Block 2 Biochem Practice Questions

d. Fructose-1,6-bisphosphate The cause of this patient's hemolytic anemia is pyruvate kinase deficiency. Pyruvate kinase is activated by fructose-1,6-bisphosphate and inhibited by alanine and ATP. Pyruvate kinase deficiency is an autosomal recessive disorder that inhibits the ability of cells to perform glycolysis. This disorder primarily manifests in red blood cells, as these cells lack mitochondria and are therefore unable to perform oxidative phosphorylation. Pyruvate kinase deficiency causes RBCs to lyse prematurely, leading to a Coombs-negative hemolytic anemia. Pyruvate kinase is one of the three major regulators of glycolysis (along with hexokinase and phosphofructokinase). It is stimulated by fructose-1,6-bisphosphate in an example of feed-forward regulation and is inhibited by alanine and ATP in an example of feedback inhibition.

A 12-year-old boy and his siblings are referred to a geneticist for evaluation of a mild but chronic hemolytic anemia that has presented with fatigue, splenomegaly, and scleral icterus. Coombs test is negative and blood smear does not show any abnormal findings. An enzymatic panel is assayed, and pyruvate kinase is found to be mutated on both alleles. The geneticist explains that pyruvate kinase functions in glycolysis and is involved in a classic example of feed-forward regulation. Which of the following metabolites is able to activate pyruvate kinase? Select one: a. Alanine b. Glyceraldehyde-3-phosphate c. ATP d. Fructose-1,6-bisphosphate e. Glucose-6-phosphate

c. Galactose-1-phosphate uridylyltransferase The child has classic galactosemia, a defect in galactose-1-phosphate uridylyltransferase. Due to the accumulation of galactose-1-phosphate, galactokinase is inhibited, and free galactose accumulates within the blood and tissues. The accumulation of galactose in the lens of the eye provides substrate for aldose reductase, converting galactose to its alcohol form (galactitol). The accumulation of galactitol leads to an osmotic imbalance across the lens, leading to cataract formation. Additionally, the increased galactose-1 phosphate, at very high levels in the liver, blocks phosphoglucomutase activity, resulting in ineffective glucose production from glycogen (phosphorylase degradation of glycogen will produce glucose-1-phosphate, but this cannot be converted to glucose-6-phosphate if phosphoglucomutase activity is inhibited). A defect in galactokinase will lead to nonclassical galactosemia, with cataract formation, but none of the feeding problems associated with classical galactosemia (associated with the accumulation of galactose-1-phosphate) are observed in nonclassical galactosemia. None of the other enzymes listed, if defi cient, will give rise to the symptoms produced, particularly cataract formation. A defect in glycogen synthase would lead to reduced glycogen levels and fasting hypoglycemia. A defect in fructokinase leads to fructosuria (fructose in the urine), but no overt symptoms of disease. The figure below indicates the pathway for galactose metabolism and the defects in classical and non classical galactosemia.

A 2-week-old newborn was brought to the pediatrician due to frequent vomiting, lethargy, and diarrhea. Family history revealed that the child never seemed to eat well, and had only been breast-fed. Physical examination revealed an enlarged liver and jaundice. The pediatrician was suspicious of an inborn error of metabolism and referred the child to an ophthalmologist for a slit lamp exam, the result of which is shown below. An enzyme that may be defective in this child is which one of the following? Select one: a. Fructokinase b. Glycogen synthase c. Galactose-1-phosphate uridylyltransferase d. Galactokinase e. Fructose-1,6-bisphosphatase

b. It is inhibited by protein kinase A activity This patient with inherited anemia and echinocytes most likely has pyruvate kinase deficiency, which is caused by a defect in glycolysis. The rate limiting enzyme of glycolysis is phosphofructokinase-1 (PFK1), which is inhibited by protein kinase A activity.The key rate-limiting step of glycolysis is mediated by the enzyme PFK1, which is allosterically stimulated by fructose-2,6-bisphosphate. The concentration of this metabolite is produced by the activity of phosphofructokinase-2 (PFK2). It is important to understand that low blood glucose leads to increased levels of glucagon relative to insulin. This increase in glucagon activates protein kinase A, which phosphorylates and inactivates PFK2. Since PFK2 is inactivated, the activity of the rate limiting enzyme PFK1 also decreases.

A 23-year-old woman is seen by her primary care physician for fatigue. She says that she has always felt a little short of breath compared to her friends; however, she did not think that it was abnormal until she started trying a new exercise regimen. On physical exam, she is found to have mild conjunctival pallor and a peripheral blood smear is obtained showing echinocytes but no intracellular accumulations. Upon further questioning, she recalls that several relatives have had similar issues with fatigue and pallor in the past. Which of the following is true about the rate limiting enzyme of the biochemical pathway that is affected by this patient's most likely condition? Select one: a. It is inhibited by fructose-2,6-bisphosphate b. It is inhibited by protein kinase A activity c. It is stimulated by citrate d. It is stimulated by ATP e. It is inhibited by AMP

d. Messenger RNA

A 24-year-old man comes to emergency department complaining of abdominal pain, vomiting, and severe watery diarrhea. He recently returned from a camping trip and admits to eating wild mushrooms that he collected in the woods. His past medical history is in significant and he takes no medications. He does not use illicit drugs. On physical examination, he is ill-appearing and jaundiced. His liver edge is soft, tender, and palpable 4 cm below the right costal margin. Laboratory tests are significant for elevated levels of liver enzymes. Synthesis of which of the following is most likely to be directly inhibited by the responsible toxin? Select one: a. Ribosomal RNA b. Transfer RNA c. DNA d. Messenger RNA e. Protein

b. Reduced glutathione Reduced glutathione. The patient has glucose-6-phosphate dehydrogenase deficiency, and his red blood cells cannot convert oxidized glutathione to reduced glutathione due to a lack of NADPH. Fava beans contain a potent oxidizing agent that will, in some patients (but not all), lead to hemolytic anemia in individuals with glucose-6-phosphate dehydrogenase deficiency; in individuals with a normal G6PDH, the oxidizing agent is handled by glutathione. The red blood cells, under these conditions, do not have a problem in regenerating NADH, NAD+, or ATP.

A 25-year-old African American male, in good health, had read about fava beans in "Silence of the Lambs." For dinner one night, the man had liver with fava beans and a nice Chianti. About 8 h after eating the beans, the man was tired and weak. Blood work showed hemolytic anemia. This patient most likely has a defect in regenerating which of the following? Select one: a. NADH b. Reduced glutathione c. ATP d. NAD+ e. Oxidized glutathione

d. Liver and kidney The major source of glucose in this patient who has been fasting for around 24 hours is gluconeogenesis, which occurs in both the liver and the kidney. The source of circulating glucose depends in large part on the physiologic state of the patient that is being examined. During the first two hours after a meal, the main source of circulating glucose is absorption of the exogenously consumed glucose by enterocytes. Between 4 and 12 hours after eating, the main source of circulating glucose becomes glycogenolysis within the liver. Finally, more than 16 hours after the last meal, the main source of circulating glucose is gluconeogenesis, which occurs in the liver as well as the kidneys. Importantly, even though muscle tissue has substantial glucose deposits, it lacks the ability to release this glucose into the blood so it does not contribute to circulating glucose.

A 26-year-old medical student who is preparing for Step 1 exams is woken up by her friend for breakfast. She realizes that she must have fallen asleep at her desk while attempting to study through the night. While walking with her friend to breakfast, she realizes that she has not eaten since breakfast the previous day. Using this as motivation to review some biochemistry, she pauses to consider what organs are responsible for allowing her to continue thinking clearly in this physiologic state. Which of the following sets of organs are associated with the major source of energy currently facilitating her cognition? Select one: a. Liver and muscle b. Liver, muscle, and kidney c. Liver only d. Liver and kidney e. Muscle only

g. B Thiamine serves as a cofactor for a number of important enzymes includingtransketolase, alpha-ketoglutarate dehydrogenase, and pyruvate dehydrogenase.Patients with alcoholism are classically deficient in thiamine and administration ofglucose to these patients without first administering a thiamine supplement resultsin the development of the acute neurological complications of thiamine deficiencyknown as Wernicke encephalopathy. Clinically Wernicke encephalopathy ischaracterized by acute confusion ophthalmoplegia and ataxia. Treatment withthiamine usually results in prompt resolution of symptoms.The bulk of ATP used for cellular processes is generated by oxidation of pyruvatethrough the tricarboxylic acid cycle (TCA cycle). The enzymes of the TCA cycle arelocated in the mitochondrion. Pyruvate generated from glycolysis is converted toacetyl C0A by the enzyme pyruvate dehydrogenase, and acetyl CoA then enters the TCA cycle

A 44-year-old homeless male develops confusion and ophthalmoplegia. Upon review of the medical record, you discover that he has been admitted to the hospital with alcohol intoxication several times before. Which of the following reactions is most likely impaired in this patient? Select one: a. D b. A c. C d. F e. E f. G g. B

a. galactosemia

A 5-week-old girl, who appeared to be healthy at birth, develops diarrhea and vomiting a few days after birth. Your current examination reveals that she has hepatomegaly, jaundice, and early cataract formation and is not meeting developmental milestones. You suspect that she has which of the following conditions? Select one: a. galactosemia b. pyloric stenosis c. Hurler syndrome d. Tay-Sachs disease e. Type I glycogenosis (von Gierke disease)

b. aconitase Fluoroacetate is a structural analog of citrate and a strong competitive inhibitor of aconitase.

A 5-year-old child was rushed to emergency room after accidentally consuming fluoroacetate, a known inhibitor of TCA cycle. Which of the following enzymes is inhibited by fluoroacetate? Select one: a. isocitrate dehydrogenase b. aconitase c. succinate dehydrogenase d. malate dehydrogenase e. citrate synthase

d. Salivary amylase Starch blockers contain a natural inhibitor of salivary amylase, which will block starch digestion in the mouth. The other amylase that digests starch, pancreatic amylase, would only be inhibited by the starch blocker if the inhibitor would survive the conditions of the acidic stomach without becoming denatured. There is no gastric amylase. Intestinal enteropeptidase activates trypsinogen, which is required for protein digestion, not starch digestion. Pancreatic lipase is required for dietary triglyceride digestion, and is not active toward starch.

A 50-year-old man has been trying to lose weight, but he enjoyed eating so much that he found it difficult to do so. He then reads about a product in the popular press, which guarantees that he can lose weight, as caloric intake due to starch ingestion will be reduced (a starch blocker). The over-the-counter product that he buys is claimed to inhibit which of the following enzymes? Select one: a. Intestinal enteropeptidase b. Pancreatic lipase c. Gastric amylase d. Salivary amylase e. Pancreatic trypsinogen

e. hereditary galactosemia

A 9-month-old child is presented to the emergency room by his parents who report that he has been vomiting and has severe diarrhea. The episodes of vomiting began when the parents started feeding their child cow's milk. The infant exhibits signs of failing to thrive. Laboratory tests show elevated blood galactose, hypergalactosuria, metabolic acidosis, albuminuria, and hyperaminoaciduria. These clinical and laboratory findings are most consistent with which of the following disorders? Select one: a. Menkes disease b. essential fructosuria c. alkaptonuria d. von Gierke disease e. hereditary galactosemia

e. ATP 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 dehy- drogenase, and malate dehydrogenase) and FADH2 at one step (catalyzed by succinate dehy- drogenase). CO2 is a product of the isocitrate and a-ketoglutarate dehydrogenase reactions.

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? Select one: a. FADH2 b. NADH c. CO2 d. GTP e. ATP

a. defective hepatic aldolase В Hereditary Fructose Intolerance.

A child born and raised in Chicago planned to spend the summer on a relative's fruit farm and help with the harvest. The summer passed uneventfully, but several days after the harvest began, the child became jaundiced and very sick. On admission to the hospital the following clinical findings were recorded: in addition to the expected hyperbilirubinemia, the patient was hypoglycemic, had a markedly elevated rise in blood fructose concentration, and was hyperlactic acidemic. Further history taking revealed that during the harvest it was customary for the family to indulge in fruit-filled meals and to snack freely on fruit while carrying out the harvest. The elevated blood fructose in this child was most likely due to which of the following? Select one: a. defective hepatic aldolase В b. defective hepatic fructokinase c. defective hepatic glucokinase d. defective hepatic fructose-1,6-bisphosphatase e. an allergic reaction to constituents in the fruit diet

c. Sucrose Glycosides are formed by condensation of the aldehyde or ketone group of a carbohydrate with a hydroxyl group of another compound. Other linked groups (aglycones) include steroids with hydroxyl groups (e.g., cardiac glycosides such as digitalis or ouabain) or other chemicals (e.g., antibiotics such as steptomycin). Sucrose (α-Dglucose-β-1 → 2-D-fructose), maltose (α-D-glucose-α-1 → 4-D-glucose), and lactose (α-D-galactose-β-1 → 4-D-glucose) are important disaccha- rides. Fructose is among several carbohydrate groups known as ketoses because it possesses a ketone group. The ketone group is at carbon 2 in fructose, and its alcohol group at carbon 1 (also at carbon 6) allows ketal formation to produce pyranose and furanose rings as with glucose. Most of the fructose found in the diet of North Americans is derived from the disaccharide sucrose (common table sugar). Sucrose is cleaved into equimolar amounts of glucose and fructose in the small intestine by the action of the pancreatic enzyme sucrase. Deficiency of sucrase can also cause chronic diarrhea. Hereditary fructose intolerance (229600) is caused by deficiency of the liver enzyme aldolase B, which hydrolyzes fructose-1-phosphate.

A child develops chronic diarrhea and liver inflammation in early infancy when the mother begins using formula that includes corn syrup. Evaluation of the child demonstrates sensitivity to fructose in the diet. Which of the following glycosides contains fructose and therefore should be avoided when feeding or treating this infant? Select one: a. Oaubain b. Streptomycin c. Sucrose d. Maltose e. Lactose

c. α-1,4-glucosidase The child has symptoms of glycogen storage disease. Glycogen is a glucose polymer with linear regions linked through the C1 aldehyde of one glucose to the C4 alcohol of the next (α-1,4-glucoside linkage). There are also branches from the linear glycogen polymer that have α-1,6-glucoside linkages. Glycogen is synthesized during times of carbohydrate and energy surplus, but must be degraded during fasting to provide energy. Separate enzymes for breakdown include phosphylases (α-1,4-glucosidases) that cleave linear regions of glycogen and debranching enzymes (α-1,6-glucosidases) that cleave branch points. Glucose-6-phosphatase is needed in the liver to liberate free glucose from glucose-6-phosphate, providing fuel for other organs. There is no glucose-6-phosphatase in muscle, and muscle glycogenolysis provides energy just for muscle with production of lactate. Deficiencies of more than eight enzymes involved in glycogenolysis, including those mentioned, can produce glycogen storage disease.

A child presents with low blood glucose (hypoglycemia), enlarged liver (hepatomegaly), and excess fat deposition in the cheeks (cherubic facies). A liver biopsy reveals excess glycogen in hepatocytes. Deficiency of which of the following enzymes might explain this phenotype? Select one: a. α-1,4-galactosidase b. α-1,6-galactosidase c. α-1,4-glucosidase d. α-1,1-glucosidase e. α-1,1-galactosidase

a. α -ketoglutarate dehydrogenase The alcoholic has become deficient in vitamin B1, thiamine, which is converted to thiamine pyrophosphate for use as a coenzyme. One of the symptoms of B1 deficiency is neurological, due to insufficient energy generation within the nervous system. B1 is required for a small number of enzymes, including transketolase, pyruvate dehydrogenase, and α-ketoglutarate dehydrogenase. By reducing the activity of the latter two enzymes, glucose oxidation to generate energy is impaired, and the ner- vous system suffers because of it.

A chronic alcoholic, while out on a binge, became very confused and forgetful. The police found the man and brought him to the emergency department. Upon examination, he displayed nystagmus and ataxia. Which enzyme is displaying reduced activity in his brain under these conditions? Select one: a. α -ketoglutarate dehydrogenase b. Succinate dehydrogenase c. Glyceraldehyde-3-phosphate dehydrogenase d. Malate dehydrogenase e. Isocitrate dehydrogenase

d. The E1 subunit of pyruvate dehydrogenase Lactic acidosis can result from a defect in an enzyme that metabolizes pyruvate (primarily pyruvate dehydrogenase and pyruvate carboxylase). The pyru- vate dehydrogenase complex consists of three major catalytic subunits, designated E1, E2, and E3. The E1 subunit is the one that binds thiamine pyrophos- phate and catalyzes the decarboxylation of pyruvate. The gene for the E1 subunit is on the X chromosome, so defects in this subunit are inherited as X-linked diseases, which primarily affects males. Since this is the second male child to have these symptoms, it is likely that the mother is a carrier for this disease. The pattern of inheritance distinguishes this diagnosis from that of an E2 or E3 deficiency. In addition, an E3 deficiency would affect more than pyruvate metab- olism, as this subunit is shared with other enzymes that catalyze oxidative decarboxylation reactions, and other metabolites would also be accumulating. Defects in citrate synthase and malate dehydrogenase would not lead to severe lactic acidosis and would not be male-specific disorders. As an example, the three subunits of α-ketoglutarate dehydrogenase are shown below.

A family that had previously had a newborn boy die of a metabolic disease has just given birth to another boy, small for gestational age, and with low Apgar scores. The child displayed spasms a few hours after birth. Blood analysis indicated extremely high levels of lactic acid. Analysis of cerebrospinal fluid showed elevated lactate and pyruvate. Hyperalaninemia was also observed. The child died within 5 days of birth. The biochemical defect in this child is most likely which of the following? Select one: a. The E3 subunit of pyruvate dehydrogenase b. The E2 subunit of pyruvate dehydrogenase c. Citrate synthase d. The E1 subunit of pyruvate dehydrogenase e. Malate dehydrogenase

e. Mutations in mitochondrial tRNA MELAS and MERRF. The two classic well-characterized diseases associated with mt-tRNA mutations are mitochondrial encephalomyopathy, lactic acidosis, and stroke-like episodes (MELAS) and myoclonic epilepsy with ragged red fibers (MERRF)

A human geneticist is studying two different families. In one family, all of the children of a mildly affected mother display myoclonic epilepsy, developmental display, and abnormal muscle biopsy (ragged red fibers). In the other family, the three children of an affected woman endure strokelike episodes and a mitochondrial myopathy. The common link between these two diseases is which of the following? Select one: a. Mutations in cytoplasmic tRNA b. Mutations in pyruvate carboxylase c. Mutations in pyruvate dehydrogenase complex d. Mutations in malate dehydrogenase e. Mutations in mitochondrial tRNA

c. Increase in intracellular AMP AMP will activate muscle glycogen phosphorylase b allosterically,allowing glycogen degradation to begin before any hormonal signal has reached the muscle. The addition of epinephrine to the muscle requires activation of adenylate cyclase to initiate glycogen degradation, and adenylate cyclase has been inactivated in this cell line. Muscle lacks glucagon receptors, so cannot respond to this hormone. An increase in intracellular calcium would lead to glycogen degradation (via activation of phosphorylase kinase b), but magnesium does not have the same effect as calcium. Increases in ADP levels will not activate glycogen phosphorylase b; the allosteric activator is specific for AMP. The table below summarizes the allosteric interactions involved in glycogen metabolism.

A muscle cell line has been developed with a nonfunctional adenylate cyclase gene. Glycogen degradation can be induced in this cell line via which of the following mechanisms? Select one: a. Increase in intracellular ADP b. Increase in intracellular magnesium c. Increase in intracellular AMP d. Addition of epinephrine e. Addition of glucagon

b. Dinitrophenol The key is the elevation in temperature. Dinitrophenol is an uncoupler of oxidation and phosphorylation in that uncouplers destroy the proton gradient across the membrane (thereby inhibiting the synthesis of ATP) without blocking the transfer of electrons through the electron transfer chain to oxygen. The energy that should have been generated in the form of a proton gradient is lost as heat, which elevates the body temperature of the affected workers. Electron flow is also enhanced in the presence of an uncoupler, so additional oxygen is required to allow the chain to continue (hence the heavy breathing). The other agents added would have stopped electron transfer totally, which would not allow for an increase in temperature, and would actually decrease the rate of breathing (since oxygen is no longer required for the nonfunctioning electron transfer chain). Atractyloside inhibits the ATP/ADP exchanger, and once there is no ADP in the mitochondrial matrix, electron flow will stop due to the inability to synthesize ATP (normal coupling). Oligomycin works in a similar mechanism in that it blocks the ATP synthase, preventing ATP synthesis, and, due to coupling, electron transfer through the chain. Rotenone blocks complex I transfer to coen- zyme Q, which significantly reduces electron flow, and will not lead to an increase in temperature.

A pair of farm workers in Mexico was spraying pesticide on crops when they both developed the following severe symptoms: heavy, labored breathing, significantly elevated temperature, and loss of consciousness. The pesticide contained an agent that interfered with oxidative phosphorylation, which most closely resembled which of the following known inhibitors? Select one: a. Cyanide b. Dinitrophenol c. Rotenone d. Oligomycin e. Atractyloside

a. Glutathione Glutathione. Glutathione is the major antioxidant in the fl uid lining the bronchial epithelium. It is essential for recovery of these tissues. Depletion of glutathione in the airway is thought to greatly increase a person's susceptibility to upper respiratory infections such as infl uenza. Glutathione is formed in the γ-glutamyl pathway, and oxidized glutathione is regenerated to reduced glutathione using NADPH produced by the HMP shunt pathway. None of the other answers (glycogen, sorbitol, pyruvate, and glucuronate) offer protection against oxidative damage. Glycogen is utilized for the storage of glucose. Pyruvate is the end product of glycolysis. Sorbitol is a product of the polyol pathway. Glucuronic acid is used for xenobiotic metabolism, in general, to increase the solubility of the xenobiotic and to prepare it for excretion.

A patient is recovering from acute respiratory distress syndrome (ARDS). Which of the following is a major antioxidant found in the fluid lining the bronchial epithelium needed in high concentration for recovery from ARDS? Select one: a. Glutathione b. Glycogen c. Glucuronic acid d. Sorbitol e. Pyruvate

b. Dinitrophenol 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. With- out 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.

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? Select one: a. Carbon monoxide b. Dinitrophenol c. Cyanide d. Rotenone e. Antimycin

b. Galactosemia The patient's symptoms and course in response to a lactose-containing formula are consistent with a diagnosis of galactosemia. Pyruvate kinase deficiency and glucose 6-phosphate dehydrogenase deficiency would manifest as anemias and are seldom seen in an infant in the case of G6PD deficiency. G6PD deficiency is usually identified by the occurrence of a hypoglycemic coma following an overnight fast but is not normally accompanied by vomiting or diarrhea. While genetic screening tests required in most states identify newborns with galactosemia, these tests may not have been performed on a child born outside the United States.

A woman returns from a yearlong trip abroad with her 2-week-old infant, whom she is breastfeeding. The child soon starts to exhibit lethargy, diarrhea, vomiting, jaundice, and an enlarged liver. The pediatrician prescribed a switch from breast milk to infant formula containing sucrose as the sole carbohydrate. The baby's symptoms resolve within a few days. Which of the following was the most likely diagnosis? Select one: a. Pyruvate kinase deficiency b. Galactosemia c. G6PD deficiency d. Hereditary fructose intolerance e. Essential fructosuria

a. Fructose 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 oxi- dase. 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.

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? Select one: a. Fructose b. Lactose c. Maltose d. Glucose e. Sorbitol

e. Succinate 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.

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? Select one: a. Pyruvate b. a-Ketoglutarate c. Malate d. Isocitrate e. Succinate

c. Pyruvate carboxylase The body's major energy source for gluconeogenesis is fatty acids, which are oxidized to acetyl-CoA, at which point acetyl-CoA enters the TCA cycle to produce ATP. Acetyl-CoA activates pyruvate carboxylase (and inhibits pyruvate dehydrogenase), a key gluconeogenic enzyme. Acetyl-CoA does not regulate any of the other enzymes listed as potential answers (PEPCK is transcriptionally regulated by CREB; Fructose-1,6-bisphosphatase is inhibited by fructose-2,6-bisphosphate; glucose-6-phosphatase is regulated by a regulatory protein; and pyruvate kinase has both allosteric and covalent controls in the liver, but none involve acetyl-CoA).

An important product of the oxidation of the body's major energy source to provide energy for gluconeogenesis regulates which of the following key gluconeogenic enzymes? Select one: a. Fructose-1,6-bisphosphatase b. PEPCK c. Pyruvate carboxylase d. Pyruvate kinase e. Glucose-6-phosphatase

d. Increase in sarcoplasmic calcium levels Increase in sarcoplasmic calcium levels. When the individual begins to run away from the alligator, muscle contraction leads to calcium release from the sarcoplasmic reticulum to the sarcoplasm. This increase in sarcoplasmic calcium binds to the calmodulin subunit of phosphorylase kinase and activates the enzymein an allosteric manner, in the absence of any covalent modifi cation. The activated phosphorylase kinase willphosphorylate and activate glycogen phosphorylase, which will initiate glycogen degradation. When epinephrinereaches the muscle, phosphorylase kinase will be fully activated via phosphorylation by PKA. The activationof glycogen degradation under these conditions is not due to a decrease in blood glucose levels, insulinbinding (insulin would not be released under these conditions), a decline in ATP levels (the AMP-activated protein kinase does not activate glycogen degradation), or lactate production, the end product of anaerobic metabolism.

An individual is taking a serene walk in the park when he spots an escaped alligator from the zoo. The individual runs away as fast as he can. Glycogen degradation is occurring to supply glycolysis with a substrate even before epinephrine has reached the muscle. This is due to which of the following? Select one: a. Insulin binding to muscle cell receptors b. Decline in ATP levels c. Sudden decrease in blood glucose levels d. Increase in sarcoplasmic calcium levels e. Lactate production

c. Providing substrate for glyceraldehyde-3-phosphate dehydrogenase Providing substrate for glyceraldehyde-3-phosphate dehydrogenase.Under anaerobic conditions, the NADH generated by the glyceraldehyde-3-phosphate dehydrogenase step accumulates. Normally, the NADH would transfer its electrons to mitochondrial NAD+, and the electrons would be donated to the electron transfer chain. However, in the absence of oxygen the electron transfer chain is not functioning. Thus, as NADH accumulates in the cytoplasm, the levels of NAD+ decrease to the point that there would be insufficient NAD+ available to allow the glyceraldehyde-3-phosphate dehydrogenase reaction to proceed, thereby inhibiting glycolysis. To prevent glycolytic inhibition, lactate dehydrogenase will convert pyruvate to lactate, regenerating NAD+ for use in glycolysis, specifically as a substrate for the glyceraldehyde-3-phosphate dehydrogenase reaction. While hexokinase is inhibited by its product glucose-6-phosphate, this allosteric effect does not explain lactate formation under anaerobic conditions.Similarly, while phosphoglyceromutase does require 2,3-bisphosphoglycerate, anaerobiosis does not increase 2,3-bisphosphoglycerate levels, nor does it alleviate the lack of NAD+ under these conditions. Pyruvate kinase is not inhibited by pyruvate (ATP and alanine are the allosteric inhibitors of this enzyme). AMP is an activator of phosphofructokinase-1; however, this activation does not relate to lactate formation under anaerobic conditions.

Anaerobiosis leads to lactate formation in muscle due to which one of the following? Select one: a. Inhibiting hexokinase by glucose-6-phosphate b. Providing 2,3-bisphosphoglycerate for the phosphoglyceromutase reaction c. Providing substrate for glyceraldehyde-3-phosphate dehydrogenase d. Inhibiting pyruvate kinase by pyruvate e. Inhibiting phosphofructokinase-1 by AMP

a. Carbon monoxide The electron transport chain shown con- tains three proton pumps linked by two mobile electron carriers. At each of these three sites (NADH-Q reductase, cytochrome reductase, and cyto- chrome oxidase) the transfer of electrons down the chain powers the pumping of protons across the inner mitochondrial membrane. The block- age of electron transfers by specific point inhibitors leads to a buildup of highly reduced carriers behind the block because of the inability to trans- fer electrons across the block. In the scheme shown, rotenone blocks step A, antimycin A blocks step B, and carbon monoxide (as well as cyanide and azide) blocks step E. Therefore a carbon monoxide inhibition leads to a highly reduced state of all of the carriers of the chain. Puromycin and chloramphenicol are inhibitors of protein synthesis and have no direct effect upon the electron transport chain.

As electrons are received and passed down the transport chain shown below, the various carriers are first reduced with acceptance of the electron and then oxidized with loss of the electron. A patient poisoned by which of the following compounds has the most highly reduced state of most of the respiratory chain carriers? Select one: a. Carbon monoxide b. Chloramphenicol c. Puromycin d. Antimycin A e. Rotenone

d. Lysosomal hydrolases Lysosomal hydrolases. In I-cell disease, the lysosomal hydrolases are mistargeted and are excreted from cells into the circulation. As the pH of the blood is above 7 and the pH optimum of these enzymes is around 5, there is no activity of the hydrolases in blood. In Hurler syndrome, a defect in the degradation of mucopolysaccharides, there is an accumulation of dermatan and heparan sulfate in the urine, but not in the blood. Short-chain dicarboxylic acids are produced with a defect in medium chain acyl-CoA dehydrogenase, and cytochrome c release into the cytoplasm of cells from mitochondria is the signal to initiate apoptosis.

Children with either I-cell disease or Hurler syndrome show very similar clinical features. One method to distinguish between the two is to find which of the following elevated in the blood? Select one: a. Short-chain dicarboxylic acids b. Heparan sulfate c. Dermatan sulfate d. Lysosomal hydrolases e. Cytochrome c

e. In the inner mitochondrial membrane

Electron transport chain carriers are located in the mitochondria: Select one: a. On the inner surface of the outer mitochondrial membrane b. On the outer surface of the outer mitochondrial membrane c. In mitochondrial matrix d. In the intermembrane space e. In the inner mitochondrial membrane

d. In the inner mitochondrial membrane.

Electron transport chain carriers are located in the mitochondria: Select one: a. On the outer surface of the outer mitochondrial membrane b. In the intermembrane space c. On the inner surface of the outer mitochondrial membrane d. In the inner mitochondrial membrane. e. In mitochondrial matrix

a. 3 The reduction of NAD+ to NADH in the TCA cycle, starting with citrate, occurs at the isocitrate dehydrogenase, α-ketoglutarate dehydrogenase, and malate dehydrogenase-catalyzed reactions, thus generating 3 moles of NADH per mole of citrate.

Energy made available for useful work from metabolism can be represented as the number of moles of ATP, reduced flavoproteins, and reduced pyridine nucleotides that are formed. How many moles of NADH are formed during the conversion of 1 mole of citrate to 1 mole of oxaloacetate and 2 moles of CO2? Select one: a. 3 b. 6 c. 12 d. 8 e. 2

e. activation of phosphorylase kinase

Epinephrine, released in response to exercise, will lead to phosphorylation events that will ultimately exert which of the following effects on hepatic glycogen metabolism? Select one: a. inhibition of glycogen synthase kinase-3 (GSK-3) b. inhibition of glycogen phosphorylase c. activation of glycogen synthase d. activation of adenylate kinase e. activation of phosphorylase kinase

e. The glycogen granules have more surface area and can be dissolved faster

Glucose is stored in plants as amylopectin granules and in animal tissues as glycogen granules. The two homopolysaccharides are similar in linkages but differ in the degree of branching and in shapes of the granules. Amylopectin usually forms large, single grains, while glycogen is usually present as smaller, clustered granules resembling bunches of grapes. How does this structure of glycogen granules benefit an animal? Select one: a. Animals would actually be better off if their starch were stored as amylopectin because it would be more resistant to chemical attack b. The glycogen granules find safety in numbers c. The glycogen granules are more resistant to oxidation d. Amylopectin granules are more stable and last longer e. The glycogen granules have more surface area and can be dissolved faster

e. 5. Given the enzymes present, only the nonoxidative reactions of the HMP shunt would take place. In order for the nonoxidative reactions to occur, the glucose-6-phosphate (G6P, labeled in the 6th position with 14C) must pass through glycolysis to produce fructose-6-phosphate (F6P, labeled in the 6th position) and glyceraldehyde-3-phosphate (labeled in the 3rd position). Transketolase will allow these two compounds to exchange carbons, which would generate erythrose-4-phosphate (E4P, labeled in the 4th position) and xylulose-5-phosphate (X5P, labeled in the 5th position). The X5P can then go to ribulose-5-phosphate (Ru5P) and ribose-5-phosphate (R5P), labeled in the fifth positions. The E4P (labeled in the 4th position) can react with another molecule of F6P (labeled in the 6th position) using transaldolase to generate sedoheptulose 7-phosphate (Se7P, labeled in the 7th position) and glyceraldehyde-3-phosphate (G3P), labeled in the 3rd position. Transketolase will then convert the Se7P and G3P to R5P and X5P, both labeled in the 5th positions.

Glucose-6-phosphate labeled in carbon 6 with 14C was added to a test tube with the enzymes phosphohexose isomerase, PFK-1, aldolase, transketolase, and transaldolase. ATP was also added to the test tube. At equilibrium, in which position would the radioactive label be found in the newly produced ribose-5-phosphate? Select one: a. 2. b. 3. c. 1. d. 4. e. 5.

c. 30 to 32 Each mole of glucose produces 2 mol of pyruvate, 2 mol of ATP, and 2 mol of NADH during glycolysis. As each pyruvate is converted to acetyl-CoA, 1 NADH is produced. Each acetyl-CoA enters the TCA cycle to produce 1 GTP, 3 NADH, and 1 FAD(2H). Therefore, the oxidation of 1 mol of glucose would yield a total of 2 mol of ATP and 2 mol of NADH from glycolytic reactions; 2 mol of NADH from the conversion of 2 mol of pyruvate to 2 mol of acetyl-CoA; and 6 mol of NADH, 2 mol of GTP, and 2 mol of FAD(2H) from the oxidation of 2 mol of acetyl- CoA in the TCA cycle. Each GTP has the same energy as 1 ATP. Each mitochondrial NADH is oxidized to produce 2.5 ATP. Each cytoplasmic NADH can generate either 1.5 ATP (using the glycerol phosphate shuttle) or 2.5 ATP (using the malate-aspartate shuttle). Each mitochondrial FAD(2H) is oxidized to produce 1.5 ATP. Therefore, the total ATP yield from the oxidation of 1 mol of glucose is 30 to 32 mol of ATP (30 mol if the glycerol phosphate shuttle is used; 32 mol if the malate-aspartate shuttle is used)

How many moles of ATP are generated by the complete aerobic oxidation of 1 mol of glucose to 6 mol of CO2? Select one: a. 60 to 64 b. 18 to 22 c. 30 to 32 d. 2 to 4 e. 10 to 12

b. Heat production The uncoupling of the normal process of proton flow across the inner mitochondrial membrane and coupling movement back into the matrix through ATP synthase releases the energy of the electrochemical proton gradient as heat.

In a study of the regulation of body weight, the administration of exogenous leptin to mice is found to increase the activity of proton leak channels located in the inner membrane of mitochondria in adipose tissue cells. Which of the following is most likely to increase as a result of the increased transport of protons through these mitochondrial leak channels? Select one: a. Mitochondrial NADH concentrations b. Heat production c. pH of the mitochondrial matrix d. Mitochondrial synthesis of ATP

b. GLUT3 In the brain, the endothelial cells of the capillaries have extremely tight junctions, and glucose must pass from the blood into the extracellular cerebrospinal fluid by GLUT 1 transporters in the endothelial cell membranes (Fig. 27.14) andthen through the basement membrane. Measurements of the overall process of glucose transport from the blood into the brain (mediated by GLUT 3 on neural cells) show a Km,app of 7 to 11 mM and a maximal velocity not much greater than the rate of glucose utilization by the brain.

In patients with suspected metastatic cancer in the brain, a useful diagnostic is positron emission tomography, PET. This technique utilizes a radioactive form of glucose that cannot be metabolized but is rapidly taken up by cancer cells due to their high rates of metabolism. Which of the following is most responsible for the rapid uptake of this compound into the tumor cells in the brain? Select one: a. PFK1 b. GLUT3 c. GLUT4 d. insulin e. GLUT2

c. succinate-CoA synthetase Succinyl-CoA synthetase catalyzes the conversion of succinyl-CoA to succinate. Simultaneously, the reaction generates GTP from GDP, utilizing the high-energy bond between CoA and succinate. This formation of GTP is referred to as substrate-level phosphorylation to distinguish it from ATP production during

In the TCA cycle, GTP is produced via a process referred to as substrate-level phosphorylation. Which of the following enzymes is involved in this process of formation of GTP from GDP? Select one: a. citrate synthase b. succinate dehydrogenase c. succinate-CoA synthetase d. malate dehydrogenase e. isocitrate dehydrogenase

e. Activation of glucose-6-phosphate dehydrogenase The answer is A: Activation of glucose-6-phosphate dehydrogenase. Glutathione reductase will utilize NADPH and reduce oxidized glutathione to reduced glutathione, generating NADP+. If Glutathione reductase is superactive, NADP+ levels accumulate, which activates glucose-6-phosphate dehydrogenase. This will lead to NADPH production via the oxidative reactions of the HMP shunt, along with ribulose-5-phosphate (Ru5P). The Ru5P will lead to increased ribose-5-phosphate production, increased 5′-phosphoribosyl 1′-pyrophosphate (PRPP) production, and increased 5′ phosphoribosyl 1′-amine levels. This eventually leads to increased purine production, in excess of what is required. Theexcess purines are converted to uric acid, and excess uric acid will lead to gout. A superactive glutathione reductase will not lead to an alteration in the activities of transketolase or transaldolase.

Individuals with a superactive glutathione reductase will develop gout. This occurs due to which of the following? Select one: a. Activation of transketolase b. Inhibition of glucose-6-phosphate dehydrogenase c. Inhibition of transketolase d. Activation of transaldolase e. Activation of glucose-6-phosphate dehydrogenase

e. Inhibiting of mitochondrial ATPase (ATP synthase) Oligomycin inhibits mitochondrial ATPase and thus prevents phosphorylation of ADP to ATP. It prevents uti- lization of energy derived from electron transport for the synthesis of ATP. Oligomycin has no effect on coupling but blocks mitochondrial phosphor- ylation so that both oxidation and phosphorylation cease in its presence.

Inhibition of the synthesis of ATP during oxidative phosphorylation by oligomycin is thought to be due to Select one: a. Blocking of the proton gradient between cytochrome c1 and cytochrome c b. Uncoupling of electron transfer between NADH and flavoprotein c. Dissociating of cytochrome c from mitochondrial membranes d. Blocking of the proton gradient between NADH-Q reductase and QH2 e. Inhibiting of mitochondrial ATPase (ATP synthase)

c. homeotic gene

Klein-Waardenburg syndrome is a single-gene disorder that includes dystopia canthorum (lateral displacement of the inner corner of the eye), impaired hearing, and pigmentary abnormalities. The gene involved is most likely to be a Select one: a. transgene b. pseudogene c. homeotic gene d. tumor suppressor gene e. proto-oncogene

e. 5'-TATTTATATTATTATTA-3' TATA box - promoter, binding transcription factors

Many different mechanisms are involved in regulation of gene expression during the differentiation of tissues and biologic processes in the multicellular organisms. One important mechanism regulates expression of different genes at the level of transcription. A DNA sequence in the human genome which is responsible for binding of multiple proteins, activating transcription is which of the following? Select one: a. 5'-CACACACACACACA-3' b. 5'-CGCGCGCGCGCGC-3 c. 5'-AAAGAGGAGAGGAG-3' d. 5'-GGGTTTTGGGTGTGT-3 e. 5'-TATTTATATTATTATTA-3'

c. Glucose --> glucose-6-phosphate MODY 2 results from a defective glucokinase. Glucokinase is an enzyme that traps glucose inside the cell by converting it to glucose-6-phophate. MODY 2 is an autosomal dominant, monogenic form of diabetes that results from any number of mutations on the glucokinase gene on chromosome 7. Normally glucokinase functions to trap glucose within the cell by converting it to glucose-6-phosphate, which does not have a transporter to exit the cell. Glucokinase allows the cell to sense the amount of glucose in the body and normally triggers insulin release. Since the defective glucokinase does not trigger the release of insulin at 90 mg/dL but works at a much higher level > 126 mg/dL, the individual experiences a chronic, mild, asymptomatic hyperglycemia.

Maturity Onset Diabetes of the Young (MODY) type 2 is a consequence of a defective pancreatic enzyme, which normally acts as a glucose sensor, resulting in a mild hyperglycemia. The hyperglycemia is especially exacerbated during pregnancy. Which of the following pathways is controlled by this enzyme? Select one: a. Fructose-6-phosphate --> fructose-1,6-biphosphate b. Glyceraldehyde-3-phosphate --> 1,3-bisphosphoglycerate c. Glucose --> glucose-6-phosphate d. Phosphoenolpyruvate --> pyruvate e. Glucose-6-phosphate --> fructose-6-phosphate

c. The high NADH/NAD+ ratio impaired gluconeogenesis

Paramedics bring a patient to the emergency department because he was found unconscious in an alleyby passers by. The man was unshaven and dishevelled, and appeared to be about 40 years old. Blood alcohol levels were found to be 0.25% and blood glucose levels 32 mg/dL. Treatment was initiated, and this enabled the man to regain consciousness, although he was still inebriated. While conscious, a history revealed that the man was a chronic alcoholic, and as far as he could remember, he had been only drinking for the past 2 weeks, with nothing to eat. Analysis of liver enzyme levels in his blood revealed normal readings. Assuming that his liver is still functioning normally, why is this patient hypoglycemic? Select one: a. Liver glycogen stores were depleted by the high NADH/NAD+ ratio b. The high NAD+ /NADH ratio impaired glycolysis c. The high NADH/NAD+ ratio impaired gluconeogenesis d. The high NAD+/NADH ratio impaired gluconeogenesis e. Liver glycogen stores were depleted by the high NAD+/NADH ratio

e. NAD+ The reduced electron carriers generated via reactions of the TCA cycle (NADH and FADH2) are subsequently reoxidized to NAD+ and FAD via the oxidative phosphorylation pathway.

Regeneration of which of the following participants in the TCA cycle is dependent on the presence of oxygen? Select one: a. ADP b. succinate c. citrate d. α-ketoglutarate e. NAD+

a. α-D-glucopyranose Hexose (six-carbon) and pentose (fivecarbon) sugars have first carbons (C1) with aldehyde groups and fifth or sixth carbons with alcohol groups. All but the initial and terminal carbons are optically active, meaning that glucose has four optically active carbons and 24 or 16 stereoisomers with the same structural formula, including galactose and mannose (epimers). By convention, the position of the hydroxyl to the right or left of the carbon next to the alcohol group (C5 in glucose) determines whether it is the D (to right) or L (to left) form. Other carbohydrates such as D-glycerol (hydroxyl on C2 to right) or D-fructose (hydroxyl on C4 to right) are named relative to D-glucose. In aqueous solutions, the C4 of hexoses (C3 of pentoses) binds to the C1 aldehyde to form a hemiacetal pyranose (six-membered) ring structure (depicted as a Hayworth formula). Less favorable for hexoses are the furanose (fivemembered) rings formed by bonding C1 aldehyde to the C3 hydroxyl. The C1 hydroxyl formed by hemiacetal formation can extend above (β) or below (α) the ring. In D-glucose, the α and β forms (anomers) are free to interconvert. When joined with other carbohydrates to form α or β polysaccharides, they are fixed. As with most stereoisomers, biological systems exhibit strong preferences for certain isomers of hexoses (e.g., glucose, fructose), pentoses (e.g., ribose), or glycosides (e.g., α-glycosides in linear glycogen and β-oriented cardiac glycosides).

Structure 2 in panel of the figure below is referred to as: Select one: a. α-D-glucopyranose b. α-D-glucofuranose c. β-D-glucopyranose d. β-L-glucofuranose e. α-D-fructofuranose

c. anomeric The carbon about which the rotation occurs in ring structures of carbohydrates is the anomeric carbon and the 2 forms are termed anomers.

