of spermatogenesis by providing them with nutrients and an environment conducive to spermatogenesis. Each Sertoli cell has a set number of spermatozoa that it can support. Sertoli cells have an additional role in forming the blood-testis barrier that provides immune privilege for the seminiferous tubules. This forms a physical barricade that keeps the developing sperm from coming into contact with cells of the immune system.\nWhile it is not clear as to all the factors involved in determining Sertoli cell number, it is apparent that their development is most critical during the first nine months of life and prior to puberty. Events that affect the development of Sertoli cells in young males can influence fertility later in life. There may also be genetic factors involved in determining the number of Sertoli cells formed in a particular male.\nThe following data were collected from two groups of men. The men in group 1 had a normal sperm count while the men in group 2 were seeking infertility treatment due to a low sperm count. Levels of follicle-stimulating hormone (FSH) and luteinizing hormone (LH) from the pituitary were measured as were the testicular hormones inhibin-B and testosterone in an attempt to determine which hormones seem to be linked to reduced sperm count.\n\\begin{tabular}{ccccc}\n\\hline Sample Group & LH & FSH & Inhibin-B & Testosterone \\\\\n\\hline 1 & $5.3 \\mathrm{IU} \/ \\mathrm{L}$ & $5.0 \\mathrm{IU} \/ \\mathrm{L}$ & $190 \\mathrm{pg} \/ \\mathrm{mL}$ & $18.0 \\mathrm{nmol} \/ \\mathrm{L}$ \\\\\n2 & $3.1 \\mathrm{IU} \/ \\mathrm{L}$ & $17.6 \\mathrm{IU} \/ \\mathrm{L}$ & $51 \\mathrm{pg} \/ \\mathrm{mL}$ & $17.8 \\mathrm{nmol} \/ \\mathrm{L}$ \\\\\n\\hline\n\\end{tabular}\nThe hormones FSH and LH are also involved in oogenesis in women. When LH surges in the ovarian cycle, what event will result?\n\nOptions:\nA. endometrium proliferation\nB. ovulation\nC. menstruation\nD. corpus luteum degradation",
  "response":"This question relies on your knowledge of the female reproductive cycle. In women, as FSH levels climb, follicles are stimulated to grow. These follicles produce estrogen, which exerts positive feedback on LH, whose levels begin to climb. When LH levels peak on the 14 th day of the cycle, the follicle ruptures, releasing the egg from the ovary and leaving the corpus luteum behind in the ovary. After ovulation, the corpus luteum secretes both estrogen and progesterone to inhibit GnRH. When GnRH is inhibited, FSH and LH will also be inhibited. Choice B correctly indicates that the LH surge causes ovulation. The correct answer is B."
 },
 {
  "instruction":"Questions 18-20 are based on the following data.\nThe generalized events in DNA replication can be depicted by the following flowchart:\nDouble helix $\\rightarrow$ Nicking of a strand $\\rightarrow$ Unwinding $\\rightarrow$ Destabilization and relief of pressure $\\rightarrow$ Initiation $\\rightarrow$ Elongation $\\rightarrow$ Closing of nicks\nBetween which two steps would the enzyme DNA polymerase be used?\n\nOptions:\nA. Double helix $\\rightarrow$ Nicking of a strand\nB. Nicking of a strand $\\rightarrow$ Unwinding\nC. Initiation $\\rightarrow$ Elongation\nD. Elongation $\\rightarrow$ Closing of nicks",
  "response":"Chain growth in DNA replication is initiated by a primase and is extended by a DNA polymerase. Thus the enzyme acting between Initiation $\\rightarrow$ Elongation is DNA polymerase. The correct answer is C."
 },
 {
  "instruction":"Osteomyelitis is a bone infection often caused by the bacterium Staphylococcus aureus, which spreads through the blood to the bones. The infection is quite common after surgeries when soft tissues become infected and the infection moves to the adjacent bones. The long bones of children are particularly susceptible as they have excellent circulation to them during the growth period. In adults, the best circulation is to the vertebrae, and these are the bones that are most likely to develop osteomyelitis.\nOsteomyelitis is treated with various antibiotics, but some patients have chronic infections that are not eradicated by antibiotic therapy due to a unique feature of S. aureus. S. aureus can be internalized by osteoblasts where the bacteria can continue to survive in the intracellular environment. Inside the osteoblasts, bacteria can induce apoptosis, killing their host cell. Surgical debridement is often required to remove the damaged and dead bone tissue created by the infection.\nA major research question deals with how an S. aureus infection induces apoptosis in the host osteoblasts. In order for apoptosis to initiate in osteoblasts, the enzyme capsase-8 must be activated. This enzyme serves as a protease that cleaves critical proteins in the cell, causing cell death. Capsase-8 is only expressed under certain circumstances. In the case of osteomyelitis, the S. aureus bacteria cause the induction of tumor necrosis factor-related apoptosis-inducing ligand (TRAIL) in the osteoblasts. Then TRAIL activates capsase-8, which triggers a series of events leading to apoptosis and death of osteoblasts.\nOsteomyelitis begins when S. aureus enters the bloodstream and moves to the bones. Under normal conditions, the nonspecific immune defense that would be activated as bacteria enter the blood would be:\n\nOptions:\nA. antibody production by plasma cells\nB. cytotoxic T-cell activation\nC. the complement system\nD. interleukin secretion\n",
  "response":"Complement proteins circulate in the blood in an inactive form. When bacteria enter the blood, the complement proteins are activated, which causes a cascade of reactions that ultimately cause lysis and death of the bacteria. Antibody production or cytotoxic T-cell activation could certainly be stimulated by bacteria in the blood, but the question stipulated that this must be a nonspecific defense and these defenses are specific. Interleukins are cytokines secreted by immune cells, and they would not directly be able to defend against bacteria in the bloodstream. The correct answer is C."
 },
 {
  "instruction":"Which of the following explanations could account for the fact that dominant alleles that cause fatal disorders are less common than recessive alleles that cause fatal disorders?\n\nOptions:\nA. Every person carrying a single fatal dominant allele dies, whereas most individuals who carry a single recessive lethal allele live and reproduce.\nB. Recessive fatal alleles must cause sterility.\nC. Dominant alleles that cause fatal disorders are more serious than recessive lethal disorders.\nD. Dominant alleles are generally less common than recessive alleles.",
  "response":"This question relies on a knowledge of dominant and recessive patterns of inheritance. Dominant alleles are expressed, even in the heterozygous individual. In this case, our example is a dominant allele that causes a fatal condition. Since every individual carrying a single allele of this type would likely not survive, we would expect the frequency of the allele to decrease over time. In contrast, disorders that follow recessive patterns of inheritance require two copies of the allele for expression. Those who are heterozygous do not express the disorder and thus would survive. The correct answer is A."
 },
 {
  "instruction":"The hormone responsible for maintaining control of water balance in the human body is:\n\nOptions:\nA. antidiuretic hormone\nB. luteinizing hormone\nC. adrenocorticotropic hormone\nD. growth hormone",
  "response":"This problem can be solved through knowledge of each of the various hormones given in the four options. Antidiuretic hormone maintains control of water balance in the human body by stimulating increased water reabsorption by the kidneys. Luteinizing hormone stimulates secretion of sex hormones by ovaries and testes. Adrenocorticotropic hormone stimulates the adrenal glands. Growth hormone regulates a wide variety of activities involved in growth and metabolism. The correct answer is A."
