BRS Biochemistry, Molecular Biology, Genetics - Questions

अब Quizwiz के साथ अपने होमवर्क और परीक्षाओं को एस करें!

Malonate is a competitive inhibitor of succinate dehydrogenase, a key enzyme in the Krebs tricarboxylic acid cycle. The presence of malonate will affect the kinetic parameters of succinate dehydrogenase in which one of the following ways? (A) Increases the apparent Km but does not affect Vmax. (B) Decreases the apparent Km but does not affect Vmax. (C) Decreases Vmax but does not affect the apparent Km. (D) Increases Vmax but does not affect the apparent Km. (E) Decreases both Vmax and Km.

A. A competitive inhibitor competes with the substrate for binding to the active site of the enzyme, in effect increasing the apparent Km (in the presence of inhibitor, it will require a higher concentration of substrate to reach ½ Vmax, as the substrate is competing with the inhibitor for binding to the active site). As the substrate concentration is increased, the substrate, by competing with the inhibitor, can overcome its inhibitory effects, and eventually the normal Vmax is reached. A noncompetitive inhibitor will decrease the Vmax without affecting the binding of substrate to the active site, so the Km is not altered under those conditions. An activator of an allosteric enzyme will decrease the apparent Km without affecting Vmax (less substrate is required to reach the maximal velocity).

A medical student has been studying for exams, and neglects to eat anything for 12 hours. At this point, the student opens a large bag of pretzels and eats every one of them in a short period. Which one of the following effects will this meal have on the student's metabolic state? (A) Liver glycogen stores will be replenished. (B) The rate of gluconeogenesis will be increased. (C) The rate at which fatty acids are converted to adipose triacylglycerols will be reduced. (D) Blood glucagon levels will increase. (E) Glucose will be oxidized to lactate by the brain and to CO2 and H2O by the red blood cells.

A. After a meal of carbohydrates (the major ingredient of pretzels), glycogen is stored in the liver and in muscle, and triacylglycerols are stored in adipose tissue. Owing to the rise in glucose level in the blood (from the carbohydrates in the pretzels), insulin is released from the pancreas and the level of glucagon in the blood decreases. Since blood glucose levels have increased, there is no longer a need for the liver to synthesize glucose, and gluconeogenesis decreases. The change in insulin-to-glucagon ratio also inhibits the breakdown of triacylglycerols and favors their synthesis. The brain oxidizes glucose to CO2 and H2O, whereas the red blood cells produce lactate from glucose, since red blood cells cannot carry out aerobic metabolism.

A 27-year-old male got lost while hiking in Yosemite National Park. He was found 8 days later. He had nothing to eat and only water to drink before being rescued. The man's brain would attempt to decrease consumption of glucose and increase consumption of ketones in order to protect the breakdown (catabolism) of which one of the following? (A) Muscle glycogen (B) Liver glycogen (C) Muscle protein (D) Red blood cells (to provide heme) (E) Adipose triacylglycerol

A. Albumin, though nonspecific, is considered the standard for assessing overall protein malnutrition. Albumin is made by the liver and is found in the blood. It acts as a nonspecific carrier of fatty acids and other hydrophobic molecules. When amino acid levels become limiting, the liver reduces its levels of protein synthesis, and a reduction in albumin levels in the circulation is an indication of liver dysfunction. Ferritin is an iron storage protein within tissues, and its circulating levels are low at all times. Creatinine is a degradation product of creatine phosphate (an energy storage molecule in muscle), and its presence in the circulation reflects the rate of creatinine clearance by the kidney. High levels of creatinine indicate a renal insufficiency. Creatine phosphokinase is a muscle enzyme that is released into circulation only when there is damage to the muscle. Blood urea nitrogen indicates the rate of amino acid metabolism to generate urea, but does not indicate protein malnutrition.

A patient seen in the ER has ingested antifreeze in a suicide attempt. Other than bicarbonate, which one of the following is the major buffer of acids to help maintain the pH in the blood within the range compatible with life? (A) Hemoglobin (B) Acetoacetate (C) β-Hydroxybutyrate (D) Phosphate (E) Collagen

A. Antifreeze contains ethylene glycol (HO-CH2-CH2-OH), and ingestion of ethylene glycol (which has a sweet taste) will lead to a metabolic acidosis due to the metabolism of ethylene glycol to glycolic and oxalic acids. As the acids form, protons are released, and bicarbonate and hemoglobin are the major buffers in the blood that will bind these protons to blunt the drop in blood pH. Acetoacetate and β-hydroxybutyrate are ketone bodies produced by the liver, and since they are both acids, their accumulation is often a cause of metabolic acidosis. Increasing their synthesis under acidotic conditions would only exacerbate the acidosis. Phosphate is an intracellular buffer, but its role is not as significant as that of either hemoglobin or bicarbonate.

An alcoholic presents with swelling and fissuring of the lips, cracking at the angles of the mouth, red eyes, and an oily, scaly rash of his scrotum. Which of the following foods would best help reverse the symptoms described in the above patient? (A) Broccoli (B) Carrots (C) Grapefruits (D) Wheat (E) Chocolate cake

A. Dark green vegetables, especially broccoli, meats, and dairy products are all high in riboflavin. Carrots are high in vitamin A, grapefruits in vitamin C, and whole grains in niacin. Chocolate cake is high in flavonoids, an antioxidant, fats, and carbohydrates

This tissue converts lactate from muscle to a fuel for other tissues

A. Exercising muscle produces lactate, which the liver can convert to glucose by gluconeogenesis. Blood glucose is oxidized by red blood cells and other tissues. Only the liver and kidney (to a small extent) can release free glucose into the circulation for use by other tissues

A 29-year-old female presents with chronic fatigue for the past 9 months. She did have mononucleosis in the past, and blood work reveals a chronic viral infection. An analysis of a liver biopsy indicated that when placed under conditions in which apoptosis should be initiated, the cells continued to grow. The viral infection was most likely caused by which one of the following? (A) Epstein-Barr virus (B) Influenza virus (C) Simian sarcoma virus (D) Polio virus (E) Herpes simplex virus

A. Infection by the Epstein-Barr virus will lead to the synthesis of a Bcl-2-like factor that antagonizes apoptosis, and allows the virus-infected cells to survive and continue producing more viruses. This factor is not present in the other viruses listed as potential answers.

A 39-year-old man is brought to the ER for a suspected suicide attempt. He has blurred vision; very dry, hot, red skin; dry mouth; urinary retention; confusion; hallucinations; loss of balance; and tachycardia. Emergency medical technicians (EMTs) found an empty bottle of amitriptyline in his apartment. The date on the bottle was just 1 week ago, yet all the pills were missing. The effects the man is experiencing is due to an inhibition of which of the following processes? (A) Muscarinic acetylcholine receptor signaling (B) Nicotinic acetylcholine receptor signaling (C) GABA signaling (D) Serotonin signaling (E) Catecholamine signaling

A. Many classes of drugs, including some antihistamines (e.g., Benadryl), some antipsychotics (e.g., olanzapine), tricyclic antidepressants (e.g., amitriptyline), and atropinelike drugs (e.g., atropine, scopolamine) have anticholinergic effects or side effects, and function as antagonists to the acetylcholine receptor. Muscarinic acetylcholine receptors act through G-protein activation, whereas nicotinic acetylcholine receptors act as an ion channel, allowing sodium to flow through the receptor once it has been activated. The drug overdose in this case is inhibiting the muscarinic receptors, which occur in the autonomic and central nervous symptoms. This is a very typical case of anticholinergic overdose, with the "classic" symptoms classified as "blind as a bat," "dry as a bone," "red as a beet," "mad as a hatter," and "hot as a hare." The overdose is not affecting the nicotinic acetylcholine receptors, or receptors for GABA, serotonin, or catecholamines.

An 18-year-old person with type 1 diabetes has not injected her insulin for 2 days. Her blood glucose is currently 600 mg/dL (normal values are 80 to 100 mg/dL). Which one of the following cells of her body can still utilize the blood glucose as an energy source? (A) Brain cells (B) Muscle cells (C) Adipose cells

A. Muscle and adipose cells require insulin to stimulate the transport of glucose into the cell, whereas the glucose transporters for the blood-brain barrier are always present, and are not responsive to insulin. Thus, the brain can always utilize the glucose in circulation, whereas muscle and adipose tissue are dependent on insulin for glucose transport into the tissue.

A bodybuilder has gained 50 pounds of muscle over the last 6 months, facilitated by both increased weight lifting and black-market pharmaceutical injection of one substance. He never experienced hypoglycemia during this time frame. How much energy is needed for the transport of the majority of the illegal substance that the bodybuilder is using for the drug to enter cells? (A) No energy is required. (B) One ATP molecule is used for each molecule of the substance transported. (C) Only a few ATP molecules are being used to open and close the channel through which many substance molecules diffuse. (D) This is an example of cotransport, in which the energy generates a sodium gradient across the membrane, and it is difficult to calculate an exact energy amount. (E) The transporter has to be phosphorylated once to allow transport to occur for many solutes.

A. No energy is needed for simple diffusion, which is the case if this is a steroid hormone. The other answer choices require ATP, which would be necessary for an active transport process, whether it be activation of a channel by phosphorylation, or generation of a gradient across the membrane for cotransport. Simple and facilitated diffusion do not require any energy sources for transport.

A bodybuilder has gained 50 pounds of muscle over the last 6 months, facilitated by both increased weight lifting and black-market pharmaceutical injection of one substance. He never experienced hypoglycemia during this time frame. The mechanism whereby the illegal substance is entering the muscle cells of the bodybuilder is most likely which one of the following? (A) Simple diffusion (B) Active transport (C) Endocytosis (D) Facilitative diffusion (E) Pinocytosis

A. The bodybuilder is injecting (most probably) testosterone, a steroid hormone, which aids in building muscle mass. Steroid hormones are lipid-soluble substances, and cross membranes by simple diffusion. The receptor for steroid hormones is present inside the cell (either the cytoplasm or nucleus), and once the steroid hormone enters the cell, it will bind to the receptor in a saturable manner. Once the concentration of the hormone inside and outside the cells is equal, transport will stop. Active transport refers to using energy to concentrate a solute against its concentration gradient, which is not the case for steroid hormone transport across the membrane. Facilitative diffusion requires a membrane-bound carrier (no energy), but as indicated previously, the carrier (receptor) for these hormones is intracellular. Steroid hormones do not enter cells through either endocytosis or pinocytosis. The fact that the bodybuilder never became hypoglycemic after taking the drug suggests that it was not insulin being injected.

A 12-year-old boy is displaying tiredness and lethargy, and is found to have a hypochromic, microcytic anemia. Microscopic examination of the boy's red blood cells demonstrated a spherical shape, rather than concave. The mutation in this child is most often found in a protein located in which part of the red blood cell? (A) The cytoskeleton (B) The nucleus (C) The mitochondria (D) The endoplasmic reticulum (E) The plasma membrane

A. The boy has hereditary spherocytosis, which is due to a mutation in a red blood cell cytoskeletal protein. The most common mutation is in spectrin, although mutations in ankyrin, band 3, and protein 4.2 can also lead to this phenotype. Owing to the mutation in the cytoskeletal protein, the membrane shape becomes spherical instead of concave. This leads to the removal of the spherical cells by the spleen, leading to both anemia and splenomegaly. Mutations in the proteins in the plasma membrane or the endoplasmic reticulum will not lead to this disorder. Red blood cells do not have a nucleus or mitochondria.

During starvation, this tissue uses amino acids to maintain blood glucose levels

A. The liver converts amino acids to blood glucose by gluconeogenesis. The other substrates for gluconeogenesis are lactate from the metabolism of glucose within the red blood cells and glycerol from the breakdown of triacylglycerol to free fatty acids and glycerol. Neither the brain, nor the skeletal muscle, nor the red blood cell can export glucose into the circulation.

A family, while on a picnic, picked some wild mushrooms to add to their picnic salad. Shortly thereafter, all the members of the family became ill, with the youngest child showing the most severe symptoms. The family is suffering these effects owing to a primary inability to accomplish which one of the following in their cells and tissues? (A) Synthesize proteins (B) Synthesize lipids (C) Synthesize DNA (D) Synthesize carbohydrates (E) Repair damage in DNA

A. The poison in poisonous mushrooms is α-amanitin, an inhibitor of eukaryotic RNA polymerases, primarily RNA polymerase II. As the family ate the mushrooms containing the poison, RNA polymerase II stopped functioning, and mRNA was no longer produced. This led to a lack of protein synthesis. There is no direct effect on the synthesis of lipids, carbohydrate, or DNA, other than replacing the required enzymes due to protein turnover. However, the net effect of α-amanitin poisoning would be to stop protein synthesis, which may then lead to a cessation of lipid or DNA synthesis. α-Amanitin has no direct effect on DNA repair.