The aldehyde and ketone moieties of the carbohydrates with 5 and 6 carbons will spontaneously react with alcohol groups present in neighboring carbons to produce ring structures. The rings can open and close allowing for different configurations of the atoms in the ring. Which of the following represents the carbon about which this rotation occurs? Select one: a. glycosidic b. hydroxy c. anomeric d. terminal e. chiral

a. Synthesis of ATP as protons flow into the mitochondrial matrix along a proton gradient that exists across the inner mitochondrial membrane The chemiosmotic hypothesis of Mitchell describes the coupling of oxidative phosphorylation and electron trans- port. The movement of electrons along the electron transport chain allows protons to be pumped from the matrix of the mitochondria to the cyto- plasmic side. The protons are pumped at three sites in the electron trans- port chain to produce a proton gradient. When protons flow back through proton channels of the asymmetrically oriented ATPase of the inner mito- chondrial membrane, ATP is synthesized.

The connection between oxidation phosphorylation and electron transport is best described by Select one: a. Synthesis of ATP as protons flow into the mitochondrial matrix along a proton gradient that exists across the inner mitochondrial membrane b. Existence of a higher pH in the cisternae of the endoplasmic reticulum than in the cytosol c. Dissociation of electron transport and oxidative phosphorylation d. Symmetric distribution of the ATPase of the inner membrane of the mitochondria e. Absence of ATPase in the inner membrane of the mitochondria

c. One high-energy bond The answer is A: One high-energy bond. For a molecule of glucose-6-phosphate (G6P) to be incorporated into glycogen, the following pathway must be utilized: G6P is converted to glucose-1-phosphate (G1P) via phosphoglucomutase, the G1P reacts with UTP to form UDPglucose via glucose-1-phosphate uridyl transferase, releasing pyrophosphate. The resultant pyrophosphate is hydrolyzed to two inorganic phosphates, with the loss of one high-energy bond. The UDPglucose then reacts with glycogen to produce a glycogen chain with one additional sugar, and UDP is released. The overall equation for these steps is: G6P + UTP + glycogenn yields UDP + 2Pi + (glycogen)n+1. These steps are outlined below:Gluose-6-phosphate → Glucose-1-phosphateGlucose-1-phosphate + UTP → UDPglucose + PPi PPi + H2O → 2 PiUDPglucose + glycogenn → Glycogenn+1 + UDP UDP + ATP → UTP + ADP Sum: Glucose-6-phosphate + ATP + glycogenn + H2O → glycogenn+1 + ADP + 2Pi

The energy required to store one molecule of glucose-6-phosphate as a portion of glycogen is which of the following? Select one: a. Three high-energy bonds b. No high-energy bonds c. One high-energy bond d. Four high-energy bonds e. Two high-energy bonds

e. Lactate inhibition of kidney tubule absorption of urate Lactate inhibition of kidney tubule absorption of urate. Patients with von Gierke disease display elevated levels of lactate, which interferes with the kidney's ability to remove uric acid from the blood and place it in the urine. This leads to hyperuricemia. The reason lactate levels are elevated is that the high glucose-6-phosphate in the cell (recall, the defect in this disorder is a lack of glucose-6-phosphatase activity) forces glycolysis forward, producing pyruvate, which is converted to lactate in order to regenerate NAD+ to allow glycolysis to continue. Glucose-6-phosphate does not inhibit glucose-6-phosphate dehydrogenase (that enzyme is regulated by the NADP+ levels), nor does it regulate a committed step of de novo purine synthesis, amidophosphoribosyl transferase (which is regulated by adenine and guanine nucleotides). Glucose-6-phosphate does stimulate glycogen synthase, but that activation does not play a role in elevated urate levels. Glucose-6-phosphate does not affect urate absorption within the kidney.

The hyperuricemia observed in patients with von Gierke disease comes about due to which of the following? Select one: a. Glucose-6-phosphate inhibition of kidney tubule absorption of urate b. Glucose-6-phosphate inhibition of glucose-6-phosphate dehydrogenase activity c. Glucose-6-phosphate stimulation of glycogen synthase d. Glucose-6-phosphate activation of amidophosphoribosyltransferase activity e. Lactate inhibition of kidney tubule absorption of urate

d. Lactate

The major metabolic product produced under normal circumstances by erythrocytes and by muscle cells during intense exercise is recycled through the liver in the Cori cycle. The metabolite is Select one: a. Glycerol b. Oxaloacetate c. Pyruvate d. Lactate e. Alanine

c. C In the citric acid cycle, the conversion of α-ketoglutarate to succinate results in decarboxylation, transfer of an H+/e− pairtoNADH+H+,andthesubstrate-levelphosphorylationofGDPtoGTP. The series of reactions involved is quite complex. First, α-ketoglutarate reacts with NAD+ + CoA to yield succinyl CoA + CO2 + NADH + H+. These reactions occur by the catalysis of the α-ketoglutarate dehydrogenase com- plex, which contains lipoamide, FAD+, and thiamine pyrophosphate as prosthetic groups. Under the action of succinyl CoA synthetase, succinyl CoA catalyzes the phosphorylation of GDP with inorganic phosphate cou- pled to the cleavage of the thioester bond of succinyl CoA. Thus, the pro- duction of succinate from α-ketoglutarate yields one substrate-level phosphorylation and the production of three ATP equivalents from NADH via oxidative phosphorylation.

Transfer of H+/e− pairs to electron transport carriers, decarboxylation, and substrate-level phosphorylation occur at some of the steps shown in the following diagram of the citric acid cycle. All three of these events occur at which step? Select one: a. E b. D c. C d. B e. A

a. A. Upon insulin release, the cAMP phosphodiesterase is activated, reducing cAMP levels in theliver, thereby leading to inactivation of protein kinase A. In addition, protein phosphatase 1 has become active anddephosphorylates the enzymes that were phosphorylated by protein kinase A. Therefore, PFK-2 is not phosphorylated,which leads to an active kinase activity and an inactive phosphatase activity (choices A, D, or E). The active kinase of PFK-2 produces more fructose-2,6-bisphosphate, leading to the activation of PFK-1 (answers A through D; combined with PFK-2 activity, now only choice A or D can be correct). Insulin stimulates preformed GLUT4 transporters in the muscle to fuse with the plasma membrane, thereby enhancing glucose transport into the muscle (choices A through C; combined with the other two columns, only choice A can be correct).

Which of the following changes in enzyme activity will occur within 1 h of a type 1 diabetic taking an injection of insulin? Select one: a. A. b. E. c. C. d. B. e. D.

c. cytochrome c oxidase Complex IV, also known as cytochrome c oxidase, contains the heme proteins known as cytochrome a and cytochrome a3, as well as copper-containing proteins in which the copper undergoes a transition from Cu+ to Cu2+ during the transfer of electrons from the heme iron in cytochrome c through the complex to molecular oxygen.

Which of the following enzymes catalyzes the reaction shown? Fe2+ + O2 + 4H+ => 2H2O + Fe2+ Select one: a. peroxidase b. cytochrome P450 c. cytochrome c oxidase d. superoxide dismutase e. catalase

e. Amylo-(1,4 → 1,6)-transglycosylase Amylo-(1,4 →1,6)-transglycosylase (also known as the branching enzyme) functions in glycogen synthesis. The other enzymes listed are involved in glycogenolysis. Deficiencies in these enzymes lead to glycogen storage with enlarged liver and hypoglycemia. Mobilization of glycogen stores to produce glucose in the liver requires phosphorolysis of the glycogen chain by the enzyme phosphorylase, which is phosphorylated by phosphorylase kinase. Also needed are the hydrolysis of α-1,6-glycosidic bonds by amylo-1,6-glucosidase (also known as the debranching enzyme) and the hydrolysis of glucose-6-phosphate derived from glucose-1-phosphate (a product of phosphorylase) by glucose-6-phosphatase to produce glucose for export into the blood.

Which of the following enzymes is associated with glycogen synthesis? Select one: a. Phosphorylase kinase b. Amylo-1,6-glucosidase c. Glucose-6-phosphatase d. Phosphorylase e. Amylo-(1,4 → 1,6)-transglycosylase

b. CO2 is consumed In the formation of phosphoenolpyruvate during gluconeogenesis, oxaloacetate is an intermediate. In the first step, catalyzed by pyruvate carboxylase, pyruvate is carboxylated with the utilization of one high-energy ATP phosphate bond: pyruvate + ATP + CO2 → oxaloacetate + ADP + Pi In the second step, catalyzed by phosphoenolpyruvate carboxykinase, a high-energy phosphate bond of GTP drives the decarboxylation of oxaloacetate: oxaloacetate + GTP → phosphoenolpyruvate + GDP + CO2 In contrast to gluconeogenesis, the formation of pyruvate from phosphoenolpyruvate during glycolysis requires only pyruvate kinase, and ATP is made.

Which of the following events occurs during formation of phosphoenolpyruvate from pyruvate during gluconeogenesis? Select one: a. Inorganic phosphate is consumed b. CO2 is consumed c. ATP is generated d. GTP is generated e. Acetyl CoA is utilized

a. Glucose-6-phosphate Glucose-6-phosphate is a pivotal compound in many pathways. Immediately upon entering cells, blood glucose is phosphorylated to glucose-6-phosphate by hexokinase in most cells and by glucokinase in the liver. Glucose may only leave a cell in the dephosphorylated form produced by glucose-6-phosphatase, which is only found in liver. Glucose-6-phosphate may be the starting point of glycolysis, glycogen synthesis, and the pentose phosphate pathway. It can be considered the end point or switching point of glycogenolysis and gluconeogenesis. Uridine-diphosphoglucose (UDP-glucose) and UDP-galactose are high-energy forms of their respective sugars that are involved in converting galactose-I-phosphate to glucose-1-phosphate (the block in galactosemia) and in donating sugar groups to polysaccharides such as glycogen or glycosaminoglycans. Fructose-6-phosphate is involved in glycolysis and gluconeogenesis.

Which of the following metabolites is involved in glycogenolysis, glycolysis, and gluconeogenesis? Select one: a. Glucose-6-phosphate b. Galactose-1-phosphate c. Uridine diphosphogalactose d. Fructose-6-phosphate e. Uridine diphosphoglucose

b. They can be associated with sulfur and nonheme iron Some monooxygenases found in liver endoplasmic reticulum require cytochrome P450. This cytochrome acts to transfer electrons between NADPH, O2, and the substrate. It can be an elec- tron acceptor from a flavoprotein. In the mitochondrial electron transport chain, flavoproteins donate electrons to coenzyme Q, which then transfers them to other cytochromes. Flavoproteins that are oxidases often react directly with molecular oxygen to form hydrogen peroxide. Flavoproteins can be NADH dehydrogenases that oxidize NADH and transfer the elec- trons to coenzyme Q. The electron transfer centers of flavoproteins in the electron transport chain contain nonheme iron and sulfur.

Which of the following statements about flavoproteins is true? Select one: a. They cannot produce hydrogen peroxide b. They can be associated with sulfur and nonheme iron c. They do not participate in oxidation of NADH dehydrogenases d. They are not oxidized by coenzyme Q e. They receive electrons from cytochrome P450 in liver mitochondria

c. phosphorylation of hepatic glycogen synthase prevents glucose incorporation into glycogen

Which of the following statements correctly defines the situation under conditions of low blood glucose? Select one: a. the activity of muscle glycolysis is inhibited by the phosphorylation of PFK-1 b. hepatic glycogen releases glucose through a reversal of the glycogen synthase reaction c. phosphorylation of hepatic glycogen synthase prevents glucose incorporation into glycogen d. a reversal of the muscle hexokinase reaction converts glucose 6-phosphate to glucose for oxidation in glycolysis e. hepatic glycogen releases glucose in response to insulin-mediated dephosphorylation of glycogen phosphorylase

d. riboflavin The conversion of succinate to fumarate is catalyzed by the enzyme succinate dehydrogenase. This enzyme requires FAD, derived from riboflavin, as a cofactor for the reaction.

Which of the following vitamins is needed for the synthesis of a cofactor required for the conversion of succinate to fumarate? Select one: a. lipoic acid b. niacin c. thiamine d. riboflavin e. pantothenic acid

d. Xylulose-5-phosphate Xylulose-5-phosphate. In order for ribose-5-phosphate to be converted to glucose-6-phosphate, the nonoxidative reactions of the HMP shunt pathway must be used (the oxidative steps are not reversible reactions). In order for this to occur, the ribose-5-phosphate is isomerized to ribulose-5-phosphate, which is then epimerized to xylulose-5-phosphate. R5P and X5P then initiate a series of reactions utilizing transketolase and transaldolase to generate fructose-6-phosphate, which can be isomerized to glucose-6-phosphate. Glyceraldehyde-3-phosphate is also formed during this series of reactions, which then goes back to fructose-6-phosphate production. Pyruvate, oxaloacetate, 1,3-bisphosphoglycerate, and 6-phosphogluconoate are not obligatory intermediates in this conversion.

Which one of the following is an obligatory intermediate in the conversion of ribose-5-phosphate to glucose-6-phosphate? Select one: a. 1,3-bisphosphoglycerate b. Oxaloacetate c. 6-phosphogluconate d. Xylulose-5-phosphate e. Pyruvate

a. It is due to the lack of a fundamental coenzyme 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.

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? Select one: a. It is due to the lack of a fundamental coenzyme b. It is the result of noncompetitive inhibition c. It cannot be effectively treated. d. It is the result of competitive inhibition e. It is the result of irreversible inhibition.

e. lactose

You are examining a patient complaining of cramping in the lower belly, bloating, gas, and diarrhea. These symptoms appear within 30 minutes to 2 hours after the ingestion of dairy products. You advise your patient to avoid foods that contain a particular sugar. Which of the following is the sugar this patient should avoid? Select one: a. fructose b. galactose c. sucrose d. glucose e. lactose

e. lactose

You are examining a patient complaining of cramping in the lower belly, bloating, gas, and diarrhea. These symptoms appear within 30 minutes to 2 hours after the ingestion of dairy products. You advise your patient to avoid foods that contain a particular sugar. Which of the following is the sugar this patient should avoid? Select one: a. galactose b. fructose c. sucrose d. glucose e. lactose

b. galactose-1-phosphate uridyltransferase

You are examining a patient who complains of gastric discomfort following the consumption of milk. Additional signs and symptoms include liver and kidney impairment as well as neural involvement. Blood work indicates an increased concentration of galactose-1 -phosphate. This patient likely has a defect in which of the following enzymes? Select one: a. fructose-1,6-bisphosphatase b. galactose-1-phosphate uridyltransferase c. glucokinase d. galactokinase e. ketohexokinase (fructokinase)

(C) Liver α-glucosidase

1. A 3-month-old infant was brought to the pediatrician due to muscle weakness (myopathy) and poor muscle tone (hypotonia). Physical exam revealed an enlarged liver and heart, and heart failure. The infant had always fed poorly, had failure to thrive, and had breathing problems. He also had trouble holding up his head. Blood work indicated early liver failure. A liver biopsy indicated that glycogen was present and of normal structure. A poten-tial defect in this child is which of the following? (A) Liver glycogen phosphorylase (B) Liver glycogen synthase (C) Liver α-glucosidase (D) Liver debranching enzyme (E) Liver branching enzyme

(B) A mercaptan The man is suffer-ing from acetaminophen poisoning. As shown below, MEOS (the microsomal ethanol oxidizing system, also named CYP2E1) will convert acetaminophen into a toxic intermediate. In a chronic alcoholic, the MEOS has been induced and is very active. The toxic inter-mediate (NAPQI) can be rendered inactive by adding glutathione to the compound for safe excretion, and glutathione is a mercaptan (a compound with a free sulfhydryl group). Individuals with acetaminophen poisoning are given N-acetyl cysteine as a mechanism to increase glutathione production. Iron and vitamin C will not aid in detoxifying the toxic intermediate. Rifampin blocks RNA polymerase in bacteria. Aspi-rin will block cyclooxygenase, but will not stimulate NAPQI excretion.

1. A 52-year-old man has had bouts of alcohol abuse in his past. During his binges, he takes acetaminophen to help control some muscle pain. He then gets very ill (nausea, vomiting, and right upper quadrant pain), and is rushed to the emergency department. A potential treatment for this patient's symptoms is to take which of the following? (A) Aspirin (B) A mercaptan (C) Rifampin (D) Iron (E) Vitamin C

(C) α-ketoglutarate dehydrogenase The alcoholic has become defi cient in vitamin B1, thiamine, which is converted to thiamine pyrophosphate for use as a coenzyme. One of the symptoms of B1 defi ciency is neurological, due to insuffi cient energy generation within the nervous system. B1 is required for a small number of enzymes, including transketolase, pyruvate dehydrogenase, and α-ketoglutarate dehydrogenase. By reducing the activity of the latter two enzymes, glucose oxidation to generate energy is impaired, and the ner-vous system suffers because of it.

1. A chronic alcoholic, while out on a binge, became very confused and forgetful. The police found the man and brought him to the emergency department. Upon exam-ination, he displayed nystagmus and ataxia. Which enzyme is displaying reduced activity in his brain under these conditions? (A) Glyceraldehyde-3-phosphate dehydrogenase (B) Isocitrate dehydrogenase (C) α-ketoglutarate dehydrogenase (D) Succinate dehydrogenase (E) Malate dehydrogenase

(D) Providing substrate for glyceraldehyde-3-phosphate dehydrogenase Under anaerobic condi-tions, the NADH generated by the glyceraldehyde-3-phosphate dehydrogenase step accumulates. Normally, the NADH would transfer its electrons to mitochondrial NAD+, and the electrons would be donated to the elec-tron transfer chain. However, in the absence of oxygen the electron transfer chain is not functioning. Thus, as NADH accumulates in the cytoplasm, the levels of NAD+ decrease to the point that there would be insuf-fi cient NAD+ available to allow the glyceraldehyde-3-phosphate dehydrogenase reaction to proceed, thereby inhibiting glycolysis. To prevent glycolytic inhibition, lactate dehydrogenase will convert pyruvate to lactate, regenerating NAD+ for use in glycolysis, specifi cally as a substrate for the glyceraldehyde-3-phosphate dehy-drogenase reaction. While hexokinase is inhibited by its product glucose-6-phosphate, this allosteric effect does not explain lactate formation under anaerobic con-ditions. Similarly, while phosphoglyceromutase does require 2,3-bisphosphoglycerate, anaerobiosis does not increase 2,3-bisphosphoglycerate levels, nor does it alle-viate the lack of NAD+ under these conditions. Pyruvate kinase is not inhibited by pyruvate (ATP and alanine are the allosteric inhibitors of this enzyme). AMP is an activator of phosphofructokinase-1; however, this acti-vation does not relate to lactate formation under anaero-bic conditions. The fi gure summarizes these key points outlined above.

1. Anaerobiosis leads to lactate formation in muscle due to which one of the following? (A) Inhibiting hexokinase by glucose-6-phosphate (B) Providing 2,3-bisphosphoglycerate for the phos-phoglyceromutase reaction (C) Inhibiting pyruvate kinase by pyruvate (D) Providing substrate for glyceraldehyde-3-phosphate dehydrogenase (E) Inhibiting phosphofructokinase-1 by AMP

(A) There is a mutation in cis with the operon Constitutive synthesis can occur by either of the two mechanisms. The fi rst is an inability to synthe-size lac repressor; the second is to have a mutation in the operator region that renders repressor binding impos-sible (an oc mutation; see the fi gure below). An inability to synthesize lac repressor can be repaired in trans; if the partial diploid contains a functional lac repressor gene, functional protein will be synthesized from the gene, which can bind to the chromosomal operator region, and regulate lac gene expression. If, however, an oc mutation occurred, the operator region is in cis with the operon and can only regulate regions of DNA that are adjacent to the operator. Thus, if a partial diploid contains a normal operator region on the extrachro-mosomal region of DNA, that operator region cannot regulate the operon on the chromosomal DNA. Thus, the constitutive mutant that was not rescued in a par-tial diploid is most likely an oc mutation. If the inducer could no longer bind to the repressor, then no expres-sion would occur, as the repressor would not leave the operator region. In addition, if this were the case, then adding normal repressor to the cell (via the partial dip-loid) should allow expression of the operon. Similarly, if inducer bound too tightly to the repressor, the introduc-tion of normal repressor should reverse the effects of the mutated repressor. The lac repressor does not contain a

1. While studying the lac operon in bacteria, a scientist isolates mutants of Escherichia coli, which always express the genes of the lac operon (constitutive synthesis). The scientist creates partial diploids of the regulatory elements of the lac operon in these mutants of E. coli. In one partial diploid, expression of the lac operon is still constitutive (synthesis of the genes is observed even in the absence of an inducer). A likely explanation for this result is which of the following? (A) There is a mutation in cis with the operon (B) There is a mutation in trans with the operon (C) Inducer can no longer bind to the repressor (D) Inducer binds too tightly to the repressor (E) The transactivation domain of the repressor is mutated

(A) Prostaglandin synthesis Eicosanoids are potent regulators of cellular function. They are derived from arachidonic acid and are metabolized by three pathways: the cyclooxygenase pathway (prostaglandins and thromboxanes), lipoxygenase pathway (leukotrienes), and the cytochrome P450 pathway (epoxides) (see the fi gure below). Nonsteroidal anti-infl ammatory drugs (NSAIDs) do not block arachidonic acid release rom the membrane (which would block all eicosanoid synthesis); however, they do interfere with the cyclooxygenase pathway. Prostaglandins affect infl ammation, thromboxanes affect formation of blood clots, and leukotrienes affect bronchoconstriction and bronchodilatation. NSAIDs block prostaglandins as one of their anti-infl ammatory mechanisms. Thus, while NSAIDS will block both prostaglandin and thromboxane synthesis, it is the blockage of prostaglandin synthesis which will block the infl ammatory symptoms.

1. You prescribe ibuprofen to help reduce your patient's infl ammation. Which of the following pathways is blocked as an anti-infl ammatory mechanism of action of nonsteroidal anti-infl ammatory drugs? (A) Prostaglandin synthesis (B) Thromboxane synthesis (C) Leukotriene synthesis (D) All eicosanoid synthesis (E) Arachidonic acid release from the membrane

(A) Pancreatic glucokinase The boy has developed MODY (maturity onset diabetes of the young), and one variant of MODY is a mutated glucoki-nase (an inheritable disorder) such that the Km for glu-cose has increased, and insulin release only occurs when hyperglycemia is present. Both an increase in ATP and NADPH are required for the pancreatic β-cell to release insulin. When pancreatic glucokinase has an increased Km for glucose, ATP levels can only increase at greater than normal levels of glucose. Thus, moderate hypergly-cemia is not suffi cient to induce insulin release. As insulin release occurs from the pancreas, liver, muscle, or intes-tinal hexokinase will not affect the process. The pancreas does not express hexokinase, only glucokinase. MODY is a monogenetic autosomal dominant disease of insulin secretion. There are at least six amino acid substitutions known in a number of different proteins. MODY1 is a mutation in the transcription factor HNF4-α:∼ MODY2 is a mutation in pancreatic glucokinase. MODY3 is a mutation in the transcription factor HNF1-α while MODY4 contains a mutation in insulin promoter factor 1. MODY5 is a mutation in another transcription factor, HNF1-β. MODY6 is a mutation in neurogenic differen-tiation factor 1. MODY is not insulin resistance. There-fore, all the other aspects of insulin resistance syndrome are not present (obesity, hypertension, and hypertrig-lyceridemia). Since MODY is autosomal dominant, it can be traced through the family tree. It was thought at one time that the patient had to be young to present with this disorder, but patients up to age 50 have been reported. It is not type 1 diabetes mellitus as no islet cell antibodies are present. Glucokinase is acting as a glucose sensor for the β-cell. A mutated, less sensitive sensor leads to mildly elevated blood glucose levels.

10. A 28-year-old male develops diabetes, as noted by con-stant, mildly elevated hyperglycemia. His father had similar symptoms at the same age as did his paternal grandmother. This patient is not obese, does not have hypertension, does not have dyslipidemia, and does not have antibodies directed against islet cells. This altera-tion in glucose homeostasis may be due to a mutation in which of the following enzymes? (A) Pancreatic glucokinase (B) Pancreatic hexokinase (C) Liver glucokinase (D) Muscle hexokinase (E) Intestinal glucokinase

(D) 22.5 moles of ATP per mole of citrate The following steps (see the fi gure on page 95) are required for the complete oxidation of citrate to car-bon dioxide and water. First, citrate goes to isocitrate, which goes to α-ketoglutarate (this last step generates carbon dioxide and NADH, which can give rise to 2.5 ATP). The α-ketoglutarate is further oxidized to suc-cinyl-CoA, plus carbon dioxide and NADH (this is the second carbon released as CO2, and another 2.5 ATP). Succinyl-CoA is converted to succinate, generating a GTP (at this point, fi ve high-energy bonds have been created, plus two carbons lost as carbon dioxide). Succinate goes to fumarate, with the generation of FADH2 (another 1.5 ATP), fumarate is converted to malate, and malate leaves the mitochondria (via the malate/aspartate shuttle) for further reactions. Once in the cytoplasm, the malate is oxidized to oxaloace-tate, generating NADH (another 2.5 ATP if the malate/aspartate shuttle is used). At this point, citrate has been converted to cytoplasmic oxaloacetate, with the generation of ten high-energy bonds and the loss of two carbons as carbon dioxide. The oxaloacetate is then converted to phosphoenolpyruvate and carbon dioxide at the expense of a high-energy bond (GTP, the phosphoenolpyruvate carboxykinase reaction). The high-energy bond is recovered in the next step, however, as PEP is converted to pyruvate, generating an ATP. Thus, at this point in our conversion, citrate has gone to pyruvate, plus three CO2, with a net yield of ten ATP (or high-energy bonds). The pyruvate re-enters the mitochondria and is oxidized to acetyl-CoA and carbon dioxide, also generating NADH (another 2.5 ATP). When this acetyl-CoA goes around the TCA cycle, two carbon dioxide molecules are produced, along with another ten high-energy bonds. The net total is therefore six carbon dioxide molecules and 22.5 high energy bonds for the complete oxidation of citrate.

10. A crazed friend of yours has gone on an orange juice, fi sh, and vitamin pill diet. He tells you that the citric acid, since it is a component of the TCA cycle, is always recy-cled and does not count toward his caloric total each day. You disagree, and inform him that citrate can, in addition to having its carbons stored as glycogen or fat for later use, produce energy for his daily metabolic needs. The energy yield for the complete oxidation of citrate to six carbon dioxides and water is which of the following? (A) 15.0 moles of ATP per mole of citrate (B) 17.5 moles of ATP per mole of citrate (C) 20.0 moles of ATP per mole of citrate (D) 22.5 moles of ATP per mole of citrate (E) 25.0 moles of ATP per mole of citrate

(D) Hydrogen peroxide In the absence of glutathione, the enzyme glutathione peroxidase will be less active due to the lowered concentration of glutathione. Glutathione peroxidase catalyzes the oxidation of two reduced glutathione molecules by hydrogen peroxide, generating oxidized glutathione and two molecules of water. As glutathione peroxidase is one mechanism whereby hydrogen peroxide levels are reduced, hydrogen peroxide would be expected to accumulate, and can then lead to radical damage of membrane proteins and lipids. Glutathione peroxidase does not require, or react with, superoxide, nitrogen dioxide, nitrous oxide, and peroxynitrate. It is possible that under these conditions, superoxide would also accumulate, due to the increase in concentration of one of the reaction products of superoxide dismutase, hydrogen peroxide. However, there is no evidence that hydrogen peroxide accumulation will inhibit the reac-tion catalyzed by superoxide dismutase.

10. A researcher has generated a cell line in which the γ-glutamyl cycle is defective, and glutathione cannot be synthesized. Which radical species might you initially expect to accumulate in this cell? (A) Superoxide (B) Nitrogen dioxide (C) Nitrous oxide (D) Hydrogen peroxide (E) Peroxynitrate

(A) Muscle glycogen phosphorylase The patient is lacking muscle glycogen phosphorylase and cannot utilize muscle glycogen for energy. This is another glycogen storage disease, type V, McArdle disease. The lack of muscle glycogen phosphorylase is why lactate production during exercise is very low. As shown in the fi gure below, there are many glycogen particles present in the muscle cells just below the sar-colemma, as the glycogen is not able to be degraded. Muscle damage also results from vigorous exercise, releasing myoglobin into the circulation, which is what leads to the reddish-brown urine after exercise. Altera-tions in liver enzymes (phosphorylase or PFK-1) would not affect exercise tolerance in the muscle. Muscle does not contain glucose-6-phosphatase, and this problem is not due to a lack of muscle GLUT4 transporters, as the muscle cannot utilize stored, internal glucose supplies.

10. An 18-year-old man visits the doctor due to exercise intolerance. His muscles become stiff or weak during exercise, and he sometimes cramps up. At times, his urine appears reddish-brown after exercise. An ischemic forearm exercise test indicates very low lactate produc-tion. A potential enzyme defect in this man is which of the following? (A) Muscle glycogen phosphorylase (B) Liver glycogen phosphorylase (C) Liver PFK-1 (D) Muscle glucose-6-phosphatase (E) Muscle GLUT4 transporters

(E) Acetyl-CoA carboxylase 2 An inactivating mutation in acetyl-CoA carboxylase would lead to an inability to produce malonyl-CoA, which regulates fatty acid oxidation through an inhibition of carnitine acyl transferase 1. As malonyl-CoA levels increase, fatty acid oxidation is reduced, and as the levels decrease, fatty acid oxidation will increase. If malonyl-CoA decarboxylase were inactivated, malonyl-CoA levels would remain elevated, and fatty acid oxidation would be inhibited. Inactivating mutations in either carnitine acyltransferase 1 or 2 would lead to an inability to oxidize fatty acids, as they would not enter the mitochondria. A defect in medium-chain acyl-CoA dehydrogenase (MCAD) would also result in reduced fatty acid oxidation, as the initial step of the oxidation spiral would be inhibited once the fatty acid had been reduced to about 10 carbons in length. The reactions catalyzed by malonyl-CoA decarboxylase and acetyl-CoA carboxylase are shown below.

10. An individual contains an inactivating mutation in a particular muscle protein, which leads to weight loss due to unregulated muscle fatty acid oxidation. Such an inactivated protein could be which of the following? (A) Malonyl-CoA decarboxylase (B) Carnitine acyl transferase I (C) Carnitine acyl transferase II (D) Medium chain acyl-CoA dehydrogenase (E) Acetyl-CoA carboxylase 2

(B) Increased opportunity for hydrogen bonding to a transacting factor A single nucleotide substitution in a promoter-proximal region has the capability of increasing hydrogen bonding to a transact-ing factor, thereby increasing the interaction between the factor and DNA and enhancing recruitment of RNA polymerase to the promoter. Since the mutation is in the promoter region, it is not involved in splicing (which occurs at exon-intron borders), it does not code for an amino acid (since exons are not in the promoter region), it is not related to sigma (which is prokaryotic specifi c), and the DNA helix does not have to be signifi -cantly melted in this area to allow transcription factor binding.

10. Theoretically, a disease could result from an increased expression of a particular gene. This can occur in eukary-otes through a single-nucleotide mutation in a promoter-proximal element. This is best explained by which one of the following? (A) More effi cient splice site recognition (B) Increased opportunity for hydrogen bonding to a transacting factor (C) Benefi cial amino acid replacement derived from the missense mutation (D) Increased amount of sigma factor binding (E) Reduced energy need to melt the DNA helix at this position

(C) Reduced glutathione The patient has glucose-6-phosphate dehydrogenase defi ciency, and his red blood cells cannot convert oxidized glutathi-one to reduced glutathione due to a lack of NADPH. Fava beans contain a potent oxidizing agent that will, in some patients (but not all), lead to hemolytic anemia in individuals with glucose-6-phosphate dehydrogenase defi ciency; in individuals with a normal G6PDH, the oxidizing agent is handled by glutathione. The red blood cells, under these conditions, do not have a problem in regenerating NADH, NAD+, or ATP.

11. A 25-year-old African American male, in good health, had read about fava beans in "Silence of the Lambs." For dinner one night, the man had liver with fava beans and a nice Chianti. About 8 h after eating the beans, the man was tired and weak. Blood work showed hemolytic anemia. This patient most likely has a defect in regener-ating which of the following? (A) NADH (B) NAD+ (C) Reduced glutathione (D) Oxidized glutathione (E) ATP

(C) Galactokinase The child has non-classical galactosemia, a defect in galactokinase. With this disorder, galactose cannot be accumulated within cells, and so it accumulates in the blood, spilling over to the urine. Because of its high level, the galactose can enter the eye and be reduced to galactitol by aldose reductase, trapping the galactitol within the eye. As galactitol accumulates, an osmotic imbalance is created, leading to cataract formation. However, since galactose-1-phosphate is not accumulating (as occurs in classical galactosemia, a defect in galactose-1-phosphate uridylyl transferase), the other effects seen with classical galac-tosemia (hypoglycemia and neurological defi cit) do not occur. The sugar that is accumulating in the urine is galactose, which contains an aldehyde, which gener-ates a positive response in a reducing test. A defect in fructokinase leads to fructosuria, a benign condition (fructose is not a substrate for aldose reductase, as it is a ketose and not an aldose). A defect in hexokinase would lead to elevated glucose levels, and can lead to sorbitol production in the lens of the eye, but the urine reducing sugar test was negative for glucose. A defect in aldolase would lead to the intracellular accumulation of metabolites, but not a great increase in circulating galactose. Refer to the fi gure in the answer to question 3 of this chapter for the pathway of galactose metabolism and the enzyme defects in both classical and nonclassi-cal galactosemia.

11. A 3-month-old girl with developing cataracts is shown to contain a reducing sugar in her urine, but the glu-cose oxidase test was negative. She has had no problems eating, and her growth curve is at the 60th percentile. Fasting blood glucose tests show normal levels of circu-lating glucose. A likely enzyme defi ciency is which of the following? (A) Fructokinase (B) Hexokinase (C) Galactokinase (D) Galactose-1-phosphate uridylyltransferase (E) Aldolase

(C) A gene encoding a transcriptional repressor has been mutated For the sake of this answer, let us assume that the overexpressed gene (named A) is located on chromosome X and the mutated gene (named B) is located on chromosome Y. The gene on chromosome Y is producing a transcriptional repressor that binds to the promoter region of the gene A on chromosome X. When the repressor is expressed, gene A transcription is reduced. The protein product of gene B is acting in trans in regulating gene A expression. A mutation that inactivates the protein product of gene B, then, would be unable to repress gene A transcription, and lead to overexpression of gene A. If the promoter for gene Ahad a TATA box mutation, one would expect reduced expression (due to an inability of the basal transcrip-tion complex to bind), rather than enhanced expres-sion. Similarly, if the locus control region of the gene is deleted, then reduced expression for gene A would be expected, not enhanced expression. If a transcrip-tional activator (gene B) suffered a missense mutation that reduced its affi nity for DNA, there would be less transcription of gene A, not enhanced expression. And, fi nally, promoter regions do not undergo splicing.

11. A human genetic condition in which too much of a gene is routinely expressed has been mapped to a locus on a different chromosome from where the gene in ques-tion is located. Which one of the following is a potential explanation for the condition? (A) The activated gene has a TATA box mutation (B) The locus control region for the gene is deleted (C) A gene encoding a transcriptional repressor has been mutated (D) A transcriptional activator sustained a missense mutation, which reduces its affi nity for DNA (E) A variant promoter region is formed owing to a splice site mutation

(D) Stimulation of glycogen synthase D Glycogen synthase D (the inactive, phosphorylated form) can be allosterically activated by glucose-6-phosphate bind-ing to the enzyme. Glucose-6-phosphate will inhibit the AMP-stimulation of muscle phosphorylase b, but does not have any allosteric effect on the other enzymes listed (PFK-1, glucose-6-phosphatase, or GLUT4 transporters) as answer choices for this problem.

11. Patients with von Gierke disease display hepatomegaly. Glycogen content in the liver is increased, relative to normal, due to which of the following effects of glucose-6-phosphate in these patients? (A) Inhibition of phosphorylase a (B) Stimulation of phosphorylase b (C) Inhibition of glycogen synthase I (D) Stimulation of glycogen synthase D (E) Inhibition of glycogen phosphorylase kinase

(C) 19 Acetoacetate will react with succinyl-CoA to produce acetoacetyl-CoA and succinate (this costs 1 GTP, as the succinate thiokinase step is skipped). The acetoacetyl-CoA is converted to two acetyl-CoA, each of which can generate 10 ATP when completely oxidized (each acetyl-CoA generates 1 GTP, 3 NADH, and 1 FADH2). The sum, then, is 20 minus the 1 lost in the CoA transferase step, for a net yield of 19 ATP

11. The net energy yield obtained (moles of ATP per mole of substrate oxidized) when acetoacetate is utilized by the nervous system as an alternative energy source is which of the following? Consider that acetoacetate must be oxidized to four molecules of carbon dioxide during the reaction sequence. (A) 17 (B) 18 (C) 19 (D) 20 (E) 21

(D) Lack of energy to the heart due to B1 deficiency The patient has thiamine defi ciency, and because of this, his heart is having trouble generating suffi cient energy to effectively pump his blood (due to a reduction in the rate of both pyruvate oxidation and TCA oxidative steps). The resultant congestive heart failure leads to edema in the lower extremities, pulmo-nary edema, and inability to participate in even mild exercise. The thiamine defi ciency has resulted from the patient's poor diet and the effect of ethanol blocking thiamine absorption from the diet. The nervous system also suffers from thiamine defi ciency, in which case, neurological signs of the defi ciency would be evident. These are not yet observed in this patient. The symp-toms observed are not due to niacin defi ciency (which are dementia, dermatitis, and diarrhea). The problem is also not due to insuffi cient energy for the kidney to appropriately fi lter the blood.

11. You have been following a patient for several years, who has recently become clinically depressed, and is eating very little and drinking alcohol very heavily. He presents to you one day with noticeable swelling of the lower legs, increased heart rate, lung congestion, and com-plaints of shortness of breath with virtually any activity. These symptoms have come about due to which of the following? (A) Lack of energy to the nervous system due to niacin defi ciency (B) Heart has trouble generating energy due to niacin defi ciency (C) Lack of energy to the nervous system due to B1 defi -ciency (D) Lack of energy to the heart due to B1 deficiency (E) Lack of TCA cycle activity in the kidneys, leading to excessive water retention

(C) Development of intimal narrowing in another artery The patient is experiencing ischemic reperfusion injury. When oxygen delivery to cardiac cells was compromised, the electron transfer chain in the mitochondria was fully reduced, as the terminal oxygen acceptor was missing. When oxygen is reintro-duced to the cell, at a high concentration, the likelihood of electron transfer from reduced coenzyme Q to oxy-gen is increased, such that the possibility of superoxide generation is increased. The superoxide produced reacts with lipids and proteins and can lead to cell death above that originally occurring from the initial heart attack. Radicals do not form during the ischemic event since oxygen is missing from the tissues. There is no effect on glycogen stores or the HMP shunt pathway under these conditions. Intimal narrowing occurs over a long time period, not over the short time period covered in this case. Injured cells do leak enzymes into the blood-stream, but these enzymes do not cause the death of other, healthy cells.

12. A 52-year-old male complained of sudden onset of left-sided chest pain radiating down his left arm. Rapid breathing, sweating, and a feeling of doom accompanied this. He was rushed to the emergency department. An angiogram revealed 90% occlusion of the left anterior descending artery (LAD) and no occlusions in any other artery. The LAD was opened by angioplasty. However, shortly after this procedure, with normal blood fl ow through the LAD, the patient's condition worsened. This was most likely due to which of the following? (A) Disruption of the HMP shunt in cardiac cells (B) Damage to healthy cells by loss of essential enzymes from cells due to membrane damage (C) Development of intimal narrowing in another artery (D) Radical-induced damage once blood fl ow was reinitiated (E) Lack of glycogen for ATP synthesis in the heart

(C) Citrate translocase Citrate translocase is required for citrate to exit the mitochondria and enter the cytoplasm in order to deliver acetyl-CoA for fatty acid biosynthesis (see the fi gure below). Acetyl-CoA, which is produced exclusively in the mitochondria, has no direct path through the inner mitochondrial membrane. However, under conditions conducive to fatty acid biosynthesis (high energy levels, and allosteric inhibition of the TCA cycle), citrate will accumulate and leave the mitochondria (see the fi gure below). Once in the cytoplasm, citrate lyase will cleave the citrate to produce acetyl-CoA and oxaloacetate. The oxaloacetate is recycled to pyruvate, producing NADPH in the process, which is also required for fatty acid biosynthesis. A defect in either carnitine acyl transferase will not affect fatty acid biosynthesis, as those enzymes are required to transport the fatty acid into the mitochondria for its oxidation. A lack of glucose-6-phosphate dehydrogenase will not interfere with fatty acid synthesis, as malic enzyme can provide suffi cient NADPH for the pathway. MCAD is involved in fatty acid oxidation and does not affect fatty acid synthesis.

12. A mouse model has been generated as an in vivo system for studying fatty acid synthesis. An inactivating mutation was created which led to the cessation of fatty acid synthesis and death to the mice. This mutation is most likely in which of the following proteins? (A) Carnitine acyl transferase I (B) Carnitine acyl transferase II (C) Citrate translocase (D) Glucose-6-phosphate dehydrogenase (E) Medium chain acyl-CoA dehydrogenase

(D) Blocking the dephosphorylation of specifi c tran-scription factors Cyclosporin A binds to the protein cyclophilin in immunocompetent lymphocytes, and this protein complex leads to the inactivation of cal-cineurin. Calcineurin, in response to increases in cyto-plasmic calcium, will activate its phosphatase activity and dephosphorylate cytoplasmic nuclear factor of activated T-lymphocytes (NF-AT), a transcription factor. When dephosphorylated NF-AT will translocate to the nucleus, it will interact with nuclear factors and bind to DNA, initiating new gene transcription. When calcineurin is inactivated via cyclosporin-cyclophilin binding, NF-AT cannot translocate to the nucleus, and cytokine synthesis by the cell (second messengers) is compromised. Immu-nocompetent cells have very low levels of calcineurin, which make them susceptible to cyclosporin treatment. Drug treatment does not directly affect the translation of cytokine genes (it is indirect because the mRNA for these genes is never made), nor does it lead to the activation of transcription of cytokine receptors. Cyclosporin does not lead to the phosphorylation of transcription factors, nor does it stimulate translation of cytokine genes.

12. A patient is taking cyclosporin A after receiving a kidney transplant. Cyclosporin A protects against organ rejec-tion by which of the following mechanisms? (A) Blocking translation of cytokine genes (B) Activating transcription of cytokine receptors (C) Stimulating the phosphorylation of transcription factors (D) Blocking the dephosphorylation of specifi c tran-scription factors (E) Stimulating translation of cytokine genes

(E) Biotin The child has biotinidase defi -ciency, which results in a functional biotin defi ciency. Biotinidase is required to remove covalently linked biotin from proteins in our diet and from proteins that have turned over within the body. An inability to do this leads to a biotin defi ciency (as most ingested biotin is linked to proteins). The hair and scalp problems have been attributed to an inability to synthesize fatty acids (as acetyl-CoA carboxylase is missing biotin). Since pyruvate carboxylase is also inoperative (due to the lack of biotin), gluconeogenesis is impaired, and ketone bodies will be synthesized by the liver to compensate for reduced glucose production. Priopionyl-CoA carboxy-lase is also impaired, leading to the elevated levels of pro-pionic acid. Since gluconeogenesis is impaired, excess pyruvate will be converted to lactate since it cannot be converted to oxaloacetate. The optic atrophy may be due to an inability to synthesize fatty acids within the neurons or a lack of energy due to reduced gluconeogenesis.

