Board review biochemistry questions

2. A 56-year-old woman with a 60-pack year history of smoking is recently found to have a large neoplastic pulmonary mass. Laboratory tests demonstrate a sodium of 127 mmol/L (normal, 135 to 145 mmol/L) and reduced urine osmolality. She likely has which of the following endocrine abnormalities? (A) Cushing disease (B) SIADH (C) Cushing syndrome (D) Acromegaly (E) Prolactinoma

2. The answer is B. This patient has a metabolic derangement consistent with SIADH secondary to a paraneoplastic syndrome associated with lung tumors. Head trauma, strokes, and intracranial tumors can disrupt the hypothalamus-pituitary axis and cause SIADH. The inappropriate release of antidiuretic hormone leads to hyponatremia and fluid overload. Cushing syndrome is a result of increased cortisol, which, when due to a pituitary adenoma secreting tumor, is referred to as Cushing disease. Acromegaly results from growth hormone excess, and a prolactinoma would cause amenorrhea and galactorrhea in a female.

41. Which glucose derivative is found in high levels in seminal fluid? (A) Sorbitol (B) Fructose (C) Glycogen (D) Glucose (E) Lactose

41. The answer is B. Fructose is the major energy source for sperm cells, and low levels contribute to infertility. Sorbitol is a precursor of fructose and is converted into fructose by sorbitol dehydrogenase. Glycogen is formed from glucose in the liver by several enzymes, including glycogen synthase. Glucose is a monosaccharide that functions as a major cellular energy source and metabolic intermediate; lactose is a disaccharide and the source of galactose. Of the sugars listed, fructose is present at the highest concentration in the seminal fluid.

6. An 11-year-old Ashkenazi Jewish girl presents with an enlarged liver and spleen, low white and red blood cell counts, bone pain, and bruising. She is diagnosed with Gaucher disease, a lysosomal storage disease. Which of the following compounds is accumulating in her lysosomes? (A) Galactocerebroside (B) Ceramide (C) Glucocerebroside (D) Sphingosine (E) GM1

6. The answer is C. Patients with Gaucher disease have a deficiency of b-glucocerebrosidase, resulting in glucocerebroside accumulation in the lysosomes of cells of the liver, spleen, and bone marrow. Galactocerebroside accumulates in Krabbe disease. Ceramide accumulation is associated with Farber disease. Sphingosine accumulation is associated with Niemann-Pick disease. GM1 accumulation is associated with generalized gangliosidosis.

6. A 7-year-old boy suffers from mental retardation and self-mutilation (e.g., biting through lip) and has an increased susceptibility to gout. These symptoms are characteristic of Lesch-Nyhan syndrome, which is due to a mutation in which of the following pathways? (A) Salvage pathway for pyrimidines (B) De novo biosynthesis of purines (C) Pathway of uric acid synthesis (D) Salvage pathway for purines (E) De novo biosynthesis of pyrimidines

6. The answer is D. Lesch-Nyhan syndrome results from a defect in HGPRT, an enzyme involved in the purine salvage pathway. HGPRT catalyzes the conversion of the free base (hypoxanthine of guanine) to a nucleotide (IMP or GMP). HGPRT is not required for de novo purine synthesis, or urate synthesis, or for either the de novo or salvage pathways of pyrimidine metabolism.

64. A 45-year-old woman presents with severe dehydration and decreased urine output. Her blood urea nitrogen level is abnormally elevated because her kidneys are not able to properly excrete urea in the urine. Which of the following is an important molecule in urea synthesis? (A) Serine (B) Glutamate (C) Proline (D) Ornithine (E) Leucine

64. The answer is D. Ornithine serves as a carrier that is regenerated in the urea cycle. Serine, glutamate, proline, and leucine are not involved in the urea cycle or in urea production.

1. Which of the following apoproteins is an activator of lipoprotein lipase? (A) Apo A (B) Apo B (C) Apo C-II (D) Apo D (E) Apo E

1. The answer is C. Apo C-II is an activator of lipoprotein lipase, Apo A is the major apolipoprotein of HDL, Apo B100 is the major apolipoprotein of LDL and VLDL (and apo B48 is the major apolipoprotein of chylomicrons), and Apo E is transferred by HDL to nascent chylomicrons and nascent VLDL to form mature forms of those particles. Apo D is unlike all the other apolipoproteins, and its role in metabolism has yet to be clearly defined.

2. For the management of deep vein thrombi, heparin is a common medication given to patients with myocardial ischemia, derived from either acute coronary syndrome or disseminated intravascular coagulation. Which of the following is true about the structure of this glycosaminoglycan? (A) It is composed with repeating disaccharide units. (B) It is composed of a short, branched carbohydrate. (C) It has NANA residues branching from a linear oligosaccharide chain. (D) Its base unit is a member of the class of sialic acids. (E) The structure is seen in ABO blood groups.

2. The answer is A. Glyscosaminoglycans are proteoglycans typified by being composed of a large number of repeating disaccharide units with a core protein and long unbranced polysaccharide chain (the lone exception is hyaluronic acid, which does not contain a core protein). Conversely, glycoproteins, for example ABO blood groups, have shorter, more branched carbohydrates and usually contain sialic acid.

6. A 23-year-old diabetic woman reports having fevers and dysuria. Physical examination reveals costovertebral tenderness, and her urinary analysis shows the presence of bacteria in her urine. Her physician suspects a complicated urinary tract infection and begins a 5-day course of ciprofloxacin. Such quinolone antibiotics inhibit which one of the following enzymes? (A) Eukaryotic topoisomerase (B) Helicase (C) Primase (D) Gyrase (E) Poly(A) polymerase

6. The answer is D. Quinolone antibiotics inhibit DNA gyrase, a prokaryotic topoisomerase, and are important drugs in the treatment of urinary tract infections. Ciprofloxacin in particular has high urinary excretion and is effective against urinary tract infections (although resistance to the drug is becoming more common). Eukaryotic topoisomerase inhibitors include etoposide, which is used in the treatment of some cancers. There are no current drugs regimens that target helicase, which unwinds DNA during replication, or primase, which creates the RNA primers during DNA replication, or poly(A) polymerase.

1. A 14-year-old African American girl with sickle cell anemia presents with extreme pain in her chest and legs. A peripheral smear shows sickling of her red blood cells, and her reticulocyte count is 6.7%. She is diagnosed with an acute pain crisis secondary to microvascular occlusion from sickled erythrocytes. Sickle cell anemia is caused by a mutation that results in one amino acid being replaced by another, which results from which of the following? (A) A missense point mutation (B) A nonsense point mutation (C) An insertion (D) A deletion (E) A frameshift mutation

1. The answer is A. A missense point mutation is the change of 1 base in the DNA that changes the ''meaning'' of a codon. In this case, the 1 base change is in an exon and leads to a glutamate codon (GAG) being changed to a valine codon (GTG). This change in primary structure of the bglobin protein alters its properties, which leads to the sickling observed under low-oxygen conditions. A nonsense point mutation is due to a change in a single base in DNA that converts a codon into a stop codon. This will result in premature termination of the growing polypeptide chain. Insertions occur when 1 base or more are added to DNA. Deletions occur when 1 base or more are removed from DNA, such as the D508 mutation, which leads to cystic fibrosis. Frameshift mutations occur when the number of bases added or deleted is not in a multiple of three, and the reading frame of the mRNA is altered

1. Acetazolamide is a carbonic anhydrase inhibitor and is used in the treatment of all the following conditions except which one? (A) Dehydration (B) Glaucoma (C) Epilepsy (D) Altitude sickness (E) Congestive heart failure

1. The answer is A. Acetazolamide is a potent carbonic anhydrase inhibitor and helps to reduce conditions of volume overload (not volume decrease, which would be brought about by dehydration). In the eye, carbonic anhydrase inhibitors lead to a decrease in the secretion of aqueous humor, which reduces intraocular pressure. In patients with epilepsy, these inhibitors block the activity of the central nervous system neuron carbonic anhydrase, which decreases excessive neuronal discharge. In the treatment of individuals with altitude sickness, the mechanism of the carbonic anhydrase inhibitor appears to be related to the acid-base effects of the drug. Patients with congestive heart failure take these inhibitors, and the effect of the inhibitors is to act as a diuretic, which helps to manage and reduce intravascular volume.

1. Allopurinol is used in the treatment of gout because of its ability to inhibit xanthine oxidase. This inhibition makes it impossible for the enzyme to degrade xanthine and hypoxanthine, which reduces the synthesis of urate, the culprit of gout. Allopurinol works through which one of the following mechanisms? (A) Suicide inhibition (B) Noncompetitive inhibition (C) Allosteric interaction with the enzyme that increases Vmax (D) Feedback inhibition (E) Subunit cooperativity

1. The answer is A. Allopurinol is a substrate for xanthine oxidase, which converts allopurinol to oxypurinol, and which binds tightly to the enzyme and is not released from the enzyme. This blocks substrate binding and further activity of the enzyme. This is an example of irreversible suicide inhibition. Such inhibition leads to a decrease of Vmax. Noncompetitive inhibition occurs when inhibitors bind to a site other than the active site, which is not the case with allopurinol. Additionally, noncompetitive inhibitors are not substrates for the enzyme, as are suicide inhibitors. There is no end-product regulation of xanthine oxidase by allopurinol, thereby ruling out feedback inhibition. Cooperativity exists when there are two or more sites for substrate binding (and in positive cooperativity it is easier for substrate to bind as the concentration of substrate is increased), but this is not observed with xanthine oxidase.

1. A teenager comes to the emergency room looking quite ill. Emergency medical services report that this appears to be a suicide attempt using rat poison. The patient is in hypovolemic shock from profuse vomiting and diarrhea. You note the strong smell of garlic from the patient. Which one of the following correctly describes an action of this toxin? (A) It inhibits pyruvate dehydrogenase. (B) It activates pyruvate dehydrogenase. (C) It reduces the concentration of pyruvate. (D) It increases glutathione production. (E) It enables gluconeogenesis to proceed.

1. The answer is A. Arsenic poisoning is a major public health hazard. Arsenic is an element found in rodenticides, herbicides, industrial chemicals, and improperly made alcohol (e.g., moonshine). Its main toxic effects are to inhibit the enzymes of glycolysis, especially pyruvate dehydrogenase. This results in a reduced flux of carbon through the TCA cycle via the inhibition of the conversion of pyruvate to acetyl CoA. The reduced citric acid cycle activity leads to a reduction in the production of reduced cofactors and a decrease in the production of cellular ATP via oxidative phosphorylation. The lack of ATP leads to a decrease in glutathione production. The lack of ATP also blocks gluconeogenesis from occuring. Because pyruvate dehydrogenase is inhibited, pyruvate will increase, not decrease, in concentration.

1. Which of the following statements is correct concerning prion disease? (A) It is a disease process in which proteins appear to be the sole pathophysiologic entity (B) It is a disease process in which a messenger RNA secondary structure appears to be the sole pathophysiologic entity (C) The disease is only found in humans (D) Effective treatments are available for the disease (E) The disease process is readily reversible

1. The answer is A. Prion diseases include a handful of diseases that affect animals (e.g., bovine spongiform encephalopathy is ''mad cow disease'') and humans (Creutzfeldt-Jakob disease). The pathogenesis is mediated primarily by a protein that can exist in two conformations: one normal, the other leading to disease. The abnormally folded proteins are resistant to degradation by the host and affect the central nervous system. The change in protein structure is not reversible. The messenger RNA in the disease state is not altered compared with the normal state. There is no effective treatment, and patients develop rapidly progressive dementia.

1. A biochemistry graduate student isolates all the enzymes of the TCA cycle and adds OAA and acetyl CoA, including the appropriate energy precursors, cofactors, and water. Which of the following will not be a direct product of his experiment? (A) ATP (B) GTP (C) NADH (D) CO2 (E) FADH2

1. The answer is A. The Krebs cycle does not directly produce ATP. The one substrate level phosphorylation reaction in the cycle generates GTP (the step catalyzed by succinate thiokinase). NADH is generated at three steps (catalyzed by isocitrate dehydrogenase, a-ketoglutarate dehydrogenase, and malate dehydrogenase) and FADH2 at one step (catalyzed by succinate dehydrogenase). CO2 is a product of the isocitrate and a-ketoglutarate dehydrogenase reactions.

1. You have a patient who is undergoing workup for a genetic disorder. You receive the results of test that was ordered and the report states, ''Delta-F508 mutation—absence of the phenylalanine at this position on the protein.'' What is true about this patient's condition? (A) The patient will have a positive sweat chloride test. (B) This patient will develop or has Duchenne muscular dystrophy. (C) There is a very high likelihood of Alzheimer disease with this mutation. (D) This patient will develop Marfan syndrome. (E) This patient has von Willebrand disease.

1. The answer is A. This patient has cystic fibrosis (CF), which is a hereditary disease that affects the ability to produce normal mucus in the exocrine glands of the lungs, liver, pancreas, and intestines. CF is most commonly caused by the DF508 mutation, which results in the deletion (D) of the amino acid phenylalanine (F) at the 508th (508) position on the CFTR protein. This is due to a 3-base deletion within the gene. Diagnosis is possible with newborn screening and confirmed by detecting high levels of salt in a patient's sweat. Duchenne muscular dystrophy is caused by a mutation, most often a large deletion, within the dystrophin gene. Although there is no specific gene yet identified that causes Alzheimer disease (AD), the presence of an abnormally folded protein, called amyloid precursor protein, and its proteolysis results in b-amyloid neurofibrillary tangles, which is a pathognomonic finding in AD. Marfan syndrome is caused by mutations in the FBN-1 gene, which creates an abnormal fibrillin-1 protein, which is normally found in the extracellular matrix. Von Willebrand disease (vWD) is the most common hereditary coagulation abnormality and is caused by a deficiency of von Willebrand factor (vWF), precluding normal platelet adhesion.

1. A 56-year-old diabetic patient with end-stage renal disease receives a kidney transplant from his son. His nephrologist is concerned about the possibility of transplant rejection and puts the patient on mycophenolic acid, which inhibits which important enzyme in the synthesis of nucleotides? (A) PRPP synthetase (B) IMP dehydrogenase (C) Adenylosuccinate synthetase (D) Ribonucleotide reductase (E) Adenylosuccinate lyase

1. The answer is B. Mycophenolic acid is a potent immunosuppressant and an inhibitor of IMP dehydrogenase, which normally converts IMP to xanthosine monophosphate. PRPP synthetase catalyses the initial step in nucleotide metabolism, forming PRPP from ATP and ribose. Adenylosuccinate synthetase and adenylosuccinate lyase are sequential enzymes in the synthesis of AMP and are not affected by mycophenolic acid.

1. Which condition can lead to life-threatening autodigestion of lipids and proteins? (A) Peptic ulcer disease (B) Pancreatitis (C) Celiac sprue (D) Crohn disease (E) Ulcerative colitis

1. The answer is B. Pancreatitis is an inflammatory process commonly caused by gallstones or alcoholism that results in pancreatic autodigestion by lipase. In advanced cases, life-threatening hypocalcemia can occur from fat saponification by lipase. Peptic ulcer diseasemay rarely lead to perforation of a viscous (gastric ulcer), but is not commonly seen in industrialized countries. Peptic ulcer disease affects the stomach. Celiac sprue is an autoimmune disorder in response to gluten, which leads to destruction of the villi on the intestinal epithelial cells. Crohn disease is another autoimmune disorder that affects various parts of the gastrointestinal system, but does not affect the pancreas. Ulcerative colitis results from ulcers in the colon, without pancreatic involvement.

1. A patient has a disease that leads to hyperexcretion of a protein in the urine. Which methodology is the fastest, easiest, and least expensive to determine the molecular weight of the native protein in the urine? (A) Ion exchange chromatography (B) Size exclusion chromatography (C) X-ray crystallography (D) NMR (E) SDS-PAGE

1. The answer is B. Size exclusion (gel) chromatography separates proteins by mass. Ion exchange chromatography involves either anion or cation exchange for separating proteins with a net positive or negative charge. No information concerning the mass of a protein results from ion exchange chromatography. NMR spectra and x-ray crystallographic diffraction techniques require laborious effort and are not fast, easy, or inexpensive. SDS-PAGE does not yield the mass of a native protein because the proteins are denatured before being separated by size.

1. A newborn is found to have fasting hypoglycemia. The nursery staff begins overnight feeds by nasogastric tube because they find that the child has consistently low blood sugars. A liver biopsy andmolecular studies demonstrate an absence of glycogen synthase. The normal function of this enzyme is to do which of the following? (A) Remove glucose residues one at a time from glycogen in the liver (B) Remove glucose residues one at a time from glycogen in muscles (C) Transfer glucose from UDP-glucose to the nonreducing end of a glycogen primer (D) Hydrolyze a-1,6 bonds of glycogen (E) Function as a glucosyl 4:6 transferase

1. The answer is C. Glycogen synthase is the first enzyme in the synthesis of glycogen. It transfers glucose from UDP-glucose to the nonreducing end of a glycogen primer and adds subsequent residues to the growing chain. The removal of glucose residues (answers A and B) during the catabolism of glycogen is mediated by glycogen phosphorylase, a deficiency of which results in Hers disease if in the liver and McArdle disease if in the muscle. Debranching enzyme hydrolyzes a-1,6 bonds of glycogen (answer D). Finally, deficiency of glucosyl 4:6 transferase (the branching enzyme) results in Andersen disease (answer E).

1. A 56-year-old woman is recently diagnosed with breast cancer. She undergoes a lumpectomy and a lymph node dissection. The biopsy results indicate that tumor cells have migrated to the lymph nodes. Her oncologist recommends chemotherapy including the agent paclitaxel (Taxol), which blocks microtubule depolymerization, thereby interfering with the cells' ability to traverse the cell cycle. During which phase of mitosis and the cell cycle would this drug be most active? (A) Cytokinesis (B) Metaphase (C) Anaphase (D) Telophase (E) Interphase

1. The answer is C. The ability of paclitaxel (Taxol) to inhibit microtubule depolymerization results in the cells being unable to complete anaphase. The cells complete metaphase, but the spindle fibers cannot shorten to allow anaphase to initiate. The replicated chromatids are unable to move to opposite ends of the cell for daughter cell formation. Metaphase occurs when the sister chromatids align on the equatorial plane. Telophase and cytokinesis are not completed if anaphase does not occur. Very few chemotherapy agents work during interphase (the portion of the cell cycle when the cell is preparing to divide).

1. A 73-year-old woman is admitted to the intensive care unit for septic shock from a urinary tract infection. The critical care fellow is concerned she may not have an appropriate stress response and orders a cosyntropin test. Which hormone does this test evaluate? (A) Oxytocin (B) Vasopressin (C) Cortisol (D) Corticotropin-releasing hormone (CRH) (E) Adrenocorticotropic hormone (ACTH)

1. The answer is C. The primary stress hormone in the body is cortisol. Normally, CRH from the hypothalamus stimulates the release of ACTH from the anterior pituitary. ACTH then acts on the adrenal gland to produce cortisol. Cosyntropin is a synthetic ACTH injected to stimulate the release of cortisol to evaluate an appropriate stress response. Thus, a cosyntropin test is measuring the release of cortisol. Oxytocin is involved in labor during birth and milk ejection afterward. Vasopressin comes from the posterior pituitary and controls overall volume status and blood pressure.

1. A 23-year-old man presents to his family physician with a painless swelling of his testicles. An ultrasound is suspicious for a neoplasm, and a biopsy confirms the presence of cancer. He is referred to an oncologist, who begins treatment with the topoisomerase inhibitor etoposide. The normal function of this enzyme is to do which of the following? (A) Repair nuclear DNA in the event of DNA damage (B) Unwind the DNA helix during replication (C) Break and rejoin the DNA helix during replication (D) Prevent the single strands of DNA from reannealing during replication (E) Synthesize RNA primers for DNA polymerase

1. The answer is C. Topoisomerase creates double-stranded breaks ahead of the replication fork to relieve the supercoiling induced by the action of helicase, which unwinds the DNA helix during replication. DNA polymerase and ligase, along with specific endonucleases, are important in repairing DNA damage. Single-stranded binding proteins prevent the separated strands from reannealing during replication. Primase is required to synthesize primers for DNA replication by DNA polymerase.

1. A 41-year-old woman presents with severe, sharp epigastric abdominal pain that radiates to her back and with nausea and vomiting. Laboratory results indicate a serum triglyceride level of 5000 mg/dL. She is diagnosed with pancreatitis, in part owing to her elevated serum triacylglycerol levels. To form triacylglycerol from diacylglycerol, which of the following compounds is also required? (A) Glycerol (B) Glycerol 3-phosphate (C) Fatty acyl CoA (D) Acetyl CoA (E) Malonyl CoA

1. The answer is C. Triacylglycerol is formed when a diacylglycerol reacts with a fatty acyl CoA. Glycerol and glycerol 3-phosphate form the backbone of the triacylglycerol. Acetyl CoA and malonyl CoA are involved in fatty acid synthesis, and not directly in triacylglycerol synthesis.

1. A 12-year-old boy presents with fatigue, polydipsia, polyuria, and polyphagia. A fingerstick glucose measurement shows a glucose level of 350mg/dL in his serum. He is diagnosed with type 1 diabetes mellitus, a disease characterized by a deficiency of insulin.Which one of the following ismost likely occurring in this patient? (A) Increased fatty acid synthesis from glucose in liver (B) Decreased conversion of fatty acids to ketone bodies (C) Increased stores of triacylglycerol in adipose tissue (D) Increased production of acetone (E) Chronic pancreatitis

1. The answer is D. A decreased insulin-to-glucagon ratio leads to a decrease in fatty acid synthesis and an increase in adipose triacylglycerol degradation, leading to fatty acid release into the circulation. The liver takes up the fatty acids, and within the mitochondria, fatty acids undergo b-oxidization. As acetyl CoA accumulates, the ketone bodies, acetoacetate and b-hydroxybutyrate, are formed and are released into the circulation. These ketone bodies are used to fuel the heart, brain, and muscle. Nonenzymatic decarboxylation of acetoacetate forms acetone, which can be smelled by some providers on the breath of patients in diabetic ketoacidosis. Because triglycerides are degraded under these conditions, there is not an increase in triglyceride storage. Pancreatitis does not result from an inability to produce insulin.

1. An 8-year-old boy of Ashkenazi Jewish descent presents with bone pain and easy bruising. His parents have no known history of serious medical ailments. He is found to have hepatosplenomegaly (enlarged liver and spleen), anemia, and an Erlenmeyer flask deformity of his distal femur. Which of the following is true for this patient? (A) The likelihood of another, to be born, sibling being affected with the disease is 50%. (B) At least one person in every generation of his pedigree is likely to be affected. (C) His disease will only affect males. (D) The probability that the patient's unaffected siblings will be heterozygous carriers is 67%. (E) His parents will become affected with the disease.

1. The answer is D. The patient is exhibiting the signs of Gaucher disease, an autosomal recessive disorder. This means that each of the child's parents are carriers of Gaucher, and they have a 25% chance of having an affected offspring (not 50%, so answer A is incorrect). In an autosomal recessive pedigree, the probability of an unaffected sibling of an affected child being a carrier is 67% (of the four genetic possibilities for a child, the one chance of being affected is eliminated, leaving three possibilities; two of those three are carrier status, and the other is noncarrier, nonaffected status). Because Gaucher is an autosomal recessive disorder, generations can be skipped in terms of an individual expressing the disease. In autosomal dominant disorders, it is likely that at least one person in every generation of the pedigree is affected (answer B is incorrect). Autosomal recessive disorders are not X-linked (the term autosomal refers to chromosomes 1 to 22, not the X and Y chromosomes, which are the sex chromosomes), so there is an equal probability of males and females being affected (answer C is incorrect). Carriers of autosomal recessive disorders do not develop the disease later in life (answer E is incorrect).

10. A 43-year-old alcoholic man has been taking the drug cimetidine for gastric reflux. His primary care physician warns that this is not a good idea given his poor liver function and decreased ability for glucuronidation. Glucuronidation involves the addition of a carbohydrate molecule that has been derived from glucose, by which of the following mechanisms? (A) Oxidation (B) Sulfation (C) Reduction (D) Phosphorylation (E) Mutarotation

10. The answer is A. Glucuronidation makes the drug more water soluble and, therefore, more easily secreted by the kidneys. Glucuronic acid is derived from glucose via oxidation of the oxygen on carbon 6 of glucose. Sulfated sugars are found in glycosaminoglycans. Reduction of glucose at carbon 1 forms sorbitol, whereas phosphorylation of glucose (usually at position 6) traps glucose within the cell and commits it to metabolism. Mutarotation occurs when a-glucose is converted to b-glucose, a process that requires passage through a straight-chain aldehyde.

Questions 1-4: Using answers choices (A) through (D) below, match the appropriate vitamin with the clinical vignette. (A) A postpartum woman from a rural Appalachian community recently gave birth to a baby boy with the aid of a midwife at home. She now brings the baby to the hospital because of continued bleeding and oozing from the umbilical stump. (B) A child with bowing of the long bones and growth in the lowest fifth percentile (C) A child with dark, purplish spots on the legs, bleeding gums and gingivitis with tooth loss, epistaxis, and profuse diarrhea (D) A child from a developing country with visual impairment and poor skin healing 1. Vitamin A (A) (B) (C) (D) 2. Vitamin C (A) (B) (C) (D) 3. Vitamin D (A) (B) (C) (D) 4. Vitamin K (A) (B) (C) (D)

1. The answer is D. Vitamin A is required for formation of the visual pigments and maintenance of epithelial tissues. The child described in this answer choice is exhibiting the symptoms of vitamin A deficiency. 2. The answer is C. Vitamin C is required for the hydroxylation of proline and lysine residues in collagen. The lack of hydroxylation leads to altered connective tissue formation. Vitamin C deficiency results in scurvy, which is depicted in the clinical vignette of answer choice C. 3. The answer is B. Vitamin D stimulates calcium uptake from the intestine and resorption from bone and urine. The lack of vitamin D will lead to rickets. The child described in the vignette has rickets as evidenced by the poor skeletal development and growth. 4. The answer is A. Vitamin K is required for the g-carboxylation of coagulation factors II, VII, IX, and X. In the absence of this modification, the clotting factors cannot bind to developing clots. Bleeding problems result from vitamin K deficiency, as described in the clinical vignette.

1. A 56-year-old man with long-standing, poorly controlled diabetes visits his primary care physician for a follow-up after a recent hospitalization. The patient experienced an episode of acute renal failure while in the hospital, and his creatinine level rose to 3.4 (normal, 0.7 to 1.5). Creatinine, a marker of kidney function, is produced from which of the following precursors? (A) Glutamine, aspartic acid, and CO2 (B) Glutamine, cysteine, and glycine (C) Serine and palmityl CoA (D) Glycine and succinyl CoA (E) Glycine, arginine, and SAM

1. The answer is E. Creatinine is formed from the cyclization of creatine phosphate, which is formed from glycine, arginine, and SAM. Glutamine, aspartic acid, and CO2 are involved in the synthesis of purines and pyrimidines. Glutamine, cysteine, and glycine form the antioxidant molecule glutathione. Serine and palmityl CoA form sphingosine. Glycine and succinyl CoA are the precursors to the formation of heme.

1. A newborn undergoes a physical examination relevant for hepatomegaly, inguinal hernia, and deformed chest (pectus carinatum). A family history ofmucopolysaccharidosis (MPS) leads you to check enzyme activities from a sample of fibroblasts. The findings were significant for decreased activity in b-glucuronidase, which is indicative of which of the following syndromes? (A) Hurler syndrome (MPS type I) (B) Morquio syndrome (MPS type IV) (C) Hunter syndrome (MPS type II) (D) Sanfilippo A syndrome (MPS type III) (E) Sly syndrome (MPS type VII)

1. The answer is E. Sly syndrome is one of the few lysosomal storage disorders with clinical manifestations in utero or at birth. The signs of coarse facial feature (gargoyle facies), mental developmental problems, and short stature can be seen in Sly syndrome as well as all the mucopolysaccharidoses. Hurler syndrome is due to a lack of a-L-iduronidase; Morquio syndrome to a lack of galactose 6-sulfatase; Hunter syndrome to a lack of iduronate sulfatase; and Sanfilippo A syndrome to a lack of heparan sulfamidase.

1. A 5-year-old mentally retarded child is seen by an ophthalmologist for ''blurry vision.'' Ocular examination demonstrates bilateral lens dislocations, and further workup is significant for osteoporosis and homocystinuria. Serum analysis would most likely show an elevation of which of the following substances? (A) Cystathionine (B) Valine (C) Phenylalanine (D) Tyrosine (E) Methionine

1. The answer is E. The child has homocystinuria, a deficiency of cystathionine b-synthase, which manifests with mental retardation, osteoporosis, and lens dislocations. This enzyme is responsible for the metabolism of sulfur-containing amino acids and normally catalyzes the conversion of homocysteine to cystathionine. When the enzyme is defective, homocysteine can dimerize via disulfide bond formation, generating homocystine. Another fate of homocysteine is remethylation to methionine, which can also accumulate in this disorder. Because cystathionine cannot be formed under these conditions, it will not accumulate. Valine, phenylalanine, and tyrosine are not associated with the defective pathway, and their blood levels remain normal.

10. A 25-year-old woman presents with a low red blood cell count, corneal opacities, and kidney insufficiency. She is diagnosed with LCAT deficiency. LCAT is involved in which of the following processes? (A) Converting cholesterol to cholesterol esters (B) The transfer of cholesterol esters from HDL to other lipoproteins (C) Endocytosis of HDL particles into hepatocytes (D) Hydrolysis of HDL (E) Decreased uptake of cholesterol by hepatocytes

10. The answer is A. LCAT converts cholesterol to cholesterol esters, which accumulate in the core of HDL. This is an important part of reverse cholesterol transport. These cholesterol esters are transferred from HDL to VLDL and LDL (via the cholesterol ester transfer protein reaction), which are then taken up by receptors in the liver. Therefore, LCAT is important for removing cholesterol from peripheral cells by inducing esterification of cholesterol, thus allowing eventual cholesterol uptake by the liver. LCAT is not involved in endocytosis of HDL into hepatocytes, in the transfer of cholesterol esters from HDL to other lipoproteins, or in hydrolysis of HDL

10. A 6-year-old child has been suffering from muscle weakness that has progressively worsened over the past 6 months. Measurement of oxygen consumption with mitochondria isolated from a muscle biopsy revealed normal rates of succinate oxidation but very poor rates of pyruvate oxidation. Assays conducted on extracts of the mitochondria revealed normal malate dehydrogenase and PDH activities. The patient may have a mutation in a mitochondrial gene encoding a subunit of which of the following? (A) Complex I (B) Complex II (C) Complex III (D) Complex IV (E) ATP synthase

10. The answer is A. PDH activity was normal, which produces NADH, but oxidation of pyruvate was impaired, which suggests a defect in the sequence of electron flow from complex I, to coenzyme Q, to complex III, then IV, then to oxygen. Because succinate oxidation was normal, a defect in complex I is strongly suspected. Succinate oxidation transfers electrons from complex II to coenzyme Q, then to complex III, on to complex IV, and finally on to molecular oxygen, generating water. Because succinate oxidation is normal, electron flow through complexes II, III, and IV is normal. This localizes the defect to electron flow within complex I. If the ATP synthase were defective, neither pyruvate or succinate oxidation would be normal because oxidation and phosphorylation are coupled. The mitochondrial genome encodes subunits of complexes I, III, and IV and the ATP synthase, but not complex II.

10. An 18-year-old man is involved in a severe motor vehicle crash. He is rushed to surgery for an emergent laparotomy. Unfortunately, he undergoes a complete pancreatectomy, which will have all of the following consequences except for which one? (A) Failure to convert pepsinogen into pepsin (B) Diabetes (C) Increased likelihood of duodenal ulcer formation (D) Decreased synthesis of trypsinogen and chymotrypsinogen (E) Steatorrhea

10. The answer is A. Patients who undergo a pancreatectomy suffer from many debilitating conditions, including diabetes due to the inability to secrete or produce insulin, increased likelihood of duodenal ulcers from the absence of bicarbonate secretion, and decreased ability to absorb fats (due to the lack of pancreatic lipase), resulting in steatorrhea. Trypsinogen and chymotrypsinogen are secreted by the pancreas and will not be present in this patient, resulting in poor protein digestion. Pepsinogen is activated to pepsin in the stomach by the release of HCl by the parietal cells, which is not affected by the removal of the pancreas

10. A 16-year-old African-American girl comes to the emergency room with complaints of painful muscle cramps. She states that she has sickle cell anemia and that she ran out of her pain medication. A complete blood count and smear rapidly confirm the diagnosis, and she is started on intravenous fluids and pain medications. The molecular defect underlying her disease is which of the following? (A) A valine rather than a glutamate at position 6 of the b-globin protein (B) A glutamate rather than a valine at position 6 of the b-globin protein (C) A valine rather than a glutamine at position 6 of the a-globin protein (D) A glutamine rather than a valine at position 6 of the a-globin protein (E) Expansion of a polyglutamine repeat within the b-globin gene

10. The answer is A. Sickle cell anemia results from a single mutation in the b-globin subunit of hemoglobin, in which a valine rather than a glutamate is present at position 6. There are other abnormalities of hemoglobin that affect the a chain as well as the b chain. Expansion of a polyglutamine repeat occurs in the huntington protein, resulting in Huntington disease, a progressive movement disorder, but is not observed in sickle cell disease.

10. A second-year medical student is working in a laboratory studying gene regulation in a mouse model of hepatocellular carcinoma. He isolates nucleic acids from the cells after exposure to a known carcinogen and has the sequences analyzed. He is surprised to find that some of the nucleotides are pseudouridine and ribothymidine. Which type of nucleic acid has the student likely isolated? (A) tRNA (B) rRNA (C) hnRNA (D) mRNA (E) snRNP

10. The answer is A. tRNAs often contain post-transcriptionally modified bases, including the conversion of uridine to pseudouridine and ribothymidine. rRNAs are found within ribosomes and do not contain these modified bases. snRNPs (small nuclear RNAs complexed with protein) are involved in splicing and also lack ribothymidine and pseudouridine. hnRNA is processed to produce mRNA, which also lacks these modified bases.

10. A 65-year-old man with a long history of uncontrolled diabetes presents to his physician after failing the driver's license renewal eye examination. Despite having abnormally high blood glucose levels and a hemoglobin A1C of 10.4, the patient wants a more specific explanation. You begin by explaining that glucose enters the lens of the eye, where it is converted to sorbitol. Which of the following converts glucose to sorbitol? (A) Hexokinase (B) Aldose reductase (C) Aldose mutase (D) Sorbitol dehydrogenase (E) Aldose oxidase

10. The answer is B. Glucose enters tissues such as nerves, kidney, and the lens of the eye by an insulin- independent mechanism. In these tissues, glucose is reduced to sorbitol by aldose reductase. The damage to these tissues is believed to be due to an osmotic effect because the sorbitol is unable to escape these tissues readily. Sorbitol dehydrogenase converts sorbitol to fructose. Hexokinase phosphorylates glucose to glucose 6-phosphate. There are no enzymes named aldose oxidase and aldose mutase.

10. A 12-year-old Jamaican boy presents with intractable vomiting, abdominal pain, and lethargy and is profoundly hypoglycemic. His symptoms are caused by Jamaican vomiting syndrome, a sickness caused by ingestion of hypoglycin, which is present in unripe ackee fruit. Hypoglycin is metabolized to a form of nonmetabolizable carnitine, which interferes with normal fatty acid oxidation. What is the primary role of carnitine? (A) Activates long-chain fatty acids in the cytosol (B) Transport of acyl groups across the inner mitochondrial membrane (C) Is converted to enoyl CoA (D) Is converted to b-hydroxyacyl CoA (E) Is involved in breakdown of even-chain, but not odd-chain, fatty acids

10. The answer is B. In the outer mitochondrial membrane, carnitine reacts with fatty acyl CoA to form fatty acyl carnitine, which can then pass to the inner mitochondrial membrane. Therefore, carnitine is important for the transport of fatty acyl CoA from the cytosol to the mitochondria and allow for b-oxidation to occur. Carnitine is not involved in activation of fatty acids or b-oxidation itself (which eliminates answer choices A, C, D, and E). As a note of interest, hypoglycin leads to inhibition of gluconeogenesis (due to a lack of fatty acid oxidation, leading to low ATP and acetyl CoA levels). Profound hypoglycemia results, which is how hypoglycin was named.