 },
 {
  "instruction":"\\section{Passage $\\mathrm{V}$}\nA student performed qualitative tests on a series of unknown chemicals in order to identify them. The unknowns included $\\mathrm{NaCl}, \\mathrm{NaHCO}_{3}$, sugar $\\left(\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}\\right), \\mathrm{MgSO}_{4}$, $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$, cornstarch, and chalk $\\left(\\mathrm{CaCO}_{3}\\right)$. These are all household chemicals. $\\mathrm{MgSO}_{4}$ is Epsom salts, $\\mathrm{NaHCO}_{3}$ is baking soda, and $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ is photographic fixer. The student was asked to identify six unknowns: A, B, C, D, E, and F.\nThe qualitative tests were based on the following:\n$>$ Solubility in water: starch is insoluble, as well as one of the other compounds. The rest are soluble in water.\n$>$ Chalk produces a gas when treated with acid.\n$>$ Starch turns blue when treated with iodine.\n$>\\mathrm{MgSO}_{4}$ produces a milky precipitate when treated with aqueous ammonia.\n$>\\mathrm{NaHCO}_{3}$ turns pink when treated with phenolphthalein.\n$>\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ decolorizes iodine solution.\n$>\\mathrm{NaCl}$ conducts electricity in solution.\n$>$ Sugar does not conduct electricity in solution.\nThe student prepared a flowchart that would aid in doing the experiments in a systematic manner and allow efficient identification of the unknowns.\n\\section{Experiment}\nThe experiments were conducted in the following order. The solubility of each chemical in water was tested first. From those results, the insoluble unknowns were tested next. The unknowns that were soluble in water were tested as aqueous solutions, rather than tested for conductivity.\n\\section{Results}\n\\begin{tabular}{ccccccl}\n\\hline Unknown & Solub & Conductivity & Acid & Phenolphthalein & $\\mathbf{N H}_{\\mathbf{3}}$ & $\\mathbf{I}_{\\mathbf{2}}$ \\\\\n\\hline A & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ \\\\\nB & $\\mathrm{N}$ & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ \\\\\nC & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ \\\\\n$\\mathrm{D}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{Y}_{\\mathrm{BLUE}}$ \\\\\nE & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ \\\\\nF & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{Y}_{\\text {COLORLESS }}$ \\\\\n\\hline\n\\end{tabular}\nFrom these results the student was able to identify all six unknowns.\nThe milky precipitate formed when $\\mathrm{MgSO}_{4}$ reacts with aqueous ammonia solution is $\\mathrm{Mg}(\\mathrm{OH})_{2}$. Enough magnesium sulfate is used to saturate a solution that, when tested with a $\\mathrm{pH}$ indicator strip, has a resulting $\\mathrm{pH}$ of 10 . What is the approximate $K_{\\mathrm{sp}}$ of magnesium hydroxide?\n\nOptions:\nA. $10^{-12}$\nB. $10^{-8}$\nC. $10^{-4}$\nD. $10^{-3}$",
  "response":"If the $\\mathrm{pH}=10$, then the $\\mathrm{pOH}$ is 4 because $\\mathrm{pH}+\\mathrm{pOH}=14$. Given the $\\mathrm{pOH}=4$, the $\\left[\\mathrm{OH}^{-}\\right]$is $1 \\times 10^{-4} \\mathrm{M}$. When the $K_{\\mathrm{sp}}$ expression is written for magnesium hydroxide, we get $K_{\\mathrm{sp}}=\\left[\\mathrm{Mg}^{2+}\\right]\\left[\\mathrm{OH}^{-}\\right]^{2}$. This then becomes $K_{\\mathrm{sp}}=[x][x]^{2}=[x]^{3} .\\left[1 \\times 10^{-4} \\mathrm{M}\\right]^{3}=1 \\times 10^{-12}$. This is close to the actual value of $1.6 \\times 10^{-12}$. The correct answer is A."
 },
 {
  "instruction":"Endothelin is a vasoactive peptide involved in a variety of processes, including the control of blood flow. Endothelin is capable of binding to two different receptors. There are receptors for endothelin located on smooth muscle (ETA) and receptors for endothelin located on vascular endothelium (ETB). When endothelin binds to ETA, it initiates vasoconstriction by narrowing blood vessels. However, when endothelin binds to ETB, vasodilation occurs due to the widening of blood vessels. The activity of the ETB receptor seems to counterregulate the ETA effects and prevents excessive vasoconstriction.\nDrug $X$ is a selective receptor antagonist for endothelin. It is designed to bind preferentially to certain endothelin receptors. Drug $\\mathrm{Y}$ is another selective receptor antagonist for endothelin. It is also designed to preferentially bind to one of the endothelin receptors. Researchers are hopeful that drugs $\\mathrm{X}$ and $\\mathrm{Y}$ will be of value in the treatment of conditions such as pulmonary arterial hypertension (PAH). In PAH, excess endothelin is produced, which causes an increase in blood pressure within the pulmonary arteries. Normal pressure in the pulmonary arteries is expected to be about $14 \\mathrm{~mm} \\mathrm{Hg}$, and pressure at or above $25 \\mathrm{~mm}$ Hg constitutes PAH. Both drugs have been tested in patients with PAH. There were five patients in each group. At the end of the study, the percent decrease in mean pulmonary artery pressure was measured for each patient and reported in the table.\n\\begin{tabular}{cccc}\n\\hline Patient & Drug X & DrugY & Placebo \\\\\n\\hline 1 & $+10 \\%$ & $-10 \\%$ & $+5 \\%$ \\\\\n2 & $+3 \\%$ & $-5 \\%$ & $0 \\%$ \\\\\n3 & $-5 \\%$ & $+2 \\%$ & $+13 \\%$ \\\\\n4 & $+2 \\%$ & $0 \\%$ & $-4 \\%$ \\\\\n5 & $0 \\%$ & $-7 \\%$ & $-2 \\%$ \\\\\nAverage decrease & $+2 \\%$ & $-4 \\%$ & $+2.4 \\%$ \\\\\n\\hline\n\\end{tabular}\nSuppose a patient suffers from chronic systolic and diastolic hypertension due to an inherited defect in an endothelin receptor. How would the kidneys attempt to alleviate the hypertension?\n\nOptions:\nA. by decreasing water absorption in the nephrons\nB. by decreasing urine output thus increasing blood volume\nC. by increasing the amount of aldosterone secreted from the adrenal glands\nD. by increasing the amount of antidiuretic hormone secreted from the pituitary gland",
  "response":"This question expects you to be able to consider how the kidneys are involved in regulating blood volume and blood pressure. Water that is reabsorbed in the nephrons returns to the bloodstream to increase plasma volume, which in turn increases blood pressure as the volume of blood increases. When the nephrons absorb more water to increase blood volume, the urine volume decreases. The two hormones aldosterone and antidiuretic hormone help increase the amount of water reabsorbed by nephrons, which in turn decreases urine volume. In order to reduce blood pressure, the kidneys can decrease the amount of water reabsorbed by the nephrons. The less water reabsorbed means there will be less blood volume and less blood pressure. This would ultimately increase urine volume. The question asks how the kidneys might try to reduce hypertension. This would mean that we want less water reabsorbed into the blood, which would decrease blood volume and blood pressure. To do this, urine volume would have to increase. The correct answer is A."
 },
 {
  "instruction":"Sperm count in human males is influenced by a variety of factors. One of the most important variables affecting sperm count is the number of Sertoli cells located in seminiferous tubules of the testes. The Sertoli cells' primary function is to nurture cells through the process of spermatogenesis by providing them with nutrients and an environment conducive to spermatogenesis. Each Sertoli cell has a set number of spermatozoa that it can support. Sertoli cells have an additional role in forming the blood-testis barrier that provides immune privilege for the seminiferous tubules. This forms a physical barricade that keeps the developing sperm from coming into contact with cells of the immune system.\nWhile it is not clear as to all the factors involved in determining Sertoli cell number, it is apparent that their development is most critical during the first nine months of life and prior to puberty. Events that affect the development of Sertoli cells in young males can influence fertility later in life. There may also be genetic factors involved in determining the number of Sertoli cells formed in a particular male.\nThe following data were collected from two groups of men. The men in group 1 had a normal sperm count while the men in group 2 were seeking infertility treatment due to a low sperm count. Levels of follicle-stimulating hormone (FSH) and luteinizing hormone (LH) from the pituitary were measured as were the testicular hormones inhibin-B and testosterone in an attempt to determine which hormones seem to be linked to reduced sperm count.\n\\begin{tabular}{ccccc}\n\\hline Sample Group & LH & FSH & Inhibin-B & Testosterone \\\\\n\\hline 1 & $5.3 \\mathrm{IU} \/ \\mathrm{L}$ & $5.0 \\mathrm{IU} \/ \\mathrm{L}$ & $190 \\mathrm{pg} \/ \\mathrm{mL}$ & $18.0 \\mathrm{nmol} \/ \\mathrm{L}$ \\\\\n2 & $3.1 \\mathrm{IU} \/ \\mathrm{L}$ & $17.6 \\mathrm{IU} \/ \\mathrm{L}$ & $51 \\mathrm{pg} \/ \\mathrm{mL}$ & $17.8 \\mathrm{nmol} \/ \\mathrm{L}$ \\\\\n\\hline\n\\end{tabular}\nFully differentiated Sertoli cells lose the ability to proliferate to produce more Sertoli cells. This would mean that they cannot perform:\n\nOptions:\nA. mitosis\nB. meiosis\nC. apoptosis\nD. binary fission",
  "response":"This question relies on your knowledge of cell division. Mitosis is the process by which differentiated cells produce identical daughter cells. The correct answer is A."