A 42-year-old woman has slowly developed an inability to keep her eyes open at the end of the day. The eyelids droop, despite her best efforts to keep them open. This does not occur first thing in the morning. Further examination shows a generalized muscle weakness as the day progresses. In addition to the answer to the previous question, a drug that may help to stabilize this condition would do which one of the following? (A) Stimulate apoptosis (B) Inhibit apoptosis (C) Stimulate cell growth (D) Inhibit cell growth (E) Induce muscle growth (F) Inhibit muscle growth

A. The woman has myasthenia gravis, which is due to an autoimmune disorder in which antibodies directed against the acetylcholine receptor block the ability of acetylcholine to stimulate the muscle cells at the neuromuscular junction. Immunosuppressants can be taken to reduce the autoantibody production. Such drugs work, in part, through the activation of the tumor necrosis factor receptor, which activates apoptosis in the cells, leading to their destruction. Inhibiting apoptosis would exacerbate the problem, as the antibody-producing cells would survive longer and continue to produce the antibodies directed against the acetylcholine receptor. Drugs affecting the muscle would not help with this disorder, as it is a problem unique to the acetylcholine receptor expressed on the muscle surface. Stimulation or inhibition of cell growth does not stop the antibody-producing cells from continuing to make antibodies, and would not be an effective drug target for this disease.

A 62-year-old male has a reddish, rough patch with white scales on the top of his ear. He does not get this treated, and 3 years later it has become an enlarged, raised lesion with a central ulcerated area that will not heal. Of the following, which is the most likely causative factor for this malignancy? (A) Creation of pyrimidine dimers (B) Creation of hydroxyl radicals (C) Oncogenic RNA virus (D) TNF receptor mutation (E) Double-strand breaks in the DNA

A. This man originally displayed an actinic keratosis that has, over the intervening 3-year period, become a squamous cell carcinoma. Actinic keratosis develops in areas of the skin that are frequently exposed to sunlight, such as the top of the ear. The frequent exposure to UV light led to the creation of pyrimidine dimers in the DNA. If the cells cannot repair the DNA damage rapidly enough, cancerous changes do occur over time. The presentation of actinic keratosis may represent a precancerous state toward squamous cell carcinoma. Removal of the actinic keratosis would have prevented the development of the tumor. Hydroxyl radicals are created by ionizing radiation such as X-rays, not by sunlight. Oncogenic RNA viruses such as HTLV-1 could cause lymphomas or leukemias, but have not been implicated in squamous cell carcinoma. TNF receptor mutations can occur in immune system cells, leading to apoptosis, which would lead to an immune defect, but not the cancer observed. UV light does not lead to the creation of double-strand breaks in DNA.

A 43-year-old female has been on a "grapefruit and potatoes" diet for several months in an effort to lose weight. She now complains of a rash covering most of her body, a large, beefy tongue, nausea and diarrhea, and some confusion Which one of the following cofactors or enzyme complexes would be most affected by this condition? (A) The concentration of NAD+ (B) The concentration of FAD (C) The concentration of coenzyme Q (D) The functioning of the FMN components of complex I (E) The functioning of the cytochrome-containing components of complex III

A. This patient has the classic symptoms of pellegra, a vitamin B3 (niacin) deficiency. NAD+ is derived from niacin. Pellagra leads to the four Ds-dermatitis, dementia, diarrhea, and death. Riboflavin is the precursor for both FAD and FMN. Coenzyme Q is synthesized from acetyl-CoA, and its levels would not be affected as much as those of NAD+. Heme is synthesized from succinyl-CoA and glycine, and a reduction in heme levels would lead to an anemia and not the symptoms as described for this patient.

A 40-year-old male is well controlled on warfarin for a factor V leiden deficiency and recurrent deep vein thrombosis. He presents today with a community-acquired pneumonia, and is placed on erythromycin. Three days later, he develops bleeding and his INR is 8.0 (indicating an increased time for blood clotting to occur, where INR is international normalized ratio). Which of the following best explains why this bleeding occurred? (A) The erythromycin inhibited cytochrome P450 (B) The erythromycin stimulated cytochrome P450 (C) The causative agent of the pneumonia inhibited vitamin K utilization (D) The causative agent of the pneumonia stimulated vitamin K utilization (E) The erythromycin inhibited mitochondrial translation (F) The erythromycin inhibited mitochondrial transcription

A. Warfarin is metabolized by a specific subset of induced p450 isozymes. The p450 system is used by cells to modify the xenobiotic (in this case the warfarin) such that it can be more easily excreted. Erythromycin, along with other macrolide antibiotics, inhibits the p450 oxidizing system, which in this case would lead to a higher blood level of warfarin and, therefore, the balance of clotting and bleeding is shifted toward excessive bleeding. A stimulation of p450 production by erythromycin would lead to a lower level of warfarin (due to increased metabolism and loss of warfarin by p450) and the potential of excessive clotting. This effect of p450 is a common drug-drug interaction. The causative agents of communityacquired pneumonia do not affect vitamin K absorption in the small intestine, or distribution throughout the body. Erythromycin does not affect mitochondrial transcription, although it may affect mitochondrial translation. Inhibition of mitochondrial protein synthesis, however, will not alter the inhibition of cytochrome p450 activity, and the increased levels of warfarin present, which may lead to increased bleeding.

The largest amount of stored energy in the body

Adipose triacylglycerols contain the largest amount of stored energy in humans, followed by protein (even though loss of too much protein will lead to death), muscle glycogen, and liver glycogen (see Table 1.1).

A young woman (5' 3" tall, 1.6 m) who has a sedentary job and does not exercise consulteda physician about her weight, which was 110 lb(50 kg). A dietary history indicates that she eatsapproximately 100 g of carbohydrate, 20 g of protein, and 40 g of fat daily. According to the woman's BMI, into what classification does her weight and height place her? (A) Underweight (B) Normal range (C) Overweight (preobese) (D) Class I obese range (E) Class II obese range

B. According to Table 1.2, a BMI of 19.5 places the woman at the lower end of the normal range. Underweight is indicated by a BMI of ,18.5; preobesity occurs above a BMI of 25, but ,30. Class I obesity is indicated by a BMI between 30 and 35, and class II obesity by a BMI between 35 and 40.

A patient is going skiing high in the Rockies, and is given acetazolamide to protect against altitude sickness. Unfortunately, the patient is also a Type 1 diabetic. He is admitted to the hospital in a worsening ketoacidosis. In which of the following cells has acetazolamide inhibited a reaction that has led to the severity of the metabolic acidosis? (A) White blood cells (B) Red blood cells (C) Lens of the eye (D) Hepatocyte (E) Muscle

B. Acetazolamide is a carbonic anhydrase inhibitor, which is found primarily in red blood cells. The red blood cells contain carbonic anhydrase that catalyzes the reaction that forms carbonic acid from CO2 and H2O. Under high-altitude conditions, the inhibition of carbonic anhydrase will lead to a decrease in blood pH, which stabilizes the deoxygenated form of hemoglobin. This is due to an increased loss of bicarbonate in the urine by the inhibition of carbonic anhydrase within the kidney. The change in pH increases oxygen delivery to the tissues, and can overcome, in part, the symptoms of altitude sickness. However, in the case of the person with Type I diabetes who begins to produce ketone bodies, the body's main compensatory mechanism to overcome the acidosis is blocked. As ketone bodies are formed and protons generated, the H+ will react with bicarbonate to form carbonic acid. Carbonic anhydrase, which catalyzes a reversible reaction, will then convert the carbonic acid to CO2 and H2O, with the CO2 being exhaled. These reactions soak up excess protons, and help to buffer against the acidosis. If, however, carbonic anhydrase has been inhibited by acetazolamide, then the bicarbonate cannot buffer the blood pH and the acidosis could become more severe. White blood cells, muscle cells, liver cells, and the lens of the eye do not contribute to the buffering of the blood, and inhibition of carbonic anhydrase in those cells would not affect the ability to overcome an acidosis.

Consider a normal 25-year-old man, about 70 kg in weight, who has been shipwrecked on a desert island, with no food available, but plenty of freshwater. Which of the following fuel stores is least likely to provide significant calories to the man? (A) Adipose triacylglycerol (B) Liver glycogen (C) Muscle glycogen (D) Muscle protein (E) Adipose triacylglycerol and liver glycogen

B. As indicated in Table 1.1, in the average (70 kg) man, adipose tissue contains 15 kg of fat or 135,000 calories (kcal). Liver glycogen contains about 0.08 kg of carbohydrate (320 calories), and muscle glycogen contains about 0.15 kg of carbohydrate (600 calories). In addition, about 6 kg of muscle protein (24,000 calories) can be used as fuel. Therefore, liver glycogen contains the fewest available calories.

A pathologist, while doing an autopsy of a patient who died from Creuzfeldt-Jakob syndrome, accidentally cut himself while examining the brain. The pathologist became very concerned for his well-being, due primarily to the possibility of which one of the following materials entering his circulation? (A) A virus (B) A protein (C) A lipid (D) A bacteria (E) A polynucleotide

B. Creuzfeldt-Jakob syndrome is a prion disorder, and the infectious agent is a protein. The altered protein forms precipitates in the brain, and shifts the equilibrium of the normal protein to that which will aggregate with the altered protein. The pathologist is concerned that the infectious protein will migrate to his brain and seed the process of aggregation with the normal prion proteins in his brain. Prion disorders are not transmitted by viruses, lipids, bacteria, or any form of nucleic acid.

A 32-year-old female has developed breast cancer. Her mother and one maternal aunt had breast cancer and her maternal grandmother had ovarian cancer. Which of the following best describes the mechanism behind this inherited problem? (A) A tumor suppressor leading to loss of apoptosis (B) A tumor suppressor leading to an inability to repair DNA (C) A tumor suppressor leading to a constitutively active MAP kinase pathway (D) An oncogene leading to loss of apoptosis (E) An oncogene leading to an inability to repair DNA (F) An oncogene leading to a constitutively active MAP kinase pathway

B. Hereditary breast cancer is due to inherited mutations in either of the tumor suppressor genes BRCA1 or BRCA2. These genes encode proteins that play important roles in DNA repair (primarily single- and double-strand breaks), and it is the loss of this function that predisposes the patient to breast and ovarian cancers. The inability to repair the breaks in the backbone leads to errors during replication, and mutations will develop that eventually lead to a loss of growth control. This is a loss-of-function disorder, which characterizes the genes involved as tumor suppressors. Inheriting one mutated copy of BRCA1 means that the other, normal copy of BRCA1 must be lost in a particular cell in order to initiate the disease (loss of heterozygosity). For breast cancer, this occurs 85% of the time (penetrance upon inheriting a BRCA1 or BRCA2 mutation). An oncogene is a dominant gene, so only one mutated copy can bring about the disease. BRCA1 or BRCA2 mutations do not directly lead to a loss of apoptosis, or to a constitutively active MAP kinase pathway.

An individual is visiting Mexico City, which is at an altitude of 7,350 feet. The person is having trouble breathing due to difficulty in getting sufficient oxygen to the tissues. Which one of the following treatments might the person try to get hemoglobin to release oxygen more readily? (A) Take a drug that initiates a metabolic alkalosis. (B) Take a drug that increases the production of BPG. (C) Hyperventilate, which will lead to decreased levels of carbon dioxide in the blood. (D) Take a drug that induces the synthesis of the γ subunits of hemoglobin. (E) Take a drug that induces the synthesis of the β subunits of hemoglobin.

B. In order for hemoglobin to release oxygen more readily, the deoxygenated state of hemoglobin needs to be stabilized. This can occur by decreasing the pH (the Bohr effect), increasing the CO2 concentration, or increasing the concentration of BPG. Fetal hemoglobin (HbF = α2γ2) has a greater affinity for O2 than does HbA (α2β2), so inducing the synthesis of the γ genes would have the opposite of the intended effect. Inducing the concentration of the β chains would not decrease oxygen binding to hemoglobin (in fact, if there is not a concurrent increase in α-gene synthesis, this may be quite detrimental to the individual, as an imbalance in the synthesis of the hemoglobin chains leads to a disorder known as thalassemia, and overall oxygen transport to the tissues would be decreased). Increased BPG would cause O2 to be more readily released. A metabolic alkalosis would raise the pH of the blood, which would stabilize the oxygenated form of hemoglobin. Reducing carbon dioxide levels in the blood through hyperventilation will also stabilize the oxygenated form of hemoglobin, and make it more difficult to deliver oxygen to the tissues.