10. An otherwise healthy 19-year-old man recovering from a respiratory infection sees his family physician. His examination is unremarkable except for a slight degree of yellow discoloration to his skin and eyes. Laboratory tests are ordered that reveal a mild increase in unconjugated bilirubin but no other abnormalities. Which of the following is the most likely diagnosis in this patient? (A) Crigler-Najjar syndrome, type I (B) Crigler-Najjar syndrome, type II (C) Gilbert syndrome (D) Lead poisoning (E) Erythropoietin deficiency

10. The answer is C. This patient has Gilbert syndrome, which is a common disorder that manifests with mild jaundice as a result of decreased bilirubin UDP-GT activity. Crigler-Najjar syndrome also results from a deficiency of the same enzyme, although it is far rarer and, in the case of type I disease, is lethal. Lead poisoning would lead to anemia, not jaundice. Erythropoietin deficiency is seen in patients with renal failure because erythropoietin is normally produced by the kidney

10. A fourth-year medical student does an international rotation in Sub-Saharan Africa. While immunizing children against polio, he sees hundreds of malnourished children in refugee camps with bloated-appearing abdomens. He learns that they are severely protein deficient because they are fed a diet of cornmeal that is provided by international relief agencies. These children likely suffer from which of the following? (A) Marasmus (B) Anorexia nervosa (C) Bulimia (D) Kwashiorkor (E) Cachexia

10. The answer is D. Protein deficiency (but not overall calorie deficiency), as in kwashiorkor, results in a deficiency of visceral proteins, including those in the blood that normally provide oncotic pressure to retain fluid within vessels. As such, patients with kwashiorkor have abdominal bloating secondary to edema. Marasmus is a deficiency of calories and protein; the children in this question are presumably receiving carbohydrate calories through the cornmeal. Anorexia nervosa and bulimia are disorders of self-induced weight loss that are found mostly in developed countries. Cachexia is weight loss associated with cancer.

10. A 75-year-old man complains of increased urinary frequency, especially at night. He has difficulty starting his stream (hesitancy) and often dribbles urine when he finishes. His urologist suspects BPH and places him on a 5-areductase inhibitor. This drug would decrease which of the following? (A) Conversion of cAMP to AMP (B) Release of calcium from the endoplasmic reticulum (C) Prostaglandin synthesis (D) Conversion of angiotensin I to angiotensin II (E) Conversion of testosterone to DHT

10. The answer is E. DHT is a stimulatory growth factor for prostate cells, and a 5-a-reductase inhibitor would decrease the conversion of testosterone to DHT. Reducing DHT levels shrinks the prostate, thereby reducing the patient's symptoms. Caffeine inhibits cAMP phosphodiesterase, which converts cAMP to AMP. IP3, derived from the cleavage of PIP2 by phospholipase C, is the signal that releases calcium from the endoplasmic reticulum to the cytoplasm. 5-a-Reductase inhibitors do not interfere with the phospholipase C pathway. Prostaglandin synthesis can be blocked by cyclooxygenase inhibitors (such as aspirin), or by lipocortin, an inhibitor of phospholipase A2. ACE inhibitors, which are used in the control of blood pressure, are used to inhibit the conversion of angiotensin I to angiotensin II.

10. Emergency medical services are called to the scene of a diabetic patient who has collapsed and is in a confused state. The patient uses an insulin pump, which appears to have malfunctioned. The patient's blood sugar is found to be 12 mg/dL, and the squad is having difficulty getting intravenous access to administer intravenous glucose. On the way to the hospital, the squad administers an intramuscular injection of glucagon. Which one of the following statements is true regarding the use of glucagon? (A) It is synthesized in the liver. (B) It inhibits gluconeogenesis. (C) It is secreted in the presence of somatostatin. (D) It is secreted in the presence of insulin. (E) It inhibits pyruvate formation and is also used in the treatment of b-blocker overdose.

10. The answer is E. Glucagon is an amino acid hormone that instructs the liver to secrete glucose, obtained via glycogenolysis and gluconeogenesis. Glucagon secretion from the pancreas (the organ that synthesizes glucagon) is inhibited by insulin and somatostatin. Glucagon is used in b-blocker overdose and exerts its beneficial effects by increasing the concentration of cAMP in the myocardium, through activation of adenylate cyclase.

Questions 5-11: Using answers (A) through (G) below, match the correct drug with its clinical effect 10. Nitroprusside and isosorbide dinitrate (A) Folic acid analog(s) that inhibit(s) dihydrofolate reductase, which leads to an inhibition of purine and deoxythymidine synthesis (B) Folate analog(s) that bind(s) specifically to bacterial dihydrofolate reductase and used in conjunction with sulfonamides (C) Histamine and basophil inhibitor(s) used to treat allergic reactions (D) Increase(s) synaptic serotonin concentration in the treatment of depression (E) Inhibitor(s) of tyrosine residue iodination (F) Replenish(es) glutathione levels during acetaminophen toxicity (G) Release(s) NO, create(s) smooth muscle relaxation, and very effective as antihypertensive

10. The answer is G. Nitroprusside and isosorbide dinitrate are compounds that decompose and release NO, which leads to smooth muscle relaxation. NO stimulates guanylate cyclase in smooth muscle cells, leading to an increase of cGMP levels. The elevated cGMP results in smooth muscle relaxation. This serves as an effective antihypertensive mechanism, and these medications are used in the treatment of hypertensive emergencies.

11. A 33-year-old triathlete is admitted to the hospital after he spent the whole day training. He looks ill and complains of diffuse weakness, fatigue, and myalgia. Laboratory tests are sent for analysis, and his lactate level is elevated, creatinine is elevated (suggesting acute renal failure), creatine kinase is 76,000, and urine tests positive for myoglobin. You determine he has rhabdomyolysis and treat him with aggressive intravenous hydration. The basis for the elevated lactate is which one of the following? (A) An increase in ATP due to the lack of oxygen for the muscle (B) An increase in NADH due to the lack of oxygen for the muscle (C) A defect in the Mform of lactate dehydrogenase (D) A defect in the H form of lactate dehydrogenase (E) A defect in the B form of muscle aldolase

11. The answer is B. Because of the intensity of the patient's training, oxygen delivery to the muscle lagged behind the need to produce ATP, so anaerobic glycolysis was providing the majority of energy (ATP) formation. This leads to elevated NADH, which is converted back to NAD+ by the lactate dehydrogenase reaction. Because the patient was exercising, ATP levels in the muscle are low, and ADP and AMP levels increase. There is no indication for a change either in the muscle (M) or in the heart (H) forms of lactate dehydrogenase. Muscle expresses the A form of aldolase, but not the B form (which is expressed in the liver).

11. A 58-year-old man develops progressive lower extremity weakness, slurring of his words, and weakness of his hands. A neurologist performs a thorough workup, confirming the diagnosis of ALS. The patient recalls that his father had similar symptoms and eventually died of respiratory failure. Some patients with the familial form of ALS have a defect in the enzyme that normally catalyzes which of the following reactions? (A) The conversion of peroxide to water and oxygen (B) The conversion of superoxide to hydrogen peroxide and water (C) The conversion of carbon tetrachloride to the CCL3Æ radical (D) The regeneration of oxidized hemoglobin (methemoglobin) (E) The conversion of carbonic acid to carbon dioxide and water

11. The answer is B. Some patients with the familial form of ALS have mutations in the enzyme SOD, which normally catalyzes the conversion of superoxide to hydrogen peroxide and water. Catalase, in turn, converts the hydrogen peroxide to water and oxygen. Carbon tetrachloride (CCl4), used in the dry cleaning business, is converted to the hepatotoxic free radical CCL3Æ by the cytochrome P-450 system. Carbonic anhydrase stimulates the conversion of carbonic acid to carbon dioxide and water. Methemoglobin reductase, which utilizes NADPH, can reduce oxidized hemoglobin (methemoglobin, in which the iron is in the +3 state) to its normal state (in which iron is in the +2 state). Glutathione can also be used to regenerate hemoglobin from methemoglobin.

11. A 65-year-old man expresses his concern about developing colon cancer to his physician. He states that his father and grandmother died of colon cancer. After a normal physical examination, negative Hemoccult tests, and unremarkable colonoscopy, the physician recommends a diet low in fat and high in fiber. The doctor explains that studies have shown a decreased risk for colon cancer with insoluble fiber consumption. Which of the following glycosidic bonds prevents humans from completely digesting fiber? (A) Glucose alfa (1---4) glucose (B) Glucose alfa (1---6) glucose (C) Glucose beta (1----4) glucose (D) Glucose alfa (1---- 2) fructose (E) Galactose beta (1---4) glucose

11. The answer is C. The glucose b (1fi4) glucose glycosidic bond, found in cellulose, a major component of fiber, cannot be digested by human enzymes. Humans can, however, digest glucose a (1fi4) glucose (starch, glycogen,maltose) bonds, galactose b (1fi4) glucose (lactose) bonds, glucose (a 1fi2) fructose (sucrose) bonds, and glucose a (1fi6) glucose (glycogen, isomaltose) bonds.

11. A young girl presents to the physician's office for a sports physical before participation in volleyball. She appears to have a perfect habitus for the sport because she is much taller than her peers and has exceptionally long arms and fingers. Auscultation of her heart reveals a midsystolic click. The physician suspects she may have Marfan syndrome, which is a defect in which of the following proteins? (A) Myosin heavy chain (B) Spectrin (C) Ankyrin (D) Fibrillin (E) Collagen

11. The answer is D. Marfan syndrome presents similarly to the case described and results from an autosomal dominant mutation in the structural protein fibrillin. Familial hypertrophic cardiomyopathy is associated with defects in myosin heavy chain. Hereditary spherocytosis is associated with mutations in spectrin or secondary defects in ankyrin. Collagen defects are seen in Ehlers-Danlos syndrome and vitamin C deficiency. Fibrillin is an essential component of microfibrils. The main organ systems affected in Marfan syndrome are the muscuoloskeletal system (arachnodactyly, dolichostenomelia, scoliosis), cardiovascular system (acute aortic dissection, mitral valve prolapse), pulmonary system (spontaneous pneumothorax), and eyes (lens subluxation, decreased nighttime vision

Questions 5-11: Using answers (A) through (G) below, match the correct drug with its clinical effect 11. N-acetylcysteine (A) Folic acid analog(s) that inhibit(s) dihydrofolate reductase, which leads to an inhibition of purine and deoxythymidine synthesis (B) Folate analog(s) that bind(s) specifically to bacterial dihydrofolate reductase and used in conjunction with sulfonamides (C) Histamine and basophil inhibitor(s) used to treat allergic reactions (D) Increase(s) synaptic serotonin concentration in the treatment of depression (E) Inhibitor(s) of tyrosine residue iodination (F) Replenish(es) glutathione levels during acetaminophen toxicity (G) Release(s) NO, create(s) smooth muscle relaxation, and very effective as antihypertensive

11. The answer is F. N-acetylcysteine is used as a precursor for glutathione synthesis. Increasing glutathione levels is essential to adequately treat acetaminophen overdose. Glutathione plays a major role in detoxifying excessive acetaminophen, and maintaining adequate levels is critical to avoid hepatotoxicity.

12. A newborn girl is delivered after her mother had an uncomplicated 9-month pregnancy. The family is concerned because their 10-year-old son has been diagnosed with cystic fibrosis and has already developed several severe pulmonary infections requiring hospitalization. They request that their pediatrician order a sodium chloride sweat test to determine whether their newborn daughter has the disease. The disease is due to a defect in which of the following? (A) A peripheral membrane protein (B) A transmembrane protein (C) Increased cholesterol content of the lipid bilayers (D) An enzyme (E) The ability to glycosylate ligands for selectins

12. The answer is B. The protein involved in cystic fibrosis is the cystic fibrosis transmembrane conductance regulator (CFTR), encoded by the CRFT gene. About 90% of cystic fibrosis patients in the American Caucasian population have a particular mutation known as Delta-F508. Delta-F508 refers to the loss of three nucleotides from the CFTR gene, at codon 508, which codes for a phenylalanine (F) residue. Thus, the protein produced is missing this critical phenylalanine in the primary structure. Individuals who inherit two copies of this mutation frequently die from respiratory failure secondary to repeat pulmonary infections and buildup of thick, tenacious mucus in the respiratory passages. There are numerous diseases associated with defects in enzymes, particularly those of key metabolic enzymes. An increased cholesterol content of lipid bilayer membranes can result in spur cell anemia. Defects in the ability to glycosylate ligands for selectins are found in the disorder leukocyte adhesion deficiency II.

12. A newborn is seen by a neonatologist for failure to thrive. The nurses in the pediatric intensive care unit note that the child has diarrhea every time he feeds. In addition, laboratory studies suggest severe hypoproteinemia. The neonatologist sends for a panel of tests, which indicate a congenital deficiency of enterokinase. This enzyme normally activates which zymogen? (A) Procolipase (B) Procarboxypeptidase (C) Trypsinogen (D) Proelastase (E) Chymotrypsinogen

12. The answer is C. Enterokinase (also called enteropeptidase) cleaves the zymogen trypsinogen to the active enzyme trypsin. Trypsinogen is also capable of autoactivation. Trypsin, in turn, cleaves the zymogens proelastase to elastase, chymotrypsinogen to chymotrypsin, procarboxypeptidase to carboxypeptidase, and procolipase to colipase.

12. A pediatric hematologist sees an 18-monthold patient with jaundice, splenomegaly, and hemolytic anemia. A blood smear indicates RBCs that are more rigid in appearance than normal, and a diagnosis of pyruvate kinase deficiency is made. Because pyruvate kinase catalyzes the last step in the glycolytic pathway, products before this step of the pathway will accumulate. Which one of the following products associated with the pathway will be made in abnormal amounts? (A) Acetyl CoA (B) Glucose (C) 2,3-Bisphosphoglycerate (D) OAA (E) Pyruvate

12. The answer is C. Pyruvate kinase converts PEP to pyruvate. Because this reaction has reduced activity, PEP accumulates, which leads to a buildup of 2-phosphoglycerate (2-PG), 3-phosphoglycerate (3-PG), and the intermediate required to convert 2-PG to 3-PG, 2,3-bisphosphoglycerate. RBCs contain no mitochondria, so pyruvate cannot be converted to acetyl CoA or OAA (and, in addition, pyruvate levels are low because of the inherited defect in pyruvate kinase). RBCs lack glucose 6-phosphatase and cannot produce glucose from glucose 6-phosphate, even though glucose 6-phosphate does accumulate in this disorder owing to the block in the glycolytic pathway.

12. A 47-year-old woman passes out at work and is brought to the emergency room. Her blood sugar is 24 mg/dL (normal, >70 mg/dL). After her sugar is corrected, she is able to tell you that she has had previous similar episodes and has been dealing with palpitations, one isolated seizure, and bouts of confusion, amnesia, and unconsciousness. Her C-peptide levels are much higher than normal. What is a possible cause of this woman's state? (A) Anorexia (B) High-protein diet (C) Exogenous administration of insulin (D) Insulin-secreting islet cell tumor (E) High-glucose diet

12. The answer is D. The pancreas synthesizes proinsulin, which is cleaved in the Golgi apparatus to form mature insulin and the C-peptide. If the patient is not eating carbohydrates, there will not be a large enough stimulus to result in a high insulin or corresponding C-peptide level. If she were self-administering pharmaceutical insulin, it would not include C-peptide; thus, her serum C-peptide level would not be elevated. Therefore, answer choice D is the most likely answer, and this patient should be evaluated for an insulinoma. Her low blood glucose is due to excessive insulin secretion resulting in the peripheral tissues removing the glucose from circulation. A high glucose diet would not lead to greatly reduced blood glucose levels.

13. A 4-year-old girl is referred to a neurologist because of progressive neurologic deficits, including increased spasticity. On further workup, a peripheral blood smear indicates nonspherocytic hemolytic anemia of Dacie type II (in vitro autohemolysis not corrected by glucose addition). A diagnosis of triose phosphate isomerase deficiency is made. On examination of muscle cells, in the initial phases of the disease, it was found there was an elevated amount of which one of the following? (A) 6-Carbon glycolytic products (B) Many 3-carbon glycolytic products (C) A component of the glycerol phosphate shuttle (D) A component of the malate-aspartate shuttle (E) Pyruvate

13. The answer is C. A defect in triose phosphate isomerase would lead to an accumulation of DHAP (the glyceraldehyde 3-phosphate produced via the aldolase reaction would be metabolized normally through the rest of the glycolytic pathway). DHAP is a component of the glycerol phosphate shuttle for transferring reducing equivalents across the mitochondrial membrane. It is not involved in the malate-aspartate shuttle. Pyruvate levels would not accumulate with this deficiency (because both pyruvate dehydrogenase and pyruvate carboxylase are working normally), nor would glucose 6-phosphate or fructose 6-phosphate accumulate in the initial phases of the disorder. Only one three-carbon product is accumulating (DHAP) in this disorder. The others can be metabolized by the active glycolytic enzymes.

13. In 1795, the British navy began to dispense limes during long sea voyages (hence the name ''limeys'' for British sailors), a measure that was largely successful in preventing scurvy. Scurvy is a condition characterized by general weakness, anemia, gum disease (gingivitis), and skin hemorrhages resulting from a lack of ascorbic acid (vitamin C) in the diet. Ascorbic acid plays a crucial role in which of the following processes in collagen synthesis? (A) Transcription (B) Glycosylation (C) Hydroxylation (D) Covalent cross-linkage (E) C-peptide cleavage

13. The answer is C. Vitamin C (ascorbic acid) is an important cofactor for the enzymes prolyl hydroxylase and lysyl hydroxylase. The hydroxylation of proline stabilizes the triple helix stricture of collagen. The hydroxyl group of lysine is often glycosylated with glucose and galactose. Cross-linkage of collagen results from the oxidation of lysine. C-peptide cleavage occurs in insulin processing

13. A 65-year-old woman is evaluated by a gastroenterologist for progressive signs and symptoms of malnutrition. Throughout her workup, she is found to have significantly decreased stomach acid. She is also found to have antibodies to gastric parietal cells, which are normally responsible for the production of acid. Why is hydrochloric acid secretion within the stomach important in digestion? (A) It stimulates the cleavage of trypsinogen to trypsin. (B) It is required for the activity of a-amylase. (C) It drives secondary active transport of amino acids. (D) It converts pepsinogen to pepsin. (E) It is required for lipid digestion.

13. The answer is D. The acidic environment of the stomach stimulates the conversion of pepsinogen to pepsin, the major proteolytic enzyme of the stomach. Patients with achlorhydria, as in this case, have deficiencies in protein digestion as well as vitamin B12 absorption (as the parietal cells, in addition to secreting HCl, also synthesize intrinsic factor). Amylase is inactivated by acid, as are lipases and other proteases. When the acidic stomach contents enter the duodenum, the pancreas is stimulated to relase bicarbonate, which aids in neutralizing the stomach acid so that the intestinal and pancreatic digestive enzymes will be active. The conversion of trypsinogen to trypsin is catalyzed by the enzyme enterokinase.

14. A social worker refers a 2-year-old child to the physician because of suspected child abuse. On physical examination, the child has blue sclerae and diminished hearing in both ears, and radiographs of the child's extremities show subacute healing long bone fractures. The child is diagnosed with osteogenesis imperfecta. This disorder results from a defect in the synthesis of which of the following proteins? (A) Transthyretin (B) Calcitonin (C) Spectrin (D) b-Mysosin heavy chain (E) Collagen

14. The answer is E. Osteogenesis imperfecta results from mutations in the gene for type I collagen. As such, the patients have multiple fractures with minimal trauma. The blue sclerae result from defects in the collagen that is found in the eye. Transthyretin forms the amyloid of familial amyloidotic neuropathies, whereas calcitonin forms the amyloid in medullary carcinoma of the thyroid. Spectrin mutations are found in hereditary spherocytosis, and b-myosin heavy-chain mutations are found in familial hypertrophic cardiomyopathy.

15. An 18-year-old boy works in the circus as a contortionist because of his hypermobile joints that allow for abnormal flexibility. He also has increased elasticity of his skin and bruises easily. This is a typical presentation of Ehlers-Danlos syndrome, resulting from a defect in collagen. Ehlers-Danlos type VI results from a lysyl oxidase deficiency. Which of the following processes will therefore be impaired in this disorder? (A) Transcription (B) Glycosylation (C) Hydroxylation (D) Covalent cross-linkage (E) Secretion into the extracellular space

15. The answer is D. The triple helix of collagen spontaneously associates into collagen fibrils, where the extracellular enzyme, lysyl oxidase, converts lysine to allysine. The newly formed residue then covalently links to the amino group of lysine on a neighboring collagen molecule, giving the fibril increased tensile strength. Lysine oxidase is not involved in the transcription, glycosylation, hydroxylation, or secretion of collagen from the cell. Lysyl hydroxylase is responsible for the hydroxylation of lysine residues in collagen. Glycosylation will occur upon the hydroxylated lysines in collagen.

16. A 24-year-old bride prepares for her wedding day. After her manicure and pedicure, her hairdresser uses rollers to create a new style for her hair. To create a ''permanent wave,'' the stylist applies thioglycollate to break apart the -S-S- bonds in the cystine units, reducing them to -SH groups. Which level of protein structure is most greatly affected by this treatment? (A) Primary structure (B) Secondary structure (C) Supersecondary structure (D) Tertiary structure (E) a-Helix formation

16. The answer is D. The tertiary structure is stabilized by covalent disulfide bonds as well as hydrophobic interactions, electrostatic interactions, and hydrogen bonds. The primary structure is composed of covalent amide bonds. The secondary, supersecondary, and a-helical structures are stabilized by noncovalent interactions. a-Helical structures are a subset of secondary structure.

17. Isoniazid is used in the treatment of tuberculosis but is is metabolized differently by different groups of patients. In comparing two groups of patients, the primary isozyme for metabolizing the drug was found to have a similar Km, but there is was a threefold difference in Vmax between the two isozymes. What are the clinical implications of the kinetic properties of these two isozymes? (A) Patients with the lower Vmax require lower doses. (B) Patients with the lower Vmax require higher doses. (C) Patients with the higher Vmax require lower doses. (D) Patients with the lower Vmax that are given higher doses have reduced toxicity. (E) Drug toxicity is not related to Vmax.

17. The answer is A. Isoniazid is acetylated in the liver in order to detoxify it and prepare it for excretion. The low Vmax enzyme acetylates the drug at a slower rate than the high Vmax enzyme (given that the Km values are the same for each enzyme). Thus, individuals with the low Vmax enzyme do not excrete the drug as rapidly as those with the high Vmax enzyme, and a lower dose of drug will be effective in those patients. If given a high dose, those patients would experience potentially toxic levels of the drug because their metabolic clearance of the drug is slow.

18. Different tissues use different isozymes to phosphorylate glucose. Both glucokinase (Km ¼ 10 mM) and hexokinase (Km ¼ 0.10 mM) convert glucose to glucose 6-phosphate. Which of the following is a correct statement based on the kinetic properties of these two enzymes? (A) Hexokinase will metabolize 100 times more glucose molecules per second than will glucokinase. (B) Hexokinase will lower the energy of activation of the reaction more than glucokinase. (C) Hexokinase will have a higher Vmax than glucokinase. (D) Hexokinase will reach Vmax at a lower glucose concentration than glucokinase. (E) Glucokinase will metabolize 100 times more glucose molecules per second than will hexokinase.

18. The answer is D. The Km is the substrate concentration at which 50% maximal velocity is reached. The Km is related to the Vmax via the Michaelis-Menton equation and is not a substitute for Vmax. Thus, because hexokinase has a lower Km for glucose than glucokinase, hexokinase will reach one-half maximal velocity at a lower substrate concentration than will glucokinase. Km is not an indicator of the energy of activation of the reaction. Because the Vmax of the enzymes is not known, one cannot determine whether answer choices A, C, and E are correct.

2. An 18-year-old woman presents with xanthomas on her eyelids and is found to have a rare genetic deficiency of lipoprotein lipase. She is diagnosed with type I hyperlipidemia. In this disorder, chylomicrons are abnormally elevated in the serum. In which cell or tissue does triacylglycerol packaging into chylomicrons occur? (A) Intestinal epithelial cell (B) Liver cell (C) Muscle cell (D) Heart cell (E) Adipose cell

2. The answer is A. Intestinal epithelial cells are the site of chylomicron formation. Dietary triacylglycerols are bound to apoproteins and other lipids to form the chylomicrons. In the liver, triacylglycerols are incorporated into VLDLs, which enter the blood. Triacylglycerols are stored in adipose tissue. The muscle, heart, and adipose cells do not package triacylglycerol into particles for export into the circulation.

2. A 2-year-old boy with Down syndrome requires intubation in the intensive care unit due to difficulty breathing. He is afflicted with congenital heart disease associated with the disease, and he dies shortly after admission. What is the most common genetic cause of Down syndrome? (A) Meiotic nondisjunction (B) Autosomal dominant inheritance (C) X-linked recessive inheritance (D) Decreased maternal age (E) Monosomy 21

2. The answer is A. Meiotic nondisjunction is the failure of chromosomes to separate and move to opposite poles of the dividing cell. This results in two copies of the same chromosome (chromosome 21 in this case) in one gamete, and nullisomy (no chromosome 21) in the other gamete. If the gamete with two copies of chromosome 21 is fertilized by a gamete with another chromosome 21, trisomy 21 will result. This is the most common cause of Down syndrome (monosomy 21 is incompatible with life). Down syndrome does not occur due to autosomal dominant inheritance. There is no dominant gene that offspring can inherit that would lead to this disorder. Because Down syndrome is due to trisomy 21, it is clearly not X-linked. The incidence of Down syndrome is essentially equal between boys and girls. Increased (not decreased) maternal age is associated with increased incidence of meiotic nondisjunction.

2. A 58-year-old woman is undergoing a myocardial infarct and is given 162 mg of aspirin, owing to the cardioprotective effects of aspirin during such an incident. Aspirin is a nonsteroidal anti-inflammatory drug that inhibits cyclooxygenase. Cyclooxygenase is required for which one of the following conversions? (A) Thromboxanes from arachidonic acid (B) Leukotrienes from arachidonic acid (C) Phospholipids from arachidonic acid (D) Arachidonic acid from linoleic acid (E) HPETEs and subsequently hydroxyeicosatetraenoic acids (HETEs) from arachidonic acid

2. The answer is A. Prostaglandins, prostacyclins, and thromboxanes are synthesized from arachidonic acid by the action of cyclooxygenase. Inhibiting cyclooxygenase decreases the synthesis of prostaglandins. Leukotriene synthesis requires the enzyme lipoxygenase. Phospholipid synthesis does not require any oxygenase reaction. The conversion of linoleic acid to arachidonic acid involves fatty acid elongation and desaturation reactions, but not the participation of cyclooxygenase. HPETE and HETE synthesis is through the leukotriene pathway, or through a cytochrome P-450-mediated pathway.

2. The major carriers of triacylglycerols are which of the following? (A) Chylomicrons and VLDL (B) IDL and LDL (C) VLDL and LDL (D) HDL and LDL (E) Chylomicrons and LDL

2. The answer is A. The major carriers of triacylglycerols are chylomicrons (synthesized in the intestine from dietary fat) and VLDL (synthesized in the liver). The triacylglycerols are digested in capillaries by lipoprotein lipase. The fatty acids that are produced are used for energy by cells or are converted back to triacylglycerols and stored. IDL and LDL are digestion products of VLDL, which have reduced amounts of triglyceride compared with VLDL. HDL has the least amount of triglyceride of any lipoprotein particle.

2. A 7-year-old child presents to the pediatrician for the development of numerous pigmented lesions on his body (cafe-au-late spots), difficulty hearing, and tinnitus. Magnetic resonance imaging confirms the presence of cerebellopontine angle acoustic neuromas, and the patient is diagnosed with neurofibromatosis-1. The defective protein in this disease normally has which of the following functions? (A) Activates the GTPase function of ras (B) Integrates signals from the extracellular environment (C) Mediates proliferation of cells during normal wound healing (D) Acts as a transcription factor (E) Is the primary regulator for the G1 to S transition of the cell cycle

2. The answer is A. The protein defective in neurofibromatosis-1 is neurofibrin, which is a GAP. GAPs function to activate the GTPase activity of ras. As the bound GTP is converted to guanosine diphosphate, ras signaling will be terminated. DCC is an example of a gene that integrates signals from the extracellular environment. DCC acts as a membrane-bound receptor, binding ligands and initiating signal transduction pathways. FGF is an example of a growth factor involved in wound healing, and aberrant expression of FGF is sometimes found in tumors. WT-1, the protein mutated in Wilms tumor, is a nuclear transcription factor. Cyclin D, which is mutated in some tumors, is the primary regulator of the G1 to S transition of the cell cycle

2. A 53-year-old man sees his family physician with concerns that his skin is ''bronzing.'' He is found to have diabetes as well as an elevated ferritin level (a sign of iron overload). The physician suspects hemochromatosis and confirms the diagnosis with genetic testing. It is found that the patient's DNA carries a mutation in which tyrosine is substituted for cysteine at position 282 (C282Y) within the HFE gene. What word below best describes this type of mutation? (A) Silent (B) Missense (C) Nonsense (D) Frameshift (E) Deletion

2. The answer is B. A missense mutation results from a nucleotide change within the DNA, which results in a change in the meaning of the codon, such that a different amino acid is placed within the protein. This is the case of the C282Y mutation in hereditary hemochromatosis. A silent mutation does not result in an amino acid change, by definition, owing to the degenerate nature of the genetic code, even though there has been a base change in the DNA (e.g., UUU and UUC both code for phenylalanine). A frameshift mutation, as in Duchenne muscular dystrophy, can result in premature termination of protein synthesis with a dysfunctional protein and requires either the insertion or deletion of 1 base or more. Deletion mutations, as in the most common mutation leading to cystic fibrosis, are due to the loss of one or more amino acids. A nonsense mutation is a single base change that converts a codon to a stop codon, leading to premature termination of protein synthesis.

2. A 75-year-old man experiences severe chest pain radiating down his left arm. He calls 911 and is transferred to the emergency room where an electrocardiogram indicates that he had a myocardial infarction. Serum levels of cardiac creatine kinase are found to be elevated. What is the biologic role of the product of this enzyme? (A) An intracellular antioxidant (B) A storage form of high-energy phosphate (C) An inhibitory neurotransmitter (D) Stimulates the release of hydrochloric acid from the stomach (E) A bactericidal product produced by macrophages

2. The answer is B. Cardiac creatine kinase phosphorylates creatine to form creatine phosphate, a source of high-energy phosphate in muscle cells. Glutathione functions as an intracellular antioxidant. GABA is an example of an inhibitory neurotransmitter. Histamine, which is derived from histidine, stimulates the release of hydrochloric acid from the stomach. Nitric oxide is one of the bactericidal substances (free radicals) produced by macrophages.

2. A 33-year-old homosexual man is recently diagnosed with HIV. His CD4+ T-cell count is dramatically decreased, and he has a high HIV viral load. He is referred to an infectious disease clinic where they begin him on a nucleoside analog. Certain nucleoside analogs inhibit DNA synthesis because they lack which of the following properties required for normal DNA polymerization? (A) A 50-phosphate (B) A 30-hydroxyl (C) A 7-methyl G modification (D) A poly(A) tail (E) A consensus sequence

2. The answer is B. DNA polymerase requires a free 30-hydroxyl group on the deoxyribonucleotide, which acts as a nucleophile to attack the 50-a-phosphate of the incoming nucleotide, which forms the phosphodiester bond. Nucleoside analogs inhibit polymerization because they lack the 30 OH for chain elongation. A 7-methyl G modification and a poly(A) tail are added to eukaryotic mRNA to stabilize the transcript, and are not found in nucleoside analogs. A consensus sequence is a DNA sequence that is found at the promoter region of genes and functions to bind various factors that regulate the transcription of the gene.

2. A 14-day-old breast-fed neonate fails to gain weight during infancy. Although concerned, the mother continues to breast-feed and wait. The infant subsequently develops cataracts, an enlarged liver, and mental retardation. Urinalysis is significant for high levels of galactose in the urine, as well as galactosemia. What food product in the baby's diet is leading to these symptoms? (A) Fructose (B) Lactose (C) Phenylalanine (D) Glucose (E) Sorbitol

2. The answer is B. This patient has classic galactosemia, resulting from the inability to process galactose once the lactose in the breast milk is cleaved to its monomers, galactose and glucose. This disease results from an autosomal recessively inherited mutation in galactose 1-phosphate uridylyltransferase. Logically, treatment is removal of lactose and galactose from the diet. The enzyme defect does not allow galactose 1-phosphate to react with UDP-glucose, leading to the accumulation of both galactose 1-phosphate and galactose. The high galactose levels lead to galactitol accumulation and cataract formation. The high galactose 1-phosphate inhibits phosphoglucomutase, which leads to the hepatomegaly.

2. A 24-year-old woman presents with diarrhea, dysphagia, jaundice, and white transverse lines on the fingernails (Mee lines). The patient is diagnosed with arsenic poisoning, which inhibits which one of the following enzymes? (A) Citrate synthase (B) Isocitrate dehydrogenase (C) Pyruvate dehydrogenase (D) Malate dehydrogenase (E) Succinate dehydrogenase

2. The answer is C. Arsenic binds the sulfhydryl groups of lipoic acid, creating an inactive 6- membered ring. Because lipoic acid is one of the three cofactors of PDH, this inactivates the enzyme's ability to synthesize acetyl CoA from pyruvate. Because a-ketoglutarate dehydrogenase also utilizes lipoic acid, its activity would also be inhibited by arsenic. Citrate synthase is not inhibited by arsenic. The other dehydrogenases listed as answer choices—isocitrate dehydrogenase, malate dehydrogenase, and succinate dehydrogenase—do not require lipoic acid and are not directly inhibited by arsenic.

2. A 47-year-old obese man complains of having to get out of bed three times a night to urinate (polyuria), being constantly thirsty (polydipsia), and eating more often (polyphagia). The patient is diagnosed with insulinresistant diabetes mellitus (type 2). If the patient's symptoms are due to a problem at the level of the glucose transporter, which one of the tissues indicated below will be most affected? (A) RBCs (B) Small intestine (C) Muscle (D) Brain (E) Liver

2. The answer is C. Both muscle and adipose tissue rely primarily on the glucose transporter GLUT-4, which requires insulin for optimal expression on the cell surface. The other glucose transporters are found on the cell surface in the absence of insulin secretion. These include GLUT-1, -2, -3, and -5. GLUT-1 is ubiquitously distributed in various tissues. GLUT-2 is present in liver and pancreatic b cells. GLUT-3 is also found in the intestine with GLUT-1. Finally, GLUT-5 functions primarily as a fructose transporter.

2. The mutation associated with Marfan syndrome is with the fibrillary protein fibrillin. What aspect of protein structure is affected in this disorder? (A) b-turn (B) b-sheet (C) Primary structure (D) Tertiary structure (E) Quaternary structure

2. The answer is C. The fibrillin mutation found in Marfan syndrome results in a defective a-helix due to an alteration in the sequence of amino acids in the protein, an altered primary structure. The tertiary and quaternary structures are not altered, nor are b-sheets or turns. Additionally, it is important to note that the main organ systems affected are the musculoskeletal system (arachnodactyly, dolichostenomelia, scoliosis), cardiovascular system (acute aortic dissection, mitral valve prolapse), pulmonary system (spontaneous pneumothorax), and eyes (lens subluxation, decreased nighttime vision).

2. A 38-year-old man gets bloated and has episodes of diarrhea after eating his favorite ice cream. It also occurs when he consumes yogurt, cheese, and other milk-containing products. The patient lacks the ability to cleave which one of the following glycosidic bonds? (A) Glucose-alfa (1----4) glucose (B) Glucose-alfa (1----2) fructose (C) Galactose-beta (1----4) glucose (D) Glucose-alfa (1---6) glucose (E) Glucose-beta (1---4) glucose

2. The answer is C. The vignette is a classic presentation of lactase deficiency. Lactase is an inestinal enzyme that cleaves the b (1----4) linkage of lactose, releasing galactose and glucose. Sucrase cleaves the glucose-a (1---2) fructose bond. Maltase cleaves glucose-a (1-----4) glucose bonds, whereas isomaltase cleaves glucose-a (1-----6) glucose bonds. Humans lack the enzymes necessary to cleave glucose b (1-----4) glucose bonds, a predominant glycosidic bond found in fiber

2. A 3-month-old child presents with vomiting and convulsions. Notable findings include hepatomegaly and hyperammonemia. A deficiency in which of the following enzymes would most likely cause an elevation of blood ammonia levels? (A) CPS II (B) Glutaminase (C) Argininosuccinate lyase (D) Asparagine synthetase (E) Urease

2. The answer is C. There are two major types of hyperammonemia: acquired and hereditary. The hereditary type can result from deficiencies of any of the five enzymes of the urea cycle, which include CPS I, ornithine transcarbamoylase, argininosuccinate synthetase, argininosuccinate lyase, and arginase. CPS II is involved in pyrimidine synthesis and utilizes glutamine as a substrate (not ammonia). Glutaminase will generate (not fix) ammonia, and therefore a loss of glutaminase activity will not increase ammonia levels. Asparagine synthetase requires glutamine as a substrate (not ammonia). Urease is a bacterial and plant enzyme that can degrade urea into ammonia; it is not present in humans.