12. An 8-month-old girl was taken to the emergency department due to the onset of sudden seizures. The child had brittle hair, with some bald spots, and skin rashes. An ophthalmologist noted optic atrophy. Urinal-ysis showed slightly elevated ketones and the presence of other organic acids (such as propionate and lactate). Treatment of this child with which of the following can successfully alleviate the problems? (A) Thiamine (B) Niacin (C) Ribofl avin (D) Carnitine (E) Biotin

(B) Lactate inhibition of kidney tubule absorption of urate Patients with von Gierke disease display elevated levels of lactate, which interferes with the kidney's ability to remove uric acid from the blood and place it in the urine. This leads to hyperuricemia. The reason lactate levels are elevated is that the high glucose-6-phosphate in the cell (recall, the defect in this disorder is a lack of glucose-6-phosphatase activity) forces glyco-lysis forward, producing pyruvate, which is converted to lactate in order to regenerate NAD+ to allow glycolysis to continue. Glucose-6-phosphate does not inhibit glucose-6-phosphate dehydrogenase (that enzyme is regulated by the NADP+ levels), nor does it regulate a committed step of de novo purine synthesis, amidophosphoribosyl transferase (which is regulated by adenine and guanine nucleotides). Glucose-6-phosphate does stimulate glyco-gen synthase D, but that activation does not play a role in elevated urate levels. Glucose-6-phosphate does not affect urate absorption within the kidney.

12. The hyperuricemia observed in patients with von Gierke disease comes about due to which of the following? (A) Glucose-6-phosphate inhibition of kidney tubule absorption of urate (B) Lactate inhibition of kidney tubule absorption of urate (C) Glucose-6-phosphate inhibition of glucose-6- phosphate dehydrogenase activity (D) Glucose-6-phosphate stimulation of glycogen syn-thase D (E) Glucose-6-phosphate activation of amidophospho-ribosyl transferase activity

(E) Phosphoglycerate kinase

12. The synthesis of one mole of glucose from two moles of lactate requires six moles of ATP. Which one of the follow-ing steps requires ATP in the gluconeogenic pathway? (A) Pyruvate kinase (B) Triosephosphate isomerase (C) Glucose-6-phosphatase (D) Fructose-1,6-bisphosphatase (E) Phosphoglycerate kinase

c. Transcription •Acetylation occurs on lysine residues near the N-termini of histones ✮ •This modification reduces the positive charge of these tails and decreases the binding affinity of histone for the negatively charged DNA • Weakens the electrostatic attraction between histones and DNA thereby loosening chromatin structure Acetyl groups are added to specific lysines by a set of different histone actetyl transferases (HATs) and removed by a set of histone deacetylase complexes(HDACs)

13 y.o. girl is brought to the doctor because of 5 month history of behavioral problems. Diagnosis of bipolar disorder is made, and treatment is started with Valproate which inhibits histone deacetylase.This drug affect which of the following processes? Select one: a. Polyadenylation b. mRNA splicing c. Transcription d. Translation. e. Post-transcriptional processing

d. Transcription •Acetylation occurs on lysine residues near the N-termini of histones ✮ •This modification reduces the positive charge of these tails and decreases the binding affinity of histone for the negatively charged DNA • Weakens the electrostatic attraction between histones and DNA thereby loosening chromatin structure Acetyl groups are added to specific lysines by a set of different histone actetyl transferases (HATs) and removed by a set of histone deacetylase complexes(HDACs)

13 y.o. girl is brought to the doctor because of 5 month history of behavioral problems. Diagnosis of bipolar disorder is made, and treatment is started with Valproate which inhibits histone deacetylase.This drug affect which of the following processes? Select one: a. mRNA splicing b. Polyadenylation c. Post-transcriptional processing d. Transcription e. Translation.

(B) Inability to induce histone acetylation Glucocorticoids bind to a cytoplasmic recep-tor and translocate to the nucleus, where they bind to glucocorticoid response elements on the DNA, leading to a complex of proteins at this site. A number of trans-activating factors recruited to the DNA contain histone acetyl transferase (HAT) activity, which leads to further unwinding of the DNA from the nucleosome, enabling gene transcription. Patients who become resistant to glucocorticoid treatment show a reduction in histone acetylation in response to the drug for a variety of rea-sons. The resistance does not appear to be due to inabil-ity of the drug to enter target cells. Once inside the cell, the drug will bind to the receptor, and dimerization of the receptor is normal. In some resistant individu-als, the dimerized receptor has trouble translocating to the nucleus, which leads to the reduction in HAT activ-ity. In other patients, however, translocation is normal, and the reduction in HAT activity appears to be due to an inability to recruit transactivating factors to the complex, which contain the HAT activity. The levels of transactivating factors are normal, but it has been hypothesized that the receptor is phosphorylated under resistance conditions, leading to an inability to attract the transactivating factors.

13. A patient has asthma, but has become resistant to glucocorticoid inhalation. A potential mechanism for this resistance is which of the following? (A) Inability of glucocorticoids to enter target cells (B) Inability to induce histone acetylation (C) Reduction of levels of transactivating factors in the nucleus (D) Cytokine induction of protein kinases (E) Increased dimerization of the glucocorticoid recepto

(D) Pyruvate carboxylase The body's major energy source for gluconeogenesis is fatty acids, which are oxidized to acetyl-CoA, at which point acetyl-CoA enters the TCA cycle to produce ATP. Acetyl-CoA activates pyruvate carboxylase (and inhibits pyruvate dehydrogenase), a key gluconeogenic enzyme. Acetyl-CoA does not regulate any of the other enzymes listed as potential answers (PEPCK is transcriptionally regu-lated by CREB; Fructose-1,6-bisphosphatase is inhibited by fructose-2,6-bisphosphate; glucose-6-phosphatase is regulated by a regulatory protein; and pyruvate kinase has both allosteric and covalent controls in the liver, but none involve acetyl-CoA).

13. An important product of the oxidation of the body's major energy source to provide energy for gluconeogen-esis regulates which of the following key gluconeogenic enzymes? (A) PEPCK (B) Fructose-1,6-bisphosphatase (C) Glucose-6-phosphatase (D) Pyruvate carboxylase (E) Pyruvate kinase

E Under the conditions described, DNA synthesis is occurring without any requirement for NADPH (such as fatty acid synthesis). Under these conditions, NADPH levels are high and glucose-6-phosphate dehydrogenase is inactive. The cell requires ribose-5-phosphate, however, for nucleotide biosynthe-sis, and this is synthesized from fructose-6-phosphate and glyceraldehyde-3-phosphate using the nonoxida-tive reactions of the pathway. Thus, both transketolase and transaldolase will be active under these conditions. PFK-1 is active as well, as the only way to generate glyc-eraldehyde-3-phosphate from a sugar precursor is via enzymes of the glycolytic pathway.

13. Consider an intestinal epithelial cell in S phase, and for which the major, active biosynthetic pathway is nucle-otide synthesis. Which one of the following best repre-sents the activity state of a series of key enzymes under these conditions?

(A) Glycogen synthase I The glucose in the sports drink will bind to liver glycogen phosphorylase a and inhibit its activity allosterically. Once the insulin sig-nal reaches the liver, phosphorylase a will be converted to the dephosphorylated phosphorylase b by activated phosphatases. There is no allosteric inhibitor for glyco-gen synthase I, or protein phosphatase 1 (which is regu-lated by protein inhibitor 1). Adenylate kinase is not regulated allosterically, and there is no allosteric inhibi-tor of phosphorylase kinase a (the nonphosphorylated form can be activated by calcium).

13. Consider the case of an athlete who has just completed a work out. At this point, the athlete consumes a sports drink, which contains a large amount of glucose, which enters the circulation. Glycogen degradation is inhib-ited in the liver under these conditions, prior to insulin release, due to allosteric inhibition of which of the fol-lowing enzymes? (A) Glycogen synthase I (B) Phosphorylase kinase a (C) Phosphorylase a (D) Protein phosphatase 1 (E) Adenylate kinase

(D) Pyridoxal phosphate Pyridoxal phosphate is required for the transamination of aspar-tate to oxaloacetate and glutamic acid to α-ketoglutarate. Both the α-keto acids are TCA cycle components, and when their levels decrease, they can be replenished through such a reaction. Niacin, ribofl avin, and lipoate are required for oxidative decarboxylation reactions, but that reaction type does not lead to a refi lling of TCA cycle intermediates. Carnitine is required to transport acyl groups into the mitochondria and is not used to transport TCA cycle intermediates from the cytoplasm to the mitochondria. Biotin would be a correct answer (for the pyruvate carboxylase reaction, to regenerate oxalo-acetate from pyruvate), but it was not offered as a choice. A typical transamination reaction is shown below.

13. The refilling of TCA cycle intermediates is frequently dependant upon which of the following cofactors? (A) Niacin (B) Riboflavin (C) Carnitine (D) Pyridoxal phosphate (E) Lipoate

B α-oxidation leads to the oxidation of the α-carbon of a branched chain fatty acid to generate an α-keto acid, which undergoes oxidative decarboxylation. This reorients the methyl groups on the branched chain fatty acid such that they are on the α-carbon, rather than the β-carbon. In this manner, the methyl groups do not interfere with the β-oxidation spiral (if the methyl group were on the β-carbon, a carbonyl group would be unable to form on that carbon, which would block further oxidation of the fatty acid). Answer choices A, C, D, and E are eliminated as requiring α-oxidation because, after one round of normal β-oxidation, the methyl group (or butyl group) will be on the α-carbon and would not interfere with the β-oxidation spiral. An overview of α-oxidation is shown below.

13. α-oxidation would be required for the complete oxidation of which of the following fatty acids?

(A) Oxidation of very long chain fatty acids The child has Zellweger's syndrome, an absence of peroxisomal enzyme activity. Of the pathways listed as answers, only the oxidation of very long chain fatty acids is a peroxisomal function. Fatty acid synthesis occurs in the cytoplasm. Acetyl-CoA oxidation takes place in the mitochondria. Glucose oxidation is a combination of glycolysis (cytoplasm) and the TCA cycle (mitochondria). Triglyceride synthesis occurs in the cytoplasm.

14. A 2-month-old infant with failure to thrive displays hepatomegaly, high levels of iron and copper in the blood, and vision problems. This child has diffi culty in carrying out which of the following types of reactions? (A) Oxidation of very long chain fatty acids (B) Synthesis of unsaturated fatty acids (C) Oxidation of acetyl-CoA (D) Oxidation of glucose (E) Synthesis of triacylgycerol

(D) Increase in intracellular AMP AMP will activate muscle glycogen phosphorylase b allosteri-cally, allowing glycogen degradation to begin before any hormonal signal has reached the muscle. The addition of epinephrine to the muscle requires activation of adenylate cyclase to initiate glycogen degradation, and adenylate cyclase has been inactivated in this cell line. Muscle lacks glucagon receptors, so cannot respond to this hormone. An increase in intracellular calcium would lead to glycogen degradation (via activation of phos-phorylase kinase b), but magnesium does not have the same effect as calcium. Increases in ADP levels will not activate glycogen phosphorylase b; the allosteric activa-tor is specifi c for AMP. The table below summarizes the allosteric interactions involved in glycogen metabolism.

14. A muscle cell line has been developed with a nonfunc-tional adenylate cyclase gene. Glycogen degradation can be induced in this cell line via which of the following mechanisms? (A) Addition of glucagon (B) Addition of epinephrine (C) Increase in intracellular magnesium (D) Increase in intracellular AMP (E) Increase in intracellular ADP

B The conditions of the cell indicate that NADPH is required for fatty acid synthesis, but there is no need for ribose-5-phosphate, as the cells are in Go phase and are not undergoing DNA synthesis (so nucleotides are not required). The HMP shunt will utilize the oxidative reactions to generate NADPH, and then the ribulose-5-phosphate produced will use the nonoxidative reactions to regenerate glucose-6-phosphate. For this to occur, transketolase, transal-dolase, glucose-6-phosphate dehydrogenase (as the major oxidative enzyme of the pathway), and fruc-tose-1,6-bisphosphatase all have to be active. These nonoxidative reactions are indicated in the fi gure on page 125.

14. A researcher is studying cultured human hepatocytes and is examining the specifi c condition in which fatty acid synthesis is activated, but the cell remains in the Go phase of the cell cycle. Under such conditions, what would be the activity state of the following enzymes?

E Under the conditions of a 48-h fast, the liver is exporting glucose, and glycolysis will be inhibited. PFK-1 activity is reduced due to a reduction of fructose-2,6-bisphosphate levels, brought about by glucagon-induced phosphorylation of PFK-2, which activates PFK-2 phosphatase activity, which converts fructose-2, 6-bisphosphate to fructose-6-phosphate. Pyruvate kinase activity, in the liver, is also reduced by phosphorylation by protein kinase A (which is activated by glucagon). As blood glucose levels have dropped during the fast, and the liver is exporting glucose, the concentration of glucose in the hepatocyte is not suffi cient for glucoki-nase (which has a high Km) to phosphorylate glucose. Glucokinase is not regulated by phosphorylation.

14. An individual with a BMI of 34 was advised by the physician to eat less and exercise more. The patient took this advice to an extreme, and has not eaten for 48 h. Which of the following best describes the patient's activity and phosphorylation state of the following key liver enzymes?

(C) By decreasing the rate of RNA polymerase binding to the promoter Transcription factors can be either positive or negative acting. In either event, to exert an effect on transcription, the factor must bind to the DNA and then either promote RNA polymerase binding to DNA (in which case, it is a positive-acting factor) or inhibit RNA polymerase binding to DNA (in which case it is a negative-acting factor). Negative acting factors can do so by blocking the binding of positively acting factors to the DNA or by blocking necessary trans-activating factors from binding to other factors already bound to the DNA. In either event, the net result is a reduction in the rate at which RNA polymerase binds to the promoter region to initiate transcription. Once initiated, the rate of phosphodiester bond formation is constant. RNA editing is a rare event, and is not used to regulate gene transcription. If HAT activity were stimulated, gene transcription would be increased, as the acetylated histones would have a reduced affi nity for DNA, and it would be easier for RNA polymerase to bind to the promoter region. Increasing enhancer bind-ing to DNA would also increase the rate at which RNA polymerase would bind to the promoter, as enhancers increase the association of positive-acting transactiva-tion factors, which would promote RNA polymerase binding to the promoter and transcription.

14. Induction of certain transcription factors leads to a decrease in the expression of certain genes. This occurs through which of the following mechanisms? (A) Decreasing the rate of RNA polymerase catalyzed phosphodiester bond formation (B) Inducing the synthesis of a protein that posttran-scriptionally edits mRNA such that translation ini-tiation is blocked (C) By decreasing the rate of RNA polymerase binding to the promoter (D) Through stimulation of proteins with HAT activity (E) Increasing enhancer binding to DNA

(C) Succinyl-CoA Barbiturates are metabo-lized via cytochrome P450 enzymes, which are induced by their substrates. The induction of synthesis requires that heme be synthesized, and the fi rst step in heme synthesis requires succinyl-CoA and glycine and occurs within the mitochondrial matrix (see the fi gure below). Thus, succinyl-CoA levels can drop in the matrix during heme synthesis, and anaplerotic reactions are required to keep the cycle going.

14. The concentration of TCA cycle intermediates can be reduced under certain conditions. Consider a patient who initiates taking barbiturates. During the initial phase of his taking this drug, which TCA cycle interme-diate is reduced in concentration? (A) Citrate (B) α-ketoglutarate (C) Succinyl-CoA (D) Fumarate (E) Oxaloacetate

(C) To reduce thromboxane synthesis Thromboxane A2 release from platelets is an essential element of forming blood clots, and aspirin will block prostaglandin, prostacyclin, and thromboxane synthesis. It is the thromboxane inhibition which reduces the risk of blood clots. Leukotrienes and lipoxins require the enzyme lipoxygenase, which is not inhibited by aspirin. These pathways are outlined below.

15. A 55-year-old man had been advised by his physician to take 81 mg of aspirin per day to reduce the risk of blood clots leading to a heart attack. The rationale for this treatment is which of the following? (A) To reduce prostaglandin synthesis (B) To reduce leukotriene synthesis (C) To reduce thromboxane synthesis (D) To increase prostacyclin synthesis (E) To increase Lipoxin synthesis

(D) An altered glycogenin with an increased Km for UDPglucose A reduction in over-all glycogen synthesis suggests that the biosynthetic pathway is defective in some step. All glycogen molecules have, at their core, a glycogenin protein molecule, which autocatalyzes the addition of six glucose residues, using UDPglucose as the carbohy-drate donor. This structure then provides the initial primer required by glycogen synthase. If the Km for UDP glucose is increased, the rate of formation of gly-cogen primers will be decreased, as the levels of UDP-glucose may not be suffi cient to allow glycogenin to self-prime. This would result in an overall reduction of glycogen levels within the cell. If a glycogen synthase had a reduced Km for UDPglucose, then the enzyme would be active at lower UDPglucose levels, and one would expect greater than normal glycogen synthesis. Phosphorylase kinase has as its substrate phosphory-lase, not glycogen, so answer B is not correct. If the uridyl transferase had a reduced Km for a substrate, it would proceed at low substrate levels and would not give the resultant phenotype. And, if phosphory-lase kinase had an increased Km for glycogen synthase, then glycogen synthase would not be inactivated as rapidly, and glycogen synthesis would be expected to continue under conditions where it should not, lead-ing to enhanced glycogen synthesis.

15. A researcher created a liver cell line that displayed very low levels of glycogen. The glycogen that was synthesized was of normal structure, but the overall lev-els of glycogen were about 5% of normal. Which of the following is a potential alteration in the cell line that would lead to these results? (A) An altered glycogen synthase with a reduced Km for UDPglucose (B) An altered phosphorylase kinase with an increased Km for glycogen (C) An altered UTPglucose-1-phosphate uridyl trans-ferase with a decreased Km for glucose-1-phosphate (D) An altered glycogenin with an increased Km for UDPglucose (E) An altered phosphorylase kinase with an increased Km for glycogen synthase

(A) Activation of glucose-6-phosphate dehydrogenase Glutathione reductase will utilize NADPH and reduce oxidized glutathione to reduced glu-tathione, generating NADP+. If Glutathione reductase is superactive, NADP+ levels accumulate, which activates glucose-6-phosphate dehydrogenase. This will lead to NADPH production via the oxidative reactions of the HMP shunt, along with ribulose-5-phosphate (Ru5P). The Ru5P will lead to increased ribose-5-phosphate pro-duction, increased 5′-phosphoribosyl 1′-pyrophosphate (PRPP) production, and increased 5′-phosphoribosyl 1′-amine levels. This eventually leads to increased purine production, in excess of what is required. The excess purines are converted to uric acid, and excess uric acid will lead to gout. A superactive glutathione reductase will not lead to an alteration in the activities of transketolase or transaldolase.

15. Individuals with a superactive glutathione reductase will develop gout. This occurs due to which of thefollowing? (A) Activation of glucose-6-phosphate dehydrogenase (B) Inhibition of glucose-6-phosphate dehydrogenase (C) Activation of transketolase (D) Activation of transaldolase (E) Inhibition of transketolase

(A) The ability to modulate the binding of positive, or negative, transacting factors to the DNA The recognition of overlapping sequences on DNA by differ-ent factors allows either positive-acting or negative-acting effects, depending on which transcription factor is pres-ent in higher concentration. The γ-globin gene is turned off after birth, and this has to do, in part, with factors bound to the promoter. When stage selector protein (SSP) is bound, γ-globin synthesis is favored, but when SP1 binds to this area instead, γ-globin synthesis is repressed. A similar story occurs at the CAAT sites; when CP1 is bound, γ-globin synthesis is favored, but when CAAT displacement protein binds, CP1 can no longer bind, and gene transcription is inhibited. Thus, γ-globin expression is regulated by the concentrations of both positive-acting and negative-acting factors available in the cell. The over-lapping binding sites were not designed for redundancy in case of mutation, to provide ribosome binding sites (the ribosomes bind to RNA produced from exons, not from the promoter region), or to provide a target for interfering RNAs (the siRNAs are targeted to mRNA). The multiple binding sites, by themselves, do not promote looping of DNA, but once transcription factors have bound to the DNA, looping may occur as the transactivating factors bind to the proteins bound to DNA.

15. Shown below is a partial map of the promoter region, and promoter-proximal region, for the γ-globin gene. The overlapping binding sites for transcription factors allow for which of the following to occur? (A) The ability to modulate the binding of positive, or negative, transacting factors to the DNA (B) The ability to reduce the risk of losing transcrip-tional control via mutation in this region (C) Promoting looping of this DNA region (D) Providing a target for interfering RNAs (E) Providing ribosome binding sites for translation ini-tiation

A Upon insulin release, the cAMP phos-phodiesterase is activated, reducing cAMP levels in the liver, thereby leading to inactivation of protein kinase A. In addition, protein phosphatase 1 has become active and dephosphorylates the enzymes that were phosphorylated by protein kinase A. Therefore, PFK-2 is not phospho-rylated, which leads to an active kinase activity and an enzymes that remove phosphates from substrates, releasing the phosphates as inorganic phosphate. They do not require, nor generate, ATP. Pyruvate kinase is not utilized for gluconeogenesis, and triose phosphate isomerase catalyzes the conversion of dihydroxyacetone phosphate and glyceraldehyde-3-phosphate, without the involvement of a high-energy phosphate bond. These are shown in the pathway below. Glycolysis and Gluconeogenesis 87inactive phosphatase activity (choices A, D, or E). The active kinase of PFK-2 produces more fructose-2,6-bis-phosphate, leading to the activation of PFK-1 (answers A through D; combined with PFK-2 activity, now only choice A or D can be correct). Insulin stimulates pre-formed GLUT4 transporters in the muscle to fuse with the plasma membrane, thereby enhancing glucose transport into the muscle (choices A through C; combined with the other two columns, only choice A can be correct)

15. Which of the following changes in enzyme activity will occur within 1 h of a type 1 diabetic taking an injection of insulin?

(D) Iron response element binding protein s shown in the fi gure below, the iron-response element binding protein (IRE-BP) binds to loops in RNA (RNA secondary structure), and in the case of ferritin mRNA, the IRE-BP blocks translation when iron levels are low. In the case of the transferrin receptor mRNA, the IRE-BP stabilizes the mRNA when iron levels are low, allowing effi cient translation of the mRNA such that transferrin receptors can be synthe-sized and placed in the membrane. If a patient had a mutated IRE-BP, which could no longer bind to its target sequence in mRNA, then the transferrin receptor mRNA would be unstable and degraded, and cells would not be able to take up iron from the circulation, leading to higher than normal iron levels in the blood yet still would result in cellular depletion of iron. However, the lack of IRE-BP binding would allow ferritin translation and synthesis, leading to high levels of ferritin inside the cell despite the low levels of intracellular iron. Ceru-loplasmin is involved in copper transport, not iron. A mutation in transferrin would lead to lower circulat-ing iron levels in the blood as the iron carrier would be mutated. Mutations in the transferrin receptor would not lead to high intracellular levels of ferritin (intracel-lular iron levels would be low, the IRE-BP would remain bound to the ferritin mRNA, and translation would be blocked). Mutations in transcobalamin would affect vitamin B12 transport, not iron.

16. A hypothetical patient was suffering from excessive free iron in the blood, yet a cellular analysis indicated low intracellular levels of iron, despite high intracellular lev-els of ferritin, and normal transferrin levels in the blood. The disease was shown to be caused by a single base change in the DNA that led to a dysfunctional protein. The mutation is likely to be in which of the following proteins? (A) Transferrin (B) Transferrin receptor (C) Transcobalamin (D) Iron response element binding protein (E) Ceruloplasmin

(E) 5 Given the enzymes present, only the nonoxidative reactions of the HMP shunt would take place. In order for the nonoxidative reactions to occur, the glucose-6-phosphate (G6P, labeled in the 6th position with 14C) must pass through glycolysis to pro-duce fructose-6-phosphate (F6P, labeled in the 6th posi-tion) and glyceraldehyde-3-phosphate (labeled in the 3rd position). Transketolase will allow these two com-pounds to exchange carbons, which would generate erythrose-4-phosphate (E4P, labeled in the 4th position) and xylulose-5-phosphate (X5P, labeled in the 5th posi-tion). The X5P can then go to ribulose-5-phosphate (Ru5P) and ribose-5-phosphate (R5P), labeled in the fi fth positions. The E4P (labeled in the 4th position) can react with another molecule of F6P (labeled in the 6th position) using transaldolase to generate sedoheptulose 7-phosphate (Se7P, labeled in the 7th position) and glyceraldehyde-3-phosphate (G3P), labeled in the 3rd position. Transketolase will then convert the Se7P and G3P to R5P and X5P, both labeled in the 5th positions. The nonoxidative reactions can be seen, schematically, in the fi gure below.

16. Glucose-6-phosphate labeled in carbon 6 with 14C was added to a test tube with the enzymes phosphohexose isomerase, PFK-1, aldolase, transketolase, and transal-dolase. ATP was also added to the test tube. At equilib-rium, in which position would the radioactive label be found in the newly produced ribose-5-phosphate? (A) 1 (B) 2 (C) 3 (D) 4 (E) 5

(D) 2-phosphoglycerate Fluoride inhibits the glycolytic enzyme enolase, which catalyzes the dehy-dration of 2-phosphoglycerate to phosphoenolpyruvate. Thus, 2-phosphoglycerate accumulates under these conditions.

16. Streptococcus mutans, found in dental plaque, produces acids from the metabolism of carbohydrates. Topical fl uoride treatment in the dental offi ce can slow the pro-duction of acids, resulting in the accumulation of which metabolite? (A) Glucose-6-phosphate (B) Fructose-1,6-bisphosphate (C) Glyceraldehyde-3-phosphate (D) 2-phosphoglycerate (E) Phosphoenolpyruvate

E

16. Ten hours into a fast, in a normal individual, which of the following best represents the activity and phospho-rylation state of a number of key enzymes within the liver?

(D) Acyl-carnitine With a carnitine defi - ciency, fatty acids cannot be added to carnitine, and acyl-carnitine would not be synthesized. With a carnitine acyl-transferase 2 defi ciency, the fatty acids are added to carnitine, but the acyl-carnitine cannot release the acyl group within the mitochondria. This will lead to an accumulation of acylcarnitine, which will lead to an accumulation in the circulation. The end result of either defi ciency is a lack of fatty acid oxidation, such that ketone body levels would be minimal under both conditions, and blood glucose levels would also be similar in either condition. Insulin release is not affected by either defi ciency, and carnitine levels, normally low, would not be signifi cantly modifi ed in either defi ciency.

16. You are examining a patient who exhibits fasting hypoglycemia and need to decide between a carnitine defi ciency and a carnitine acyltransferase 2 defi ciency as the possible cause. You order a blood test to specifi cally examine the levels of which one of the following? (A) Glucose (B) Ketone bodies (C) Insulin (D) Acyl-carnitine (E) Carnitine

(C) Decrease in hydrogen bonding between the tran-scription factor and DNA The necessary tran-scription factor is not binding as well to the DNA, leading to reduced effi ciency of transcription. The side chain of aspartic acid can participate in hydrogen bond-ing with DNA bases, but the side chain of valine (com-pletely carbon-hydrogen bonds) cannot (see the fi gure below). Transcription factors do not bind to DNA via ionic interactions. An increase in hydrogen bonding between the factor and DNA would lead to enhanced transcription (not reduced transcription). An inability of the transcription factor to bind to DNA would lead to no transcription of the gene; since there is a low level of expression, binding of this factor to the DNA must be occurring, although the affi nity for the factor and DNA has been reduced.

17. A cell is producing a certain transcription factor that contains a single point mutation, such that N is con-verted to V. Circular dichroism experiments show that this altered factor has the same secondary structure as the nonmutated factor. Under normal conditions, serum stimulation of quiescent cultures shows strong induction of fi ve genes. When quiescent cells harbor-ing the mutant transcription factor are exposed to serum, the level of expression of those fi ve genes stays at basal levels (there is no increase in mRNA produc-tion). This fi nding is most likely due to which of the following? (A) Loss of ionic interactions between the transcription factor and DNA (B) Increase in hydrogen bonding between transcrip-tion factor and DNA (C) Decrease in hydrogen bonding between the tran-scription factor and DNA (D) Increase in ionic interactions between the transcrip-tion factors and DNA (E) Inability of the transcription factor to bind to the DNA

(B) Phosphoglucomutase For this woman to synthesize lactose, she needs to synthesize the pre-cursors UDPgalactose and glucose, both of which are available from glucose. Glucose is converted to glucose-6-phosphate by hexokinase in the breast, and then phosphoglucomutase will convert this to glucose-1-phosphate (G1P). The G1P will react with UTP in the glucose-1-phosphate uridyl transferase reaction, producing UDPglucose. The C4 epimerase will then produce UDPgalactose from UDPglucose. The UDPgalactose then condenses with free glucose (using lactose synthase) to produce lactose and UDP. The other enzymes listed as answers are not required to produce lactose from the single precursor glucose. Fructokinase is unique for fructose metabolism. Aldolase is a glyco-lytic enzyme, which is defi cient in hereditary fructose intolerance. Phosphohexose isomerase coverts glucose-6-phosphate to fructose-6-phosphate, which is not required for lactose synthesis. Classical galactosemia (severe, type 1) is a defi cit of galactose-1-phosphate uridyl transferase. Patients cannot metabolize galactose, and the accumulating galactose-1-phosphate interferes with glycogen degradation. Nonclassical galactosemia (type 2) is a defi cit in galactokinase, such that galactose cannot be phosphorylated. The complications in type 1 galactosemia due to the accumulation of galactose-1-phosphate are not seen in type 2 galactosemia. In either case, the missing enzymes are not required for the syn-thesis of lactose. See the fi gure below for both the path-way of lactose synthesis, and the defects in classical and nonclassical galactosemia.

17. A woman with nonclassical galactosemia is consider-ing becoming pregnant and is concerned that she will be unable to synthesize lactose in order to breast-feed her child. Her physician, who recalls her biochemistry, tells her this should not be a problem, and that she will be able to synthesize lactose at the appropriate time. This is true due to the presence of which of the following? (A) Galactose-1-phosphate uridyl transferase (B) Phosphoglucomutase (C) Fructokinase (D) Aldolase (E) Phosphohexose isomerase

(E) Na+, K+ ATPase

17. After eating a meal containing carbohydrates, the mono-saccharides must be absorbed from the intestinal lumen. This transport is dependent on which of the following enzymes? (A) Na+/H+ antiporter (B) Glucose-6-phosphate dehydrogenase (C) Hexokinase (D) Chloride transporter (E) Na+, K+ ATPase

(C) COX-2 is specifi cally induced during infl ammation COX-2 is induced during infl ammatory conditions, while COX-1 is constitutively expressed. Thus, when an injury occurs, and an immune response is mounted at the site of injury, COX-2 is induced in those cells to produce second messengers that play a role in mediating the pain response. Specifi cally inhibiting the COX-2 isozyme will block the production of those second messengers, without affecting the normal function of COX-1. Inhibiting COX-1 may reduce the frequency of heart attacks, and inhibiting COX-2 will block prostaglandin production via the cylco-oxygenase. Recent data suggests that certain drugs that specifi cally block COX-2 have unwanted side effects, such as an increase in heart attacks.

17. Inhibitors specifi c for cyclooxygenase 2 (COX-2) were deemed more effi cacious for certain conditions than inhibitors which blocked both COX-1 and COX-2 activities. This is due to which of the following? (A) Inhibiting COX-1 increased the frequency of heart attacks (B) Inhibiting COX-2 did not alter prostaglandin production (C) COX-2 is specifi cally induced during infl ammation (D) Specifi cally inhibiting COX-2 reduces the rate of heart attacks (E) COX-1 is inducible and only expressed during wound repair, while COX-2 is expressed constitutively

(B) Glucose-6-phosphatase deficiency The HMP shunt can have increased activity under two conditions, one being an increase in the cofactor NADP+levels and the other being an increase in the substrate levels (glucose-6-phosphate). The only enzyme listed, which when defective would lead to an increase in either glucose-6-phosphate or NADPH, is glucose-6-phos-phatase. A defi ciency in glycogen phosphorylase would not produce glucose-1-phosphate; thus, there would not be an increase in the HMP shunt under these conditions. A defi ciency in fructose-1,6-bisphosphatase defi ciency would impair gluconeogenesis and would not lead to the synthesis of glucose-6-phosphate. Defi ciencies in either pyruvate carboxylase or pyruvate dehydrogenase would lead to pyruvate accumulation and NAD+ accumulation, but not NADP+ or glucose-6-phosphate accumulation.

17. Which one of the following disorders would lead to increased activity of the HMP shunt pathway? (A) Glycogen phosphorylase defi ciency (B) Glucose-6-phosphatase deficiency (C) Fructose-1,6-bisphosphatase defi ciency (D) Pyruvate kinase defi ciency (E) Pyruvate dehydrogenase defi ciency

(B) Inhibition of a detoxifying cytochrome P450 system The patient is suffering from the potential side effects of statins, namely, muscle damage and pain. This may occur due to an inhibition of coenzyme Q synthesis (which requires a product derived from mevalonic acid) and a lack of energy generation in the muscle. The reason this comes about is that statins are detoxifi ed through a cytochrome P450 system, and the particular system that works on statins is inhibited by grapefruit juice. Thus, in the presence of grapefruit juice, the effective intracellular levels of statins are higher than in the absence of the juice, due to the decreased rate of destruction. The artifi cially induced higher levels of statins then lead to muscle damage. Statins inhibit the conversion of HMG-CoA to mevalonic acid (catalyzed by HMG-CoA reductase). Thus, answers indicating that mevalonic acid levels are increasing cannot be correct; they are reduced in the presence of a statin. Statins do not bring aboutcalcium effl ux from the sarcoplasmic reticulum.

18. A 45-year-old man was diagnosed with hypercholes-terolemia, for which he was prescribed a statin. After a month on medications, the patient decided to adopt a healthier life style and replaced eggs and coffee for breakfast with fruit juices and whole-wheat toast. Within 2 weeks of changing his diet, the man developed severe muscle pain in his arms and shoulders. The muscle pain could be the result of which of the following? (A) Induction of a detoxifying cytochrome P450 system (B) Inhibition of a detoxifying cytochrome P450 system (C) Increased mevalonate inhibiting actin/myosin inter-actions (D) Increased mevalonate stabilizing actin/myosin inter-actions (E) HMG-CoA stimulation of calcium effl ux from the sarcoplasmic reticulum

(E) Reduced ability to form malonyl-CoA Biotinidase is required to remove covalentlybound biotin from proteins, which is how most of the biotin in our diet is received. In the absence of biotinidase, individuals can become functionally biotindefi cient, due to the lack of free biotin in the body (as compared to being covalently bound to proteins). The formation of malonyl-CoA, via acetyl-CoA carboxylase, requires biotin as a required cofactor (see the fi gure below). Citrate lyase, malic enzyme, acetyl transacylase (an activity of fatty acid synthase) and acyl carrier protein (another component of fatty acid synthase) do not require biotin for their activity

18. An individual with a biotinidase defi ciency was shown to produce fatty acids at a greatly reduced rate (in the absence of supplements) as compared to someone who did not have the defi ciency. This is due to which of the following? (A) Low activity of citrate lyase (B) Reduced activity of malic enzyme (C) Reduced activity of acetyl transacylase (D) Defective acyl carrier protein (E) Reduced ability to form malonyl-CoA

B When region B is deleted from the promoter region, there is an increase in overall expression of the reporter gene (the second and last lines of the fi gure indicate this). Removal of any other region leads to a decrease in expression, indicating that positive- acting transcription factors bind to those regions of DNA.

18. Indicate which region (as designated by the letters A, B, C, D, and E) binds an inhibitory transcription factor.

(A) Enhanced production of fructose-2,6-bisphosphate When phosphorylated, heart PFK-2 is activated to produce more fructose-2,6-bisphosphate to stimulate heart PFK-1 and to increase the glycolytic rate of the heart. Phosphorylation of heart PFK-2 can be accomplished through the AMP-activated protein kinase (when the heart is having trouble generating energy) or in response to insulin (indicating that high levels of glucose are available for use). Phosphorylation of heart PFK-2 does not affect its transcription or turnover rate, and also does not affect the degradation of fructose-1,6-bisphosphate.

18. Skeletal muscle PFK-2 is not regulated by phospho-rylation, but heart muscle PFK-2 is. In the heart, phosphorylation of PFK-2 leads to what effect? (A) Enhanced production of fructose-2,6-bisphosphate (B) Reduced production of fructose-2,6-bisphosphate (C) Degradation of fructose-1,6-bisphosphate (D) Increased turnover of PFK-2 (E) Increased transcription of PFK-2

(A) One high-energy bond For a mole-cule of glucose-6-phosphate (G6P) to be incorporated into glycogen, the following pathway must be utilized: G6P is converted to glucose-1-phosphate (G1P) via phosphoglucomutase, the G1P reacts with UTP to form UDPglucose via glucose-1-phosphate uridyl transferase, releasing pyrophosphate. The resultant pyrophosphate is hydrolyzed to two inorganic phosphates, with the loss of one high-energy bond. The UDPglucose then reacts with glycogen to produce a glycogen chain with one additional sugar, and UDP is released. The overall equa-tion for these steps is: G6P + UTP + glycogenn yields UDP + 2Pi + (glycogen)n+1. These steps are outlined below:Gluose-6-phosphate → Glucose-1-phosphateGlucose-1-phosphate + UTP → UDPglucose + PPiPPi + H2O → 2 PiUDPglucose + glycogenn→ Glycogenn+1+ UDPUDP + ATP → UTP + ADPSum: Glucose-6-phosphate + ATP+ glycogenn+ H2O → glycogenn+1+ ADP + 2Pi

18. The energy required to store one molecule of glucose-6-phosphate as a portion of glycogen is which of the following? (A) One high-energy bond (B) Two high-energy bonds (C) Three high-energy bonds (D) Four high-energy bonds (E) No high-energy bonds

(E) Glutathione Glutathione is the major antioxidant in the fl uid lining the bronchial epithelium. It is essential for recovery of these tissues. Depletion of glutathione in the airway is thought to greatly increase a person's susceptibility to upper respiratory infec-tions such as infl uenza. Glutathione is formed in the γ-glutamyl pathway, and oxidized glutathione is regen-erated to reduced glutathione using NADPH produced by the HMP shunt pathway. None of the other answers (glycogen, sorbitol, pyruvate, and glucuronate) offer protection against oxidative damage. Glycogen is utilized for the storage of glucose. Pyruvate is the end product of glycolysis. Sorbitol is a product of the polyol pathway. Glucuronic acid is used for xenobiotic metabolism, in general, to increase the solubility of the xenobiotic and to prepare it for excretion.

19. A patient is recovering from acute respiratory distress syndrome (ARDS). Which of the following is a major antioxidant found in the fl uid lining the bronchial epithelium needed in high concentration for recovery from ARDS? (A) Glucuronic acid (B) Pyruvate (C) Sorbitol (D) Glycogen (E) Glutathione

(A) α-ketoglutarate dehydrogenase In order for citrate to be converted to glycogen, the citrate must fi rst be converted to oxaloacetate in the TCA cycle (which requires the participation of α-ketoglutarate dehydrogenase). From oxaloacetate, PEP carboxyki-nase will convert this to PEP, which will go through the gluconeogenic pathway up to glucose-6-phosphate. From there, G1P is produced, then UDPglucose, and fi nally incorporation of the glucose into glycogen. Pyru-vate carboxylase, while being a gluconeogenic enzyme, converts pyruvate to OAA, which is not required in this series of reactions. PFK-1 and pyruvate kinase are irreversible enzymes of glycolysis and are not used in the gluconeogenic pathway. Glucose-6-phosphatase removes the phosphate from G6P, which is not required when glycogen is being synthesized. See the fi gure above for the pathways.

19. An individual has been eating a large number of oranges during the winter months to protect against getting a cold. The excess carbons of citrate can be used to produce glycogen in the liver. Which one of the following liver enzymes is required for this conver-sion to occur? (A) α-ketoglutarate dehydrogenase (B) Pyruvate carboxylase (C) Pyruvate kinase (D) PFK-1 (E) Glucose-6-phosphatase

(B) Activation of pyruvate carboxylase Fatty acid oxidation increases the levels of acetyl-CoA within the mitochondrial matrix, and acetyl-CoA is a potent activator of pyruvate carboxylase, a key gluconeogenic enzyme (it will convert pyruvate to oxaloacetate, a necessary fi rst step to bypass the irreversible pyruvate kinase reaction). Acetyl-CoA cannot be used to synthesize net glucose, so it is not an effective precursor of glucose production. Acetyl-CoA does not activate PEP carboxykinase (that enzyme is transcriptionally controlled), nor does it affect pyruvate kinase (a cytoplasmic enzyme). PFK-2 is not regulated by acetyl-CoA (phosphorylation by protein kinase A is the key regulator effect for PFK-2 in the liver)

19. Liver fatty acid oxidation leads to an enhancement of gluconeogenesis via which of the following? (A) Generation of precursors for glucose synthesis (B) Activation of pyruvate carboxylase (C) Activation of phosphoenolpyruvate carboxykinase (D) Inhibition of pyruvate kinase (E) Inhibition of PFK-2

C When region C is removed from the DNA, there is a tremendous reduction in the expres-sion of the reporter gene, indicating that this region of the DNA binds important positively-acting transcription factors for the expression of the reporter gene.

19. Referring to the fi gure above, which region binds stimu-latory transcription factors?

(D) Blood glucose and insulin levels measured while he was symptomatic

19. Your 20-year-old male patient received a medical discharge from the US Army. He has had multiple episodes of lightheadedness, sweating, fatigue, tremor, and intense hunger. He had one seizure. During two of these episodes, his blood glucose was 40 mg/dL. Which of the following tests could help you diagnose his problem? (A) Fasting blood glucose (B) HbA1c (C) Noncontrast CT scan of the abdomen (D) Blood glucose and insulin levels measured while he was symptomatic (E) Determining the presence of islet cell antibodies

b. There are 64 possible combinations of 3 nucleotides, each combination specifying a specific amino acid or serving as a termination signal. The "genetic code" uses 64 possible combinations of threenucleotide "words" (codons) each specifying one of 20 amino acids or three stop signals that catalyze translational termination when present in the transcribed messenger RNA (answer b correct, answers a, c, and d incorrect). Single nucleotide changes in genes (point mutations) can thus change the codon in transcribed mRNA and produce single amino acid changes in the translated protein. Amino acid changes in proteins implied by altered enzyme activity or altered electrophoretic migration (eg, the abnormal Z form of AAT) can now be predicted by DNA sequencing of the encoding gene to demonstrate specific altered codons (eg, the GAG to AAG change in the AAT gene that changed the 342nd amino acid from lysine to glutamine. Mutations are best described by reference to amino acid position, since the linear correspondence of codons in DNA and of amino acids in protein domains is interrupted by the presence of introns in DNA (answer e is incorrect). Accumulation of abnormal AAT protein in liver or other cells can be detected by periodic acid Schiff (PAS) staining that recognized carbohydrate groups (also recognizing glycogen). The accumulation in liver to cause early disease is different from lung damage (emphysema) that occurs later due to lack of AAT protection.