 },
 {
  "instruction":"The DNA doubles and chromosomes replicate during which phase of the cell cycle?\n\nOptions:\nA. $\\mathrm{G} 1$\nB. metaphase\nC. $\\mathrm{S}$\nD. G2",
  "response":"The majority of the cell cycle is spent in interphase, which consists of three stages: G1, S, and G2. In the G1 stage of interphase, the cell organelles are doubled, and materials required for DNA synthesis are accumulated for the onset of cell division. The $\\mathrm{S}$ stage of interphase is the stage of the cell cycle where the amount of DNA doubles with the replication of chromosomes. In the G2 stage of interphase, which follows the synthesis of DNA, proteins required for the next cell division are synthesized. Metaphase is the stage in mitosis characterized by the precise lineup of the chromosomes along the equatorial plane. The correct answer is C."
 },
 {
  "instruction":"A geneticist allows a cell to replicate in the presence of radioactive nucleotides. Which of the following would occur?\n\nOptions:\nA. The DNA in one of the daughter cells would be radioactive but not in the other daughter cell.\nB. The mRNA made by the daughter cells would be radioactive.\nC. The DNA in each of the daughter cells would be radioactive.\nD. The DNA would not be radioactive in either of the daughter cells. MCAT",
  "response":"This question relies on your knowledge of semiconservative DNA replication where a single double helix is separated in order for both strands of the double helix to serve as templates for the production of new strands. The correct answer is C."
 },
 {
  "instruction":"Leukemia is a term that describes a wide variety of blood cell cancers. Most leukemia cases involve elevated leukocyte counts. In acute cases of leukemia, immature leukocytes proliferate quickly, which leads to many abnormal immature cells in the bone marrow. This inhibits the ability of the bone marrow to produce healthy new leukocytes. This form of the cancer progresses rapidly, and the cells metastasize easily as they move into the bloodstream from the bone marrow. It can be fatal within weeks or months if not treated promptly. In contrast, chronic cases of leukemia typically involve the proliferation of abnormal mature leukocytes. This form of the disease progresses more slowly, and treatment need not always be immediate.\nOnce classified as acute or chronic, leukemia can be further subdivided into lymphocytic or myelogenous. This distinction depends on which type of leukocyte is abnormal in the patient. In lymphocytic leukemia, the lymphoid cells, such as the lymphocytes, are abnormal. In the myelogenous form, the myeloid cells, such as basophils, neutrophils, and eosinophils, are affected.\nChemotherapy drugs are often used in the management of leukemia cases. There are numerous categories of these drugs, and they have varied mechanisms of action. In treatment, several drugs are used in combination because some of the drugs are cell-cycle specific while others are not. Alkylating agents are capable of disrupting the function of DNA. Tubulin-binding agents are designed to interfere with microtubule formation. Other drugs are designed to interfere with DNA polymerase, DNA repair mechanisms, or protein synthesis, or to act as purine analogs. Many of these drugs cannot enter into the brain when injected or ingested, so if the cancer has spread to\nthis area, special methods must be used to introduce the drugs to the brain.\nA variety of risk factors exist for leukemia. Some of these factors include exposure to radiation, certain viral infections, and inherited tendencies. What do all of these factors have in common?\n\nOptions:\nA. All involve problems with protein translation in the cells.\nB. All can cause an increase in the production of cellular growth factors that enhance cell proliferation.\nC. All cause changes to the DNA that can lead to the production of altered proteins that can influence the development of cancer.\nD. All cause problems with transcription in the cells.",
  "response":"Of the items listed in the question, the thing they all have in common is that they are mutagens or involve some genetic change. These mutations to the DNA can lead to the production of altered proteins that can be associated with cancer. While some cancers are associated with changes in cellular growth factors, this would not be the case for all cancers. The translation of proteins is directly from the mRNA, which is derived from the DNA in transcription. There would have to be a change in the DNA for the translation of proteins to be affected. The correct answer is C."
 },
 {
  "instruction":"\\section{\nIn October 2012, Felix Baumgartner attempted a world-record sky-diving jump from just over 128,000 ft (39,014 m) above the Earth's surface. He was fitted with a customized, pressurized spacesuit that brought his total weight to $260 \\mathrm{lbs}$ ( $1156 \\mathrm{~N}$ ). The spacesuit was designed to collect and transmit important monitoring, communications, and tracking information throughout the jump. Baumgartner then ascended to the jumping altitude housed within a pressurized capsule of weight 2900 lbs ( $12900 \\mathrm{~N})$ attached to a $30 \\times 10^{6}$ cubic feet $\\left(8.5 \\times 10^{8} \\mathrm{~L}\\right)$ helium-filled weather balloon. The weather balloon with the attached capsule ascended at a rate of $1000 \\mathrm{ft} \/ \\mathrm{min}(5.1 \\mathrm{~m} \/ \\mathrm{s})$, reaching the jumping altitude in approximately two hours.\nOnce the capsule reached the jumping altitude, the door to the capsule was opened and Baumgartner stepped out and fell forward, beginning his free-fall descent. He continued falling until he reached a maximum speed of $834 \\mathrm{mph}(373 \\mathrm{~m} \/ \\mathrm{s})$, which exceeded the speed of sound. This speed was Baumgartner's terminal velocity-the velocity attained in which he was no longer being accelerated downward due to gravity. Once he reached this speed, Baumgartner deployed his parachute and glided safely to Earth's surface.The distance traversed by Mr. Baumgartner before he reached maximum velocity was approximately:\n\nOptions:\nA. $4800 \\mathrm{~m}$\nB. $5480 \\mathrm{~m}$\nC. $6350 \\mathrm{~m}$\nD. $7100 \\mathrm{~m}$",
  "response":"The initial velocity $\\left(v_{i}\\right)$ of Felix Baumgartner in the vertical direction is 0 , his final velocity $\\left(v_{f}\\right)$ is $373 \\mathrm{~m} \/ \\mathrm{s}$, and the acceleration due to gravity is $9.8 \\mathrm{~m} \/ \\mathrm{s}^{2}$ (although in reality it is probably slightly lower because of the significantly large jumping altitude). The equation of motion that can be used to determine the distance traversed by Baumgartner is: The correct answer is D."
 },
 {
  "instruction":"Genetically modified (GM) foods contain DNA from more than one source. The goal of creating GM foods is to modify the organisms to express new and useful traits. A few examples involve engineering crops to be resistant to bacterial or viral infections, to be resistant to herbicides, to ripen more slowly, to survive extreme temperature changes, to contain vaccines, or to create crops that are more nutritious.\nOne GM food that has received media attention over the years is golden rice. This GM product was first introduced in 2000 and was engineered to provide carotene that is a precursor to vitamin $\\mathrm{A}$. The carotene in the rice provides the golden color. Vitamin A deficiency is a major problem in underdeveloped parts of the world, killing more than one million people a year, most of whom are children. This vitamin is critical for the development and maintenance of epithelial tissues, for vision, and for growth.\nRice plants normally produce carotene in the inedible parts of the plant. To be able to access and use the carotene, it was necessary to engineer the rice to produce it in the edible grains. In order to achieve this goal, two genes from daffodil plants, psy and $l y c$, and a bacterial gene called $c r t 1$ were added to the rice genome. These genes engineered a pathway that produces lycopene, which is eventually converted to carotene. Each of these genes was placed under the control of a grain-specific promoter so that the products were produced in the rice grains.\nWhile golden rice was produced to provide carotene, which the human body can convert into the essential vitamin A, the rice provided much less carotene than was anticipated. Golden rice 2, a modified version released in 2005, contains a much larger amount of carotene and, as such, is a better source of the vitamin. Field tests were conducted in 2009, and human consumption trials were completed in 2012. However, neither version has been grown in large-scale quantities at this time.\nThe engineering of golden rice to have several exogenous genes that can be induced under the control of a single promoter would MOST resemble:\n\nOptions:\nA. an operon system\nB. hybridized DNA\nC. a vector\nD. Okazaki fragments",
  "response":"A series of several genes under the control of a single promoter is most similar to the inducible operon system in bacteria. Okazaki fragments are produced during semiconservative replication on the lagging strand of DNA. Strands of DNA that are hybridized are complementary to each other. A vector is used to carry DNA from one source into another. The correct answer is A."