2. A 10-year-old boy presents with vomiting, sweating, drooling, and a decreased heart rate. His friends state that he was in a corn field when it was sprayed by a crop duster. The chemical being sprayed was an organophosphate derivitive that covalently binds to acetylcholinesterase and inactivates the enzyme. What type of inhibition is being displayed? (A) Competitive (B) Noncompetitive (C) Irreversible (D) Feedback (E) Allosteric

2. The answer is C. This is an example of irreversible inhibition because a covalent bond has been formed between the inhibitor and the required serine at the active site of the enzyme. This enzyme can only be reactivated if that covalent bond is hydrolyzed, which is unlikely. Both competitive and noncompetitive binding are reversible because the inhibitor is not covalently linked to the enzyme. Allosteric inhibitors also bind to enzymes via noncovalent forces, and feedback inhibition refers to the normal regulation of a pathway by an end product of the pathway.

2. A physician evaluates a 42-year-old patient for fatigue. The patient is found to have an elevated white blood cell count and an enlarged spleen. A referral to an oncologist results in a diagnosis of chronic myelogenous leukemia. Treatment with hydroxyurea, a ribonucleotide reductase inhibitor, is begun. The normal function of ribonucleotide reductase is to catalyze which one of the following reactions? (A) Form PRPP from adenosine diphosphate (ADP) and ribose (B) Convert xanthine to uric acid (C) Form carbamoyl phosphate from glutamine, CO2, and two ATP molecules (D) Convert ADP to dADP (E) Convert guanosine to guanine and ribose 1-phosphate

2. The answer is D. Ribonucleotide reductase converts nucleoside diphosphates (NDPs) to deoxynucleoside diphosphates (dNDPs) for their use in DNA synthesis (an example is ADP to dADP). PRPP synthetase forms PRPP from ATP and ribose. Xanthine oxidase converts xanthine to uric acid. Carbamoyl phosphate is synthesized by carbamoyl phosphate synthetase II from glutamate, CO2, and two ATP molecules (carbamoyl phosphate synthetase I utilizes ammonia, carbon dioxide, and two ATP molecules as substrates). Purine nucleoside phosphorylase degrades guanosine to guanine and ribose 1-phosphate (as well as inosine to hypoxanthine and ribose 1- phosphate).

23. A 33-year-old nonalcoholic man is referred to a gastroenterologist for the management of a newly diagnosed a1-antitrypsin deficiency from a liver biopsy. This enzyme normally inhibits the enzyme elastase but also will inhibit trypsin. Trypsin will cleave proteins at which of the following sites? (A) The carboxyl side of arginine or lysine (B) The carboxyl side of aromatic amino acids (C) The carboxyl side of uncharged amino acids (D) Aromatic amino acids found at the carboxy terminus (E) Basic amino acids found at the carboxy terminus

23. The answer is A. Trypsin normally cleaves a peptide on the carboxyl side of arginine or lysine. Chymotrypsin has the ability to cleave a peptide on the carboxyl side of aromatic amino acids. Elastase is actually the major target of a1-antitrypsin, and elastase cleaves peptides on the carboxyl side of uncharged amino acids. Carboxypeptidase A cleaves aromatic amino acids from the carboxy terminus, whereas carboxypeptidase B cleaves basic amino acids from the carboxy terminus.

2. An SNP in the coding region of a tumorsuppressor gene was recently discovered. Heterozygous individuals synthesize both the normal and mutant forms of this gene. The normal form codes for a leucine at a particular position within the protein; when the SNP is present, that codon is altered to code for an arginine. Although the SNP was identified at the nucleotide level, you wish to separate and characterize 10 mg of each of the proteins. Assuming you had a mixture of the two forms of the protein, what initial approach would you use to separate these two forms? (A) Mass spectrometry (B) X-ray crystallography (C) NMR (D) Ion exchange chromatography (E) Affinity purification on a KLF-6 polyclonal antibody column

2. The answer is D. The key is isolation as well as detection. The change of leucine to arginine alters the charge properties of the protein (leucine has no charge on the side chain, whereas arginine has a +1 charge on the side chain at pH 7). This suggests that ion exchange chromatography, at the appropriate pH, can separate the two forms of the protein. Mass spectrometry can differentiate the masses but cannot obtain separated material for further analyses. Likewise, NMR and Xray crystallography might detect a difference, but would not provide material for further studies. A polyclonal antibody would not differentiate the two forms, although a monoclonal antibody, if designed properly, might separate the forms. The polyclonal antibody would likely bind to both forms of the protein with equal affinity.

2. A newborn is experiencing failure to thrive. On physical examination, organomegaly is appreciated owing to accumulation of glycogen in the lysosomes of several organs, including the heart, muscle, and liver. You diagnose the condition as Pompe disease. Which one of the following biochemical deficits is seen in this disorder? (A) A deficiency of glycogenin (B) Loss of a-1,6-glucosidase activity (C) Loss of glucose 6-phosphatase activity (D) Loss of muscle glycogen phosphorylase activity (E) Loss of a lysosomal glucosidase activity

2. The answer is E. Pompe disease results from a deficiency of a-1,4-glucosidase (acid maltase), halting the release of glucose from glycogen found in lysosomes. McArdle syndrome is caused by a deficiency in muscle glycogen phosphorylase (answer D), which also cleaves the same a-1,4- glycosidic bond but, instead, presents with muscle symptoms, such as weakness and cramps. The difference between Pompe and McArdle symptoms is that Pompe disease results in a loss of lysosomal function, but not so McArdle disease. Glycogenin initiates glycogen synthesis, and therefore, a deficiency would result in a decrease in glycogen storage. An a-1,6-glucosidase deficiency results in the inability to liberate the 1,6 branch points of glycogen as seen in Cori disease, but does not have lysosomal involvement. Glucose 6-phosphatase deficiency, or von Gierke disease, results in hypoglycemia, hepatomegaly, hyperlipidemia, hyperuricemia, gouty arthritis, nephrolithiasis, and chronic renal failure, without lysosomal involvement.

2. Sickle cell disease results in abnormal hemoglobin formation because of a point mutation in DNA that leads to the insertion of which amino acid into b-globin? (A) Glutamate (B) Glutamic acid (C) Tyrosine (D) Serine (E) Valine

2. The answer is E. Sickle cell anemia is caused by a point mutation in DNA, which leads to glutamic acid at position 6 of the b-chain of globin being replaced with the hydrophobic amino acid valine. This mutation of the b-globin gene causes the polymerization of hemoglobin under low oxygen conditions, distorting the red blood cells into an inelastic, sickle shape. The most lifethreatening manifestations of sickling, or a ''sickle crisis,'' are aplastic crisis, splenic sequestration, vaso-occlusive crisis, and acute chest syndrome.

20. You are in the laboratory and are trying to understand the interaction between lovastatin and HMG-CoA reductase. Analyses reveal that the Vmax of the reductase is unchanged in the absence or presence of the statin, yet the Km is increased. Lovastatin is most likely functioning as which of the following? (A) Competitive inhibitor (B) Noncompetitive inhibitor (C) Feedback inhibitor (D) Allosteric activator (E) Irreversible inhibitior

20. The answer is A. A competitive inhibitor will compete for binding with the substrate to the enzyme. Thus, in kinetic terms, competitive inhibitors increase the apparent Km (because it takes increased substrate to displace the inhibitor from the active site), but do not alter the maximal velocity (because at high enough substrate concentrations, the enzyme is saturated with substrate). A noncompetitive, as well as irreversible, inhibitor reduces Vmax without affecting Km. Feedback inhibition refers to an end product of a pathway inhibiting a previous step of the pathway; lovastatin is not a product or intermediate of the cholesterol biosynthetic pathway. Because the Km has been increased in the presence of lovastatin, the drug is not acting as an activator, which usually reduces the Km, which allows the reaction to proceed at lower substrate concentrations

21. A 19-year-old woman is running on a treadmill. Exercise increases her respiratory quotient. Which of the following statements is correct concerning the respiratory quotient? (A) It is higher for fat than carbohydrate. (B) It is higher for protein than carbohydrate. (C) It is the ratio of CO2 produced to O2 utilized during substrate oxidation. (D) It is increased in exercise due to enhanced fat metabolism. (E) It is increased in exercise due to enhanced amino acid metabolism.

21. The answer is C. The respiratory quotient (R.Q.) is the ratio of CO2 produced to O2 used (CO2/ O2) by a tissue in oxidation of a foodstuff. The R.Q. for fat is 0.7, 0.8 for protein, and 1 for carbohydrate. R.Q. is increased in exercise owing to enhanced carbohydrate metabolism (glycogen degradation producing glucose for energy needs) compared with fat metabolism (fatty acid oxidation) in the resting state.

22. A 25-year-old man is intubated in the intensive care unit. He is being treated for an overwhelming infection. Through a gastric tube, he is being fed proteins that are broken down to amino acids. When his dietary nitrogen intake exceeds his excreted nitrogen, this is most accurately called which of the following? (A) Negative nitrogen balance (B) Positive nitrogen balance (C) Nitrogen balance (D) Biosynthesis (E) An anabolic state

22. The answer is B. Positive nitrogen balance describes the state when dietary nitrogen exceeds excreted nitrogen. Negative nitrogen balance occurs when dietary nitrogen is less than excreted nitrogen. Nitrogen balance occurs when dietary nitrogen equals excreted nitrogen. Anabolism describes biosynthetic pathways, which require energy. Catabolism describes degradative pathways, some of which yield energy. The patient is most likely in an anabolic, biosynthetic state because of increased antibody synthesis, which requires the use of amino acids. However, the most accurate term used to describe this state is positive nitrogen balance.

24. A 27-year-old woman with end-stage cystic fibrosis has lost the function of her pancreas. Absorption of which of the following vitamins will not be affected by the loss of pancreatic exocrine function? (A) Vitamin C (B) Vitamin A (C) Vitamin D (D) Vitamin E (E) Vitamin K

24. The answer is A. Vitamin C is the only water-soluble vitamin listed and not affected by this patient's pancreatic insufficiency. Vitamins A, D, E, and K are the fat-soluble vitamins that the body needs. Absorption of these is not possible without pancreatic lipase and bile salt secretion, in order to emulsify the lipids and to allow the vitamins to be absorbed by the intestinal epithelial cells.

25. A 23-year-old man develops steatorrhea, weight loss, and a bloody diarrhea. He notes that the diarrhea is worse when he eats breads or cereals. A gastroenterologist performs a biopsy during a colonoscopy, which reveals celiac disease. This disorder is most directly due to which of the following? (A) Excess lipids in the feces (B) Deficiency of enterokinase (C) Defective transport of the amino acid cystine (D) A defect in the transport of neutral amino acids (E) Hypersensitivity to the protein gluten

25. The answer is E. Celiac disease, or nontropical sprue, results from hypersensitivity to the grain protein gluten. Biopsy of the intestine demonstrates destruction of the absorptive cells of the gut because gluten stimulates an inflammatory response in the gut. Excess lipids in the feces does result in steatorrhea, which is an effect, not a cause, of celiac disease. Defective transport of cystine in the kidney leads to kidney stones, a condition known as cystinuria. Defects in the transport of neutral amino acids results in Hartnup disease. Enterokinase activates pancreatic zymogens in the intestinal lumen so that digestion can occur. In the absence of enterokinase, protein and fat digestion would be minimal. However, a loss of enterokinase does not lead to destruction of the gut epithelial cells, as occurs in celiac disease

26. A patient comes for an annual check-up with his pediatrician. He has pyruvate kinase deficiency with the typical manifestations, including occasional anemia, a severe episode requiring a blood transfusion, and some symmetric growth delay. Which of the following statements is correct concerning this disorder? (A) The defect primarily affects erythrocytes and results in cell membrane alterations and splenic sequestration. (B) Pyruvate kinase deficiency is an autosomal dominant disorder. (C) Pyruvate kinase deficiency is the most common cause of enzyme deficiency-induced hemolytic anemia. (D) Clinical manifestations are the result of decreased plasma bilirubin. (E) In severe forms, erythrocytes must rely on aerobic mechanisms to create adenosine triphosphate (ATP).

26. The answer is A. Pyruvate kinase deficiency is an autosomal recessive disorder that primarily affects erythrocytes because they rely only on glycolysis to create ATP. With decreased ATP production, the cell is unable to maintain the biconcave cell membrane shape, and the cell is sequestered in the spleen for destruction. Moreover, with an impaired glycolytic pathway, depending on the severity of the enzyme deficiency, patients may have mild to serious manifestations. Hemolysis of defective erythrocytes results in release of bilirubin, so bilirubin levels increase, do not decrease, under these conditions. Pyruvate kinase deficiency is second to glucose 6-phosphate dehydrogenase deficiency for enzyme deficiency-induced hemolytic anemia.

27. An 84-year-old woman is in the medical intensive care unit because of septic shock. Her underlying source is pneumonia, and she is doing poorly because of advanced metabolic acidosis. Her laboratory results reveal a lactate of 12.3 (>5 indicates a severe acidosis, normal lactate is 1 mM), and the values are increasing with each assay. Which of the following statements is true concerning high lactate states? A) It reflects a state in which there is a high NADH/NAD+ ratio. (B) When lactate is high, respiratory compensation is seen with a decrease in the respiratory rate. (C) High lactate indicates an increased responsiveness to catecholamines. (D) Lactate cannot be converted to glucose. (E) Treatment for high lactate includes decreased ventilation.

27. The answer is A. Lactic acidosis is a common clinical entity and the result of tissue hypoperfusion and oxygenation. On a molecular level, this results in a high NADH/NAD+ ratio and causes shunting to anaerobic metabolism and the accumulation of lactate. In advanced cases as seen in this patient with sepsis, impending cardiopulmonary collapse is a major concern because of decreased responsiveness to catecholamines to maintain adequate blood pressures and respiratory failure from persistent respiratory compensation (tachypnea and increased minute ventilation). Mechanical ventilation, exogenous administration of catecholamines, and correction of the acidosis/volume deficits are the mainstays of treatment.

28. A 58-year-old man is taken to the emergency room after being poisoned with arsenic. Arsenic has several harmful effects, including forming an inactivating complex with a-lipoic acid in pyruvate dehydrogenase. As a result of this poisoning, how many ATP equivalents will the oxidation of glucose yield within the muscle, under aerobic conditions? (A) 1 (B) 2 (C) 4 (D) 5 (E) 6

28. The answer is D. With the arrest of the pyruvate dehydrogenase complex, glucose will only provide energy gains from glycolysis. Because the glycerol 3-phosphate shuttle is the major shuttle system in muscle, the two molecules of cytoplasmic NADH produced will yield three ATP equivalents. When added to the two ATP molecules produced at the substrate level, a total of five ATP equivalents will result

29. A child presents with developmental delay of verbal milestones, intermittent ataxia, poor muscle tone, abnormal eye movements, and seizures. The mother notes worsening of these symptoms with exertion. You are concerned that the child has pyruvate dehydrogenase complex deficiency. Which one of the following is true about this disorder? (A) The patient is unable to form acetyl coenzyme A (CoA) from carbohydrate precursors. (B) Clinical sequelae are related to an excessive production of citrate. (C) Underlying neuropathology is usually observed in individuals whose onset is in childhood. (D) Presentation and progression are predictable and uniform once the diagnosis is confirmed. (E) An absence of compensatory anaerobic metabolism is seen under low oxygen conditions.

29. The answer is A. Pyruvate dehydrogenase complex deficiency (PDCD) is one of the most common neurodegenerative disorders and is caused by abnormal mitochondrial metabolism. Specifically, an inability to convert pyruvate to acetyl CoA leads to decreased downstream synthesis of TCA substrates (e.g., citrate) and cellular conversion from aerobic metabolism to the anaerobic/ lactate pathway. Childhood-onset (as opposed to in utero) forms of this disorder result in intermittent periods of decompensation, albeit normal neurologic development. There is significant variability in the severity of the disease based on how defective the enzyme is, resulting in variable phenotypes as extreme as death within 1 year to survival to adulthood. The default pathway without aerobic metabolism (lack of oxygen) is anaerobic metabolism.

3. A 4-day-old infant develops severe jaundice and is transferred to the neonatal intensive care unit for aggressive phototherapy. He is found to have a complete loss of UDP-GT activity. The loss of this enzyme activity leads to which of the following disorders? (A) Crigler-Najjar syndrome (B) Gilbert syndrome (C) Dubin-Johnson syndrome (D) Hereditary orotic aciduria (E) Gout

3. The answer is A. Crigler-Najjar syndrome, type I, results from a complete lack of UDP-GT activity and is a lethal condition. Gilbert syndrome is a mild defect in bilirubin conjugation that is usually asymptomatic, although it is due to a subtle defect in the same enzyme. Dubin-Johnson syndrome is a transport defect in bilirubin and does not involve its conjugation with glucuronic acid. Hereditary orotic aciduria results from a defect in pyrimidine synthesis. Gout results from hyperuricemia, not hyperbilirubinemia.

3. A 53-year-old man is referred to a neurologist because he is beginning to develop spasticlike movements in his lower limbs. Magnetic resonance imaging of the head shows loss of mass in the caudate nucleus, and a presumptive diagnosis of Huntington disease is made. The genetic basis of this disease is best described as which of the following? (A) Triple repeat expansion (B) Nucleotide deletion (C) Point mutation (D) Transposition of genetic material (E) DNA methylation

3. The answer is A. Huntington disease is a neurodegenerative disorder that results from a triplet repeat expansion (CAG) within the Huntington gene. The most common mutation leading to cystic fibrosis is an example of a deletion (D508, a loss of three nucleotides). Point mutations may cause disease when they occur either in coding or noncoding regions, such as the point mutation in the a1-antitrypsin coding sequence or mutations in introns of certain types of b-thalassemia. Transposition is a phenomenon of transposons moving about the genome, but they are not responsible for Huntington disease. DNA methylation leads to the genetic phenomena of imprinting.

3. Nerve agents are toxic substances used in chemical warfare that add covalently to the enzyme acetylcholinesterase, thereby preventing the enzyme from hydrolyzing acetylcholine. This results in persistent acetylcholine in the synapse and continual muscle contractions. Which of the following substances can directly reactivate acetylcholinesterase after exposure to such a toxin? (A) Pralidoxime chloride (B) Atropine (C) Scopolamine (D) Ipratropium (E) Diphenhydramine

3. The answer is A. Pralidoxime chloride (2-PAM) is the nerve agent antidote that reactivates the poisoned acetylcholinesterase. Nerve agents (e.g., Sarin, VE) inhibit acetylcholinesterase by phosphorylation of the active site serine hydroxyl group of the enzyme. 2-PAM reactivates the cholinesterase by removing the phosphoryl group that is bound to the serine hydroxyl side chain and creates inactivated organophosphate and pralidoxime that undergo rapid metabolism and removal from the synapse. Atropine is a competitive antagonist for the muscarinic acetylcholine receptor and decreases the effects of acetylcholine. Although it is used in nerve agent poisoning, it does not directly reactivate the poisoned acetylcholinesterase. Scopolamine is an anticholinergic competitive anatagonist at muscarinic (M1) acetylcholine receptors used to treat nausea and eye conditions that require mydriasis (e.g., iritis, uveitis). Scopolamine does not reactivate inhibited acetylcholinesterase. Ipratropium is a nonselective anticholinergic that blocks muscarinic receptors in the lung and decreases bronchoconstriction and mucus secretion, but has no direct effect on acetylcholinesterase. Diphenhydramine is an antihistamine, which has no effects on acetylcholinesterase.

3. A 2-week-old child underwent complex congenital heart malformation repair. The cardiothoracic surgeon accompanies the patient back from the operating room and tells the pediatric intensive care unit staff that the ASD (atrial septal defect) and VSD (ventricular septal defect) were successfully repaired. However, the thoracic duct was accidentally cut, and daily echocardiograms will be needed to evaluate for cor pulmonale (alterations in the right ventricle of the heart). Which one of the following statements is true concerning fat metabolism in this patient? (A) The thoracic duct carries a substantial volume of lymph and triglycerides from the enteric circulation to the venous system. (B) Triacylglycerol is primarily stored in the liver. (C) In the intestinal cell, glucose is converted to triacylglycerol by phosphatidic acid. (D) In adipose cells, triacylglycerol is converted to VLDL. (E) Somatostatin has no role in the treatment of chylothorax in children.

3. The answer is A. The thoracic duct carries lymph and triglyceride from the enteric circulation to the venous system. Chylothorax is the accumulation of chylous fluid from a compromised thoracic duct. Nontraumatic causes (e.g., malignant erosion) or traumatic causes (e.g., blunt trauma, cardiothoracic surgery) result in the slow accumulation of a milky fluid rich in triglycerides in the chest cavity. Treatment is medical (somatostatin in children) and surgical (percutaneous drainage or thoracostomy tube drainage). Triglyceride is primarily stored in the adipose cells. Intestinal cells do not produce triglyceride from glucose; these cells pass glucose directly into the circulation. VLDL is produced by the liver, not adipose tissue.

3. A 67-year-old man suffers from congestive heart failure. He is taking digoxin, an effective chronotrope and inotrope, which is an ether that contains a sugar component (glycol) and a nonsugar (aglycone) component attached via oxygen. Digoxin would be best classified as which of the following? (A) Glycoprotein (B) Glycoside (C) Oligosaccharide (D) Glucosteroid (E) Thioester

3. The answer is B. Digoxin is a medication that can improve the contraction of the heart. It is a drug that has been around for centuries and is made from the foxglove plant. A glycoside is an ether containing a sugar component (glycol) and a nonsugar (aglycone) component attached via oxygen or nitrogen bond; hydrolysis of a glycoside yields one or more sugars. A glycoprotein contains sugars attached via glycosidic linkage to amino acid side chains of the protein. An oligosaccharide is the linkage of a number of sugars in glycosidic bonds. A glucosteroid is a type of steroid hormone. A thioester linkage contains a sulfur bonded to a carbon, which has a carbonyl group also attached to it.

3. The cardioprotective effects of aspirin occur due to the inhibition of the synthesis of which one of the following? (A) PGF2a (B) PGE2 (C) TXA2 (D) PGA2 (E) PGI2

3. The answer is C. Even though inhibition of cyclooxygenase leads to a decrease in synthesis of all the answers listed (PGF2a, PGE2, TXA2, PGA2, and PGI2), it is the inhibition of thromboxane A2 that is cardioprotective. TXA2 is a potent vasoconstrictor and a stimulator of platelet aggregation. The stimulation of platelet aggregation initiates thrombus formation at sites of vascular injury as well as in the vicinity of a ruptured atherosclerotic plaque in the lumen of vessels. Inhibition of TXA2 synthesis reduces the risk for thrombus formation and occlusion of a vascular vessel.

3. A 23-year-old woman develops a tumor of the soft tissue of her arm (sarcoma) and is being evaluated by an oncologist. The physician learns that her older brother had leukemia when he was 12 years old, her grandmother died of a brain tumor, and her aunt has breast cancer. With this clustering of tumors, the oncologist suspects Li-Fraumeni syndrome and orders molecular studies on the patient's tumor for which of the following genes? (A) TNF (B) p53 (C) WT-1 (D) Rb (E) RET

3. The answer is B. Li-Fraumeni syndrome is a familial DNA repair syndrome caused by mutation in the tumor-suppressor gene p53. The p53 gene is the most common genetic alteration in cancer, and patients who inherit a mutated copy of the p53 gene have an increased predisposition to soft tissue neoplasms, leukemias, central nervous system tumors, and breast cancer. TNF is a proapoptotic molecule. Mutations in TNF do not lead to Li-Fraumeni syndrome. WT- 1 is mutated in Wilms tumor, the most common malignancy of the kidney in children. The function of WT-1 is still controversial. Patients with hereditary Rb mutations develop retinoblastoma or osteosarcoma. Rb is important in regulating the E2F family of transcription factors. RET is mutated in patients, some with thyroid tumors and some with MEN. RET is a receptor tyrosine kinase

Questions 3-7: Using answers choices (A) through (E) below, match the clinical vignette with the appropriate defective or deficient enzyme. (A) Glucose 6-phosphate dehydrogenase (B) Galactose 1-phosphate uridylyltransferase (C) N-acetylglucosamine 1-phosphate transferase (D) Galactokinase (E) Fructokinase 3. An immigrant 4-month-old child who did not receive prenatal or antenatal care appears ill. The child presents with hepatomegaly, jaundice, hypoglycemia, and convulsions. Blood work shows an elevated white blood cell count of 14,000, 16% bands, and blood cultures that grow out Escherichia coli, signifying a bacterial infection.

3. The answer is B. This child is exhibiting the signs of classic galactosemia, a lack of galactose 1-phosphate udriyltransferase. The lack of this enzyme leads to hypoglycemia due to galactose 1-phosphate inhibition of phosphoglucomutase, and jaundice due to the liver's inability to conjugate bilirubin. Sepsis is a common complication of untreated galactose 1-phosphate uridylyltransferase deficiency.

3. A 55-year-old man suffers from cirrhosis of the liver. He has been admitted to the hospital several times for hepatic encephalopathy. His damaged liver has compromised his ability to detoxify ammonia. Which of the following amino acids can be used to fix ammonia and thus transport and store ammonia in a nontoxic form? (A) Aspartate (B) Glutamate (C) Serine (D) Cysteine (E) Histidine

3. The answer is B. Three enzymes can fix ammonia into an organic molecule: glutamate dehydrogenase (a-ketoglutarate plus ammonia yield glutamate), CPS I (carbon dioxide, ammonia, and two ATP molecules yield carbamoyl phosphate), and glutamine synthetase (glutamate plus ammonia plus ATP yield glutamine). Thus, of the answer choices provided, glutamate is the correct answer. Aspartate is the precursor for asparagine synthesis, but glutamine is the nitrogen donor in that reaction, not ammonia. Serine, cysteine, and histidine are not utilized for nitrogen transport.

3. A second-year medical student decides to do research in a nutrition laboratory that is studying the effects of caffeine on cellular metabolism. Caffeine inhibits cAMP phosphodiesterase. If caffeine were added to liver cells, in the presence of glucagon, which of the following enzymes would be phosphorylated and inactivated? (A) Phosphorylase kinase (B) Pyruvate kinase (C) Phosphorylase (D) Protein kinase A (E) Calmodulin

3. The answer is B. Under the conditions of the experiment, cAMP levels would be elevated by glucagon treatment and would remain elevated owing to the presence of caffeine. This leads to constant activation of protein kinase A (PKA). PKA will phosphorylate pyruvate kinase, leading to its inactivation. Phosphorylase kinase and phosphorylase are activated by phosphorylation. PKA is not regulated by phosphorylation but by dissociation of inhibitory subunits that bind to cAMP. Calmodulin is a calcium-binding protein that serves as a subunit of phosphorylase kinase, but is not phosphorylated by PKA.

3. A 42-year-old man presents to the emergency room and receives a neurologic consult for a slowly progressing chorea with some rigidity and seizures. These episodes have occurred in the past, and a deceased mother had choreic neurologic episodes and dementia. You suspect the patient might have Huntington disease, and the hospital has a laboratory equipped to perform RFLP analyses. Which of the following techniques is required to carry out RFLP analysis? (A) Western blot (B) Northern blot (C) Southern blot (D) X-ray crystallography (E) Mass spectrometry

3. The answer is C. Because RFLP analyses involve DNA digested with restriction enzymes, and the fragments resolved by gel electrophoresis, a Southern blot needs to be performed to complete the RFLP study. A Western blot is performed on proteins separated by size, and a Northern blot is performed on RNA, also usually separated by size. Mass spectrometry could be used on the isolated protein, as could x-ray crystallography, but it is DNA that is being probed in this question.

3. A 43-year-old woman, who has recently had difficulty getting up and out of a chair and finds going up stairs to be troublesome, is referred to a rheumatologist for her progressive weakness. A full rheumatologic workup reveals anti-Jo-1 antibodies, indicating that the patient likely has polymyositis. The physician explains to the patient that she has autoantibodies directed against histidinyl-tRNA synthetase (the Jo-1 antigen). Which of the following best describes a property of this protein? (A) To form an aminoacyl-tRNA synthetase complex in the absence of energy (B) To initiate transcription by interacting with the 30S subunit of the ribosome (C) To recognize and covalently link a particular amino acid and a particular tRNA for that amino acid (D) To bind puromycin, which terminates protein synthesis (E) To covalently link amino acids to the 50 end of a corresponding tRNA

3. The answer is C. Each aminoacyl-tRNA synthetase recognizes a single amino acid and a particular tRNA specific for that amino acid. The enzyme then covalently links the two, forming an aminoacyl-tRNA, in a reaction requiring the hydrolysis of ATP. The amino acid is added to the 30 end of the tRNA. The initiation of transcription requires the interaction of the small ribosomal subunit and an initiator methionyl-tRNA, not an aminoacyl tRNA synthetase. Puromycin inhibits translation by binding to the ribosome, not to the aminoacyl tRNA synthetase.

3. A 65-year-old man with a history of type 2 diabetes is complaining of blurred vision and numbness in his toes. Laboratory results were significant for increased blood urea nitrogen (BUN) and creatinine, indicative of renal failure. Laboratory work also revealed an HbA1C of 9.0. One of the mechanisms for the damage responsible for the man's symptoms is the nonenzymatic covalent bonds formed between glucose and structural proteins. How would this reaction best be classified? (A) Acylation (B) Carboxylation (C) Glycation (D) Hydroxylation (E) Esterification

3. The answer is C. Glycation refers to the reaction of the aldehyde group of glucose reacting with the amino groups of protein, forming an amide linkage. The increased rate of glycation of collagen during hyperglycemia is implicated in the development of complications of diabetes, such as blindness and renal and vascular disease. Clotting factors are often carboxylated; histones can be acylated. Collagen is a prominent example of a protein that is hydroxylated during its production

3. A couple of African American descent gives birth to a boy after an otherwise uneventful pregnancy. The child is exceptionally fairskinned and has almost white hair. Further examination reveals red pupils. A postnatal screen is likely to confirm the deficiency of which of the following enzymes in the child? (A) Peroxidase (B) Inducible nitric oxide synthase (iNOS) (C) Glutathione reductase (D) Tyrosinase (E) Phenylalanine hydroxylase

3. The answer is D. Albinism results from a defect in the melanocyte isozyme of tyrosinase, which is required for the conversion of tyrosine to dihydroxyphenylalanine, on the pathway to melanin. Peroxidase is important in the formation of thyroid hormone. Glutathione reductase is an NADPH-requiring enzyme involved in regenerating oxidized glutathione. Phenylalanine hydroxylase converts phenylalanine to tyrosine; the absence of this enzyme leads to phenylketonuria.

3. By what mechanism does digoxin help patients with congestive heart failure? (A) Stimulates beta-1 receptors on myocytes (B) Decreases intracellular myocyte calcium levels (C) Activates myocyte Na+/K+ ATPase (D) Increases intracellular myocyte calcium levels (E) Stimulates alfa-1 receptors on myocytes

3. The answer is D. Digoxin increases intracellular myocyte calcium levels, thereby increasing myocyte contractility. This occurs primarily by digoxin's ability to inhibit the extracellular a subunit of the myocyte's Na+/K+ ATPase pump. This pump normally transports three sodium ions out of the cell in exchange for two potassium ions entering the cell, at the expense of ATP. By inhibiting the Na+/K+ ATPase, intracellular Na+ levels increase, reducing the rate of calcium extrusion from the cell (the Na+/Ca2+ exchange pump that pumps sodium out of the cell requires a high extracellular Na+ concentration, which cannot be maintained owing to the inhibition of the Na+/K+ ATPase). More intracellular calcium increases the length of phase 4 and 0 action potentials, effectively increasing the contractility of the heart. b- and a-adrenergic receptors belong to a class of G-protein-coupled receptors that are targeted by catecholamines. These include— with varying specificity to the b and a receptors—epinephrine, norepinephrine, phenylephrine, and albuterol, to name a few, and are not used in the standard management of congestive heart failure.

3. In which one of the following scenarios would one expect to observe an increase in liver fructose 2,6-bisphosphate levels? (A) After the release of epinephrine (B) In an individual who had just finished running a marathon (C) In a patient exhibiting diabetic ketoacidosis (D) After the consumption of a large bowl of ice cream (E) In a patient with kwashiorkor

3. The answer is D. F-2,6-BP serves an intracellular signal that indicates glucose is abundant. It is a potent activator of PFK-1, the major regulated step of glycolysis. Its levels are increased in the liver when insulin is released; the levels drop when glucagon or epinephrine is released. A marathon runner would have high epinephrine levels and low F-2,6-BP in the liver. A patient in diabetic ketoacidosis lacks insulin. A patient with kwashiorkor also exhibits reduced insulin levels.

3. A 30-year-old woman presents to her obstetrician for her 6-month pregnancy examination. The patient has already had one son born with Edwards syndrome, who died after 1 week of life. She is concerned about the fate of the current pregnancy because after her first child was born she was seen by a clinical geneticist who told her that she is a mosaic for trisomy 18. Which of the following statements is true for this patient? (A) She will develop severe trisomy 18 later in her life. (B) Her baby will definitely have Edwards syndrome if the child is a boy. (C) Her baby will not have trisomy if the child is a girl. (D) Trisomy 18 in her son was likely due to some of her primary oocytes being 45,XX, 18 (E) Some of her primary oocytes are 47,XX, +18.

3. The answer is E. A mosaic individual originated from a single cell line within the zygote, but because of a mitotic event during development, two separate cell lines are created. In this case, the patient originated as 46,XX cells, but a mitotic disjunction (mosaicism) event resulted in 45,XX, 18 cells (monosomy 18 cells typically do not survive) and 47,XX, +18 cells. She would therefore be a mosaic of 46,XX cells and 47,XX, +18 cells. Severity of disease in a mosaic depends on how many of her cells are 47,XX, +18, but because this patient is healthy, she will not develop symptoms later in life. Transmission of trisomy 18 from a female mosaic to her child occurs from a 47,XX, +18 primary oocyte from the mosaic, which generates an egg with two copies of chromosome 18. Transmission will not likely occur from a mosaic to her child if all of her primary oocytes are 46,XX (normal). Trisomy 18 is not determined by gender and can occur in both males and females. One cannot predict the occurrence of trisomy based on the X and Y chromosomes. Trisomy 18 in her son was likely due to at least some of her primary oocytes being 47,XX, +18, not 45,XX, 18. If the latter were correct, there would be a chance of monosomy 18, which is incompatible with life.

3. A 3-year-old boy presents to the pediatric clinic with the symptoms of hypotonia, lactic acidosis, and seizures. After an extensive workup, he is diagnosed with PDHC deficiency, an X-linked recessive disorder. Which one of the following cofactors is not required by this enzyme to convert pyruvate to acetyl CoA? (A) Thiamine (B) Lipoic acid (C) Pantothenate (D) Niacin (E) Ascorbic acid

3. The answer is E. Ascorbic acid is a required cofactor for the hydroxylation reactions that occur in collagen formation. It has no role in TCA cycle reactions. The PDHC utilizes the other cofactors listed as answers. The conversion of pyruvate to acetyl CoA begins with the decarboxylation of pyruvate, which is bound to the cofactor thiamine pyrophosphate (TPP). The next reaction of the complex is the transfer of the 2-carbon acetyl group from acetyl TPP to lipoic acid, the covalently bound coenzyme of lipoyl transacetylase. The enzyme dihydrolipoyl dehydrogenase, with FAD as a cofactor, catalyzes the oxidation of the two sulfhydryl groups of lipoic acid. The final activity of the PDHC is the transfer of reducing equivalents from the FADH2 of dihydrolipoyl dehydrogenase to NAD+, forming FAD and NADH.

3. A 40-year-old Hispanic woman with a body mass index of 34 presents with acute right upper quadrant pain, nausea, and vomiting after eating a meal rich in lipids. She is diagnosed with having cholelithiasis and is placed on a bile salt analog that is used to inhibit the formation of cholesterol gallstones. Which of the following is an example of a bile salt? (A) HMG-CoA (B) Mevalonate (C) Squalene (D) Lanosterol (E) Chenocholic acid

3. The answer is E. Chenocholic acid is an example of a bile salt. HMG-CoA, mevalonate, squalene, and lanosterol are intermediates in cholesterol synthesis. Bile salts are synthesized from cholesterol in the liver, are stored in the gallbladder, and are released to facilitate lipid digestion in the intestines.