195. A 6-year-old boy of Scandinavian descent presents with recurrent vomiting, fatigue, and weight loss. Physical examination reveals scleral icterus (yellow whites of eyes) and tenderness over the right upper quadrant of the abdomen. Laboratory studies reveal elevated levels of serum glutamic oxaloacetic transaminase (AST—reflecting liver or heart cell death) and bilirubin (reflecting icterus or jaundice), but no hepatitis A or B viral antigens. A diagnosis of nonviral hepatitis is made, and a family history reveals that the father and his sister suffer from early-onset emphysema, and that the paternal grandmother died from this illness. A suspected diagnosis of α1 antitrypsin (AAT-MIM*104700) deficiency is confirmed by finding the child has only the Z form of AAT on protein electrophoresis, confirmed by DNA sequencing showing homozygous AAG sequences at the 342nd codon (expected to encode lysine) rather than normal GAG (expected to encode glutamine). Which of the following statements about the "genetic code" explain this molecular diagnosis? a. There are 64 codons, each of which can encode several different amino acids. b. There are 64 possible combinations of 3 nucleotides, each combination specifying a specific amino acid or serving as a termination signal. c. There are 16 possible combinations of 2 nucleotides, allowing the first two nucleotides of each codon to specify unique amino acids and rendering the third nucleotide as irrelevant (degenerate). d. Information is stored as sets of trinucleotide repeats called codons, accounting for the evolution of complex genes from simpler ones. e. The 5′ to 3′ sequence of 3-base pair codons within genes is exactly matched by the amino to carboxy sequence of amino acids within the translated protein.

c. -CCA-CCC-TAG-GTT-CAG Insertion (correct answer c) or deletion (incorrect answer d) because no termination codon generated of nucleotides shifts the reading frame unless the change is a multiple of three (incorrect answer e). Frameshifts may create unintended stop codons as in correct answer c and are typically more severe than p. Point mutations with substitutions are named by their position in the protein, that is, P 21A (choice b). The protein change P 21A could also be denoted by the corresponding change in the DNA reading frame, that is, C 63A. Note that the single base substitution of C65A for incorrect answer a does not result in an amino acid change. Deletions may be prefixed by the letter delta, as with ΔF25 (incorrect answer e).

196. A 3-year-old Caucasian boy has grown along the third percentile for age, despite his parents being tall. He also has required treatment for chronic asthma, and his pediatrician notes unusual curvature of his knees and wrists typical of rickets. Testing confirms rickets by x-ray and deficiency of vitamin D, a fat-soluble vitamin that can become deficient with inadequate sunlight or intestinal malabsorption. The latter possibility, together with the asthma and growth delay, suggests a diagnosis of cystic fibrosis (MIM*219700). DNA testing is now very accurate, examining over 100 possible mutations including the 3 bp deletion that removes a codon for phenylalanine (ΔF508). The DNA sequence shown below is the sense strand from a coding region known to be a mutational "hot spot" for a gene. It encodes amino acids 21 to 25. Given the genetic and amino acid codes CCC = proline (P), GCC = alanine (A), TTC = phenylalanine (F), and TAG = stop codon, which of the following sequences is a frameshift mutation that causes termination of the encoded protein 5′-CCC-CCT-AGG-TTC-AGG-3′? a. -CCA-CCT-AGG-TTC-AGG b.-GCC-CCT-AGG-TTC-AGG c. -CCA-CCC-TAG-GTT-CAG d. -CCC-CTA-GGT-TCA-GG e. -CCC-CCT-AGG-AGG-

b. Inhibition of mRNA synthesis Three RNA polymerases are responsible for RNA transcription in mammalian cells, RNA polymerase I for ribosomal RNA, RNA polymerase II for messenger RNA, and RNA polymerase III for transfer and small RNAs (5s, μRNAs). Mammalian RNA polymerase II is highly sensitive to the mushroom toxin α-amanitin, and thus synthesis of mRNA (correct answer b) rather than other RNA species (incorrect answers a, c-e) will be inhibited by α-amanitin (even experienced mushroom gatherers can confuse the toxic Amanita species with edible varieties). Inhibition of RNA polymerase II-catalyzed HnRNA/mRNA transcription seems to have the most dramatic effects in liver, perhaps, because of its active role in protein synthesis. The liver damage caused by α-amanitin ingestion cannot be treated, requiring regeneration in milder ingestions and transplant in severe ones.

197. A 5-year-old Caucasian girl returns from a mushroom hunt in Northern Michigan and begins vomiting, then shows tremors, extreme anxiety, and finally disorientation—not recognizing her parents. She is taken to the emergency room where jaundice and delirium are noted and followed by liver function tests showing elevated ammonia, liver enzymes, and bilirubin. A toxicologist is called, who suggests poisoning from Amanita phalloides, a mushroom containing the toxin α-amanitin that inhibits RNA polymerase II. The child's liver failure is most likely due to which of the following? a. Inhibition of tRNA and protein synthesis b. Inhibition of mRNA synthesis c. Inhibition of small RNA synthesis d. Inhibition of ribosomal RNA synthesis e. Inhibition of mitochondrial RNA synthesis

c. 550 bp-decreased amount of mRNA The size of the primary β-globin RNA transcript will include the 5′- and 3′-untranslated regions (50 plus 50 bp) and the exons encoding protein domains (450 amino acids requires 150 coding bp), producing a processed (mature) mRNA of 550 bp (correct answer c). Introns (900 bp) would be included in the 1450 bp primary or HnRNA (answer a incorrect). RNA processing removes the 150 and 750 bp introns, yielding a mature mRNA of 550 bp (excluding answers b, d, and e that incorrectly include an intron or exclude untranslated regions that remain in the mRNA). Thalassemias are caused by mutations that alter the ratio of α- and β-globin, with β-globin being reduced in β-thalassemia. The altered ratios reduce hemoglobin (α2β2) synthesis, leading to severe anemia and extramedullary hematopoiesis in liver in an attempt at compensation (causing liver enlargement or hepatomegaly). β-thalassemia requires mutations affecting both β-globin alleles to be severe, while α- thalassemia requires mutations affecting three of the four α-globin alleles to be severe (hemoglobin H disease). Exons are the coding portions of genes and consist of trinucleotide codons that guide the placement of specific amino acids into protein. Introns are the noncoding portions of genes that may function in evolution to provide "shuffling" of exons to produce new proteins. The primary RNA transcript contains both exons and introns, but the latter are removed by RNA splicing. The 5′ (upstream) and 3′ (downstream) untranslated RNA regions remain in the mature RNA, and are thought to regulate RNA transport or translation. A polyA tail is added to the primary transcript after transcription, which facilitates transport and processing from the nucleus. The discovery of introns complicated Mendel's idea of the gene as the smallest hereditary unit; a modern definition might be the colinear sequence of exons, introns, and adjacent regulatory sequences that accomplish protein expression.

198. A 9-month-old Italian American boy presents with tachypnea and respiratory distress. His physical examination reveals a large spleen and his blood counts reveal severe anemia with targeted red blood cells on smear. A diagnosis of β-thalassemia (MIM*141900) is suggested and analysis of gene structure and expression is begun. As shown in the diagram above Questions 241 to 248, the β-globin gene contains three exons that encode a protein of 150 amino acids, separated by introns of 150 and 750 bp. The β-globin messenger RNA (mRNA) has 5′- and 3′-untranslated regions of 50 nucleotides. Choose the size of the mature β-globin messenger RNA and its amount in this boy with β-thalassemia: a. 1450 bp-normal amount of mRNA b. 700 bp-decreased amount of mRNA c. 550 bp-decreased amount of mRNA d. 500 bp-increased amount of mRNA e. 450 bp-decreased amount of mRNA

c. A small RNA that binds the promoter and enhances transcription The small size and alkaline lability of the factor suggests it is a small RNA (correct answer c) rather than mRNA (>100 to thousands of bp—incorrect answer b) or tRNA (75-90 bp—incorrect answer a). Some simple RNAs can self-catalyze their own splicing, but such mechanisms are less important during the complex processing of mammalian mRNAs (incorrect answer e). A transposon is a mobile DNA element that usually inhibits gene expression (incorrect answer d). Small or mciroRNAs ranging from 21 to 30 nucleotides in length are now recognized in all eucaryotes and include more than 1000 species in humans. Micro RNAs may stimulate or interfere with gene transcription by binding to chromatin or DNA sequence elements. Regulation by microRNAs is complex, since many have double-stranded or hairpin regions that are cleaved and activated by specific nucleases such as Dicer, Drosha, or Pasha in flies. Silencing or interfering microRNAs are an important tool for genetic analysis in that they allow gene knockout to define their effects; they also are being examined as therapeutic agents.

199. A "factor" is studied that stimulates a gene controlling differentiation of human immune stem cells into B-cells that fight bacterial infection. Some diseases such as Bruton agammaglobulinemia (MIM*300300) involve an inability to produce B-cells. The factor can hybridize to DNA, is hydrolyzed by alkali treatment, and migrates as a 2 to 30 bp species on electrophoresis. The differentiation gene is not stimulated by the factor if a 10 bp promoter element near the initiation site for transcription is removed. Which of the following factors is most likely? a. A transfer RNA that recognizes a codon within the promoter element b. An mRNA that is translated to produce a stimulatory transcription factor c. A small RNA that binds the promoter and enhances transcription d. A transposon that recognizes the promoter element and inserts to activate the gene e. An RNA catalyst that is essential for mRNA splicing

(D) Induction of drug-metabolizing enzymes Barbiturates are xenobiotics, and the body induces specifi c cytochrome P450 systems to help detoxify and excrete the barbiturates. When the man fi rst begins taking the drug, a low concentration of drug is suffi cient to exert a physiological effect, as the drug detoxifying system has not yet been induced. As the detoxifying system is induced, however, higher concen-trations of drug are required to have the same effect, as the rate of excretion of the drug is increased as the detoxifi cation system is induced. The "tolerance" to drugs, in this case, is not due to downregulation of drug receptors or decreased absorption of the drug from the stomach. There is no induction of target gene expres-sion, leading to enhanced drug action, nor are oppos-ing neurotransmitters expressed. The tolerance comes about due to enhanced inactivation of the drug due to the induction of drug-metabolizing enzymes.

2. A 34-year-old man was prescribed barbiturates 6 months ago for a seizure disorder. However, with time, the phy-sician has had to increase his daily dosage to maintain the same therapeutic drug level. This is due to which of the following? (A) Downregulation of drug receptors on the cell surface (B) Decreased absorption of the drug from the stomach (C) Increased synthesis of opposing neurotransmitters (D) Induction of drug-metabolizing enzymes (E) Induction of targeted enzyme synthesis

(C) Muscle PFK-1 The child has a form of glycogen storage disease known as type VII, Tarui dis-ease, which is a lack of muscle phosphofructokinase 1 (PFK-1) activity. The lack of muscle PFK-1 means that glycolysis is impaired, so anaerobic activities are signifi -cantly curtailed in such individuals. Slow, aerobic activi-ties, which can be powered by fatty acid oxidation, are normal in such children. Strenuous activity will lead to muscle damage and weakness due to this block in glycolysis. Glucose-6-phosphatase is only found in the liver (and to a small extent, the kidney), and a lack of such activity would lead to fasting hypoglycemia, but would not affect muscle glycolytic activity. A defect in liver PFK-1 activity would not affect muscle glycolysis. A defect in liver glycogen phosphorylase would also lead to fasting hypoglycemia, but would not alter the rate of muscle glycolysis, or lactate formation from that pathway.

2. A 7-year-old boy is brought to the pediatrician due to severe exercise intolerance. In gym class, the boy has trouble with anaerobic activities. Laboratory tests showed a lack of lactate production under such conditions. The boy was eventually found to have a mutation in which one of the following enzymes? (A) Liver glycogen phosphorylase (B) Liver PFK-1 (C) Muscle PFK-1 (D) Muscle glucose-6-phosphatase (E) Liver glucose-6-phosphatase

(C) To induce synthesis of a functional protein The patient has sickle cell anemia, and hydroxyurea treatment is designed to activate tran-scription of the γ-globin chain, which is normally only expressed during development (fetal hemoglobin). When expressed, the γ-globin gene will form func-tional hemoglobin tetramers with the α-globin chains, thereby reducing the effects of the mutated β-globin chain. Hydroxyurea does not bind to hemoglobin and denature it; it does not reduce the synthesis of the β-chains, nor does it alter oxygen levels in the blood or 2,3-bisphosphoglycerate levels in the eryth-rocyte. The γ-chain is normally turned off at birth as part of the hemoglobin-switching pathway (see the fi gure on page 49). The challenge to scientists at this time is to understand how to reactivate γ-chain syn-thesis in patients with both sickle cell disease and β-thalassemias.

2. An African American patient has displayed vaso-occlusive episodes for most of his life. The incidents are more prevalent under conditions in which blood oxygen levels are low, such as during exercise or taking trips to locations at high altitudes. The patient has been placed on hydroxyurea. The rationale behind this treatment is which of the following? (A) To prevent vaso-occlusive episodes through hydroxy-urea induced protein degradation (B) To reduce synthesis of a defective protein (C) To induce synthesis of a functional protein (D) To enhance oxygen levels in the blood (E) To activate the enzyme that produces 2,3-bisphos-phoglycerate

(C) 3 moles of ATP When glycogen produces glucose via the action of glycogen phosphory-lase, glucose-1-phosphate is produced. As this is con-verted to two molecules of pyruvate, four moles of ATP are generated and one is utilized at the PFK-1 step for the net production of three moles of ATP. Two moles of NADH are also produced, but those are utilized by lac-tate dehydrogenase to reduce pyruvate to lactate (anaer-obic conditions) such that NAD+ can be regenerated for the glyceraldehyde-3-phosphate dehydrogenase step. A small amount of free glucose will be released from glycogen by the debranching enzyme (about 5% of the total); for that glucose, the net yield is two moles of ATP (since hexokinase has to phosphorylate the free glucose to glucose-6-phosphate), but since the majority of glu-cose released is in the form of glucose-1-phosphate, three moles of ATP is the better answer. These reactions are outlined below.

2. In muscle, under anaerobic conditions, the net synthesis of ATP starting from one mole of glucose derived from muscle glycogen is which one of the following? (A) 1 mole of ATP (B) 2 moles of ATP (C) 3 moles of ATP (D) 4 moles of ATP (E) 5 moles of ATP

(C) 10

2. The energy yield from the complete oxidation of acetyl-CoA to carbon dioxide is which of the following in terms of high-energy bonds formed? (A) 6 (B) 8 (C) 10 (D) 12 (E) 14

(B) Lipoxygenase Montelukast is a leukotriene blocker. Leukotrienes are formed through the lipoxygenase pathway and affect bronchoconstriction and allergy pathways (see the fi gure in answer to question 1). The cyclooxygenase pathway produces prostaglandins and thromboxanes. The P450 pathway produces epoxides. The Cori cycle is related to gluconeogenesis (lactate transfer from the muscle to the liver), while the TCA cycle is utilized to oxidize acetyl-CoA to CO2 and H2 O.

2. You have an asthmatic patient who is already on an inhaled steroid and albuterol, but is still having diffi - culty. You add montelukast to her regimen. Montelukast (Singulair) specifi cally blocks the product of which of the following metabolic pathways? (A) Cyclooxygenase (B) Lipoxygenase (C) P450 (D) Cori cycle (E) TCA cycle

(D) Fatty acid transport into the mitochondria The man had eaten the unripe fruit of the ackee tree (from Jamaica). The unripened fruit contains the toxin hypoglycin A, which will interfere with carnitine's ability to transport acyl-carnitine groups across the inner mitochondrial membrane. This leads to a complete shutdown of fatty acid oxidation in all tissues in the affected individual, leading to severe hypoglycemia. Hypoglycin has no effect on fatty acid release from the adipocyte, or fatty acid entry into liver cells. Fatty acid oxidation is not directly inhibited, nor does this toxin directly inhibit the complexes of the electron transport chain and the proton-translocating ATPase.

20. A 35-year-old man in New York city, originally from Jamaica, purchased an illegally imported fruit from a street vendor and, within 4 h of eating the fruit, began vomiting severely. When brought to the emergency department the man was severely dehydrated and exhibited several seizures. The toxic effects of the fruit were interfering with which of the following? (A) Fatty acid release from the adipocyte (B) Fatty acid entry into the liver cell (C) Fatty acid activation (D) Fatty acid transport into the mitochondria (E) Oxidative phosphorylation

(A) Cytochrome synthesis The woman has a form of porphyria, an inborn error in the biosyn-thesis of heme. The buildup of heme intermediates is what generates the symptoms observed. Barbiturates are degraded through the cytochrome P450 series of enzymes, which are induced by the drug. The enzymes require heme, so heme synthesis is also induced, and whenever heme synthesis is induced, the toxic inter-mediates get accumulated. The end product of heme degradation is bilirubin, and if excessive bilirubin accumulates, jaundice is the result. Both dolichol and ubiquinone synthesis are from isoprenoids, and not related to the heme pathway.

20. A woman developed the following symptoms after tak-ing certain drugs such as barbiturates. The symptoms included severe pain in the abdomen, hallucinations, disorientation, and a reddish tint to the urine. These symptoms appeared due to the induction of genes involved in which of the following pathways? (A) Cytochrome synthesis (B) Cytochrome degradation (C) Ubiquinone biosynthesis (D) Ubiquinone degradation (E) Dolichol synthesis

(B) A high Km for glucokinase The liver expresses glucokinase, which has a high Km for glucose, particularly as compared to the Km for hexokinase. This means that glucokinase will only phosphorylate glucose when the intrahepatic glucose concentrations are high, and the intrahepatic levels of glucose only reach these levels after a meal. Under normal, fasting conditions, the concentration of blood glucose is lower than the Km for glucokinase, and very little phosphorylation of glu-cose will occur. PFK-1 is not a phosphorylated enzyme, and glucagon does not stimulate an increase in glucose transporters in the liver.

20. Under conditions of hypoglycemia, the liver is not uti-lizing glucose as an energy source due to which of the following? (A) A low Km for glucokinase (B) A high Km for glucokinase (C) An inhibited, phosphorylated PFK-1 (D) An activated, phosphorylated PFK-1 (E) A reduction of glucose transporters in the membrane

(C) γ-Glutamyl cycle Glutathione is pro-duced via the γ-glutamyl cycle; the HMP shunt pathway provides the NADPH that allows oxidized glutathione to be converted to reduced glutathione. The other path-ways listed (TCA cycle, glycolysis, HMP shunt, and the polyol pathway) do not provide for glutathione synthe-sis. The TCA cycle is designed to oxidize acetyl-CoA to carbon dioxide and water. Glycolysis is the entry point of sugars into metabolism. The HMP shunt pathway gen-erates fi ve-carbon sugars and NADPH, and the polyol pathway generates sugar alcohols. The γ-glutamyl cycle is shown in the fi gure below.

20. Which of the following biochemical pathways produces the antioxidant referred to in the previous question? (A) TCA cycle (B) Glycolysis (C) γ-Glutamyl cycle (D) HMP shunt (E) Polyol pathway

(B) He is not depleting glycogen stores prior to loading Marathon runners deplete their stores of glycogen during a race and need to catabolize other sources for energy to continue running. In the vernacular of the sport, when all the glycogen stores are exhausted, the runner "hits the wall." This is usu-ally somewhere around mile 20. Research has shown that proper "carb loading" prior to a race can increase body stores of glycogen and increase performance. Though it is a small increase (1% to 2%), it has been documented repeatedly in research studies even in highly trained athletes. Therefore, it is not a myth. To properly carbohydrate load, one must deplete gly-cogen stores with very vigorous exercise about 2 to 3 days prior to a race. This stimulates glycogen syn-thase which increases glycogen stores over the next 2 to 3 days before it returns to baseline levels. This is a critical step in the process of "overbuilding" gly-cogen stores. This is the step the patient is not doing properly. Vigorous exercise cannot then be continued during the 2 to 3 days of glycogen building or the glycogen stores will be utilized. Pancakes, potatoes, brown rice, and pasta are excellent sources of simple carbohydrates.

20. Your patient is a marathon runner and has visited your offi ce to ask you about carbohydrate loading to increase his performance during a race. For a full week prior to a race, he eats three meals a day of pancakes, potatoes, brown rice, and pasta and does not exercise at all. He has not noticed any success with this regimen. Which of the following answers best explains why he is getting no benefi t from his "carb loading"? (A) Carbohydrate loading is a myth (B) He is not depleting glycogen stores prior to loading (C) He is not on the carbohydrate loading diet long enough prior to the race (D) He is eating the incorrect foods for carbohydrate loading (E) He is too highly trained as an athlete for anything to increase his performance

c. The PAH mutation has disrupted spliceosome activity and caused PKU. The splicing of messenger RNA is carried out by spliceosomes— complexes of small ribonucleoprotein particles (snRNPs) and messenger RNA precursor. The presence of the GT to AT mutation in both PAH alleles, together with the larger (incompletely spliced) mRNA precursor, indicates a homozygote affected with PKU. Self-splicing by RNA catalysis (ribozymes) occurs in simpler organisms and some ribosomal RNA precursors, but not in spliceosome-mediated processing of mammalian mRNAs (incorrect answers a, b). Spliceosomes use snRNPs to recognize the 5′ splice site and the 3′ splice site followed by excision of the intronic RNA; RNA polymerase or helicase are not involved because it is the RNA within the spliceosome that catalyzes intron excision (incorrect answers d, e). The indicated splicing mutation was the first molecular change demonstrated in PKU, a necessary step in diagnosis because PAH enzyme activity cannot be measured in blood leucocytes or in cultured fibroblasts. Although PKU caused by phenylalanine hydoxylase can be treated using phenylalanine-deficient formula, the diet is demanding in that most food protein is minimized, aspartame (a phenylalanine derivative) minimized, and tyrosine balance (becoming essential without dietary phenylalanine as its precursor) is critical. Thus, some parents do choose prenatal diagnosis, especially for severe forms of PKU reflecting defective biopterin cofactor binding.

200. A 3-week-old Caucasian girl has returned for follow-up after her initial newborn metabolic screen showed elevated levels of phenylalanine, causing concern for phenylketonuria (PKU-MIM*261600). The second screen performed at age 2 weeks showed a phenylalanine level of 30 mg/dL with upper limits of normal being 6 mg/dL. Molecular characterization demonstrated a single-base change GT to AT at an intron-exon junction in both of her phenylalanine hydroxylase (PAH) alleles and an unusually large PAH mRNA. Which of the following most accurately describes these test results? a. The PAH mutation has disrupted RNA catalysis, producing a PKU carrier. b. The PAH mutation has disrupted RNA self-splicing and caused PKU. c. The PAH mutation has disrupted spliceosome activity and caused PKU. d. The PAH mutation has disrupted action of RNA polymerase and caused PKU. e. The PAH mutation has disrupted RNA helicase, producing a PKU carrier.

d. 5′ ends of mRNA unprotected by caps The primary transcripts of all eukaryotic mRNAs are capped at the 5′ ends as opposed to prokaryotic RNAs or eukaryotic tRNAs and rRNAs (incorrect answers b, c). Spliceosomes accomplish RNA splicing and excision of intronic sequences from mammalian mRNA (incorrect answer a), and poly(A) is added to the 3′ end (incorrect answer e). The cap is composed of 7-methylguanylate attached by a pyrophosphate linkage to the 5′ end. The cap protects the 5′ ends of mRNAs from nucleases and phosphatases and is essential for the recognition of eukaryotic mRNAs in the protein-synthesizing system

201. Antibiotics that selectively inhibit bacterial growth exploit differences between bacterial and mammalian cells. Design of a drug that would selectively inhibit bacterial RNA synthesis would target which of the following? a. Spliceosome components b. 3′ ends of tRNA unprotected by caps c. 3′ ends of mRNA unprotected by caps d. 5′ ends of mRNA unprotected by caps e. Poly(A) addition to the 5′ end

d. Ribosomal RNA The nucleolus, an organelle unique to eukaryotic cells, is the site where RNA polymerase I transcribes the multiple, tandemly repeated genes for ribosomal RNAs to give 45S primary transcripts that form the 28S and 18S ribosomal RNAs. The ribosomal RNA (rRNA) gene clusters can undergo amplification through unequal crossing over, accounting for individual variability in the size of these clusters, located at the ends of the five pairs of acrocentric chromosomes (numbers 13-15, 21-22). These variably sized clusters form stalks and "satellites" on the acrocentric chromosomes when visualized at metaphase by karyotyping, and their associated rRNA gives a distinctive dark color when incubated with silver stain. The rRNA gene clusters (rDNA) associate within nucleoli at interphase and comprise nucleolar organizing regions (NOR) on chromosomes visualized at metaphase. Enzymatic modification and cleavage remove spacer regions to yield 28S, 18S, and 5.8S ribosomal RNA. The 5S subunit is synthesized by RNA polymerase III in the nucleoplasm rather than in the nucleolus. Ribosomal proteins combine with the ribosomal subunits to assemble into a 60S subunit containing the 5S, 5.8S, and 28S RNAs and a 40S subunit containing the 18S RNA. Combined, the two subunits produce a functional eukaryotic ribosome with a sedimentation coefficient of 80S. Messenger RNAs are synthesized by RNA polymerase II in the nucleoplasm, while 5S, tRNA, and small RNAs are synthesized by RNA polymerase III in the cytoplasm.

202. A 36-year-old female elects to have amniocentesis with her third pregnancy because of "advanced" maternal age—her family history and that of her 38-year-old husband are normal. The fetal chromosomes show a normal number of 46, but one chromosome 21 has a large satellite region. The couple is called back for parental chromosomes and is extremely anxious that their fetus may have Down syndrome. The same enlarged satellite is seen on one chromosome 21 of the father, and silver staining shows this amplified DNA to be part of the nucleolar organizing region (NOR). The counselor explains that these NOR regions are seen on chromosomes 13 and 15 and 21 and 22 (acrocentric chromosomes) and are transcriptionally active in the nucleolus. The molecules associated with these NOR would most likely be which of the following? a. Messenger RNA used to synthesize histone proteins b. Small silencing RNA (siRNA) c. Poly(A) RNA d. Ribosomal RNA e. Transfer RNA

c. Several different genes may produce identical mRNA molecules. Although precursor heterogenous RNA may undergo alternative processing to encode several different proteins, mature mRNAs are highly processed (incorrect answers d, e) to encode one specific protein (incorrect answer b). A collinear 3-nucleotide to amino acid code is employed (incorrect answer a). About 30% of the DNA of humans and other mammals consists of repeated sequences. Repetitive DNA includes numerous families of genes like those for histones. Some families of repeated genes make identical mRNA molecules, suggesting that their multiple gene copies are needed to make adequate amounts of protein. Although many genes in bacteria produce a polycistronic mRNA that encodes several different peptides, all mRNAs in mammals encode a single peptide and are monocistronic. In addition, RNA is initially transcribed from protein-encoding genes as larger molecules called heterogenous nuclear RNA (hnRNA). These immature hnRNA molecules must be spliced to remove introns and chemically modified with 5′ caps and 3′-poly(A) sequences before reaching the cytoplasm as functional mRNA. The initial hnRNA transcript is colinear with its encoded protein within exons, but not within introns. Mature mRNAs also have 5′- and 3′-untranslated regions that are not colinear with the encoded peptide.

203. Which one of the following statements best describes the synthesis of mammalian messenger RNA (mRNA)? a. There is colinearity of the RNA nucleotides transcribed from a gene and the amino acids encoded by each nucleotide. b. Each mRNA often encodes several different proteins. c. Several different genes may produce identical mRNA molecules. d. The RNA sequence transcribed from a gene is virtually identical to the mRNA that exits from nucleus to cytoplasm. e. Mammalian mRNA undergoes minimal modification during its maturation.

e. Western blot using antibody to the amino-terminal portion of apoB protein Northern blotting to detect RNA and Western blotting to detect protein were named as puns on the name of Ed Southern, who developed his cognate method for electrophoretic separation and hybridization probe detection of DNA segments. Changes in protein size due to RNA editing (two forms of mRNA from the same gene) would be detected by Western blotting (incorrect answers b, c). Specific antibodies to apoB protein are needed to visualize the protein after Western blotting (incorrect answers a, d). Specific antibodies are usually raised by injection into rabbits and purification of antibody from rabbit serum. Antibodies are labeled with radioactive iodine or fluorescent dyes, then floated over the Western blot membrane to bind and identify specific peptides. Antibody to the aminoterminus region of apoB protein is needed to identify.

204. A 35-year-old African American male has noted deteriorating skills in activities such as golf and tennis, lately noting difficulties with handwriting. Evaluation reveals poor tandem walking and finger-to-nose coordination, and a peripheral smear shows shrunken and spiny red blood cells called acanthocytes. Serum lipoprotein electrophoresis shows very low density lipoproteins (VLDL) and low-density lipoproteins (LDL), suggesting a diagnosis of abetalipoproteinemia (MIM*200100). Which of the following is the best method to demonstrate yield of different apolipoprotein B (apoB) protein sizes through RNA editing? a. Western blot using apoB DNA as probe b. Northern blot using antibody to apoB protein as probe c. Southern blot using the proximal apoB gene segment as probe d. Western blot using apoB cDNA as probe e. Western blot using antibody to the amino-terminal portion of apoB protein

b. Loading of the small ribosomal subunit with initiation factors, messenger RNA, and initiation aminoacyl-tRNA Despite the more complex initiation factors loading the small ribosomal subunit (correct answer b), protein synthesis in prokaryotes and eukaryotes is quite similar. The ribosomal subunits are odd numbers 30S and 50S in prokaryotes and even numbers 40S and 60S in eukaryotes. At the start of translation, initiation factors, mRNA, and initiation aminoacyltRNA bind to the dissociated small ribosomal subunit (answers a, c-e thus incorrect). Only after the small ribosomal subunit is primed with mRNA and initiation aminoacyl-tRNA does the large ribosomal subunit bind to it. Once this happens, elongation factors bring the first aminoacyl-tRNA of the nascent protein to the A site. This is the step inhibited by tetracycline. Then, peptidyl transferase forges a peptide bond between the initiation amino acid and the first amino acid of the forming peptide. The now uncharged initiation tRNA leaves the P site, and the peptidyl-tRNA from the A site moves to the now vacant P site with the two amino acids attached. The ribosome advances three bases to read the next codon, and the process repeats. Macrolide antibiotics (eg, azithromycin, Zithromax, and erythromycin) bind to the bacterial 50S ribosome and inhibit this translocation step. When the stop signal is reached after the complete polypeptide has been synthesized, releasing factors bind to the stop signal, causing peptidyl transferase to hydrolyze the bond that joins the polypeptide at the A site to the tRNA. Factors prevent the reassociation of ribosomal subunits in the absence of new initiation complex.

205. A 26-year-old male medical student contracts a skin infection from an elderly patient and notes an expanding circular red lesion on his thigh at dinner that night. He places a warm washrag on the lesion at bedtime and awakens in early morning feeling hot with shortness of breath. After reluctantly going to the health clinic, he faints during the registration process and awakens in the intensive care unit. An infectious disease specialist tells him he has methicillin-resistant Staphylococcus aureus with toxic shock syndrome (low blood pressure) and adult respiratory distress syndrome (swelling of alveolar membranes with hypoxemia). The attending wishes to use a new antibiotic that acts on the earliest step of protein synthesis. Which of the following steps would most likely be targeted? a. Peptide bond formation b. Loading of the small ribosomal subunit with initiation factors, messenger RNA, and initiation aminoacyl-tRNA c. Sliding of the ribosome three bases forward to read a new codon d. Binding of the small ribosomal subunit to the large ribosomal subunit e. Elongation factors deliver aminoacyl-tRNA to the A site of the small ribosomal subunit

a. Disruption of signal peptide function with accumulation of AAT protein in the Golgi apparatus Protein synthesis occurs in the cytoplasm, on groups of free ribosomes called polysomes, and on ribosomes associated with membranes, termed the rough endoplasmic reticulum. However, proteins destined for secretion have a signal peptide targeting their ribosomes to the endoplasmic reticulum and the synthesized protein is extruded into the ER lumen. Normal AAT synthesis would involve signal peptide targeting to the ER, secretion from ER lumen to plasma membrane and the bloodstream, and subsequent inhibition of damaging serum proteases (incorrect answer c). Mutations causing AAT deficiency could disrupt signal peptide function and rough ER targeting with decreased serum AAT levels (incorrect answers b, d, and e).

206. A 15-year-old Caucasian adolescent of Scandinavian descent has suffered from mild liver disease thought to be related to receiving parenteral nutrition as a premature infant. Recently she has developed shortness of breath, and has decreased oxygen saturation measured at rest. Emphysema is suspected that with liver disease suggests α1-antitrypsin (AAT) deficiency (MIM*107400), confirmed by protein electrophoresis that shows only mutant protein. Given that normal AAT blocks serum proteases, which of the following is the most likely mechanism for AAT disease? a. Disruption of signal peptide function with accumulation of AAT protein in the Golgi apparatus b. Disruption of signal peptide function with increased amounts of AAT in plasma c. Disruption of AAT protease function d. Decreased targeting of AAT mRNA to smooth ER ribosomes e. Increased targeting of AAT mRNA to rough ER ribosomes

b. Insertion of 1 bp, 25% The insertion or deletion of nucleotides other than multiples of three will change the reading frame by which the linear nucleotides of RNA parsed three at a time (codons) into the linear amino acids of protein (eliminates answers a, c). Substitution with a different nucleotide will usually produce missense (change in amino acid) or silent (no change in amino acid) mutations that have milder effects on protein function (incorrect answers d, e—rare substitutions could produce a terminating codon and truncated protein). The inserted base pair or similar nonsense mutation would have to be present in both β-globin alleles of the deceased child in order to cause disease, and thus both parents (assuming correct paternity) would be heterozygotes or carriers. The next child of carrier parents would have a 1/4 or 25% chance to be affected (1/2 chance for the abnormal allele in each parental gamete) and could be given the option of preimplantation/prenatal diagnosis (the 25% risk eliminates answers a, e). A shift in the reading frame by one or two positions will change the sense of ensuing nucleotides, causing all subsequent codons to be read out of frame to produce a completely different (nonsense) amino acid sequence. Often the shift in frame will introduce a stop codon after the frameshift and produce a truncated protein. Whether the amino acid sequence is garbled or truncated, a frameshift mutation will disrupt production of normal protein. In the case of frameshift mutations in the β-globin gene, production of β-globin protein will be reduced as with transcription or RNA processing mutations, causing another form of β-thalassemia when both β-globin gene copies are ineffective. Missense mutations may produce a protein with variant structure and function (eg, as in a hemoglobinopathy), while silent mutations will have no effect on protein structure/function and may be called polymorphisms (harmless genetic variations).

207. A couple request genetic counseling because their first child had severe β-thalassemia (MIM*141900) and expired at age 1 year from graftversus- host reaction after bone marrow transplant. DNA analysis of the child demonstrated a promoter mutation on one copy of the β-globin gene and a mutation in the middle of exon 2 on the other copy. What is the most likely type of mutation in exon 2 that would lead to β-thalassemia (ie, decreased production of β-globin peptide from both β-globin gene copies), and what is the couple's recurrence risk for their next child to be affected? a. Deletion of 3 bp, 50% b. Insertion of 1 bp, 25% c. Insertion of 3 bp, 25% d. Missense mutation, 25% e. Silent mutation, virtually zero recurrence risk

c. Two ATP is required for the esterification of amino acids to their corresponding tRNAs. This reaction is catalyzed by the class of enzymes known as aminoacyl-tRNA synthetases. Each one of these enzymes is specific for one tRNA and its corresponding amino acid: amino acid + tRNA + ATP → aminoacyl-tRNA + AMP + PP i As with most ATP hydrolysis reactions that release pyrophosphate, pyrophosphatase quickly hydrolyzes the product to Pi, which makes the reaction essentially irreversible. Since ATP is hydrolyzed to AMP and PPi during the reaction, the equivalent of two high-energy phosphate bonds is utilized.

208. How many high-energy phosphate-bond equivalents are utilized in the process of activation of amino acids for protein synthesis? a. Zero b. One c. Two d. Three e. Four

b. Peptidyl transferase During the course of protein synthesis on a ribosome, peptidyl transferase catalyzes the formation of peptide bonds. However, when a stop codon such as UAA, UGA, or UAG is reached, aminoacyl-tRNA does not bind to the A site of a ribosome. One of the proteins, known as a release factor, binds to the specific trinucleotide sequence present. This binding of the release factor activates peptidyl transferase to hydrolyze the bond between the polypeptide and the tRNA occupying the P site. Thus, instead of forming a peptide bond, peptidyl transferase catalyzes the hydrolytic step that leads to the release of newly synthesized proteins. Following release of the polypeptide, the ribosome dissociates into its major subunits.

209. The hydrolytic step leading to the release of a polypeptide chain from a ribosome is catalyzed by which of the following? a. Dissociation of ribosomes b. Peptidyl transferase c. Release factors d. Stop codons e. UAA

c. 5′-CUAUAGGUAAUCACUG-3′ The template strand refers to the DNA strand that is transcribed into RNA. As for DNA, RNA is synthesized in the 5′ to 3′ direction, so the template strand shown in the question should be rewritten in the 3′ to 5′ direction (eg, 3′-CAGTGA . . ., giving a 5′ to 3′ RNA transcript of 5′-GUCACU . . ., correct answer e). The DNA complementary to the template strand is known as the coding strand and will have the same sequence as the RNA transcript, except that U replaces T in RNA. This coding strand sequence is used for DNA sequence databases so potential genes can be identified by coding sequences that would yield stretches of amino acids and thus encode proteins. Incorrect answers b and d show DNA sequences without U, while incorrect answers a and c have the wrong sequence orientation.

210. The sequence of the template DNA strand is 5′-GATATCCATTAGTGAC-3′. What is the sequence of the RNA produced? a. 5′-CAGUGAUUACCUAUAG-3′ b. 5′-CTATAGGTAATCACTG-3′ c. 5′-CUAUAGGUAAUCACUG-3′ d. 5′-GTCACTAATGGATATC-3′ e. 5′-GUCACUAAUGGAUAUC-3′

d. They are found both free in the cytoplasm and bound to membranes. The two subunits of ribosomes are composed of proteins and rRNA. Ribosomes are found in the cytoplasm, in mitochondria, and bound to the endoplasmic reticulum. Transcription refers to the synthesis of RNA complementary to a DNA template and has nothing immediately to do with ribosomes. Cells with high levels of protein synthesis such as erythroid precursors (globins) or plasma cells (immunoglobulins) have abundant ribosomes.

211. Which of the following statements about ribosomes is true? a. They are composed of RNA, DNA, and protein. b. They are composed of three subunits of unequal size. c. They are bound together so tightly they cannot dissociate under physiologic conditions. d. They are found both free in the cytoplasm and bound to membranes. e. They are an integral part of transcription.

c. Translocation of tRNA-nascent protein complex from A to P sites Two molecules of GTP are used in the formation of each peptide bond on the ribosome. In the elongation cycle, binding of aminoacyl- tRNA delivered by EF-2 to the A site requires hydrolysis of one GTP. Peptide bond formation then occurs. Translocation of the nascent peptide chain on tRNA to the P site requires hydrolysis of a second GTP. The activation of amino acids with aminoacyl-tRNA synthetase requires hydrolysis of ATP to AMP plus PPi.

212. Guanosine triphosphate (GTP) is required by which of the following steps in protein synthesis? a. Aminoacyl-tRNA synthetase activation of amino acids b. Attachment of ribosomes to endoplasmic reticulum c. Translocation of tRNA-nascent protein complex from A to P sites d. Attachment of mRNA to ribosomes e. Attachment of signal recognition protein to ribosomes

a. Inhibiting translocation by binding to 50S ribosomal subunits Virulent strains of bacteria such as Mycoplasma pneumoniae, various Legionella species, and Bordetella pertussis cause atypical community- acquired pneumonias that often be successfully treated with erythromycin (or related antibiotics such as azithromycin [Zithromax]). The mechanism of action of erythromycin is to specifically bind the 50S subunit of bacterial ribosomes (answers b-e incorrect). Under normal conditions, after mRNA attaches to the initiation site of the 30S subunit, the 50S subunit binds to the 30S complex and forms the 70S complex that allows protein chain elongation to go forward. Elongation is prevented in the presence of erythromycin.

213. A 50-year-old African American male develops cough and fever for 3 days while attending a convention, then returns to his home city for medical evaluation. Chest x-rays reveal diffuse (interstitial) pneumonia, and the male's poor oxygen saturation and severe cough mandate hospitalization. His physician calls a hospital in the convention town and hears that over a dozen conventioneers received ER treatment for similar symptoms. The physician suspects mycoplasma or pertussis infection similar to Legionnaires disease and prescribes erythromycin. Erythromycin, azithromycin (Zithromax), and clarithromycin (Biaxin) are macrolide antibiotics that inhibit protein synthesis in certain bacteria by doing which of the following? a. Inhibiting translocation by binding to 50S ribosomal subunits b. Acting as an analogue of mRNA c. Causing premature chain termination d. Inhibiting initiation e. Mimicking mRNA binding

d. Blocks elongation of proteins by inactivating elongation factor 2 (EF-2 or translocase) The gene that produces the deadly toxin of Corynebacterium diphtheriae comes from a lysogenic phage that grows in the bacteria. Prior to immunization, diphtheria was a frequent cause of death in children. The protein toxin produced by this bacterium inhibits protein synthesis by ADP-ribosylation of elongation factor 2 (EF-2 or translocase). Diphtheria toxin is a single protein composed of two portions (A and B). The B portion enables the A portion to translocate across a cell membrane into the cytoplasm. The A portion catalyzes the transfer of the adenosine diphosphate ribose unit of NAD to a nitrogen atom of the diphthamide ring of EF-2, thereby blocking translocation. Diphthamide is an unusual amino acid residue of EF-2.

214. An immigrant family from rural Mexico brings their 3-month-old son to the emergency room because of whistling inspiration (stridor) and high fever. The child's physician is perplexed because the throat examination shows a gray membrane almost occluding the larynx. A senior physician recognizes diphtheria, now rare in immunized populations. The child is intubated, antitoxin is administered, and antibiotic therapy is initiated. Diphtheria toxin is often lethal in unimmunized persons because it does which of the following? a. Inhibits initiation of protein synthesis by preventing the binding of GTP to the 40S ribosomal subunit b. Binds to the signal recognition particle receptor on the cytoplasmic face of the endoplasmic reticulum receptor c. Shuts off signal peptidase d. Blocks elongation of proteins by inactivating elongation factor 2 (EF-2 or translocase) e. Causes deletions of amino acid by speeding up the movement of peptidyl-tRNA from the A site to the P site

a. A specific tRNA and a specific amino acid Aminoacyl-tRNA synthetases are responsible for charging a tRNA with the appropriate amino acid for translation. Charging a tRNA is a two-step reaction. In the first step, the enzyme forms an aminoacyl-AMP enzyme complex in a reaction that requires one ATP. In the second step, the activated amino acid is attached to the appropriate tRNA, and the enzyme and AMP are released.

215. Aminoacyl-tRNA synthetases must be capable of recognizing which of the following? a. A specific tRNA and a specific amino acid b. A specific rRNA and a specific amino acid c. A specific tRNA and the 40S ribosomal subunit d. A specific amino acid and the 40S ribosomal subunit e. A specific amino acid and the 60S ribosomal subunit

d. Pancreatic mitochondria Prokaryotic ribosomes have a sedimentation coefficient of 70S, and are composed of 50S and 30S subunits (odd numbers). Eukaryotic cytoplasmic ribosomes, either free or bound to the endoplasmic reticulum, are larger—60S and 40S subunits that associate to an 80S ribosome (even numbers). Nuclear ribosomes are attached to the endoplasmic reticulum of the nuclear membrane. Ribosomes in chloroplasts and mitochondria of eucaryotic cells are more similar to prokaryotic ribosomes than to eukaryotic cytosolic ribosomes. Like bacterial ribosomes, chloroplast and mitochondrial ribosomes use a formylated tRNA and are sensitive to bacterial protein synthesis inhibitors. Antibiotics that target bacterial protein synthesis may interfere with human mitochondrial function at high doses.