A person with Type 1 diabetes ran out of her prescription insulin and has not been able to inject insulin for the past 3 days. An overproduction of which of the following could cause a metabolic acidosis? (A) Hemoglobin (B) Ketone bodies (C) HCl (D) Bicarbonate

B. Ketone bodies are weak acids. In diabetic ketoacidosis, the liver produces ketone bodies, which will reduce the brain's dependency on glucose as its sole energy source. This is due to the lack of insulin, and the liver switching to starvation mode owing to the constant signaling by glucagon. Hemoglobin in the red blood cells and bicarbonate, both in the red blood cells and the plasma, are two of the body's major buffers, and their overproduction would not lead to an acidosis. HCl overproduction within the stomach might lead to duodenal ulcers or gastroesophageal reflux, but not to an overall metabolic acidosis, as the protons do not find their way into the circulation. A loss of chloride, if severe enough, could produce a metabolic alkalosis, but not an acidosis.

An experimental drug has been added to a eukaryotic cell, and while the drug was designed to interfere with a membrane transport process, the investigators found that in cells treated with the drug the lysosomes quickly turn into inclusion bodies. None of the material directed to the lysosomes for removal was being digested in the lysosome, and remained intact inside the organelle. An analysis of lysosomal contents in drug-treated cells indicated that the full complement of lysosomal enzymes were present in the organelle. Assuming that the drug is targeting just one protein, which one of the following proteins is most likely the target? (A) An outer membrane protein that allows the lysosomal membrane to become permeable to small molecules (B) A proton-translocating ATPase in the lysosomal membrane (C) A chloride pump in the lysosomal membrane (D) The enzyme that adds mannose- 6-phosphate to lysosomal enzymes (E) The mannose-6-phosphate receptor

B. Lysosomes contain a single membrane (so there is no outer membrane, such as in mitochondria) that contains a proton-translocating ATPase. The ATPase will concentrate protons inside of the lysosome, at the expense of ATP hydrolysis, to acidify the intraorganelle pH such that the lysosomal enzymes will be active. If the intravesicular pH cannot be lowered, the digestive enzymes will be inactive, and no digestion will take place. Targeting a chloride pump in the lysosomal membrane will not affect the activity of the lysosomal enzymes. If the lysosomal enzymes were not marked with a mannose-6-phosphate residue in the Golgi apparatus, they would not be able to bind to the mannose-6-phosphate receptor to be targeted to the lysosomes. If the drug altered either of those proteins (the enzyme responsible for adding the mannose-6-phosphate or the mannose-6-phosphate receptor), then the lysosomal enzymes would not be in the lysosomes, which is not the case. Such a drug would bring about the symptoms of I-cell disease.

A 45-year-old woman has been admitted to a substance abuse center for her alcoholism. As a first attempt to curb the patient's drinking, she is given a drug that will lead to an elevation of which one of the following metabolites if she drinks alcohol? (A) Acetic acid (B) Acetaldehyde (C) Ethanol (D) Carbon dioxide (E) Carbon monoxide

B. One treatment for chronic alcoholism is to inhibit the enzyme aldehyde dehydrogenase, which would lead to the accumulation of acetaldehyde if ethanol has been imbibed. Ethanol metabolism, at the first step, converts ethanol to acetaldehyde (the enzyme is alcohol dehydrogenase). Aldehyde dehydrogenase then converts the acetaldehyde to acetic acid, which is eventually converted to acetyl-CoA. The accumulation of acetaldehyde is what initiates the symptoms associated with a hangover, such as headache and nausea. The theory behind the treatment is that if the individual drinks alcohol while on the drug, the buildup of acetaldehyde will make the person feel very uncomfortable, and will lead to a reduction, or cessation, ofdrinking alcohol. Inhibiting aldehyde dehydrogenase will not lead to elevations of acetic acid, or ethanol, carbon dioxide, or carbon monoxide.

A patient is brought to the emergency room after being found by search and rescue teams. He was mountain climbing, got caught in a sudden snowstorm, and had to survive in a cave. He had no food for 6 days. In adapting to these conditions, which metabolic process has increased rather than decreased? (A) The brain's use of glucose (B) Muscle's use of ketone bodies (C) The red blood cells' use of glucose (D) The brain's use of ketone bodies (E) The red blood cells' use of ketone bodies (F) Muscle's use of glucose

B. Red blood cells lack mitochondria, so they can use only glucose for fuel (fatty acids and ketone bodies require mitochondrial proteins for their oxidative pathways). The brain can also use ketone bodies, along with glucose. The liver can use glucose, fatty acids, and amino acids as energy sources. The heart can use glucose, fatty acids, amino acids, and lactic acid as potential energy sources.

A 50-year-old male with a "pot belly" and a strong family history of heart attacks is going to his physician for advice on how to lose weight. He weighs 220 lb (100 kg) and is about 6' tall (1.85 m). His lifestyle can be best described as sedentary. What is this patient's BMI? (A) 24 (B) 29 (C) 31 (D) 36 (E) 40

B. The BMI is equal to kg/m2, which in this case is equal to 100/1.852, which is about 29.

After a fast of a few days, ketone bodies become an important fuel

B. The brain begins to use ketone bodies when levels start to rise after 3 to 5 days of fasting. Normally, the brain will use only glucose as a fuel (most fatty acids cannot cross the blood-brain barrier to be metabolized by the brain), but when ketone bodies are elevated in the blood, they can enter the brain and be used for energy.

On a routine newborn exam, it is noted that the red reflex is absent in one eye. An MRI shows a tumor blocking the retina. Regulation at which phase of the cell cycle would be affected by the mutation that leads to this tumor? (A) G0 to G1 (B) G1 to S (C) S to G2 (D) G2 to M (E) M to G1 (F) G1 to G0

B. The child has hereditary retinoblastoma, which is due to an inherited mutation in the rb gene. As the rb gene is a tumor suppressor gene, once loss of heterozygosity occurs, the function of rb in the cell cycle is lost. Rb helps to regulate the E2F family of transcription factors. Once cyclin D is synthesized, and activates a pair of CDKs, rb protein is phosphorylated, which causes it to leave a complex with the E2F factors. The removal of rb from the protein complex activates the E2F proteins, which initiate new gene transcription to allow the cell to transition to the S phase of the cell cycle. In the absence of any functional rb gene product, the transition to S phase is unregulated, and occurs continuously, leading to tumor growth. The rb gene product is not required for any other checkpoints in the cell cycle.

An 18-year-old college freshman shares a dorm room with three roommates. One of his roommates has been diagnosed with meningococcal meningitis, caused by the bacteria Neisseria meningitidis. The other three roommates are isolated, and treated twice a day with an antibiotic as prophylaxis against this organism, as none of them had received the meningococcal vaccine prior to enrollment. They are told that this antibiotic can give a reddish discoloration of their urine or tears. The reason this drug is effective in killing the bacteria is which one of the following? (A) DNA synthesis is inhibited. (B) RNA synthesis is inhibited. (C) The process of protein synthesis is inhibited. (D) The bacterial membrane becomes leaky. (E) ATP generation is reduced.

B. The drug given to prevent Neisseria infection (used prophylaxically), which is common in crowded conditions such as freshman dormitory rooms or military barracks, is rifampin. Rifampin inhibits RNA polymerase, and also exhibits a red color. Loss of rifampin in the urine or tears would give a reddish tint to those fluids. Rifampin does not interfere with DNA synthesis, the bacterial membrane, the process of protein synthesis, or ATP generation by the bacteria.

A family has been using an additional propane heater in their enclosed apartment during the winter months. One morning, a family member is difficult to awake, and when awake, complains of a splitting headache and being very tired. His mucous membranes are also a cherry red color. These symptoms are the result of which one of the following? (A) Increased oxygen delivery to the tissues (B) Decreased oxygen delivery to the tissues (C) Increased blood flow to the brain (D) Decreased blood flow to the brain (E) Decreased oxygen affinity to hemoglobin

B. The family member is exhibiting the symptoms of carbon monoxide (CO) poisoning. CO will bind to hemoglobin, with a higher affinity than oxygen, and decrease oxygen delivery to the tissues. In addition to competing with oxygen for binding to hemoglobin, CO, once bound to hemoglobin, shifts the oxygen-binding curve to the left, stabilizing the "R" state, or oxygenated state, which makes it more difficult for oxygen to be released from hemoglobin in the tissues. Thus, in the presence of CO, oxygen affinity for hemoglobin is actually increased. CO poisoning does not affect the blood flow to the brain.

A scientist is studying a novel hepatocyte cell line that cannot produce a nucleolus when the cells are grown at 42°C. When examining cells that have been at 42°C for 96 hours, the scientist finds that the incorporation of 14C-leucine into proteins is greatly reduced as compared to cells grown at 35°C. This is most likely due to which one of the following at the nonpermissive temperature? (A) Lack of charged tRNA molecules (B) Inability to form peptide bonds during protein synthesis (C) Lack of initiation factors (D) Inability to form mature mRNA (E) Lack of GTP needed for protein synthesis

B. The nucleolus is the site within the nucleus at which rRNA is produced, and ribosomal subunits assembled. In the absence of nucleoli, mature ribosome content within the cell will decrease, which will lead to an overall reduction in protein synthesis. One of the functions of the mature ribosomal complex is to catalyze the formation of peptide bonds, using the enzymatic activity within the large ribosomal subunit rRNA (23S in prokaryotes, and 28S in eukaryotes). In nucleoli, rRNA genes are transcribed to produce the 45S rRNA precursor, which is trimmed, modified, and complexed with proteins to form ribosomal subunits. The synthesis of tRNA does not require the nucleolus, and charging reactions occur in the cytoplasm, so it is unlikely that the levels of charged tRNAs will be reduced. The capping, splicing, and polyadenylation of mRNA does not require the nucleolus (or ribosomes), and would proceed normally at the nonpermissive temperature. The only reason for initiation factors to be reduced is if their turnover number is high, and new proteins need to be synthesized to replenish the eIF pool. The lack of nucleoli, for 96 hours, should not affect the ability of the cell to produce energy in the form of ATP and GTP.

A 45-year-old man presents with blood in his stool. Workup reveals a stage 3 (Dukes 3) colon carcinoma, with multiple polyps within the colon. A family history reveals that his father and grandfather both had colon cancer in their fifth decade of life. A potential initiating activating event in the development of this tumor is which one of the following? (A) Loss of β-catenin activity (B) Activation of β-catenin activity (C) Loss of transcription factor myc activity (D) Activation of Bcl-2 activity (E) Loss of cyclin expression (F) Gain-of-cyclin expression

B. The patient has hereditary colon cancer, specifically adenomatous polyposis coli, which presents in the fourth or fifth decade of life, with multiple polyps lining the lumen of the colon. The defective protein is APC, which regulates β-catenin activity. The loss of APC function leads to inappropriately activated β-catenin, which is a transcription factor and can stimulate the expression of myc and cyclin D1, to promote cell growth. The inappropriate expression of myc and cyclin D1 due to the loss of APC is an initiating event in tumorigenesis. The APC mutation does not affect BCl-2 activity, which is an antiapoptotic activity. While the loss of APC activity will lead to a gain-of-cyclin expression, that gain is due to the activation of β-catenin, which would be the initiating event for tumor formation.

A 23-year-old male presents to the ER with a fracture of his humerus, sustained in what appeared to be a minor fall. He has a history of multiple fractures after a seemingly minor trauma. He also has "sky blue" sclera and an aortic regurg murmur. His underlying problem is most likely due to a mutation in which one of the following proteins? (A) Fibrillin (B) Type 1 collagen (C) Type IV collagen (D) α1-Antitrypsin (E) β-Myosin heavy chain

B. The patient is exhibiting the signs of osteogenesis imperfecta, brittle bones, as exemplified by various mutations in type 1 collagen, the building blocks of the bones. The aortic regurgitation murmur is also due to a lack of type 1 collagen in the extracellular matrix of the aorta. Mutations in fibrillin give rise to Marfan syndrome, which does exhibit long bones, but not brittle or easily broken bones. Marfan syndrome would also be associated with lens dislocation, which is not occurring in this patient. Mutations in type IV collagen would lead to Alport syndrome, not brittle bones, and there is no mention of kidney/urine problems with the patient. A defect in α1-antitrypsin would lead to emphysema (not brittle bones), and a mutation in β-myosin heavy chain would lead to hypertrophic cardiomyopathy, not brittle bones.