3. A 75-year-old woman with osteoporosis complains of back pain. A computed tomographic scan of her back confirms a compression fracture of the L3 vertebra. The attending physician begins treating the patient with morphine for pain control. Morphine is an analgesic that works similarly to which of the following endogenously produced substances? (A) ACTH (B) POMC (C) Lipotropin (D) MSH (E) Endorphin

3. The answer is E. Morphine and other opioids are agonists of the endorphin receptors and mimic the action of naturally occurring endorphins. Endorphins result from the proteolytic cleavage of POMC. The two main peptides produced from POMC are ACTH, which stimulates cortisol production, and lipotropin. Lipotropin is further processed to MSH and the endorphins.

30. A 2-month-old presents to the pediatrician for follow-up care. The child appears abnormal, and the mother states that the baby was small for gestational age and is developing much differently than her other children. The child has difficulty sleeping at night and is frequently short of breath during the day. In addition, the mother states that the child exhibits abnormal eye movements and does not respond to visual stimuli. On examination, the baby has poor muscle tone, has microcephaly, and is hyperventilating. You suspect a defect in pyruvate carboxylase. Which of the following is true about this disorder? (A) It is an X-linked disorder. (B) Clinical sequelae are related to an excessive production of citrate. (C) Morbidity and mortality can be corrected with early medical intervention. (D) The patient has an inability to form oxaloacetate from carbohydrates. (E) Compensatory anaerobic metabolism is absent under low oxygen conditions.

30. The answer is D. Pyruvate carboxylase deficiency is a rare, difficult to manage disorder that affects neonates and prevents the conversion of pyruvate to oxaloacetate. This forces glucose metabolism down the anaerobic pathway and results in the accumulation of lactic acid. As fatty acid oxidation generates acetyl CoA, pyruvate dehydrogenase activity in the mitochondria will be inhibited. In the presence of a deficient pyruvate carboxylase, pyruvate begins to accumulate and is converted to lactate in order for glycolysis to continue. The inability to generate oxaloacetate results in a slowing of the TCA cycle and inefficient energy production. Decreased ATP production severely affects the developing nervous system, which has high energy needs. The result is progressive motor pathway degeneration (e.g., hypotonia), decreased cerebellar functioning (e.g., ataxia) and respiratory compensation (i.e., respiratory alkalosis via hyperventilation to correct the lactic acid metabolic acidosis).

31. An elderly homeless man is brought in by EMS because he was found unresponsive in a cold basement. It's the middle of winter, and according to bystanders, he recently resorted to a wood stove to provide him some relief from the cold. On examination, he exhibits sinus tachycardia (a value of 124, with normal between 60 and 100), displays a pulse oximetry reading of 100% oxygen saturation, is disoriented to person/place/time, and is yelling that his head hurts. You obtain laboratory tests, including an arterial blood gas with cooximetry, and his carbon monoxide level is 23%. What is the pathophysiology of carbon monoxide toxicity? (A) Binds to and inhibits ATP synthase activity (B) Binds to and inhibits the cytochrome oxidase complex (C) Uncouples oxidative phosphorylation (D) Binds to and inhibits the ATP-ADP antiportor (E) Complexes with NADH dehydrogenase and inhibits its activity

31. The answer is B. Carbon monoxide (and cyanide) combine with cytochrome oxidase and block the transfer of electrons to oxygen in the final step of the electron transport chain (complex III also contains cytochromes, which can bind carbon monoxide). Patients typically present with intentional exposure (e.g., sitting in a running car in an enclosed area) or have accidental exposures to carbon monoxide fumes during cold weather. Patients present with alterations of conciousness, normal oxygen saturations, and the pathognomonic ''cherry-red skin.'' Carbon monoxide does not act as an uncoupler, nor does it inhibit ATP/ADP exchange across the inner mitochondrial membrane.

32. Which statement is true of glycogen in humans? (A) Glycogen is a minor form of glucose storage. (B) Linkages between glucose residues are a-1,4 and a-1,6 at branch points. (C) Glycogen degradation produces glucose 3-phosphate as its major product. (D) Glycogen stores in muscle are used to directly form GTP. (E) Glycogen is formed from fatty acids or glucose.

32. The answer is B. Glycogen is a major form of glucose storage and is formed from glucose. The structure of glycogen consists of a-1,4 linkages and a-1,6 linkages at branch points. Glycogen degradation produces glucose 1-phosphate and, at the branch points, free glucose. In the liver, glycogen degradation is used to help maintain proper blood glucose levels during exercise or fasting; in the muscle, glycogen is used to generate ATP to fuel muscle contraction. The degradation of glycogen does not lead directly to the formation of GTP.

33. A 3-year-old child presents to the pediatrician for failure to thrive. A workup including an ultrasound of his liver shows cirrhosis. A biopsy of the liver demonstrates a deficiency of glucosyl 4:6 transferase. Which of the following is the most likely glycogen storage disease (GSD) that affects this child? (A) Type I: von Gierke disease (B) Type II: Pompe disease (C) Type III: Cori disease (D) Type IV: Andersen disease (E) Type V: McArdle disease

33. The answer is D. Andersen disease (type IV glycogen storage disease [GSD]) results from a deficiency of glucosyl 4:6 transferase (the branching enzyme, also known as transglucosidase). This enzyme is responsible for forming branches in glycogen. The other answer choices are all deficiencies in enzymes responsible for the degradation of glycogen: for example, glucose 6-phosphatase deficiency in von Gierke disease; maltase deficiency in Pompe disease; glycogen debrancher deficiency in Cori disease; and muscle glycogen phosphorylase deficiency in McArdle disease.

34. An infant was brought into the emergency room after her parents witnessed her having a seizure. The child's blood glucose was 28 mmol/L. After a thorough workup, a glycogen storage disease (GSD) is suspected, and a muscle biopsy is significant for the accumulation of dextrin, a form of glycogen in which the branches only contain a few glucose molecules. Which of the following GSDs is most likely the cause of the hypoglycemia and subsequent seizure? (A) Type I: von Gierke disease (B) Type II: Pompe disease (C) Type III: Cori disease (D) Type IV: Andersen disease (E) Type V: McArdle disease

34. The answer is C. Of the glycogen storage diseases, the presence of dextrin is unique to Cori disease, a deficiency of debranching enzyme activity. Hypoglycemic seizures may occur in the first decade of life. Long-term morbidity arises from hepatic disease and progressive muscle weakness. Von Gierke disease, a deficiency of glucose 6-phosphatase, and Pompe disease, a deficiency of acid a-glucosidase (also known as lysosomal maltase), both result in excessive glycogen with normal structure and cardiomyopathy. McArdle disease also results in excessive glycogen with normal structure, but the deficient muscle phosphorylase results in symptoms of muscle cramps and myoglobinuria. Andersen disease results from the deficiency of the branching enzyme, transglucosidase (glucosyl 4:6 transferase), which is found in all tissue. Because of the abnormal glycogen structure, hepatic deposition (preciptiation) may occur and result in severe cirrhosis, hepatic failure, or neuromuscular failure. It also can present as abnormal liver function tests in its mildest presentation.

35. A 24-year-old student is training for the track and field events at her college. She presents to her physician with complaints of severe muscle cramps and weakness when training. Muscle biopsy demonstrates glycogen accumulation, liver biopsy is unremarkable, and laboratory tests indicate a deficiency of myophosphorylase (muscle glycogen phosphorylase). Which of the following is the most likely diagnosis? (A) Andersen disease (B) Cori disease (C) McArdle disease (D) von Gierke disease (E) Hers disease

35. The answer is C. McArdle disease (type V GSD) is due to a defect specific to muscle phosphorylase, with normal expression and activity of liver phosphorylase. A presentation involvingmuscle failure during demands such as exercise is typical.Many affected individuals also experience myoglobinuria due to rhabdomyolysis. Rhabdomyolysis, with concomitant spillage ofmyoglobin into the bloodstream, can result in serious renal damage. The diagnosis can bemade by observing gross blood in the urine, but lacking red blood cells on microscopic examination. Andersen disease (lack of branching enzyme) primarily affects the liver and skeletal muscle. Cori disease is a deficiency of debranching enzyme. Lastly, von Gierke and Hers diseases primarily target the liver. Von Gierke disease is the lack of glucose 6-phosphatase, whereas Hers disease is a lack of liver glycogen phosphorylase activity (with normalmuscle glycogen phosphorylase activity).

36. Which statement regarding glucagon is true? (A) A high-carbohydrate meal will cause blood glucagon levels to increase. (B) A high-protein meal will cause blood glucagon levels to decrease. (C) A high-carbohydrate meal will cause blood glucagon levels to decrease. (D) Tumors of the b cells of the pancreas are called glucagonomas. (E) Glucagon secretion is decreased with cholecystokinin.

36. The answer is C. Increased glucose levels in the blood will decrease pancreatic a-cell secretion of glucagon. Other factors that decrease glucagon secretion are insulin and somatostatin. A tumor of the a cells, leading to constant glucagon release, is a glucagonoma. Tumors of the b cells of the pancreas are insulinomas. Factors that increase glucagon secretion include decreased plasma glucose, increased plasma amino acids, catecholamines, and cholecystokinin, a hormone released from the gut when stomach contents enter the small intestine.

37. A 56-year-old, newly diagnosed type 2 diabetic patient fails an initial attempt at controlling her diabetes with dietary measures alone. She follows up with her family physician, who starts her on a sulfonylurea. This drug works through which of the following mechanisms? (A) Stimulating the production of GLUT-2 (B) Stimulating the synthesis of new insulin (C) Antagonizing the effects of arginine on pancreatic b cells (D) Inhibiting the release of glucagon (E) Stimulating the release of preformed insulin

37. The answer is E. Sulfonylureas stimulate the release of preformed insulin from pancreatic islets (via an alteration in membrane potential) and have been important drugs in the management of diabetes. Sulfonylureas have the same action on b cells as does arginine, in that preformed insulin is released. Neither sulfonylurea nor arginine is capable of stimulating the production of new insulin. No agent listed directly inhibits glucagon secretion. The production of GLUT-2 is insulin independent and would not be altered with sulfonylurea treatment.

38. A 4-month-old boy presents with frequent episodes of weakness, accompanied by sweating and feelings of dizziness. Physical examination is remarkable for palpably enlarged liver and kidneys. Laboratory tests reveal hypoglycemia and lactic acidemia. The patient is diagnosed with an enzyme deficiency of glucose 6-phosphatase, which is normally only expressed in which of the following? (A) Liver and muscle (B) Liver and brain (C) Liver and kidney (D) Erythrocytes (E) Liver and adipose tissue

38. The answer is C. Type 1 glycogen storage disease, von Gierke disease, results from the deficiency of glucose 6-phosphatase. The deficiency blocks the release of glucose from glycogen stores and also obstructs glucose synthesis in the last step of gluconeogenesis. Thus, glucose is only available from the diet, resulting in severe hypoglycemia when fasting. Although muscle is a major storage area for glycogen, the glucose 6-phosphate is converted to glucose in the liver before it can be used by muscle as an energy source. Brain and erythrocytes depend on glucose in the serum for their energy source, whereas adipose tissue uses fatty acids entering the Kreb cycle as their main source of energy. The kidney also contains glucose 6-phosphatase activity, and will contribute to gluconeogenesis during starvation conditions.

39. A 14-year-old high school girl who is extremely conscious about her appearance has gone a full day without eating. She hopes by the day of her school dance to fit into a dress she intentionally bought a size too small. Which of the following organs contributes to the glucose that is being synthesized through gluconeogenesis? (A) Spleen (B) Red blood cells (C) Skeletal muscle (D) Liver (E) Brain

39. The answer is D. The two organs that produce all of the enzymes necessary for gluconeogenesis are the liver and kidneys. Although the kidneys only supply 10% of the newly formed glucose, their participation takes on a major role in starvation. Mature red blood cells lack a nucleus and, therefore, are unable to transcribe the messenger RNA (mRNA) needed to translate and synthesize the needed enzymes for gluconeogenesis. Skeletal muscle and brain tissue also lack the ability to transcribe and translate the gene for glucose 6-phosphatase and must rely on the blood glucose supplied by the diet, gluconeogenesis, and glycogenolysis for their needed energy requirements.

Questions 3-7: Using answers choices (A) through (E) below, match the clinical vignette with the appropriate defective or deficient enzyme. (A) Glucose 6-phosphate dehydrogenase (B) Galactose 1-phosphate uridylyltransferase (C) N-acetylglucosamine 1-phosphate transferase (D) Galactokinase (E) Fructokinase 4. A native of East Africa presents with jaundice and splenomegaly after eating fava beans. A blood smear reveals hemolysis.

4. The answer is A. The patient has a defect in glucose 6-phosphate dehydrogenase (G6PDH). A lack of G6PDH leads to a reduced ability to generate NADPH, which is required to regenerate the protective form (the reduced form) of glutathione. This is particularly evident in red blood cells (RBCs) because they lack mitochondria and can only generate NADPH through G6PDH. In the presence of strong oxidizing agents (which are present in fava beans), the RBC cannot regenerate reduced glutathione, which protects the membrane from oxidative damage. This results in RBC membrane damage and cell lysis results. The jaundice results from the excess heme (from hemoglobin) released from the RBCs being converted to bilirubin and overwhelming the conjugation system of the liver

4. In a patient with severe chronic obstructive pulmonary disease (COPD), COPD ''flares'' are common and result in an inability to ventilate. This leads to the accumulation of carbon dioxide in the body, a primary respiratory acidosis. Of the following mechanisms, which is the most important one for the management of acid-base status? (A) CO2 + H2O --------- H2CO3 -------- H+ + HCO3 (B) H+ + NH3 -------- NH4 + (C) CH3COOH ----------- CH3COO + H+ (D) H2O -------- H+ + OH (E) CH3CHOHCOOH --------- CH3CHOHCOO+ H+

4. The answer is A. The primary conversion of carbon dioxide into a soluble form that can be expired (and thereby removed from the body) is through mechanism (A). Mechanism (B) is seen in metabolic acidosis as the kidney tries to excrete hydrogen protons via NH4+. An accumulation of acetic acid during metabolic acidosis (e.g., diabetic ketoacidosis, DKA) can result in the reaction seen in mechanism (C). Mechanism (D) is the simple equilibrium reaction of water into its conjugate acid and base. Mechanism (E) is the breakdown reaction of b-hydroxybutyric acid, a ''ketone'' product seen in DKA.

4. A 50-year-old woman presents with spleen enlargement and a decreased red blood cell count. She is diagnosed with chronic myelogenous leukemia (CML), a blood cell cancer in which a portion of chromosome 9 is aberrantly attached to chromosome 22. CML thus serves as a paradigm for other blood cell cancers, as an example of which of the following? (A) Transgenic expression (B) Reciprocal gene translocation (C) Robertsonian translocation (D) Gene knockout (E) Single gene defect disease

4. The answer is B. Gene translocation in CML refers to the reciprocal gene translocation between chromosomes 9 and 22. Gene translocation occurs in other blood cell cancers as well. Transgenic expression refers to intentional expression of a gene, as opposed to expression of that gene due to spontaneous mutation as occurs in CML. Creation of the Philadelphia chromosome is not due to transgenic expression. Robertsonian translocation is a type of nonreciprocal chromosomal translocation that only occurs with the 5 acrocentric chromosomes (chromosomes with centromeres near their ends, which includes chromosomes 13, 14, 15, 21, and 22). Nonhomologous acrocentric chromosomes can break at their centromeres, and their long arms can join at a single centromere. For example, the long q arm of 21 can fuse to a q arm of chromosome 14 at a single centromere. The short p arm of 21 and the short p arm of chromosome 14 fuse and are lost after a few cell divisions. Gene knockout refers to intentional genetic engineering in which a gene is eliminated or ''knocked out'' of a genome. This does not occur in CML. A single gene defect disease refers to a disease in which a mutation or deficiency of a single gene determines disease diagnosis and manifestation. It does not apply to creation of a new gene by translocation of genetic material between chromosomes

4. Which amino acid is a major neurotransmitter in the brain? (A) Tyrosine (B) Glutamate (C) Trytophan (D) Serine (E) Hisitidine

4. The answer is B. Glutamate functions as the most important and abundant excitatory neurotransmitter in the brain. It is released from the presynaptic membrane and interacts with postsynaptic glutamate receptors such as the NMDA (N-methyl-D-aspartate) receptor. Antagonists of NMDA, such as ketamine, are used clinically to provide dissociative anesthesia in children.

4. An 8-year-old boy presents with orangecolored tonsils, a very low HDL level, and an enlarged liver and spleen and is diagnosed with Tangier disease. Which of the following statements best describes HDL? (A) It is produced in skeletal muscle. (B) It scavenges cholesterol from cell membranes. (C) Its major protein is apo E. (D) It is formed when VLDL is digested by lipoprotein lipase. (E) It activates ACAT.

4. The answer is B. HDL scavenges cholesterol from cell membranes and lipoproteins. HDL is produced in the liver (not muscle), and its major apoprotein is apo A. IDL is formed when VLDL is digested by lipoprotein lipase. Cholesterol activates ACAT, which converts cholesterol to cholesterol esters for storage in cells.

4. A 37-year-old immigrant from Thailand develops fevers, night sweats, weight loss, and a blood-tinged cough. He present to the emergency room, where an infectious disease doctor is consulted and immediately prescribes a multidrug regimen that includes rifampin. Rifampin inhibits which one of the following types of enzymes? (A) DNA-dependent DNA polymerase (B) DNA-dependent RNA polymerase (C) RNA-dependent DNA polymerase (D) RNA-dependent RNA polymerase (E) Reverse transcriptase

4. The answer is B. Rifampin is an important agent in a multidrug regimen for tuberculosis and works by inhibiting the b subunit of the bacterial DNA-dependent RNA polymerase (RNA polymerase). A DNA-dependent DNA polymerase, such as bacterial Pol III, directs DNA replication, whereas bacterial Pol I is involved with DNA repair and lagging strand DNA synthesis. An example of a RNA-dependent DNA polymerase is reverse transcriptase, found in retroviruses. Only certain RNA viruses, such as poliovirus, code for RNA-dependent RNA polymerases.

4. A 40-year-old woman has rheumatoid arthritis, a crippling disease causing severe pain and deformation in the joints of the fingers. She is prescribed prednisone, a steroid that exerts its beneficial effects through anti-inflammatory pathways. What is the mechanism of steroidal anti-inflammatory agents? (A) Prevent conversion of arachidonic acid to epoxides (B) Inhibit phospholipase A2 (C) Promote activation of prostacyclins (D) Degrade thromboxanes (E) Promote leukotriene formation from HPETEs

4. The answer is B. Steroids such as cortisone and prednisone inhibit phospholipase A2, which cleaves arachidonic acid from membrane phospholipids. In the absence of free arachidonic acid, the formation of prostaglandins and leukotrienes is reduced. Because these molecules are mediating the ''pain'' response, a reduction in their synthesis results in a feeling of less pain for the affected individual. Steroids do not prevent the conversion of arachidonic acid to epoxides, activate prostaglandins, degrade thromboxane, or stimulate leukotriene production.

4. A patient you have been treating for Gaucher disease, a lysosomal storage disease, comes into your office wanting to have her relatives tested to determine whether they are carriers of the disorder. The most predominant b-glucosidase mutation, N370S, is expressed in this family. Wanting to do this in the most expeditious and least costly manner would require the use of which one of the following techniques? (A) SNP assessment (B) DNA sequencing (C) Allele-specific oligonucleotide probe hybridization (D) RLFP analysis (E) DNA fingerprinting

4. The answer is C. Because the mutation is known and is due to a single nucleotide change, allelespecific oligonucleotides can be synthesized that differentiate between the normal and disease gene. By appropriately labeling the probes, and using stringent hybridization conditions, it can be determined which probe (or probes) will hybridize to sample DNA in a simple experiment. The experiment is to spot sample DNA on nitrocellulose and then probe the sample with the two available probes. If both probes bind to the sample, the patient is a carrier. DNA sequencing, RFLP analysis, and DNA fingerprinting involve electrophoresis on a gel-based material, which will require more time and expense. SNP assessment does not require gel electrophoresis, but the assays are more time-consuming than a simple hybridization experiment.

4. A 28-year-old professional cyclist has been training for an opportunity to race in the Tour de France. His coach strongly suggests that he consume carbohydrates after each of his workouts to ensure that his muscle glycogen storage can endure the 28-day race. The activity of muscle glycogen synthase in resting muscles is increased by the action of which of the following? (A) Epinephrine (B) Glucagon (C) Insulin (D) Phosphorylation (E) Fasting and starvation

4. The answer is C. Glycogen synthesis occurs at times of rest and when the energy needs of the cells are being met. Of the hormones influencing the storage of glucose, insulin promotes the synthesis of energy stores through the dephosphorylation and activation of glycogen synthase. In fact, a helpful generalization is that glucagon and epinephrine typically mobilize energy stores through the activation of enzymes via direct phosphorylation, whereas insulin accomplishes the opposite. Fasting and starvation will not result in an increase in muscle glycogen stores.

4. A 35-year-old woman is found to have breast cancer. A core biopsy is sent for molecular studies to guide treatment, and the pathology report indicates that the tumor is HER2/neu positive. Which of the following correctly describes HER2? (A) A tumor-suppressor gene (B) An anti-apoptotic gene (C) A growth factor receptor (D) A steroid hormone (E) A cell cycle regulator

4. The answer is C. HER2/neu is a growth factor receptor that is sometimes overexpressed in breast cancer, in which case it is considered to be a marker of poor prognosis. Patients who overexpress this protein can be treated with trastuzumab (Herceptin), which is a monoclonal antibody that blocks growth-promoting signals via the HER2 receptor. The p53 gene is a tumor-suppressor gene, whereas bcl-2 is an antiapoptotic gene. Estrogen is an example of a steroid hormone. Estrogen acts as a growth factor for breast tumors that overexpress the estrogen receptor. Cyclin D is an example of a cell cycle regulator that is aberrantly expressed in mantle cell lymphoma and other tumors.

4. A 24-year-old man presents to your clinic with several concerning symptoms. He states that he has uncontrollable movements called chorea, occasional stiffness, slurring of speech, difficulty planning his day and balancing his checkbook, and bouts of anxiety and crying spells. He also professes that this has been noted in some relatives on his mother's side. What is true about the nature of the molecular mutation of this disorder? (A) A point mutation in a single gene (B) A nucleotide deletion in a single gene (C) A triplet repeat expansion within a gene (D) A frameshift mutation within a gene, creating a truncated protein (E) A chromosomal deletion of many bases that covers many genes

4. The answer is C. This patient has all the classic signs and symptoms of Huntington disease. Huntington disease is an autosomal dominant disorder that involves the huntington gene. This gene encodes a sequence of repeating trinucleotides, which gives rise to a polyglutamine stretch in the protein. In certain individuals, those that express Huntington disease, this trinucleotide repeat is greatly expanded, and the stretch of polyglutamine in the protein is enlarged, leading to a dysfunctional protein that, over time, leads to altered neuronal function. Other diseases that result from trinucleotide repeat expansion include spinobulbar muscular atrophy, spinocerebellar ataxia, fragile X syndrome, Friedreich ataxia, and myotonic dystrophy. Huntington disease is not due to a single nucleotide change, a frameshift mutation, or a deletion event within the gene.

4. A 27-year-old, semiprofessional tennis player seeks advice from a hospital-based nutritionist concerning his diet supplements. His coach had given him amino acid supplements consisting of phenylalanine and tyrosine. The rationale was that these neurotransmitter precursors would ''help his brain focus'' on his game. In reality, excess phenylalanine will be metabolized to provide energy. Phenylalanine will enter the TCA cycle as which one of the following TCA cycle intermediates? (A) Oxaloacetate (B) Citrate (C) a-Ketoglutarate (D) Fumarate (E) Succinyl CoA

4. The answer is D. Although it is true (see Chapter 13) that tyrosine and phenylalanine are precursors for neurotransmitter synthesis, excess amino acid intake will lead to their degradation. Phenylalanine is converted directly to tyrosine and, through homogentisic acid, enters the TCA cycle as fumarate. Aspartate and asparagine enter through oxaloacetate; glutamate directly feeds into a-ketoglutarate; and valine, threonine, isoleucine, and methionine enter via propionyl CoA to succinyl CoA.

4. A 3-year-old boy is brought to the emergency room with abdominal pain, mental status changes, and fatigue. On history, the physician finds that the patient lives in an older house and has been sucking on the paint chips that have crumbled in the windowsills, making the doctor suspicious for lead poisoning. Lead typically interferes with which of the following enzymes? (A) Cytochrome oxidase (B) Protoporphyrinogen oxidase (C) UMP synthase (D) d-ALA dehydratase (E) Porphobilinogen deaminase

4. The answer is D. Lead inhibits hemoglobin synthesis by inhibiting d-ALA dehydratase. Cytochrome oxidase is a component of complex IV of the electron transport chain, which is inhibited by cyanide and carbon monoxide, but not by lead. Protoporphyrinogen oxidase is deficient in variegate porphyria, and is not sensitive to lead. UMP synthase is defective in the genetic condition of hereditary orotic aciduria. UMP synthase is not sensitive to the presence of lead. Patients with acute intermittent porphyria have a deficiency of porphobilinogen deaminase. Lead has no effect on this enzyme.

4. Which statement is true concerning the intestinal brush border membrane? (A) Amylase is only found in the brush border. (B) Disaccharides cross the brush border. (C) Insulin is required for the uptake of glucose. (D) Fructose requires a sodium-independent monosaccharide transporter. (E) Minimal carbohydrate digestion occurs here because most occurs in the mouth and stomach.

4. The answer is D. The brush border of the jejunum is one of the main sites of carbohydrate digestion through the action of maltase, pancreatic amylase, lactase, and other disaccharidases and oligosaccharidases. Initial digestion of carbohydrate occurs in the mouth (using salivary amylase), but once the food enters the stomach, the reduction in pH inactivates salivary amylase. Once digested into monosaccharides in the intestine, absorption occurs through specific transporters that are sodium dependent (SGLT-1 for the transport of glucose) and independent (GLUT-5 for the transport of fructose). GLUT-4, the insulin-responsive transporter, is not expressed in the intestinal epithelial cells. Once galactose, glucose, and fructose are in the intestinal mucosal cells, GLUT-2 transports these monosaccharides into the portal circulation. Disaccharides are not transported across the intestinal epithelial cell membrane.

4. A 58-year-old woman with breast cancer presents with confusion, headaches, and persistent nausea. To evaluate for metastases to the brain, the oncologist orders a positron emission tomography (PET) scan, which covalently links a radioactive isotope to glucose and reveals highly active areas in the body (e.g., a tumor). Which of the following proteins enables the tracer to remain in the cell? (A) Insulin (B) GLUT-4 (C) GLUT-1 (D) Hexokinase (E) PFK-1

4. The answer is D. The conversion of glucose to glucose 6-phosphate by glucokinase and hexokinase traps the labeled glucose within the cell. Glucose is transported into the brain through either the insulin-sensitive GLUT-4 transporter or the insulin-independent GLUT-1 transporter. However, until the molecule is phosphorylated, it can be transported out of the cell and down its concentration gradient. Insulin accelerates the rate of glucose transport into cells via GLUT-4 transporters, but does not phosphorylate the glucose molecule to trap it in the cell. PFK-1 is the first committed step of glycolysis, but the conversion to glucose 6-phosphate, and trapping of glucose, occurs before the PFK-1 step.

4. An 18-year-old obese woman maintains a sedentary lifestyle and eats a high-fat, highcarbohydrate diet. Maintenance of this diet and lifestyle has led to lipogenesis and obesity. Which of the following statements correctly describes an aspect of lipogenesis? (A) The primary source of carbons for fatty acid synthesis is glycerol. (B) Fatty acids are synthesized from acetyl CoA in the mitochondria. (C) Fatty acid synthesis and esterification to glycerol to form triacylglycerols occurs primarily in muscle cells. (D) The fatty acyl chain on the fatty acid synthase complex is elongated two carbons at a time. (E) NADP+, which is important for fatty acid synthesis, is produced by the pentose phosphate pathway.

4. The answer is D. The primary source of carbons for fatty acid synthesis is dietary carbohydrate. Fatty acids are synthesized from acetyl CoA in the hepatocyte cytosol, and esterification to glycerol to form triacylglycerols also occurs primarily in the liver. The fatty acyl chain on the fatty acid synthase complex is elongated two carbons at a time. With each two-carbon addition to the elongating chain, the b-keto group is reduced in a reaction that requires NADPH. NADPH is a reducing equivalent produced by the pentose phosphate pathway and the malic enzyme. NADP+ is a product of fatty acid biosynthesis, not a substrate

4. A 40-year-old woman complains of decreased energy, significant weight gain, and cold intolerance. She is seen by her family physician, who has laboratory tests done that indicate she has a decreased level of thyroid hormone. Which of the following amino acids is iodinated in mature thyroid hormone? (A) Serine (B) Threonine (C) Tryptophan (D) Tyrosine (E) Phenylalanine

4. The answer is D. Thyroid hormone is an iodinated molecule produced by several complex reactions in the follicular cells of the thyroid and the colloid. Certain tyrosine residues are iodinated in the precursor protein thyroglobulin. The side chains of serine, threonine, tryptophan, or phenylalanine are not substrates for iodination in thyroglobulin.

4. A 54-year-old man presents to his family physician with a 3- to 4-week history of a nonproductive cough and a low-grade fever. The physician suspects an atypical pneumonia from Mycoplasma pneumoniae, and decides to treat the patient empirically with erythromycin. Which of the following best describes the mechanism of action of erythromycin? (A) Binds to the 50 cap of mRNA and prevents translation (B) Binds to the 30S ribosomal subunit and blocks peptidyl transferase activity (C) Binds to the Shine-Dalgarno sequence of mRNA and prevents translation (D) Binds to and inhibits initiation factor IF-1 from binding to the initiation complex (E) Binds to the 50S ribosomal subunit and blocks translocation during translation

4. The answer is E. Erythromycin, as well as clindamycin, binds to the 50S ribosomal subunit of bacteria and blocks translocation, thereby inhibiting translation. Aminoglycosides, like gentamicin, tetracycline, and streptomycin, bind to the 30S ribosomal subunit to elicit their effects. The ribosomes bind to the 50 cap of the eukaryotic mRNA (prokaryotic mRNA lacks a cap). The Shine- Dalgarno sequence in prokaryotic mRNA allows proper binding of the mRNA to the 30S ribosomal subunit; erythromycin does not bind to this sequence. IF-1, IF-2, and IF-3 are bacterial initiation factors required for the formation of the initiation complex, although they are not the targets of any antibiotic.

4. A 13-year-old boy has developed polydipsia, polyuria, polyphagia, and weight loss over the past few weeks. He was brought to the emergency room by his parents because he woke up this morning very lethargic. His blood glucose was found to be 600 mg/dL, and he was immediately placed on an insulin drip. Insulin works primarily by which one of the following mechanisms? (A) Activating adenylate cyclase (B) Binding to an intracellular receptor (C) Activating caspases (D) Producing cGMP (E) Causing phosphorylation of tyrosine residues

4. The answer is E. Insulin binds to extracellular receptors, promoting dimerization and subsequent phosphorylation of tyrosine residues on the intracellular portion of the receptor. This leads to increased tyrosine phosphorylation of other proteins, which mediate the intracellular effects of insulin action, including glucose uptake and storage. Hormones like epinephrine bind to G protein-coupled receptors with activation of adenylate cyclase. Insulin's primary effect is not achieved via activation of adenylate cyclase. Steroid hormones are lipid soluble and diffuse through the cell membrane, binding intracellular receptors that activate gene transcription. Caspases are involved in apoptosis, a process that is antagonized by insulin. ANP activates guanylate cyclase, which produces cGMP.

4. You are called to the emergency room (ER) to admit a patient to the medicine service. The patient appears malnourished and suffers from alcoholism. These are chronic issues, but there is an acute change that resulted in him being brought to the ER by the life squad. The patient is exhibiting some ataxia and increased confusion, and has new-onset short-term memory loss. Besides eliciting the above on examination, you note that he also has a lateral rectus muscle palsy. Which one of the following statements is correct concerning this patient's condition? (A) It is the result of irreversible inhibition. (B) It cannot be effectively treated. (C) It is the result of noncompetitive inhibition. (D) It is the result of competitive inhibition. (E) It is due to the lack of a fundamental coenzyme.

4. The answer is E. This patient has the classic symptoms of Wernicke encephalopathy, which results from an inadequate intake or absorption of thiamine. The patient is thus thiamine deficient. Within the United States, the condition is most often observed in chronic alcoholics with poor diets. Thiamine pyrophosphate is a required coenzyme for the oxidative decarboxylation of pyruvate and a-ketoglutarate during energy metabolism. The absence of thiamine leads to reduced energy production by all organs and tissues. Treatment includes intravenous thiamine, magnesium, and glucose and is reversible in the acute setting. The reduction in neuronal energy metabolism is not based on a type of enzyme inhibition (such as competitive, noncompetitive, or irreversible), but on the lack of a required cofactor for two enzymes

40. A newborn infant is found to have persistent hypoglycemia despite increased feeding intervals. The child is also irritable with a moderate degree of hepatomegaly. He is found to have normal levels of muscular fructose 1,6- bisphosphatase but decreased levels of the hepatic isoform. Which of the following statements is true of fructose 1,6-bisphosphatase? (A) Its synthesis is induced by adenosine monophosphate (AMP). (B) Its synthesis is induced by insulin. (C) It is inhibited by fructose 2,6-bisphosphate. (D) Its synthesis is induced in the fed state. (E) Its synthesis is inhibited during fasting.

40. The answer is C. Fructose 1,6-bisphosphatase is an important regulatory step in gluconeogenesis. Its activity is inhibited by fructose 2,6-bisphosphate. It is also inhibited by AMP. In addition, its synthesis is induced during the fasting state but not in the fed state. Insulin has no direct effect on this critical regulatory enzyme. Because insulin is released in the fed state, it will not lead to increased synthesis of this enzyme.

42. A 23-year-old woman gives birth to a healthy baby and plans on breast-feeding. Prolactin is an important hormone for the synthesis of milk in the breast because it stimulates the synthesis of a-lactalbumin. The function of this protein is to do which of the following? (A) Convert galactose to galactitol (B) Lower the Km of galactosyl transferase for glucose (C) Add galactose for the glycosylation of proteins (D) Reduce oxidized glutathione (E) Form fructose from sorbitol

42. The answer is B. Prolactin, released from the anterior pituitary, stimulates the synthesis of alactalbumin, which reduces the Km of galactosyl transferase for the substrate glucose. Glucose is then joined to UDP-galactose, thereby forming lactose. The same enzyme, galactosyl transferase, is also responsible for the glycosylation of proteins. Aldose reductase converts galactose to galactitol. Glutathione reductase is required to reduce glutathione, in a reaction requiring NADPH. Sorbitol is converted to fructose by the action of sorbitol dehydrogenase.

43. A 5-year-old child presents with Hurler syndrome, which is characterized by dwarfism, hunchback, coarse facies, mental retardation, clouding of the cornea, and sensorineural deafness. The patient also has organomegaly due to the accumulation of which of the following? (A) Glucocerebroside (B) Sphingolipids (C) Heparan sulfate and dermatan sulfate (D) Glycogen (E) Galactose 1-phosphate

43. The answer is C. Hurler syndrome results from a deficiency of a-L-iduronidase. Children with this illness have progressive mental retardation and have an average life span of 10 years. Death is usually due to accumulation of the glycosaminoglycans, such as heparan sulfate and dermatan sulfate, in the arteries, leading to arterial damage and ischemia. A sphingolipidosis would lead to the accumulation of glucocerebroside (Gaucher disease) or other sphingolipids. A defect in liver phosphorylase (Hers disease) would lead to an accumulation of liver glycogen. A defect in galactose 1-phosphate uridylyl transferase (classic galactosemia) would lead to an accumulation of galactose 1-phosphate.

44. A 75-year-old man presents to his doctor with significant knee pain. He is a former plumber who spent years on his knees and is also an avid tennis player. X-rays demonstrate osteoarthritis, and the patient is considering knee replacement surgery. Which synovial fluid component is important to provide ''cushioning'' in joints? (A) Glycosaminoglycans (B) N-acetylgalactosamine (C) Dolichol phosphates (D) Sphingolipids (E) Collagen

44. The answer is A. Glycosaminoglycans are found in synovial fluid and the vitreous humor of the eye. Negatively charged uronic acid and sulfate groups contained within the glycosaminoglycans create a very hydrophilic environment, and water strongly associates with these molecules. Compression from outside forces, such as axial loading during walking, lead to water molecules being ''squeezed'' out, providing a lubricant effect. When the outside force is removed, the negatively charged glycosaminoglycans repair each other, and water reenters the structure, restoring the cushion to the joint. N-acetylgalactosamine, by itself, dolichol phosphates, collagen, or sphingolipids would not behave as the glycosaminoglycans do.