 },
 {
  "instruction":"Venous blood coming from the head area in humans returns to the heart through which major vessel?\n\nOptions:\nA. superior vena cava\nB. aorta\nC. hepatic portal vein\nD. carotid artery",
  "response":"The superior vena cava is the vein that carries deoxygenated blood from the upper half of the body to the heart into its right atrium. The aorta is the largest and longest of the arteries of the human body that transports oxygenated blood directly from the left ventricle to all other organs and tissues in the body. The hepatic portal vein is the major vein that transports blood from the digestive organs into the liver. The carotid artery is one of two types of arteries (the other being vertebral) that feed blood from the heart and into the head\/ brain area. Answers and Explanations No passage MCAT The correct answer is A."
 },
 {
  "instruction":"The kidneys are responsible for filtering certain components of blood and producing urine from them. In an adult, the kidneys are typically capable of processing about 1 to $1.5$ liters of water per hour depending on the individual and conditions within the body. In a condition known as water intoxication, a person ingests more water (usually 3 or more liters in a single sitting) in a shorter period of time than the kidneys can process. In a child, the amount of water required to induce water intoxication is much less. Of the options listed, what would be the best treatment for someone suffering from water intoxication?\n\nOptions:\nA. administering a saline solution that is more concentrated than the cells\nB. administering an isotonic solution\nC. administering a solution low in solute concentration\nD. performing dialysis to remove excess water from the plasma",
  "response":"In this case, administering a solution that is hypertonic to the cells would help. A hypertonic extracellular solution would draw water out of the swollen cells to counteract the effects of water intoxication. An isotonic solution would not help. A solution low in solute concentration would be high in water relative to the cells making the solution hypotonic, which would cause more water to enter the cells. A saline solution would be high in solutes and hypertonic to the cells. The correct answer is A."
 },
 {
  "instruction":"\\section{\nIn October 2012, Felix Baumgartner attempted a world-record sky-diving jump from just over 128,000 ft (39,014 m) above the Earth's surface. He was fitted with a customized, pressurized spacesuit that brought his total weight to $260 \\mathrm{lbs}$ ( $1156 \\mathrm{~N}$ ). The spacesuit was designed to collect and transmit important monitoring, communications, and tracking information throughout the jump. Baumgartner then ascended to the jumping altitude housed within a pressurized capsule of weight 2900 lbs ( $12900 \\mathrm{~N})$ attached to a $30 \\times 10^{6}$ cubic feet $\\left(8.5 \\times 10^{8} \\mathrm{~L}\\right)$ helium-filled weather balloon. The weather balloon with the attached capsule ascended at a rate of $1000 \\mathrm{ft} \/ \\mathrm{min}(5.1 \\mathrm{~m} \/ \\mathrm{s})$, reaching the jumping altitude in approximately two hours.\nOnce the capsule reached the jumping altitude, the door to the capsule was opened and Baumgartner stepped out and fell forward, beginning his free-fall descent. He continued falling until he reached a maximum speed of $834 \\mathrm{mph}(373 \\mathrm{~m} \/ \\mathrm{s})$, which exceeded the speed of sound. This speed was Baumgartner's terminal velocity-the velocity attained in which he was no longer being accelerated downward due to gravity. Once he reached this speed, Baumgartner deployed his parachute and glided safely to Earth's surface.The drag force, $F$ drag, can be expressed in terms of the terminal velocity, $v t$, as $F$ drag $=C v_{t}^{2}$, where $C$ is a drag constant. The value of the drag constant for Mr. Baumgartner is:\n\nOptions:\nA. $0.008 \\mathrm{~kg} \/ \\mathrm{m}$\nB. $0.680 \\mathrm{~kg} \/ \\mathrm{m}$\nC. $1.320 \\mathrm{~kg} \/ \\mathrm{m}$\nD. $5.640 \\mathrm{~kg} \/ \\mathrm{m}$",
  "response":"When an object in free fall reaches terminal velocity, it is in equilibrium. In other words, the drag force acting upward becomes equal in magnitude yet opposite in direction to the object's weight, or: The correct answer is A."
 },
 {
  "instruction":"Iodine-131, a radioisotope typically employed in the diagnosis of thyroid disorders, has a mass number, $A$, of 131 and an atomic number, $Z$, of 53 . The number of neutrons, $N$, of iodine-131 is:\n\nOptions:\nA. 131\nB. 78\nC. 53\nD. 25",
  "response":"The number of neutrons is the mass number $A$ minus the atomic number $Z$. Thus, the number of neutrons of iodine-131 is $N=A-Z=$ $131-53=78$ The correct answer is B."
 },
 {
  "instruction":"Osteomyelitis is a bone infection often caused by the bacterium Staphylococcus aureus, which spreads through the blood to the bones. The infection is quite common after surgeries when soft tissues become infected and the infection moves to the adjacent bones. The long bones of children are particularly susceptible as they have excellent circulation to them during the growth period. In adults, the best circulation is to the vertebrae, and these are the bones that are most likely to develop osteomyelitis.\nOsteomyelitis is treated with various antibiotics, but some patients have chronic infections that are not eradicated by antibiotic therapy due to a unique feature of S. aureus. S. aureus can be internalized by osteoblasts where the bacteria can continue to survive in the intracellular environment. Inside the osteoblasts, bacteria can induce apoptosis, killing their host cell. Surgical debridement is often required to remove the damaged and dead bone tissue created by the infection.\nA major research question deals with how an S. aureus infection induces apoptosis in the host osteoblasts. In order for apoptosis to initiate in osteoblasts, the enzyme capsase-8 must be activated. This enzyme serves as a protease that cleaves critical proteins in the cell, causing cell death. Capsase-8 is only expressed under certain circumstances. In the case of osteomyelitis, the S. aureus bacteria cause the induction of tumor necrosis factor-related apoptosis-inducing ligand (TRAIL) in the osteoblasts. Then TRAIL activates capsase-8, which triggers a series of events leading to apoptosis and death of osteoblasts.\nOsteoblasts are unusual in that they actively engulf $S$. aureus, which survives in the osteoblast and eventually triggers apoptosis. In engulfing the bacteria, the osteoblasts resemble the normal activities of which type of leukocyte?\n\nOptions:\nA. basophils\nB. lymphocytes\nC. eosinophils\nD. macrophages",
  "response":"The leukocytes that engulf microbes such as bacteria are macrophages. This process is termed phagocytosis. Basophils secrete histamine, which is involved in the initiation of the inflammatory response. Lymphocytes are involved in specific immune system defenses. Eosinophils are involved in the defense response to antigens from unique sources such as parasites. The correct answer is D."
 },
 {
  "instruction":"Passage $\\mathrm{V}$\nFor many years, researchers have been interested in developing edible vaccines that would provide protection against various bacterial and viral diseases. An edible vaccine is a genetically modified organism (GMO) that is created when genes from the bacteria or virus being vaccinated against are inserted into plant cells. As the plants grow, they produce proteins from the bacteria or virus that the researchers are attempting to vaccinate against. When a person eats the plant, the viral or bacteria proteins serve as antigens, which initiates a primary immune response.\nSome of the plants that are of interest in edible vaccine development are bananas, potatoes, tomatoes, corn, rice, tobacco, and soybeans. While early research focused on growing crops that produced edible products such as bananas, the most recent research is focused on extracting material from the leaves and seeds of the plants, dehydrating it, and putting it into capsules that can be swallowed. This allows for a more predictable dosage of the vaccine and eliminates concerns that the crops containing the vaccine would be accidentally eaten by individuals who are not in need of the vaccine. The term edible vaccine has now been replaced with the term plantderived oral vaccine.\nThe development of plant-derived oral vaccines solves many problems associated with injectable vaccines. Because plant-derived oral vaccines are grown in crops, there is no problem with storage or refrigeration that is required of most traditional vaccines. These plant-derived vaccines also require no special training to administer. Additionally, when a vaccine is taken orally, it provides mucosal immunity due to the production of IgA antibodies by the immune system-in addition to systemic immunity. Since 90 percent of diseases enter the body through a mucus membrane, a vaccine that coerces mucosal immunity is of great benefit in the prevention of many diseases.\nA traditional vaccine that is administered either intramuscularly or subcutaneously would be expected to cause the formation of which type of antibodies initially following administration?\n\nOptions:\nA. $\\operatorname{IgE}$\nB. IgG\nC. IgM\nD. IgA",
  "response":"IgM antibodies are produced first in response to exposure to a new antigen. Later, IgG antibodies will be produced. The correct answer is C."