A 25-year-old female presents with intense fear whenever she has to drive her car through a tunnel. She feels faint, sweats profusely, has palpitations, and hyperventilates. She is prescribed diazepam to reduce her symptoms. The type of chemical messenger enhanced by this treatment is best described as which one of the following? (A) Neuropeptide (B) Biogenic amine (C) Large-molecule neurotransmitter (D) Cytokine (E) G-protein

B. The patient is experiencing an anxiety disorder and panic attacks. These symptoms are often treated with psychotherapy and benzodiazepams. Patients with anxiety disorders have low gamma-aminobutyric acid (GABA). Benzodiazepams, such as diazepam, increase the efficiency of the synaptic transmission of GABA, helping to make any existing GABA more efficacious. This is through the drug binding to GABA receptors such that when GABA binds to the receptor, the response to GABA is greater than in the absence of the drug (one effect is to leave chloride channels open for greater periods of time in response to GABA, thereby depolarizing the membrane and sending an inhibitory signal). GABA is the chief inhibitory neurotransmitter. GABA is a biogenic amine or "small-molecule" neurotransmitter, and is derived from the decarboxylation of glutamate. Neuropeptides are the other type of chemical messenger secreted by the nervous system, and are usually small peptides. Cytokines are small protein messengers of the immune system. G-proteins aid in transmitting the signals induced by proteins that bind to heptahelical receptors (such as the epinephrine or glucagon receptors). GABA does not transmit its signal through a G-protein.

A 23-year-old female patient presents to the ER with a feeling of being unable to catch her breath, light-headedness, and "tingling" of her fingers, toes, and around her mouth. This happens whenever she drives through a tunnel, and that is what set off this episode. Which of the following arterial blood pHs would be most consistent with her diagnosis? (A) 8.10 (B) 7.55 (C) 7.15 (D) 6.40 (E) 6.10

B. The patient is having a panic attack (due to driving in tunnels) and is hyperventilating, causing an acute respiratory alkalosis. The loss of CO2 pushes the carbonic anhydrase reaction in the direction of CO2 production, which reduces the proton concentration (and thereby raising the pH). The patient would lose consciousness with a more severe attack. A respiratory alkalosis is usually mild as compared to a metabolic alkalosis. For this reason, the pH increase is smaller (7.55 is more likely than a pH of 8.10, which could occur via a metabolic alkalosis). The other choices given, 7.15, 6.40, and 6.10, are all lower than physiologic pH (7.4), and would be considered acidosis, instead of alkalosis. An example of a metabolic alkalosis is hypokalemia, a reduction in normal potassium values. Owing to low serum potassium levels, potassium leaves the cells and is replaced by protons from the circulation. The loss of protons from the blood leads to the alkalosis.

A 50-year-old male with a "pot belly" and a strong family history of heart attacks is going to his physician for advice on how to lose weight. He weighs 220 lb (100 kg) and is about 6' tall (1.85 m). His lifestyle can be best described as sedentary. For which of the following disease processes is this patient at higher risk? (A) Diabetes mellitus, type 1 (B) Insulin resistance syndrome (C) Gaucher disease (D) Low blood pressure (E) Sickle cell disease

B. The patient's weight, age, and activity put him at higher risk for insulin resistance syndrome. The entire syndrome includes hypertension, diabetes mellitus (type 2), decreased high-density lipoprotein levels, increased triglyceride levels, increased urate, increased levels of plasminogen activator inhibitor 1, nonalcoholic fatty liver, central obesity, and polycystic ovary syndrome (PCOS) (in females). Insulin resistance syndrome leads to early atherosclerosis throughout the entire body. The patient is not at increased risk for diabetes mellitus, type 1, as that is the result of an autoimmune condition that destroys the β cells of the pancreas such that insulin can no longer be produced. The lifestyle exhibited by the patient has not been linked to autoimmune disorders. Gaucher disease is a disorder of the enzyme β-glucocerebrosidase, and is an autosomal recessive disorder. Since this disease is an inherited disorder, the patient's lifestyle does not increase his risk of having this disease. The patient's increasing weight might lead to increased blood pressure, but not to reduced blood pressure. Sickle cell disease is another autosomal recessive disorder leading to an altered β-globin gene product, and like Gaucher disease, it is an inherited disorder that is not altered by the patient's lifestyle.

Which one of the following is the amino acid in hemoglobin that accepts H+ and allows hemoglobin to act as a buffer to acids? (A) Alanine (B) Histidine (C) Serine (D) Threonine (E) Aspartate

B. The side chain of histidine has a pKa of 6.0, which, of all amino acid side chains, is the one closest to physiologic pH. The local environment of the protein can raise this pKa value closer to 7 such that the histidine side chains within hemoglobin will be the major groups that accept and donate protons when hemoglobin acts as a buffer. The alanine side chain (a methyl group) cannot accept or donate protons. The pKa for the side chains of serine or threonine are above 10.0, so at physiologic pH these side chains are always protonated, and cannot act as a binding site for excess protons generated during an acidotic event. The pKa for the side chain of aspartate is about 4.0, so at physiologic pH that group is always deprotonated, and will not accept protons generated during an acidotic event.

A young woman (5' 3" tall, 1.6 m) who has a sedentary job and does not exercise consulteda physician about her weight, which was 110 lb(50 kg). A dietary history indicates that she eatsapproximately 100 g of carbohydrate, 20 g of protein, and 40 g of fat daily. What is the woman's approximate DEE in calories (kilocalories) per day at this weight? (A) 1,200 (B) 1,560 (C) 1,800 (D) 2,640 (E) 3,432

B. This woman's DEE is 1,560 calories (kcal). DEE equals BMR plus physical activity. Her weight is 110 lb/2.2 5 50 kg. Her BMR (about 24 kcal/kg) is 50 kg 3 24 5 1,200 kcal/day. She is sedentary and needs only 360 additional kcal (30% of her BMR) to support her physical activity. Therefore, she needs 1,200 1 360 5 1,560 kcal each day.

A 42-year-old man is placed on a twodrug regimen to prevent the activation of the tuberculosis bacteria, as his tuberculin skin test (PPD) was positive, but he shows no clinical signs of tuberculosis, and his chest X-ray is negative. One of the drug's mechanism of action is to inhibit which one of the following enzymes? (A) DNA polymerase (B) RNA polymerase (C) Peptidyl transferase (D) Initiation factor 1 of protein synthesis (IF-1) (E) Telomerase

B. Two drugs are utilized for latent tuberculosis: isoniazid and rifampin. Isoniazid works by blocking the synthesis of mycolic acid, a necessary component of the cell wall of the bacteria that leads to tuberculosis. Rifampin works by inhibiting bacterial RNA polymerase, and blocking the synthesis of new proteins. Neither drug affects DNA polymerase or peptidyl transferase (chloramphenicol is the antibiotic that inhibits bacterial peptidyl transferase activity). Rifampin also has no effect on IF-1.

A 43-year-old female has been on a "grapefruit and potatoes" diet for several months in an effort to lose weight. She now complains of a rash covering most of her body, a large, beefy tongue, nausea and diarrhea, and some confusion. To reverse the symptoms described in the patient, a diet high in which one of the following should be recommended? (A) Green, leafy vegetables (B) Whole grains and meat (C) Citrus fruits (D) Orange and yellow vegetables (E) Chocolate cake

B. While green, leafy vegetables are rich in other B vitamins, whole grains, meats, fish, and liver are the best sources of niacin. Citrus fruits are high in vitamin C. Orange and yellow vegetables are high in vitamin A. Chocolate cake is high in flavonoids, an antioxidant, fats, and carbohydrates.

Which one of the following is a common metabolic feature of patients with anorexia nervosa, untreated type 1 DM, hyperthyroidism, and nontropical sprue? (A) A high BMR (B) Elevated insulin levels in the blood (C) Loss of weight (D) Malabsorption of nutrients (E) Low levels of ketone bodies in the blood

C. All of these patients will lose weight—the anorexic patients because of insufficient calories in the diet, the patients with type 1 DM because of low insulin levels that result in the excretion of glucose and ketone bodies in the urine, those with hyperthyroidism because of an increased BMR, and those with nontropical sprue because of decreased absorption of food from the gut. The untreated diabetic patients will have high ketone levels because of low insulin. Ketone levels may be elevated in anorexia and also in sprue, due to a reduction in levels of gluconeogenic precursors. An increased BMR would be observed only in hyperthyroidism. Nutrient malabsorption would occur only in nontropical sprue and anorexia.

A young woman (5' 3" tall, 1.6 m) who has a sedentary job and does not exercise consulteda physician about her weight, which was 110 lb(50 kg). A dietary history indicates that she eatsapproximately 100 g of carbohydrate, 20 g of protein, and 40 g of fat daily. On the basis of the woman's current weight, diet, and sedentary lifestyle, which one of the following does the physician correctly recommend that she should undertake? (A) Increase her exercise level (B) Decrease her protein intake (C) Increase her caloric intake (D) Decrease her fat intake to ,30% of her total calories (E) Decrease her caloric intake

C. Because her caloric intake (840 kcal/day) is less than her expenditure (1,560 kcal/day), the woman is losing weight. She needs to increase her caloric intake. Exercise would cause her to lose more weight. She is probably in negative nitrogen balance because her protein intake is low (0.8 g/kg/day is recommended). Although her fat intake is 43% of her total calories and recommended levels are, 30%, she should increase her total calories by increasing her carbohydrate and protein intake rather than decreasing her fat intake.

A patient is taking omeprazole for gastric reflux disease. Omeprazole contains a free sulfydryl group that is critical for its mechanism of action. This drug will most likely act in which one of the following ways? (A) Reduce an existing sulfhydryl on the intestinal proton pump. (B) Form a disulfide bond with a methionine on the gastric proton pump. (C) Form a disulfide bond with a cysteine on the gastric proton pump. (D) Reduce an existing sulfhydryl group on the gastric proton pump. (E) Form a disulfide bond with a cysteine on the intestinal proton pump.

C. Free sulfhydryl groups can form disulfide bonds with a cysteine side chain, which may then interfere with the functioning of the protein. Gastric reflux disease is caused by a failure of the gastroesophageal sphincter, allowing stomach contents to travel up the esophagus. Since there is acid in these contents, damage to the esophagus can result. Inhibiting the gastric proton pump, thereby increasing the pH of the stomach contents, would reduce the damage that occurs during reflux due to the increased pH. Inhibiting an intestinal proton pump would not address gastric reflux disease. Methionine residues, though they contain a sulfur atom, do not form disulfide bonds. Free sulfhydryl groups are already reduced, so the drug cannot reduce one further.

A 27-year-old male got lost while hiking in Yosemite National Park. He was found 8 days later. He had nothing to eat and only water to drink before being rescued. Which one of the following lab tests should be run on the patient to determine whether he is suffering from overall protein malnutrition? (A) Albumin (B) Blood urea nitrogen (BUN) (C) Creatinine (D) Ferritin (E) Creatine phosphokinase (CPK)

C. In an attempt to save muscle tissue (amino acids used for gluconeogenesis), the brain in starvation mode will utilize ketone bodies for a portion of its energy needs. Liver glycogen stores would be depleted under the conditions described. Heme is not used for energy production, and produces bilirubin when degraded, which cannot be used to generate energy or ketone bodies. Muscle glycogen cannot contribute to blood glucose levels, as muscle tissue lacks the enzyme that allows free glucose to be produced within the muscle.

A 50-year-old female has shortness of breath, cough, and fever for 3 days. She lives with her husband and has no medical problems. Her pulse ox in the office is 89 and her pulse rate is 110. She is admitted for treatment of community-acquired pneumonia, and her intravenous (IV) antibiotic treatment includes levofloxacin. A mutation in which bacterial enzyme would be required for levofloxacin resistance to be observed? (A) DNA primase (B) DNA polymerase III (C) DNA gyrase (D) DNA ligase (E) DNA polymerase I

C. Levofloxacin is a member of the quinolone family of antibiotics that inhibits bacterial topoisomerases, primarily DNA gyrase (etoposide is the drug that inhibits eukaryotic topoisomerases). Without gyrase activity, the DNA of the bacterial chromosome cannot be unwound properly, and DNA replication would cease, leading to the death of the bacteria. The quinolone family of antibiotics does not directly affect DNA polymerases, DNA ligase, or DNA primase.