45. A 58-year-old man with a 30-year history of heavy drinking presents with confusion, unstable gait, and nystagmus. On the Mini Mental State Examination, he scores 21/30, and a diagnosis of Wernicke-Korsakoff syndrome is made. Which of the patient's enzymes is working at a reduced activity level due to the patient's condition? (A) Transaldolase (B) Phosphopentose isomerase (C) Transketolase (D) Phosphopentose epimerase (E) Glucose 6-phosphate dehydrogenase

45. The answer is C. Wernicke-Korsakoff syndrome results from a deficiency of thiamine due to any condition resulting in a poor nutritional intake. Heavy, long-term alcohol use is the most common association with Wernicke-Korsakoff syndrome. Thiamine is converted to thiamine pyrophosphate, which serves as a cofactor for several enzymes that function in glucose utilization. These enzymes include transketolase, pyruvate dehydrogenase, and a-ketoglutarate dehydrogenase. None of the other enzymes listed require thiamine pyrophosphate as a cofactor.

46. A 24-year-old man is brought to the emergency room with multiple gunshot wounds. He is rushed immediately to the operating room because of intractable hypotension. His blood type was determined to be B negative, and blood products were ordered. Which of the following carbohydrate units will be found on the reducing end of his red blood cell surface proteins? (A) N-acetylgalactosamine (B) N-acetylglucosamine (C) Galactose (D) Glucose (E) Fructose

46. The answer is C. The human ABO blood types arise from differences in the oligosaccharide content of glycoproteins and glycolipids on the surface of blood cells.Differences in a single sugar on the nonreducing end result in blood types A, B, or O. In all three blood types, a generic oligosaccharide consisting of fucose, galactose, andN-acetylglucosamine is attached to the red blood cell surface proteins. In the absence of an additional sugar, the result is type O.When galactose or N-acetylgalactosamine is covalently bound to the galactose, the types are considered B and A, respectively.

47. A 2-year-old child is brought for evaluation for developmental delay, mental retardation, and abnormal bleeding. After a thorough workup by a pediatric geneticist, the child is found to have a congenital disorder of N-linked oligosaccharides. Which of the following substances is involved in the synthesis of N-linked glycoproteins? (A) a-Mannosidase (B) a-Lactalbumin (C) Thiamine pyrophosphate (D) a-Fucosidase (E) Dolichol phosphate

47. The answer is E. Dolichol phosphate is a long-chain alcohol associated with the endoplasmic reticulum that transfers branched polysaccharide chains to proteins during the synthesis of N-linked glycoproteins. Both a-mannosidase and a-fucosidase are enzymes involved in the lysosomal degradation of glycoproteins. a-Lactalbumin is involved in the synthesis of lactose. Thiamine pyrophosphate is a cofactor for transketolase in the pentose phosphate pathway, and also for enzymes that catalyze oxidative decarboxylation reactions (such as pyruvate dehydrogenase).

48. What enzyme deficiency leads to classic galactosemia? (A) Lactase (B) Glucose 6-phosphate dehydrogenase (C) Galactokinase deficiency (D) Galactose 1-phosphate uridyltransferase (E) UDP-glucose epimerase

48. The answer is D. Classic galactosemia is a rare metabolic disorder that results in an inability to metabolize galactose and results in significant long-term effects (e.g., cirrhosis, renal failure, cataracts, and brain damage) if exposed to galactose. There are three forms of galactosemia, with classic galactosemia being the most severe manifestation. Classic galactosemia is caused by a deficiency of galactose 1-phosphate uridyltransferase. Galactokinase deficiency and UDP-galactose epimerase deficiency are two other, less severe forms of galactosemia. Treatment of classic galactosemia includes early detection (typically from mandated newborn screening) and elimination of dietary galactose. Lactase deficiency is seen in lactose-intolerant individuals and is an acquired deficiency of lactase that does not lead to significant long-term sequelae. Glucose 6-phosphate dehydrogenase deficiency is a defect in the hexose monophosphate shunt pathway that leads to hemolytic anemia in the presence of strong oxidizing agents.

49. Which statement is true regarding pancreatitis? (A) Grey-Turner and Cullen signs are common findings. (B) Surgery is a mainstay of treatment. (C) Gallstones and scorpion bites are not risk factors. (D) Advanced pancreatitis leads to hypercalcemia. (E) Hypertriglyceridemia is a cause of pancreatitis, especially with levels higher than 1000 mg/dL.

49. The answer is E. Pancreatitis is an inflammatory process of the pancreas with significant morbidity and mortality. Gallstones and alcohol consumption are the most common causes, but other causes exist, including trauma, steroids, mumps, autoimmune causes, scorpion and snake venom, endoscopic retrograde cholangiopancreatography (ERCP), and certain drugs (e.g., azathioprine). Prognosis depends on the degree of pancreatitis. Ranson's criteria are helpful to stage the severity with hypocalcemia (not hypercalcemia), an ominous sign of pancreatic parenchymal saponification. Treatment is largely supportive: nothing by mouth (NPO), intravenous fluids, nasogastric tube, and pain control. Only in advanced cases with evidence of phlegmon on CT are surgical intervention and antibiotics indicated. Grey-Turner (bruising of the flanks) and Cullen signs (bruising around the umbilicus) are not associated with pancreatitis.

5. Which is the best technique to separate oxygenated normal hemoglobin A (HbA) from oxygenated sickle cell hemoglobin (HbS), assuming no protein aggregation? (A) Native gel electrophoresis (B) SDS-PAGE (C) Gel filtration (D) Affinity chromatography with a C-terminal antibody (E) Ultracentrifugation

5. The answer is A. Because the difference between HbA and HbS is the sixth amino acid of the Hb b-chain (an E6V mutation), the difference in mass is very small, and separation based on charge will be able to differentiate between the two forms of hemoglobin. Native gel electrophoresis can accomplish this. SDS-PAGE blankets the protein with a uniform negative charge that will mask the inherent difference. Gel filtration and ultracentrifugation are not sensitive enough, and a C-terminal antibody will detect both forms because the difference between the two forms is manifest at the amino-terminal end of the protein.

5. An infant is born prematurely at 28 weeks and increasingly has significant difficulty breathing, taking rapid breaths with intercostal retractions. The child soon becomes cyanotic. He is diagnosed with respiratory distress syndrome due to a deficiency of surfactant. Which of the following is the phospholipid in highest concentration in surfactant? (A) Dipalmitoyl phosphatidylcholine (B) Dipalmitoyl phosphatidylethanolamine (C) Dipalmitoyl phosphatidylglycerol (D) Dipalmitoyl phosphatidylinositol (E) Dipalmitoyl phosphatidylserine

5. The answer is A. Dipalmitoyl phosphatidylcholine (DPPC), also called lecithin, is the major phospholipid in surfactant. Surfactant is a protein and lipid mixture that is responsible for lowering surface tension at the alveolar air-fluid interface. DPPC contains a glycerol backbone, palmitic acid, esterified at positions 1 and 2, and phosphocholine esterified at position 3. The other phospholipids suggested as answers are present in nonappreciable levels in surfactant.

5. A 57-year-old alcoholic man with chronic pancreatitis is admitted to the hospital for treatment. The absorption of which one of the following vitamins may be affected with pancreatitis? (A) Vitamin B12 (cobalamin) (B) Folic acid (C) Vitamin B2 (riboflavin) (D) Vitamin B6 (pyridoxine) (E) Vitamin D

5. The answer is E. Although alcoholics are often malnourished with various nutrient deficiencies, pancreatitis will affect the role of the exocrine pancreas in the absorption of fat-soluble vitamins. Vitamin D is the only vitamin listed that is fat soluble; the other vitamins listed are water soluble. With the lack of pancreatic lipase dietary fat cannot be digested, and the fat-soluble vitamins cannot be released from the lipids. The vitamins thus pass through the intestine, nonabsorbed. The other fat-soluble vitamins are vitamins E, K, and A.

5. A 43-year-old man presents to the neurologist with headache and double vision. He also complains of a milky discharge from his breast. An MRI of his head confirms the diagnosis of a prolactinoma. Which of the following substances could inhibit the release of prolactin from the pituitary? (A) Dopamine (B) Caffeine (C) Endorphins (D) Renin (E) PGF2a

5. The answer is A. Dopamine normally inhibits the release of prolactin from the anterior pituitary. When there is damage to the dopamine-producing neurons of the hypothalamus, or drugs inhibiting dopamine release are given, prolactin levels increase, resulting in galactorrhea. Caffeine inhibits phosphodiesterase and, therefore, the conversion of cAMP to AMP. Caffeine does not affect prolactin release. Endorphins are produced by proteolytic processing of POMC from the anterior pituitary and do not play a role in prolactin release. Renin is produced by the kidney and cleaves angiotensinogen to angiotensin I. PGF2a is important in stimulating uterine contractions during delivery but does not play a role in regulating prolactin levels.

5. A 27-year-old man is seen by his physician for a week-long cough, sore throat, and difficulty swallowing. He is diagnosed with diphtheria, which has reactivated because of waning immunity. One way in which diphtheria toxin leads to cell death is through the inhibition of eEF-2. Which statement best explains the function of eEF-2? (A) It is required for the translocation of peptidyl- tRNA during translation. (B) It is required for the initiation of protein synthesis. (C) It is the agent that binds to, and is inactivated by, chloramphenicol. (D) It functions as a peptidyl transferase. (E) It is analogous to the prokaryotic factor eIF-1.

5. The answer is A. Eukaryotic elongation factor 2 (eEF-2) is required for eukaryotic translation in that it facilitates the translocation of peptidyl-tRNA along the mRNA. The elongation reaction requires GTP hydrolysis. The corresponding elongation factor in bacteria is EF-G. Diphtheria toxin ADP-ribosylates eEF-2, leading to its inactivation. Initiation of eukaryotic protein synthesis requires several initiation factors, designated eIF-1, eIF-2, etc. Peptidyl transferase is an rRNA (within the large ribosomal subunit) involved in formation of the peptide bond between the amino acid groups within the A and P sites of the ribosome. This activity is not affected by the toxin. Chloramphenicol inhibits prokaryotic peptidyl transferase. Prokaryotes are not affected by diphtheria toxin because they do not contain eEF-2.

5. An 8-year-old girl presents with polydipsia, polyuria, and fatigue. A urinalysis is significant for glucose. To differentiate between type 1 diabetes mellitus and maturity-onset diabetes of the young (MODY), an assay is run to identify one of the six proteins responsible for MODY. Results reveal a missense mutation in exon 7 of the glucokinase gene establishing MODY2. Which of the following is a significant characteristic of glucokinase? (A) The Km is above the fasting concentration of glucose in the blood (B) It is expressed in many tissues (C) Its activity is stimulated in response to fructose 2,6-bisphosphate (D) Its activity is inhibited by glucose 6-phosphate (E) It is expressed only in muscle

5. The answer is A. Glucokinase and hexokinase catalyze the same reactions; however, they differ in their kinetic properties. Glucokinase has a low-affinity, high Km, allowing substantial catalytic activity only at high glucose concentration, such as after a meal, whereas hexokinase has a highaffinity, low Km, and is active at very low glucose concentrations. The Km of glucokinase is above that of the fasting concentration of glucose in the blood. Glucokinase also escapes any local regulation, although regulation at the level of transcription is influenced via the hormones insulin and glucagon. Glucose 6-phosphate and fructose 2,6-bisphosphate do not regulate glucokinase activity, although glucose 6-phosphate is an allosteric inhibitor of hexokinase activity. Glucokinase is only found in the liver and pancreas and to a small extent in the brain. It is not found in muscle.

5. A 40-year-old man presents with chest pain that radiates to his left jaw and shoulder. He is diagnosed with a myocardial infarction and is prescribed a statin medication. Statins are competitive inhibitors of HMG-CoA reductase, which converts HMG-CoA to which of the following? (A) Mevalonate (B) Isopentenyl pyrophosphate (C) Geranyl pyrophosphate (D) Farnesyl pyrophosphate (E) Cholesterol

5. The answer is A. HMG-CoA reductase converts HMG-CoA to mevalonate, using two NADPH molecules. Isopentenyl pyrophosphate, geranyl pyrophosphate, and farnesyl pyrophosphate are intermediates in cholesterol synthesis. Mevalonate is required to synthesize isopentenyl pyrophosphate, which leads to geranyl and farnesyl pyrophosphate production.

5. Two couples present to the emergency room with severe nausea, vomiting, and diarrhea. One of the patients admits that she had a dinner party and served a salad containing mushrooms she had picked during a hike in the forest earlier that day. Inhibition of which enzyme or process explains the clinical manifestations of a-amanitin poisoning seen in these patients? (A) RNA polymerase II (B) RNA polymerase I (C) RNA splicing (D) RNA polyadenylation (E) RNA polymerase III

5. The answer is A. The cellular toxin a-amanitin from the Amanita phalloides mushroom (death cap mushroom) specifically inhibits RNA polymerase II, the enzyme required for mRNA synthesis. Loss of transcriptional activity results in severe gastrointestinal symptoms, liver toxicity, and sometimes death. Eukaryotic RNA polymerase I is responsible for the synthesis of rRNA, whereas RNA polymerase III produces tRNA. Small ribonucleoproteins are required for RNA splicing. RNA poly(A) polymerase normally adds stretches of adenine residues to the 30 end of mRNA. The only enzyme inhibited by a-amanitin is RNA polymerase II.

5. Influenza virus results in more than 500,000 deaths worldwide annually. Influenza A contains an eight-piece segmented negative-sense RNA genome. Two important proteins encoded by this genome are HA (hemagglutinin) and NA (neuraminidase). The HA protein directly binds to which host cell epithelial component? (A) Sialic acid (B) Cerebrosides (C) Cytokine receptors (D) Serine-threonine kinase receptors (E) Uronic acid subgroups

5. The answer is A. The influenza virus enters the epithelial host cell by binding to sialic acid residues found on the cell surface. Sialic acid is a modified sugar residue. Cerebrosides are glycolipids synthesized from ceramide and a UDP-sugar; a common one is glucocerebroside. Cytokine receptors (which work through the JAK kinase and STAT transcription factors) and serine- threonine kinase receptors are two types of receptors involved in signal transduction, which is initiated after a chemical messenger (e.g., hormone, neurotransmitter, or cytokine) binds to the receptor on the plasma membrane. Uronic acid is an oxidized sugar and is a component of proteoglycans. Uronic acid is not usually found as a part of glycoproteins, as sialic acid is.

5. While on an aid mission to central Africa with Doctors Without Borders, a physician encounters a 23-year-old man who appears to have metastatic liver cancer. The grain storage facility outside his village was found to be contaminated with aflatoxin B1. In addition to this carcinogen, which of the following might act as a cocarcinogen in the development of this patient's cancer? (A) HTLV-1 (B) HBV (C) Asbestos (D) Vinyl chloride (E) Aniline dyes

5. The answer is B. Hepatocellular carcinoma is the most common cancer worldwide. HBV is a major contributor to the development of this cancer, by acting as a cocarcinogen with aflatoxin B1. HTLV-1, a retrovirus related to HIV, is associated with a form of leukemia and lymphoma. Asbestos, used as an insulator, is closely associated with a tumor of the pleural linings of the lung, a mesothelioma. Vinyl chloride is associated with a rare form of liver cancer, angiosarcoma. Finally, aniline dyes, used in the clothing industry, have been associated with bladder cancer. There is no evidence to suggest that HTLV-1, asbestos, vinyl chloride, or aniline dyes contribute to aflatoxin B1-induced liver cancer.

5. A 45-year-old man presents with multiple gunshot wounds to the abdomen requiring an emergent laparotomy, jejunectomy, and colectomy. After surgery, he is placed on intravenous nutrition (i.e., TPN). Which of the following compounds should be a component of TPN? (A) Palmitate (B) Linoleate (C) Phosphatidic acid (D) Glycerol (E) Glucose

5. The answer is B. Linoleate and a-linolenate are the essential fatty acids required in the human diet. Palmitate (C16:0) is synthesized by the fatty acid synthase complex. Phosphatidic acid is an intermediate in triacylglycerol synthesis, which is formed using glycerol as a precursor in the liver and using glucose as a precursor in adipose tissue. It can be synthesized without the need for an essential fatty acid.

5. A competitive reversible inhibitor such as physostigmine is used to treat glaucoma and myasthenia gravis and to reverse anticholinergic syndrome. Based on this, which one of the following statements is true concerning the clinical implications of using physostigmine? (A) Use of the drug will decrease the Km of the targeted enzyme. (B) An overdose of physostigmine can typically be reversed. (C) Physostigmine will increase the Vmax of the targeted enzyme. (D) Physostigmine will decrease the Vmax of the targeted enzyme. (E) Physostigmine is unable to cross the bloodbrain barrier.

5. The answer is B. Physostigmine is both a naturally occurring substance (Calabar bean) and a chemically synthesized substance that is a competitive reversible inhibitor of acetylcholinesterase. The drug easily crosses the blood-brain barrier. By definition, a competitive reversible inhibitor acts at the catalytic site with the substrate and competes with substrate binding to the enzyme. Thus, the effects of the inhibitor can be overcome by addition of the substrate, leading to an effective reversal of drug overdose. This is a reversible inhibition, so the Vmax is unchanged because if sufficient substrate is added, the effects of the inhibitor can be overcome. With a competitive inhibitor, the Km is increased because more substrate is needed to reach (1/2)Vmax

5. A 31-year-old Asian woman presents with a chief complaint of ''difficulty eating dairy products.'' The patient clarifies her symptoms and states that she develops bloating, abdominal pain, diarrhea, and excessive flatulence. Which one of the following statements is true about this patient's condition? (A) She lacks b1-----4 endoglucosidase activity. (B) She cannot obtain galactose and glucose from a disaccharide. (C) She has an inability to emulsify dietary lipids. (D) She is exhibiting an autosomal dominant condition. (E) Her associated diarrhea is of an infectious etiology.

5. The answer is B. This patient has lactose intolerance, which is a reduced ability to metabolize lactose (a disaccharide) into galactose and glucose. This problem usually manifests after adulthood. The loss of lactase activity is not due to an autosomal dominant disorder; in fact, lactase persistence (high levels of lactase throughout life) is inherited in an autosomal dominant fashion. Absence of b1-----4 endoglucosidases is innate to all humans, making digestion of dietary fiber (i.e., cellulose) not possible. An inability to emulsify dietary lipids will lead to steatorrhea, which may have some overlap with this patient's symptoms, but is not the unifying diagnosis given her chief complaint. Diarrhea in patients with lactose intolerance is an osmotic diarrhea from the remnant colonic lactose that is not entirely fermented by normal gut flora

Questions 5-11: Using answers (A) through (G) below, match the correct drug with its clinical effect. 5. Trimethoprim. (A) Folic acid analog(s) that inhibit(s) dihydrofolate reductase, which leads to an inhibition of purine and deoxythymidine synthesis (B) Folate analog(s) that bind(s) specifically to bacterial dihydrofolate reductase and used in conjunction with sulfonamides (C) Histamine and basophil inhibitor(s) used to treat allergic reactions (D) Increase(s) synaptic serotonin concentration in the treatment of depression (E) Inhibitor(s) of tyrosine residue iodination (F) Replenish(es) glutathione levels during acetaminophen toxicity (G) Release(s) NO, create(s) smooth muscle relaxation, and very effective as antihypertensive

5. The answer is B. Trimethoprim is a folate analog that binds specifically to bacterial dihydrofolate reductase and is used in conjunction with sulfonamides, which block folate synthesis. A common formulation is trimethoprim-sulfamethoxazole (Bactrim).

5. A 57-year-oldman with a long history of alcohol abuse comes to the emergency room with symptoms of confusion and heptomegaly on examination. The patient also has a flapping tremor at the wrist (asterixis). He is diagnosed with hepatic encephalopathy, which can be partially treated with a diet of branched-chain amino acids. Which of the following sets of amino acids would you suggest? (A) Tryptophan, phenylalanine, tyrosine (B) Aspartate, glutamate, asparagine (C) Valine, leucine, isoleucine (D) Glycine, alanine, serine (E) Methionine, proline, cysteine

5. The answer is C. The branched-chain amino acids make up the three essential amino acids Lleucine, L-isoleucine, and L-valine. These amino acids are found in proteins of all life forms. Dietary sources of the branched-chain amino acids are principally derived from animal and vegetable proteins. None of the other amino acids suggested as answers falls into the category of branched-chain amino acids

63. In the nursery, an infant with blond hair and blue eyes is noted to have a mousy odor to his urine upon diaper changes. As is mandated by the state, all infants are screened for multiple inborn errors of metabolism, and he is found to have phenylketonuria (PKU). The cells of this child are unable to convert phenylalanine to which one of the following? (A) Tyrosine (B) Serine (C) Glycine (D) Cysteine (E) Alanine

63. The answer is A. Phenylalanine hydroxylase converts phenylalanine to tyrosine. A defect in this enzyme results in an accumulation of phenylketones in the urine, which gives urine its characteristic odor. PKU is treated by restriction of phenylalanine in the diet. Phenylalanine is not a precursor for serine, glycine, cysteine, or alanine synthesis

5. A child is noted to have recurrent respiratory infections that necessitate hospitalization. His laboratory tests demonstrate a decrease in T cells, B cells, and NK cells. He has decreased levels of circulating antibodies and is diagnosed with severe combined immunodeficiency. The enzyme that is defective in this disorder is important in which one of the following processes? (A) The conversion of ribonucleotides to deoxyribonucleotides (B) The de novo synthesis of AMP (C) The degradation of adenosine (D) The de novo synthesis of UMP (E) The conversion of dUMP to dTMP

5. The answer is C. The enzyme deficiency in severe combined immunodeficiency disease (SCID) is likely adenosine deaminase, which normally degrades adenosine to inosine. The conversion of ribonucleotides to deoxyribonucleotides is performed by ribonucleotide reductase. AMP is formed from IMP through the action of adenylosuccinate synthetase, followed by the action of adenylosuccinate. UMP synthase is an important enzyme in the formation of UMP and, subsequently, cytidine triphosphate (CTP) and thymidine triphosphate (TTP). The conversion of dUMP to dTMP is mediated by thymidylate synthase. A second enzyme deficiency that can lead to SCID is the loss of a cytokine receptor subunit (the interleukin-2 receptor g-chain). This is an X-linked form of SCID owing to the inability of immune cells to proliferate in response to interleukin- 2.

5. A 2-week-old infant presents to your rural family medicine clinic and appears ill; he is febrile and jaundiced and has extensive, reddened skin. According to his mother, it appears that since the delivery of the baby at home he has always ''carried a fever.'' You are able to send off some laboratory tests and are surprised to see that he has significant neutropenia and hypergammaglobulinemia. What is the most likely diagnosis? (A) DiGeorge syndrome (B) Severe combined immunodeficiency disease (C) Chediak-Higashi syndrome (D) Wiskott-Aldrich syndrome (E) Myeloperoxidase deficiency

5. The answer is C. This patient has Chediak-Higashi syndrome, a primary immunodeficiency disorder of phagocytic vesicles in which the lysosomes are unable to destroy bacteria. Primary immunodeficiency disorders are grouped into humoral/antibody/B-cell disorders (e.g., hypogammaglobulinemia, common variable immunodeficiency, leukocyte adhesion deficiency), cellmediated/ T-cell disorders (e.g., DiGeorge syndrome), combined humoral and cell-mediated deficiencies (e.g., severe combined immunodeficiency, Wiskott-Aldrich syndrome), complement deficiencies (e.g., angioedema), and phagocyte dysfunction (e.g., Chediak-Higashi syndrome). All these disorders affect children early in life and are associated with significant morbidity and mortality.

5. Which one of the following occurs in an individual who is rested and has fasted for 12 hours? (A) Gluconeogenesis is the major process by which blood glucose is maintained. (B) Adenylate cyclase has been inactivated in liver. (C) Liver glycogen stores have been depleted. (D) Glycogen phosphorylase, pyruvate kinase, and glycogen synthase are phosphorylated in the liver. (E) Glycogen synthase has been activated in liver.

5. The answer is D. After 12 hours of fasting, liver glycogen stores are still substantial (liver glycogen stores are not depleted until after about 30 hours of fasting). Glycogenolysis is stimulated by glucagon, which activates adenylate cyclase. The cAMP generated by adenylate cyclase activates protein kinase A, which phosphorylates glycogen phosphorylase kinase, pyruvate kinase (PK), and glycogen synthase. As a result, glycogen phosphorylase is activated, whereas glycogen synthase and PK are inactivated. Gluconeogenesis does not become the major process for maintaining blood glucose until fasting has occurred for 18 to 20 hours.

5. A 2-year-old girl was seen in the emergency room for vomiting and tremors. Laboratory tests revealed a plasma ammonium ion concentration of 195 mM(normal, 11- to 50 mM) and serum elevation of arginine. Two days later, after stabilization, ammonia and arginine levels were normal. You conclude that this patient may have a defect in which of the following enzymes? (A) CPS I (B) CPS II (C) Ornithine transcarbamoylase (D) Arginase (E) Argininosuccinate lyase

5. The answer is D. Arginase deficiency, the least common of the urea cycle defects, presents with episodic increases in serum ammonia and arginine, leading to the observed symptoms of vomiting and tremors. A defect in CPS I leads to constant hyperammonemia, without elevated arginine. A defect in CPS II would interfere with pyrimidine synthesis and does not alter blood ammonia levels. An ornithine transcarbamoylase deficiency leads to hyperammonemia and orotic aciduria. An argininosuccinate lyase deficiency leads to elevated argininosuccinate, not elevated arginine levels.

5. A patient had large deposits of liver glycogen, which, after an overnight fast, contained shorter than normal branches. A defective form of which of the following could cause this abnormality? (A) Glycogen phosphorylase (B) Glucagon receptor (C) Glycogenin (D) Amylo-1,6-glucosidase (a-glucosidase) (E) Amylo-4,6-transferase (4:6 transferase)

5. The answer is D. If, after fasting, the branches were shorter than normal, phosphorylase must be functional and capable of being activated by glucagon. The branching enzyme (the 4:6 transferase) must be normal because branches are present. The protein glycogenin must be present in order for large amounts of glycogen to be synthesized and deposited. The defect most likely is in the debranching enzyme (which contains an a-1,6-glucosidase). If the debrancher is defective, phosphorylase would break the glycogen down to within four glucose residues of the branch points, but complete degradation would not occur. Therefore, short branches would be present in the glycogen. If the short branches contain only one glucose unit, the defect is in the a-1,6- glucosidase activity of the debrancher. If they contain four glucose units, the defect is in the 4:4 transferase activity of the debrancher

5. A 3-year-old boy presents to the emergency room after having a generalized tonic-clonic seizure. The child has a history of epilepsy, ataxia, and lactic acidosis. When questioned, the parents state that their child was born with a rare metabolic disease, pyruvate carboxylase deficiency. Which one of the following metabolites is this child unable to produce effectively? (A) Pyruvate (B) Alanine (C) Acetyl CoA (D) Oxaloacetate (E) Acetoacetate

5. The answer is D. Pyruvate carboxylase is a biotin-dependent mitochondrial enzyme that converts pyruvate to OAA. In the absence of pyruvate carboxylase activity, OAA can only be produced from the transamination of aspartate, so OAA levels cannot be replenished effectively. Pyruvate production is normal, and as pyruvate accumulates, lactate is formed. Alanine is also formed from pyruvate via a transamination reaction. The PDHC will convert pyruvate to acetyl CoA. Acetoacetate is a ketone body, which is made from acetyl CoA.

Questions 3-7: Using answers choices (A) through (E) below, match the clinical vignette with the appropriate defective or deficient enzyme. (A) Glucose 6-phosphate dehydrogenase (B) Galactose 1-phosphate uridylyltransferase (C) N-acetylglucosamine 1-phosphate transferase (D) Galactokinase (E) Fructokinase 5. A healthy, well-appearing child with no pertinent medical history has 3+ glucose on a urine dipstick.

5. The answer is E. The child has a fructokinase deficiency. A deficiency in fructokinase will not allow fructose to be metabolized (fructose cannot be converted to fructose 1-phosphate), resulting in elevated serum fructose after eating a meal containing sucrose or fructose. The elevated fructose leads to no metabolic problems, but does enter the urine for excretion. Because fructose is a reducing sugar (as is glucose), it will react positively in a glucose dipstick test. A more specific glucose oxidase test would need to be run to demonstrate that the positive result on the dipstick test was or was not due to elevated glucose.

50. A 53-year-old man presents with xanthomas under his eyes, hepatomegaly, and a triglyceride level almost 8 times the upper limit of normal at 1500 mg/dL. He is diagnosed with type V hyperlipidemia. In which tissue are triacylglycerols primarily synthesized? (A) Skeletal muscle (B) Heart muscle (C) Liver (D) Spleen (E) Blood cells

50. The answer is C. Triacylglycerols are formed primarily in the liver, but they can also be generated in adipose tissue and intestinal cells. In the liver, they are packaged in very-low-density lipoprotein (VLDL) and are secreted into the blood. Triacylglycerols are stored in adipose tissue. Skeletal muscle does not produce triacylglycerol, nor do the spleen or red blood cells. Type V hyperlipidemia results from elevated levels of VLDL and chylomicrons.

51. A 50-year-old alcoholic man presents with a ruddy face, distended abdomen, and enlarged, fatty liver. Both the liver and adipose tissue can synthesize triglyceride. Which of the following reactions is liver specific for triglyceride synthesis? (A) Glucose to DHAP (B) Glycerol to glycerol 3-phosphate (C) DHAP to glycerol 3-phosphate (D) Phosphatidic acid to diacylglycerol (E) Fatty acid to fatty acyl CoA

51. The answer is B. The liver can use glycerol to produce glycerol 3-phosphate (G-3-P) by a reaction that requires ATP and is catalyzed by glycerol kinase. The liver can also produce G-3-P by converting glucose through glycolysis to DHAP, which is reduced to G-3-P. Adipose tissue lacks glycerol kinase and thus cannot generate G-3-P from glycerol; therefore, it must convert glucose to DHAP, which is then reduced to G-3-P. Both tissues can convert glucose to DHAP, DHAP to glycerol 3-phosphate, phosphatidic acid to diacylgylcerol, and the activation of free fatty acids to fatty acyl CoA.

52. A 40-year-old woman presents with pain in her legs that is elicited upon walking and relieved by rest. Imaging reveals that diffuse atherosclerosis is causing her leg pain. She is found to have no functional apoprotein C-II. Which of the following will be elevated in this patient's blood? (A) Cholesterol esters (B) Chylomicrons (C) LDL (D) HDL (E) Cholesterol

52. The answer is B. Apoprotein C-II is transferred from HDL to nascent chylomicrons and nascent VLDL. Apoprotein C-II activates lipoprotein lipase, which hydrolyzes the triacylglycerols of chylomicrons and VLDL to fatty acids and glycerol. In the absence of C-II activity, chylomicrons and VLDL levels will be increased because the triglyceride cannot be removed from those particles. LDL contains the highest content of cholesterol and its esters, and its levels are not increased in this disorder (LDL is derived from IDL, which is derived from VLDL as the triglyceride content of the particle is reduced). Triglycerides are also elevated because they are the major component of chylomicrons and VLDL.

53. A 5-year-old boy presents with altered mental status, heart failure, and muscle weakness. His serum levels of ketones and glucose are abnormally low. He is diagnosed with primary carnitine deficiency. What is the primary problem with this disorder? (A) Activation of fatty acids (B) Impaired transport of long-chain fatty acids (C) b-Oxidation (D) o-Oxidation (E) a-Oxidation

53. The answer is B. Primary carnitine deficiency is caused by a deficiency in the plasma membrane carnitine transporter leading to urinary excretion of carnitine. Cells cannot transport carnitine from the circulation across the plasma membrane and into the cell. This depletion of intracellular carnitine impairs transport of long-chain fatty acids into the mitochondria. Without a route of entry into the mitochondria, fatty acid oxidation is blocked. The lack of energy generation from fatty acid oxidation leads to hypoglycemia and low ketones. Primary carnitine deficiency does not directly affect fatty acyl activation, b-oxidation, o-oxidation, or a-oxidation.

54. An infant presents with difficulty moving his limbs, facial abnormalities, and seizures. His blood level of very-long-chain fatty acids is abnormally elevated, and he is diagnosed with Zellweger syndrome. Which of the following is true concerning the oxidation of very-longchain fatty acids? (A) Initially occurs in mitochondria (B) Oxidation occurs at the a-carbon (C) Produces acetyl CoA (D) Generates no reduced cofactors (E) Degraded three carbons at a time

54. The answer is C. Oxidation of very-long-chain fatty acids initally occurs in peroxisomes and generates hydrogen peroxide and NADH as reduced cofactors. Oxidation occurs at the a-carbon, and produces acetyl CoA (oxidation occurs two carbons at a time, not three). The last steps of oxidation occur in the mitochondria.

55. A 38-year-old woman is found to have an obstructing gallstone in the common bile duct. What is clinical implication of this obstruction? (A) Increased formation of chylomicrons (B) Increased recycling of bile salts (C) Increased excretion of bile salts (D) Increased conjugation of bile salts (E) Increased excretion of fat in the feces

55. The answer is E. Obstruction of bile in the biliary tree will prevent the recycling and excretion of bile salts. Reduced excretion of bile salts leads to reduced digestion of fats and decreased formation of chylomicrons. As a consequence, ingested fat would not be digested and would remain in the chyme and ultimately in the feces, leading to steatorrhea. The conjugation of bile salts would also decrease because their recycling is also blocked.

7. A 32-year-old woman presents to the physician with extreme fatigue and vague neurologic complaints. On examination, it is found that she has decreased positional and vibrational sense, and her complete blood count reveals a megaloblastic anemia. She relates a history of gastric resection 4 years ago for severe stomach ulcers. Which vitamin deficiency does this represent? (A) Vitamin C (B) Vitamin D (C) Vitamin K (D) Vitamin B12 (E) Folate

7. The answer is D. Both folate and vitamin B12 deficiency lead to a megaloblastic anemia secondary to a reduction in DNA synthesis. Only vitamin B12 deficiency causes neurologic dysfunction associated with damage to the dorsal spinal columns. The history of gastric resection is consistent with a deficiency of intrinsic factor required for reabsorption of vitamin B12 in the terminal ileum. Vitamin C, D, and K deficiencies will not lead to megaloblastic anemia.

56. A 60-year-old woman had an episode of chest pain radiating to her left arm, which was diagnosed as a myocardial infarction. She was prescribed a statin medication. How does statin treatment lower cholesterol and LDL levels? (A) It inhibits the rate-limiting step in cholesterol biosynthesis. (B) It increases synthesis of bile salts to digest cholesterol. (C) It increases the serum level of HDL. (D) It inhibits formation of LDL from IDL. (E) It inhibits synthesis of LDL receptors.

56. The answer is A. HMG-CoA reductase is the rate-limiting enzyme in cholesterol biosynthesis, and statins block HMG-CoA reductase activity. Inhibition of HMG-CoA reductase results in decreased intracellular cholesterol levels. Decreased intracellular cholesterol levels result in an increased expression of LDL receptors on the surface of hepatocytes. This results in increased intracellular uptake of serum LDL, leading to decreased serum LDL levels. Inhibition of HMGCoA reductase does not directly affect synthesis of bile salts, HDL, or IDL.

57. A 40-year-old man presents with severe claudication secondary to atherosclerotic plaques in the arteries of his legs. His laboratory tests indicate high levels of cholesterol and LDL. Which of the following is the precursor of LDL? (A) IDL (B) Cholesterol (C) Cholesterol esters (D) HDL (E) Chylomicrons

57. The answer is A. IDL is degraded to form LDL. Cholesterol is converted to cholesterol esters for storage in cells and HDL particles. HDL does not form LDL (LDL is derived from IDL, which is derived from VLDL). Chylomicrons are synthesized by intestinal epithelial cells

58. Which statement correctly describes a role of HDL? (A) It is the largest of the lipoproteins, and it eliminates cholesterol esters and triacylglycerol from the serum. (B) It carries triglyceride from tissue to tissue. (C) It transports free fatty acids throughout the body. (D) It serves as a shuttle for cholesterol from the periphery to the liver for metabolism. (E) It carries amino acids from tissue to tissue.

58. The answer is D. HDL is the densest, smallest lipoprotein with a low triglycerol content and high protein content and contains primarily ApoA1. HDL transfers cholesterol ester to other lipoproteins in exchange for various lipids, which it shuttles back to the liver. Within the liver, the cholesterol esters are degraded by lysosomal enzymes. HDL does not carry triglyceride to any great extent, fatty acids, or amino acids throughout the body. The triglyceride in HDL arises via the cholesterol ester transfer protein (CETP) reaction, in which HDL exchanges a cholesterol ester for triglyceride with a VLDL particle.