 },
 {
  "instruction":"A certain type of cell makes antibodies that are secreted from the cell. It is possible to track the path of these as they leave the cell by labeling them with radioactive isotopes. Which of the following might be the path of antibodies from where they are made to the cell membrane?\n\nOptions:\nA. Golgi complex to lysosomes to cell membrane\nB. rough endoplasmic reticulum to Golgi complex to cell membrane\nC. nucleus to Golgi complex to cell membrane\nD. smooth endoplasmic reticulum to lysosomes to cell membrane",
  "response":"Antibodies are proteins that are secreted from the cell. Proteins would be produced and labeled by the rough endoplasmic reticulum. From there, they would be sorted by the Golgi complex and finally directed to the cell membrane for secretion. The correct answer is B."
 },
 {
  "instruction":"Passage $\\mathrm{V}$\nFor many years, researchers have been interested in developing edible vaccines that would provide protection against various bacterial and viral diseases. An edible vaccine is a genetically modified organism (GMO) that is created when genes from the bacteria or virus being vaccinated against are inserted into plant cells. As the plants grow, they produce proteins from the bacteria or virus that the researchers are attempting to vaccinate against. When a person eats the plant, the viral or bacteria proteins serve as antigens, which initiates a primary immune response.\nSome of the plants that are of interest in edible vaccine development are bananas, potatoes, tomatoes, corn, rice, tobacco, and soybeans. While early research focused on growing crops that produced edible products such as bananas, the most recent research is focused on extracting material from the leaves and seeds of the plants, dehydrating it, and putting it into capsules that can be swallowed. This allows for a more predictable dosage of the vaccine and eliminates concerns that the crops containing the vaccine would be accidentally eaten by individuals who are not in need of the vaccine. The term edible vaccine has now been replaced with the term plantderived oral vaccine.\nThe development of plant-derived oral vaccines solves many problems associated with injectable vaccines. Because plant-derived oral vaccines are grown in crops, there is no problem with storage or refrigeration that is required of most traditional vaccines. These plant-derived vaccines also require no special training to administer. Additionally, when a vaccine is taken orally, it provides mucosal immunity due to the production of IgA antibodies by the immune system-in addition to systemic immunity. Since 90 percent of diseases enter the body through a mucus membrane, a vaccine that coerces mucosal immunity is of great benefit in the prevention of many diseases.\nThe goal of any type of vaccination is to induce active immunity without people getting ill from the disease that they are being vaccinated against. In order to have long-term effectiveness, the vaccine must cause the development of which type of cells in the immune system?\n\nOptions:\nA. cytotoxic T cells\nB. helper T cells\nC. plasma cells\nD. memory cells",
  "response":"Once activated, most cells in the immune system do not live long. Activated cytotoxic T cells and plasma cells are inactivated once the antigen that they have been designed to fight has been eliminated. The basis of long-term immune system protection is the development of memory T and memory B cells (choice D). These memory cells are produced upon primary exposure to a specific antigen and have the ability to survive for years within the body. Should the same antigen return, memory cells quickly proliferate into cytotoxic T cells and plasma cells to fight the antigen. The correct answer is D."
 },
 {
  "instruction":"Leukemia is a term that describes a wide variety of blood cell cancers. Most leukemia cases involve elevated leukocyte counts. In acute cases of leukemia, immature leukocytes proliferate quickly, which leads to many abnormal immature cells in the bone marrow. This inhibits the ability of the bone marrow to produce healthy new leukocytes. This form of the cancer progresses rapidly, and the cells metastasize easily as they move into the bloodstream from the bone marrow. It can be fatal within weeks or months if not treated promptly. In contrast, chronic cases of leukemia typically involve the proliferation of abnormal mature leukocytes. This form of the disease progresses more slowly, and treatment need not always be immediate.\nOnce classified as acute or chronic, leukemia can be further subdivided into lymphocytic or myelogenous. This distinction depends on which type of leukocyte is abnormal in the patient. In lymphocytic leukemia, the lymphoid cells, such as the lymphocytes, are abnormal. In the myelogenous form, the myeloid cells, such as basophils, neutrophils, and eosinophils, are affected.\nChemotherapy drugs are often used in the management of leukemia cases. There are numerous categories of these drugs, and they have varied mechanisms of action. In treatment, several drugs are used in combination because some of the drugs are cell-cycle specific while others are not. Alkylating agents are capable of disrupting the function of DNA. Tubulin-binding agents are designed to interfere with microtubule formation. Other drugs are designed to interfere with DNA polymerase, DNA repair mechanisms, or protein synthesis, or to act as purine analogs. Many of these drugs cannot enter into the brain when injected or ingested, so if the cancer has spread to\nthis area, special methods must be used to introduce the drugs to the brain.\nWhat would be the BEST explanation as to why people with chronic forms of leukemia can live longer untreated than those with acute cases of leukemia?\n\nOptions:\nA. Chronic cases involve types of leukocytes that have less important functions in the body.\nB. Chronic cases of leukemia involve abnormal cell division of mature cells, so there are still some immature cells that can proliferate into normal mature cells.\nC. The immune system is less likely to recognize and destroy abnormal immature leukocytes than it would be to recognize and destroy abnormal mature leukocytes.\nD. all of the above",
  "response":"The passage explains that in acute cases of leukemia, Passage I immature leukocytes are involved. This implies that these immature cells are abnormal and may not proliferate into normal mature leukocytes. The passage also indicates that in chronic cases of leukemia, mature leukocytes are affected so that it is possible that some unaffected immature cells may proliferate into functional mature cells. If there are some functional leukocytes, the person would be likely to survive longer without treatment. The correct answer is B."
 },
 {
  "instruction":"The process of quorum sensing is observed in a variety of species. It involves a system of communication between organisms that is dependent on population density. Bacteria are notorious for using quorum sensing (QS) via the secretion of secreted signaling molecules, called autoinducers (AI), to coordinate their gene expression once a threshold density of population is reached. Not only can bacteria communicate via QS within their species, but they can also cross-talk through QS between different species. In order to participate in quorum sensing, bacteria require a way to produce an autoinducer and a way to respond to autoinducers via binding to specific receptors.\nOne example of how bacteria use quorum sensing to their advantage is to regulate the gene expression of virulence factors. During infection, if a small number of bacteria were to express their virulence factors, this would largely have no impact on the host. Instead, bacteria wait until a quorum is reached in terms of population size, indicating that the population is large enough to express virulence factors. These virulence factors relate to such abilities as escaping the immune response, invasion, and spreading within the host.\nWe know that we are facing a current healthcare crisis related to the loss of effectiveness of many traditional types of antibiotics based on bacterial resistance. This provides a major impetus for identifying novel ways to target bacterial pathogens that avoid the use of antibiotics. Several chemical compounds and molecules have been identified as facilitators of quorum sensing inhibition. Typically, these factors can serve as signal-targeting enzymes or small-molecule inhibitors of signal synthases and receptors.\nTargets of traditional antibiotics include all of the following EXCEPT:\n\nOptions:\nA. ribosomes\nB. metabolic processes\nC. the cell wall\nD. the nucleus",
  "response":"Antibiotics must target structures that are unique to prokaryotic cells. Of the choices listed, the nucleus is a structure lacking in prokaryotic cells and therefore could not serve as a target for antibiotics. The correct answer is D."