A 40-year-old tobacco farmer is seen in the ER with bradycardia, profuse sweating, vomiting, increased salivation, and blurred vision.He was spraying his field with malathion when the hose ruptured and he was covered with the malathion. Which of the following types of inhibition of enzymes does this poisoning represent? (A) Competitive (B) Noncompetitive (C) Irreversible (D) Reversible (E) The drug does not work by inhibiting enzyme activity.

C. Malathion is an organophosphate that inhibits the action of acetylcholinesterase in an irreversible manner. It is one of the most common causes of poisoning worldwide. Malathion forms an irreversible covalent bond between the inhibitor and the active site serine side chain of the enzyme. Without acetylcholinesterase, acetylcholine accumulates in the neuromuscular junction and causes the symptoms described in the case. Both competitive and noncompetitive inhibition are reversible forms of inhibition, and their mechanism of action does not apply to malathion.

Ketone bodies are used as a fuel after an overnight fast

C. Skeletal muscle oxidizes ketone bodies, which are synthesized in the liver from fatty acids derived from adipose tissue. As the fast continues, the muscle will switch to oxidizing fatty acids, which allows ketone body levels to rise such that the brain will begin using them as an energy source.

A 50-year-old male with a "pot belly" and a strong family history of heart attacks is going to his physician for advice on how to lose weight. He weighs 220 lb (100 kg) and is about 6' tall (1.85 m). His lifestyle can be best described as sedentary. How many kilocalories per day would the patient need to maintain this weight? (A) 2,400 (B) 2,620 (C) 3,120 (D) 3,620 (E) 3,950

C. The DEE is equal to the BMR plus physical activity factor. For the patient in question, the BMR 5 24 kcal/kg/day 3 100 kg, or 2,400 kcal/day. Since the patient is sedentary, the activity level is 30% that of the BMR, or 720 kcal/day. The overall daily needs are therefore 2,400 1 720 kcal/day, or 3,120 kcal/day. If the patient consumes ,3,000 kcal/day, or increases his physical activity level, then weight loss would result.

A 15-month-old girl has been given an MMR immunization. Which of the following chemical messengers is responsible for the body's ability to mount an immune response to this vaccination? (A) Neuropeptides (B) Biogenic amines (C) Steroid hormones (D) Cytokines (E) Amino acids

D. Cytokines are the messengers of the immune system. Neuropeptides and biogenic amines (small-molecule neurotransmitters) are messengers of the nervous system. Steroid hormones are messengers of the endocrine system. Amino acids (such as glycine and glutamate) can act as mediators within the nervous system. Once the shot is given, immune cells secrete cytokines to induce the synthesis of antibodies against the antigens injected into the girl.

A 10-year-old boy, small for his age in both height and weight with a calculated, projected adult height of less than 5 feet, is photophobic, and develops a "butterfly" rash over his nose and cheeks if exposed to the sun. He has a high-pitched voice, large nose, prominent ears, and has had multiple pneumonias in his childhood. An examination of fibroblasts from this patient demonstrated an increased sister chromatid exchange rate during mitosis as compared to cells from a normal child. The defective enzymatic activity in this child can be traced to which one of the following activities? (A) A DNA polymerase (B) An RNA polymerase (C) A helicase (D) An exonuclease (E) An endonuclease

C. The child has Bloom's syndrome, a DNA synthesis defect due to a defective DNA helicase. The defective helicase leads to an increased mutation rate in the cells, through an unknown mechanism. Cells derived from patients with Bloom's syndrome display a significant increase in recombination events between homologous chromosomes as compared to normal cells (increased sister chromatid exchange rate). Mutations in the helicase increase genomic instability; the normal Bloom's protein suppresses sister chromatid exchange, and helps to maintain genomic stability. Bloom's syndrome is not due to a mutation in either DNA or RNA polymerase, an exonuclease, or an endonuclease.

A physician working in a refugee camp in Africa notices a fair number of children with emaciated arms and legs, yet a large protruding stomach and abdomen. An analysis of the children's blood would show significantly reduced levels of which one of the following as compared with those in a healthy child? (A) Glucose (B) Ketone bodies (C) Albumin (D) Fatty acids (E) Glycogen

C. The children are exhibiting the effects of kwashiorkor, a disorder resulting from adequate calorie intake but insufficient calories from protein. This results in the liver producing less serum albumin (due to the lack of essential amino acids), which affects the osmotic balance of the blood and the fluid in the interstitial spaces. Owing to the reduction in osmotic pressure of the blood, water leaves the blood and enters the interstitial spaces, producing edema in the children (which leads to the expanded abdomen). The children are degrading muscle protein to allow the synthesis of new protein (due to a lack of essential amino acids), and this leads to the wasting of the arms and legs of children with this disorder. The children will exhibit normal or slightly elevated levels of ketone bodies and fatty acids in the blood, as the diet is calorie sufficient. Glycogen levels may only be slightly reduced (since the diet is calorie sufficient), but glycogen is not found in the blood. Glucose levels will be only slightly reduced, as gluconeogenesis will keep glucose levels near normal.

An environmentalist attempted to live in a desolate forest for 6 months, but had to cut his experiment short when he began to suffer from bleeding gums, some teeth falling out, and red spots on the thighs and legs. This individual is suffering from an inability to properly synthesize which one of the following proteins? (A) Myoglobin (B) Hemoglobin (C) Collagen (D) Insulin (E) Fibrillin

C. The environmentalist is suffering from scurvy, a deficiency of vitamin C. The hydroxylation of proline and lysine residues in collagen requires vitamin C and oxygen. In the absence of vitamin C, the collagen formed cannot be appropriately stabilized (owing, in part, to reduced hydrogen bonding between subunits due to the lack of hydroxyproline) and is easily degraded, leading to the bleeding gums and loss of teeth. Globin synthesis might be indirectly affected because absorption of iron from the intestine is stimulated by vitamin C, but globin is not modified through a hydroxylation reaction. Iron is involved in heme synthesis, which regulates globin synthesis. Insulin and fibrillin synthesis are not dependent on vitamin C (lack of insulin will lead to diabetes, and mutations in fibrillin lead to Marfan syndrome).

A 12-year-old boy is admitted to the hospital in ketoacidosis with a blood glucose level of 700 mg/dL (normal fasting levels are between 80 and 100 mg/dL). The boy is shown to have no detectable C-peptide upon further testing. The child is treated appropriately such that the glucose levels have been reduced, and he does not become dehydrated. Once glucose is transported into his cells, which organelle is responsible for generating energy from the oxidation of glucose to carbon dioxide and water? (A) Lysosome (B) Golgi complex (C) Mitochondria (D) Nucleus (E) Peroxisome

C. The mitochondria are organelles of fuel oxidation and ATP generation. Lysosomes contain hydrolytic enzymes that degrade proteins and other large molecules. The Golgi form vesicles for transport of molecules to the plasma membrane and for secretion. The nucleus carries out gene replication and transcription of DNA.

A 28-year-old man presents to the ER with a large amount of blood and protein in his urine. He has had a sensorineural hearing loss since his teen years and has misshaped lenses (anterior lenticonus). The physician is suspicious of a genetic disorder that may lead to eventual kidney failure. If this is the case, the patient most likely has a mutation in which one of the following proteins? (A) Spectrin (B) α1-Antitrypsin (C) Collagen (D) Fibrillin (E) β-Myosin heavy chain

C. The patient has Alport syndrome, a mutation in type IV collagen that alters the basement membrane composition of kidney glomeruli. In the absence of a functional basement membrane, the kidneys have difficulty in properly filtering waste products from blood into the urine, and both blood and proteins can enter the urine. Type IV collagen is also important for hearing (it is found in the inner ear) and for the eye. Type IV collagen forms a meshlike structure, which is different from the rodlike structures found in type I collagen, and is found in almost all basement membrane structures. Given sufficient time, the alteration in the basement membrane in the glomeruli will lead to their destruction, and loss of kidney function. A mutation in α1-antitrypsin will lead to emphysema, mutations in spectrin can lead to hereditary spherocytosis, mutations in fibrillin lead to Marfan syndrome, and mutations in β-myosin heavy chain can lead to FHC.

A 3-year-old boy presents with 3 days of a low-grade fever, joint pain, and a "lacy-" appearing rash on his arms and legs. His rash began on his face and he appeared to have "slapped cheeks." The chemical messenger that caused the symptoms (vasodilatation presenting clinically as a "rash") can be classified as which one of the following? (A) Cytokine (B) Neuropeptide (C) Eicosanoid (D) Steroid hormone (E) Amino acid

C. The patient has Fifth disease, a viral illness caused by parvovirus B19. The "slapped cheek" appearance of this rash is very distinctive. The eicosanoids control cellular function in response to injury (in this case, a viral infection). In response to the viral infection, vascular endothelial cells will secrete prostaglandins that act on smooth muscle cells to cause vasodilation, which leads to the reddish appearance of the infected individual. The release of eicosanoids may also be responsible for the fever that sometimes accompanies Fifth disease. Neuropeptides, cytokines, steroid hormones, or amino acids are not responsible for the vasodilation that occurs in this disease.

A thin, emaciated 25-year-old male presents with purple plaques and nodules on his face and arms, coughing, and shortness of breath. In order to diagnose the cause of his problems most efficiently, you would order which one of the following types of tests? (A) Southern blot (B) Northern blot (C) Western blot (D) Sanger technique (E) Southwestern blot

C. The patient has Kaposi's sarcoma and AIDS. The causative agent is HIV, an RNA virus. The Western blot technique is used to identify whether a specific blood sample contains antibodies that will bind to HIV-specific proteins. The HIV proteins are run through a gel, transferred to filter paper, and probed using the sera from the patient. If the patient has antibodies to the HIV proteins, then a positive result will be obtained. A Southern blot is used to identify the DNA, and in this case it is easier to check for the presence of anti-HIV antibodies in the patient's sera. A Northern blot would check for viral RNA, but it is more efficient, and reliable, due to the low levels of viral RNA, to check for anti-HIV proteins instead. The Sanger technique identifies a portion of the DNA chain through sequencing the bases in the DNA, and is not used for determining the HIV status. A Southwestern blot is used to detect DNA binding to proteins, and would not be applicable for AIDS testing.

A 47-year-old woman, who has been on kidney dialysis for the past 7 years, has developed jaundice, fatigue, nausea, a lowgrade fever, and abdominal pain. A physical examination indicates a larger-than-normal liver, and blood work demonstrates elevated levels of aspartate aminotransferase (AST) and alanine aminotransferase (ALT). The physician places the patient on two drugs, one of which is a nucleoside analog, geared to inhibit DNA and RNA syntheses. The primary function of the other drug is to do which one of the following? (A) Inhibit DNA repair in infected cells (B) Enhance the rate of the elongation phase of protein synthesis (C) Reduce the rate of initiation of protein synthesis (D) Inhibit ribosome formation (E) Promote ribosome formation

C. The patient has an acute version of hepatitis C infection, which primarily affects the liver and its function. The two-drug treatment for hepatitis C is ribavirin and modified interferon (it is modified so it is more stable). The interferon works by activating a kinase (protein kinase R) that phosphorylates a key initiation factor for protein synthesis, thereby inhibiting the factor from participating in protein synthesis. This leads to a reduction in protein synthesis, and reduced replication of the virus infecting the cells. Interferon does not inhibit DNA repair, enhance the elongation phase of protein synthesis, or affect ribosome formation.

A 30-year-old female presents with a 15-pound weight loss over 1 month, heat intolerance, tachycardia, tremor, bilateral exophthalmos, and a mass in the anterior neck. The hormone overproduced in this condition requires which one of the following? (A) Arachidonic acid (B) Cholesterol (C) Tyrosine (D) Tryptophan (E) Glutamate

C. The patient has hyperthyroidism, or Grave disease, an overproduction of thyroid hormone. Thyroid hormone is derived, in part, from tyrosine, which is iodinated to produce the active forms of thyroid hormone, T3 and T4. Cholesterol is a precursor to steroid hormones. Arachidonic acid is a precursor to eicosanoids (prostaglandins). Tryptophan is a necessity in the production of serotonin, and glutamate is needed to produce GABA. The symptoms described do not occur if there is an overproduction of steroid hormones, eicosanoids, serontonin, or GABA.