216. Ribosomes similar to those of bacteria are found in which of the following? a. Cardiac muscle cytoplasm b. Liver endoplasmic reticulum c. Neuronal cytoplasm d. Pancreatic mitochondria e. Plant nuclei

b. Acceleration of protein synthesis by phosphorylating initiation inhibitors The initiation of protein synthesis is a multistep process that includes several protein factors—eIF-4E and eIF-4G proteins that bind the mRNA cap, eIF-4A and eIF-4B proteins that bind to the 5′ end of mRNA and reduce its secondary structure. The eIF-3 proteins bind to the prior factors, linking them to the 40S ribosomal subunit and scanning for suitable AUG codons that will initiate peptide bond formation. This is usually the 5′-most AUG, but others may be selected by virtue of Kozak consensus sequences that surround it (GCC-AUG-UGG). Other protein factors regulate the rate of protein synthesis initiation by acting on eIF-4E, including those such as BP1 that bind and inactivate eIF-4E unless it is phosphorylated. Insulin and other growth factors act to phosphorylate BP-1 at several sites, releasing eIF-4E and stimulating initiation and protein synthesis. Although stimulation of protein synthesis by insulin is one plausible factor in increased fetal growth, other factors undoubtedly contribute to this complex process. Once the EF factors, 40S subunit, and relaxed mRNA are associated (as a 43S preinitiation complex), it binds the 60S ribosomal subunit and its peptidyl transferase that is a component of the 28S rRNA—an example of a ribozyme. In translation, the P site of the ribosome is occupied by the peptidyl-tRNA, which has the growing peptide chain attached. The appropriate aminoacyl-tRNA enters the A site and the carboxyl group of the peptidyl tRNA undergoes nucleophilic attack by the α-amino group of the aminoacyl-tRNA, as catalyzed by peptidyl transferase. The peptide chain is transferred to the tRNA in the A site, which is subsequently displaced to the P site, freeing the A site to bind the next charged tRNA.

217. Endogenous insulin is ineffective in type 2 diabetes mellitus (MIM*222100), leading to high blood glucose levels (hyperglycemia) with gradual effects on blood vessels in eye, kidney, and skin. Pregnant women with diabetes mellitus present high glucose loads to their fetus; this stimulates production of fetal insulin (hyperinsulinemia) and causes rapid fetal growth with large birth weight (macrosomia). The stimulation of fetal growth by insulin may be correlated with its effect on gene expression, which is which of the following? a. Stimulation of mRNA production by enhancing 5′ capping b. Acceleration of protein synthesis by phosphorylating initiation inhibitors c. Acceleration of protein synthesis by phosphorylating the 40S ribosomal subunit d. Acceleration of protein synthesis by phosphorylating proteinase K e. Stimulation of mRNA production by enhancing RNA splicing

b. Enhancer Eukaryotic promoters are much more complex than bacterial promoters. The most common bacterial promoter elements are the -10 and -35 sequences, which are the recognition sites for RNA polymerase binding. In addition, bacterial promoters may have operators (repressor binding sites) and activator binding sites. Eukaryotic promoters have a core region that may contain a TATA box (which is bound by TBP, the TATA binding protein), an initiator sequence, and downstream promoter elements. Many promoters lack one or more of these elements. In addition, eukaryotic promoters have upstream elements such as GC boxes and CCAAT boxes that bind specific transcription factors (Sp1 and CTF, respectively). Enhancers are elements that increase or enhance the rate of transcription initiation from a promoter. Enhancers can be upstream or downstream of the transcription start site and can exert their effect from hundreds or thousands of bases away. In addition, enhancers are orientation-independent (ie, whether their recognition sequence is on the Watson or Crick strand).

218. Which of the following eukaryotic promoter/regulatory elements has the most variable position with respect to the start site of transcription? a. Downstream promoter element (DPE) b. Enhancer c. Initiator sequence d. Operator e. TATA box

e. Signal recognition particle Signal recognition particles (SRPs) recognize the signal sequence on the N-terminal end of proteins destined for the lumen of the endoplasmic reticulum (ER). SRP binding arrests translation and an SRP receptor facilitates import of the nascent protein into the ER lumen. A signal peptidase removes the signal sequence from the protein, which may remain in the membrane or be routed for secretion. Common to both eukaryotic and prokaryotic protein synthesis is the requirement for ATP to activate amino acids. The activated aminoacyl-tRNAs then interact with ribosomes carrying mRNA. Peptidyl transferase catalyzes the formation of peptide bonds between the free amino group of activated aminoacyl-tRNA on the A site of the ribosome and the esterified carboxyl group of the peptidyl- rRNA on the P site; the liberated rRNA remains on the P site.

219. Which of the following is required for certain types of eucaryotic protein synthesis, but not for prokaryotic protein synthesis? a. GTP b. Messenger RNA c. Ribosomal RNA d. Peptidyl transferase e. Signal recognition particle

e. Missense mutation Missense mutations, which cause the substitution of one amino acid for another, may significantly alter the function of the resultant protein without altering the size of DNA restriction fragments detected by Southern blotting. In this case, Northern blot results would most likely also be normal. Single-base changes may also result in nonsense mutations. Large insertions or deletions in the exon or coding regions of the gene alter the Southern blot pattern and usually ablate the activity of one gene copy. In the case of an autosomal locus like that for ornithine aminotransferase, the homologous allele remains active and gives 50% enzyme activity (heterozygote or carrier range with a normal phenotype). Similar effects on enzyme activity would be predicted from complete gene deletions at one locus, while duplication might produce 150% or 50% of normal enzyme activity depending on the status of promoter sites.

221. Gyrate atrophy (MIM*258870) is a rare autosomal recessive genetic disorder caused by a deficiency of ornithine aminotransferase. Affected individuals experience progressive chorioretinal degeneration with vision and neurologic defects. The gene for ornithine aminotransferase has been cloned, its structure has been determined, and mutations in affected individuals have been extensively studied. Which of the following mutations best fits with test results showing normal Southern blots with probes from all ornithine aminotransferase exons but absent enzymatic activity? a. Duplication of entire gene b. Two-kb deletion in coding region of gene c. Two-kb insertion in coding region of gene d. Deletion of entire gene e. Missense mutation

(B) Galactose-1-phosphate uridylyltransferase The child has classic galactosemia, a defect in galactose-1-phosphate uridylyltransferase. Due to the accumulation of galactose-1-phosphate, galac-tokinase is inhibited, and free galactose accumulates within the blood and tissues. The accumulation of galactose in the lens of the eye provides substrate for aldose reductase, converting galactose to its alco-hol form (galactitol). The accumulation of galactitol leads to an osmotic imbalance across the lens, lead-ing to cataract formation. Additionally, the increased galactose-1-phosphate, at very high levels in the liver, blocks phosphoglucomutase activity, result-ing in ineffective glucose production from glycogen (phosphorylase degradation of glycogen will produce glucose-1-phosphate, but this cannot be converted to glucose-6-phosphate if phosphoglucomutase activ-ity is inhibited). A defect in galactokinase will lead to nonclassical galactosemia, with cataract formation, but none of the feeding problems associated with clas-sical galactosemia (associated with the accumulation of galactose-1-phosphate) are observed in nonclassi-cal galactosemia. None of the other enzymes listed, if defi cient, will give rise to the symptoms produced, particularly cataract formation. A defect in glycogen synthase would lead to reduced glycogen levels and fasting hypoglycemia. A defect in fructokinase leads to fructosuria (fructose in the urine), but no overt symptoms of disease. The fi gure below indicates the pathway for galactose metabolism and the defects in classical and nonclassical galactosemia.

3. A 2-week-old newborn was brought to the pediatri-cian due to frequent vomiting, lethargy, and diarrhea. Family history revealed that the child never seemed to eat well, and had only been breast-fed. Physical examination revealed an enlarged liver and jaundice. The pediatrician was suspicious of an inborn error of metabolism and referred the child to an ophthalmolo-gist for a slitlamp exam, the result of which is shown (A) Fructose-1,6-bisphosphatase (B) Galactose-1-phosphate uridylyltransferase (C) Galactokinase (D) Glycogen synthase (E) Fructokinase

(C) Fructose-1-phosphate inhibition of glycogen phos-phorylase The child has hereditary fructose intolerance, a defect in aldolase B activity in the liver. This leads to an accumulation of fructose-1-phosphate in the liver (and, as fructokinase has a high Vmax, a large amount of fructose-1-phosphate accumu-lates). At high levels, fructose-1-phosphate, through similarity in structure to glucose-1-phosphate, inhibits glycogen phosphorylase activity, leading to hypoglyce-mia (glycogen degradation is inhibited when blood glu-cose levels drop). The fructose is derived from the fruit juices introduced to the child's diet. Fructose does not inhibit debranching enzyme, and fructose-6-phosphate has no effect on glycogen phosphorylase (recall, one of the products of the glycogen phosphorylase reaction is glucose-1-phosphate, not glucose-6-phosphate). Galac-tose is found in lactose, which, while present in milk, is not found in fruit juice.

3. A 3-month-old infant, when switched to a formula diet plus fruit juices, begins to vomit and displays severe hypoglycemia after eating. Removal of the fruit juices from the diet seemed to reduce the severity of the symptoms. At the pediatrician's offi ce, an inborn error of metabolism was considered, which could explain the hypoglycemia. Which explanation is most likely? (A) Fructose inhibition of the debranching enzyme (B) Galactose-1-phosphate inhibition of glycogen phos-phorylase (C) Fructose-1-phosphate inhibition of glycogen phos-phorylase (D) Fructose-6-phosphate inhibition of glycogen phos-phorylase (E) Galactose inhibition of aldolase

(C) Increased gamma chain synthesis The sibling has, in addition to sickle cell disease, hereditary persistence of fetal hemoglobin (HPFH). Individuals with HPFH express the γ-globin chain throughout their life, and it can be at high levels. Since this child is express-ing both the HbS protein and the γ-protein, some nor-mal fetal hemoglobin can be formed in this child, with reduced levels of HbS formed. This reduces the level of sickling and allows oxygen delivery to the tissues. Thus, the HPFH protects against the effects of homozygous HbS expression, and the sibling shows few, if any, symp-toms of his HbS mutations. Alterations in the expression of the δ-chain have not been observed. Reduced α-chain synthesis would lead to an anemia, and there is no effec-tive way to reduce the synthesis of HbS chain. Increased ζ-synthesis has also not been observed.

3. A hematologist is studying an African American family as one of the children was recently diagnosed with sickle cell disease. His sibling shows no symptoms of the disease, although genetic tests showed homozygos-ity for the HbS gene. An analysis of his red blood cells is likely to show which of the following? (A) Reduced alpha chain synthesis (B) Reduced sickle chain synthesis (C) Increased gamma chain synthesis (D) Increased zeta chain synthesis (E) Increased delta chain synthesis

(A) Coconut/palm oil is a saturated fat Saturated fats do not liquefy until a much higher temperature than that at which monounsaturated or polyunsaturated fats do (the melting temperature for saturated fats is greater than that for unsaturated fats). Conversely, saturated fats are solids at a higher temperature than unsaturated fats and cannot exist in a liquid form at a lower temperature. Since the oil of a plant is its "lifeblood," at a lower temperature, a saturated oil would solidify and the plant would die. Saturated oil plants cannot survive in a temperate climate (Kansas) and need a tropical climate of warm temperatures all year round. Only polyunsaturated oil plants can survive in a temperate climate (corn, fl ax, wheat, and canola). Monounsaturated oils need a warmer climate, but not as warm as the tropics (olive, peanut). Knowing where a plant grows gives a large clue as to whether the oil will be saturated, monounsaturated, or polyunsaturated. The difference in oil content between plants appears to be an evolutionary process. Kansas soil is very rich and supports growth of most plants.

3. Coconut palm tress cannot survive growing outdoors in Kansas. Which of the following is the best explanation for this fi nding? (A) Coconut/palm oil is a saturated fat (B) Coconut/palm oil is a monounsaturated fat (C) Coconut/palm oil is a polyunsaturated fat (D) Kansas soil is not sandy enough to support growth (E) Kansas soil is too rocky to support growth

(B) Ethanol inhibition of a cytochrome P450 system Ethanol inhibits the drug detoxifying system for barbiturates; thus, in the presence of ethanol, the high levels of barbiturates being taken (due to the toler-ance) are now toxic (the system that breaks down the drug has been inhibited). Ethanol does not increase absorption of the drug from the digestive tract, nor does acetaldehyde, ethanol's oxidation product, react with barbiturates. Barbiturate action is not affected by energy production (acetyl-CoA). The ethanol inhibition of cytochrome P450 systems is also not due to ethanol's dehydration effect.

3. Considering the patient in question 2, one night, the patient consumes a large amount of alcohol. He continues to take his usual dose of seizure medication. He dies that night in his sleep. This is due to which of the following? (A) Ethanol stimulating barbiturate absorption by the stomach (B) Ethanol inhibition of a cytochrome P450 system (C) Acetaldehyde reacting with the drug, creating a toxic compound (D) Acetyl-CoA production leads to enhanced energy pro-duction, which synergizes with barbiturate action (E) Ethanol's dehydration effect leads to toxic concen-trations of the seizure medication in the blood

(D) Ethanol's carbons are lost as carbon dioxide before a gluconeogenic precursor can be generated Ethanol is converted to acetaldehyde, which is further oxidized to acetic acid and is then activated to acetyl-CoA. The acetyl-CoA enters the TCA cycle to gen-erate energy, and two carbons are lost for each turn of the cycle as CO2. Thus, ethanol cannot provide carbons for the net synthesis of glucose. Ethanol is not converted. to acetone, nor is it directly lost in the urine. Ethanol is primarily oxidized in the liver, and its carbons can-not be used for the biosynthesis of lysine, which is an essential amino acid for humans. Ethanol oxidation is outlined in the fi gure below.

3. Ethanol ingestion is incapable of supplying carbons for gluconeogenesis. This is due to which of the fol-lowing? (A) Ethanol is converted to acetone, and the carbons are lost during exhalation (B) Ethanol is lost directly in the urine (C) Ethanol cannot enter the liver, where gluconeogen-esis predominantly occurs (D) Ethanol's carbons are lost as carbon dioxide before a gluconeogenic precursor can be generated (E) Ethanol is converted to lysine, which is strictly a ketogenic amino acid

(D) Glucose-6-phosphatase The child has Von Gierke disease, glycogen storage disease type I, a lack of glucose-6-phosphatase. In such a disorder, glucose-6-phosphate, whether produced from glycogen degradation or gluconeogenesis, cannot be dephospho-rylated for glucose export, and the liver cannot main-tain blood glucose levels. The small amount of glucose which is exported (10% of the expected) is derived from the activity of debranching enzyme, which hydrolyzes an α-1,6-glucose linkage, which produces free glu-cose. The hepatomegaly arises due to excess glycogen in the liver (glucose-6-phosphate will activate glycogensynthase D), as does the increase in kidney size. A pic-ture of a 25-month-old untreated child with this disor-der is shown below. A lack of glycogen synthase would not lead to hepatomegaly, while a lack of branching enzyme leads to a different glycogen storage disease, with very different symptoms. A lack of debranching activity would not lead to hepatomegaly and would allow more glucose release than is observed through the normal action of glycogen phosphorylase. A defect in fructose-1,6-bisphosphatase would impair gluconeo-genesis, but should not affect the ability of glycogen to be degraded to raise blood glucose levels.

4. A 6-month-old infant was brought to the pediatri-cian due to fussiness and a tender abdomen. The child seemed to do well until the time between feeding was increased to more than 3 h. The baby always seemed hungry and irritable if not fed frequently. Upon exami-nation, hepatomegaly and enlarged kidneys were noted, and blood work showed fasting hypoglycemia. Subse-quent laboratory analysis demonstrated that in response to a glucagon challenge, only about 10% of the normal amount of glucose was released into circulation, which signifi cantly contributed to the fasting hypoglycemia. Which enzyme defect in the patient is the most likely? (A) Glycogen synthase (B) Branching enzyme (C) Debranching enzyme (D) Glucose-6-phosphatase (E) Fructose-1,6-bisphosphatase

(C) Transketolase The patient is experi-encing the symptoms of vitamin B1 (thiamine) defi ciency. Ethanol blocks thiamine absorption from the gut, so in the United States, one will usually only see a B1 defi -ciency in chronic alcoholics. One assay for B1 defi ciency is to measure transketolase activity (which requires B1as an essential cofactor) in the presence and absence of added B1. If the activity level increases when B1 is added, a vitamin defi ciency is assumed. None of the other enzymes listed (transaldolase, aldolase, β-ketothiolase, and acetylcholine synthase) require B1 as a cofactor, and, thus, could not be used as a measure of B1 levels. A reaction catalyzed by transketolase is shown below (note the breakage of a carbon-carbon bond, and then the synthesis of a carbon-carbon bond to generate the product of the reaction).

4. A chronic alcoholic presents to the emergency department with nystagmus, peripheral edema, pulmonary edema, ataxia, and mental confusion. The physician orders a test to determine if there is a vitamin defi ciency. An enzyme used for such a test can be which of the following? (A) Transaldolase (B) Aldolase (C) Transketolase (D) β-ketothiolase (E) Acetylcholine synthase

(A) The E1 subunit of pyruvate dehydrogenase Lactic acidosis can result from a defect in an enzyme that metabolizes pyruvate (primarily pyruvate dehydrogenase and pyruvate carboxylase). The pyru-vate dehydrogenase complex consists of three major catalytic subunits, designated E1, E2, and E3. The E1 subunit is the one that binds thiamine pyrophos-phate and catalyzes the decarboxylation of pyruvate. The gene for the E1 subunit is on the X chromosome, so defects in this subunit are inherited as X-linked diseases, which primarily affects males. Since this is the second male child to have these symptoms, it is likely that the mother is a carrier for this disease. The pattern of inheritance distinguishes this diagnosis from that of an E2 or E3 defi ciency. In addition, an E3 defi ciency would affect more than pyruvate metab-olism, as this subunit is shared with other enzymes that catalyze oxidative decarboxylation reactions, and other metabolites would also be accumulating. Defects in citrate synthase and malate dehydrogenase would not lead to severe lactic acidosis and would not be male-specifi c disorders. As an example, the three subunits of α-ketoglutarate dehydrogenase are shown below.

4. A family that had previously had a newborn boy die of a metabolic disease has just given birth to another boy, small for gestational age, and with low Apgar scores. The child displayed spasms a few hours after birth. Blood analysis indicated extremely high levels of lactic acid. Analysis of cerebrospinal fl uid showed elevated lactate and pyruvate. Hyperalaninemia was also observed. The child died within 5 days of birth. The biochemical defect in this child is most likely which of the following? (A) The E1 subunit of pyruvate dehydrogenase (B) The E2 subunit of pyruvate dehydrogenase (C) The E3 subunit of pyruvate dehydrogenase (D) Citrate synthase (E) Malate dehydrogenase

(C) Deletions in the locus control region of the β-globin gene cluster The girl is expressing HPFH (hereditary persistence of fetal hemoglobin). This can come about by deletions on the locus control region of the β-globin gene cluster (since fetal hemo-globin is α2γ2, and adult hemoglobin is α2β2, mutations in the locus control region of the α-gene cluster will not affect fetal hemoglobin synthesis). A general loss of transcription factors would not lead to increased tran-scription of the γ-chains, nor would a general increase in all transcription factor expression in the cell. While inappropriate looping may help to lead to γ-globin gene expression, the looping needs to be modulated by tran-scription factors for gene expression to occur.

4. A healthy teenage girl has come to her pediatrician for a presports physical. Results of hemoglobin electrophore-sis indicated an elevation of fetal hemoglobin. This can come about via which of the following mechanisms? (A) Overall increased expression of all transcription factors (B) Overall reduced expression of all transcription factors (C) Deletions in the locus control region of the β-globin gene cluster (D) Deletions in the locus control region of the α-globin gene cluster (E) Inappropriate looping of chromosomal DNA, allow-ing transcription of previously inaccessible genes to occur

(C) Acyl-CoA dehydrogenase The acylCoA dehydrogenases catalyze the fi rst step of the fatty acid oxidation spiral in that these enzymes create a carbon-carbon double bond between carbons 2 and 3 of the fatty acyl-CoA, generating an FADH2 in the process. The FADH2 then donates its electrons to the electron transfer fl avoprotein (ETF), which then transfers the electrons to coenzyme Q (via the ETF:CoQ oxidoreductase). A lack of the oxidoreductase activity will lead to an accumulation of mitochondrial FADH2, depleting FAD levels, and reducing the activity of the acyl-CoA dehydrogenases. The lack of FAD does not directly inhibit the β-ketothiolase or enoyl-CoA dehydrogenase steps, nor does it affect the activity of the carnitine acyltransferases. The fi gure below shows the normal transport of electrons from FADH2 to coenzyme Q when the FADH2 is generated by the acyl-CoA dehydrogenases.

4. An inactivating mutation in the ETF:CoQ oxidoreductase will lead to an initial inhibition of which of the following enzymes in fatty acid oxidation? (A) Carnitine acyltransferase I (B) Carnitine acyltransferase II (C) Acyl-CoA dehydrogenase (D) Enoyl-CoA dehydrogenase (E) β-keto thiolase

(E) Medium chain acyl-CoA dehydrogenase The child has MCAD (medium-chain acyl-CoA dehydrogenase) defi ciency, an inability to completely oxidize fatty acids to carbon dioxide and water. With an MCAD defi ciency, gluconeogenesis is impaired due to a lack of energy from fatty acid oxidation, and an inability to fully activate pyruvate carboxylase, as acetyl-CoA activates pyruvate carboxylase, and acetyl-CoA production from fatty acid oxidation is greatly reduced. In an attempt to generate more energy, medium-chain fatty acids are oxidized at the ω ends to generate the dicarboxylic acids seen in the question (see the fi gure below for an overview of ω oxidation). The fi nding of such metabolites (dicarboxylic acids) in the blood is diagnostic for MCAD defi ciency. If there were mutations in any aspect of carnitine metabolism, there would be no oxidation of fatty acids (the fatty acids would not be able to enter the mitochondria), and the dicarboxylic acids (which are byproducts of fatty acid metabolism) would not be observed. Similarly, a mutation in the fatty acyl-CoA synthetase (the activating enzyme, converting a free fatty acid to an acyl-CoA) would also result in a lack of fatty acid oxidation, as fatty acids are not able to enter the mitochondria in their free (nonactivated) form.

5. A 3-month-old child had her fi rst ear infection and was feeding poorly due to the ear pain. One morning the parents found the child in a nonresponsive state and rushed her to the emergency department. A blood glucose level was 45 mg/dL, and upon receiving intravenous glucose the child became responsive. Further blood analysis displayed the absence of ketone bodies, normal levels of acyl- carnitine, and the presence of the following unusual carboxylic acids shown below. The enzymatic defect in this child is most likely in which of the following enzymes? (A) Fatty acyl-CoA synthetase (B) Carnitine translocase (C) Carnitine acyltransferase I (D) Carnitine acyltransferase II (E) Medium chain acyl-CoA dehydrogenase

(C) The E3 subunit of pyruvate dehydrogenase The child is defective in a variety of oxidative decarboxylation reactions (pyruvate dehy-drogenase, leading to a buildup of lactate and pyruvate; α-ketoglutarate dehydrogenase, leading to the buildup of α-ketoglutarate; and branched-chain α-ketoacid dehydrogenase, leading to a buildup of many of the other metabolites). Enzymes, which catalyze oxidative decarboxylation reactions, contain three catalytic sub-units, E1, E2, and E3 (see the fi gure in the answer to the previous question). E3 subunit, which contains the dihydrolipoyl dehydrogenase activity, is common among these enzymes. Thus, a mutation in E3 would render all of these enzymes inoperable, leading to a buildup of the α-ketoacid precursors. Defects in citrate synthase or malate dehydrogenase would not lead to the buildup of these α-ketoacids.

5. A 3-month-old girl developed lactic acidemia. Blood analysis also indicated elevated levels of pyru-vate, α-ketoglutarate, and branched-chain amino acids. A urinalysis showed elevated levels of lactate, pyruvate, α-hydroxyisovalerate, α-ketoglutarate, and α-hydroxybutyrate. A likely mutation in which of the following proteins would lead to this clinical fi nding? (A) The E1 subunit of pyruvate dehydrogenase (B) The E2 subunit of pyruvate dehydrogenase (C) The E3 subunit of pyruvate dehydrogenase (D) Citrate synthase (E) Malate dehydrogenase

(E) Branching enzyme

5. A 4-month-old infant is seen by the pediatrician for fail-ure to thrive. Examination shows distinct hepatosple-nomegaly. Lab results show elevated transaminases and bilirubin, suggestive of liver failure. The boy dies shortly thereafter, and upon autopsy, precipitated carbohydrate was found throughout the liver. The boy most likely had a mutation in which of the following enzymes? (A) Glycogen phosphorylase (B) Debranching enzyme (C) Glycogen synthase (D) β-glucosidase (E) Branching enzyme

(C) Fructose-6-phosphate and glyceraldehyde-3-phos-phate In the absence of NADP+, the oxidative steps of the HMP shunt pathway are nonfunctional, so only the nonoxidative steps will occur. In addition, PFK-1 has been made nonfunctional, such that glyceraldehyde-3-phosphate (G3P) cannot be produced from either fructose-6-phosphate (F6P) or glucose-6-phosphate (G6P). In order to generate ribose-5-phosphate (R5P) under these conditions, both F6P and G3P need to be provided. These two substrates will react, using transketolase as a substrate, to generate erythrose-4-phosphate (E4P) and xylulose-5-phosphate (X5P, step 1 in the fi gure below). The X5P will be epimerized to ribulose-5-phosphate (Ru5P, step 2 in the fi gure below), and then isomerized to R5P (step 3 in the fi gure below). Glucose-6-phosphate can-not be used as a substrate because it cannot be con-verted to G3P (due to the block in PFK-1). Pyruvate cannot be used as a substrate in extracts of red blood cells because such cells do not have pyruvate carboxy-lase, so the pyruvate cannot be converted to either F6Por G3P.

5. A researcher is studying the HMP shunt pathway in extracts of red blood cells, in the absence of NADP+, and in which PFK-1 has been chemically inactivated. Which carbon substrates are required to generate ribose-5-phosphate in this system? (A) Glucose-6-phosphate and sedoheptulose-7-phos-phate (B) Glucose-6-phosphate and glyceraldehyde-3-phosphate (C) Fructose-6-phosphate and glyceraldehyde-3-phos-phate (D) Fructose-6-phosphate and pyruvate (E) Glucose-6-phosphate and pyruvate

(B) Inhibition of apolipoprotein B translation The cholesterol tag on the dsRNA allowed cells to take up the dsRNA, which was processed by intracellular ribonucleases to make a specifi c silenc-ing RNA for the apolipoprotein B mRNA. Binding of the processed dsRNA to the apoB mRNA will lead to either the destruction of the mRNA or the blockage of translation of the mRNA. In either event, there will be a reduction in apoB translation such that cells can no longer produce apoB100 or apoB48. The dsRNA does not affect the transcription of the apoB gene, nor does it interfere with apoB folding once it becomes transcribed and translated. The dsRNA does not affect the turnover of the apoB protein, nor does it edit the apoB mRNA (other systems in the cell will do that).

5. In a study with mice exhibiting hypercholesterolemia, cholesterol was affi xed to double-stranded RNA, which targeted the dsRNA to enter cells through cholesterol diffusion through the plasma membrane. The dsRNA was targeted to bind to mRNA that encoded the apoli-poprotein B gene and resulted in a lowering of circulat-ing cholesterol levels. This result occurs due to which of the following? (A) Inhibition of apolipoprotein B transcription (B) Inhibition of apolipoprotein B translation (C) Inhibition of apolipoprotein B folding (D) Enhanced degradation of apolipoprotein B (E) RNA editing of the apolipoprotein B mRNA

(C) Mutations in mitochondrial tRNA Both families are suffering from mitochondrial dis-eases. Family 1 has MERRF (myoclonic epilepsy with ragged red fi bers) while family 2 has MELAS (mito-chondrial myopathy, encephalopathy, lactic acidosis, and stroke). Both disorders are due to mutations in a mitochondrially encoded tRNA. MERRF is a muta-tion in tRNAlys, whereas MELAS has a mutation in a tRNAleu gene. In both cases, the tRNA mutations inter-fere with protein synthesis within the mitochondria, leading to a reduction of functional proteins necessary for various aspects of oxidative phosphorylation. These disorders are not due to mutations in nuclear encoded genes (which eliminates all of the other answers).

6. A human geneticist is studying two different families. In one family, all of the children of a mildly affected mother display myoclonic epilepsy, developmental display, and abnormal muscle biopsy (ragged red fi bers). In the other family, the three children of an affected woman endure strokelike episodes and a mitochondrial myopathy. The common link between these two diseases is which of the following? (A) Mutations in pyruvate dehydrogenase complex (B) Mutations in cytoplasmic tRNA (C) Mutations in mitochondrial tRNA (D) Mutations in malate dehydrogenase (E) Mutations in pyruvate carboxylase

(C) CREB CREB (cyclic AMP response element binding protein) is a transcription factor that is activated by protein kinase A and that regulates, in part, the expression of phosphoenolpyruvate carboxykinase (PEPCK), a necessary protein for gluconeogenesis. Since the patient is anorexic, her blood glucose levels are being maintained primarily by gluconeogenesis, and the enzymes for that pathway need to be upregulated. The release of glucagon and epinephrine, both of which would be elevated in this patient, leads to the activation of protein kinase A and an increase in gene transcription for those genes regulated by CREB. Under these conditions, protein synthesis will be limited, so factors necessary for protein synthesis would not be generally activated (eIF4, eEF2, and ribosomal protein S6). Neither glucagon nor epineph-rine works through steroid hormone receptors (they both utilize serpentine receptors on the cell membrane).

6. A woman with a BMI of 16.5 visits her family physician because she always feels tired. The history indicates that the woman is always on a diet, exercises over 3 h/day, and perceives herself as fat. Blood work indicates that her glucose levels are only slightly below normal under fasting conditions. The patient's ability to maintain her blood glucose levels near normal results, in part, from activation of which of the following proteins? (A) eIF4 (B) eEF2 (C) CREB (D) Ribosomal subunit S6 (E) Steroid hormone receptor7

(C) Adenylate cyclase If adenylate cyclase is defective, glucagon cannot initiate the activation of glycogenolysis and inhibition of glycolysis in the liver (cAMP levels will not increase, and PKA will stay inac-tive). Under such conditions, only the allosteric effec-tors in liver will be active, and there is no activator of glycogen phosphorylase b. When the hypoglycemia is severe enough, epinephrine release, working through its α-receptors, will activate phospholipase C, leading to calcium release. The increased calcium can activate phosphorylase kinase, which will activate phosphory-lase, but fasting hypoglycemia will still occur. Defects in liver PFK-1 or glucokinase will not affect glycogenoly-sis or gluconeogenesis. Defects in liver galactokinase or fructokinase will not allow for metabolism of galactose or fructose, but do not affect the ability of the liver to degrade glycogen, or perform gluconeogenesis from other precursors.

6. An inactivating mutation in which of the following pro-teins can lead to fasting hypoglycemia? (A) Liver PFK-1 (B) Liver glucokinase (C) Adenylate cyclase (D) Galactokinase (E) Fructokinase

(C) Insuffi cient energy for gluconeogenesis Defects in fatty acid oxidation deprive the liver of energy when fatty acids are the major energy source (such as during exercise, or a fast). Because of this, there is insuffi cient energy to synthesize glucose from gluconeogenic precursors (it requires 6 moles of ATP to convert 2 moles of pyruvate to 1 mole of glucose). Acyl-carnitines and dicarboxylic acids have no effect on the enzymes of gluconeogenesis, nor do they hinder the ability of the red blood cell to utilize glucose through the glycolytic pathway. Additionally, acetyl-CoA levels are low due to the lack of complete fatty acid oxidation and pyruvate carboxylase, a key gluconeogenic enzyme, is not fully activated. This also contributes to the reduced gluconeogenesis observed in patients with MCAD defective.

6. Regarding the child described in question 5, why were fasting blood glucose levels so low? (A) Acyl-carnitine inhibition of gluconeogenesis (B) Dicarboxylic acid inhibition of gluconeogenesis (C) Insuffi cient energy for gluconeogenesis (D) Dicarboxylic acid inhibition of glycogen phosphorylase (E) Reduction of red blood cell production of lactate for gluconeogenesis

(D) Xylulose-5-phosphate In order for ribose-5-phosphate to be converted to glucose-6-phos-phate, the nonoxidative reactions of the HMP shunt pathway must be used (the oxidative steps are not revers-ible reactions). In order for this to occur, the ribose-5-phosphate is isomerized to ribulose-5-phosphate, which is then epimerized to xylulose-5-phosphate (steps 1 and 2 in the fi gure on page 123). R5P and X5P then initi-ate a series of reactions utilizing transketolase (step 3 in the fi gure on page 123) and transaldolase (step 4 in the fi gure on page 123) to generate fructose-6-phosphate, which can be isomerized to glucose-6-phosphate (step 5 in the fi gure on page 123). Glyceraldehyde-3-phosphate is also formed during this series of reactions, which then goes back to fructose-6-phosphate production. Pyruvate, oxaloacetate, 1,3-bisphosphoglycerate, and 6-phosphogluconoate are not obligatory intermediates in this conversion.

6. Which one of the following is an obligatory intermediate in the conversion of ribose-5-phosphate to glucose-6-phosphate? (A) Pyruvate (B) 1,3-bisphosphoglycerate (C) Oxaloacetate (D) Xylulose-5-phosphate (E) 6-phosphogluconate

a. Sucrose Glycosides are formed by condensation of the aldehyde or ketone group of a carbohydrate with a hydroxyl group of another compound. Other linked groups (aglycones) include steroids with hydroxyl groups (eg, cardiac glycosides such as digitalis or ouabain) or other chemicals (eg, antibiotics such as streptomycin). Sucrose (α-D-glucose-β-1 → 2- D-fructose), maltose (α-D-glucose-α-1 → 4- D-glucose), and lactose (α-D-galactose-β-1 → 4-D-glucose) are important disaccharides. Fructose is among several carbohydrate groups known as ketoses because it possesses a ketone group. The ketone group is at carbon 2 in fructose, and its alcohol group at carbon 1 (also at carbon 6) allows ketal formation to produce pyranose and furanose rings as with glucose. Most of the fructose found in the diet of North Americans is derived from the disaccharide sucrose (common table sugar). Sucrose is cleaved into equimolar amounts of glucose and fructose in the small intestine by the action of the pancreatic enzyme sucrase. Deficiency of sucrase can also cause chronic diarrhea. Hereditary fructose intolerance (MIM*229600) is caused by deficiency of the liver enzyme aldolase B, which hydrolyzes fructose 1-phosphate.

66. Diarrhea from infection or malnutrition is the world's most prevalent killer of children. A 2-month-old Caucasian girl develops chronic diarrhea and liver inflammation in early infancy when the mother begins using formula that includes corn syrup. Evaluation of the child demonstrates sensitivity to fructose in the diet. Which of the following glycosides contains fructose and therefore should be avoided when feeding or treating this infant? a. Sucrose b. Ouabain c. Lactose d. Maltose e. Streptomycin

c. Cellulose Cellulose, the most abundant compound known, is the structural fiber of plants and bacterial walls. It is a polysaccharide, consisting of chains of glucose residues linked by β1 → 4 bonds. Since humans do not have intestinal hydrolases that attack β1 → 4 linkages, cellulose cannot be digested but forms an important source of "bulk" in the diet. Lactose is a disaccharide of glucose and galactose found in milk. Amylose is an unbranched polymer of glucose residues in β1 → 4 linkages. Glycogen is a branched polymer of glucose with both β1 → 4 and β1 → 6 linkages. Maltose is a disaccharide of glucose, which is usually the breakdown product of amylose.

67. Which of the following carbohydrates would be most abundant in the diet of strict vegetarians? a. Amylose b. Lactose c. Cellulose d. Maltose e. Glycogen

b. NAD+ In humans, ethanol is cleared from the body by oxidation catalyzed by two NAD+-linked enzymes: alcohol dehydrogenase and acetaldehyde dehydrogenase (eliminating answers a, c-e). These enzymes act mainly in the liver to convert alcohol to acetaldehyde and acetate, respectively. In chronic alcoholics, alcohol dehydrogenase may be elevated somewhat. The NADH level is significantly increased in the liver during oxidation of alcohol, owing to the consumption of NAD +. This leads to a swamping of the normal means of regenerating NAD +. Thus, NAD + becomes the rate-limiting factor in oxidation of excess alcohol.

68. Alcohol abuse affects over 13% of adults in the United States, costing more health care dollars ($184 billion annually) than cancer ($107 billion) or obesity ($100 billion) in some studies. In 2001, 47% of those between ages 12 and 20 years reported drinking, 30% of these to binge drinking in the last month. Chronic alcoholics require more ethanol than do nondrinkers to become intoxicated because of a higher level of a specific enzyme. However, independent of specific enzyme levels, the availability of what other substance is rate-limiting in the clearance of ethanol? a. NADH b. NAD+ c. FADH d. FAD + e. NADPH

d. Decreased ATP generation and increased glucose utilization on changing from aerobic to anaerobic metabolism The exposure of tissues to chronic hypoxia makes them rely more on anaerobic metabolism for the generation of energy as ATP and other high-energy phosphates. Most tissues except for red blood cells can metabolize glucose under anaerobic or aerobic conditions (red blood cells do not have mitochondria for electron transport and must rely on other tissues to generate glucose back from lactate). In most tissues, a switch from aerobic to anaerobic metabolism greatly increases glucose utilization and decreases energy production. (Increased glucose utilization under anaerobic conditions in bacteria is known as the Pasteur effect after its discoverer.) Under aerobic conditions, the cell can produce a net gain, in moles of ATP formed per mole of glucose utilized that can be as high as 18 times that produced under anaerobic conditions. Thus, the cell generates more energy and requires less glucose under aerobic conditions. Such increased ATP concentrations, together with the release of citrate from the citric acid cycle under aerobic conditions, allosterically inhibit the key regulatory enzyme of the glycolytic pathway, phosphofructokinase. Decreased phosphofructokinase activity decreases metabolism of glucose by glycolysis.

69. In lung diseases such as emphysema or chronic bronchitis, there is chronic hypoxia that is particularly obvious in vascular tissues such as the lips or nail beds (cyanosis). Certain genetic diseases such as α1-antitrypsin deficiency (MIM*107400) predispose to emphysema, as do environmental exposures such as cigarette smoking or asbestos. Poorly perfused areas exposed to chronic hypoxia have decreased metabolic energy for tissue maintenance and repair. Which of the following is an important reason for this? a. Increased hexokinase activity owing to increased oxidative phosphorylation b. Increased ethanol formation from pyruvate on changing from anaerobic to aerobic metabolism c. Increased glucose utilization via the pentose phosphate pathway on changing from anaerobic to aerobic metabolism d. Decreased ATP generation and increased glucose utilization on changing from aerobic to anaerobic metabolism e. Decreased respiratory quotient on changing from carbohydrate to fat as the major metabolic fuel

(B) Oxidative damage to red blood cell membranes The man has glucose-6-phosphate dehydrogenase defi ciency and is incapable of regen-erating reduced glutathione to protect red blood cell membranes from oxidative damage. In the presence of a strong oxidizing agent (the new drug the patient was taking), the red cell membranes undergo oxidative damage and the red cell bursts, leading to hemolytic anemia. This is all due to a lack of protective glutathi-one in the membrane. As the red cell lacks a nucleus, the cell cannot induce new gene synthesis. The drug the patient was taking does not induce ion pores in red cell membranes or inhibit the HMP shunt pathway. It also does not cause oxidative damage to bone marrow. The drugs to avoid while prescribing for a patient with a G6PDH defi ciency include primaquine, dapsone, nitro-furantoin, and sulfonylurea. The reduced (Panel A) and oxidized (Panel B) forms of glutathione are indicated to the side.

7. A 23-year-old man of Mediterranean descent was recently prescribed ciprofl oxacin to treat a urinary tract infection. After 2 days on the drug, the patient was feel-ing worse, and weak, and went to the emergency depart-ment. He was found to have hemolytic anemia. This most likely resulted due to which of the following? (A) Induction of red blood cell cytochrome P450s, lead-ing to membrane damage (B) Oxidative damage to red blood cell membranes (C) Drug induced ion pores in the red blood cell membrane (D) Drug induced inhibition of the HMP shunt pathway (E) Oxidative damage to bone marrow, interfering with red blood cell production

(E) Carnitine transporter The child has a mutation in the enzyme which transports carnitine into liver and muscle cells, leading to a primary carnitine defi ciency. The carnitine stays in the blood and is eventually lost in the urine (the same carnitine transporter is required to recover the carnitine from the urine in the kidney). Since the liver is carnitine defi cient, ketone body production is minimal at all times, even during a fast (thus, the lack of baseline ketone bodies in the circulation under these conditions). Fatty acids will rise in circulation, as they cannot be stored in the cells as acylCoA. The liver shows evidence of triglyceride formation as the acyl-CoA cannot be degraded, and acyl-CoA accumulates within the cytoplasm, leading to triglyceride formation. A defect in carnitine acyl transferase 1 would lead to elevated levels of carnitine in the circulation. A defect in carnitine acyltransferase II would lead to elevated levels of acyl-carnitine in the circulation (since the acyl group cannot be removed from the carnitine). The lack of circulating dicarboxylic acids indicates that the defect is not in MCAD (medium-chain acyl-CoA dehydrogenase). A defect in hormone sensitive lipase would show a decrease in free fatty acid levels, rather than the increase observed in the patient.

7. A 6-month-old child presents to the physician in a hypotonic state. The child has previously had a number of hypoglycemic episodes, at which times blood glucose levels were between 25 and 50 mg/dl. Blood work shows normal levels of ketone bodies (not elevated) during hypoglycemic episodes. Carnitine levels in the blood were, however, below normal. Free fatty acid levels were elevated in the blood, however acyl-carnitine levels were normal. Dicarboxylic acid levels were non-detectable in the blood. A liver biopsy shows elevated levels of triglyceride. A likely enzymatic defect is which of the following? (A) Carnitine acyltransferase I (B) Carnitine acyltransferase II (C) Medium chain acyl-CoA dehydrogenase (D) Hormone sensitive lipase (E) Carnitine transporter

(A) Impaired gene transcription Indi-viduals with Li-Fraumeni syndrome inherit a mutated copy of p53, the product of the TP53 gene (on chromo-some 17). p53 is a transcription factor whose major job is to monitor the health of DNA; if DNA alterations are found, p53, acting as a transcription factor, will initiate new gene transcription to arrest the cell cycle until the DNA damage is repaired and to also induce genes neces-sary for DNA repair. If the DNA cannot be repaired, p53will initiate gene transcription leading to cellular death (apoptosis). In the absence of p53 activity, damaged DNA will be replicated, which increases the probability of errors, eventually causing a mutation that leads to a can-cer. Thus, the initial inactivating event is impaired gene transcription by p53, which is the trigger for all other events that follow. Mutations in p53 do not lead, directly, to enhanced gene transcription (this may occur as a result of secondary mutations, but not directly from the mutations in p53) or to alterations in protein synthesis. p53 mutations also do not alter chromosome structure.

7. A long-standing patient of yours has developed multi-ple tumor types during his life (41 years old). You have diagnosed him as having a specifi c syndrome involving p53. The multiple cancers that result from this syndrome are primarily due to which of the following initial direct effects of the inherited mutation? (A) Impaired gene transcription (B) Enhanced gene transcription (C) Impaired protein synthesis (D) Enhanced protein synthesis (E) Altered chromosomal structure

(D) Additional thiamine in the diet Leigh syndrome can result from a defi ciency of pyruvate dehydrogenase (PDH) activity, leading to lactic acido-sis. In some cases, the enzyme has a reduced affi nity for thiamine pyrophosphate, a required cofactor for the enzyme. Adding thiamine to the diet may over-come this deficiency by raising the concentration of thiamine pyrophosphate such that it will bind to the altered enzyme. Increasing the carbohydrate in the diet will make the disease worse, as more pyru-vate would be generated due to the increase in the glycolytic rate. Vitamin B6 does not play a role in gly-colysis or the PDH reaction. Lipoamide is a required cofactor for the PDH reaction, so reducing lipoam-ide would have an adverse effect on the activity of PDH. Decreasing the fat content of the diet may be harmful, particularly if the calories are replaced as carbohydrate.