 },
 {
  "instruction":"The process of quorum sensing is observed in a variety of species. It involves a system of communication between organisms that is dependent on population density. Bacteria are notorious for using quorum sensing (QS) via the secretion of secreted signaling molecules, called autoinducers (AI), to coordinate their gene expression once a threshold density of population is reached. Not only can bacteria communicate via QS within their species, but they can also cross-talk through QS between different species. In order to participate in quorum sensing, bacteria require a way to produce an autoinducer and a way to respond to autoinducers via binding to specific receptors.\nOne example of how bacteria use quorum sensing to their advantage is to regulate the gene expression of virulence factors. During infection, if a small number of bacteria were to express their virulence factors, this would largely have no impact on the host. Instead, bacteria wait until a quorum is reached in terms of population size, indicating that the population is large enough to express virulence factors. These virulence factors relate to such abilities as escaping the immune response, invasion, and spreading within the host.\nWe know that we are facing a current healthcare crisis related to the loss of effectiveness of many traditional types of antibiotics based on bacterial resistance. This provides a major impetus for identifying novel ways to target bacterial pathogens that avoid the use of antibiotics. Several chemical compounds and molecules have been identified as facilitators of quorum sensing inhibition. Typically, these factors can serve as signal-targeting enzymes or small-molecule inhibitors of signal synthases and receptors.\nN-Acyl homoserine lactones (N-AHLs) are one type of autoinducer produced by certain species of bacteria. Different types of bacteria can produce N-AHLs with unique molecular structures by making minor changes to the core structure of the molecule. Given the structure of the molecule, what would be the MOST likely way they could do this?\n\nOptions:\nA. by altering the $\\mathrm{R}$ group side chains\nB. by altering fatty acid chains\nC. by glycosylating the core structure\nD. by methylating the core structure",
  "response":"This question relies on your knowledge of the structure of organic molecules. The N-AHLs in this question should be recognized as some derivative of an amino acid. Given that information, the most likely way to alter them would be by altering their R group side chains. The correct answer is A."
 },
 {
  "instruction":"Genetically modified (GM) foods contain DNA from more than one source. The goal of creating GM foods is to modify the organisms to express new and useful traits. A few examples involve engineering crops to be resistant to bacterial or viral infections, to be resistant to herbicides, to ripen more slowly, to survive extreme temperature changes, to contain vaccines, or to create crops that are more nutritious.\nOne GM food that has received media attention over the years is golden rice. This GM product was first introduced in 2000 and was engineered to provide carotene that is a precursor to vitamin $\\mathrm{A}$. The carotene in the rice provides the golden color. Vitamin A deficiency is a major problem in underdeveloped parts of the world, killing more than one million people a year, most of whom are children. This vitamin is critical for the development and maintenance of epithelial tissues, for vision, and for growth.\nRice plants normally produce carotene in the inedible parts of the plant. To be able to access and use the carotene, it was necessary to engineer the rice to produce it in the edible grains. In order to achieve this goal, two genes from daffodil plants, psy and $l y c$, and a bacterial gene called $c r t 1$ were added to the rice genome. These genes engineered a pathway that produces lycopene, which is eventually converted to carotene. Each of these genes was placed under the control of a grain-specific promoter so that the products were produced in the rice grains.\nWhile golden rice was produced to provide carotene, which the human body can convert into the essential vitamin A, the rice provided much less carotene than was anticipated. Golden rice 2, a modified version released in 2005, contains a much larger amount of carotene and, as such, is a better source of the vitamin. Field tests were conducted in 2009, and human consumption trials were completed in 2012. However, neither version has been grown in large-scale quantities at this time.\nGenetically modified organisms, typically plants, are being developed as a means of administering vaccines. If these transgenic plants are to be successful as vaccines, they will need to:\n\nOptions:\nA. stimulate inflammation in the recipient\nB. express antigens to stimulate adaptive defenses in the recipient\nC. contain antibodies to the organism being vaccinated against\nD. contain memory cells against the organism being vaccinated against",
  "response":"Vaccines are designed to work by stimulating the primary immune response of the recipient. This requires the immune system to recognize an antigen and to mount a specific immune response. Ultimately, this leads to the production of memory cells that can survive for long periods of time. When the system is challenged by the actual antigen, different types of memory cells can quickly proliferate into plasma cells and cytotoxic $\\mathrm{T}$ cells in order to destroy the antigen. The correct answer is B."
 },
 {
  "instruction":"According to the bystander effect, which is MOST likely to be true in an emergency?\n\nOptions:\nA. the greater the number of people present, the slower the response because of increased diffusion of responsibility\nB. the greater the number of people present, the slower the response because of decreased diffusion of responsibility\nC. the greater the number of people present, the faster the response because of increased diffusion of responsibility\nD. the greater the number of people present, the faster the response because of decreased diffusion of responsibility",
  "response":"The bystander effect is characterized by slowness in response due to diffusion of responsibility. The more people who are present, the less an individual is likely to feel responsible to respond to the emergency because of the collective presence of other individuals. The correct answer is A."
 },
 {
  "instruction":"Which of the following statements BEST explains why your body prefers to perform aerobic respiration as opposed to anaerobic respiration?\n\nOptions:\nA. Anaerobic respiration does not allow for NADH to be recycled to NAD+.\nB. Aerobic respiration requires less of an adenosine triphosphate (ATP) investment than anaerobic respiration does.\nC. Aerobic respiration produces far more ATP than anaerobic respiration.\nD. Aerobic respiration is easier for cells to perform than anaerobic respiration.",
  "response":"The primary difference between aerobic and anaerobic No Passage cellular respiration is the amount of ATP generated. Anaerobic respiration generates far less ATP per glucose molecule than does aerobic respiration. The correct answer is C."
 },
 {
  "instruction":"Prokaryotic and eukaryotic cells generally have which of the following structures in common?\n\nOptions:\nA. a membrane-bound nucleus\nB. a cell wall made of cellulose\nC. ribosomes\nD. linear chromosomes made of DNA",
  "response":"Prokaryotic cells lack membrane-bound organelles including a nucleus. Their DNA is circular and does not exist as linear chromosomes. Prokaryotic cells can have a cell wall, but it is typically made of peptidoglycan. The only item on this list that is common between prokaryotic and eukaryotic cells is that they both contain ribosomes needed for protein synthesis within the cell. The correct answer is C."
 },
 {
  "instruction":"When two solutions that differ in solute concentration are placed on either side of a semipermeable membrane and osmosis is allowed to occur, which of the following will be observed?\n\nOptions:\nA. Water will move from an area of low solute concentration to an area of high solute concentration.\nB. The solute will move from an area of high concentration to an area of low concentration.\nC. There will be no net movement of water.\nD. Water will move from an area of high solute concentration to an area of low solute concentration.",
  "response":"Osmosis is a specialized form of diffusion, or passive transport. During osmosis, water always moves from the side of the membrane that has more water to the side of the membrane that has less water. The only thing that moves by osmosis is water. This means that choice B can be eliminated because solutes are not moving by osmosis. If there is a concentration difference across the membrane, there will be water movement, which eliminates choice $\\mathrm{C}$. When osmosis occurs, water moves from the side of the membrane that has more water to the side that has less water. This means that the side of the membrane with more water has less solute (and is therefore less concentrated) than the side of the membrane with less water and more solute (being more concentrated). Based on the choices provided, choice A is the most appropriate. The correct answer is A."
 },
 {
  "instruction":"Questions 18-20 are based on the following data.\nThe generalized events in DNA replication can be depicted by the following flowchart:\nDouble helix $\\rightarrow$ Nicking of a strand $\\rightarrow$ Unwinding $\\rightarrow$ Destabilization and relief of pressure $\\rightarrow$ Initiation $\\rightarrow$ Elongation $\\rightarrow$ Closing of nicks\nBetween which two steps would the enzyme DNA ligase be used?\n\nOptions:\nA. Double helix $\\rightarrow$ Nicking of a strand\nB. Nicking of a strand $\\rightarrow$ Unwinding\nC. Initiation $\\rightarrow$ Elongation\nD. Elongation $\\rightarrow$ Closing of nicks",
  "response":"Abutting segments of unjoined DNA can be annealed by ligases. Thus the enzyme acting between Elongation $\\rightarrow$ Closing of nicks is $D N A$ ligase. The correct answer is D."