A 21-year-old patient is being evaluated for a major depressive disorder. During the interview, he admits to having several episodes in the past of feeling "on top of the world," able to function very well with only 4 hours of sleep per night, maxing out his credit cards (a very unusual characteristic for him), and "indiscriminate" sexual encounters with multiple partners. The elemental medication most commonly used to treat this patient's disorder will lead to the accumulation of which one of the following compounds? (A) Inositol (B) Phosphatidylinositol (C) Inositol phosphate (D) Inositol bisphosphate (E) Inositol trisphosphate (F) Diacylglycerol

C. The patient has presented with classic symptoms of bipolar disorder. Lithium is a first-line treatment of bipolar disorder whose mechanism of action is to interrupt the PI cycle by blocking the action of inositol monophosphatases, the enzyme that converts inositiol phosphate to free inositol, such that phosphatidylinositol can be resynthesized from CDP-diacylglycerol and inositol. Through an interruption in the cycle, the key PI-cycle second messengers cannot be continually generated, leading to a reduction in signaling capabilities.

A young black man was brought to the emergency room (ER) due to severe pain throughout his body. He had been exercising vigorously when the pain started. He has had such episodes about twice a year for the past 10 years. An analysis of the blood shows a reduced blood cell count (anemia), and odd- looking red blood cells that were no longer concave and looked like an elongated sausage. The type of mutation leading to this disorder is best described as which one of the following? (A) Insertion (B) Deletion (C) Missense (D) Nonsense (E) Silent

C. The patient has sickle cell anemia. Sickle cell anemia arises owing to a substitution of valine (GTG) for glutamate (GAG). This is the definition of a missense mutation (one amino acid replaced by another), and since only one amino acid is replaced, it is also a point mutation. A nonsense mutation is a point mutation that converts a codon to a stop codon and premature termination of the growing peptide chain. A silent mutation is the result of a DNA change that does not change the amino acid sequence of the protein. Since this is a singlenucleotide substitution, it is not due to an insertion or deletion of genetic material

A 50-year-old male with a "pot belly" and a strong family history of heart attacks is going to his physician for advice on how to lose weight. He weighs 220 lb (100 kg) and is about 6' tall (1.85 m). His lifestyle can be best described as sedentary. Into which of the following categories does his BMI place him? (A) Underweight (B) Healthy (C) Overweight (preobese) (D) Obese (class I) (E) Obese (class II)

C. The patient is in the overweight (preobese) category with a BMI of 29. As indicated in Table 1.2, a BMI of ,18.5 is the underweight category, a BMI between 18.5 and 24.9 is the healthy range, a BMI between 25 and 30 is the overweight (preobese) category, and any BMI of 30 or above is considered the obese range. Class I obese is between 30 and 35, whereas class II obese is between 35 and 40. Class III obesity, or morbidly obese, is the classification for individuals with a BMI of 40 or higher

Your diabetic patient is using the short-acting insulin lispro to control his blood glucose levels. Lispro is a synthetic insulin formed by reversing the lysine and proline residues on the C-terminal end of the B-chain. This allows for more rapid absorption of insulin from the injection site. The engineering of this drug is an example of which of the following technologies? (A) Polymorphism (B) DNA fingerprinting (C) Site-directed mutagenesis (D) Repressor binding to a promoter (E) PCR

C. This is a prime example of recombinant DNA technology to manufacture a very useful treatment for human diseases, by using site-directed mutagenesis. As the amino acid and DNA sequences of mature insulin is known, the engineering of lispro required that a proline codon be converted to a lysine codon, and the adjacent lysine codon converted to a proline codon. A polymorphism refers to the differences in DNA sequences amongst individuals in a population at a particular location within the genome. A polymorphism would not result in the synthesis of lispro insulin. DNA fingerprinting is used to identify unknown DNA samples by comparing the polymorphisms present in the sample DNA to a particular individual's DNA. Other than identical twins, everyone's DNA is different, and can be distinguished by fingerprinting using genetic polymorphisms. Repressor binding to a promoter is part of gene regulation in prokaryotypes, and would not be useful in generating the genetic changes necessary to produce lispro insulin. The PCR can amplify a particular DNA segment between two known segments of DNA, but the technique does not lead to the alteration of amino acid sequence between lispro and normal insulin.

The liver enzyme glucokinase catalyzes the phosphorylation of glucose to glucose 6-phosphate. The value of Km for glucose is about 7 mM. Blood glucose is 5 mM under fasting conditions, and can rise in the liver to 20 mM after a high-carbohydrate meal. Therefore, if a person who is fasting eats a high-carbohydrate meal, the velocity of the glucokinase reaction will change in which one of the following ways? (A) Remain at <50% Vmax (B) Remain above 80% Vmax (C) Increase from <50% Vmax to >50% Vmax (D) Decrease from >50% Vmax to <50% Vmax (E) Remain at Vmax

C. This problem is best solved using the Michaelis-Menton equation and comparing the velocity (as a function of maximal velocity) under fasting and nonfasting conditions. During fasting, [S] = 5 mM, and the Km is 7 mM; so v = (5 × Vmax)/(7 + 5) = 42% Vmax. In the fed state, [S] = 20 mM, and the Km is 7 mM; so v = (20 × Vmax)/(7 + 20) = 74% Vmax. Glucokinase is more active in the fed than in the fasting state, and the velocity will increase from <50% Vmax to >50% Vmax.

A 72-year-old man acquired a bacterial infection in the hospital while recuperating from a hip replacement surgery. The staph infection was resistant to a large number of antibiotics, such as amoxicillin, methicillin, and vancomycin, and was very difficult to treat. The bacteria acquired its antibiotic resistance owing to which one of the following? Choose the one best answer. (A) Spontaneous mutations in existing genes (B) Large deletions of the chromosome (C) Transposon activity (D) Loss of energy production (E) Alterations in the membrane structure

C. Transposons have the ability to move DNA elements from one piece of DNA to another, including antibiotic-resistance genes from R-plasmids to the host chromosome. Thus, over time, as a bacteria obtains plasmids with antibiotic-resistance genes on them, the transposons can move the gene to the bacterial chromosome so it is always expressed by the cell, and no longer requires the plasmid for antibiotic resistance. Alterations in the membrane structure do not occur, nor do large deletions of the bacterial chromosome (antibiotic-resistance genes are not normal components of the bacterial chromosome). Antibiotic resistance is neither due to a loss of energy production, nor to spontaneous mutations in existing genes, as the bacteria do not encode genes that may confer antibiotic resistance to begin with.

A vegan has been eating low-quality vegetable protein for many years, and is now exhibiting a negative nitrogen balance. This may be occurring due to a lack of which one of the following in his/her diet? (A) Linoleic acid (B) Starch (C) Serine (D) Lysine (E) Linolenic acid

D. A negative nitrogen balance will result from a diet deficient in one essential amino acid, or in a very diseased state. Linoleic and linolenic acids are the essential fatty acids in the diet, and a lack of these fatty acids will not affect nitrogen balance. Starch is a glucose polymer, and the lack of starch will not affect nitrogen balance. Lysine is an essential amino acid, whereas serine can be synthesized from a derivative of glucose. Lack of lysine in the diet will lead to a negative nitrogen balance as existing protein is degraded to provide lysine for new protein synthesis.

The shipwrecked man described in the previous question will have most of his fuel stored as triacylglycerol instead of protein in muscle due to triacylglycerol stores containing which of the following as compared to protein stores? (A) More calories and more water (B) Less calories and less water (C) Less calories and more water (D) More calories and less water (E) Equal calories and less water

D. Adipose tissue contains more calories (kilocalories) and less water than does muscle protein. Triacylglycerol stored in adipose tissue contains 9 kcal/g, and adipose tissue has about 15% water. Muscle protein contains 4 kcal/g and has about 80% water.

After a stressful week of exams, a medical student sleeps for 15 hours, then rests in bed for an hour before getting up for the day. Under these conditions, which one of the following statements concerning the student's metabolic state would be correct? (A) Liver glycogen stores are completely depleted. (B) Liver gluconeogenesis has not yet been activated. (C) Muscle glycogen stores are contributing to the maintenance of blood glucose levels. (D) Fatty acids are being released from adipose triacylglycerol stores. (E) The liver is producing and oxidizing ketone bodies to CO2 and H2O.

D. During fasting, fatty acids are released from adipose tissue and oxidized by other cells. Liver glycogen is not depleted until about 30 hours of fasting. After an overnight fast, both glycogenolysis and gluconeogenesis by the liver help maintain blood glucose levels. Muscle glycogen stores are not used to maintain blood glucose levels. The liver produces ketone bodies but does not oxidize them, but under the conditions described in this question, ketone body formation would be minimal.

A 27-year-old male got lost while hiking in Yosemite National Park. He was found 8 days later. He had nothing to eat and only water to drink before being rescued. Which one of the following is an essential nutrient that he has not received over the last 8 days? (A) Lactic acid (B) Oleic acid (C) Steric acid (D) EPA (E) Palmitic acid

D. Eicosapentaenoic acid (EPA, a 20-carbon fatty acid containing five double bonds) can be derived from an essential fatty acid found in fish oils (linolenic acid), and is a precursor of eicosanoids (prostaglandins, leukotrienes, and thromboxanes). EPA is also ingested from fish oils. Lactic acid is produced from muscle and red blood cells, and is not an essential nutrient. Palmitic acid (a fatty acid containing 16 carbons, with no double bonds), oleic acid (a fatty acid containing 18 carbons, with one double bond), and stearic acid (a fatty acid containing 18 carbons, with no double bonds) can all be synthesized by the mammalian liver through the normal pathway of fatty acid synthesis.

A baby did well with no discernable problems until 3 months of age when he began having cyanotic spells and later had a myocardial infarction. An autopsy revealed a congenital defect, where the left main coronary artery arose from the pulmonary artery instead of the aorta. Of the following, which is the most likely reason that he showed no symptoms until 3 months of age? (A) He had abnormal α chains of hemoglobin. (B) He had abnormal β chains of hemoglobin. (C) He had abnormal γ chains of hemoglobin. (D) HbF has a lower affinity for BPG. (E) HbF has a higher affinity for BPG.

D. Fetal hemoglobin (HbF) is composed of two α subunits and two γ subunits. It has a lower affinity for BPG and therefore a higher affinity for oxygen. The congenital defect in the child has led to a reduced oxygenation of the blood supplying the heart (the end organ of the coronary arteries). The coronary arteries normally receive oxygenated blood from the aorta, but in this case, the left main coronary artery is supplying the left ventricle with deoxygenated blood from the pulmonary artery. However, as the child matures and begins producing HbA instead of HbF, there is insufficient oxygen being delivered to the aorta, leading to a myocardial infarction due to the lack of oxygen reaching the heart. As the HbF was replaced with HbA (two α subunits and two β subunits), the lower affinity for oxygen became manifest since the pulmonary artery blood is lower in oxygen than the left ventricle, leading to a myocardial infarction from the lack of O2 delivered to the myocardium. If the child had been born with abnormal γ chains, the deficiency would have been manifest at birth, and not first appeared at 3 months of age.

A 27-year-old male got lost while hiking in Yosemite National Park. He was found 8 days later. He had nothing to eat and only water to drink before being rescued. Which one of the following would be his primary source of carbons for maintaining blood glucose levels when he was found? (A) Liver glycogen (B) Muscle glycogen (C) Fatty acids (D) Triacylglycerol (E) Ketone bodies

D. In the starvation state, muscle decreases the use of ketone bodies, causing an elevation of ketone bodies in the bloodstream. The brain uses the ketone bodies for energy and uses less glucose, which decreases the need for gluconeogenesis, thus sparing muscle protein degradation to provide the precursors for gluconeogenesis. Red blood cells cannot use ketone bodies and must utilize glucose. Therefore, the use of glucose by red blood cells would be unchanged under these conditions.

Unbeknownst to its owners, a cow recently sacrificed for meat production had mad cow disease. The precipitating event in the cow's brain that led to this disease is which one of the following? (A) Altered gene expression (B) Infection of the brain with a virus (C) Proteolytic cleavage of an existing brain protein (D) An altered secondary and tertiary structures for an existing brain protein (E) Loss of the nuclear membrane

D. Mad cow disease is a prion disorder, in which a misfolded prion protein in the brain forms aggregrates and precipitates, interfering with normal brain function. Prions can adopt a "stable" conformation, which consists primarily of α-helices, and an aggregation-prone conformation, which consists primarily of β-sheets. Once in the aggregation-prone conformation, the protein aggregates, shifting the equilibrium between structure forms toward the aggregation- prone form. This feeds the aggregation-prone form until the precipitated protein begins to interfere with brain function, and will eventually lead to death. Prion disorders are not dueto altered gene expression, viruses, proteolytic cleavage of a prion protein, or to the loss of the nuclear membrane.