7. A toddler has been diagnosed with a mild case of Leigh syndrome. One possible treatment is which of the fol-lowing? (A) Increased carbohydrate diet (B) Additional B6 in the diet (C) Decreased lipoamide in the diet (D) Additional thiamine in the diet (E) Decreased fat diet

1,000,000 One active PKA can acti-vate in 1 s 100 molecules of phosphorylase kinase. Each phosphorylase kinase can, in 1 s activate 100 molecules of glycogen phosphorylase (so at this point we have 100 times 100 active molecules of phosphorylase, or 10,000 active phosphorylase molecules). Each active phosphorylase molecule can release 100 glucose residues per second from glycogen, and since there are 10,000 active phosphorylase molecules, 1,000,000 molecules of glucose are released per second once a single molecule of PKA has been activated. This is an example of cascade amplifi cation, in which an increase in activity of just one molecule at the top of the cascade can result in a large response further down the cascade

7. If the turnover number of all enzymes involved in gly-cogen metabolic regulation and activity is 100 reac-tions per second, how many glucose molecules could be removed from glycogen in 1 s upon activation of one molecule of protein kinase A (PKA)? (A) 100 (B) 1,000 (C) 10,000 (D) 100,000 (E) 1,000,000

c. Acetaldehyde The principal pathway for hepatic metabolism of ethanol is thought to be oxidation to acetaldehyde in the cytoplasm by alcohol dehydrogenase. Acetaldehyde is then oxidized, chiefly by acetaldehyde dehydrogenase within the mitochondrion, to yield acetate. Acetone, methanol, hydrogen peroxide, and glycerol do not appear in this biodegradation pathway, eliminating answers a-b, d-e. The genetic variations of acetaldehyde dehydrogenase have few phenotypic effects aside from sensitivity to alcoholic beverages and are extremely common in the affected populations. These characteristics qualify acetaldehyde dehydrogenase variation as an example of enzyme polymorphism. Alcohol sensitivity is less of a factor in alcohol overdose or poisoning, where large intake overwhelms the liver's capacity to metabolize about ounce per hour. Symptoms of coma and impending respiratory arrest mandate emergency room treatment to support respiration, avoid aspiration, reverse hypoglycemia (with IV dextrose), and restore depleted thiamine.

70. An 18-year-old Korean college student attends a fraternity party and is embarrassed because he becomes flushed and sick to his stomach after his first drink of alcohol. He learns that this reaction is due to genetic variation in some Asians and Native Americans that affects metabolism of a metabolite of alcohol. Which of the following is the variably degraded metabolite? a. Methanol b. Acetone c. Acetaldehyde d. Hydrogen peroxide e. Glycerol

b. Glycogen and lactate to glucose in liver Glycolysis in muscle produces lactate, which must be converted to glucose by liver or kidney via the Cori cycle (incorrect answers a, c, and e). Defects in liver glycogen metabolism therefore impair glucose 6-phosphate production or gluconeogenesis (alanine is also a substrate for gluconeogenesis— incorrect answer d) with resulting hypoglycemia and liver glycogen storage with or without toxicity (cirrhosis). Defects in muscle glycogen metabolism impair contraction (cramps, fatigue) with decreased serum lactate production during exercise and muscle glycogen accumulation (progressive weakness and atrophy). Glucose 1-phosphate is the first intermediate in the conversion of glycogen to glucose. The enzyme glycogen phosphorylase catalyzes this first step. The second intermediate, glucose 6-phosphate, is subsequently converted to glucose by the enzyme glucose-6-phosphatase. This enzyme is found only in the liver and kidney; thus, these are the only tissues able to break down glycogen for use by other tissues. In tissues such as muscle, glycogen can be broken down to glucose 6-phosphate but can only be used in the cell in which it was produced.

71. A 6-month-old Caucasian girl is hospitalized for evaluation of short stature, enlarged liver, and intermittent lethargy/irritability that, on a recent emergency room visit, was accompanied by low blood glucose (hypoglycemia). In the same ward is a 6-year-old African American boy who is being evaluated for severe muscle cramping that raised suspicions for sickle cell disease. The family history is unremarkable for these children; each has a normal sibling and normal parents. Both children are given diagnoses of glycogen storage disease with the infant girl having type I (MIM*232200) affecting liver and the older boy type V (McArdle disease, MIM*232600) affecting mainly muscle. Which of the following conversions explains the difference in these presentations? a. Glycogen to lactate in liver b. Glycogen and lactate to glucose in liver c. Glycogen to glucose in muscle d. Glycogen to alanine in muscle e. Glycogen to glucose 6-phosphate in liver

d. Ribose and NADPH Glucose-6-phosphate dehydrogenase (G6PD) is the first enzyme of the pentose phosphate pathway, a pathway that metabolizes glucose to produce ribose and NADPH (eliminating answers a-c, e). Its deficiency (MIM*305900) is the most common enzymopathy, affecting 400 million people worldwide. It contrasts with glycolysis in its use of NADP rather than NAD for oxidation, its production of carbon dioxide, its production of pentoses (ribose, ribulose, xylulose), and its production of the highenergy compound PRPP (5-phosphoribosyl-1-pyrophosphate) rather than ATP. Production of NADPH by the pentose phosphate pathway is crucial for reduction of glutathione, which in turn removes hydrogen peroxide via glutathione peroxidase. Erythrocytes are particularly susceptible to hydrogen peroxide accumulation, which oxidizes red cell membranes and produces hemolysis. Stresses such as newborn adjustment, infection, or certain drugs can increase red cell hemolysis in G6PD-deficient individuals, leading to severe anemia, jaundice, plugging of renal tubules with released hemoglobin, renal failure, heart failure, and death. Since the locus encoding G6PD is on the X chromosome, the deficiency exhibits X-linked recessive inheritance with severe affliction in males and transmission through asymptomatic female carriers. Ribose 5-phosphate produced by the pentose phosphate pathway is an important precursor for ribonucleotide synthesis, but alternative routes from fructose 6-phosphate allow ribose synthesis in tissues without the complete cohort of pentose phosphate enzymes or with G6PD deficiency. The complete pentose phosphate pathway is active in liver, adipose tissue, adrenal cortex, thyroid, erythrocytes, testis, and lactating mammary gland. Skeletal muscle has only low levels of some of the enzymes of the pathway but is still able to synthesize ribose through fructose 6-phosphate.

72. A 3-day-old boy from an Iraqi refugee family exhibits severe tissue swelling (edema), rapid heart rate (tachycardia), enlargement of liver and spleen (hepatosplenomegaly), and jaundice of the eyes and skin. The couple had two prior female infants who are alive and well, and the wife relates that she lost a brother in infancy with severe hemolysis induced after a viral infection. Blood tests show hemoglobin of 5 g/dL (normal neonate mean 18.5) with reticulocyte count of 15% (normal 1.8-4.6). Exchange transfusion is attempted but the child suffers cardiac arrest. Postmortem assay of red cell enzymes confirms the suspected diagnosis of glucose- 6-phosphate dehydrogenase deficiency (MIM*305900), implying defective synthesis of which of the following compounds? a. Deoxyribose and NADP b. Glucose and lactate c. Lactose and NADPH d. Ribose and NADPH e. Sucrose and NAD

c. Fructose has a specific kinase in liver that allows bypass of phosphofructokinase. A special pathway for fructose metabolism (a specific fructokinase plus aldolase B and triokinase) is present in liver, kidney, and small intestine (incorrect answers a, b, d, and e). Foods high in sucrose (glucose-fructose) such as syrups, beverages, or diabetic substitutes yield high concentrations of fructose in the portal vein. Fructose is catabolized more rapidly than glucose by its specific fructokinase, bypassing hexokinase that is regulated by fasting and insulin. While providing a fuel for glycolysis, fructose also increases fatty acid, VLDL, and cholesterol-LDL production that are also side effects of diabetes mellitus due to the necessary shift to fat oxidation when intracellular glucose is less available. Recent data also suggest that fructose in soft drinks and other foods promotes insulin resistance and type II diabetes. For these reasons, the American Diabetic Association (www.diabetes. org) now suggests avoidance of fructose with the exception of naturally occurring fructose in fruits. Fructose metabolism begins when fructokinase catalyzes the phosphorylation of fructose to fructose 1-phosphate, which is then split to D-glyceraldehyde and dihydroxyacetone by aldolase B. Triokinase converts D-glyceraldehyde to glyceraldehyde 3-phosphate, which can be metabolized further by glycolysis or be condensed with dihydroxyacetone phosphate by adolase to form fructose 1,6-diphosphate, glucose 6-phosphate, and glucose through gluconeogenesis.

73. Which of the following best explains why fructose was formerly recommended for patients with diabetes mellitus? a. Fructose is a better substrate for hexokinase. b. Fructose stimulates residual insulin release. c. Fructose has a specific kinase in liver that allows bypass of phosphofructokinase. d. Fructose is phosphorylated and cleaved to triose phosphates, which cannot be used for gluconeogenesis. e. Hexokinase phosphorylates fructose in extrahepatic tissues, and its activity will not be affected by high glucose concentrations in diabetes.

b. cAMP-dependent protein kinase Glycogen synthesis and breakdown (glycogenolysis) are accomplished by separate pathways rather than reversible reactions. Glycogen synthase is active when dephosphorylated; glycogen phosphorylase is active when phosphorylated by a cyclic AMP-dependent protein kinase (incorrect answers a, c-e). These enzyme phosphorylations and dephosphorylations integrate glycogen synthesis/breakdown with food and glucose availability (refer to Fig. 8 in the High-Yield Facts). Glycogen storage diseases are a group of inherited enzyme deficiencies that cause accumulation of glycogen in liver, heart, or muscle. Glucose is the primary source of energy for most cells and excess glucose is stored as glycogen. Glycogen provides for short-term high-energy consumption in muscle and is an emergency energy supply for the brain. Glycogen stored in the liver can be converted back to glucose for release into the blood stream for use by other tissues. Deficiency of adenylyl kinase or cAMPdependent protein kinase can alter glycogenesis/glycogenolysis regulation and produce glycogen storage (refer to Table 3, High-Yield Facts).

74. A 2-month-old Caucasian girl is evaluated for hypoglycemia and lactic acidosis and noted to have an enlarged liver. Biopsy reveals stored glycogen, and a glycogen storage disease is suspected. Assay of usual glycogen enzyme deficiencies in the liver specimen is normal, so the metabolic consultant recommends assay of rarer enzyme deficiencies that influence glycogen metabolism. These enzymes would most likely include which of the following? a. Hexokinase b. cAMP-dependent protein kinase c. Glucose-6-phosphate dehydrogenase d. Phosphofructokinase e. Fructose-1,6-diphosphatase

a. Linear α1 → 4 linkages with branching α1 → 6 linkages Normal glycogen is composed of glucose residues joined in straight chains by β1 → 4 linkages. At 4- to 10-residue intervals, a branch of β1 → 4 linkages is initiated at a β1 → 6 linkage (incorrect answers b-e). Glycogen particles can contain up to 60,000 glucose residues. In the absence of the debrancher enzyme, glycogen can be degraded only to the branch points, inhibiting release of glucose into the serum and causing glycogen storage. As noted in High-Yield Facts, Table 3, Forbes/Cori or type 3 glycogen storage disease (MIM*232400) involves deficiency of debranching enzyme.

75. An 11-month-old infant with hypoglycemia and a palpable liver is evaluated for possible glycogen storage disease. The parents have immigrated from Russia, and report that the child's older brother was diagnosed with a "debrancher" enzyme deficiency with similar glycogen storage. This diagnosis would imply accumulation of glycogen with which type of glucose linkages? a. Linear α1 → 4 linkages with branching α1 → 6 linkages b. Linear α1 → 6 linkages with branching β1 → 4 linkages c. Linear β1 → 4 linkages only d. Linear β1 → 6 linkages only e. Branching β1 → 6 linkages only

e. Xylulose It is important to differentiate glucosuria due to diabetes mellitus or renal tubular problems from other sugars in the urine, like galactose in galactosemia or fructose/xylulose in essential fructosuria (all these are reducing sugars that are positive with Clinitest but only glucose is positive with the glucose oxidase reagent strip Dextrostix—the test is nonspecific like dipstick tests for hemoglobin/myoglobin). The uronic acid pathway, like the pentose phosphate pathway, provides an alternate fate for glucose without generating ATP. Glucose 6-phosphate is converted to glucose 1-phosphate and reacted with UTP to form the higher energy compound UDP-glucose. UDP-glucose is converted to UDP-glucuronic acid that is a precursor for glucuronide units in proteoglycan polymers. Unused glucuronic acid is converted to xylulose (incorrect answers a-d) and then to xylitol by a xylulose reductase, the enzyme deficiency in essential pentosuria (MIM*260800). In this "disease," which is better called a trait, excess xylulose is excreted into urine but causes no pathology. Pentoses (5-carbon sugars) are important in the pentose phosphate and uronic acid pathways, providing ribose for nucleic acid metabolism. The other sugars listed as options except for lactose are 6-carbon hexoses.

76. A 7-year-old obese Hispanic girl presents with dehydration after 3 days of vomiting and diarrhea. Her parents mention that a sibling was diagnosed with a type of diabetes that spilled sugar into the urine but did not need treatment. Urine reagent strip test for reducing sugars is strongly positive. The physician obtains a blood glucose level that is normal and a urine glucose oxidase test on the urine is also negative for glucose. Further analysis of the urine reveals a small amount of fructose and a large amount of an unidentified pentose that is most likely which of the following? a. Galactose b. Glucose c. Lactose d. Mannose e. Xylulose

a. Fructose 6-Carbon hexoses with a C1 aldehyde group and four asymmetric carbons can generate 16 isomeric forms including ketoses with a C2 ketone group (eg, fructose—incorrect answers b-e) and aldoses with a C1 aldehyde group (glucose, galactose, and mannose). All are reducing sugars that will give a positive reducing substance reaction by urine dipstick, including fructose from the sucrose in table sugar. Glucofuranose and glucopyranose are ring structures of glucose, with the majority of glucose in solution in the glucopyranose form.

77. A 2-week-old Caucasian boy returns to his pediatrician for evaluation of increased jaundice during his nursery stay and to follow up on a positive newborn screen for galactosemia (MIM*230400). Clinical assessment suggests the child's jaundice has resolved and he has a good weight gain with no feeding concerns. The pediatrician obtains a repeat newborn screen, but recalls that certain carbohydrates can be recognized as reducing substances in urine by the Clinitest reaction that produces a green color. The urine reagent strip test is positive and physician suggests a switch to nonlactose formula until the nurse mentions that she spilled the urine and performed the test on a tabletop where sugar for coffee had been spilled. Among the following C6 isomers of glucose, which is a ketose and reducing substance? a. Fructose b. Galactose c. Glucofuranose d. Glucopyranose e. Mannose

a. Fructose can bypass phosphofructokinase using a fructokinase in liver. Fructose is taken in by humans as sucrose, sucrose-containing syrups, and the free sugar. In liver, fructose is phosphorylated to fructose 1-phosphate by liver fructokinase, allowing it to bypass the ATP-regulated phosphofructokinase, yield glyceraldehyde and dihydroxyacetone phosphate by aldolase cleavage, and increase triglyceride/lipid biosynthesis (incorrect answers b-d). Fructose phosphorylation can also deplete liver cell ATP, lessening its inhibition of adenine nucleotide degradation and increasing uric acid that is the problem in gout. In adipocytes, fructose can be alternatively phosphorylated by hexokinase to fructose 6-phosphate. However, this reaction is competitively inhibited by appreciable amounts of glucose, as it is in other tissues.

78. Which of the following explains why individuals with hyperlipidemia and/or gout should minimize their intake of sucrose and high-fructose syrups? a. Fructose can bypass phosphofructokinase using a fructokinase in liver. b. Fructose can be phosphorylated by hexokinase in liver cells. c. Fructose is converted to UDP-fructose in liver. d. Fructose is ultimately converted to galactose in liver. e. Fructose can be phosphorylated by hexokinase in adipose cells.

c. Stool glucose after hydrolysis to break up cellulose β1 → 4 bonds Glucose (glucopyranose) residues in cellulose are linked by β1 → 4 bonds in straight chains that humans cannot hydrolyze because they do not possess an enzyme to carry out this function. Cellulose is a structural constituent of plants that is insoluble and provides a source of fiber in the diet. Humans do have intestinal lactase to cleave galactose β1 → 4 glucose (lactose) bonds, maltase to cleave glucose β1 → 4 glucose (maltose) bonds, and sucrose to cleave glucose β1 → 4 fructose (sucrose) bonds. N-acetylcysteamine therapy must be started within 8 hours of acetaminophen ingestion to prevent and within 16 hours to ameliorate liver toxicity, a severe potentially lethal result if over 10 times the therapeutic dose of 5 mg/kg acetaminophen is ingested.

79. A 14-year-old Caucasian adolescent with past history of drug/alcohol abuse is brought to the emergency room after she cannot be aroused for school; her parents think that a bottle of acetaminophen (Tylenol) in their medicine cabinet has fewer tablets than they remember. The ER physician notes that cellulose is listed as the solid binder in the particular brand of acetaminophen. Since liver-protective N-acetylcysteamine therapy must be started within 16 hours of potential toxicity, the physician suggests that a rapid test for cellulose ingestion would be helpful in ascertaining if the child took the acetaminophen. Which of the following tests would be most informative regarding cellulose ingestion? a. Serum glucose to reflect intestinal digestion and absorption of cellulose b. Serum glucose after hydrolysis to break up cellulose β1 → 4 bonds c. Stool glucose after hydrolysis to break up cellulose β1 → 4 bonds d. Stool glucose to reflect intestinal digestion of ingested cellulose e. Stool glucose to reflect intestinal digestion of β1 → 4 bonds distal to β1 → 6 branching points in cellulose

(A) Amplifi cation of the dihydrofolate reductase (DHFR) gene Methotrexate resistance most often occurs due to amplifi cation of the gene for DHFR, the target for methotrexate treatment. Through overpro-duction of DHFR, there is suffi cient enzyme available to overcome the effects of the drug given to the patient. Resistance does not come about by altering the rate of entry of the drug into the cell, by inactivating DHFR, or by inducing an enzyme that can degrade methotrexate.

8. A patient has been prescribed methotrexate for che-motherapy. After an initial success in reducing tumor growth, the tumor resumes its rapid growth. One poten-tial mechanism for this is which of the following? (A) Amplifi cation of the dihydrofolate reductase (DHFR) gene (B) Amplifi cation of the gene for degrading metho trexate (C) Reduced transcription of the gene allowing metho-trexate entry into the cell (D) Increased transcription of the gene allowing metho-trexate effl ux from the cell (E) An inactivating mutation in the gene for DHFR

(D) Malate to coenzyme Q Complex I accepts electrons from NADH, and will transfer them to coenzyme Q. Malate dehydrogenase will convert malate to oxaloacetate, generating NADH in the process. The NADH will then donate electrons to complex I to ini-tiate electron transfer. Succinate donates electrons at complex II (via succinate dehydrogenase, a component of complex II), which donates to coenzyme Q, thereby bypassing complex I. Cytochrome c transfers electrons from complex III to complex IV. Once electrons are car-ried by coenzyme Q, complex I is no longer required for electron transfer to oxygen. These transfers are outlined in the fi gure below.

8. A patient was diagnosed with a mitochondrial DNA mutation that led to reduced complex I activity. This patient would have diffi culties in which of the following electron transfers? (A) Succinate to complex III (B) Cytochrome c to complex IV (C) Coenzyme Q to complex III (D) Malate to coenzyme Q (E) Coenzyme Q to oxygen

(B) Increase in sarcoplasmic calcium levels When the individual begins to run away from the alligator, muscle contraction leads to calcium release from the sarcoplasmic reticulum to the sarcoplasm. This increase in sarcoplasmic calcium binds to the calmodulin subunit of phosphorylase kinase and activates the enzyme in an allosteric manner, in the absence of any covalent modifi cation. The activated phosphorylase kinase will phosphorylate and activate glycogen phosphorylase, which will initiate glycogen degradation. When epi-nephrine reaches the muscle, phosphorylase kinase will be fully activated via phosphorylation by PKA. The acti-vation of glycogen degradation under these conditions is not due to a decrease in blood glucose levels, insu-lin binding (insulin would not be released under these conditions), a decline in ATP levels (the AMP-activated protein kinase does not activate glycogen degradation), or lactate production, the end product of anaerobic metabolism. The fi gure above shows the stimulation of glycogen degradation, working through calcium activation of the calmodulin subunit of phosphorylase kinase.

8. An individual is taking a serene walk in the park when he spots an escaped alligator from the zoo. The individ-ual runs away as fast as he can. Glycogen degradation is occurring to supply glycolysis with a substrate even before epinephrine has reached the muscle. This is due to which of the following? (A) Sudden decrease in blood glucose levels (B) Increase in sarcoplasmic calcium levels (C) Insulin binding to muscle cell receptors (D) Decline in ATP levels (E) Lactate production

(B) Acyl-carnitine Primary carnitine defi - ciency is a lack of carnitine within the cell (such as a mutation in the carnitine transporter); secondary carnitine defi ciency occurs when the carnitine is sequestered in the form of acyl-carnitine (the carnitine cannot be removed from the acyl group, such as a defect in carnitine acyl transferase 2). Thus, elevated levels of acylcarnitine would be expected in a secondary carnitine defi ciency, but not in a primary carnitine defi ciency. In both types of carnitine defi ciencies, fatty acid oxidation is signifi cantly reduced, so the levels of ketone bodies, glucose, lactate, and fatty acids would be similar under both conditions.

8. Carnitine defi ciency can occur in a number of ways. Secondary carnitine defi ciency can be distinguished from primary carnitine defi ciency by measuring which of the following in the blood? (A) Fatty acids (B) Acyl-carnitine (C) Lactic acid (D) Glucose (E) Ketone bodies

(D) Glucose-6-phosphate dehydrogenase Given the demographics of the patient's ancestry (and the need for obtaining an accurate history), and the fact that the patient is a male, the patient may have glucose-6-phosphate dehydrogenase defi ciency (an X-linked disorder). If a person with this enzyme defi -ciency is given primaquine, which is a strong oxidiz-ing agent, hemolytic anemia is likely to develop. If a physician suspects that a patient may have such an enzymatic defi ciency, it is imperative to check before prescribing strong oxidizing agents to the patient, or prescribe another antimalarial prophylaxis that is not a strong oxidizing agent (such as tetracycline). If individuals were defi cient in transketolase, pyruvate dehydrogenase, α-ketoglutarate dehydrogenase, or glyceraldehyde-3-phosphate dehydrogenase, red cell lysis would not occur. One should also recall that the red cells lack mitochondria, so these cells do not contain pyruvate dehydrogenase or α-ketoglutarate dehydrogenase.

8. You are seeing a male patient of African American descent, whose grandparents live in a chloroquine resis-tant malaria belt in Africa. He wants to visit his grand-parents, and you want to give him primaquine as a malaria prophylaxis, but before you do so, you should test the patient for which of the following nonsymptom-atic enzymatic defi ciencies? (A) Transketolase (B) Pyruvate dehydrogenase (C) α-Ketoglutarate dehydrogenase (D) Glucose-6-phosphate dehydrogenase (E) Glyceraldehyde-3-phosphate dehydrogenase

a. Glucokinase Glucokinase promotes uptake of large amounts of glucose by the liver while the other enzymes are present in pathways active after glucose uptake (incorrect answers b-e). At normal glucose levels, the liver produces glucose from glycogen, but as glucose levels rise after feeding, the liver stops converting glycogen and instead takes up glucose. Insulin also plays a role in regulating blood glucose levels. Pancreatic β-cells produce insulin in response to hyperglycemia. Glucose uptake by the β-cells and phosphorylation by glucokinase stimulate secretion of insulin, which enhances glucose transport into adipose tissues and muscle and thus lowers blood glucose levels.

80. A frequent presentation in the newborn period is transient hypoglycemia as the child adapts to separation from maternal glucose controls. Blood glucose is generally maintained at concentrations of 4.5 to 5.5 mmol/L but may rise to 6.5 to 7.2 mmol/L after feeding or decrease to 3.3 to 3.9 mmol/L in the fasting state. Which of the following enzymes plays an important role in regulating blood glucose levels after feeding? a. Glucokinase b. Glucose-6-phosphatase c. Phosphofructokinase d. Pyruvate kinase e. Glucose-6-phosphate dehydrogenase

a. CO2 is consumed. Glucagon will stimulate gluconeogenesis and glycogenolysis in liver by increasing cAMP concentrations, thus raising blood glucose levels. The first step of gluconeogenesis is catalyzed by pyruvate carboxylase with consumption of carbon dioxide and utilization of one high-energy ATP phosphate bond (incorrect answers b-d): pyruvate + ATP + CO 2 → oxaloacetate + ADP + P i Acetyl-CoA activates pyruvate carboxylase and inhibits pyruvate kinase activity of glycolysis but is not utilized in the reactions (incorrect answer c). The second step of gluconeogenesis is catalyzed by phosphoenolpyruvate carboxykinase using energy from GTP, which is produced by succinate thiokinase of the citric acid cycle in liver and kidney (rather than ATP in most other tissues): oxaloacetate + GTP → phosphoenolpyruvate + GDP + CO 2 Two high-energy phosphates are thus employed for pyruvate to glucose conversion by gluconeogenesis that circumvents irreversible glucose to phosphoenolpyruvate to pyruvate reactions of glycolysis.

81. A 5-year-old African American girl is brought to the emergency room with tonic-clonic seizures. Laboratory studies are drawn as anticonvulsants given do not completely stop the seizures. A blood glucose of 30 mg/dL (normal 60-100) is found, prompting administration of 25% dextrose intravenously and 1 mg of glucagon intramuscularly. Which of the following events will occur in response to glucagon? a. CO2 is consumed. b. Inorganic phosphate is consumed. c. Acetyl-CoA is utilized. d. ATP is generated. e. GTP is generated.

b. α-Ketoglutarate → succinate During conversion of α-ketoglutarate to succinate, a molecule of GTP is synthesized from GDP and phosphate using energy from the hydrolysis of succinyl-CoA to produce succinate and CoA (incorrect answers a, c-e). This constitutes substrate-level phosphorylation, and, in contrast to oxidative phosphorylation, this is the only reaction in the citric acid cycle that directly yields a high-energy phosphate bond. The sequence of reactions from α-ketoglutarate to succinate is catalyzed by the α-ketoglutarate dehydrogenase complex and succinate thiokinase, respectively. α-ketoglutarate + NAD + + acetyl-CoA → succinyl-CoA + CO 2 + NADH succinyl-CoA + P i + GDP → succinate + GTP + acetyl-CoA

82. A 6-month-old Caucasian boy has exhibited somewhat slow growth but becomes very ill after contracting influenza from an older sibling. A plasma lactate level of 55 mg/dL is found that suggests underlying mitochondrial disease, and mitochondrial DNA studies confirm a large deletion that affects several oxidative-phosphorylation complexes. Exacerbation of routine illness by the child's reduced energy reserves would most affect which of the following reactions of the citric acid cycle? a. Citrate→α-ketoglutarate b. α-Ketoglutarate → succinate c. Succinate → fumarate d. Fumarate → malate e. Malate → oxaloacetate

b. Her insulin increased glucose import into extrahepatic tissues using various glucose transporters. Liver cells are permeable to glucose while extrahepatic tissues require insulin for glucose entry, reflecting different glucose transporters (GLUT) in different tissues (incorrect answers a, d, and e). Liver hexokinase has a low Km for glucose and acts at a constant rate, while glucokinase has a higher Km for glucose and promotes glucose uptake at high concentrations as found in the portal vein after meals (incorrect answer c). The liver releases glucose at normal serum glucose concentrations but takes up glucose at high serum glucose concentrations. Insulin and glucagon act in opposing fashion to regulate serum glucose concentration. Insulin, secreted by the pancreatic β-cell in response to internal increases in glucose, ATP, and calcium influx, increases glucose uptake by muscle and adipose cells by recruiting glucose transporters to their plasma membranes. Glucagon, secreted by the pancreatic α-cells, stimulates cyclic AMP synthesis with increased gluconeogenesis and glycogenolysis to increase serum glucose concentrations.

83. A 45-year-old African American female is found unconscious at her desk and rushed to the emergency room via ambulance. She is a known diabetic, and an injectable insulin pen was found on her desk. Serum glucose was less than 20 g/dL as read by glucometer on the way to the hospital. Which of the following statements reflects the pathogenesis of her hypoglycemia? a. Her liver was suddenly permeable to glucose after added insulin. b. Her insulin increased glucose import into extrahepatic tissues using various glucose transporters. c. Her insulin stimulated liver glucokinase, which phosphorylates glucose only at low glucose concentrations. d. Her extrahepatic tissues became permeable to glucose through the opposing action of glucagon. e. Insulin stimulated glucose uptake into her liver even when the serum glucose became low.

d. NADPH and ribose The pentose phosphate cycle does not produce ATP, but instead produces ribose and NADPH. NADP + is the hydrogen acceptor instead of NAD +, as in glycolysis. In the oxidative phase of the pentose phosphate pathway, NADPH is generated by glucose-6-phosphate dehydrogenase. NADPH is also generated by 6-phosphogluconate dehydrogenase. Ribose is generated in the nonoxidative phase.

84. Which of the following two compounds are the primary products of the pentose phosphate pathway? a. NAD+ and ribose b. NADH and ribose c. NADP+ and ribose d. NADPH and ribose e. NAD+ and glucose f. NADH and glucose g. NADP+ and glucose h. NADPH and glucose

a. Glucokinase Glucokinase catalyzes the conversion of glucose to glucose 6-phosphate in the energy, requiring first step of glycolysis. ATP is also required in the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate by phosphofructokinase. ATP is generated in the conversion of 1,3-bisphosphoglycerate to 3-phosphoglycerate by phosphoglycerate kinase and in the conversion of phosphoenolpyruvate to pyruvate by pyruvate kinase (incorrect answers c-d). No energy is required when lactate dehydrogenase converts pyruvate to lactate (a molecule of NADH is converted to NAD +− incorrect answer b) or when phosphohexose isomerase converts glucose 6-phosphate to fructose 6-phosphate (incorrect answer e).

85. Which of the following is an energy-requiring step of glycolysis? a. Glucokinase b. Lactate dehydrogenase c. Phosphoglycerate kinase d. Pyruvate kinase e. Phosphohexose isomerase

b. Glycerol and alanine Gluconeogenesis refers to the pathway for converting noncarbohydrate precursors to glucose (incorrect answers a, d). Glycerol, lactate, propionate, and certain amino acids such as alanine are all substrates for gluconeogenesis (substrates will undergo chemical conversion to form carbohydrates rather than serving as regulatory factors, incorrect answers a, d). Biotin is a cofactor for the first step (pyruvate carboxylase) of gluconeogenesis, while conversion of succinyl-CoA to acetyl-CoA provides GTP for the second step (phosphoenolpyruvate carboxykinase—incorrect answers c, e). Glycogen is a glucose storage molecule that can readily be converted in the liver back to glucose for maintenance of blood glucose levels between meals.

86. Which of the following are primary substrates for gluconeogenesis? a. Galactose and fructose b. Glycerol and alanine c. Acetyl-CoA and succinyl-CoA d. Sucrose and lactose e. GTP and biotin

c. α-1,4-glucosidase The child has symptoms of glycogen storage disease. Glycogen is a glucose polymer with linear regions linked through the C1 aldehyde of one glucose to the C4 alcohol of the next (α-1,4-glucoside linkage). There are also branches from the linear glycogen polymer that have α-1,6-glucoside linkages. Glycogen is synthesized during times of carbohydrate and energy surplus, but must be degraded during fasting to provide energy. Separate enzymes for breakdown include phosphorylases (α-1,4-glucosidases) that cleave linear regions of glycogen and debranching enzymes (α-1,6- glucosidases) that cleave branch points. Glucose-6-phosphatase is needed in the liver to liberate free glucose from glucose 6-phosphate, providing fuel for other organs. There is no glucose-6-phosphatase in muscle, and muscle glycogenolysis provides energy just for muscle with production of lactate. Deficiencies of more than eight enzymes involved in glycogenolysis, including those mentioned, can produce glycogen storage disease.

87. A 3-month-old Caucasian girl presents with low blood glucose (hypoglycemia), enlarged liver (hepatomegaly), and excess fat deposition in the cheeks (cherubic facies). A liver biopsy reveals excess glycogen in hepatocytes. Deficiency of which of the following enzymes best explains this phenotype? a. α-1,1-glucosidase b. α-1,1-galactosidase c. α-1,4-glucosidase d. α-1,4-galactosidase e. α-1,6-galactosidase

b. To stimulate gluconeogenesis when tissue resistance leads to excess insulin Glucagon responds to decreases in blood glucose by increasing liver cAMP, gluconeogenesis, and glycogenolysis. High cAMP levels activate cAMPdependent protein kinases that inactivate glycolytic enzymes pyruvate kinase and phosphofructokinase and activate glycogen phosphorylase by phosphorylation (incorrect answers a, c, and d). Obesity most commonly causes type II diabetes mellitus due to tissue resistance rather than decreased insulin secretion by the pancreas (incorrect answer e). Early stages of diabetes mellitus may be accompanied by periodic hypoglycemia due to high insulin levels trying to overcome tissue resistance or to periodic bursts of insulin secretion from failing pancreatic β-cells. Glucagon antagonizes insulin by increasing gluconeogenesis and glycogenolysis, both by increased action of cAMP kinases and through allosteric regulation of enzymes—for example, reciprocal activation of pyruvate carboxylase (gluconeogenesis) and inactivation of pyruvate dehydrogenase (glycolysis) by acetyl-CoA. Glucagon also inhibits lipolysis.

88. An 18-year-old Hispanic female is evaluated for fainting spells that tend to occur when she has not eaten due to illness or work. Her body mass index [weight in kg /(height in m) 2] is over 29 (<26 desirable) and the initial history makes anxiety or blood pressure changes unlikely. Several random blood samples demonstrate lower glucose than normal and a glucose tolerance test is abnormal, showing a slow decrease in administered glucose that is suggestive of diabetes mellitus. Glucagon proves effective in helping her fainting spells, and a dietician places her on a moderate, low-carbohydrate diet, with monitoring of glucose by glucometer and use of glucagon when symptomatic. What is the role of glucagon in this situation? a. To stimulate the citric acid cycle when tissue resistance leads to excess insulin b. To stimulate gluconeogenesis when tissue resistance leads to excess insulin c. To stimulate glycolysis in the presence of decreased insulin secretion d. To stimulate the pentose phosphate pathway in the presence of decreased insulin secretion e. To stimulate glycogenolysis in the presence of decreased insulin secretion

b. UDP-glucose UDPglucose is the high-energy compound from which glycogen is synthesized, made by condensing glucose-1-P and UTP using UDP-glucose pyrophosphorylase (incorrect answers a, c-e). Glucose is rapidly converted to glucose 6-phosphate by glucokinase in liver, its high Km favoring the high concentrations of glucose in the portal vein (route from glucose-absorbing intestinal veins to the liver). Glucose 6-phosphate is converted to glucose 1-phosphate and activated to form UDP-glucose, a molecule that can arise from UDP-galactose through lactose absorption or follow the glucuronic acid pathway to produce ascorbic acid (vitamin C) and building blocks for proteoglycans. UDP-glucose is added to terminal glucose units of glycogen by glycogen synthase in the form of α1 → 4 linkages. To increase the solubility of glycogen and to increase the number of terminal residues, glycogen- branching enzyme transfers blocks of ∼7 glucose residues to branch points via α1→ 6 linkages. Absence of the branching enzyme, deficient in type IV glycogen storage disease (MIM*232500), produces a less soluble glycogen with fewer termini, thus decreasing glucose release (short stature from less energy, large liver with stored glycogen) and provoking reaction to the abnormal glycogen to cause liver cell death (scarring and cirrhosis).

89. A 2-year-old African American boy is evaluated for short stature (height below the 3rd percentile with weight and head circumference at the 20th percentile for age), irritability between meals accompanied by low glucose levels, enlarged liver, and easy bruisability. Liver biopsy reveals excess glycogen and deficiency of the α1 → 6 branching enzyme is documented for a diagnosis of type IV glycogen storage disease (MIM*232500). This disorder causes variable liver and neuromuscular symptoms but can be lethal due to progressive liver scarring (cirrhosis) because of the presence of abnormal glycogen. Conventional therapy with continuous glucose supply (eg, nocturnal cornstarch by tube) to minimize the need for glycogen breakdown is often not effective, and additional agents to minimize synthesis of abnormal glycogen have been considered. Inhibition of glycogen synthesis might target which of the following molecules? a. UDP-glucose 1-phosphate b. UDP-glucose c. UDP-glucose 6-phosphate d. Glucose 6-phosphate e. Glucose 1-phosphate

(D) Increased translation of the transferrin receptor mRNA Since the patient has an iron defi ciency, leading to a microcytic anemia, cells will upregulate their mechanism for acquiring iron, which is through the transferrin receptor. Ferritin is the iron storage protein within cells, and if intracellular iron levels are low, there is no need to upregulate the syn-thesis of ferritin (its synthesis is actually downregulated under these conditions). The iron travels in the circu-lation bound to transferrin, so increasing the number of transferrin receptors on the cell surface will enable a more effi cient transport of iron and transferrin into the cells. The regulation of transferrin receptor synthe-sis is at the level of translation, as is the regulation of ferritin synthesis. Thus, cells under these conditions will increase their translation of the transferrin receptor mRNA. This translational regulation is shown below.

9. A 16-year-old girl has been losing weight and feeling lethargic over the past 4 months and is taken to the physician by her parents. During the history, the par-ents expressed concern that their daughter had seemed to eat very little during the day, a claim denied by the patient. Laboratory results indicated an iron defi ciency and a microcytic anemia. The cells of the patient have adapted to the iron defi ciency in which one of the following ways? (A) Increased transcription of ferritin mRNA (B) Reduced transcription of the transferrin receptor mRNA (C) Increased translation of the ferritin mRNA (D) Increased translation of the transferrin receptor mRNA (E) Increased degradation of the transferrin receptor mRNA

(E) Dinitrophenol The key is the elevation in temperature. Dinitrophenol is an uncoupler of oxi-dation and phosphorylation in that uncouplers destroy the proton gradient across the membrane (thereby inhibiting the synthesis of ATP) without blocking the transfer of electrons through the electron transfer chain to oxygen. The energy that should have been generated in the form of a proton gradient is lost as heat, which elevates the body temperature of the affected workers. Electron fl ow is also enhanced in the presence of an uncoupler, so additional oxygen is required to allow the chain to continue (hence the heavy breathing). The other agents added would have stopped electron trans-fer totally, which would not allow for an increase in temperature, and would actually decrease the rate of breathing (since oxygen is no longer required for the nonfunctioning electron transfer chain). Atractyloside inhibits the ATP/ADP exchanger, and once there is no ADP in the mitochondrial matrix, electron fl ow will stop due to the inability to synthesize ATP (normal coupling). Oligomycin works in a similar mechanism in that it blocks the ATP synthase, preventing ATP syn-thesis, and, due to coupling, electron transfer through the chain. Rotenone blocks complex I transfer to coen-zyme Q, which signifi cantly reduces electron fl ow, and will not lead to an increase in temperature.

9. A pair of farm workers in Mexico was spraying pesti-cide on crops when they both developed the following severe symptoms: heavy, labored breathing, signifi cantly elevated temperature, and loss of consciousness. The pesticide contained an agent that interfered with oxi-dative phosphorylation, which most closely resembled which of the following known inhibitors? (A) Oligomycin( B) Atractyloside (C) Cyanide (D) Rotenone (E) Dinitrophenol

(A) Superoxide dismutase The patient is experiencing the symptoms of familial ALS. A muta-tion in superoxide dismutase 1 (SOD1) in humans has been linked to the development of familial ALS through an unknown mechanism. Familial ALS only constitutes between 5% and 10% of all ALS cases diagnosed. The disease process, when SOD1 is mutated, is not linked to a loss of enzymatic activity, although the SOD1 may have been mutated such that it will produce other radical spe-cies and is no longer specifi c for superoxide. A second model proposes a misfolding problem similar to prion disease. For more information on such models see Nature Med. 2000;6:1320-1321 and Ann Neurol. 2007 Dec;62(6):553-559. None of the other enzymes listed (catalase, myeloperoxidase, NO synthase, and tyrosine hydroxylase) have been linked to the development of ALS, or an ALSlike disease. The reaction catalyzed by SOD1 is shown below.

9. A patient has an insidious and steadily progressing neu-rologic disorder that, after several years, results in wast-ing and paralysis of the muscles of the limbs and trunk, loss of ability to speak, and swallowing diffi culties. His paternal uncle had the same disease. A mutation in which enzyme may lead to these symptoms? (A) Superoxide dismutase (B) Catalase (C) Myeloperoxidase (D) NO synthase (E) Tyrosine hydroxylase

(D) Increase in intracellular AMP levels As AMP levels increase in the muscle due to the need for ATP for muscle contraction, and the activity of the ade-nylate kinase reaction, the AMP-activated protein kinase is turned on. One of the effects of the AMP-activated protein kinase is to increase the number of GLUT4 transporters in the muscle membrane, in a process simi-lar to the action of insulin. This enables muscle to take up glucose effi ciently from the circulation when inter-nal energy levels are low. The ability of the muscle to take up glucose under these conditions is not due to an increase in epinephrine levels, an increase in sarcoplas-mic calcium levels, or insulin binding to muscle cells. Under conditions as described in the question, insu-lin will not be present in the circulation to bind to the muscle cells. As the muscle does not contain glucagon receptors, there is no effect on muscle when glucagon is present in the circulation.

9. As the individual in the previous question continues to run from the alligator, the muscle begins to import glu-cose from the circulation. This occurs due to which of the following? (A) Insulin binding to muscle cells (B) Epinephrine binding to muscle cells (C) Glucagon binding to muscle cells (D) Increase in intracellular AMP levels (E) Increase in intracellular calcium levels

(C) cisΔ9 C18:1 An 18-carbon fatty acid will generate an additional acetyl-CoA, one NADH, and one FADH2 as compared to a 16-carbon fatty acid. Thus, the addition of two carbons will add 14 additional ATP to the overall energy yield (10 ATP per acetyl-CoA, 2.5 for NADH, and 1.5 for FADH2). An unsaturation at an odd carbon position will require the use of an isomerase during oxidation, and this will result in the loss of generation of 1 FADH2; an unsaturation at an even carbon position will require the use of the 2,4 dienoyl-CoA reductase, and this will result in the loss of generation of 1 NADPH. Thus, an unsaturation at an odd position results in the loss of 1.5 ATP, while an unsaturation at an even position results in the loss of 2.5 ATP. Thus, in comparing two 18-carbon fatty acids, one with an unsaturation at position 9, and the other at position 6, the fatty acid with the double bond at position 9 will yield one more ATP than the fatty acid with the unsaturation at position 6. An overview of the fatty acid oxidation spiral is shown above, along with the reactions required for the oxidation of unsaturated fatty acids.

9. Which one of the following fatty acids will generate the largest amount of ATP upon complete oxidation to carbon dioxide and water? (A) C16:0 (B) cisΔ9 C16:1 (C) cisΔ9 C18:1 (D) cisΔ6 C18:1 (E) cisΔ9, Δ12 C18:2

d. New glucose molecules are added to the C1 aldehyde group of chain termini, forming a hemiacetal. Glycogen is a highly branched polymer of α-d-glucose residues joined by α1 → 4-glycosidic linkage. Under the influence of glycogen synthase, the C4 alcohol of a new glucose is added to the C1 aldehyde group of the chain terminus. The branched chains occur about every 10 residues and are joined in α1 → 6-glycosidic linkages. Large amounts of glycogen are stored as 100- to 400-. granules in the cytoplasm of liver and muscle cells. The enzymes responsible for making or breaking the α1 → 4-glycosidic bonds are contained within the granules. Thus, glycogen is a readily mobilized form of glucose.