59. A 15-year-old girl presents with severe menstrual cramping, caused by increased prostaglandin production. Prostaglandins are synthesized most directly from which of the following? (A) Arachidonic acid (B) Glucose (C) Acetyl CoA (D) Oleic acid (E) Leukotrienes

59. The answer is A. Prostaglandins can be synthesized from arachidonic acid (which requires the essential fatty acid, linoleate, for its synthesis). They cannot be directly synthesized from glucose, acetyl CoA, or oleic acid. Leukotrienes are also derived from arachidonic acid. Prostaglandins are not derived from leukotrienes.

6. A 23-year-old woman has come to your clinic because of a lump in her breast that she palpated on breast self-examination. Family history reveals that her mother and her aunt both had breast and ovarian cancer. Given this presentation, you suspect the patient may have a hereditary mutation. Which one of the following genes, when mutated, is implicated in breast and ovarian cancer? (A) BRCA-1 (B) MLH-1 (C) Ataxia telangiectasia (AT) (D) p53 (E) Rb

6. The answer is A. BRCA-1 is a gene implicated in breast and ovarian cancer (as is BRCA-2). About 5% of breast cancers are due to mutations in this gene, and a strong family history of both breast and ovarian cancer is highly suspicious, especially in such a young patient. Mutations in MLH-1, a gene involved in DNA repair, are correlated with hereditary nonpolyposis colon cancer (HNPCC). Patients with AT mutations demonstrate an increase in leukemias and lymphomas. Loss of p53 function is found in many tumor types (not just breast and ovarian cancers). Mutations in p53, associated with breast cancer, signal an aggressive cancer type, with a poor prognosis for recovery. Mutations in Rb lead to retinoblastoma and lung tumors, but not to tumors of the breast and ovary.

6. A 45-year-old woman presents with oily, foul-smelling stool, which appears to be due to an obstruction of the bile duct. Which of the following statements correctly describes bile salts? (A) They can act as detergents, aiding in lipid digestion. (B) They are stored in the intestines. (C) Ninety-five percent of bile salts are excreted in the feces, and 5% are recycled back to the liver. (D) Bile salts are synthesized in the intestines. (E) Squalene and lanosterol are examples of bile salts.

6. The answer is A. Bile salts, synthesized in the liver and stored in the gallbladder, act as detergents, aiding in lipid digestion. The bile salts are secreted into the intestine in response to cholecystokinin. An inadequate concentration of bile salts in the intestines can lead to oily, foulsmelling stool with a high fat content, a condition known as steatorrhea. Ninety-five percent of bile salts are recycled back to the liver, and 5% are excreted in the feces. Squalene and lanosterol are intermediary compounds in cholesterol synthesis, not bile acid synthesis.

6. Which one of the following statements is correct concerning the formation of muscle lactate during exercise? (A) Lactate formation occurs when the NADH/ NAD+ ratio is high. (B) The liver preferentially converts lactate into carbon dioxide and water. (C) The heart preferentially converts lactate into glucose. (D) Lactate formation is less likely to be found in the eye, testes, and RBCs than in other tissues. (E) The intracellular pH is typically increased when lactate is produced.

6. The answer is A. Lactate formation occurs in a high NADH/NAD+ state. The NADH has been generated by the glyceraldehyde 3-phosphate dehydrogenase reaction, and NAD+ needs to be regenerated, under anaerobic conditions, for glycolysis to continue. The formation of lactate, an acid, results in a drop in pH. Lactate formation commonly occurs in poorly vascularized tissues (e.g., eye, renal medulla, testes) or tissues without mitochondria (e.g., RBCs). Lactate, once formed in muscle, diffuses into the bloodstream and is used by other tissues. In the liver, lacate is converted to glucose, whereas in the heart, lactate is preferentially oxidized to provide energy

Questions 5-11: Using answers (A) through (G) below, match the correct drug with its clinical effect 6. Methotrexate (A) Folic acid analog(s) that inhibit(s) dihydrofolate reductase, which leads to an inhibition of purine and deoxythymidine synthesis (B) Folate analog(s) that bind(s) specifically to bacterial dihydrofolate reductase and used in conjunction with sulfonamides (C) Histamine and basophil inhibitor(s) used to treat allergic reactions (D) Increase(s) synaptic serotonin concentration in the treatment of depression (E) Inhibitor(s) of tyrosine residue iodination (F) Replenish(es) glutathione levels during acetaminophen toxicity (G) Release(s) NO, create(s) smooth muscle relaxation, and very effective as antihypertensive

6. The answer is A. Methotrexate is a structural folic acid analog that inhibits dihydrofolate reductase and blocks de novo purine and deoxythymidine synthesis. This reduces cell proliferation and is used in the treatment of cancer and rheumatoid arthritis.

6. A 32-year-old bodybuilder has decided to go on a diet consisting of only egg whites to ensure optimal protein for muscle growth. After a few weeks, he notices decreased energy and is found to be hypoglycemic. A nutritionist tells the patient that he most likely has a functional biotin deficiency. Which of the following enzymes is unable to catalyze a key step in synthesizing glucose from pyruvate? (A) Pyruvate carboxylase (B) Phosphoenolpyruvate carboxykinase (C) Fructose 1,6-bisphosphatase (D) Glucose 6-phosphatase (E) Phosphoglycerate kinase

6. The answer is A. Pyruvate carboxylase requires the cofactor biotin to catalyze the irreversible carboxylation of pyruvate to OAA. Although the conversion of OAA to phosphoenolpyruvate is also irreversible and requires energy in the form of GTP, the enzyme catalyzing this step, PEPCK, does not require a cofactor. As with pyruvate carboxylase and PEPCK, F-1,6-bisphosphatase and glucose 6-phosphatase are used to bypass the irreversible steps of glycolysis. Neither of those enzymes requires biotin, which is used exclusively for carboxylation reactions. Phosphoglycerate kinase catalyzes a reversible reaction in which ATP is generated from 1,3-bisphosphoglycerate, or in which ATP is utilized to generate 1,3-bisphosphoglycerate during gluconeogenesis. Because no carboxylation reaction is involved in the phosphoglycerate reaction, biotin is not a required cofactor for that enzyme.

6. A 54-year-old Native American living on an Indian reservation in southwest Arizona is brought into the clinic by a family member. They are concerned because of impaired memory, diarrhea, and a rash on the face, neck, and dorsum of the hands. Which vitamin deficiency do these symptoms represent? (A) Niacin (B) Cobalamin (C) Folic acid (D) Vitamin C (E) Vitamin E

6. The answer is A. The patient presents with the classic presentation of pellagra, or niacin deficiency, with diarrhea, dementia, and dermatitis. Niacin is synthesized from the essential amino acid tryptophan, which is particularly deficient in corn-based diets. However, niacin is still required in the diet because the amount of niacin derived from tryptophan is insufficient for daily needs. A cobalamin deficiency (vitamin B12) will lead to megaloblastic anemia along with methylmalonic acidemia. Folic acid deficiency often manifests with megaloblastic anemia. Vitamin C deficiency results in scurvy. Vitamin E deficiency is rare and can result in neurologic symptoms.

6. A young infant, who was nourished with a synthetic formula, was found to have a serum and urine sugar compound that yielded a positive reducing-sugar test but was negative when measured with glucose oxidase. Treatment of the urine and serum with acid to cleave glycosidic bonds did not increase the amount of reducing sugar measured. Which of the following compounds is most likely to be present in this infant's urine and serum? (A) Glucose (B) Fructose (C) Sorbitol (D) Maltose (E) Lactose

6. The answer is B. Fructose gives a positive result in a reducing-sugar test and a negative result in a glucose oxidase test. Glucose would yield a positive test result with the enzyme glucose oxidase. Sorbitol has no aldehyde or ketone group and, thus, is not a reducing sugar and cannot be oxidized in the reducing-sugar test. Maltose and lactose are disaccharides that undergo acid hydrolysis, which doubles the amount of reducing sugar. Because fructose is a monosaccharide, acid would have no effect on the amount of reducing sugar present.

6. MELAS is a mitochondrial disorder characterized by mitochondrial encephalopathy, lactic acidosis, and strokelike episodes. If a cell were to contain 100% nonfunctional mitochondria, what would be the net ATP yield that would be produced from 1 mole of glucose? (A) 1 mole (B) 2 moles (C) 4 moles (D) 8 moles (E) 0 moles

6. The answer is B. Mitochondria contain both the TCA cycle and the ETC. Therefore, diseases that compromise and fully reduce mitochondrial activity will result in energy production solely through glycolysis. This net energy production per mole of glucose is 2 moles of ATP and 2 moles of NADH, which are unable to undergo further oxidation via mitochondrial oxidative phosphorylation. The NADH will reduce pyruvate to form lactate such that glycolysis can continue. The ''L'' in MELAS stands for lactic acidosis.

6. A 28-year-old man complains of a rash on his sun-exposed skin, diarrhea, and loss of balance. His urinalysis results are significant for an increase in neutral amino acids, and he is diagnosed with Hartnup disease. Which one of the following amino acids would have been found in the urine sample? (A) Lysine (B) Phenylalanine (C) Arginine (D) Histidine (E) Glutamate

6. The answer is B. Phenylalanine is the only amino acid listed as an answer choice that that does not have an ionizable side chain. The symptoms of dermatitis (rash), diarrhea, and dementia are the ''3 Ds'' of pellagra. Pellagra is a congregate of symptoms resulting from vitamin B3 (niacin) deficiency. In Hartnup disease, the transport of neutral amino acids into intestinal epithelial cells is compromised, and this includes tryptophan. Because tryptophan is a precursor for niacin synthesis, if the patient has a low niacin uptake, the symptoms of pellagra can manifest.

65. Which statement is true regarding the excretion of nitrogenous products? (A) Ninety percent of urea excretion occurs by the liver. (B) Ninety percent of urea excretion occurs by the kidneys. (C) Correctly measuring the kidney's functional ability to excrete nitrogenous wastes is difficult. (D) There is no link between the urea cycle and the TCA cycle. (E) Liver urea synthesis is not inducible.

65. The answer is B. Ninety percent of urea excretion occurs by the kidney (about 30 g/day), and the kidney's functional ability to excrete nitrogenous wastes is measured by the blood urea nitrogen (BUN). The liver is responsible for urea synthesis, and its synthesis is variable and dependent on protein intake. A key relationship exists between the urea cycle and the TCA cycle: one of the urea nitrogens supplied to the urea cycle as aspartic acid is formed from the TCA cycle intermediate oxaloacetic acid. The TCA cycle intermediates are formed by cytoplasmic isozymes of the TCA cycle enzymes because these reactions occur in the cytoplasm.

6. A patient coming into an outpatient clinic for metabolic blood work has a portion of his blood subjected to ion exchange chromatography on a carboxymethyl cellulose column. The patient's serum profile indicates less protein is binding than in a normal blood sample. How might this be interpreted? (A) Some serum proteins of the patient are deficient in sialic acid residues. (B) Some serum proteins of the patient are deficient in glycosaminoglycan side chains. (C) The patient's serum contains more negatively charged proteins than normal. (D) Some serum proteins are deficient in a substituent identical to diethylaminoethyl cellulose. (E) The patient has a silent mutation in a hemoglobin variant.

6. The answer is C. A carboxymethyl cellulose column contains a negative charge and binds positively charged proteins. If fewer proteins in the patient's blood are binding to the column, the patient's blood contains either fewer positively charged proteins or an increased amount of negatively charged proteins compared with the normal case. Both sialic acid and glycosaminoglycans are negatively charged. Reducing the levels of both of these agents would increase protein binding to the column. Diethylaminoethyl cellulose is not found on proteins. A silent mutation does not change the charge distribution of a protein.

6. A 50-year-old man presents with personality changes and an uncontrolled choreic (dancing) movement of one arm. His mother has Huntington disease, which is an autosomal dominant disease characterized by an abnormally long trinucleotide repeat sequence in a specific area on one chromosome. He would like to know if he also has Huntington disease. Which of the following techniques would be most precise in detecting this genetic abnormality? (A) Karyotyping (B) Phenotype characterization (C) DNA sequencing (D) Amniocentesis (E) Chorionic villus sampling

6. The answer is C. DNA sequencing is the most precise way of detecting small genetic changes in a specific area on one chromosome. Karyotyping enables visualization of chromosomes under light microscopy. Chromosomes are arrested in metaphase and are differentiated according to size and centromere location. This technique is useful for chromosomal abnormalities such as gene translocations but usually is not precise enough to detect small genetic changes on the scale of several to tens of nucleotides, as would occur with an expansion of a trinucleotide repeat. Characterizing the patient's phenotype or clinical manifestation is helpful for diagnosis, but it will not detect a patient's genetic abnormality. Amniocentesis and chorionic villus sampling are helpful for detecting fetal genetic abnormalities but are not useful for detecting genetic abnormalities in adults.

6. A 16-year-old girl presents with extreme slenderness. Her body weight is 35% below expected. She feels as though she is obese and severely restricts her food intake. She is diagnosed with anorexia nervosa. In this patient, breakdown of fatty acids is required to provide energy. Before being oxidized, fatty acids are activated in the cytosol to form which of the following? (A) ATP (B) CoA (C) Fatty acyl CoA (D) Carnitine (E) Malonyl CoA

6. The answer is C. Long-chain fatty acids are activated, in a reaction requiring ATP and CoA, to a fatty acyl CoA. Carnitine reacts with fatty acyl CoA, forming fatty acyl carnitine, in order to transport the fatty acid across the mitochondrial membrane. Malonyl CoA is an intermediate in fatty acid synthesis.

Questions 3-7: Using answers choices (A) through (E) below, match the clinical vignette with the appropriate defective or deficient enzyme. (A) Glucose 6-phosphate dehydrogenase (B) Galactose 1-phosphate uridylyltransferase (C) N-acetylglucosamine 1-phosphate transferase (D) Galactokinase (E) Fructokinase 6. A 15-month-old child has obvious signs of developmental delay. This includes an inability to roll over and poor head control. Gingival hypertrophy, coarse facial features, and an umbilical and aortic insufficiency are also noted in this child on physical examination.

6. The answer is C. The child has the symptoms of I-cell disease, which is a defect in N-acetylglucosamine 1-phosphate transferase. This results in an inability to appropriately tag lysosomal enzymes so that they can be routed to the lysosomes. Instead, they are secreted from the cell. This results in a loss of lysosomal function and a severe lysosomal storage disease.

6. A 23-year-old Golden Gloves boxing contender presents with assorted metabolic disorders, most notably ketosis. During the history and physical examination, he describes his training regimen, which is modeled after the Rocky films and involves consuming a dozen raw eggs a day for protein. Raw eggs contain a 70-kD protein called avidin, with an extremely high affinity for a cofactor required by propionyl CoA carboxylase, pyruvate carboxylase, and acetyl CoA carboxylase. The patient is functionally deficient in which one of the following cofactors? (A) Tetrahydrobiopterin (B) Tetrahydrofolate (C) Biotin (D) Methylcobalamin (E) Pyridoxal phosphate

6. The answer is C. The cofactor required for propionyl CoA carboxylase, pyruvate carboxylase, and acetyl-CoA carboxylase is biotin. Avidin binds extremely tightly (hence the name avidin) to biotin, which can then no longer be used by these enzymes as a cofactor. Loss of pyruvate carboxylase activity reduces gluconeogenesis, so hypoglycemia and ketosis will result. Avidin does not bind to tetrahydrobiopterin, tetrahydrofolate, B12, or pyridoxal phosphate

6. A 50-year-old woman complains of feeling warm all of the time. Her eyes appear as though they are ''bulging out of their sockets'' (proptosis). She sees a family physician to evaluate her condition. Laboratory tests demonstrate a decreased level of TSH. Which of the following would you expect in this patient? (A) Reduced blood pressure (B) Weight gain (C) Increased basal metabolic rate (D) Reduced heart rate (E) Excess sleep

6. The answer is C. This patient presents with the symptoms and laboratory results consistent with hyperthyroidism. Thyroid hormone increases the basal metabolic rate, and if excess thyroid hormone is released, weight loss and an increase in body temperature will result. Thyroid hormone release is regulated by TSH (thyrotropin) secretion from the pituitary gland. Low TSH would occur by feedback inhibition of its release by thyroid hormone. The other answer choices, such as reduced blood pressure, weight gain, and excess sleep, are all observed with hypothyroidism, not hyperthyroidism.

7. A 43-year-old man comes to the emergency room with a headache and blurred vision. He complains that his wedding ring no longer fits him, and that his favorite hat no longer fits on his head. His wife feels that his nose has become wider, and he is diagnosed with acromegaly. Which of the following metabolic effects would you expect in this patient? (A) Decreased protein synthesis (B) Inhibition of gluconeogenesis (C) Inhibition of lipolysis (D) Increased protein synthesis (E) Gigantism

7. The answer is D. Acromegaly results from a growth hormone-producing tumor of the anterior pituitary. GH is an important metabolic regulator, stimulating primarily anabolic pathways. As such, increased GH levels result in increased protein synthesis, increased gluconeogenesis, and increased lipolysis. GH excess in children before the closing of the growth plates of bones results in gigantism, but would not have this effect in a grown man.

6. A 64-year-old man complains of an acute onset of unilateral eye pain and reduction in visual acuity. On physical examination, you notice conjunctival injection (eye redness) and a middilated and nonreactive pupil. Funduscopic examination reveals cupping of the optic disc. Recognizing the signs and symptoms as glaucoma, you administer the medication acetazolamide to decrease the production of aqueous fluid and lower the intraocular pressure. Acetazolamide is a noncompetitive inhibitor of carbonic anhydrase and, therefore, will lead to which of the following concerning the kinetic constants of carbonic anhydrase? (A) An increase in the apparent Km (B) A decrease in the apparent Km (C) An increase in Vmax (D) A decrease in Vmax (E) A decrease in both the apparent Km and Vmax

6. The answer is D. A noncompetitive inhibitor binds to the enzyme at a site different than where the substrate binds. Therefore, the number of enzymes capable of catalyzing the reaction is decreased, resulting in a decrease in Vmax. Because the inhibitor binds at a site distinct from that of the substrate, increasing the substrate concentration cannot overcome the effect of the inhibitor. Consequently, Km remains unchanged, and the inhibitor does not interfere with substrate binding.

6. A sprinter is trying to optimize his performance. He has calculated that, even under anaerobic conditions, his glycogen stores will supply him with enough energy to last the race. What would the energy difference be between using glucose from a dietary source versus relying solely on glucose from glycogen stores as fuel for his race? (A) Dietary would give 1 more mole of ATP/ glucose (B) Dietary would give 2 more moles of ATP/ glucose (C) Dietary would give the same ATP/glucose (D) Dietary would give 1 less mole of ATP/ glucose (E) Dietary would give 2 less moles of ATP/ glucose

6. The answer is D. Dietary glucose would enter the blood as free glucose, be transported into the muscle cell, and then be metabolized through the glycolytic pathway, producing lactate. The net yield would be 2 moles of ATP per mole of glucose metabolized (there is a loss of 2 moles of ATP for the hexokinase and phosphofructokinase-1 steps, and a gain of 4 moles of ATP from the phosphoglycerate kinase and pyruvate kinase steps). Glucose derived from glycogen, however, is primarily released as glucose 1-phosphate (not free glucose), so the utilization of ATP at the hexokinase step is bypassed, resulting in a net yield of 3 moles of ATP per 1 mole of glucose residues released from glycogen. This calculation ignores the 10% of glucose residues in glycogen that are released as free glucose by the action of the debranching enzyme.

6. A third-year medical student joins a laboratory that studies gene regulation. The laboratory uses bacteria to study gene expression and metabolic regulation after exposure to toxic compounds. The goal is to generalize the observations seen in prokaryotic cells in response to the toxins and to compare the response to eukaryotic cells. Which of the following statements is true regarding prokaryotic and eukaryotic genomes? (A) They are both diploid. (B) They both organize and compact their DNA with histones. (C) They both have short repetitive DNA sequences throughout their genome. (D) They both organize their genes into operons. (E) They both use the same genetic code to convert codons to amino acids.

6. The answer is E. Although there are some differences in mitochondria, the genetic code is nearly universal. Only eukaryotes have a diploid genome, which is organized into chromatin with the aid of highly positively charged histones. Only eukaryotes have numerous short repetitive DNA sequences throughout their genome, like Alu sequences, which are found in primates and humans. Lastly, prokaryotes, but not eukaryotes, organize transcriptional units as operons, multiple gene products obtained from a single mRNA.

6. A 55-year-old man has been attempting to lose weight using a low-carbohydrate diet. After 2 months of little success, he confides to his son that he adds sugar to his coffee in the morning and after dinner. He tells him that only some of the sugar will be absorbed and should not be the cause of his limited success. The son, a medical student, states that glucose is almost completely absorbed from the gut. What type of transport mechanism does glucose utilize for gastrointestinal absorption? (A) Passive (B) Facilitated (C) Active (D) Passive and facilitated (E) Active and facilitated

6. The answer is E. Glucose absorption in the small intestine occurs via two types of transport. The first is facilitated transport (a form of passive transport in which molecules move down a concentration gradient with the assistance of transport proteins). The second is active transport (requires energy, usually in the form of ATP, to allow the substrate to move against a concentration gradient), in which sodium ions are carried along with glucose into cells. An Na+-K+ ATPase pumps Na+ into the blood, which drives the active transport of glucose into the intestinal epithelial cell. Glucose moves down a concentration gradient to enter the blood from the intestinal epithelial cell.

60. An 18-year-old girl presents with a lack of secondary sexual characteristics, such as ageappropriate pubic hair growth and breast development. Her history reveals she has never started menses. She is diagnosed with 17-ahydroxylase deficiency. The levels of various steroid hormones in her serum are found to be abnormal. Steroid hormones are most directly derived from which one of the following? (A) Acetyl CoA (B) Cholesterol (C) Fatty acids (D) Glucose (E) Oleic acid

60. The answer is B. Steroid hormones, such as progesterone, testosterone, 17-b-estradiol, cortisol, and aldosterone, are most directly formed from cholesterol. Cholesterol forms pregnenolone by cleavage of its side chain. Acetyl CoA is the precursor for cholesterol biosynthesis. Fatty acids (including oleic acid) are not utilized in the de novo synthesis of cholesterol, or steroid hormones.

61. A 52-year-old man takes a daily 81-mg aspirin for cardioprotective purposes. Aspirin will inactivate a key enzyme in platelets for which length of time? (A) 24 hours (B) Permanently (C) 12 hours (D) 48 hours (E) 72 hours

61. The answer is B. Aspirin inhibits the enzyme cyclooxygenase within platelets and it is a permanent inactivation. Aspirin will form a covalent linkage with the enzyme at the active site, leading to irreversible inhibition. Inhibition of cyclooxygenase blocks the synthesis of prostaglandins and thromboxane A2.

62. A 1-year-old infant's arms and legs have become spastic and rigid. Analysis shows an abnormally low level of sphingomyelinase activity, causing accumulation of sphingomyelin. Which of the following substrates combine to directly form sphingomyelin? (A) Serine and palmitoyl CoA (B) Fatty acyl CoA and sphingosine (C) Palmitoyl CoA and ceramide (D) Phosphatidylcholine and ceramide (E) UDP-galactose and ceramide

62. The answer is D. Patients with Niemann-Pick disease have a deficiency of sphingomyelinase activity. Sphingomyelin is formed when phosphatidylcholine reacts with ceramide. Serine and palmitoyl CoA condense to form a precursor of sphingosine. Once sphingosine is formed, a fatty acyl CoA combines with sphingosine and forms ceramide. UDP-galactose and ceramide combine to form galactocerebroside.

66. A 16-year-old girl is found by her parents on the bathroom floor unconscious. There is an empty bottle of acetaminophen in the toilet. She is rushed to the hospital, where she is given several doses of N-acetylcysteine. Acetaminophen overdose is potentially life-threatening because it depletes cellular stores of which substance? (A) Nitric oxide (B) Histamine (C) Creatine (D) Glutathione (E) Serotonin

66. The answer is D. Glutathione plays an important role in detoxifying acetaminophen. As cellular stores of glutathione are depleted and hepatocytes are damaged, jaundice, encephalopathy, and accumulation of toxic substances occur. Nitric oxide is an important vasodilator. Histamine mediates HCl release from the stomach as well as bronchoconstriction of the lungs. Creatine is a storage form of high-energy phosphate when phosphorylated. Serotonin is a neurotransmitter involved in mood and depression. N-acetylcysteine can be used as a precursor for glutathione (g-glutamyl-cysteinyl-glycine) biosynthesis.

67. A 43-year-old man with a long history of poorly controlled hypertension presents to the emergency room with a severe headache. His blood pressure is found to be dramatically elevated at 250/148mm Hg. Which of the following products, derived from amino acids, can be employed to treat his hypertension? (A) Melanin (B) Nitric oxide (C) GABA (D) Dopamine (E) Serotonin

67. The answer is B. Nitric oxide, also referred to as endothelium-derived relaxing factor (EDRF), relaxes the smooth muscle of blood vessels and thus lowers blood pressure. Nitric oxide is derived from arginine. Melanin is the major skin pigment that is derived from tyrosine. GABA is an inhibitory neurotransmitter (derived from glutamic acid), whereas both dopamine and serotonin (dopamine from tyrosine, and serotonin from tryptophan) are normally stimulatory neurotransmitters involved in affecting mood.

68. A 56-year-old woman develops diarrhea, flushing, wheezing, and a heart murmur. A computed tomography (CT) scan of the abdomen demonstrates a mass in the ileum along with multiple metastatic liver lesions. A biopsy reveals a diagnosis of a carcinoid tumor and displays elevated levels of the serotonin metabolite, 5-hydroxyindole acetic acid (5-HIAA). Serotonin is normally produced from which amino acid? (A) Tyrosine (B) Arginine (C) Histidine (D) Glycine (E) Tryptophan

68. The answer is E. Serotonin, which is overproduced in carcinoid syndrome, is an indolamine neurotransmitter derived from tryptophan. Tyrosine is the precursor for catecholamine neurotransmitters, including dopamine, norepinephrine, and epinephrine. Nitric oxide, an important vasodilator, is derived from arginine. Histidine forms histamine, which is an important inflammatory mediator. Glycine can function as an inhibitory neurotransmitter.

69. A 59-year-old woman develops a shuffling gait and a pill-rolling tremor. She is referred to a neurologist for evaluation. After a thorough workup, a diagnosis of Parkinson disease is made, and the patient is placed on a monoamine oxidase inhibitor (MAOI). The drug is given to Parkinson patients to decrease the degradation of which of the following? (A) Serotonin (B) Nicotinamide (C) 5-HIAA (D) Endothelium-derived relaxation factor (EDRF) (E) Dopamine

69. The answer is E. Parkinson disease results from a relative deficiency of dopamine. Monoamine oxidase (MAO) and catecholamine O-methyltransferase (COMT) are enzymes that degrade catecholamines such as dopamine, epinephrine, and norepinephrine. MAOs can also degrade serotonin, resulting ultimately in the formation of 5-HIAA. MAO inhibitors will allow these neurotransmitters to remain elevated for extended periods of time. Nicotinamide can be synthesized from tryptophan. EDRF is a short-acting substance that rapidly degrades spontaneously.

7. Which one of the following ailments, seen by an emergency room physician, is most likely caused by enzyme denaturation? (A) A 34-year-old man diagnosed with a gastrinoma complaining of diarrhea for 2 weeks (B) A 58-year-old man with chest pain and shortness of breath with increased activity (C) An 18-year-old boy presenting with a sore throat and fever of 101 F; he has small minimally tender anterior cervical lymph nodes and a red pharynx (D) An 18-month-old boy with a 4-day history of symptoms of an upper respiratory infection presenting with fever, irritability, and pulling at his left ear for the past 24 hours (E) A 48-year-old woman complaining of knee pain after twisting her leg playing tennis

7. The answer is A. Factors that cause protein unfolding include heat, chemical denaturants, and changes in pH. A gastrinoma is a neuroendocrine tumor that secretes excessive gastrin, resulting in increased gastric acid secretion. This, in turn, results in a paradoxical acidic environment in the duodenum and denaturation of the pancreatic digestive enzymes. The diarrhea is a result of an osmotic pull owing to the undigested nutrients' inability to be absorbed in the gut. Although a fever (choice C) included an increase in temperature, most proteins are denatured above 50 C, a temperature well above the normal body temperature of 37 C. The other choices (potential heart attack, choice B; an ear infection, choice D; and a sore knee, choice E) are not initially a result of enzyme denaturation.

7. A 27-year-old firefighter is brought to the emergency room after being exposed to smoke during a training exercise. He looks ill and has labored breathing. He is clutching his head and exhibits an altered mental status. On examination, you note that he appears red, and his pulse oximetry reads 100%. You suspect carbon monoxide toxicity. What is true of the oxygen saturation curve during carbon monoxide toxicity? (A) The oxygen saturation curve is shifted to the left. (B) The oxygen saturation curve is shifted to the right. (C) The effect of carbon monoxide on hemoglobin is similar to that of having increased levels of 2,3 bisphosphoglycerate. (D) The effect of carbon monoxide on hemoglobin is similar to that of a low pH state. (E) The effect of carbon monoxide on hemoglobin is similar to that of an increased temperature state.

7. The answer is A. Oxygen saturation curves relate the saturation of hemoglobin with oxygen for a given partial pressure of oxygen. If carbon monoxide binds to one of the subunits of hemoglobin, the affinity of the other subunits for oxygen is increased (due to the cooperative nature of oxygen binding to hemoglobin). This shifts the oxygen binding curve to the left. Because the oxygen now has a higher affinity for hemoglobin, it is more difficult for hemoglobin to release oxygen to the tissues, leading to hypoxia despite oxygen being bound to hemoglobin. Conditions that shift the curve to the right allow oxygen to be released more readily: low pH, increased PCO2, increased temperature, presence of 2,3-bisphosphoglycerate, and absence of carbon monoxide. In other words, hemoglobin will release oxygen in states that allow for normal binding of oxygen and increased oxygen demands by tissues.

7. A 63-year-old man presents to the emergency room with lower gastrointestinal bleeding, bloating, and cramping. He states that this has happened before and resolves in a day or two. On examination, he has a low-grade fever, left lower quadrant tenderness to palpation, and guaiac-positive stools. Blood work reveals leukocytosis and a normal hematocrit. A computed tomography scan is ordered with oral and intravenous contrast and reveals numerous out-pouchings in his distal colon. Which of the following is true about this condition? (A) Polysaccharides rich in b1,4 bonds of cellulose are a mainstay of treatment. (B) This patient has a digestive enzyme deficiency. (C) A biopsy of the lesions would reveal a malignant process. (D) This patient has gastroenteritis from a heat-stable protein. (E) These signs and symptoms are consistent with Clostridium difficile colitis

7. The answer is A. This patient has diverticulosis, which is out-pouchings on the colonic wall secondary to weakness of the muscle layers that are most often found in the sigmoid colon. Sometimes these out-pouchings can harbor stool and become infected, resulting in diverticulitis. Colonoscopy confirms the diagnosis, although computed tomographic scanning is the most available modality in the acute setting. Although the out-pouchings are notmalignant, colonoscopy may identifymalignant lesions. These symptoms are not consistent with gastroenteritis, and in the absence of diarrhea, recent antibiotic use, or hospitalization, Clostridium difficile infection is quite unlikely. Consuming fiber (such as foods containing cellulose) will increase stool bulk and moisture, which will reduce travel time through the colon, reducing the risk for diverticulitis.

7. A 4-month-old infant presents with muscular weakness that is progressing to paralysis. Examination of the back of the eye shows a cherryred spot on the macula. An abnormally low level of hexosaminidase A is present, causing deposition of certain gangliosides in neurons. The accumulating material in this disorder is which of the following? (A) GM1 (B) GM2 (C) GM3 (D) GD3 (E) GT3

7. The answer is B. This patient has either Tay-Sachs or Sandoff disease. Patients with these diseases have a deficiency of hexosaminidase A (Tay-Sachs), or hexosaminidase A and B (Sandhoff) activity, resulting in the buildup of GM2 in neurons, which can result in neurodegeneration and early death. Hexosaminidase A (which is composed of proteins encoded by the HexA and HexB genes) removes N-acetylgalactosamine from GM2, to form GM3. TheM series of gangliosides contain 1 sialic acid residue; the D series contain 2 sialic acid residues, and the T series contain 3 sialic acid residues.

7. After surgical resection of part of her small intestine, a 40-year-old woman presents with chronic foul-smelling diarrhea and weight loss. She is diagnosed with short bowel syndrome. In this syndrome, fat cannot be properly absorbed, so long-chain fatty acids are mobilized from adipose tissue to generate energy for cell survival. The initiating substrate for fatty acid oxidation is which of the following? (A) Long-chain fatty acid (B) Fatty acyl carnitine (C) Fatty acyl CoA (D) b-Hydroxyacyl CoA (E) Acetyl CoA

7. The answer is C. Fatty acyl CoA undergoes b-oxidation in a spiral involving four steps. Longchain fatty acids are released from adipose cells and must be activated and transported into mitochondria for oxidation. Fatty acyl CoA reacts with carnitine, forming fatty acyl carnitine, which crosses the inner mitochondrial membrane. The acyl group is then transferred back to CoA, forming fatty acyl CoA in the mitochondrial matrix. Subsequent reactions convert the fatty acyl CoA to trans2 fatty enoyl CoA, b-hydroxy acyl CoA, and keto acyl CoA. The end product of fatty acid oxidation is acetyl CoA, which is oxidized via the TCA cycle and oxidative phosphorylation to produce carbon dioxide, water, and ATP.

7. A 36-year-old woman is training for her first marathon, and her coach has her keeping a pace that allows her to stay below her anaerobic threshold. By avoiding anaerobic muscle glycolysis, the pyruvate produced in the muscle does not accumulate because it is converted to which one of the following? (A) Ethanol (B) Lactic acid (C) Acetyl CoA (D) Alanine (E) OAA

7. The answer is C. Glycolysis is dependent on NAD+ (for the glyceraldehyde 3-phosphate dehydrogenase reaction) as a substrate for the pathway to continue to metabolize glucose. Under aerobic conditions, NAD+ is generated via the electron transport chain. Under anaerobic conditions, an oxygen deficit limits the electron transport chain, and NAD+ is generated by the conversion of pyruvate to lactate in mammals and to ethanol in yeast and some microorganisms. When oxygen is not limiting, the pyruvate is converted to acetyl CoA to generate energy via the TCA cycle and oxidative phosphorylation. Because acetyl CoA levels are low during exercise, pyruvate carboxylase is not active, and OAA will not be formed.

7. A 46-year-old longshoreman is brought to the emergency room with muscle twitching, spasms, and difficulty swallowing. Apparently, the cargo container he was unloading from Indonesia contained the rodenticide strychnine. Strychnine interferes with one of the functions of which of the following nonpolar, aliphatic amino acids? (A) Tyrosine (B) Asparagine (C) Glycine (D) Glutamate (E) Lysine

7. The answer is C. Strychnine interferes with neurotransmission by the amino acid glycine. Glycine is the only nonpolar amino acid listed as an answer choice. Tyrosine is a polar aromatic amino acid. Asparagine and glutamine are polar, uncharged amino acids. Lysine is an example of a positively charged amino acid.

7. A woman is pregnant with her fourth child. She has had an uneventful pregnancy and is offered amniocentesis. Cytogenetic analysis performed on amniotic fluid reveals 47,XY, +21.Which of the following statements is true of amniocentesis? (A) The procedure can be performed at 4 to 10 weeks (B) The procedure is unlikely to reveal a cytogenetic abnormality (C) Risk includes miscarriage and spontaneous abortion (D) A biopsy of chorionic villus undergoes DNA analysis (E) The procedure is offered to women younger than 35 years.

7. The answer is C. The primary risk for amniocentesis is miscarriage. Spontaneous abortion can occur if the procedure is done earlier than the recommended time period, which is typically between 14 and 20 weeks's gestation, not as early as 4 to 10 weeks. Amniocentesis is a reliable and accurate procedure for obtaining fetal cells for the detection of cytogenetic abnormalities in the fetus. A biopsy of the chorionic villus is taken for chorionic villus sampling, which is usually performed earlier (at 9 to 12 weeks) than amniocentesis. This is a different procedure than amniocentesis. Amniocentesis is routinely offered to women older than 35 years because advanced maternal age increases the risk for incidence of cytogenetic fetal abnormalities. It is not routinely offered to women younger than 35 years.