Fatty acids are not a significant fuel source at any time

D. Oxidation of fatty acids occurs in mitochondria. Red blood cells lack mitochondria and therefore cannot use fatty acids. The brain will not transport most fatty acids across the blood-brain barrier (the essential fatty acids are a notable exception). Therefore, the brain cannot use fatty acids as an energy source. The brain does, however, synthesize its own fatty acids, and will oxidize those fatty acids when appropriate. Red blood cells can never use fatty acids as an energy source due to their lack of mitochondria.

A young woman (5' 3" tall, 1.6 m) who has a sedentary job and does not exercise consulteda physician about her weight, which was 110 lb(50 kg). A dietary history indicates that she eatsapproximately 100 g of carbohydrate, 20 g of protein, and 40 g of fat daily. What is this woman's BMI? (A) 16.5 (B) 17.5 (C) 18.5 (D) 19.5 (E) 20.5

D. The BMI is calculated by dividing the weight of the individual (in kilograms) by the square of the height of the individual (in meters). For this woman, BMI 5 50/1.62 5 19.5

A 4-year-old boy has had multiple episodes of pneumonia, steatorrhea, and has fallen off his normal growth curve. A sweat test was positive for chloride ions. The reason this boy is at risk for repeat episodes of pneumonia is which one of the following? (A) Elastase destruction of lung cells (B) Defective α1-antitrypsin activity (C) Excessive water in the lungs (D) Dried mucus accumulation in the lungs (E) Loss of lung cells due to a defect in DNA repair

D. The boy is exhibiting the symptoms of cystic fibrosis, which is due to a mutation in the CFTR. The CFTR is required for chloride transport across the membrane, is activated by phosphorylation by the cAMP-activated protein kinase, and when activated allows chloride to flow down its electrochemical gradient. A defective CFTR also alters the ion composition of mucus, reducing its ability to absorb water through osmosis, leading to the drying of mucus in various ducts and tissues, including the lung cells. The lung cells normally secrete a thin, watery mucus designed to trap small particles, which are moved through the lung so they can be swallowed or removed by coughing. When water cannot leave the lung cells, the mucus dries out, leading to pulmonary dysfunction due to clogged bronchi.

A 5-year-old boy begins to regress in terms of developmental milestones, particularly neurologically. Shortly thereafter, the child enters a coma, and dies 2 years into the coma. Upon autopsy, the myelin sheath in the brain was found to be abnormal, as it contained a large quantity of very long-chain fatty acids in its phospholipids. The adrenal glands were also abnormal in appearance. The child, at the molecular level, had inherited a mutation that led to an inability to catalyze reactions that occur in which one of the following intracellular organelles? (A) Lysosomes (B) Nucleus (C) Mitochondria (D) Peroxisomes (E) Golgi apparatus (F) Nucleolus

D. The child has the symptoms of X-linked adrenoleukodystrophy, which is an X-linked disorder with a mutation in the ABCD1 gene. The ABCD1 gene is required for the transport of very long-chain fatty acids into the peroxisome for catabolism. In the absence of this activity, the very long-chain fatty acids accumulate, become incorporated into phospholipids, and alter the structure of myelin, leading to the neurological disorders observed. The lysosomes, nucleus, and Golgi apparatus are not involved in very long-chain fatty acid oxidation. The nucleolus is found in the nucleus and is the site of ribosome formation. Mitochondria oxidize fatty acids, but not when they are very long-chain fatty acids (greater than 20 carbons). In those cases, the initial steps of oxidation occur in the peroxisome, and when the chain length has been reduced, the partially oxidized fatty acid is transferred to the mitochondria to finish the oxidation of the compound.

A 27-year-old male got lost while hiking in Yosemite National Park. He was found 8 days later. He had nothing to eat and only water to drink before being rescued. Which cell can only use glucose for energy needs? (A) Brain (B) Red blood cells (C) Hepatocyte (D) Heart (E) Muscle

D. The glycerol component of triacylglycerol would be the major contributor of carbons for gluconeogenesis among the answer choices provided. Substrates for hepatic gluconeogenesis are lactate (from red blood cells), amino acids (from muscle), and glycerol (from adipose tissue). Fatty acids would be used for energy, but the carbons of fatty acids cannot be used for the net synthesis of glucose. Hepatic glycogen stores are exhausted about 30 hours after the initiation of the fast, and muscle glycogen stores contribute only to muscle energy needs and not to the maintenance of blood glucose levels.

A 32-year-old male is on a weight-maintenance diet, so he does not want to lose or gain any weight. Which amino acid must be present in the diet so the patient does not go into a negative nitrogen balance? (A) Alanine (B) Arginine (C) Glycine (D) Threonine (E) Serine

D. The lack of one essential amino acid will lead to a negative nitrogen balance due to increased protein degradation to supply that amino acid for the ongoing protein synthesis. Of the amino acids listed, only threonine is an essential amino acid (alanine can be synthesized from pyruvate [which can be derived from glucose], arginine is produced in the urea cycle using aspartic acid and the amino acid ornithine, glycine is derived from serine, and serine is derived from 3-phosphoglycerate, which can be produced from glucose).

A young black man was brought to the emergency room (ER) owing to severe pain throughout his body. He had been exercising vigorously when the pain started. He has had such episodes about twice a year for the past 10 years. An analysis of the blood shows a reduced blood cell count (anemia), and oddlooking red blood cells that were no longer concave and looked like an elongated sausage. An underlying cause in the change of shape of these cells is which one of the following? (A) Increased ionic interactions between hemoglobin molecules in the oxygenated state. (B) Increased ionic interactions between hemoglobin molecules in the deoxygenated state. (C) Increased hydrophobic interactions between hemoglobin molecules in the oxygenated state. (D) Increased hydrophobic interactions between hemoglobin molecules in the deoxygenated state. (E) Increased phosphorylation of hemoglobin molecules in the oxygenated state. (F) Increased phosphorylation of hemoglobin molecules in the deoxygenated state.

D. The man has sickle cell disease, and his hemoglobin consists of mutated β chains, along with normal α chains. The glutamate at position 6 in the β chains of HbA is replaced by valine in HbS. Valine contains a hydrophobic side chain, whereas glutamate contains an acidic side chain. Under low oxygenation conditions (such as vigorous exercise), the HbS molecules will polymerize owing to hydrophobic interactions between the valine on the β chain and a hydrophobic patch on another HbS molecule. Under well-oxygenated conditions, the valine in the β chain is not exposed on the surface of the molecule, and it cannot form an interaction with the hydrophobic patch on another hemoglobin molecule. Once the HbS polymerizes, it forms a rigid rod within the red blood cells, which deforms the cell and gives it the "sickle" appearance. Once sickled, the red blood cells cannot easily deform and pass through narrow capillaries, leading to loss of oxygen to certain areas of the body, which is what leads to the pain experienced by the patient. The sickling is not due to increased or decreased ionic interactions between HbS molecules, or to phosphorylation of the HbS monomers.

A 53-year-old man, who has been smoking for the past 35 years at a two-pack-a-day rate, visits his physician for a cough that will not go away, and for difficulty in breathing. A chest X-ray rules out cancer, but does display an increased anterior-posterior (AP) diameter, flattened diaphragms, and "air trapping." The patient is told that his condition will not improve, and that he needs to stop smoking to stop the progression of the disease. At the molecular level, this disease is due to which one of the following? (A) Enhanced trypsin activity in the lung (B) Decreased trypsin activity in the lung (C) Enhanced α1-antitrypsin activity in the lung (D) Decreased α1-antitrypsin activity in the lung (E) Enhanced reduction of sulfhydryl groups in the lung (F) Decreased reduction of sulfhydryl groups in the lung

D. The man has the symptoms of emphysema, due to destruction of lung cells by the protease elastase. Neutrophils in the lung accidentally release elastase as they engulf and destroy inhaled bacteria and other particles, and normally α1-antitrypsin would bind to the elastase and inhibit its activity. In a long-term smoker, however, products from the cigarette smoke oxidize an essential methionine side chain in α1-antitrypsin, rendering it inactive. Thus, over time, noninhibited elastase has been destroying lung tissue until the lung no longer functions properly. Even though the inhibitor will block trypsin activity, the lung damage is the result of increased elastase activity, not trypsin activity. Sulfhydryl groups are not being affected, rather a sulfur in methionine is the target of the cigarette smoke.

A 33-year-old man had a screening colonoscopy, and was diagnosed with a right- sided, mucinous colon cancer, with no other lesions or polyps seen. The reason he had a colonoscopy at such an early age is that his father and paternal uncle had colon cancers diagnosed by age 40. His paternal grandmother had ovarian and uterine cancers. A likely defect in the patient is a reduction in the ability to carry out which one of the following processes? (A) Removal of thymine dimers from the DNA (B) Inability to remove the base U from DNA (C) Loss of DNA ligase activity (D) Inability to correct mismatched bases in newly synthesized DNA (E) Inability to form a solenoid structure from individual nucleosomes

D. The patient has HNPCC, which is due to specific mutations in proteins involved in mismatch repair (mutations in at least four different proteins have been identified that lead to HNPCC). Mismatch repair is not involved in thymine dimer removal, nor base excision repair (the removal of uracil from DNA). HNPCC does not involve a defective DNA ligase, nor does the disease result in defective DNA packaging (solenoid formation) in the nucleus.

A person is diagnosed with group A streptococcal bacteremia. One of the body's major defenses in this type of disease is for eosinophils to phagocytize the bacteria. Once internalized, the bacteria are destroyed by fusing the phagosome with a particular intracellular organelle. Which one of the following would destroy the activity of that organelle such that the bacteria would not be incapacitated? (A) Inhibiting sodium-potassium ATPase activity (B) Interfering with mitochondrial protein synthesis (C) Blocking transport through nuclear pores (D) Inhibiting a proton-translocating ATPase (E) Inhibiting a calcium-activated ATPase

D. The phagosomes fuse with lysosomes, where the acidity and digestive enzymes within the lysosomes destroy the contents of the phagosome (in this case, the bacteria within the phagosome). The digestive enzymes have a pH optimum of 5.5, which is maintained within the lysosome through the actions of a proton pump (a proton-translocating ATPase activity). The nucleus and mitochondria are not involved in the lysosomal digestion of phagosome contents. Blocking either the sodium ATPase, potassium ATPase, or a calcium-activated ATPase will greatly affect other organs, but will not affect the ability of the lysosome to degrade its contents.

A young woman (5' 3" tall, 1.6 m) who has a sedentary job and does not exercise consulteda physician about her weight, which was 110 lb(50 kg). A dietary history indicates that she eatsapproximately 100 g of carbohydrate, 20 g of protein, and 40 g of fat daily. How many calories (kcal) does this woman consume each day? (A) 1,440 (B) 1,340 (C) 940 (D) 840 (E) 640

D. The woman consumes 400 calories (kcal) of carbohydrate (100 g 3 4 kcal/g), 80 calories of protein (20 3 4), and 360 calories of fat (40 3 9) for a total of 840 calories daily.

A 42-year-old woman has slowly developed an inability to keep her eyes open at the end of the day. The eyelids droop, despite her best efforts to keep them open. This does not occur first thing in the morning. Further examination shows a generalized muscle weakness as the day progresses. A drug that may help to stabilize this condition would do which one of the following? (A) Stimulate the production of immune cells (B) Stimulate the production of epinephrine (C) Inhibit the production of acetylcholine (D) Inhibit acetylcholinesterase (E) Stimulate catechol-O-transferase

D. The woman has myasthenia gravis, which is due to autoantibodies directed against the acetylcholine receptor. As such, acetylcholine stimulation of muscle cells is decreased, owing to a reduced number of functional acetylcholine receptors at the neuromuscular junction. One way to treat this condition is to inhibit acetylcholinesterase, the enzyme that degrades acetylcholine at the neuromuscular junction. By keeping the levels of acetylcholine high at the junction, there is a greater probability that the receptors that are active are occupied, and the signal from the neuron can be transmitted. Inhibiting the production of acetylcholine would exacerbate the problem, as would stimulating the production of immune cells (more autoantibodies would potentially be generated). Epinephrine is not involved at the neuromuscular junction, and stimulation of catechol-O-transferase is a mechanism to inhibit the action of catecholamines in nonneuronal tissues, and does not contribute to the progression of myasthenia gravis.