90. Which of the following statements best describes the structure of glycogen? a. Glycogen is a copolymer of glucose and galactose. b. There are more branch residues than residues in straight chains. c. Branch points contain α1-4 glycosidic linkages. d. New glucose molecules are added to the C1 aldehyde group of chain termini, forming a hemiacetal. e. The monosaccharide residues alternate between D- and L-glucose.

c. Increased normal glycogen with deficient phosphorylase Glycogen storage diseases are caused by abnormal glycogen breakdown with hepatic (hypoglycemia, hepatomegaly, and short stature) or muscle (exercise fatigue, cramping) symptoms. The excess glycogen (incorrect answers d, e) usually reflects defective glycogenolytic enzymes or their phosphorylating kinases but some defects of glycogen synthesis (like the type IV with deficient branching enzyme) or alterations of regulatory enzymes like phosphofructokinase can lead to decreased glycogen mobilization (answers d, e incorrect). Types V (phosphorylase deficiency) and VII (phosphofructokinase deficiency of Tarui disease) are due to enzymes with major effects on muscle while others (I, III-VIII) affect enzymes with major effects on liver (some like Type IV can affect both). The compromised phosphorylation of muscle glycogen characteristic of McArdle disease compels the muscles to rely on auxiliary energy sources such as free fatty acids and ambient glucose.

91. A 25-year-old African American male notices increased fatigue and decreased performance when playing outfield for his office softball team. After a game in hot weather with many fielding chances, he has severe muscle pains and cramps that keep him awake and notices that his urine is dark that night. His physician suspects McArdle disease (type V glycogen storage disease—MIM*232600) and has him run on a treadmill breathing air with reduced oxygen. The man develops severe cramps after this ischemic exercise, further supporting the presumptive diagnosis. A diagnostic muscle biopsy is likely to show which of the following? a. Increased normal glycogen with deficient hexokinase b. Increased normal glycogen with deficient glycogen synthase c. Increased normal glycogen with deficient phosphorylase d. Decreased muscle glycogen with decreased glycogen synthase e. Decreased muscle glycogen with decreased debranching enzyme

a. Fructose, fructose 1-phosphate Fructose is a 6-carbon hexose with a ketone group (ketose), while ribulose and xylulose have 5 carbons (incorrect answers d, e) and glucose and galactose have aldehyde rather than ketone groups (incorrect answers b, c). Fructose is the substrate for fructokinase that is deficient in asymptomatic fructosuria (MIM*229800), while fructose 1-phosphate is the substrate for aldolase B that is deficient in hereditary fructose intolerance (MIM*229600) with vomiting, lactic acidosis, and, with continued fructose ingestion, failure to thrive and hepatic disease. Glucose isomers can have different orientations of the H and OH groups on the carbon atom adjacent to the CH 2OH group (D- vs L-isomers), and most of the monosaccharides in mammals are in the D form but differ in their configuration of the H and OH groups at the 2, 3, and 4 carbons (epimers). Mannose and galactose are the most biologically important epimers of glucose.

92. The 10-year experience of a biochemical genetics laboratory includes five patients who had a positive urine reagent dipstick reaction for reducing sugars. Two of these were young females evaluated for vomiting and failure to thrive, while three were a boy and two girls, who were studied as part of a control population. Mass spectrometry profiles had shown a urine chemical with six carbons and a ketone group. Which of the following metabolites satisfy these chemical criteria and accumulate in specific inborn errors? a. Fructose, fructose 1-phosphate b. Galactose, galactose 1-phosphate c. Glucose, glucose 1-phosphate d. Ribulose, ribose 1-phosphate e. Xylulose, xylose 1-phosphate

e. Normal liver uptake reflecting a higher Km for liver glucokinase Less glucose uptake by liver after postprandial glucose levels fall is normal (incorrect answers a-c) and does not reflect depleted glycogen or starvation from the patient's likely bulimia (thin tooth enamel and decay from frequent induced vomiting). Most peripheral tissues such as muscle contain hexokinase with a low Km and high affinity for glucose that brings it rapidly into cells. After a meal, glucose absorbed from the intestine will produce high levels in serum that decreases rapidly as extrahepatic tissues import and convert glucose to glucose-6-phophate with hexokinase. In contrast, liver has a different hexokinase enzyme called glucokinase (incorrect answer d) that has a high Km and low affinity, removing glucose from the portal vein only when it is at high concentrations (shortly after a meal). This ensures that the liver will sequester glucose only when it is in excess and not needed by tissues such as brain or heart, storing the excess glucose as glycogen for later use.

93. A 17-year-old Caucasian adolescent is evaluated for fainting spells and found to have a glucose level of 54 mg/dL (normal 75-105) one hour after eating. Her body mass index (BMI) is 17 (average 23-25), and physical examination reveals pallor, halitosis, and discolored teeth with multiple caries. An eating disorder is suspected, and she enlists in a research study that includes behavioral and dietary therapy along with metabolic studies. Administration of deuterium-labeled glucose shows initial rapid entry into liver and peripheral muscle followed by continued muscle but minimal liver intake. Which of the following accurately interprets these results? a. Abnormal liver uptake reflecting starvation b. Abnormal liver uptake reflecting latent diabetes mellitus c. Abnormal liver uptake reflecting prior glycogen depletion d. Normal liver uptake reflecting a lower Km for liver hexokinase e. Normal liver uptake reflecting a higher Km for liver glucokinase

a. 1-reaction A, 2-reactions A and B to C, 3-reactions C to D and E Glucose is converted to glucose 6-phosphate (G6P) by hexokinase (reaction A in figure below Question 94) in most tissues. However, liver that has a glucokinase with higher Km that ensures glucose import and conversion to glycogen occurs only during periods of glucose excess, that is after meals (incorrect answers d, e—see prior Answer 93 for more detail). Hexo- or glucokinase (reaction A in figure) and phosphofructokinase converting fructose 6-phosphate to fructose 1,6-bisphosphate (reaction B-C in figure) both are endergonic reactions that require energy and are coupled with exergonic ATP to ADP hydrolysis to allow conversion (incorrect answers b-e). This energy deficit is made up later in glycolysis by the exergonic phosphoglycerate kinase and pyruvate kinase reactions that generate ATP (four per molecule of glucose), entering glycolysis and giving a net yield of 2 ATP for each conversion of glucose to pyruvate. Glyceraldehyde 3-phosphate (substrate D in figure) is produced from fructose 1,6-diphosphate (substrate C) or from dihydroxyacetone phosphate by phosphotriose isomerase (reaction E) and reduced to yield NADH (incorrect answer c).

94. The figure below depicts the initial steps of glycolysis with enzyme reactions A and E, and substrates B, C, and D. Which of the following options depicts the correct reaction sequence: 1—a reaction regulating glucose levels, 2—endergonic reaction(s), requiring ATP to ADP coupling, and 3— reaction(s), producing a substrate that will be reduced to produce NADH? a. 1-reaction A, 2-reactions A and B to C, 3-reactions C to D and E b. 1-reaction A, 2-reactions A and C to D, 3-reactions C to D and E c. 1-reaction A, 2-reaction B to C, 3-reactions B to C and E d. 1-reaction E, 2-reaction A, 3-reactions C to D and E e. 1-reaction E, 2-reaction E, 3-reactions C to D and E

d. Phosphoenolpyruvate to pyruvate The indicated reactions are part of the glycolytic pathway and only the pyruvate kinase reaction (option d) is exergonic—produces energy that is coupled to produce ATP from ADP and P i. Less effective energy production in muscle implied by the girl's fatigue would affect this reaction less than others that are endergonic or energy neutral (incorrect answers a-c, e). Major defects affecting muscle glycogen degradation or mitochondrial electron transport will produce weakness/developmental delay at young ages or exercise intolerance (muscle aches and spasms) at later ages. Milder defects produce more subtle symptoms that can manifest as early fatigue or poor performance in high intensity or endurance sports. It is interesting to speculate if mild mutations enhancing muscle energy pathways will be found in exceptional athletes, but the only supplements that are proven to work are those that increase muscle mass (steroids, growth hormone) rather than "energy drinks" or supplements such as creatine.

95. A 14-year-old African American adolescent competes well in short distance events as she tries out for the track team but does poorly at longer distances. Evaluation at a sports institute shows she has elevation of blood lactate after longer periods on a treadmill and suggests a defect in muscle energy metabolism. Although the girl's parents decline further expensive evaluation, which of the following reactions of muscle metabolism would be least affected after she has run a long distance? a. Glucose 6-phosphate to fructose 6-phosphate b. Glucose to glucose 6-phosphate c. Fructose 6-phosphate to fructose 1,6-diphosphate d. Phosphoenolpyruvate to pyruvate e. Pyruvate to lactate

e. Glyceraldehyde-3-phosphate dehydrogenase Using NAD+ in an oxidation-reduction reaction, inorganic phosphate is added to glyceraldehyde 3-phosphate by the enzyme glyceraldehyde-3-phosphate dehydrogenase to form 1,3-diphosphoglycerate. Other enzymes (incorrect answers a, b, and d) add high-energy phosphate bonds to substrates of glycolysis. Hexokinase—or, in the case of the liver, glucokinase—adds phosphate from ATP to glucose to form glucose 6-phosphate. Strictly speaking, this is not always considered a step of the glycolytic pathway. Phosphofructokinase uses ATP to convert fructose 6-phosphate to fructose 1,6-phosphate, while phosphoglycerate kinase and pyruvate kinase transfer substrate high-energy phosphate groups to ADP to form ATP. Triose phosphate isomerase interconverts dihydroxyacetone phosphate and glyceraldehyde 3-phosphate without phosphorylation (incorrect answer c).

96. Which of the following enzymes catalyzes high-energy phosphorylation of substrates during glycolysis? a. Pyruvate kinase b. Phosphoglycerate kinase c. Triose phosphate isomerase d. Aldolase e. Glyceraldehyde-3-phosphate dehydrogenase

b. Reaction B The conversion of glucose to glucose 6-phosphate (reaction B in figure) is different in liver and muscle, while others are similar (incorrect answers a, c-e). In muscle and most other tissues, hexokinase regulates the conversionof glucose to glucose 6-phosphate. When the major regulatory enzyme of glycolysis, phosphofructose kinase, is turned off, the level of fructose 6-phosphate increases and, in turn, the level of glucose 6-phosphate rises because it is in equilibrium with fructose 6-phosphate. Hexokinase is inhibited by glucose 6-phosphate. However, in the liver, glucose is phosphorylated even when glucose 6-phosphate levels are high because the enzyme regulating transformation of glucose into glucose 6-phosphate is glucokinase. Glucokinase is not inhibited by glucose 6-phosphate in the liver. Although hexokinase has a low Km for glucose and is capable of acting on low levels of blood glucose, glucokinase has a high Km for glucose and is effective only when glucose is abundant. Therefore, when blood glucose levels are low, muscle, brain, and other tissues are capable of taking up and phosphorylating glucose, whereas the liver is not. When blood glucose is abundant, glucokinase in the liver phosphorylates glucose and provides glucose 6-phosphate for the synthesis and storage of glucose as glycogen.

97. Which reaction in the figure below occurs in both muscle and liver but has substantially different qualities in the two? a. Reaction A b. Reaction B c. Reaction C d. Reaction D e. Reaction E

F Hexokinase Patients with essential fructosuria have fructokinase deficiency and thus cannot convert fructose to fructose-1-phosphate. Hexokinase (found in adipocytes and myocytes) becomes the primary enzyme responsible for fructose metabolism and converts it to fructose-6-phosphate, which can be used for glycolysis or converted to glucose-6-phosphate by glucose-6-phosphate isomerase. Fructose that is not converted by hexokinase is excreted in urine, which leads to the presence of reducing substances.

A 12-month-old boy is brought to the physician by his mother for a well-child examination. He was delivered at term after an uncomplicated pregnancy. His mother says he is breastfeeding well. He is at the 50th percentilefor height and 65th percentile for weight. Physical examination shows no abnormalities. Urinalysis shows 3+ reducing substances. Compared to a healthy infant, giving this patient apple juice to drink will result in increased activity of which of the following enzymes? A Aldolase B B Galactokinase C Galactose-1-phosphate uridyltransferase D Fructokinase E α-1,6-glucosidase F Hexokinase

D Myophosphorylase deficiency Myophosphorylase deficiency (McArdle disease) is a glycogen storage disease that manifests with myalgias, rhabdomyolysis, myoglobinuria, and early fatigue from exercise due to the inability to break down glycogen for use as energy in muscle tissues. Treatment often involves aerobic exercise programs and nutritional supplementation, although evidence of benefit is limited.

A 15-year-old boy comes to the physician because of severe muscle cramps and pain for 3 months. He first noticed these symptoms while attending tryouts for the high school football team. Since then, he becomes easily fatigued and has severe muscle pain and swelling after 10 minutes of playing. However, after a brief period of rest, the symptoms improve, and he is able to return to the game. Two days ago, he had an episode of reddish-brown urine after playing football. There is no family history of serious illness. He appears healthy. Vital signs are within normal limits. Physical and neurological examinations show no abnormalities. Serum creatine kinase concentration is 333 U/L. Urinalysis shows: Which of the following is the most likely cause of this patient's symptoms? A Medium-chain acyl-CoA dehydrogenase deficiency B Thyroid hormone deficiency C Dystrophin gene mutation D Myophosphorylase deficiency E Acid maltase deficiency F CTG repeat in the DMPK gene

c. α-L-iduronidase The two major groups of lysosomal storage disease are sphingolipidoses and mucopolysaccharidoses. An absence of α-L-iduronidase, as in Hurler's syndrome and Scheie's syndrome, leads to accumulations of dermatan sulfate and heparan sulfate. Scheie's syndrome is less severe, with corneal clouding, joint degeneration, and increased heart disease. Hurler's syndrome has the same symptoms plus mental and physical retardation leading to early death. The later onset in this child is compatible with a diagnosis of Scheie's syndrome. Note that Hurler's and Scheie's syndromes result from mutations at the same locus—hence their identical McKusick numbers. The reasons for the differences in disease severity are unknown. All of the other enzyme deficiencies listed lead to the lack of proper breakdown of sphingolipids and their accumulation as gangliosides, glucocerebrosides, and sphingomyelins. Symptoms of lipidoses may include organ enlargement, mental retardation,m and early death.

A 15-year-old boy has a long history of school problems and is labeled as hyperactive. His tissues are puffy, giving his face a "coarse" appearance. His IQ tests have declined recently and are now markedly below normal. Laboratory studies demonstrate normal amounts of sphingolipids in fibroblast cultures with increased amounts of glycosaminoglycans in urine. Which of the following enzyme deficiencies might explain the boy's phenotype? Select one: a. β-gangliosidase A b. α-galactocerebrosidase c. α-L-iduronidase d. Hexosaminidase A e. Glucocerebrosidase

a. CPT-II deficiency The most likely diagnosis in this case is CPT-II deficiency, although this is apparently a fairly mild case. The patient's muscle weakness and "brown urine" (myoglobinuria) are characteristic of this disorder. CPT-I deficiency would most likely manifest as liver dysfunction. A secondary form of carnitine deficiency due to exogenous factors such as malnutrition, infection, or dialysis, is unlikely. MCAD ordinarilymanifests within the first 3-5 years of life. The patient's normal stature is inconsistent with Marfan syndrome, which is characterized by tall stature and very long bones in the extremities.

A 19-year-old man complains of "brown urine" and pain in the muscles of his arms andlegs experienced while playing touch football. He has had several episodes of muscle painduring exercise, but he had not noticed darkening of his urine afterward. The pain usuallyresolved overnight. Physical examination reveals a well-fed male of normal stature. Reflexesand range of motion in all arms and legs are normal, but there is some paraparesis(weakness), especially in his right leg. A muscle biopsy is taken and sent for specializedtesting. The patient is sent home with a recommendation to take a dietary carnitine supplement.Which of the following is the most likely diagnosis? Select one: a. CPT-II deficiency b. Carnitine deficiency c. Marfan syndrome d. MCAD deficiency e. CPT-I deficiency

D Glucose-6-phosphatase Glucose-6-phosphatase deficiency is the underlying cause of von Gierke disease (type a), which typically manifests with failure to thrive, lethargy, hepatomegaly, severe fasting hypoglycemia (causing hypoglycemic seizures), hyperlipidemia, and lactic acidosis. Further characteristic findings include hyperuricemia and facial fat deposition (sometimes referred to as a "doll-like" face). Like other GSDs, von Gierke disease can be managed effectively with diet modifications (e.g., uncooked corn starch, glucose preparations), which aim to prevent hypoglycemia and muscle symptoms.

A 2-month-old boy is brought to the emergency department 25 minutes after having a seizure. He has had multiple seizures during the past week. His mother also notes that he has become lethargic and has had a weak cry for the past month. He was born at 37 weeks' gestation. He is at the 20th percentile for height and 15th percentile for weight. His temperature is 36.7°C (98.0°F), respirations are 50/min, and pulse is 140/min. Examination shows a soft and nontender abdomen. The liver is palpated 4 cm below the right costal margin; there is no splenomegaly. Serum studies show: A deficiency of which of the following enzymes is the most likely cause of this patient's symptoms? A deficiency of which of the following enzymes is the most likely cause of this patient's symptoms? A Glycogen branching enzyme B Galactose-1-phosphate uridyltransferase C Fructokinase D Glucose-6-phosphatase E Acid maltase

a. Oxidation of very long chain fatty acids The child has Zellweger's syndrome, an absence of peroxisomal enzyme activity. Of the pathways listed as answers, only the oxidation of very long chain fatty acids is a peroxisomal function. Fatty acid synthesis occurs in the cytoplasm. Acetyl-CoA oxidation takes place in the mitochondria. Glucose oxidation is a combination of glycolysis (cytoplasm) and the TCA cycle (mitochondria). Triglyceride synthesis occurs in the cytoplasm.

A 2-month-old infant with failure to thrive displays hepatomegaly, high levels of iron and copper in the blood, and vision problems. This child has difficulty in carrying out which of the following types of reactions? Select one: a. Oxidation of very long chain fatty acids b. Oxidation of glucose c. Oxidation of acetyl-CoA d. Synthesis of triacylgycerol e. Synthesis of unsaturated fatty acids

A Lysosomal phosphatase Lysosomal phosphatase and other lysosomal enzymes accumulate in the serum of patients with I-cell disease. In this condition, the Golgi apparatus fails to phosphorylate mannose residues to mannose-6-phosphate on glycoproteins. Since phosphorylated mannose is a marker for exportation to the lysosome, I-cell disease results in inappropriate extracellular secretion of proteins intended for the lysosome.

A 2-year-old boy is brought to the pediatrician for evaluation of recurrent falls and clumsiness. He can only walk with support, and his vocabulary consists of approximately 20 words. He cannot hold objects or stack blocks. Physical examination shows coarse facial features and scoliosis. A holosystolic murmur is heard at the apex. The abdomen is distended, and both the liver and spleen tips are palpable. Fundoscopic examination shows cloudy corneas. Laboratory studies show complete inactivity of the N-acetylglucosaminyl-1-phosphotransferase enzyme. This condition is most likely associated with an accumulation of which of the following substances in the patient's serum? A Lysosomal phosphatase B Heparan sulfate C Ceramide trihexoside D Sphingomyelin E Glucocerebroside

E Aggressive behavior Hunter syndrome is an X-linked recessive lysosomal storage disease caused by a deficiency in iduronate-2-sulfatase and is associated with aggressive behavior, developmental delay, cognitive impairment, carpal tunnel syndrome, and hepatosplenomegaly. Unlike Hurler syndrome(mucopolysaccharidosis type I), it is not associated with corneal clouding.

A 2-year-old boy with recurrent ear infections is brought to the pediatrician for a follow-up examination. He can walk with support and his vocabulary consists of approximately 50 words. His maternal uncle died in childhood from an unknown disease. Physical examination shows coarse facial features with an enlarged tongue. The abdomen is distended and both the liver and spleen tip are palpable. Laboratory studies show elevated total urinary glycosaminoglycan levels and an absence of plasma iduronate-2-sulfatase. Which of the following additional findings is most likely in this patient? A Optic atrophy B Leukopenia and thrombocytopenia C Cherry-red macula D Corneal clouding E Aggressive behavior

e. Splicing

A 2-year-old child with b-thalassemia has required frequent transfusions throughout his life because of anemia. A peripheral blood smear demonstrates microcytic, hypochromic red cells. After further genetic testing, the child's spleen is removed to lessen the need for transfusions. A segment of the suspected gene in this patient is removed and sequenced, and the corresponding RNA is experimentally reconstructed (shown below), demonstrating a mutation at the 5' end of the segment, as shown below. This change is most likely to affect which of the following processes? Select one: a. Capping b. Transcription c. Polyadenylation d. Translation e. Splicing

b. MCAD This patient appears to have suffered brain damage and died of severe hypoglycemia coupled with hyperammonemia. Deficiencies of pyruvate dehydrogenase or pyruvate carboxylase would produce psychomotor retardation due to major disruption of carbohydrate metabolism. But this patient's tests reveal a key finding—the presence of medium-chain (C6-C12) fatty acylcarnitine species in her blood. This isdiagnostic of MCAD deficiency, an impairment of metabolism of these fats and theiraccumulation to toxic levels. There has been speculation that MCAD deficiency andother undiagnosed metabolic disorders may be responsible for a significant proportionof sudden infant death syndrome (SIDS) cases. MCAD deficiency is now being testedas a component of mandatory newborn screening in many states.

A 21-month-old girl is hospitalized with a suspected gastrointestinal virus. She is vomiting and lethargic. Physical examination reveals poor muscle tone, guarding, and some cyanosis. Blood is drawn for chemistry and complete blood count, and an intravenous line is ordered for administration of glucose and electrolytes. Before this work is completed, the patient suffers a seizure and lapses into a coma. She dies 3 days later, despite intravenous treatments to stabilize her blood sugar. The original blood sample taken on admissionreveals severe hypoglycemia and hyperammonemia. An acylcarnitine profile of her blood indicates the presence of significant C6-C10 species. An evaluation of this patient's liver would reveal deficiency of which of the following enzyme activities? Select one: a. Pyruvate dehydrogenase b. MCAD c. CPT-II d. Pyruvate carboxylase e. CPT-I

e. Pyruvate kinase deficiency This clinical presentation represents biliary colic secondary to a hemolytic normocytic anemia. The presence of echinocytes makes this most consistent with a diagnosis of pyruvate kinase deficiency. Pyruvate kinase is the final regulated enzyme in the pathway of glycolysis and is responsible for converting phosphoenolpyruvate to pyruvate through substrate level phosphorylation of ADP to ATP. Genetic deficiency of this enzyme therefore inhibits the pathway of glycolysis and disrupts the production of ATP. This inhibition is particularly significant in red blood cells because these cells do not possess mitochondria and are completely dependent on glycolysis for energy production. Pyruvate kinase deficiency therefore causes a hemolytic anemia presenting in two phases, a younger and more severe variety that leads to early failure to thrive, and an older young adult presentation that is milder until decompensation due to gallstones or stress. The presentation of this patient is immediately suggestive of anemia due to the symptom of fatigue and the physical exam findings of conjunctival pallor and pale skin. The most common way to further differentiate anemias is based on MCV and this case is found to be a normocytic (between 80-100) anemia. Normocytic anemias can usually distinguished by characteristic blood smear findings as presented in the incorrect answers section. Almost all normocytic anemias are also hemolytic anemias leading to the accumulation of bilirubin gallstones so the right upper quadrant pain is not crucial in this case but can be important if distinguishing hemolytic from nonhemolytic anemias. Important associations with the diagnosis of pyruvate kinase deficiency are increased 2,3 bisphosphoglycerate and increased intermediate metabolites of glycolysis. The relevant mechanistic association is with decreased ATP production leading to decreased Na/K pump activity and subsequent cellular osmotic and morphological changes. Figure shows echinocytes that are often found in pyruvate kinase deficiency. The morphology change is the result of low levels of ATP not being able to power the Na/K pump sufficiently and leading to osmotic water shifts. Incorrect Answers: G6PD presents with a normocytic hemolytic anemia that is characterized by heinz bodies due to oxidized hemoglobin remnants. Hereditary spherocytosis is caused by mutations in red blood cell cytoskeletal proteins such as spectrin and ankyrin and is characterized by spherocytes. Hemoglobin C is marked by hemoglobin beta chain mutations and is characterized by intracellular crystals of aggregated beta chain. Iron deficiency causes a microcytic anemia with MCV <80.

A 22-year-old male presents to his primary care physician complaining of several instances of postprandial right upper quadrant pain that he describes as being "squeezing" in nature. He also says that he also has general fatigue and was easily winded in gym when he was in school. Physical exam reveals conjunctival pallor and generally pale skin. Based on clinical suspicion blood work is obtained showing a hematocrit of 34% (normal range 41%-50%). Mean corpuscular volume is found to be 87 fl (normal range 80 fl - 100 fl). A peripheral blood smear is also obtained and the results are shown. Which of the following is the most likely cause of this patient's syndrome? Select one: a. Iron deficiency b. Hemoglobin C c. Glucose-6-phosphate dehydrogenase deficiency d. Hereditary spherocytosis e. Pyruvate kinase deficiency

b. Oxidative damage to red blood cell membranes Oxidative damage to red blood cell membranes. The man has glucose-6-phosphate dehydrogenase deficiency and is incapable of regenerating reduced glutathione to protect red blood cell membranes from oxidative damage. In the presence of a strong oxidizing agent (the new drug the patient was taking), the red cell membranes undergo oxidative damage and the red cell bursts, leading to hemolytic anemia. This is all due to a lack of protective glutathione in the membrane. As the red cell lacks a nucleus, the cell cannot induce new gene synthesis. The drug thepatient was taking does not induce ion pores in red cell membranes or inhibit the HMP shunt pathway. It also does not cause oxidative damage to bone marrow. The drugs to avoid while prescribing for a patient with a G6PDH deficiency include primaquine, dapsone, nitrofurantoin, and sulfonylurea.

A 23-year-old man of Mediterranean descent was recently prescribed ciprofloxacin to treat a urinary tract infection. After 2 days on the drug, the patient was feeling worse, and weak, and went to the emergency department. He was found to have hemolytic anemia. This most likely resulted due to which of the following? Select one: a. Drug induced inhibition of the HMP shunt pathway b. Oxidative damage to red blood cell membranes c. Oxidative damage to bone marrow, interfering with red blood cell production d. Drug induced ion pores in the red blood cell membrane e. Induction of red blood cell cytochrome P450s, leading to membrane damage

c. Propionyl-CoA carboxylase This newborn has elevated levels of hydroxypropionic acid and symptoms suggestive of propionic acidemia. Propionyl-CoA carboxylase is deficient in this patient. Propionic acidemia is an autosomal recessive disorder caused by a deficiency of propionyl-CoA carboxylase in the propionic acid cycle. This deficiency results in the inability to metabolize VOMIT via the propionic acid cycle:-Valine-Odd-chain fatty acids (proprionic acid)-Methionine-Isoleucine-ThreoninePatients present with ketoacidosis and have elevated blood levels of propionic acid, methyl citrate, and hydroxypropionic acid. Other defects of propionic acidemia include methylmalonyl-CoA mutase deficiency which results in elevated methylmalonic acid levels.

A 3-day-old female infant presents with poor feeding, lethargy, vomiting after feeding, and seizures. Labs revealed ketoacidosis and elevated hydroxypropionic acid (from propionc acid) levels. Upon administration of parenteral glucose and protein devoid of valine, leucine, methionine, and threonine, and carnitine, the infant began to recover. Which of the following enzymes is most likely deficient in this infant? Select one: a. Phenylalanine hydroxylase b. Branched-chain ketoacid dehydrogenase c. Propionyl-CoA carboxylase d. Cystathionine synthase e. Homogentisate oxidase

e. Hemophilia A The infant has glucose-6-phosphate dehydrogenase deficiency (G6PD) and is experiencing a hemolytic anemic crisis due to the TMP-SMX the mother is taking, which in excreted in breast milk. G6PD deficiency has a similar mode of inheritance as hemophilia A, which is also X-linked recessive. G6PD deficiency is an X-linked recessive disease caused by a deficiency of the enzyme resulting in episodic anemia crisis. G6PD is an enzyme in the pentose phosphate shunt pathway that helps produces NADPH which is further used to regenerate glutathione, a molecule involved in neutralizing oxygen radicals and hydrogen peroxide. In the absence of G6PD, no NADPH is produced. This results in decreased glutathione production, leaving the erythrocytes vulnerable to damage by free radicals and lysis. Offending agents such as fava beans, dapsone, sulfonamides, quinine, and primaquine increase oxidative stress and will precipitate hemolytic anemia crises.

A 3-month-old African American infant presents to the hospital with 2 days of fever, "coke"-colored urine, and jaundice. The pregnancy was uneventful except the infant was found to have hyperbilirubinemia that was treated with phototherapy. The mother explains that she breastfeeds her child and recently was treated herself for a UTI with trimethoprim-sulfamethoxazole (TMP-SMX). Which of the following diseases is similarly inherited as the disease experienced by the child? Select one: a. Marfan syndrome b. Beta thalassemia c. Sickle cell anemia d. Rett syndrome e. Hemophilia A

b. Medium chain acyl-CoA dehydrogenase Medium chain acyl-CoA dehydrogenase. The child has MCAD (medium-chain acyl-CoA dehydrogenase) deficiency, an inability to completely oxidize fatty acids to carbon dioxide and water. With an MCAD deficiency, gluconeogenesis is impaired due to a lack of energy from fatty acid oxidation, and an inability to fully activate pyruvate carboxylase, as acetyl-CoA activates pyruvate carboxylase, and acetyl-CoA production from fatty acid oxidation is greatly reduced. In an attempt to generate more energy, medium-chain fatty acids are oxidized at the ω ends to generate the dicarboxylic acids seen in the question (see the figure below for an overview of ω oxidation). The finding of such metabolites (dicarboxylic acids) in the blood is diagnostic for MCAD deficiency. If there were mutations in any aspect of carnitine metabolism, there would be no oxidation of fatty acids (the fatty acids would not be able to enter the mitochondria), and the dicarboxylic acids (which are byproducts of fatty acid metabolism) would not be observed. Similarly, a mutation in the fatty acyl-CoA synthetase (the activating enzyme, converting a free fatty acid to an acyl-CoA) would also result in a lack of fatty acid oxidation, as fatty acids are not able to enter the mitochondria in their free (nonactivated) form.

A 3-month-old child had her fi rst ear infection and was feeding poorly due to the ear pain. One morning the parents found the child in a nonresponsive state and rushed her to the emergency department. A blood glucose level was 45 mg/dL, and upon receiving intravenous glucose the child became responsive. Further blood analysis displayed the absence of ketone bodies, normal levels of acyl-carnitine, and the presence of the following unusual carboxylic acids shown below. The enzymatic defect in this child is most likely in which of the following enzymes? Select one: a. Fatty acyl-CoA synthetase b. Medium chain acyl-CoA dehydrogenase c. Carnitine acyltransferase I d. Carnitine translocase e. Carnitine acyltransferase II

d. Insufficient energy for gluconeogenesis Defects in fatty acid oxidation deprive the liver of energy when fatty acids are the major energy source (such as during exercise, or a fast). Because of this, there is insufficient energy to synthesize glucose from gluconeogenic precursors (it requires 6 moles of ATP to convert 2 moles of pyruvate to 1 mole of glucose). Acyl-carnitines and dicarboxylic acids have no effect on the enzymes of gluconeogenesis, nor do they hinder the ability of the red blood cell to utilize glucose through the glycolytic pathway. Additionally, acetyl-CoA levels are low due to the lack of complete fatty acid oxidation and pyruvate carboxylase, a key gluconeogenic enzyme, is not fully activated. This also contributes to the reduced gluconeogenesis observed in patients with MCAD defective.

A 3-month-old child had her fi rst ear infection and was feeding poorly due to the ear pain. One morning the parents found the child in a nonresponsive state and rushed her to the emergency department. A blood glucose level was 45 mg/dL, and upon receiving intravenous glucose the child became responsive. Further blood analysis displayed the absence of ketone bodies, normal levels of acyl-carnitine, and the presence of the following unusual carboxylic acids shown below. Why were fasting blood glucose levels so low? Select one: a. Acyl-carnitine inhibition of gluconeogenesis b. Reduction of red blood cell production of lactate for gluconeogenesis c. Dicarboxylic acid inhibition of gluconeogenesis d. Insufficient energy for gluconeogenesis e. Dicarboxylic acid inhibition of glycogen phosphorylase

c. Liver α-glucosidase The infant has Pompe disease, a loss of liver α-glucosidase activity. This is glycogen storage disease II. The fi nding of normal glycogen structure eliminates liver debranching and branching activities as being defi cient. The missing enzyme is a lysosomal enzyme, and nondegraded glycogen accumulates in the lysosome, interfering with lysosomal function (hence, a lysosomal storage disease). The malfunctioning of the lysosomes is what leads to the muscle and liver problems. A defect in glycogen phosphorylase (liver) would lead to fasting hypoglycemia, and an enlarged liver, but not the muscle problems exhibited by the child. A defect in glycogen synthase would also lead to fasting hypoglycemia, but would not lead to severe muscle and liver disease. Additionally, in an individual with a defect in glycogen synthase, glycogen would not be found in the liver biopsy since it could not be formed.

A 3-month-old infant was brought to the pediatrician due to muscle weakness (myopathy) and poor muscle tone (hypotonia). Physical exam revealed an enlarged liver and heart, and heart failure. The infant had always fed poorly, had failure to thrive, and had breathing problems. He also had trouble holding up his head. Blood work indicated early liver failure. A liver biopsy indicated that glycogen was present and of normal structure. A potential defect in this child is which of the following? Select one: a. Liver glycogen synthase b. Liver debranching enzyme c. Liver α-glucosidase d. Liver branching enzyme e. Liver glycogen phosphorylase

e. Aldolase B The disorder is hereditary fructose intolerance, with a reduced ability to convert fructose-1-phosphate to dihydroxyacetone phosphate and glyceraldehyde. The specifi c defect is in aldolase B, with its activity reduced by as much as 85%. This problem is only evident when sucrose is introduced into the diet, and fructose enters the liver. The accumulation of fructose-1-phosphate, due to the reduced aldolase activity, leads to a constellation of physiological problems resulting in nausea, vomiting, and hypoglycemia. Elimination of fructose from the diet will reverse the symptoms. Galactokinase is needed for galactose metabolism; since the patient digests milk normally galactokinase activity is not altered. Similarly, glucose metabolism is not adversely affected (milk contains lactose, which is split into glucose and galactose), indicating that hexokinase and glucokinase activities are normal. The defect in aldolase B will hinder glycolysis, but the liver also contains aldolase C activity (this isozyme will not split fructose-1-phosphate), whichenables glucose metabolism to be very close to normal. A deficiency in fructokinase will lead to an accumulation of fructose (not fructose-1-phosphate), which is released into the urine (fructosuria), but does not lead to the physiological symptoms exhibited by the patient.

A 3-month-old girl is brought to the pediatrician due to fussiness and lethargy. According to the parents, the baby was just fine until the mother needed to return to work, and the baby was being switched from breast milk to baby foods, formula, and fruit juices. At that time, the child cried while feeding, sometimes vomited, and had been lethargic. The baby's appetite seemed to have worsened. The parents thought that if only formula was used, the baby was better, but they really could not remember. Which possible enzyme defect might lead to this case presentation? Select one: a. Hexokinase b. Glucokinase c. Galactokinase d. Fructokinase e. Aldolase B

e. Galactose-1-phosphate This infant who vomits after feeding and has early development of bilateral cataracts most likely has classic galactosemia, which results in accumulation of galactose-1-phosphate (Ga-1-P) inside cells due to deficiency of the Ga-1-P uridyltransferase. Early onset genetic disorders involving carbohydrate metabolism can be due to either abnormal processing of galactose or fructose. Early development of cataracts is the key distinguishing feature of galactose disorders because excess galactose is processed into galactitol in the lens of the eye and subsequently results in clouding. Additional symptoms such as fatigue, vomiting, mental retardation, and failure to thrive implicate galactose-1-phosphate (Ga-1-P) uridyltransferase deficiency. These symptoms arise because Ga-1-P acts as a phosphate sink meaning that accumulation of the molecule occupies so much phosphate that other phosphate dependent pathways in the cell are inhibited.b

A 3-week-old male is brought to the emergency department because of increasing lethargy. He was born at home without prenatal care or neonatal screening and appeared to be normal at birth. Despite this, his parents noticed that he would vomit after breastfeeding. He then progressively became more lethargic and began to have a few episodes of diarrhea after feeding. His parents do not recall any significant family history and neither of his siblings have had similar symptoms. Upon presentation, the infant is found to be generally unresponsive with mild hepatomegaly. Physical exam further reveals signs of clouding in the lenses of his eyes bilaterally. The levels of which of the following metabolites will be most dramatically elevated in this patient? Select one: a. Fructose-1-phosphate b. Fructose c. Galactose d. Lactose e. Galactose-1-phosphate

E Subsarcolemmal accumulation of mitochondria This patient's lactic acidosis and diffuse muscle weakness in combination with the family history of CNSdisease and childhood death suggest a form of mitochondrial myopathy, most likely MELAS syndrome. Subsarcolemmal and intermyofibrillar accumulation of mitochondria in muscle fibers is a hallmark of mitochondrial diseases, which are characterized by defective oxidative phosphorylation and subsequent lack of energy. The compensatory proliferation of mitochondria presents as characteristic ragged-red fibers on Gomori trichrome stain.

A 3-year-old boy is brought to the emergency department because of nausea and vomiting for 1 day. His maternal uncle had a seizure disorder and died in childhood. He appears fatigued. Respirations are 32/min. Examination shows diffuse weakness in the extremities. Serum studies show a low pH, elevated lactate concentration, and normal blood glucose. A metabolic condition characterized by a defect in oxidative phosphorylation is suspected. Microscopic examination of a muscle biopsy specimen of this patient is most likely to show which of the following findings? A Fibrofatty replacement of normal muscle fibers B Muscle atrophy with perimysial inflammation C Intermyofibrillar accumulation of glycogen D Endomysial inflammation with T-cell infiltration E Subsarcolemmal accumulation of mitochondria

e. Ascorbic acid 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 cofac- tors 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 cova- lently 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 de- hydrogenase to NAD+, forming FAD and NADH.

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? Select one: a. Lipoic acid b. Pantothenate c. Thiamine d. Niacin e. Ascorbic acid

e. Glucose-6-phosphate-dehydrogenase deficiency Normocytic anemia, back pain and hemoglobinuria following oxidant stresses (dapsone therapy in this vignette) in a patient of Mediterranean descent are highly suggestive of Glucose-6-phosphate-dehydrogenase (G6PD) deficiency. G6PD deficiency is an X-linked recessive disorder most commonly found in Greeks, Italians, and African Americans. The deficiency of G6PD reduces the synthesis of the reduced form of NADPH, which subsequently decreases glutathione (GSH) levels. Because GSH neutralizes hydrogen peroxide, the absence of GSH in G6PD deficiency allows peroxidase to oxidize hemoglobin. Anemia is often precipitated by oxidant stresses that induce hemolysis. Infections are the most common cause, followed by drugs including dapsone, antimalarials, and sulfonamide antibiotics.

A 32-year-old Greek man presents with a rash consistent with dermatitis herpetiformis and is treated with a course of oral dapsone. Over the following several days he becomes increasingly fatigued and experiences persistent back pain. Complete blood count reveals a normocytic anemia, and urinalysis shows hemoglobinuria. Results from a peripheral blood smear are shown in Figure. Which of the following is most likely responsible for this patient's symptoms? Select one: a. Pyruvate kinase deficiency b. Cold immune hemolytic anemia c. Hereditary spherocytosis d. Paroxysmal nocturnal hemoglobinuria e. Glucose-6-phosphate-dehydrogenase deficiency

a. erythrocyte protection against membrane oxidation

A 32-year-old man of Mediterranean descent is brought to the physician because of a 2-day history of fatigue and shortness of breath. He recently had a urinary tract infection treated with trimethoprim-sulfamethoxazole. His pulse is 100/minute. Physical examination shows pallor and scleral icterus. Laboratory studies show a decreased erythrocyte count and an increased serum bilirubin concentration that is mainly indirect This patient's anemia is most likely due to failure of which of the following? Select one: a. erythrocyte protection against membrane oxidation b. erythrocyte production of lactic acid c. reticulocyte maturation d. reticulocyte synthesis of heme e. bone marrow production of reticulocytes

c. Fatty acid transport into the mitochondria Fatty acid transport into the mitochondria. The man had eaten the unripe fruit of the ackee tree (from Jamaica). The unripened fruit contains the toxin hypoglycin A, which will interfere with carnitine's ability to transport acyl-carnitine groups across the inner mitochondrial membrane. This leads to a complete shutdown of fatty acid oxidation in all tissues in the affected individual, leading to severe hypoglycemia. Hypoglycin has no effect on fatty acid release from the adipocyte, or fatty acid entry into liver cells. Fatty acid oxidation is not directly inhibited, nor does this toxin directly inhibit the complexes of the electron transport chain and the proton-translocating ATPase.

A 35-year-old man in New York city, originally from Jamaica, purchased an illegally imported fruit from a street vendor and, within 4 h of eating the fruit, began vomiting severely. When brought to the emergency department the man was severely dehydrated and exhibited several seizures. The toxic effects of the fruit were interfering with which of the following? Select one: a. Oxidative phosphorylation b. Fatty acid entry into the liver cell c. Fatty acid transport into the mitochondria d. Fatty acid activation e. Fatty acid release from the adipocyte

c. Acetoacetate This patient appears to be suffering from diabetic ketoacidosis inducedby his failure to take his insulin on schedule. Although patients with diabetesmay have elevated levels of both protein and erythrocytes in urine, depending on thedegree of renal impairment, the best answer in this case is acetoacetate, a ketone bodythat should be highly elevated in his urine. The very low level of insulin has allowedglucagon action to run unchecked in stimulating fuel production by his adipose tissueand liver—increased gluconeogenesis, lipolysis, and ketogenesis. Urinary elevationsof lactate and pyruvate are characteristic of several metabolic disorders other thandiabetes.

A 35-year-old man is brought to the emergency department in a confused and semicomatosestate following a motor vehicle accident. His wife explains that he has type 1 diabetes mellitus. They were at a party earlier in the evening and both of them had two or three drinks. She is unsure whether he took his insulin before they left for the party. Physical examination reveals peripheral cyanosis and dehydration. While you are checking his abdomen, the patient doubles over and vomits. A fruity odor is detectable on his breath. A spot glucose reveals severe hyperglycemia.Testing of the patient's urine would likely reveal abnormally high levels of which of the following? Select one: a. Hemoglobin b. Lactate c. Acetoacetate d. Protein e. Pyruvate

F Acetyl coenzyme A Acetyl coenzyme A activates pyruvate carboxylase, a mitochondrial-specific enzyme that catalyzes oxaloacetate production. Oxaloacetate is converted to phosphoenolpyruvate (PEP) in the cytosol by the GTP-dependent enzyme PEP carboxykinase. PEP then undergoes essentially a reversed process of glycolysis, which results in glucose formation. Partitioning of the gluconeogenic pathway between the cytosol and mitochondria allows for greater control and regulation of gluconeogenesis.