7. A 32-year-old woman receives anesthesia in preparation for a laparoscopic cholecystectomy. The anesthesiologist notices a subtle twitch of the masseter muscle in the jaw, followed by sinus tachycardia and an increase of the endexpiratory CO2. He immediately recognizes the early signs of malignant hyperthermia and administers dantrolene. Dantrolene is a muscle relaxant that acts specifically on skeletal muscle by interfering with the release of calcium from the sarcoplasmic reticulum. Which of the following enzymes would be affected by this action? (A) Phosphoglucomutase (B) Glucokinase (C) Glycogen synthase (D) Glycogen phosphorylase kinase (E) Glucosyl 4:6 transferase

7. The answer is D. Glycogen regulation in skeletal muscle versus the liver is matched well to the functions of the muscle and liver. Muscle glycogen functions as storage for mechanical energy needs, whereas liver glycogen functions to maintain blood glucose levels. With regard to regulation, when a motor neuron stimulates the release of calcium from the sarcoplasmic reticulum, the calcium binds to calmodulin and activates glycogen phosphorylase kinase, which in turn activates glycogen phosphorylase. Thus, a reduction in calcium release from the sarcoplasmic reticulum would lead to inactivation of glycogen phosphorylase kinase. None of the other enzymes listed contains a calmodulin subunit, and they would not be regulated by calcium. Glucokinase (in the liver) converts glucose to glucose 6-phosphate. Phosphoglucomutase converts glucose 6-phosphate to glucose 1-phosphate. Glycogen synthase is activated via phosphorylation. Finally, glucosyl 4:6 transferase is the enzyme that creates branches in glycogen.

7. A 58-year-old man is awakened by a throbbing ache in his great toe. He has suffered these symptoms before, usually after indulging in a rich meal. On examination, he is noted to have a greatly inflamed great toe; also of note are several small nodules on the antihelix of his ear. Inhibition of which of the following proteins might prevent further occurrences of this man's ailments? (A) Carbamoyl phosphate synthetase II (B) HGPRT (C) PRPP synthetase (D) Xanthine oxidase (E) Orotate phosphoribosyl transferase

7. The answer is D. Gout is caused by either the increased production or reduced excretion of uric acid, leading to the deposition of urate crystals. Allopurinol, a xanthine oxidase inhibitor, decreases the production of urate from hypoxanthine and xanthine. Carbamoyl phosphate synthetase II is an enzyme in pyrimidine biosynthesis and is not involved in urate formation. HGPRT is an enzyme in the pathway for purine salvage. Inhibition of HGPRT activity would increase urate production. Orotate phosphoribosyl transferase is important in the synthesis of pyrimidines. PRPP synthetase is an important enzyme in the biosynthesis of purines; loss of its activity would reduce urate production

7. A 4-year-old child is referred by the pediatrician to a pediatric neurologist after presenting with myoclonic seizures and lactic acidosis. The neurologist orders a muscle biopsy, and the pathology returns with the appearance of ''ragged red fibers.'' The parents are informed that the child has MERFF syndrome, a mitochondrial DNA (mtDNA) disorder. Which of the following statements best explains mtDNA? (A) It is inherited equally from both parents. (B) It is replicated with increased fidelity with respect to nuclear DNA. (C) It shares the same genetic code as nuclear DNA. (D) It is a double-stranded circular DNA. (E) It encodes all the proteins necessary for the electron transport chain.

7. The answer is D. Mitochondrial DNA is a small double-stranded circular DNA. It is inherited exclusively from the mother because the ovum contributes the cytoplasm to the zygote, which contains all the mitochondria for the zygote. The mitochondrial DNA replication machinery is less evolved than nuclear DNA and replicates with an error rate that is 5 to 10 times greater than that of the nuclear genome. The genetic code within mitochondria is slightly different than that of the nuclear genome. Finally, mitochondrial DNA encodes only 13 of the numerous subunits of the electron transport chain. The others are encoded by the nuclear genome.

7. A 33-year-old, obese man with an impressive family history of type 2 diabetes is concerned he may develop the disease as well. During a health maintenance examination, his family physician orders several laboratory tests to evaluate the patient. Which of the following results would lead to a diagnosis of diabetes? (A) A single random glucose level of 190 mg/dL (B) The presence of a reducing sugar in his urine (C) A single fasting blood glucose level of 160 mg/dL (D) A 2-hour oral glucose tolerance test with a blood glucose level of 210 mg/dL (E) A single fasting blood glucose level of 110 mg/dL

7. The answer is D. Of all the test values, the one that renders a diagnosis of diabetes in a single episode is a 2-hour oral glucose tolerance test yielding a blood glucose level of 200 mg/dL at the end of the test. A single random glucose level of more than 200 mg/dL (not 190 mg/dL) with symptoms of diabetes would confirm the diagnosis. Guidelines concerning fasting glucose levels indicate that to diagnose, diabetes fasting blood glucose levels of more than 126 mg/dL need to be observed on at least two occasions. Two fasting blood glucose levels between 100 and 125 mg/dL indicate impaired glucose tolerance, or what has been called prediabetes. The presence of a reducing sugar in the urine is not sufficient criteria for diabetes because patients with benign fructosuria would also be positive in such a test and not necessarily glucose intolerant (diabetic).

7. A new test is developed that can nonradioactively ''label'' compounds in the human body. As a physician with a background in the new field of metabolomics, you assess a 21-year-old with classic PKU. The patient is fed phenylalanine with a label in the phenyl ring, and a 24-hour urine sample is collected. Which of the following compounds would you expect to contain the greatest amount of label in this urine sample? (A) Tyrosine (B) Tryptophan (C) Epinephrine (D) Phenylketone (E) Acetate

7. The answer is D. PKU results from a defect in phenylalanine hydroxylase, resulting in a block in the conversion of phenylalanine to tyrosine. Phenylalanine accumulates in cells and is converted to phenylketones, which enter the blood and urine. Tyrosine is the product whose formation is blocked, and epinephrine, a product of tyrosine, would not be synthesized, so it would not contain a ''label.'' Acetate and tryptophan are not derived from labeled phenylalanine in a patient with PKU, so those compounds would not contain the label.

80. A 30-year-old woman presents to an infertility clinic with her husband stating that she and her husband have been unsuccessfully trying to have a baby for 2 years. Over the next few months, her physician runs some tests, which show that she may not be ovulating. Which of the following is responsible for ovulation? (A) Increased FSH (B) Increased estradiol (C) LH surge (D) Increased progesterone (E) Increased progesterone and estradiol

80. The answer is C. A surge of LH at the midpoint of the menstrual cycle stimulates the egg to leave the follicle. Initially, FSH acts on immature follicles to promote maturation. Estradiol causes the endometrium to thicken and vascularize in preparation for implantation. After ovulation, the residual follicle secretes both progesterone and estradiol. Progesterone also causes the endometrium to thicken and vascularize

Questions 3-7: Using answers choices (A) through (E) below, match the clinical vignette with the appropriate defective or deficient enzyme. (A) Glucose 6-phosphate dehydrogenase (B) Galactose 1-phosphate uridylyltransferase (C) N-acetylglucosamine 1-phosphate transferase (D) Galactokinase (E) Fructokinase 7. A 12-month-old, otherwise healthy male has cataracts and galactosemia.

7. The answer is D. The child has a galactokinase deficiency. Galactose cannot be converted to galactose 1-phosphate, so galactose accumulates whenever lactose is present in the diet. The elevated galactose is converted to the sugar alcohol galactitol in the lens of the eye by aldose reductase, which leads to cataract formation. This is a less severe disorder than galactose 1-phosphate uridylyltransferase deficiency because the toxic metabolite galactose 1-phosphate does not accumulate. Galactokinase deficiency is considered the nonclassic form of galactosemia.

7. A scientist is studying oxidative phosphorylation in intact, carefully isolated mitochondria. Upon adding an oxidizable substrate, such as pyruvate, a constant rate of oxygen utilization is noted. The scientist then adds a compound that greatly enhances the rate of oxygen consumption. This compound is most likely which one of the following? (A) Rotenone (B) Carbon monoxide (C) Antimycin (D) Cyanide (E) Dinitrophenol

7. The answer is E. An uncoupler was added to the mitochondria to greatly increase the rate of oxygen consumption. Dinitrophenol will allow free proton diffusion across the inner mitochondrial membrane, thereby dissipating the proton gradient and preventing ATP synthesis. Without an existing proton gradient to ''push'' against, electron flow through the electron transport chain is accelerated, resulting in enhanced oxygen consumption. Rotenone inhibits electron transfer from complex I to coenzyme Q. Carbon monoxide and cyanide block complex IV from accepting electrons. Antimycin blocks electron flow from complex III. Since electron flow is blocked using rotenone, carbon monoxide, cyanide, or antimycin, oxygen uptake will cease.

7. A PCR assay needs to be developed to determine the HIV status of a newborn in the pediatric intensive care unit whose mother is HIV positive. Which set of primers should be used for the assay? (A) The primers should consist of antiparallel complements of two parts of a noninfected human genome. (B) The primers should be designed so that, after annealing with potential infective DNA, the 50 end of primer 1 would ''face'' the 30 end of primer 2. (C) The primers should be synthesized so that, after annealing with potential infective DNA, the 50 end of both primers ''face'' each other. (D) The primers should be designed to be synthesized with dideoxynucleotides to allow sequencing of the mutation. (E) The primers should be designed with identical sequences to those in the HIV genome and must bind to DNA in a complementary, antiparallel manner.

7. The answer is E. For the development of a PCR assay, one requires primers designed to hybridize to the specific HIV target (choice A is incorrect) that are complementary and antiparallel (choices B and C are excluded) and that require deoxyribonucleotides (not dideoxynucleotides, so choice D is incorrect) because one wants continuous DNA synthesis to occur. If one examines appropriately designed primers, the 30 ends face each other after hybridization to their respective templates.

7. A 55-year-old woman presents with crushing substernal chest pain and shortness of breath. A coronary artery is occluded owing to an atherosclerotic plaque, and a high myocardial infarct is diagnosed. High serum HDL levels are protective against the development of atherosclerosis because HDL does which of the following? (A) Inhibits cholesterol production by the liver (B) Inhibits HMG-CoA reductase (C) Increases VLDL production (D) Increases LDL production (E) Brings cholesterol esters back to the liver

7. The answer is E. HDL is known as the ''good'' lipoprotein particle because HDL scavenges cholesterol from the periphery (from cell membranes and from other lipoproteins) and brings cholesterol esters back to the liver, where they can be converted to and excreted as bile salts. This is known as reverse cholesterol transport. HDL does not inhibit cholesterol production by the liver. Statin medications inhibit HMG-CoA reductase activity, not HDL production. Increasing VLDL or LDL will facilitate the development of atherosclerosis, not protect against its development.

Questions 5-11: Using answers (A) through (G) below, match the correct drug with its clinical effect 7. Propylthiouracil and methimazole (A) Folic acid analog(s) that inhibit(s) dihydrofolate reductase, which leads to an inhibition of purine and deoxythymidine synthesis (B) Folate analog(s) that bind(s) specifically to bacterial dihydrofolate reductase and used in conjunction with sulfonamides (C) Histamine and basophil inhibitor(s) used to treat allergic reactions (D) Increase(s) synaptic serotonin concentration in the treatment of depression (E) Inhibitor(s) of tyrosine residue iodination (F) Replenish(es) glutathione levels during acetaminophen toxicity (G) Release(s) NO, create(s) smooth muscle relaxation, and very effective as antihypertensive

7. The answer is E. Propylthiouracil and methimazole are used in the treatment of hyperthyroidism by inhibiting tyrosine residue iodination. This blocks the synthesis of mature and active thyroid hormone, alleviating hyperthyroidism.

7. A 23-year-old man is seen in the gastroenterologist's office for a referral concerning a family history of a1-antitrypsin deficiency. The physician arranges for the interventional radiologist to perform a liver biopsy, which will allow a determination of the extent of accumulation of nonsecreted protein in the hepatocyte. Which of the following statements is true concerning proteins like a1-antitrypsin that are normally secreted from the cell? (A) They are synthesized on ribosomes attached to the smooth endoplasmic reticulum. (B) They contain a hydrophilic signal sequence. (C) The signal sequence is found on the C terminus of the protein. (D) Glycosylation takes place only in the endoplasmic reticulum. (E) Proteins travel from the endoplasmic reticulum to the Golgi apparatus and are ultimately secreted by exocytosis.

7. The answer is E. Proteins destined to be secreted from the cell travel to the cell membrane via the endoplasmic reticulum and the Golgi apparatus. Such proteins are synthesized on ribosomes attached to the RER. The proteins enter the endoplasmic reticulum lumen with the aid of a hydrophobic signal sequence at the N terminus of the protein. Glycosylation takes place in both the endoplasmic reticulum and the Golgi before sorting and sending the protein to its final destination.

7. A medical student is assigned to a patient in the intensive care unit. A review of the patient's medications shows that he is taking a proton pump inhibitor (PPI). This class of drugs inhibits the production of which of the following major acids produced by the body? (A) Phosphoric acid (B) Sulfuric acid (C) Lactic acid (D) beta-Hydroxybutyric acid (E) Hydrochloric acid

7. The answer is E. The proton pump inhibitors, such as omeprazole, inhibit the H+-K+ ATPase, which is responsible for the production of hydrochloric acid by the gastric parietal cells. Many patients are given these medications in the hospital to prevent the development of gastric ulcers. Phosphoric acid and sulfuric acid are important acids that are byproducts of normal metabolism. Lactic acid is yet another product of metabolism, primarily anaerobic glycolysis. Beta-Hydroxybutyric acid is a ketone body that results from lipid metabolism.

70. A 12-year-old boy develops convulsions and is referred to a neurologist. After running an electroencephalogram (EEG), it is determined that the child has epilepsy. He is started on a benzodiazepine, which stimulates GABA activity. GABA is derived from its precursor amino acid via which one of the following types of reactions? (A) Deamination (B) Decarboxylation (C) Hydroxylation (D) Iodination (E) Methylation

70. The answer is B. Decarboxylation of glutamate results in the formation of the inhibitory neurotransmitter GABA. This is the only reaction type that converts glutamate to GABA. Hydroxylation of tyrosine forms dopa. Thyroid hormone requires iodination of tyrosine molecules. Methylation of norepinephrine forms the adrenal hormone epinephrine. A deaminated metabolite of catecholamines is vanillylmandelic acid, which is excreted in the urine.

71. A 63-year-old woman reports a long history of joint pain. Her fingers are severely deformed secondary to rheumatoid arthritis. Upon visiting a rheumatologist, she is started on methotrexate. This drug inhibits which of the following conversions? (A) Dopamine to norepinephrine (B) Tyrosine to dopa (C) Dihydrofolate to tetrahydrofolate (D) Histamine to formiminoglutamate (FIGLU) (E) Norepinephrine to vanillylmandelic acid

71. The answer is C. Methotrexate inhibits the enzyme dihydrofolate reductase, which is the enzyme that converts dihydrofolate to tetrahydrofolate. Dopamine b-hydroxylase converts dopamine to norepinephrine. Tyrosine hydroxylase is the enzyme that converts tyrosine to dopa. Histamine is degraded to formiminoglutamate (FIGLU). Norepinephrine is deaminated and methylated by the sequential action of monoamine oxidase (MAO) and catechol O-methyl transferase (COMT).

72. A 5-year-old boy presents with mental retardation and self-mutilation. Blood tests show an elevated uric acid level. He is diagnosed with Lesch-Nyhan syndrome, a disease caused by a defect in hypoxanthine-guanine phosphoribosyl transferase (HGPRT). HGPRT is most significant in which of the following pathways? (A) Purine synthesis (B) Purine salvage (C) Purine degradation (D) Pyrimidine synthesis (E) Pyrimidine degradation

72. The answer is B. Hypoxanthine-guanine phosphoribosyl transferase (HGPRT) and adenine phosphoribosyl transferase (APRT) are enzymes associated with the purine-salvage pathway. Purine bases can be salvaged by reacting with PRPP to re-form nucleotides. If purine bases cannot be salvaged because of a defect in one of these enzymes, purines will instead be converted to uric acid, which will rise in the blood, and can lead to gout. The enzyme is not involved in pyrimidine metabolism.

73. An infant presents with recurrent infections and a markedly decreased lymphocyte count. Elevated levels of adenosine and deoxyadenosine are found in the serum. X-ray shows the virtual absence of a thymic shadow. The defective reaction in this disease is which one of the following? (A) AMP to adenosine (B) Adenosine to inosine (C) Guanosine to guanine and ribose 1-phosphate (D) Inosine to hypoxanthine and ribose 1-phosphate (E) Hypoxanthine to xanthine

73. The answer is B. The child has severe combined immunodeficiency disease (SCID) caused by a deficiency in adenosine deaminase (ADA) activity. ADA converts adenosine to inosine, as well as deoxyadenosine to deoxyinosine. AMP is degraded to adenosine by removal of a phosphate by 50-nucleotidase. There are two reactions catalyzed by purine nucleoside phosphorylase (PNP): (1) guanosine is converted to guanine and ribose 1-phosphate, and (2) inosine is converted to hypoxanthine and ribose 1-phosphate. Like ADA deficiency, PNP deficiency also results in SCID. However, in PNP deficiency, adenosine and deoxyadenosine do not accumulate. Hypoxanthine is converted to xanthine by xanthine oxidase in the purine degradative pathway. Deoxyadenosine accumulation leads to the inhibition of ribonucleotide reductase and inhibition of immune cell maturation.

74. An infant presents with developmental delay, muscle weakness, and anemia. Urine analysis reveals a high level of excreted orotic acid. Ammonia levels are normal in this patient. In which of the following pathways is this disease manifest? (A) Purine synthesis (B) Purine degradation (C) Purine salvage (D) Pyrimidine synthesis (E) Pyrimidine degradation

74. The answer is D. The patient has hereditary orotic aciduria, a defect in the pyrimidine biosynthetic pathway. The defective enzyme is UMP synthase, which converts orotic acid to OMP, and then OMP to UMP. In this disorder, orotic acid accumulates and is released into the blood and then into the urine. A defect in the urea cycle (ornithine transcarbamoylase deficiency) will also lead to orotic aciduria, but is also associated with hyperammonemia. Orotic acid is not formed during purine synthesis or degradation, nor in pyrimidine degradation.

75. An infant presents with neonatal jaundice. After several weeks, the jaundice becomes more exaggerated. The patient has an enzyme deficiency that inhibits conjugation of bilirubin. Which of the following reacts with bilirubin in the conjugation reaction? (A) Vitamin C (B) Iron (C) Ceruloplasmin (D) Porphyrin ring (E) UDP-glucuronate

75. The answer is E. UDP-glucuronate reacts with bilirubin to form bilirubin monoglucuronide. This reaction is catalyzed by bilirubin uridine diphosphate glucuronyl transferase (UDP-GT). Vitamin C increases uptake of iron in the intestinal tract. Iron is in the center of the porphyrin ring in heme. Ceruloplasmin is involved in the oxidation of iron. None of these other factors interact with bilirubin.

76. A 26-year-old woman meets with her family physician to discuss family planning. She is interested in starting a family soon and is looking for advice on what nutritional supplements would be beneficial during pregnancy. The physician suggests which two of the following supplements as being the most important for the health of the fetus? (A) Selenium and vitamin K (B) Copper and riboflavin (C) Iron and folate (D) Vitamin C and vitamin D (E) Vitamin A and biotin

76. The answer is C. Pregnancy is a time of increased metabolic demand, and two of the most important supplements are iron, to prevent anemia, and folate, to prevent neurotubule defects in the developing fetus. Copper and seleniumare trace elements that are rarely deficient. Riboflavin is often found in grain products. Vitamin C and vitamin D are often obtained appropriately from the diet. Vitamin A derivatives are often teratogenic and, therefore, should be avoided during pregnancy. Vitamin K deficiency is common in newborns, and they are often supplemented at birth.

77. An intern is scrubbing in to a complicated surgery that is anticipated to last 15 hours. In preparation, the intern has not eaten or drunk anything for the past 15 hours so that he will not have to go to the bathroom in the middle of the surgery. After 30 hours of fasting, which of the following is most important for the maintenance of normal blood glucose levels? (A) Glycogenolysis (B) Gluconeogenesis (C) Triacylglycerol synthesis (D) Increased insulin release (E) Decreased muscle protein breakdown

77. The answer is B. About 2 to 3 hours after a meal, the liver maintains normal blood glucose by glycogenolysis. Within 30 hours, liver glycogen stores are depleted, leaving gluconeogenesis as the primary process for maintaining normal blood glucose. Ketone bodies are generated, triacylglycerols are broken down, and muscle protein breakdown increases. In the fed state, insulin increases; in the fasting state, glucagon increases, while insulin decreases

78. Which of the following steroid hormones is secreted in response to angiotensin II? (A) Cortisol (B) Aldosterone (C) Both cortisol and aldosterone (D) Neither cortisol nor aldosterone (E) Testosterone

78. The answer is B. Angiotensin II stimulates the synthesis and secretion of aldosterone from the adrenal cortex, but not cortisol or testosterone. Aldosterone is a hormone that instructs the kidneys to retain sodium and excrete potassium.

79. Which of the following stimulates the production of progesterone by the corpus luteum? (A) LH (B) PRL (C) TSH (D) GH (E) FSH

79. The answer is A. Luteinizing hormone (LH) stimulates the corpus luteum to produce progesterone. Prolactin (PRL) stimulates lactogenesis in the breast. Thyroid-stimulating hormone (TSH) is synthesized in the pituitary gland and stimulates the release of T3 and T4 from the thyroid gland. Growth hormone is released from the pituitary and does not affect the corpus luteum. Follicle-stimulating hormone (FSH) stimulates the growth of ovarian follicles. It is not involved in progesterone formation.

8. As an experiment, a high school professor convinces each of his students to put a soda cracker in their mouth and not swallow it. After several minutes, some of the students report a sweet taste. Which of the following enzymes is responsible for this phenomenon? (A) Amylase (B) Sucrase (C) Lactase (D) Maltase (E) Isomaltase

8. The answer is A. Carbohydrate digestion is initiated in the mouth through the action of salivary a-amylase. This enzyme will digest starch, releasing glucose residues. The salivary amylase is rendered inactive in the stomach because of the acidic environment, being replaced by pancreatic a-amylase, which is secreted into the duodenum. The enzyme specifically cleaves the a-1,4 bonds between the glucosyl residues of starch, resulting in glucose polysaccharides of varying numbers of residues. The other glycosidases (sucrase, lactase, maltase, and isomaltase) are associated with enzyme complexes on the brush border of enterocytes of the intestine and would not lead to glucose production in the mouth.

8. A 62-year-old, obese man complains of polydipsia (increased drinking), polyuria (increased urination), and fatigue. A glucose tolerance test confirms the diagnosis of diabetes. He is placed on metformin, which works by which of the following mechanisms? (A) Inhibiting hepatic gluconeogenesis (B) Increasing glucagon levels (C) Increasing cellular responsiveness to circulating insulin (D) Stimulating the release of preformed insulin (E) Replacing the need for endogenous insulin

8. The answer is A. Metformin, a biguanide, is beneficial in the treatment of type 2 diabetes because it inhibits hepatic gluconeogenesis, which is often increased in patients with type 2 diabetes. No known agent to treat diabetes directly affects the secretion of glucagon. Thiazolidinediones are used in the treatment of diabetes because they increase cellular responsiveness to insulin. Sulfonylureas stimulate the release of preformed insulin. None of these agents completely replaces the need for exogenous insulin in patients with insulin-dependent diabetes.

8. Proton pump inhibitors are a mainstay in the treatment of peptic ulcer disease and inhibit the gastric hydrogen ATPase. ATPases are in a class of enzymes that catalyze the hydrolysis of a high energy bond in adenosine triphosphate (ATP) to form adenosine diphosphate (ADP) and a free phosphate ion. The hydrogen ATPase in the gastric mucosal parietal cell utilizes this energy to exchange one hydrogen ion from the cytoplasm for one extracellular potassium ion. What type of transport is this enzyme catalyzing? (A) Antiport coupled transport (B) Symport coupled transport (C) Facilitated diffusion (D) Simple diffusion (E) Osmosis

8. The answer is A. The action of the gastic hydrogen ATPase is in antiport coupled transport: the exchange between hydrogen for potassium is driven by the energy released by the conversion of ATP to ADP. Symport coupled transport, although a form of active transport as well, results in the passage of molecules together across a membrane, such as the glucose-Na+ cotransporter. Facilitated and simple diffusion are passive mechanisms for the transfer of a molecule across a membrane. The driving force for passive and simple diffusion relies primarily on the concentration gradient of the molecule across the membrane and requires no energy. Facilitated diffusion uses a carrier protein to transfer the molecule across the membrane, whereas simple diffusion does not require a carrier. Osmosis is the diffusion of a solvent (usually water in biologic systems) across a semipermeable membrane in response to a difference in solute concentration across the membrane

8. During a medical rotation, a medical student volunteered for a respiratory physiology examination that determines basal metabolic rate and the respiratory quotient. She followed the protocol for a resting individual in the postabsorptive state. Which of the following amino acids would be found in the highest concentration in serum? (A) Alanine and glutamine (B) Arginine and ornithine (C) Glutamate and aspartate (D) Branched chain amino acid (E) Hydrophobic amino acids

8. The answer is A. The postabsorptive state refers to after a meal, at a point at which excess amino acids for the meal are being degraded, with the carbons used for either glycogen or fatty acid synthesis and the nitrogen being used for urea synthesis. Amino acids, which carry nitrogen to the liver from outlying tissues, include alanine and glutamine. None of the other amino acids listed as potential answers (arginine, ornithine, aspartate, glutamate, branched-chain amino acids, hydrophobic amino acids) are utilized as nitrogen carriers in the body and would not be elevated in the blood in the postabsorptive state.

8. A male infant with 3-b-hydroxylase deficiency is born with ambiguous genitalia and severe salt wasting from lack of androgens and aldosterone, respectively. Testosterone, a major androgen, is produced by which of the following reactions? (A) Oxidation of the A ring of pregnenolone (B) Removal of the side chain of the D ring of progesterone (C) Aromatization of the A ring of estradiol (D) Cleavage of the side chain of progesterone (E) Oxidation of aldosterone

8. The answer is B. 3-b-Hydroxylase deficiency is a disease resulting in decreased production of aldosterone, cortisol, and androgens (3-b-hydroxylase is required for production of all three types of steroids). Male infants manifest with ambiguous genitalia (owing to lack of androgens and testosterone), and both males and females show salt wasting (owing to lack of aldosterone). Testosterone is produced only by the removal of the side chain of the D ring of progesterone.

8. A 75-year-old chronic alcoholic man presents to the emergency room after being found unconscious on the floor of his home. On examination, he is found to have a distended abdomen consistent with ascites.Which of the following functions of the liver has been compromised to lead to the finding of abdominal ascites? (A) Lipid metabolism (B) Albumin synthesis (C) Estrogen metabolism (D) Alcohol detoxification (E) Decreased production of coagulation factors

8. The answer is B. Cirrhosis results in several complications, including ascites, caused by decreased oncotic pressure causing extravasation of intravascular fluid. Albumin is the primary protein that maintains oncotic pressure within the vessels. Albumin is synthesized by the liver, and in malnourished or chronic disease states, such as chronic alcoholism, the liver synthesizes reduced levels of albumin, and hypoalbuminemia results. Decreased production of coagulation factors leads to bleeding problems, not ascites production. Ascites production is not due to problems in lipid metabolism, estrogen metabolism, or alcohol detoxification.

8. A 53-year-old, previously successful man recently lost his job and is under investigation for racketeering. His wife returns home to find him slumped over the steering wheel of his idling car in the closed garage. He is unresponsive and has a cherry color to his lips and cheeks. Which of the following is inhibited by the carbon monoxide in the car's exhaust fumes? (A) Complex I of the ETC (B) Cytochrome oxidase (C) The ATP-ADP antiporter (D) The F0 component of the F0-F1 ATPase (E) The F1 component of the F0-F1 ATPase

8. The answer is B. In addition to binding the iron in hemoglobin and impairing oxygen transport, carbon monoxide also terminates cellular respiration by inhibiting cytochrome oxidase, which contains a heme iron. Amytal, a barbiturate, inhibits complex I of the ETC. There are no ironcontaining cytochromes in complex I because complex I contains proteins with iron-sulfur centers. The ATP-ADP antiporter is inhibited by the plant toxin atractyloside. The F0 component of the F0-F1 ATPase is inhibited by the drug oligomycin. There is no inhibitor for the F1 component of the proton-translocating ATPase.

8. One colony of bacteria is split into two Petri plates: one plate with growth medium containing glucose and all 20 amino acids and one medium with one sugar (lactose) and one nitrogen source (NH4 +). Which of the following statements is correct concerning the cells growing in the second medium? (A) cAMP levels will be lower than cells growing in the presence of glucose. (B) CAP protein (cAMP-binding protein) will be bound to the lac promoter. (C) The lac repressor will be bound to the lac operator. (D) RNA polymerase will not bind to the trp promoter. (E) Attenuation of transcription of the trp operon will increase.

8. The answer is B. In the absence of glucose and the presence of lactose (growth medium #2), the lac repressor will be inactive, cAMP levels will rise, and the cAMP-CAP complex will bind to the lac promoter, stimulating transcription of the operon. Tryptophan levels in the cell will be low; thus, the repressor for the trp operon will be inactive, and the operon will be transcribed by RNA polymerase. Attenuation of transcription of this operon will decrease.

8. A 59-year-old man presents with nephrotic syndrome. Immunoelectrophoresis detects a monoclonal immunoglobulin G (IgG) l subtype in his serum and free l light chains in his urine. A renal biopsy shows amyloidosis. Although several different proteins are precursors to amyloid deposition, all amyloid fibrils share an identical secondary structure that is which of the following? (A) a-Helix (B) b-Pleated sheet (C) Triple helix (D) Helix-turn-helix (E) Leucine zipper

8. The answer is B. Regardless of the type of amyloid disease, the pathogenesis is related to the accumulation of b-pleated protein. In the case of multiple myeloma, it is the accumulation of immunoglobulin light chains in the kidney and heart. a-Helical proteins include native fibrillary proteins. The triple helix is a unique structure found in collagen. Helix-turn-helix and leucine zippers are supersecondary structures that are often found in transcription factors, like homeobox proteins (helix-turn-helix).

8. A teenager presents to the emergency room (ER) 2 hours after an overdose of salicylic acid (aspirin). In an attempt to alkalinize the urine, the ER physician administers sodium bicarbonate, which results in salicylate anions becoming trapped within the renal tubule and preventing diffusion of the charged salicylate across the renal epithelium into the patient's systemic circulation. What is the ratio of secreted salicylate to trapped salicylate if the renal dialysate has a pH of 6, and the pKa of salicylic acid is about 3? (A) 10,000:1 (B) 1000:1 (C) 100:1 (D) 1:100 (E) 1:1000

8. The answer is B. Salicylate is excreted when it is uncharged (the AH form) and is trapped in the renal tubule when in charged form (A).

8. A 3-year-old boy in good health began having generalized seizures consisting of a sudden turning of the head to the left, tonic posturing of the left arm, and loss of awareness for 1 to 2 minutes. The patient was successfully treated with the anticonvulsant phenytoin (dilantin). Dilantin is a substrate that binds to and is metabolized by an enzyme in the liver. Which one of the following statements best describes the relationship between an enzyme, substrate, and product? (A) Enzyme-product complexes enhance substrate binding. (B) All the active sites of the enzyme are saturated with substrate at high substrate concentrations. (C) At high substrate concentrations, substrate- substrate interactions interfere with enzyme activity. (D) At low substrate concentrations, none of the enzyme is found in the ES complex. (E) Significant product formation results in activation of the reaction.

8. The answer is B. The rate of an enzyme-catalyzed reaction will generally increase exponentially with respect to substrate concentration until the substrate concentration exhausts the catalytic sites of the enzyme population. Once this occurs, the rate of reaction remains the same regardless of an increase of substrate because all enzymes are saturated (Vmax has been achieved). Substrate cannot bind to enzyme-product complexes because the substrate binding sites are occupied by product. Substrate-substrate interactions are the same regardless of concentration of substrate, and such interactions do not affect enzyme activity. An ES complex can form at low substrate concentration as well as at high substrate concentration. Product formation does not stimulate enzyme activity and can slow down the reaction rate.

Questions 5-11: Using answers (A) through (G) below, match the correct drug with its clinical effect 8. Diphenhydramine (A) Folic acid analog(s) that inhibit(s) dihydrofolate reductase, which leads to an inhibition of purine and deoxythymidine synthesis (B) Folate analog(s) that bind(s) specifically to bacterial dihydrofolate reductase and used in conjunction with sulfonamides (C) Histamine and basophil inhibitor(s) used to treat allergic reactions (D) Increase(s) synaptic serotonin concentration in the treatment of depression (E) Inhibitor(s) of tyrosine residue iodination (F) Replenish(es) glutathione levels during acetaminophen toxicity (G) Release(s) NO, create(s) smooth muscle relaxation, and very effective as antihypertensive

8. The answer is C. Diphenhydramine blocks the effect of histamine at H1 receptor sites and results in the reduction of smooth muscle contraction. It also prevents histamine release and mast cell degranulation. This drug is a mainstay in the treatment of allergic reactions.

8. A 19-year-old, African American male military recruit is about to be sent to Iraq on his assignment. In preparation for his tour of duty, he is given a prophylactic dose of primaquine to prevent malaria. Several days after he begins taking the drug, he develops fatigue and hemolytic anemia. Which of the following proteins is likely deficient? (A) Fructokinase (B) Aldolase B (C) Glucose 6-phosphate dehydrogenase (D) Galactokinase (E) Galactosyl transferase

8. The answer is C. Drugs that cause oxidative stress, like primaquine and sulfa-containing drugs, result in hemolytic disease in patients with G6PDH deficiency. In the presence of strong oxidizing agents, patients lacking G6PDH cannot adequately regenerate reduced glutathione in the red blood cells, which ultimately leads to membrane damage and lysis of the cells. Deficiency of fructokinase is a benign disorder. Deficiency of aldolase B leads to hereditary fructose intolerance. Galactokinase deficiency leads to galactosemia, a slightly milder form than is seen with galactose 1-phosphate uridylyltransferase deficiency. Galactosyl transferase is important in the glycosylation of proteins as well as in the metabolism of substances like bilirubin.

8. A 30-year-old man presents with weakness in his right upper and lower extremities. He is diagnosed with an acute middle cerebral artery stroke secondary to atherosclerosis. Genetic studies show that he has familial hypercholesterolemia, type II, a disorder caused by a deficiency of LDL receptors. Which of the following statements best describes patients with type II familial hypercholesterolemia? (A) After LDL binds to the LDL receptor, the LDL is degraded extracellularly. (B) The number of LDL receptors on the surface of hepatocytes increases. (C) Cholesterol synthesis by hepatocytes increases. (D) Excessive cholesterol is released by LDL. (E) The cholesterol level in the serum decreases.

8. The answer is C. Familial hypercholesterolemia, type II, is due to a mutation in the LDL receptor, which, in a variety of mechanisms, prevents endocytosis of the receptor with bound LDL. Cholesterol synthesis by hepatocytes increases in patients with this disorder because HMG-CoA reductase is not properly inhibited. Normally, after LDL binds to the LDL receptor, the receptor- LDL complex is internalized, the LDL is degraded intracellularly, and the cholesterol released from the LDL migrates to the cytosol. The increase in intracellular cholesterol inhibits HMG-CoA reductase, resulting in decreased cholesterol synthesis by hepatocytes. Because the LDL receptor is deficient in patients with this disease, LDL cannot be taken up by hepatocytes (to release cholesterol), and an extremely high serum cholesterol level results, both from increased cellular synthesis and release of VLDL, and high levels of LDL particles in the circulation.

8. A17-year-old boy receives a pediatric consult for an upper respiratory infection with jaundice. History reveals he frequently becomes jaundiced when he hasminor infections. The patient's history containsno evidence of consanguinity.His father had a fewmild cases of hyperbilirubinemia, and his paternal grandmother had episodic fatigue with mild scleral icterus. Based onthe family history,what is themost plausible inheritance pattern of this hyperbilirubinemia? (A) X-linked dominant (B) X-linked recessive (C) Autosomal dominant (D) Autosomal recessive (E) Mitochondrial

8. The answer is C. Inheritance of an autosomal dominant trait exhibits the following characteristics: transmission from father to son, variable expressivity, a recurrence rate risk of 50%, potential of reduced penetrance, and a significant spontaneous mutation rate. X-linked dominant transmission is excluded on the basis of male-to-male inheritance, as is a mitochondrial inheritance pattern. X-linked recessive transmission is excluded because females as well as males are affected in this family. Assuming no consanguinity, autosomal recessive transmission is unlikely because there is transmission of the disorder through three generations of the family.

8. A 34-year-old man of Italian descent is seen for a yearly physical examination. He has no complaints and is in good health. However, he does relay a family history of anemia, and a complete blood count demonstrates a mild anemia; the physician suspects thalassemia minor in the patient. Thalassemia is often due to an alteration in RNA splicing, which is an essential part of mRNA processing in eukaryotes. Which of the following is a correct statement concerning mRNA processing? (A) Poly(A) RNA is the initial transcript produced, which is subsequently spliced to mRNA. (B) The coding region of the gene is found within introns. (C) The coding regions of the gene are found within exons. (D) All human genes require splicing of introns. (E) The 50 end of the intron contains the nucleotides AG followed by an invariant GU.