 },
 {
  "instruction":"The kidneys are responsible for filtering certain components of blood and producing urine from them. In an adult, the kidneys are typically capable of processing about 1 to $1.5$ liters of water per hour depending on the individual and conditions within the body. In a condition known as water intoxication, a person ingests more water (usually 3 or more liters in a single sitting) in a shorter period of time than the kidneys can process. In a child, the amount of water required to induce water intoxication is much less.  The regulation of water and sodium levels in the nephrons is adjusted in two regions of the kidneys where water and sodium can be reabsorbed from the nephron and returned to circulation. Those regions are the:\n\nOptions:\nA. proximal convoluted tubule and distal convoluted tubule\nB. the loop of Henle and the distal convoluted tubule\nC. the distal convoluted tubule and the collecting duct\nD. the proximal convoluted tubule and the loop of Henle",
  "response":"In the proximal convoluted tubule, the primary events are the reabsorption of nutrients and water. There are two regions in the nephron where large amounts of sodium and water are reabsorbed. In the loop of Henle, the countercurrent multiplier system is used, which leads to the reabsorption of salts (sodium and chloride) and water. In the distal convoluted tubule, the hormone aldosterone influences the reabsorption of salts and water followed by osmosis. The correct answer is B."
 },
 {
  "instruction":"\\section{\nIn October 2012, Felix Baumgartner attempted a world-record sky-diving jump from just over 128,000 ft (39,014 m) above the Earth's surface. He was fitted with a customized, pressurized spacesuit that brought his total weight to $260 \\mathrm{lbs}$ ( $1156 \\mathrm{~N}$ ). The spacesuit was designed to collect and transmit important monitoring, communications, and tracking information throughout the jump. Baumgartner then ascended to the jumping altitude housed within a pressurized capsule of weight 2900 lbs ( $12900 \\mathrm{~N})$ attached to a $30 \\times 10^{6}$ cubic feet $\\left(8.5 \\times 10^{8} \\mathrm{~L}\\right)$ helium-filled weather balloon. The weather balloon with the attached capsule ascended at a rate of $1000 \\mathrm{ft} \/ \\mathrm{min}(5.1 \\mathrm{~m} \/ \\mathrm{s})$, reaching the jumping altitude in approximately two hours.\nOnce the capsule reached the jumping altitude, the door to the capsule was opened and Baumgartner stepped out and fell forward, beginning his free-fall descent. He continued falling until he reached a maximum speed of $834 \\mathrm{mph}(373 \\mathrm{~m} \/ \\mathrm{s})$, which exceeded the speed of sound. This speed was Baumgartner's terminal velocity-the velocity attained in which he was no longer being accelerated downward due to gravity. Once he reached this speed, Baumgartner deployed his parachute and glided safely to Earth's surface.Assuming the density of helium is $0.1785 \\mathrm{~g} \/ \\mathrm{cm}^{3}$, the weight of the helium used to inflate the balloon was:\n\nOptions:\nA. $1.5 \\times 10^{8} \\mathrm{~N}$\nB. $1.5 \\times 10^{7} \\mathrm{~N}$\nC. $1.5 \\times 10^{6} \\mathrm{~N}$\nD. $1.5 \\times 10^{5} \\mathrm{~N}$",
  "response":"Using the relationship The correct answer is C."
 },
 {
  "instruction":"Genetically modified (GM) foods contain DNA from more than one source. The goal of creating GM foods is to modify the organisms to express new and useful traits. A few examples involve engineering crops to be resistant to bacterial or viral infections, to be resistant to herbicides, to ripen more slowly, to survive extreme temperature changes, to contain vaccines, or to create crops that are more nutritious.\nOne GM food that has received media attention over the years is golden rice. This GM product was first introduced in 2000 and was engineered to provide carotene that is a precursor to vitamin $\\mathrm{A}$. The carotene in the rice provides the golden color. Vitamin A deficiency is a major problem in underdeveloped parts of the world, killing more than one million people a year, most of whom are children. This vitamin is critical for the development and maintenance of epithelial tissues, for vision, and for growth.\nRice plants normally produce carotene in the inedible parts of the plant. To be able to access and use the carotene, it was necessary to engineer the rice to produce it in the edible grains. In order to achieve this goal, two genes from daffodil plants, psy and $l y c$, and a bacterial gene called $c r t 1$ were added to the rice genome. These genes engineered a pathway that produces lycopene, which is eventually converted to carotene. Each of these genes was placed under the control of a grain-specific promoter so that the products were produced in the rice grains.\nWhile golden rice was produced to provide carotene, which the human body can convert into the essential vitamin A, the rice provided much less carotene than was anticipated. Golden rice 2, a modified version released in 2005, contains a much larger amount of carotene and, as such, is a better source of the vitamin. Field tests were conducted in 2009, and human consumption trials were completed in 2012. However, neither version has been grown in large-scale quantities at this time.\nWhen preparing GM organisms, it can be easier to get a eukaryotic cell to express bacterial genes than it is to get a bacterial cell to express eukaryotic genes. What might be the MOST reasonable explanation for this?\n\nOptions:\nA. Bacteria are more simplistic organisms, and it is more difficult to get them to express complex eukaryotic genes.\nB. The ribosomes of bacteria are different from eukaryotic ribosomes, and they have difficulty translating eukaryotic mRNA.\nC. Eukaryotic genes contain introns that must be spliced out prior to translation in order to make a functional protein. Bacteria lack the ability to perform RNA splicing.\nD. Bacterial cells recognize eukaryotic DNA as foreign and use nucleases to destroy it before it can be expressed.\n",
  "response":"Bacterial ribosomes are in fact different from eukaryotic ribosomes. However, the genetic code is universal so there should be no problem translating the proteins. When eukaryotic DNA is introduced to bacterial cells, it is contained in a vector such as a plasmid that is a normal component of many bacterial cells. We know that eukaryotic genes contain introns and prokaryotic genes do not. If a eukaryotic gene is placed in a prokaryotic cell, the introns cannot be removed and the mRNA produced will contain introns. When translated, the intended protein will not be produced. The correct answer is C."
 },
 {
  "instruction":"Questions 18-20 are based on the following data.\nThe generalized events in DNA replication can be depicted by the following flowchart:\nDouble helix $\\rightarrow$ Nicking of a strand $\\rightarrow$ Unwinding $\\rightarrow$ Destabilization and relief of pressure $\\rightarrow$ Initiation $\\rightarrow$ Elongation $\\rightarrow$ Closing of nicks\nBetween which two steps would the enzyme topoisomerase be used?\n\nOptions:\nA. Double helix $\\rightarrow$ Nicking of a strand\nB. Nicking of a strand $\\rightarrow$ Unwinding\nC. Initiation $\\rightarrow$ Elongation\nD. Elongation $\\rightarrow$ Closing of nicks",
  "response":"Topoisomerases are enzymes that cut one of the DNA strands so that it can begin to unwind and serve to relieve pressure in the coil caused by its unwinding. Thus the enzyme acting between Double helix $\\rightarrow$ Nicking of a strand is topoisomerase. The correct answer is A."