An alcoholic presents with swelling and fissuring of the lips, cracking at the angles of the mouth, red eyes, and an oily, scaly rash of his scrotum. Which one of the following cofactors of enzyme complexes would be most affected by this condition? (A) The concentration of NAD+ (B) The concentration of NADP+ (C) The concentration of coenzyme Q (D) The functioning of the FMN components of complex I (E) The functioning of the cytochrome-containing components of complex III

D. This patient has vitamin B2 (riboflavin) deficiency, ariboflavinosis, as indicated by the symptoms displayed by him. Both FAD and FMN require vitamin B2 to be produced. NAD+ and NADP+ are derived from niacin. Coenzyme Q is derived from acetyl-CoA, and vitamin B2 is not needed in the synthesis of the heme ring, which is derived from succinyl- CoA and glycine.

A new antibiotic has been developed that shows a strong affinity for attacking amino acids with a specific orientation in space. In order for it to work well in humans, the antibiotic must be effective against amino acids in which one of the following configurations? (A) R-configuration (B) L-configuration (C) Aromatic ring configuration (D) Polypeptide chain configuration (E) D-configuration

E. Amino acids in humans are in the L-configuration (except glycine which is neither L nor D), whereas bacterial amino acids can be in either the L- or D-configuration. An antibiotic would need to be effective against bacterial proteins and not human proteins, so developing an antibiotic that recognizes proteins or polypeptides that contain D amino acids would only be effective against bacterial products. All amino acids are in polypeptide chains, and phenylalanine, tyrosine, and tryptophan are amino acids that contain aromatic rings, and are present in both bacteria and humans. The R and S nomenclature is not commonly used in biochemistry to describe the configuration of amino acids.

A 12-year-old boy is admitted to the hospital in ketoacidosis with a blood glucose level of 700 mg/dL (normal fasting levels are between 80 and 100 mg/dL). The boy is shown to have no detectable C-peptide upon further testing. A potential reason for the elevated blood glucose is which one of the following? (A) A lack of a sodium gradient across cellular membranes (B) A lack of a calcium gradient across cellular membranes (C) A lack of a chloride gradient across cellular membranes (D) A reduced number of glucose transport molecules in the brain membrane (E) A reduced number of glucose transport molecules in the muscle membrane (F) A reduce number of glucose transport molecules in the liver membrane

E. The boy has Type 1 diabetes, and is producing no insulin. One of insulin's effects is to stimulate the translocation of GLUT4 transporters from internal vesicles to the plasma membrane of muscle and fat cells. The increase in glucose transport molecules on the cell surface is important for rapidly reducing blood glucose levels. The GLUT4 transporter is for facilitative diffusion, and is not dependent on an ion gradient across the membrane for effective transport, as are the glucose transporters in the small intestine.

A newborn has found to be very photophobic, and his skin burns even with minimal exposure to sunlight, eventually forming skin blisters. Neither parent exhibits this trait, although both are prone to burning when in the sun for a short period of time. As the child grows, he is found to be at average height and weight for his age, and is progressing normally along the developmental guidelines. He is, however, kept inside at all times, and is carefully wrapped if he has to leave the house. Fibroblasts isolated from this child are grown in culture, and in an experiment, exposed to UV light. An analysis of the fibroblast DNA will demonstrate which one of the following? (A) A preponderance of apurinic sites and apyrimidinic sites (B) An increase in sister chromatid exchange rate (C) A preponderance of abnormal base pairs in the DNA (D) Loss of telomeres within the DNA (E) An increase in cross-linked bases within the strands of DNA

E. The child has XP, a defect in nucleotide excision repair such that thymine dimers, created by exposure to UV light, cannot be removed from the DNA. XP will not affect the repair of apurinic or apyrimidinic sites (sites missing just the base from DNA, which requires the AP endonuclease for repair). An increase in sister chromatid exchange rates is a finding in Bloom's syndrome, which is a defect in a helicase required for both DNA and RNA syntheses. Patients with Bloom's syndrome are small for their age, unlike those with XP who follow normal developmental milestones. XP does not result in unusual base pairs in DNA, rather the formation of thymine dimers between the adjacent T residues in one strand of DNA. These T residues are still complementary to the A residues in the other strand. XP does not affect the ability of telomerase to extend the ends of the linear chromosomes in the cell.

A 4-year-old boy displays a failure to thrive, extreme sensitivity to the sun, hearing loss, severe tooth decay, pigmentary retinopathy, and premature aging. An analysis of fibroblasts from the boy demonstrated extensive DNA damage in cells trying to grow, but minimal damage in quiescent cells, which have a greatly reduced rate of transcription as compared to the growing cells. This child most likely has a defect in which one of the following processes? (A) Repair of thymine dimers (B) Base excision repair (C) Nucleotide excision repair (D) Mismatch repair (E) Transcription-coupled DNA repair

E. The child is exhibiting the symptoms of Cockayne syndrome, which is due to a defect in transcription-coupled DNA repair. During transcription of genes, if the RNA polymerase notices DNA damage, transcription will stop while the transcription-coupled DNA repair mechanism will correct the DNA damage. This syndrome can be due to mutations in either the ERCC6 or ERCC8 gene, and the protein products of both the genes are involved in repairing the DNA of actively transcribed genes. The key to answering the question is the amount of DNA damage in growing cells (which are transcriptionally active) versus the damage in quiescent cells (which express fewer genes). The symptoms described are also unique to individuals with this disorder. The repair of thymine dimers and the processes of base excision repair, nucleotide excision repair, and mismatch repair are all functional in individuals with this disorder.

A 15-year-old boy was diagnosed with skin cancer. He had always been sensitive to sunlight, and had remained indoors for most of his life. An analysis of his DNA, from isolated fibroblasts, indicated an increased level of thymine dimers when the cells were exposed to UV light. The boy developed a skin tumor owing to an increased mutation rate, which was caused by which one of the following? (A) A lack of DNA primase activity (B) Decreased recombination during mitosis (C) Increased recombination during mitosis (D) Loss of base excision repair activity (E) Loss of nucleotide excision repair activity

E. The damage to the DNA caused by UV light (pyrimidine dimers) can be repaired by the nucleotide excision repair pathway. In some cases, the missing enzyme is a repair endonuclease. The boy has XP, as determined by the increase in thymine dimers in his DNA after exposure to UV light. Since the dimers cannot be repaired, the DNA polymerase will "guess" when replication occurs across the dimers, increasing the mutation rate of the cells. Eventually, a mutation occurs in a gene that regulates cell proliferation, and a cancer results. An increase or decrease in mutation rate is not related to the rate of recombination during mitosis, nor to a lack of DNA primase activity (which would lead to reduced DNA synthesis, not inaccurate DNA synthesis). Base excision repair is normal in patients with XP.

A woman has been complaining of a burning sensation when urinating, and a urine culture demonstrated a bacterial infection. The physician placed the woman on ciprofloxacin. Ciprofloxacin will be effective in eliminating the bacteria because it interferes with which one of the following processes? (A) mRNA splicing (B) Initiation of protein synthesis (C) Elongation of protein synthesis (D) Nucleotide excision repair (E) DNA replication

E. The quinolone family of antibiotics (which includes ciprofloxacin) inhibits DNA gyrase, a prokaryotic-specific topoisomerase involved in unwinding the DNA strands for replication to occur. In the absence of gyrase activity, there would be no DNA replication, and the bacteria would not be able to proliferate. The quinolones do not affect eukaryotic topoisomerases. Splicing of hnRNA only occurs in eukaryotic cells. The gyrase is neither involved in nucleotide excision repair, nor in any aspect of protein synthesis.

A 16-year-old male high school student was playing basketball for his school when he collapsed on court and could not be resuscitated. An autopsy demonstrated increased thickness of the intraventricular septum and left ventricular wall. These findings could be explained by a mutation in which one of the following proteins? (A) Spectrin (B) α1-Antitrypsin (C) Collagen (D) Fibrillin (E) β-Myosin heavy chain

E. The student has died from FHC, a thickening of the left ventricle of the heart muscle due to a mutation in β-myosin heavy chain. The exact reason for the hypertrophy, which can be caused by mutations in a variety of sarcomeric proteins, is still unknown. None of the other proteins suggested as answers are muscle sarcomeric proteins. Spectrin is a red blood cell protein, and is not found in the heart. α1-Antitrypsin is a circulating protein synthesized by the liver, and in its absence, emphysema will develop. Collagen is the major structural protein of the body, but there are no mutations in collagen that lead to a greatly hypertrophied heart muscle. A lack of fibrillin leads to Marfan syndrome, which can present with defects in heart valves and the aorta, but not a heart muscle greatly increased in size.

Lysosomal hydrolases are targeted to the lysosome by the addition of a carbohydrate residue to the protein. An inability to add this carbohydrate leads to a disease in which the lysosomal hydrolases are treated as secreted proteins, and are exported from the cell, rather than taken to the lysosomes. The secreted proteins will have which one of the following effects on the cells and proteins in the circulation? (A) The blood cells will have their membrane proteins digested. (B) The blood cells will have their carbohydrates on the cell surface removed. (C) The blood cell membranes will become leaky, leading to the death of the blood cells. (D) Circulating proteins will be degraded, whereas the blood cells will be protected against the enzymes. (E) Circulating proteins will be targeted to the spleen for removal. (F) There will be no effect on the proteins and cells in the circulation.

F. Most lysosomal hydrolases have their highest activity near an acidic pH of 5.5 (pH optimum) and little activity in a neutral or basic environment. The intralysosomal pH is maintained near 5.5 by vesicular ATPases, which actively pump protons into the lysosome. The cytosol and other cellular components have a pH near 7.2, and are therefore protected from escaped hydrolases. The pH of the blood is maintained between 7.2 and 7.4, so the escaped lysosomal enzymes will have no activity at that pH, and will not affect the proteins and cells in the circulation. I-cell disease results from the inability to appropriately target lysosomal proteins, and it is a lysosomal storage disease.

The primary source of carbon for maintaining blood glucose levels during an overnight fast

Liver glycogenolysis is the major process for maintaining blood glucose levels after an overnight fast. The muscle cannot export glucose to contribute to the maintenance of blood glucose levels, and fatty acid carbons cannot be utilized for the net synthesis of glucose.

The energy source reserved for strenuous muscular activity

Muscle glycogen is used for energy during exercise. The glycogen is degraded to a form of glucose that can enter metabolic pathways for energy generation. Because exercise is strenuous, muscle requires large amounts of energy, and this can be generated at the fastest rate by converting muscle glycogen to pathway precursors within the muscle. Liver glycogen will produce glucose that enters the circulation. Once in the circulation, the muscle can take up that glucose and use it to generate energy; however, the rate of energy generation from liver-derived glucose is much slower than that from muscle-derived glucose.

A 38-year-old homeless man who has not received any medical care in the last 20 years presents with 2 days of shortness of breath, chills, fever, drooling, painful swallowing, and a "croupy" cough. A physical examination reveals a bluish discoloration of his skin and a tough, gray membrane adhered to his pharynx. The underlying mechanism through which this disease affects normal cells is which one of the following? (A) DNA synthesis is inhibited in the target cells. (B) RNA synthesis is inhibited in the target cells. (C) The process of protein synthesis is inhibited in the target cells. (D) The plasma membrane becomes leaky in the target cells. (E) ATP generation is reduced in the target cells.

The man has contracted diphtheria, and needs the diphtheria antitoxin and then antibiotics to remove the offending organism, diptheria, as the patient has not received medical care over the past 20 years, he has also missed his diphtheria vaccine, which should be received every 10 years. Diphtheria toxin blocks eukaryotic protein synthesis by phosphorylating an initiation factor, which inhibits protein synthesis in the cells. The toxin does not directly affect DNA or RNA synthesis, nor does it, as a primary target, reduce ATP production by the mitochondria or allow the plasma membrane to become leaky.

The major precursor of urea in the urine

The nitrogen in amino acids derived from protein is converted to urea and excreted in the urine. Uric acid, another excretion product that contains nitrogen, is derived from purine bases (found in nucleic acids), not from protein.


संबंधित स्टडी सेट्स

Chapter 10: Designing Organization Structure

View Set

Neuroscience (Questions - Combined Content)

View Set

APES Unit 8 exam multiple choice

View Set