8. The answer is C. The coding regions of genes are located within exons. Most, but not all, human genes contain introns, which are spliced out. The splice point at the 50 end of an intron usually has the sequence GU (the 30 end of the exon adjacent to this intron has an invariant AG). At the 30 end, an invariant AG is frequently followed by GU, which are the first two nucleotides at the 50 end of the next exon. Poly(A) RNA is added post-transcriptionally to a mRNA transcript, at the 30 end, before export from the nucleus.

8. A patient presents with dizziness, fatigue, and tremors. A fingerstick test indicates a blood glucose of 36 mmol/L. Of the allosteric activators of glycolysis in the liver, which one of the following is the most important in allowing the liver to maintain a normal blood glucose level? (A) Citrate (B) ATP (C) Fructose 2,6-bisphosphate (D) Glucose 6-phosphate (E) Acetyl CoA

8. The answer is C. The major regulated step of glycolysis is the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate, catalyzed by the enzyme PFK-1. PFK1 is activated by both F- 2,6-BP and AMP and inhibited by ATP and citrate. The modulation of F-2,6-BP levels in the liver is controlled by the insulin-to-glucagon ratio in the blood, which is tied to the regulation of PFK- 2, the enzyme that both produces and degrades F-2,6-BP. Glucose 6-phosphate acts by negative feedback inhibition on hexokinase (an enzyme not present in liver), whereas acetyl CoA is an inhibitor of the pyruvate dehydrogenase reaction.

8. A patient is referred to you by her obstetrician for genetic counseling for an apparent chromosomal microdeletion. To be able to test the heritability of this disorder by RFLP analyses, what is required? (A) SDS gel electrophoresis to detect the gene product of normal and diseased individuals (B) Mutations flanking restriction sites so the entire gene can be sequenced (C) A unique single-stranded oligonucleotide that can be amplified by successive rounds of heating, cooling, and annealing (D) Restriction sites closely linked to the mutation causing the disease (E) Antibodies used with ribonucleotide probes for detection of the altered trait

8. The answer is D. Because a chromosomal microdeletion and subsequent RFLP analysis are based on DNA, the optimal detection is by a Southern blot after digestion with a restriction enzyme that cleaves on sites surrounding the location of the deletion. If a deletion is present, the restriction fragment will be smaller than if the deletion is not present. SDS-PAGE or antibodies will not detect the disorder at the DNA level. Gene sequencing is not required to detect microdeletions. A single-stranded oligonucleotide cannot be amplified by PCR.

8. A 23-year-old woman is referred to an endocrinologist for weight gain, especially around the waist. She also has striae over the abdomen and a rounded appearance to her face. She is found to have Cushing disease. Which of the following would most likely be found in this patient, compared with someone who does not have Cushing disease? (A) Increased synthesis of immunoglobulins (B) Increased protein synthesis (C) Inhibition of lipolysis (D) Increased gluconeogenesis (E) Reduced liver glycogen stores

8. The answer is D. Cushing disease leads to excessive cortisol release due to an adenoma in the pituitary gland, leading to the release of ACTH, which stimulates cortisol release from the adrenal gland. Excess levels of cortisol lead to a variety of debilitating symptoms, which include central obesity, a ''buffalo hump,'' a round face (often referred to as a ''moon face''), excessive sweating, dilation of capillaries (telangiectasia), thinning of the skin with purple or red striae, hirsutism, sexual dysfunction, and mental changes. Cortisol is a ''stress'' hormone and is preparing the tissues for survival during the stressful period. This includes an increase in lipolysis and gluconeogenesis, and glycogen storage. Steroids, including cortisol, suppress the immune response and are often administered exogenously to control autoimmune diseases. Thus, immunoglobulin synthesis would not be increased. Protein synthesis is also not increased in response to cortisol.

8. An 8-year-old boy sees a dermatologist because he has developed vesicles and bullae on his face and arms that appeared after a week-long trip to Florida. His father has a similar condition. A diagnosis of porphyria cutanea tarda is confirmed by finding elevated levels of porphyrins in his serum, urine, and stool. His disease is due to a deficiency of which of the following enzymes? (A) d-ALA dehydratase (B) Porphobilinogen deaminase (C) Uroporphyrinogen III cosynthase (D) Ferrochelatase (E) Uroporphyrinogen decarboxylase

8. The answer is E. Porphyria cutanea tarda is the most common of the porphyrias and results from a deficiency of uroporphyrinogen decarboxylase. Deficiency of d-ALA dehydratase results in d-ALA dehydratase porphyria. Acute intermittent porphyria is due to a deficiency of porphobilinogen deaminase (also known as hydroxymethylbilane synthase). Deficiency of uroporphyrinogen III cosynthase results in congenital erythropoietic porphyria. Finally, ferrochelatase deficiency results in erythropoietic protoporphyria.

8. An infant is born with a high forehead, abnormal eye folds, and deformed ear lobes and shows little muscle tone and movement. After multiple tests, he is diagnosed with Zellweger syndrome, a disorder caused by peroxisome malformation. What type of fatty acid would you expect to accumulate in patients with Zellweger syndrome? (A) Short-chain fatty acids (B) Acetyl CoA (C) Dicarboxylic acids (D) Long-chain fatty acids (E) Very-long-chain fatty acids

8. The answer is E. Very-long-chain fatty acids are initially oxidized in peroxisomes, generating hydrogen peroxide, NADH, and acetyl CoA. Once the fatty acids have been shortened to about 8 to 10 carbons in length, they are transferred to the mitochondria to finish their oxidation via traditional b-oxidation. Thus, very-long-chain fatty acids will accumulate with this peroxisomal disorder. Short-chain and long-chain fatty acids are oxidized within the mitochondria via b-oxidation. Acetyl CoA will not accumulate with a peroxisomal disorder because it will also be oxidized in the mitochondria. Dicarboxylic acids accumulate when there is a defect in mitochondrial b-oxidation, and o-oxidation begins to play a larger role in generating energy.

81. One colony of bacteria is split into two Petri plates: one plate with growth medium containing glucose and all 20 amino acids and the other plate with growth medium with one sugar (lactose) and one nitrogen source (NH4+). Which one of the following statements is correct concerning cells grown with only lactose and one nitrogen source? (A) cAMP levels will be lower than cells grown in the presence of glucose and all 20 amino acids. (B) Catabolite-activator protein (CAP, or cAMP-binding protein) will be bound to the lac promoter. (C) The lac repressor will be bound to the lac operator. (D) RNA polymerase will not bind to the trp promoter. (E) Attenuation of transcription of the trp operon will increase compared with cells grown in the presence of glucose and all 20 amino acids.

81. The answer is B. In the absence of glucose and the presence of lactose, the lac repressor will be inactive, cAMP levels will rise, and the CAP protein-cAMP complex will bind to the lac promoter, stimulating transcription of the operon. In the absence of amino acids, tryptophan levels in the cell will be low; thus, the repressor for the trp operon will be inactive, and the operon will be transcribed by RNA polymerase. Attenuation of transcription of this operon will decrease because of the reduced tryptophan levels.

82. A 75-year-old man is brought to the emergency room after being hit by a car. He suffered an open femur fracture that required orthopedic surgery. After several days, a wound infection sets in with a new bacterial organism that is resistant to all antibiotics currently available. The effort toward creating new antibiotics requires targeting of characteristics exclusive to bacteria and not to humans. Which of the following statements is exclusively true of prokaryotes but not eukaryotes? (A) Transcription and translation are coupled. (B) Most cells are diploid. (C) Each gene has its own promoter. (D) DNA is complexed with histones. (E) Genes contain introns.

82. The answer is A. In prokaryotes, transcription and translation occur concurrently and are coupled. Eukaryotes contain a membrane bound nuclei, so transcription is separated from translation. In eukaryotes, most cells are diploid, and each gene has its own promoter, DNA is complexed with histones, and genes contain introns. These characteristics are not true of prokaryotes.

83. Eukaryotic secreted proteins are synthesized on ribosomes attached to the rough endoplasmic reticulum, whereas cytoplasmic proteins are synthesized on cytoplasmic (free) ribosomes. Which one of the following statements does not accurately describe an aspect of secreted protein synthesis? (A) They are sorted in the endoplasmic reticulum and Golgi apparatus. (B) A signal sequence is present at the N terminus. (C) A hydrophilic signal sequence is required. (D) After entering the endoplasmic reticulum, the signal sequence is removed by signal peptidase. (E) The protein is secreted via exocytosis.

83. The answer is C. A hydrophobic signal sequence (not hydrophilic) at the N terminus of a secretory protein allows the nascent protein to pass into the lumen of the rough endoplasmic reticulum (RER). The signal sequence is then cleaved, such that the protein may be further processed within the RER and the Golgi apparatus in preparation for exocytosis.

84. A 47-year-old man, with no known family history of cancer, develops changes in his bowel habits, including pencil-caliber stools with occasional bleeding. A colonoscopy and biopsy confirm the diagnosis of adenocarcinoma of the colon. Furthermore, the tumor is found to express a mutation in the ras protein. Which of the following correctly describes the ras protein? (A) A nonreceptor tyrosine kinase (B) A nuclear transcription factor (C) A polypeptide growth factor (D) A receptor tyrosine kinase (E) A guanosine triphosphate (GTP)-binding protein

84. The answer is E. The ras protein is a guanosine triphosphate (GTP)-binding protein, which is turned on when it is bound to GTP and shut off when it hydrolyzes GTP to guanosine diphosphate (GDP). Mutations that destroy this hydrolytic activity are among the most common in cancer, resulting in continuous growth-promoting signals. The abl and src genes are two nonreceptor tyrosine kinases, whereas the epidermal growth factor receptor is a receptor tyrosine kinase. The myc gene is an example of a nuclear transcription factor. Fibroblast growth factor is a typical example of a polypeptide growth factor implicated in cancer.

85. A 53-year-old man presents to the physician because he is fatigued and ''not feeling himself.'' The doctor orders a routine set of tests, which demonstrates a white blood cell count of 85,000/mL (normal, 3000-10,000/mL). Molecular studies suggest that he has chronic myelogenous leukemia (CML). Which of the following translocations is associated with CML? (A) t(8;14) (B) t(14;18) (C) t(11;14) (D) t(9;22) (E) t(15;17)

85. The answer is D. Chronic myelogenous leukemia is associated with t(9;22), a translocation between chromosomes 9 and 22. The resulting chromosome is known as the Philadelphia chromosome, in which the bcr gene from chromosome 22 is fused to the abl gene on chromosome 9. The aberrant bcr-abl protein is thus created and results in an unregulated tyrosine kinase being activated in the cell. There are several other important chromosomal translocations associated with blood and lymph cancers, including t(8;14) found in Burkitt lymphoma, t(14;18) found in follicular lymphoma, t(11;14) found in mantle cell lymphoma, and finally, t(15;17) found in acute myelogenous leukemia, M3 variant.

86. A 22-year-old woman, who has had numerous episodes of unprotected intercourse since the age of 12 years, visits a gynecologist for her first Pap smear. The results return positive for atypical cells, indicative of human papilloma virus (HPV) infection. Which of the following correctly describes the effects of HPV on cell growth? (A) The viral E7 protein degrades cellular p53. (B) The virus causes insertional inactivation of critical genes. (C) The viral E6 protein perturbs the normal function of cellular Rb. (D) The viral LMP-1 protein prevents the expression of cellular bcl-2. (E) The viral E6 protein degrades cellular p53.

86. The answer is E. HPV infection causes cervical cancer and is acquired as a sexually transmitted disease. Its oncogenic potential is related to the viral E6 protein, which disrupts cellular growth by degrading cellular p53. The viral E7 protein perturbs the normal function of Rb. Retroviruses can cause cancer because they can cause insertional inactivation of key growth-controlling genes. The Ebstein-Barr virus (EBV) LMP-1 protein results in cancer through its ability to prevent the expression of bcl-2. HPV does not insert into the host genome (it replicates autonomously), nor does it express the LMP-1 protein.

87. Which of the following conditions is caused by a trinucleotide repeat? (A) Hemophilia A (B) Prader-Willi syndrome (C) Fragile X syndrome (D) Angelman syndrome (E) Leber hereditary optic neuropathy

87. The answer is C. Fragile X syndrome is the second leading inherited cause of mental retardation and is caused by trinucleotide repeats of CGG (up to 200) within the FMR-1 gene on the X chromosome. Some other examples of trinucleotide repeat disorders include Huntington disease, spinocerebellar ataxia, Friedreich ataxia, and myotonic dystrophy. Hemophilia A is an X-linked recessive disease that leads to a deficiency of clotting factor VII. Uniparental disomy leads to specific genetic abnormalities: paternal transmission of the del(15)(q11q13) deletion results in Prader-Willi syndrome; maternal transmission of this deletion results in Angelman syndrome. This is the classic example of genetic imprinting. Because all the mitochondria in the developing zygote originate from the ovum, defects in the mitochondrial genome are exclusively maternally transmitted, as seen in Leber hereditary optic neuropathy (LHON). Most often, this disorder results from one of three distinct point mutations in the mitochondrial DNA (mtDNA).

88. Which of the following statements is correct concerning robertsonian translocations? (A) Form the basis for genetic imprinting (B) Failure of the homologous chromosomes to separate during meoisis (C) Manifest clinically as cri du chat syndrome (D) When a chromosome experiences two internal breaks with a subsequent rejoining of the internal fragment in the reverse orientation (E) A variant translocation between two acrocentric chromosomes, whereby two long arms are joined at the centromere with the loss of the two short arms

88. The answer is E. Robertsonian translocations are variant translocations between two acrocentric chromosomes, whereby two long arms are joined at the centromere with the loss of the two short arms. Parent-specific expression is an example of genetic imprinting and is seen in Prader- Willi syndrome and Angelman syndrome. Both disorders result from a chromosome deletion (del(15)q11q13). Inheriting the deletion from the father results in Prader-Willi syndrome; inheriting the deletion from the mother results in Angelman syndrome. Failure of the homologous chromosomes to separate during meoisis is an example of a nondisjunction. Cri du chat syndrome is a rare genetic disorder caused by loss of genetic material on the short arm of chromosome 5. Inversions occur when a chromosome experiences two internal breaks with a subsequent rejoining of the internal fragment in the reverse orientation.

89. Which statement is true regarding mendelian genetics? (A) The inheritance pattern of one trait is dependent on the inheritance of another trait. (B) Independent assortment occurs during anaphase II of meiosis after crossover has occurred. (C) The offspring inherit one allele at each locus from each of the parents. (D) The two alleles for each characteristic segregate during fertilization. (E) In autosomal recessive inheritance, 25% of offspring are carriers.

89. The answer is C. The offspring inherit one allele at each locus from each of the parents. The other answers are incorrect. The inheritance pattern of one trait is independent and does not influence the inheritance of another trait (A), independent assortment occurs during anaphase I of meiosis after crossover has occurred (B), the two alleles for each characteristic segregate during gamete production (D), and in autosomal recessive inheritance, 50% of offspring are carriers (E).

9. A 50-year-old man undergoes genetic testing for hemochromatosis, an autosomal recessive disease characterized by abnormally elevated serum iron levels leading to organ toxicity. He is positive for a genetic mutation and is diagnosed with the disease. However, he never develops signs of elevated serum iron levels or organ toxicity. Which of the following terms best describes this patient's disease? (A) Low penetrance (B) Low expressivity (C) Low mosaicism (D) Low lyonization (E) Low mendelian inheritance

9. The answer is A. Low penetrance indicates that a patient carrying the disease allele does not manifest the traits that are typical for that disease, whereas low expressivity indicates that a patient carrying the disease allele shows a decreased degree of expression of the traits typical for that disease, but traits or signs are still detectable (and none are detectable in our patient). Mosaicism indicates that some of the patient's cells are positive for a specific genetic abnormality, and some of the patient's cells are negative for the same genetic abnormality. This is not the case for our patient. Lyonization refers to X-inactivation in females, which could manifest as expression of an X-linked recessive disease in heterozygous females. This is a rare occurrence, and our patient is a male. Mendelian inheritance refers to inheritance of genes from parents to their offspring, but this term does not describe clinical manifestations of a genetically inherited disease.

9. A 40-year-old woman presents with an LDL serum level of 400 (recommended level is <130), and a triglyceride level of 170 (recommended level is <150). She is diagnosed with type II familial hypercholesterolemia. In this disorder, a mutated LDL receptor is formed, such that it cannot bind to LDL. Which of the following would result? (A) Cellular HMG-CoA reductase activity is not inhibited. (B) The triglycerides in chylomicrons cannot be degraded. (C) The VLDL level in the serum increases. (D) The HDL level in the serum increases. (E) The VLDL cannot be converted to IDL.

9. The answer is A. Normally, LDL, after binding to its receptor and internalization into the cell, is digested by lysosomal enzymes to release cholesterol. High levels of intracellular cholesterol inhibit HMG-CoA reductase, and cholesterol synthesis decreases in hepatocytes. In familial hypercholesterolemia type II, LDL cannot be taken up into hepatocytes. Because HMG-CoA reductase is not properly inhibited, excessive cholesterol levels result. Triglyceride digestion in chylomicrons, the level of VLDL, the level of HDL, and conversion of VLDL to IDL are not affected in type II hypercholesterolemia.

9. A 4-month-old infant presents with a seizure. His mother reports that her infant has been irritable and lethargic over the past several days. The infant is found to be profoundly hypoglycemic and have low ketones. Short-chain dicarboxylic acids are found to be elevated in the serum. The most likely enzyme deficiency is which of the following? (A) Medium-chain acyl CoA dehydrogenase (MCAD) (B) Carnitine acyltransferase I (C) Hormone-sensitive lipase (D) Pyruvate carboxylase (E) Fatty acyl CoA synthetase

9. The answer is A. The infant has MCAD deficiency. The child can only partially oxidize fatty acids (to the 6- to 10-carbon stage), leading to reduced energy generation and low acetyl CoA levels. The low acetyl CoA reduces gluconeogenesis because pyruvate carboxylase cannot be fully activated. The reduced energy also contributes to the reduced levels of gluconeogenesis because that pathway requires energy to proceed. The dicarboxylic acids result from o-oxidation of the medium-chain acyl CoAs, to try and generate more energy. Defects in CAT I or hormonesensitive lipase would result in a complete lack of fatty acid oxidation, and the dicarboxylic acids would not be observed. A defect in pyruvate carboxylase, although negatively affecting gluconeogenesis, would not affect fatty acid oxidation.

9. A man visits the emergency room with signs of severe dehydration: he is thirsty, has decreased skin turgor, is tachycardic, and is somnolent. According to the patient, he abruptly began to suffer from diarrhea this morning, which he describes as multiple bouts of watery diarrhea. You suspect Vibrio cholerae and begin volume resuscitation. By what biochemical mechanism does Vibrio lead to these clinical manifestations? (A) Adenylate cyclase activation (B) Phospholipase C activation (C) Phosphodiesterase activation (D) Tyrosine kinase receptor activation (E) Serine-threonine kinase activation.

9. The answer is A. Vibrio cholerae is a comma-shaped gram-negative flagellated rod that is carried in water, food, and, classically, shellfish. This noninvasive organism requires large inoculums and secretes a toxin (AB5), which ADP-ribosylates a Gas protein, which in turn keeps adenylate cyclase active and increases cAMP levels. The high cAMP causes secretion of chloride ions and decreases sodium absorption, which leads to extensive osmotic losses of water into the intestinal lumen. This leads to ''rice-water stools'' in which the ''rice'' is the desquamated intestinal mucosa and the water is the osmotically lost water. Phospholipase C activation occurs in, for example, IgE activation and results in histamine secretion from mast cell degranulation. Phosphodiesterase activation is seen, as an example, when light interacts with rhodopsin, enabling vision. Tyrosine kinase receptor activation is part of the insulin and other growth factor signaling pathways. Serine-threonine kinase activation is also an intracellular signaling pathway that controls several cell functions, including apoptosis. The TGF-b receptor acts through a serine- threonine kinase activity.

9. A 45-year-old alcoholic man walks into the emergency room with a clumsy, wide-based gait and appears confused. He has pronounced nystagmus, and laboratory tests are significant for a metabolic acidosis and a serum blood alcohol level of 0.13. This patient should most probably be treated with IV fluids containing which of the following? (A) Thiamine (B) Riboflavin (C) Niacin (D) Pantothenic acid (E) Biotin

9. The answer is A. Wernicke encephalopathy, with the classic triad of ataxia, confusion, and ophthalmoplegia (and nystagmus), is due to thiamine deficiency. Thiamine is an essential coenzyme in carbohydrate metabolism, including the pentose-phosphate pathway (transketolase) and the TCA cycle (pyruvate dehydrogenase and a-ketoglutarate dehydrogenase). Riboflavin deficiency is possible in malnourished alcoholics, causing cheilosis, glossitis, and corneal changes. Niacin deficiency causes diarrhea, dementia, and dermatitis. Deficiencies of pantothenic acid and biotin are rare, although a biotin deficiency will lead to hypoglycemia and mild ketosis.

9. A 34-year-old woman presents with central obesity, relatively thin extremities, and purple stria on her abdomen. Further workup reveals an excessive serum cortisol level and a blood sugar level of 258 mg/dL. Which of the following is the most likely cause of her hyperglycemia? (A) A pancreatic adenoma secreting adrenocorticotropic hormone (ACTH) and glucagon (B) Glucocorticoid-enhanced transcription of PEPCK (C) Increased substrates for gluconeogenesis due to excess fatty acid degradation (D) Cortisol inhibition of insulin secretion (E) Excess consumption of processed carbohydrates

9. The answer is B. Cushing syndrome results in increased circulating glucocorticoids, primarily cortisol. Glucocorticoids bind to cytosolic receptor proteins that traverse the nuclear envelope and bind to specific sequences in the PEPCK gene and cause an increase in PEPCK gene expression. PEPCK RNA is translated into PEPCK protein in the cytosol. The increased PEPCK protein levels then catalyze the formation of PEP from OAA, which is an initiating step of gluconeogenesis, resulting in hyperglycemia due to the inappropriate stimulation of gluconeogenesis. Tumors of the pancreas more commonly produce insulin or glucagon. There are documented cases of pancreatic tumors secreting ACTH, but this is much less likely than another form of Cushing syndrome, as described by the patient's signs and symptoms. Fatty acid degradation produces primarily acetyl CoA, which cannot be used to synthesize net glucose. Cortisol actually stimulates insulin secretion, although, paradoxically, it decreases the tissues' sensitivity to the hormone. Excess consumption of carbohydrates, on a short-term basis, does not lead to hyperglycemia because insulin release will lead to glucose uptake by the tissues and a return of blood glucose levels to a normal value.

9. Gene transcription rates and mRNA levels were determined for an enzyme that is induced by glucocorticoids. Compared with untreated levels, glucocorticoid treatment caused a 10-fold increase in the gene transcription rate and a 20-fold increase in both mRNA levels and enzyme activity. These data indicate that a primary effect of glucocorticoid in this assay is to decrease which of the following? (A) The activity of RNA polymerase II (B) The rate of mRNA translation (C) The ability of nucleases to act on mRNA (D) The rate of binding of ribosomes to mRNA (E) The activity of RNA polymerase III

9. The answer is C. If the rate of degradation of mRNA is not altered by glucocorticoids, the increase in mRNA levels should reflect the increase in transcription rate (10-fold in this case). Because the increase in mRNA level (20-fold) is greater than the increase in transcription rates (10-fold), the glucocorticoids must also be increasing mRNA stability (i.e., decreasing the rate of degradation by nucleases). The activity of RNA polymerase II is increased (transcription is increased), and the rate of translation (the binding of ribosomes to mRNA) is increased (the enzyme activity is increased).

9. A morbidly obese woman decides to see her physician to begin a weight loss program. He tells her that diet and exercise play an essential role in her program. She is concerned that she does not have the time to devote to exercise and wants to know if there is any pharmacologic treatment for her. The physician decides to start her on orlistat, which will directly inhibit which one of the following steps in fat digestion? (A) Bile salt formation (B) Micelle formation (C) Pancreatic and gastric lipase (D) Absorption of free fatty acids (E) Chylomicron formation

9. The answer is C. Orlistat inhibits pancreatic and gastric lipase, thereby preventing the release of free fatty acids from triacylglycerol in the intestinal lumen. The absorption of fatty acids by the intestinal epithelial cells is thus blocked because free fatty acids are not being produced from the triacylglycerol of the diet. The normal sequence of triacylglycerol digestion is that triacylglycerols are emulsified by bile salts from the gallbladder. These micelles are then acted on by lipases, leading to the liberation of fatty acids. The fatty acids are absorbed by enterocytes, which then reassemble triacylglcerol for packaging into chylomicrons

9. A 24-year-old woman complains of intermittent right upper quadrant pain that extends to the inferior tip of her scapula. An ultrasound confirms your suspicion of cholelithiasis, and the patient undergoes cholecystectomy. Analysis indicates gallstones containing bilirubin. Measurement of metabolic intermediates, such as 2,3- bisphosphoglycerate and glucose 6-phosphate, are elevated in her serum. A deficiency of which of the following enzymes most likely led to her pigmented gallstones and release of these metabolites into the blood? (A) Glucose 6-phosphate dehydrogenase (B) PFK-1 (C) Pyruvate kinase (D) Pyruvate dehydrogenase (E) Pyruvate carboxylase

9. The answer is C. Pyruvate kinase deficiency is an autosomal recessive disease that causes hemolytic anemia of varying degrees depending on the amount of pyruvate kinase activity lost. The RBCs of affected individuals have a significantly reduced ability to make ATP. The lack of ATP impairs the cells' ability to achieve osmotic balance via ion pumping. Because of the loss of osmotic balance, the cells lyse readily. The heme released is converted to bilirubin in excess, which accumulates in the gallbladder, leading to gallstone formation. The presence of 2,3- bisphosphoglycerate and glucose 6-phosphate indicates a pyruvate kinase deficiency as opposed to the more prevalent glucose 6-phosphate dehydrogenase deficiency. The defect in pyruvate kinase leads to PEP accumulation, which is converted back to 3-phosphoglycerate, leading to increased 2,3-bisphosphoglycerate production. The RBCs lack mitochondria and thus do not contain pyruvate dehydrogenase or pyruvate carboxylase. A PFK-1 deficiency would not lead to increased 2,3-bisphosphoglycerate in the serum.

9. The patient described in question 8, who has multiple myeloma, has not responded to numerous treatments, and his disease is progressing. He sees his oncologist, who wants to start him on the drug bortezomib. Bortezomib inhibits the proteasome from degrading proteins. Which class of intracellular proteins will not be specifically degraded as a result of taking this drug? (A) Proteins with PEST sequences (B) Amyloid proteins (C) Polyubiquitinated proteins (D) Immunoglobulin light chains (E) Immunoglobulin heavy chains

9. The answer is C. The proteasome normally degrades proteins that have been polyubiquitinated. As such, in the presence of bortezomib, polyubiquitinlated proteins will accumulate within cells, leading to a selective adverse effect on the cancer cells (myeloma cells) because these are the cells growing most rapidly. Proteins with PEST sequences are rapidly degraded by nonspecific intracellular proteases. Although the immunoglobulin light chains are forming the amyloid proteins in this disease, these structures are difficult to degrade, such that inhibiting the proteasome has no effect on the degradation of the amyloid proteins. Immunoglobulin heavy chains are not accumulating in this disorder

9. A 56-year-old woman with no known medical conditions presents to the emergency room with pain in the upper arm. She denies any trauma; however, a fracture of the humerus is found on radiograph. She is found to have an elevated PTH level. Which of the following statements best describes PTH? (A) It lowers serum calcium. (B) It directly promotes the absorption of calcium from the intestine. (C) It stimulates the conversion of vitamin D to the active form. (D) It promotes the reabsorption of phosphate from the kidney. (E) It promotes the excretion of calcium from the kidney.

9. The answer is C. This patient likely has a PTH-producing tumor of the parathyroid. PTH is an important regulator of calcium homeostasis. It stimulates the conversion of vitamin D to the active form, which, in turn, promotes the absorption of calcium from the intestine as well as reabsorption of calcium (not phosphate) from the kidney filtrate. PTH functions to increase serum calcium by liberating calcium phosphate from bone, making it weaker and prone to fracture. PTH prevents accumulation of phosphate by promoting its excretion from the kidneys. PTH does not promote the excretion of calcium from the kidney.

9. A 17-year-old woman who recently began taking birth control pills presents to the emergency room with cramping abdominal pain, anxiety, paranoia, and hallucinations. A surgical evaluation, including ultrasound and computed tomography scan, fails to demonstrate an acute abdominal process. A urinalysis reveals an increase in urine porphyrins. Which of the following is the most likely? (A) Congenital erythropoietic porphyria (B) Variegate porphyria (C) Porphyria cutanea tarda (D) Acute intermittent porphyria (E) Erythropoietic protoporphyria

9. The answer is D. Acute intermittent porphyria is an autosomal dominant disease resulting from the deficiency of porphobilinogen deaminase (also known as hydroxymethylbilane synthase). Often these intermittent attacks are provoked by drugs such as gonadal steroids, barbiturates, or alcohol. These drugs are metabolized by cytochrome P-450 systems, which contain heme. The presence of these drugs induces cytochrome P-450 synthesis (which includes an increase in heme biosynthesis). The induction of heme biosynthesis is the event that leads to an accumulation of the toxic intermediate. The other choices, including congenital erythropoietic porphyria, porphyria cutanea tarda, variegate porphyria, and erythropoietic protoporphyria, are considered erythropoietic porphyries, which are characterized by photosensitivity and rarely exhibit abdominal pain.

9. A patient has come in for an HIV test. This test is run in two phases. The first test is an ELISA as a screen, and if two positive test results occur by ELISA, the second test will be run. The second test is a confirmatory Western blot. What do the ELISA and Western blots measure in their respective assays for HIV? (A) The ELISA is measuring the presence of HIV antigen in the sera, whereas the Western blot is measuring the presence of antibodies to HIV proteins in the sera. (B) The ELISA is measuring the presence of antibodies to HIV proteins in the sera only, whereas the Western blot is measuring the presence of HIV antigens in the sera. (C) The ELISA is measuring the presence of HIV antigen in the sera, whereas the Western blot is measuring the presence of HIV antigen in the sera as well. (D) The ELISA is measuring the presence of antibodies to HIV proteins in the sera only, whereas the Western blot is also measuring the presence of antibodies to HIV proteins in the sera. (E) The ELISA measures the presence of antibodies directed against human leukocyte antigen (HLA) molecules to HIV, whereas the Western blot measures levels of free, circulating virus in the sera of the patient.

9. The answer is D. Both an ELISA and Western blot detect host antibodies against HIV antigens. Neither detects the virus or antigens of the virus. Anti-HLA antibodies may be present, but the HIV test is not designed to measure them.

Questions 5-11: Using answers (A) through (G) below, match the correct drug with its clinical effect 9. Fluoxetine (A) Folic acid analog(s) that inhibit(s) dihydrofolate reductase, which leads to an inhibition of purine and deoxythymidine synthesis (B) Folate analog(s) that bind(s) specifically to bacterial dihydrofolate reductase and used in conjunction with sulfonamides (C) Histamine and basophil inhibitor(s) used to treat allergic reactions (D) Increase(s) synaptic serotonin concentration in the treatment of depression (E) Inhibitor(s) of tyrosine residue iodination (F) Replenish(es) glutathione levels during acetaminophen toxicity (G) Release(s) NO, create(s) smooth muscle relaxation, and very effective as antihypertensive

9. The answer is D. Fluoxetine (Prozac) is a selective serotonin reuptake inhibitor that increases the extracellular serotonin concentration in the synapse. It is effective in the treatment of depression.

9. A 76-year-old bedridden nursing home resident begins to develop swelling of her left leg. A venous Doppler ultrasound is ordered and shows an obstructive deep vein thrombosis extending from her left common femoral vein to her popliteal vein with limited blood flow. The patient is immediately started on heparin to further prevent the clot from enlarging. Heparin is an example of which of the following? (A) Sphingolipid (B) Cerebroside (C) Ganglioside (D) Glycosaminoglycan (E) Prostaglandin

9. The answer is D. Heparin is an example of a glycosaminoglycan, a long repeating chain of disaccharide units attached to a core protein. The sugar residues of heparin are sulfated. Cerebrosides and gangliosides are both examples of sphingolipids derived from the lipid ceramide. Prostaglandins are derived from polyunsaturated fatty acids, an example of which is arachidonic acid.

9. A 3-year-old girl presents with developmental delay and growth failure. The physical examination is remarkable for coarse facial features, craniofacial abnormalities, gingival hyperplasia, prominent epicanthal fold, and macroglossia. The patient was diagnosed with I-cell diseases. Lysosomal proteins are mistargeted in this disorder. Rather than being targeted to the cell's lysosomes, lysosomal proteins in this disease are found in which of the following? (A) In the endoplasmic reticulum (ER) (B) In the Golgi apparatus (C) In the mitochondria (D) Exported from the cell (E) In the cytoplasm

9. The answer is D. If lysosomal proteins are not appropriately tagged with mannose 6-phosphate in the ER and Golgi apparatus, the proteins will be exported from the cell. The lysosomal proteins do not contain the appropriate targeting signals to be sent to the ER, Golgi apparatus, or mitochondria. Because these enzymes are synthesized onmembrane-bound ribosomes (the rough ER), they will not be found in the cytoplasm (cytoplasmic proteins are synthesized on cytoplasmic ribosomes). Although the child's physical abnormalities are similar to other storage diseases, gingival hyperplasia is a unique clinical feature to I-cell disease.

9. An 8-year-old boy is seen by an ophthalmologist for vision difficulties, and the physician notices a slowing of the boy's eye movements. The ophthalmologist finds ophthalmoplegia and pigmentary retinopathy and suspects the child has Kearns-Sayre syndrome. Assuming that the defect in this disorder is due to a mutation in complex II of the ETC, electron transfer from which substrate would be impaired? (A) Malate (B) a-Ketoglutarate (C) Isocitrate (D) Succinate (E) Pyruvate

9. The answer is D. Succinate feeds electrons into complex II, which, in this case, would be impaired. The other substrates listed, malate, a-ketoglutarate, isocitrate, and pyruvate, all feed their electrons via NADH through complex I. Malate dehydrogenase, a-ketoglutarate dehydrogenase, isocitrate dehydrogenase, and PDH all generate NADH during the course of the reactions that they catalyze.

9. A 2-year-old girl is failing to meet age-appropriate milestones, including a progressive difficulty in walking. An abnormally low level of arylsulfatase A is found in her cells, causing accumulation of sulfated glycolipids in neurons. Unfortunately, she dies 5 years later. Which one of the following is the most likely diagnosis for this disorder? (A) Fabry disease (B) Gaucher disease (C) Niemann-Pick disease (D) Tay-Sachs disease (E) Metachromatic leukodystrophy

9. The answer is E. Metachromatic leukodystrophy is due to a deficiency in arylsulfatase A, a lysosomal enzyme that degrades sulfated glycolipids. These sulfatide compounds accumulate in neural tissue, causing demyelination of central nervous system and peripheral nerves, with resultant loss of cognitive and motor functions. Fabry disease is a result of a deficiency in a-galactosidase A; Niemann-Pick is a result of a deficiency in sphingomyelinase; Gaucher disease is a result of a deficiency in glucocerebrosidase; and Tay-Sachs is a result of a deficiency in hexosaminidase A.

9. A 4-year-old child, on a well-child checkup, is found to have a large flank mass. Computed tomography demonstrates a large mass arising from the kidney, and a subsequent biopsy reveals a diagnosis of Wilms tumor. A pediatric oncologist starts chemotherapy including the transcription inhibitor actinomycin D. Which of the following statements is correct regarding transcription regulation in bacteria? (A) All mRNAs are monocistronic. (B) RNA polymerase requires a primer. (C) The RNA chain grows in the 30 to 50 direction. (D) Rho factor is critical for initiation of RNA synthesis. (E) The TATA box contains a consensus sequence for the binding of RNA polymerase.

9. The answer is E. RNA polymerase binds to specific consensus sequences on DNA, with the TATA box being an example of such a sequence. The TATA box is also called the Hogness box in eukaryotic cells and the Pribnow box in prokaryotes. Prokaryotic cells produce both monocistronic and polycistronic mRNA, whereas eukaryotic cells only produce monocistronic mRNA. RNA polymerase, unlike DNA polymerase, does not require a primer. RNA polymerase synthesizes RNA in the 50 to 30 direction, reading the DNA template in the 30 to 50 direction. The rho factor (r) is critical for rho-dependent termination, whereas the sigma factor (s) aids in bacterial transcriptional initiation.