 },
 {
  "instruction":"\\section{Passage $\\mathrm{V}$}\nA student performed qualitative tests on a series of unknown chemicals in order to identify them. The unknowns included $\\mathrm{NaCl}, \\mathrm{NaHCO}_{3}$, sugar $\\left(\\mathrm{C}_{12} \\mathrm{H}_{22} \\mathrm{O}_{11}\\right), \\mathrm{MgSO}_{4}$, $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$, cornstarch, and chalk $\\left(\\mathrm{CaCO}_{3}\\right)$. These are all household chemicals. $\\mathrm{MgSO}_{4}$ is Epsom salts, $\\mathrm{NaHCO}_{3}$ is baking soda, and $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ is photographic fixer. The student was asked to identify six unknowns: A, B, C, D, E, and F.\nThe qualitative tests were based on the following:\n$>$ Solubility in water: starch is insoluble, as well as one of the other compounds. The rest are soluble in water.\n$>$ Chalk produces a gas when treated with acid.\n$>$ Starch turns blue when treated with iodine.\n$>\\mathrm{MgSO}_{4}$ produces a milky precipitate when treated with aqueous ammonia.\n$>\\mathrm{NaHCO}_{3}$ turns pink when treated with phenolphthalein.\n$>\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$ decolorizes iodine solution.\n$>\\mathrm{NaCl}$ conducts electricity in solution.\n$>$ Sugar does not conduct electricity in solution.\nThe student prepared a flowchart that would aid in doing the experiments in a systematic manner and allow efficient identification of the unknowns.\n\\section{Experiment}\nThe experiments were conducted in the following order. The solubility of each chemical in water was tested first. From those results, the insoluble unknowns were tested next. The unknowns that were soluble in water were tested as aqueous solutions, rather than tested for conductivity.\n\\section{Results}\n\\begin{tabular}{ccccccl}\n\\hline Unknown & Solub & Conductivity & Acid & Phenolphthalein & $\\mathbf{N H}_{\\mathbf{3}}$ & $\\mathbf{I}_{\\mathbf{2}}$ \\\\\n\\hline A & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ \\\\\nB & $\\mathrm{N}$ & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ \\\\\nC & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ \\\\\n$\\mathrm{D}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{Y}_{\\mathrm{BLUE}}$ \\\\\nE & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ \\\\\nF & $\\mathrm{Y}$ & $\\mathrm{Y}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{N}$ & $\\mathrm{Y}_{\\text {COLORLESS }}$ \\\\\n\\hline\n\\end{tabular}\nFrom these results the student was able to identify all six unknowns.\nUnknown A was determined to be:\n\nOptions:\nA. $\\mathrm{CaCO}_{3}$\nB. $\\mathrm{NaCl}$\nC. $\\mathrm{Na}_{2} \\mathrm{~S}_{2} \\mathrm{O}_{3}$\nD. starch",
  "response":"Unknown A is soluble in water and has negative results for all of the tests except for conductivity, so it must be $\\mathrm{NaCl}$. The correct answer is B."
 },
 {
  "instruction":"Endothelin is a vasoactive peptide involved in a variety of processes, including the control of blood flow. Endothelin is capable of binding to two different receptors. There are receptors for endothelin located on smooth muscle (ETA) and receptors for endothelin located on vascular endothelium (ETB). When endothelin binds to ETA, it initiates vasoconstriction by narrowing blood vessels. However, when endothelin binds to ETB, vasodilation occurs due to the widening of blood vessels. The activity of the ETB receptor seems to counterregulate the ETA effects and prevents excessive vasoconstriction.\nDrug $X$ is a selective receptor antagonist for endothelin. It is designed to bind preferentially to certain endothelin receptors. Drug $\\mathrm{Y}$ is another selective receptor antagonist for endothelin. It is also designed to preferentially bind to one of the endothelin receptors. Researchers are hopeful that drugs $\\mathrm{X}$ and $\\mathrm{Y}$ will be of value in the treatment of conditions such as pulmonary arterial hypertension (PAH). In PAH, excess endothelin is produced, which causes an increase in blood pressure within the pulmonary arteries. Normal pressure in the pulmonary arteries is expected to be about $14 \\mathrm{~mm} \\mathrm{Hg}$, and pressure at or above $25 \\mathrm{~mm}$ Hg constitutes PAH. Both drugs have been tested in patients with PAH. There were five patients in each group. At the end of the study, the percent decrease in mean pulmonary artery pressure was measured for each patient and reported in the table.\n\\begin{tabular}{cccc}\n\\hline Patient & Drug X & DrugY & Placebo \\\\\n\\hline 1 & $+10 \\%$ & $-10 \\%$ & $+5 \\%$ \\\\\n2 & $+3 \\%$ & $-5 \\%$ & $0 \\%$ \\\\\n3 & $-5 \\%$ & $+2 \\%$ & $+13 \\%$ \\\\\n4 & $+2 \\%$ & $0 \\%$ & $-4 \\%$ \\\\\n5 & $0 \\%$ & $-7 \\%$ & $-2 \\%$ \\\\\nAverage decrease & $+2 \\%$ & $-4 \\%$ & $+2.4 \\%$ \\\\\n\\hline\n\\end{tabular}\nIf a patient has PAH, the heart can become overly stressed which can ultimately lead to heart failure. Which area of the heart is MOST likely to initially be affected by an increased blood pressure in the pulmonary arteries?\n\nOptions:\nA. the right and left atria\nB. the right and left ventricles\nC. the right ventricle\nD. the left ventricle\n",
  "response":"To answer this question, you must be familiar with the anatomy of the heart. The right side of the heart is the pulmonary circuit while the left side is the systemic circuit. Deoxygenated blood enters the right atrium of the heart via the superior and inferior vena cava. Blood from the right atrium passes through the tricuspid valve to the right ventricle and then leaves the heart through the pulmonary semilunar valve. Blood is carried via the pulmonary arteries to the lungs where gas exchange occurs. After gas exchange occurs, oxygenated blood returns to the heart via pulmonary veins to enter the left atrium. From there blood passes through the bicuspid valve into the left ventricle. Blood leaves the left side of the heart by passing through the aortic valve into the aorta. If there was a pressure increase in the pulmonary arteries, the right ventricle of the heart would be most likely affected as it is the last chamber that blood is located in before it moves toward the lungs in the pulmonary arteries. The correct answer is C."
 },
 {
  "instruction":"Endothelin is a vasoactive peptide involved in a variety of processes, including the control of blood flow. Endothelin is capable of binding to two different receptors. There are receptors for endothelin located on smooth muscle (ETA) and receptors for endothelin located on vascular endothelium (ETB). When endothelin binds to ETA, it initiates vasoconstriction by narrowing blood vessels. However, when endothelin binds to ETB, vasodilation occurs due to the widening of blood vessels. The activity of the ETB receptor seems to counterregulate the ETA effects and prevents excessive vasoconstriction.\nDrug $X$ is a selective receptor antagonist for endothelin. It is designed to bind preferentially to certain endothelin receptors. Drug $\\mathrm{Y}$ is another selective receptor antagonist for endothelin. It is also designed to preferentially bind to one of the endothelin receptors. Researchers are hopeful that drugs $\\mathrm{X}$ and $\\mathrm{Y}$ will be of value in the treatment of conditions such as pulmonary arterial hypertension (PAH). In PAH, excess endothelin is produced, which causes an increase in blood pressure within the pulmonary arteries. Normal pressure in the pulmonary arteries is expected to be about $14 \\mathrm{~mm} \\mathrm{Hg}$, and pressure at or above $25 \\mathrm{~mm}$ Hg constitutes PAH. Both drugs have been tested in patients with PAH. There were five patients in each group. At the end of the study, the percent decrease in mean pulmonary artery pressure was measured for each patient and reported in the table.\n\\begin{tabular}{cccc}\n\\hline Patient & Drug X & DrugY & Placebo \\\\\n\\hline 1 & $+10 \\%$ & $-10 \\%$ & $+5 \\%$ \\\\\n2 & $+3 \\%$ & $-5 \\%$ & $0 \\%$ \\\\\n3 & $-5 \\%$ & $+2 \\%$ & $+13 \\%$ \\\\\n4 & $+2 \\%$ & $0 \\%$ & $-4 \\%$ \\\\\n5 & $0 \\%$ & $-7 \\%$ & $-2 \\%$ \\\\\nAverage decrease & $+2 \\%$ & $-4 \\%$ & $+2.4 \\%$ \\\\\n\\hline\n\\end{tabular}\nIn addition to regulation of vessel diameter, ETA is known to have other roles in the body. When activated by a specific type of endothelin, ETA can synergize with growth factors to cause rapid cell proliferation. Based on this information, it is MOST likely that ETA can be involved with:\n\nOptions:\nA. an increase in metabolism\nB. the development of cancer\nC. immune system hypersensitivities\nD. increased stress to the kidneys and liver",
  "response":"The question suggests that ETA can be activated by certain types of endothelin, which causes ETA to synergize with growth factors to increase cell proliferation rates. Growth factors are one of many mechanisms used to regulate the cell cycle. Of the choices listed, A, C, and D can be eliminated as they have nothing to do with growth factors and cell proliferation. The only reasonable answer is choice B since one of the many characteristics of cancer is increased cell proliferation. The correct answer is B."
 }
]