A 36-year-old woman is fasting prior to a religious ceremony. Her only oral intake in the last 36 hours has been small amounts of water. A metabolic enzyme that plays an important role in maintaining normal blood glucosein this patient is located exclusively within the mitochondria. An increase in which of the following substances is most likely to increase the activity of this enzyme? A Insulin B Adenosine monophosphate C Glucagon D Oxidized nicotinamide adenine dinucleotide E Citrate F Acetyl coenzyme A

a. Glucose-6-phosphatase This patient is presenting with symptoms that are characteristic of type I glycogen storage disease (GSD), or von Gierke disease, which is a defect in glucose-6-phosphatase. Von Gierke disease is due to an autosomal recessive deficiency in glucose-6-phosphatase, an enzyme involved in producing glucose from glycogenolysis and gluconeogenesis. Type 1 GSD classically presents around 3-6 months of age. These patients tend to have severe hypoglycemia, lactic acidosis, hyperuricemia, and hyperlipidemia. On physical exam, they also classically have a "doll-like face" from fat deposition, as well as thin limbs and a protuberant abdomen from enlarged kidneys and liver. A glucagon stimulation test is used to differentiate between different types of glycogen storage disease. For this test, the patient fasts overnight, after which intramuscular glucagon is administered and blood glucose and plasma lactate are recorded in response. A defect in glucose-6-phosphatase results in impaired glycogenolysis; thus, the patient would not experience a rise in plasma glucose in response to this test.

A 4-month-old boy is brought to his pediatrician for a well-child visit. His parents have noticed that he has had poor growth compared to his older siblings. The boy was delivered vaginally after a normal pregnancy. His temperature is 98.8°F (37.1°C), blood pressure is 98/68 mmHg, pulse is 88/min, and respirations are 20/min. On exam, his abdomen appears protuberant, and the boy appears to have abnormally enlarged cheeks. A finger stick reveals that the patient's fasting blood glucose is 50 mg/dL. On further laboratory testing, the patient is found to have elevated blood lactate levels, as well as no response to a glucagon stimulation test. What enzymatic defect is most likely present? Select one: a. Glucose-6-phosphatase b. Alpha-1,6-glucosidase c. Glycogen phosphorylase d. Glycogen synthase e. Alpha-1,4-glucosidase

A A The deficient enzyme in von Gierke disease is glucose-6-phosphatase, which catalyzes the final steps of glycogenolysis and gluconeogenesis. Loss of this enzyme means the liver can no longer release glucose from its abundant glycogen stores, and it loses the ability to control blood sugar. Patients therefore present early in life (3-6 months) with severe fasting hypoglycemia, hepatomegaly, severe hyperlipidemia, hyperuricemia, and lactic acidosis. Treatment is focused on preventing hypoglycemia through frequent glucose supplementation (oral cornstarch), a diet of complex carbohydrates, and avoidance of other sugars that require this enzyme for metabolism into glucose (lactose, fructose, galactose, sucrose).

A 4-month-old boy is brought to the emergency department 30 minutes after having a generalized seizure. His mother reports that he has become lethargic and has had a weak cry for the past 3 weeks. He is at the 20th percentile for height and 15th percentilefor weight. Examination shows prominent cheeks and a distended, non-tender abdomen. The liver is palpated 5 cm below the right costal margin and there are palpable masses in both flanks. His serum glucose is 37 mg/dL, cholesterol is 253 mg/dL, uric acid is 9.5 mg/dL, and lactate is 4.3 mEq/L (N = 0.5-2.2). An enzyme responsible for which of the following reactions is most likely deficient in this patient? A A B B C C D D E E F F

d. Medium chain acyl-CoA dehydrogenase deficiency This patient is presenting with hypoketotic hypoglycemia, which is consistent with medium chain acyl-CoA dehydrogenase deficiency. Medium chain acyl-CoA dehydrogenase deficiency presents with hypoketotic hypoglycemia early in post-natal life with an accumulation of C8-C10 carnitines in the blood. Acetyl-CoA is a positive regulator of gluconeogenesis and decreased acetyl-CoA leads to decreased fasting glucose levels. Treatment includes dietary modification along with avoidance of fasting beyond 4-5 hours.

A 4-month-old girl is seen for ongoing lethargy and vomiting. She was born to a 31-year-old G2P2 mother with a history of hypertension. She has had 7 episodes of non-bloody, non-bilious vomiting and 3 wet diapers over the last 24 hours. Laboratory results are shown below. Serum: (Norma)Na+: 132 mEq/L (136-145 mEq/L)Cl-: 100 mEq/L (95-105 mEq/L)K+: 3.2 mEq/L (3.5-5.0 mEq/L)HCO3-: 27 mEq/L (22-28 mEq/L)BUN: 13 mg/dL (7-18 mg/dL)Glucose: 30 mg/dL (70-110 mg/dL)Lactate: 2 mmol/L (0.67-1.8 mmol/L)Urine ketones: < 20 mg/dL (Small: <20 mg/dL. Moderate: 30 to 40 mg/dL. Large: >80 mg/dL.)Which of the following is the most likely diagnosis? Select one: a. Glucocerebrosidase deficiency b. Galactose-1-phosphate uridyltransferase deficiency c. Sphingomyelinase deficiency d. Medium chain acyl-CoA dehydrogenase deficiency e. Glucose-6-phosphatase deficiency

c. Galactosylceramide The child has Krabbe disease, a mutation in a galactosidase, which cannot remove galactose from galactose cerebroside (an inability to break the bond between galactose and ceramide). The buildup of galactose-ceramide leads to the neuronal and muscle damage seen in the child. An inability to degrade a sulfatide would lead to metachromatic leukodystropy, which has very different symptoms than Krabbe disease. An inability to degrade glucosylceramide leads to Gaucher disease, again, with a very different disease progression than that seen with Krabbe disease. A defect in the degradation of sphingomyelin leads to Niemann-Pick disease, with a different set of symptoms than that seen with Krabbe disease. A defect in degrading ceramide leads to Farber disease, a defect in ceramidase. Farber disease is similar to Krabbe disease, but often presents with hepatomegaly and splenomegaly.

A 4-month-old infant is brought to the pediatrician for a variety of problems. The child is frequently irritable, small for age, vomits frequently, and displays hypotonia, as well as hyperesthesia (auditory, tactile, and visual). Liver and spleen size are normal. As the child ages, his condition worsens, with rapid psychomotor deterioration, seizures, and blindness. This disorder is caused by an accumulation of which of the following in neuronal lysosomes? Select one: a. Sulfatide b. Glucosylceramide c. Galactosylceramide d. Ceramide e. Sphingomyelin

e. Branching enzyme Branching enzyme. The child has a lack of branching enzyme activity, another glycogen storage disease, type IV (Andersen disease). In this case, the glycogen produced is a long, straight chain amylopectin, which has limited solubility, and precipitates in the liver (recall, the liver has the highest concentration of glycogen of all tissues). This leads to early liver failure (thus, the high bilirubin and transaminases in the serum) and death if a liver transplant is not performed. Defects in any of the other enzymes listed would lead to a different clinical scenario. Lack of glycogen phosphorylase or synthase, within the liver, would lead to fasting hypoglycemia, but not liver failure. Lack of these enzymes in the muscle would lead to exercise intolerance but would not affect blood glucose levels. Lack of α-glucosidase is Pompe disease, which also leads to an early death, but is due to the lack of a lysosomal enzyme, and there is no glycogen precipitation within the body of the liver. A lack of debranching activity is glycogen storage disease III, but would also lead to fasting hypoglycemia, without glycogen precipitation within the liver

A 4-month-old infant is seen by the pediatrician for failure to thrive. Examination shows distinct hepatosplenomegaly.Lab results show elevated transaminases and bilirubin, suggestive of liver failure. The boy dies shortly thereafter, and upon autopsy, precipitated carbohydrate was found throughout the liver. The boy most likely had a mutation in which of the following enzymes? Select one: a. Glycogen phosphorylase b. β-glucosidase c. Debranching enzyme d. Glycogen synthase e. Branching enzyme

e. Deficiency of CD-55 and CD-59 cell membrane proteins This patient has paroxysmal nocturnal hemoglobinuria (PNH) (hemolytic anemia, pancytopenia, and dark urine in the evenings/mornings (positive for hemoglobin and hemosiderin)), in which a deficiency of hematopoietic cell membrane proteins, specifically CD-55 and CD-59 is seen. The lack of the CD-55 (decay accelerating factor) and CD-59 (membrane inhibitor of reactive lysis) membrane proteins on red blood cells leads to an uncontrolled activation of complement. In turn, this unchecked, activated complement cascade results in intravascular lysis/destruction of RBCs. The presentation of PNH includes hemolytic anemia, pancytopenia, and venous thrombosis. Although hemolysis occurs throughout the day, the dramatic darkening in the color of urine in the evenings/early morning is due to the accumulation and concentration of the urine overnight. Diagnosis is confirmed by flow cytometry showing absence of CD55 and CD59.

A 42-year-old male presents to his primary care physician with complaints of fatigue and occasionally darkened urine over the past 3 months. Upon further questioning, the patient reveals that he has regularly had dark, 'cola-colored' urine when he has urinated at night or early in the morning. However, when he urinates during the day, it appears a much lighter yellow color. Laboratory work-up is initiated and is significant for a hemoglobin of 10.1 g/dL, elevated LDH, platelet count of 101,000/uL, and leukopenia. Urinalysis, taken from an early morning void, reveals brown, tea-colored urine with hemoglobinuria and elevated levels of hemosiderin. Which of the following is responsible for this patient's presentation? Select one: a. Autosomal dominant deficiency of spectrin protein in the RBC membrane b. Autosomal recessive deficiency of platelet Glycoprotein IIb/IIIa receptor c. Deficiency of C1 esterase-inhibitor d. Presence of a temperature-dependent IgG autoantibody e. Deficiency of CD-55 and CD-59 cell membrane proteins

E Lactase deficiency Lactase deficiency is the most likely cause of this patient's prolonged symptoms of abdominal discomfort, diarrhea, and bloating. Secondary lactase deficiency can be due to underlying disorders of the small intestine that result in mucosal damage, e.g., viral gastroenteritis. Lactase is found distally in the intestinal villi, which are particularly affected by mucosal damage. Lactose intolerance results in the passing of undigested, osmotically active lactose to the large intestine, where it binds water and is degraded by the native flora of the intestine, causing increased gas (flatulence) and short‑chain fatty acidformation (diarrhea).

A 43-year-old man comes to the physician because of a 2-week history of nonbloody diarrhea, abdominal discomfort, and bloating. When the symptoms began, several of his coworkers had similar symptoms but only for about 3 days. Abdominal examination shows diffuse tenderness with no guarding or rebound. Stool sampling reveals a decreased stool pH. Which of the following is the most likely underlying cause of this patient's prolonged symptoms? A Intestinal type 1 helper T cells B Anti-endomysial antibodies C Heat-labile toxin D Bacterial superinfection E Lactase deficiency

e. Western blot

A 43-year-old man is evaluated for progressive neuropsychiatric symptoms. A year ago, he began feeling depressed and having hallucinations. Five months later, he developed intermittent paresthesias and progressively worsening choreiform movements, myoclonus, and ataxia. These symptoms have not improved despite multiple hospitalizations; an extensive workup has been unrevealing. The patient is a slaughterhouse worker with extensive exposure to bovine offal. As part of the evaluation for prion disease, a tissue sample digested with protease is processed via gel electrophoresis and transferred to filter paper. Antibodies to a specific prion protein are added to the filter. Next, a marked protein that combines with the antibody-protein complex is used to determine whether the test is positive. Which of the following best describes this test? Select one: a. Southern blot b. Microarray c. Southwestern blot d. Northern blot e. Western blot

a. Coenzyme Q Coenzyme Q. Coenzyme Q is derived from isoprene units, which are produced in the pathway of cholesterol biosynthesis, after the HMG-CoA reductase step. If HMG-CoA reductase is inhibited (as it is by statins), then the production of the isoprenes is also reduced, and both Coenzyme Q and dolichol levels could become limiting. The biosynthesis of hemoglobin, ketone bodies, glycogen, or dihydrobiopterin is not dependent on isoprene units.

A 44-year-old man displayed elevated cholesterol levels and was prescribed a statin to reduce such levels. Statin treatment has the potential to interfere with the synthesis of which of the following? Select one: a. Coenzyme Q b. Hemoglobin c. Dihydrobiopterin d. Ketone bodies e. Glycogen

e. transketolase

A 50-year-old alcoholic male presents with pain, numbness, tingling, and weakness in his feet. He is diagnosed with thiamine deficiency. Thiamine is required as a cofactor for which of the following enzymes? Select one: a. gluconolactone hydrolase b. transaldolase c. 6-phosphogluconate dehydrogenase d. G6PDH e. transketolase

B Activation of phosphorylase kinase The activation of glycogen phosphorylase kinase in hepatocytes and myocytes is necessary for mediating the catabolism of glycogen to glucose (e.g., glycogenolysis). The whole process is initiated by the binding of glucagon to the glucagon receptor (liver) or epinephrine to the β adrenergic receptor (liver and muscle). This results in the stimulation of adenylate cyclase to increase conversion of ATP to cAMP,which leads to the activation of protein kinase A. Protein kinase A subsequently phosphorylatesglycogen phosphorylase kinase, which then activates glycogen phosphorylase. The enzymatic activity of glycogen phosphorylase is the rate-limiting step of glycogenolysis.

A 52-year-old man undergoes an exercise stress test for a 1-week history of squeezing substernal chest pain that is aggravated by exercise and relieved by rest. During the test, there is a substantial increase in the breakdown of glycogen in the muscle cells. Which of the following changes best explains this intracellular finding? A Decrease in protein kinase A B Activation of phosphorylase kinase C Increase in glucose-6-phosphate D Activation of protein phosphatase E Inactivation of glycogen synthase kinase

b. Cytochrome oxidase 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 iron- containing cytochromes in complex I because complex I contains proteins with iron-sulfur cen- ters. 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.

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? Select one: a. The F0 component of the F0-F1 ATPase b. Cytochrome oxidase c. The F1 component of the F0-F1 ATPase d. Complex I of the ETC e. The ATP-ADP antiporter

e. To reduce thromboxane synthesis To reduce thromboxane synthesis. Thromboxane A2 release from platelets is an essential element of forming blood clots, and aspirin will block prostaglandin, prostacyclin, and thromboxane synthesis. It is the thromboxane inhibition which reduces the risk of blood clots. Leukotrienes and lipoxins require the enzyme lipoxygenase, which is not inhibited by aspirin.

A 55-year-old man had been advised by his physician to take 81 mg of aspirin per day to reduce the risk of blood clots leading to a heart attack. The rationale for this treatment is which of the following? Select one: a. To reduce leukotriene synthesis b. To increase Lipoxin synthesis c. To reduce prostaglandin synthesis d. To increase prostacyclin synthesis e. To reduce thromboxane synthesis

D Renal cortex The renal cortex is the primary extrahepatic site of gluconeogenesis. It responds to the same hypoglycemic cues as the liver, including glucagon, glucocorticoids (cortisol), and acidosis(typically from starvation ketosis). Another important site is the intestinal epithelium, and recent evidence indicates that astrocytes of the central nervous system perform gluconeogenesis as well.

A 55-year-old man with alcoholic cirrhosis is admitted to the hospital for a scheduled liver transplantation. The physician asked the patient to stop eating 24 hours prior to the surgery. At the time of surgery, which of the following structures directly contributes to preventing hypoglycemia by releasing glucose into the blood? A Red blood cells B Skeletal muscle C Skin D Renal cortex E Adrenal gland

C Avoidance of milk feeds The most common form of galactosemia is galactose-1-phosphate uridyltransferase deficiency, which leads to the accumulation of galactose-1-phosphate and galactitol. The severe manifestations of galactosemia are a direct result of these toxic metabolites. Infantile cataracts are caused by the accumulation of galactitol in the eye lens. Newborns typically become symptomatic after the initiation of breastfeeding or formula feeding. The condition can be diagnosed by the presence of increased levels of galactose in the urine or increased levels of galactose-1-phosphate in the blood. Patients with classic galactosemia must avoid food containing lactose (e.g., cow's milk, breast milk).

A 6-day-old female newborn is brought to the physician because of yellowing of her eyes and body, vomiting, and poor feeding for 3 days. She has had diarrhea for the past 2 days. She was born at 38 weeks' gestation and the antenatal period was uncomplicated. She appears lethargic. Vital signs are within normal limits. Examination shows jaundice of the skin and conjunctivae. Bilateral cataracts are present. The abdomen is soft and nontender. The liver is palpated 4 cm below the right costal margin; there is no splenomegaly. Muscle tone is decreased in all extremities. Serum glucose concentration is 37 mg/dL. Which of the following is the most appropriate recommendation to prevent long-term complications of this illness? A Phototherapy B Frequent glucose feeds C Avoidance of milk feeds D Thiamine therapy E Levothyroxine therapy

D Aldolase B Aldolase B deficiency causes hereditary fructose intolerance, which manifests with vomiting, poor feeding, hypoglycemia, jaundice, and hepatomegaly. Aldolase B normally catalyzes the breakdown of fructose-1-phosphate to glyceraldehyde and dihydroxyacetone phosphate. Symptoms manifest around the time of weaning, as food containing fructose and/or sucrose food is added to the diet at that time. Treatment of this condition involves a lifelong fructose-free and sucrose-free diet.

A 6-month-old boy is brought to the emergency department by his mother because of recurrent vomiting and yellowing of his eyes. The mother says that he has been eating poorly since she started weaning him off of breast milk 5 days ago. At this time, mashed vegetables and fruits were added to his diet. Examination shows scleral jaundice and dry mucous membranes. The tip of the liver is palpable 4 cm below the right costal margin. His serum glucose concentration is 47 mg/dL, serum alanine aminotransferase is 55 U/L, and serum aspartate aminotransferase is 66 U/L. Which of the following enzymes is most likely deficient? A Galactokinase B Debranching enzyme C Galactose-1 phosphate uridyltransferase D Aldolase B E Fructokinase F Glucose-6-phosphatase

e. Glucosylceramide The child has a form of Gaucher disease, which is a defect in a glucosidase which removes glucose from glucosylceramide. The accumulation of glucosylcerebroside in the lysosomes leads to the observed symptoms. Defects in degrading galactosylceramide lead to Krabbe disease, which does not result in hepatomegaly and splenomegaly. A defect in degrading sulfatide leads to metachromatic leukodystrophy, which has different symptoms than what the child is experiencing. A defect in the degradation of sphingomyelin leads to Niemann-Pick disease, with a different set of symptoms than that seen with Gaucher disease. A defect in degrading ceramide leads to Farber disease, a defect in ceramidase, with more severe symptoms than those observed in Gaucher disease.

A 6-month-old boy is brought to the pediatrician due to a large stomach. The doctor noticed splenomegaly, with no pain. The boy was always tired and had anemia. The boy also has thrombocytopenia and bruises easily. X-rays show a deformity of the distal femur, as shown below. This disorder is caused by an accumulation of which of the following in macrophage lysosomes? Select one: a. Ceramide b. Galactosylceramide c. Sulfatide d. Sphingomyelin e. Glucosylceramide

C C The enzyme that activates galactose to galactose-1-phosphate is galactokinase (GALK). Congenital GALK deficiency leads to pathologically increased conversion of galactose to galactitol by aldose reductase and manifests with elevated blood galactose, galactosuria, and early-onset cataracts (due to accumulation of galactitol in the lens). Compared to the more common classic galactosemia, GALK deficiency is less severe and does not result in liver, kidney, or brain damage. Also, cognitive and motor development remains unaffected, as in this patient. Treatment includes a lifelong lactose-free and galactose-free diet.

A 6-month-old boy is brought to the physician for the evaluation of developmental delay. His mother reports that he babbles and turns his attention to her when she calls his name but she is concerned because he has never smiled and does not seem to recognize her face. He is exclusively breastfed and at the 60th percentile for height and 85th percentile for weight. Ophthalmological examination shows a white reflex in both eyesand a failure to track the penlight. During the examination, he sits without support and grasps toys that are placed in his hand. A reducing substance test result of the urine is positive, and a glucose oxidase test result is negative. This patient is most likely deficient in an enzyme that catalyzes which of the following labeled reactions? A A B B C C D D E E

b. Carnitine transporter Carnitine transporter. The child has a mutation in the enzyme which transports carnitine into liver and muscle cells, leading to a primary carnitine deficiency. The carnitine stays in the blood and is eventually lost in the urine (the same carnitine transporter is required to recover the carnitine from the urine in the kidney). Since the liver is carnitine defi cient, ketone body production is minimal at all times, even during a fast (thus, the lack of baseline ketone bodies in the circulation under these conditions). Fatty acids will rise in circulation, as they cannot be stored in the cells as acyl-CoA. The liver shows evidence of triglyceride formation as the acyl-CoA cannot be degraded, and acyl-CoA accumulates within the cytoplasm, leading to triglyceride formation. A defect in carnitine acyl transferase 1 would lead to elevated levels of carnitine in the circulation. A defect in carnitine acyltransferase II would lead to elevated levels of acyl-carnitine in the circulation (since the acyl group cannot be removed from the carnitine). The lack of circulating dicarboxylic acids indicates that the defect is not in MCAD (medium-chain acyl-CoA dehydrogenase). A defect in hormone sensitive lipase would show a decrease in free fatty acid levels, rather than the increase observed in the patient.

A 6-month-old child presents to the physician in a hypotonic state. The child has previously had a number of hypoglycemic episodes, at which times blood glucose levels were between 25 and 50 mg/dl. Blood work shows normal levels of ketone bodies (not elevated) during hypoglycemic episodes. Carnitine levels in the blood were, however, below normal. Free fatty acid levels were elevated in the blood, however acyl-carnitine levels were normal. Dicarboxylic acid levels were non-detectable in the blood. A liver biopsy shows elevated levels of triglyceride. A likely enzymatic defect is which of the following? Select one: a. Carnitine acyltransferase II b. Carnitine transporter c. Medium chain acyl-CoA dehydrogenase d. Carnitine acyltransferase I e. Hormone sensitive lipase

e. Glucose-6-phosphatase Glucose-6-phosphatase. The child has Von Gierke disease, glycogen storage disease type I, a lack of glucose-6-phosphatase. In such a disorder, glucose-6-phosphate, whether produced from glycogen degradation or gluconeogenesis, cannot be dephosphorylated for glucose export, and the liver cannot maintain blood glucose levels. The small amount of glucose which is exported (10% of the expected) is derived from the activity of debranching enzyme, which hydrolyzes an α-1,6-glucose linkage, which produces free glucose. The hepatomegaly arises due to excess glycogen in the liver (glucose-6-phosphate will activate glycogen synthase D), as does the increase in kidney size. A picture of a 25-month-old untreated child with this disorder is shown below. A lack of glycogen synthase would not lead to hepatomegaly, while a lack of branching enzyme leads to a different glycogen storage disease, with very different symptoms. A lack of debranching activity would not lead to hepatomegaly and would allow more glucose release than is observed through the normal action of glycogen phosphorylase. A defect in fructose-1,6-bisphosphatase would impair gluconeogenesis,but should not affect the ability of glycogen to be degraded to raise blood glucose levels.

A 6-month-old infant was brought to the pediatrician due to fussiness and a tender abdomen. The child seemed to do well until the time between feeding was increased to more than 3 h. The baby always seemed hungry and irritable if not fed frequently. Upon examination, hepatomegaly and enlarged kidneys were noted, and blood work showed fasting hypoglycemia. Subsequent laboratory analysis demonstrated that in response to a glucagon challenge, only about 10% of the normal amount of glucose was released into circulation, which significantly contributed to the fasting hypoglycemia. Which enzyme defect in the patient is the most likely? Select one: a. Debranching enzyme b. Fructose-1,6-bisphosphatase c. Glycogen synthase d. Branching enzyme e. Glucose-6-phosphatase

a. His hepatic NADH/NAD+ ratio is high The metabolism of alcohol involves the generation of NADH, leading to a high NADH/NAD+ ratio. Alcohol metabolism is a two-step process involving the transformation of alcohol to acetaldehyde (catalyzed by alcohol dehydrogenase, and then subsequent conversion of the acetaldehyde to acetic acid by aldehyde dehydrogenase. Both of the these reactions are oxidation reactions and result in the generation of NADH, leading to an elevated NADH/NAD+ level. The elevated NADH/NAD+ ratio favors glycolytic reactions and decreased hepatic gluconeogenesis, contributing to the hypoglycemia that chronic alcohol users are often found to have.

A 65-year-old homeless man with a history of hospitalization for alcohol intoxication is brought in confused. His serum glucose is 39mg/dl. Which of the following is likely true? Select one: a. His hepatic NADH/NAD+ ratio is high b. Hepatic gluconeogenesis is elevated c. The next step in management is glucose repletion d. He has decreased activity of alcohol dehydrogenase e. He has also been using cocaine

C Lysosomal acid maltase A defect in lysosomal acid maltase (also referred to as lysosomal acid alpha-glucosidase) is responsible for Pompe disease (i.e., GSD-II). Without this enzyme, the conversion of glycogen to glucose in the lysosome cannot be completed and glycogen buildup becomes toxic to numerous cells, predominantly skeletal and cardiac muscle cells. Notably, the condition does not manifest with hypoglycemia or metabolic acidosis because glycolysis and gluconeogenesis do not depend on lysosomal glycogenolysis. This patient has infantile-onset Pompe disease, which is the most severe form of this condition. It is characterized by rapidly progressing symptoms and has an average life expectancy of one year if left untreated. Enzyme replacement therapy (e.g., with recombinant human acid alpha-glucosidase) should be initiated as soon as possible in all patients with infantile-onset Pompe disease.

A 7-month-old boy is brought to the physician because of a 2-month history of fatigue, weakness, and difficulty feeding. He was delivered at term to a 32-year-old woman. He is not able to sit upright on his own. He is at the 75th percentile for height and 25th percentile for weight. His temperature is 37.7°C (99.8°F), blood pressure is 110/68 mm Hg, pulse is 150/min, and respirations are 50/min. His tongue protrudes beyond the teeth. Neck veins are distended. Crackles are heard at both lung bases. Cardiac examination shows an S3 gallop. The liver is palpated 2 cm below the right costal margin. Neurologic examination shows profound weakness in proximal and distal muscles of the upper and lower extremities. He has 2+ reflexes bilaterally. A chest x-ray shows cardiomegaly. Serum glucose is 95 mg/dL. Which of the following is the enzyme most likely to be defective in this patient? A Muscle glycogen phosphorylase B Very-long-chain acyl-CoA dehydrogenase C Lysosomal acid maltase D Glucose-6-phosphatase E Glycogen debranching enzyme

b. Reduced effectiveness of carboxylation reactions, leading to reduced glucose production Reduced effectiveness of carboxylation reactions, leading to reduced glucose production. Raw eggs contain a potent binding partner to biotin, avidin, which, while bound to biotin, blocks biotin's participation in carboxylation reactions. This leads to reduced activity of pyruvate carboxylase, a necessary step in many gluconeogenic pathways, thereby leading to a reduced ability of the liver to properly maintain blood glucose levels. As oxaloacetate levels drop due to the need of oxaloacetate for gluconeogenesis, acetyl-CoA derived from fatty acid oxidation increases, leading to ketone body formation. Avidin does not affect NAD or FAD levels, nor does it interfere with coenzyme A or vitamin C.

A 7-year-old girl, who lives on a farm, started to have shaking and sweating episodes. Upon physical examination, she was found to be hypoglycemic under fasting conditions (fasting blood glucose was 50 mg/dL) and positive for ketones in her blood and urine. Her growth curve is normal. Further analyses showed no other metabolic abnormalities. Probing further into her history, in the absence of her parents, revealed that one of her chores was to collect eggs from the chicken coop every morning, and she had gotten into the habit of eating one or two raw eggs every morning. This had been going on for the past 6 weeks or so. A reasonable explanation for her laboratory results is which one of the following? Select one: a. Reduced effectiveness of protein hydroxylation, leading to reduced enzymatic activity and reduced glucose production b. Reduced effectiveness of carboxylation reactions, leading to reduced glucose production c. Reduced effectiveness of acyl activation, leading to reduced glucose production d. Reduced levels of electron donors in her system, leading to reduced glucose production e. Reduced levels of electron acceptors in her system, leading to reduced glucose production

E Glycogen debrancher Cori disease is an autosomal recessive glycogen storage disorder resulting from a deficiency of glycogen debranching enzyme, which performs 2 functions as α-1,6-glucosidase and 4-α-D-glucanotransferase. It presents with hypotonia, muscle wasting, mild hypoglycemia, and ketosis, as observed in this patien

A 9-month-old girl is brought to the physician because of a 1-month history of poor feeding and irritability. She is at the 15th percentile for height and 5th percentile for weight. Examination shows hypotonia and wasting of skeletal muscles. Cardiopulmonary examination shows no abnormalities. There is hepatomegaly. Her serum glucose is 61 mg/dL, creatine kinase is 100 U/L, and lactic acid is within the reference range. Urine ketone bodies are elevated. Which of the following enzymes is most likely deficient in this patient? A Muscle phosphorylase B Acid alpha-glucosidase C Glucose-6-phosphatase D Glucocerebrosidase E Glycogen debrancher

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 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 3' ends face each other after hybridization to their respective templates.

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? Select one: a. The primers should consist of antiparallel complements of two parts of a noninfected human genome. b. The primers should be synthesized so that, after annealing with potential infective DNA, the 5' end of both primers "face" each other c. The primers should be designed to be synthesized with dideoxynucleotides to allow sequencing of the mutation d. The primers should be designed so that, after annealing with potential infective DNA, the 5' end of primer 1 would "face" the 3' end of primer 2 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

d. produce mannose-6-phosphate modifications in lysosomal enzymes Enzymes that are destined for the lysosomes (lysosomal enzymes) are directed there by a specific carbohydrate modification. During transit through the Golgi apparatus a residue of N-acetylglucosamine-1-phosphate is added to carbon 6 of one or more specific mannose residues that have been incorporated into these enzymes. The N-acetylglucosamine is activated by coupling to UDP and is transferred by an N-acetylglucosamine phosphotransferase yielding N-acetylglucosamine-1-phosphate-6-mannoseprotein. A second reaction removes the N-acetylglucosamine leaving mannose residues phosphorylated in the sixth position. A specific mannose-6-phosphate receptor is present in the membranes of the Golgi apparatus. Binding of mannose-6-phosphate to this receptor targets proteins to the lysosomes. Defects in the proper targeting of glycoproteins to the lysosomes can also lead to clinical complications. Deficiencies in N-acetylglucosamine phosphotransferase lead to the formation of dense inclusion bodies in fibroblasts. Two disorders related to deficiencies in the targeting of lysosomal enzymes are termed I-cell disease (mucolipidosis II) and pseudo-Hurler polydystrophy (mucolipidosis III). I-cell disease is characterized by severe psychomotor retardation, skeletal abnormalities, coarse facial features, painful restricted joint movement, and early mortality. Pseudo-Hurler polydystrophy is less severe; it progresses more slowly, and afflicted individuals live to adulthood.

A child aged 5 months was referred to a specialist. The child had been born with dislocated hips and a coarse featured face. He had been suffering repeated upper respiratory tract infections and did not seem to be developing his motor abilities, Clinical examination revealed hyperplasia of the gums, restriction of joint mobility and hepatosplenomegaly. On listening to the heart a mitral valve murmur could be detected. Further investigation involved cell culture of the child's fibroblasts obtained from a skin biopsy. Examination of the fibroblasts under the microscope revealed the presence of numerous intracellular inclusions, which on electron microscopy were revealed to be large lysosomes. Biochemical analysis showed decreased levels of the lysosomal hydro lase B-glucuronidase within the fibroblasts, but elevated levels of this enzyme within the culture medium. These symptoms arise from a defect in the targeting of lysosomal enzymes due to an inability to carry out which of the following processes? Select one: a. remove mannose-6-phosphates from lysosomal enzymes prior to their transport to the lysosomes b. synthesize the mannose-6-phosphate receptor found in lysosomes c. recycle the lysosomal receptor for mannose-6-phosphate present on lysosomal enzymes d. produce mannose-6-phosphate modifications in lysosomal enzymes e. transport mannose-6-phosphate receptors to lysosomes

b. N-acetylgalactosamine N-acetylgalactosamine. Tay-Sachs is a defect in hexosaminidase A, which removes the terminal N-acetylgalactosamine residue from ganglioside GM2, producing the free sugar and GM3. Hexosaminidase A does not cleave glucose, ceramide, sphingosine, or the fatty acyl component of ceramide from GM2; it is specific for N-acetylgalactosamine.

A child has been diagnosed with Tay-Sachs disease, in which a particular lipid accumulates within the lysosomes. The component of this lipid which cannot be removed in the lysosome is which of the following? Select one: a. Glucose b. N-acetylgalactosamine c. Ceramide d. Sphingosine e. Fatty acid

b. Transaldolase Transketolas. The patient is experiencing the symptoms of vitamin B1 (thiamine) defi ciency. Ethanol blocks thiamine absorption from the gut, so in the United States, one will usually only see a B1 deficiency in chronic alcoholics. One assay for B1 deficiency is to measure transketolase activity (which requires B1 as an essential cofactor) in the presence and absence of added B1. If the activity level increases when B1 is added, a vitamin defi ciency is assumed. None of the other enzymes listed (transaldolase, aldolase, β-ketothiolase, and acetylcholine synthase) require B1 as a cofactor, and, thus, could not be used as a measure of B1 levels. A reaction catalyzed by transketolase is shown below (note the breakage of a carbon-carbon bond, and then the synthesis of a carbon-carbon bond to generate the product of the reaction).

A chronic alcoholic presents to the emergency department with nystagmus, peripheral edema, pulmonary edema, ataxia, and mental confusion. The physician orders a test to determine if there is a vitamin deficiency. An enzyme used for such a test can be which of the following? Select one: a. β-ketothiolase b. Transaldolase c. Acetylcholine synthase d. Transketolase e. Aldolase

e. Dinitrophenol All of the poisons shown affect either electron transport or oxidative phosphorylation. Dinitrophenol is unique in that it disconnects the ordinarily tight coupling of electron transport and phosphorylation. In its presence, electron transport continues normally with no oxidative phosphorylation occurring. Instead, heat energy is gen- erated. The same principle is utilized in a well-controlled way by brown fat to generate heat in newborn humans and cold-adapted mammals. The bio- logical uncoupler in brown fat is a protein called thermogenin. Barbitu- rates, the antibiotic piericidin A, the fish poison rotenone, dimercaprol, and cyanide all act by inhibiting the electron transport chain at some point.

A comatose laboratory technician is rushed into the emergency room. She dies while you are examining her. Her most dramatic symptomis that her body is literally hot to your touch, indicating an extremely high fever. You learn that her lab has been working on metabolic inhibitors and that there is a high likelihood that she accidentally ingested one. Which one of the following is the most likely culprit? Select one: a. Dimercaprol b. Cyanide c. Barbiturates d. Piericidin A e. Dinitrophenol

c. TGTACA

A graduate student is running a laboratory experiment on Escherichia coli. For the experiment, she must insert a segment of human DNA into the circular genome of the bacteria. To accomplish this, she utilizes a restriction enzyme, which is known for its ability to cut DNA at palindromic sequences. Which of the following is a palindromic sequence of DNA? Select one: a. TATAAG b. TAGGAT c. TGTACA d. CGTCGC e. CAGAGC

c. Exhibits epithelial specific expression A northern blot is used to determine the level of RNA in a given sample. Comparisons of RNA levels in samples obtained from different cell types or tissues, from cells treated under different conditions, or from different species are examples of comparisons that can be made. Samples are separated by running an electric current through an agarose gel, transferring the separated sample to filter paper, and hybridization done with a specific RNA or DNA probe labeled to allow detection. 32P-labeled nucleotides are often used as probes and the hybridization detected by autoradiography. The hybridization can be done under high or low stringency conditions. Low stringency conditions allow for more mismatches and are often used to detect similar but not identical sequences. In the image shown, the experiment was carried out under high stringency conditions and the sample from the 2 epithelial cell types shows the strongest hybridization signal. A reasonable conclusion is that this protein exhibits epithelial cell specific expression. No conclusions can be drawn as to the mechanism of increased expression in these cells. If the experiment was repeated under low stringency, more RNA species would hybridize with the probe and more bands would likely be visible on the autoradiograph for epithelial cells and fibroblasts. The additional bands would represent similar but not identical sequences. High stringency requires the highest amount of base pairing for hybridization to occur. High stringency conditions consist of low salt concentrations, high temperature, and high concentrations of formamide while low stringency conditions consist of high salt concentrations, low temperature, and low concentrations of formamide.

A membrane protein was discovered that was important in cell-cell recognition. In order to learn more about the expression of this protein a Northern blot was done using RNA obtained from gingival epithelial cells, oral fibroblasts, and buccal epithelial cells. An RNA probe specific for the N-terminal region of this membrane protein was prepared, radio labeled, and hybridization carried out under conditions of high stringency. The autoradiograph shown in the image was obtained (refer to the image). What is the most logical conclusion about the protein based on the data shown in the image? Select one: a. Expression in epithelial cells results from gene amplification b. Is specifically expressed only in oral tissues c. Exhibits epithelial specific expression d. In epithelial cells and fibroblasts have identical amino acid compositions e. Exhibits fibroblast specific expression

d. CTTCTCCACAGG The complementary probe will be antiparallel to the coding strand of the mutant allele, with all sequences written 5' → 3'.

A missense mutation in codon 6 of the Sickle cell anemia is caused by B-globin gene. A man with sickle cell disease and his phenotypically normal wife request genetic testing because they are concerned about the risk for their unborn child. DNA samples from the man and the woman and from fetal cells obtained by amniocentesis are analyzed using the PCR to amplify exon 1 of the B-globin gene. Which 12-base nucleotide sequence was most likely used as a specific probe complementary to the coding strand of the sickle cell allele? Select one: a. CCTCACCTCAGG b. CTTCTCCCTCAGG c. GGACACCTCTTC d. CTTCTCCACAGG e. CCTGTGGAGAAG

c. Citrate translocase Citrate translocase. Citrate translocase is required for citrate to exit the mitochondria and enter the cytoplasm in order to deliver acetyl-CoA for fatty acid biosynthesis (see the fi gure below). Acetyl-CoA, which is produced exclusively in the mitochondria, has no direct path through the inner mitochondrial membrane. However, under conditions conducive to fatty acid biosynthesis (high energy levels, and allosteric inhibition of the TCA cycle), citrate will accumulate and leave the mitochondria. Once in the cytoplasm, citrate lyase will cleave the citrate to produce acetyl-CoA and oxaloacetate. The oxaloacetate is recycled to pyruvate, producing NADPH in the process, which is also required for fatty acid biosynthesis. A defect in either carnitine acyl transferase will not affect fatty acid biosynthesis, as those enzymes are required to transport the fatty acid into the mitochondria for its oxidation. A lack of glucose-6-phosphate dehydrogenase will not interfere with fatty acid synthesis, as malic enzyme can provide suffi cient NADPH for the pathway. MCAD is involved in fatty acid oxidation and does not affect fatty acid synthesis.

A mouse model has been generated as an in vivo system for studying fatty acid synthesis. An inactivating mutation was created which led to the cessation of fatty acid synthesis and death to the mice. This mutation is most likely in which of the following proteins? Select one: a. Glucose-6-phosphate dehydrogenase b. Carnitine acyl transferase I c. Citrate translocase d. Medium chain acyl-CoA dehydrogenase e. Carnitine acyl transferase II

c. Northern blot

A mutation in a non-coding sequence is believed to affect expression of the gene coding for a specific fetal enzyme. Liver and bone marrow cells from the fetus and his parents are obtained. Which of the following is the best method to determine if this gene is being transcribed in cultures of isolated cells? Select one: a. Southern blot b. Western blot c. Northern blot d. Southwestern blot e. ELISA

e. Peroxisomes This child appears to have died of a form of Zellweger syndrome. The key findings supporting this conclusion include the dysmorphic skeletal features, hepatomegaly, elevated blood copper and, most importantly, the accumulation of VLCFAs in tissues. Zellweger syndrome is a disorder of peroxisome biogenesis, and cells ofaffected individuals have very small or absent peroxisomes. The other major peroxisomaldisorder involving accumulation of VLCFAs, X-linked adrenoleukodystrophy, does not normally manifest during the neonatal period and is not associated with skeletal abnormalities. Further, peroxisomes are of normal size and appearance in the cells of patients with X-linked adrenoleukodystrophy.

A newborn baby boy is unconscious after having suffered a seizure. A variety of dysmorphic facial features are evident, including a high forehead, a flat occiput, large fontanelles, and a high arched palate. All reflexes are depressed. There is hepatomegaly consistent with impaired liver function revealed by blood chemistry. Testing also reveals high levels of copper in the blood, but adrenal function is within normal limits. Despite all interventions, the infant dies within a week of birth. Autopsy reveals an accumulation of VLCFAs in tissue samples of the liver and kidneys.Microscopic examination of tissues from this patient would likely indicate an absence of which of the following cellular components? Select one: a. Lipid droplets b. Mitochondria c. Endoplasmic reticulum d. Lysosomes e. Peroxisomes

(A) Fasting blood glucose A pyruvate carboxylase defi ciency will impair gluconeogenesis from lactate and pyruvate, thereby leading to fasting hypogly-cemia more easily than a pyruvate dehydrogenase defi -ciency (which will primarily affect the ability to generate energy from carbohydrates). Alanine amino transferase activity in the blood is a measure of liver damage, which would not distinguish between the two possibilities. Free fatty acid levels would be the same under both conditions, during fasting conditions, as would insulin and glucagon levels.

A newborn displays lethargy and crying episodes. Blood analysis indicates lactic acidosis and hyperalaninemia. In order to distinguish between a pyruvate dehydro-genase complex defi ciency and a pyruvate carboxylase defi ciency, one can measure which of the following in the blood? (A) Fasting blood glucose (B) Alanine aminotransferase activity (C) Free fatty acids levels when fasting (D) Insulin levels when fasting (E) Glucagon levels when fasting

c. DNA microarray

A newborn girl in the nursery is being evaluated for the first time. The pediatrician notes multiple anomalies including a cleft palate, microcephaly, anisocoria, and an anteriorly displaced anus. The genetics team is consulted, and recommends performing a test in which DNA samples from the patient's cells are added to a glass chip that contains thousands of sample genes. Once the DNA sample is added, the glass chip measures the amount of hybridization of each gene.The described test used by the geneticists is most consistent with which of the following? Select one: a. Next generation DNA sequencing b. Northern blot c. DNA microarray d. RNA interference e. Western blot

e. Restriction sites closely linked to the mutation causing the disease 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.

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? Select one: a. Antibodies used with ribonucleotide probes for detection of the altered trait b. Mutations flanking restriction sites so the entire gene can be sequenced c. gel electrophoresis to detect the gene product of normal and diseased individuals d. A unique single-stranded oligonucleotide that can be amplified by successive rounds of heating, cooling, and annealing e. Restriction sites closely linked to the mutation causing the disease

b. Allele-speciflc oligonucleotide probe hybridization

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 P-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? Select one: a. RLFP analysis b. Allele-speciflc oligonucleotide probe hybridization c. DNA fingerprinting d. DNA sequencing e. SNP assessment

c. Acetylation of core histones

A pharmacologist employed by a pharmaceutical company is investigating the mechanism of action of a new drug that significantly inhibits the division of tumor cells obtained from patients with acute myelogenous leukemia. He has determined that the drug serves as a potent inactivator of chromatin modifying activity that up-regulates the expression of a cluster of oncogenes in these tumor cells. Which type of chromatin-modifying activity is most likely stimulated by the enzyme target of this drug? Select one: a. Binding of histone Hl to nucleosomes b. Deacetylation of core histone H4 c. Acetylation of core histones d. Methylation o f cytosine bases in DNA e. Deamination of cytosine bases in DNA

c. Acetylation of core histones

A pharmacologist employed by a pharmaceutical company is investigating the mechanism of action of a new drug that significantly inhibits the division of tumor cells obtained from patients with acute myelogenous leukemia. He has determined that the drug serves as a potent inactivator of chromatin modifying activity that up-regulates the expression of a cluster of oncogenes in these tumor cells. Which type of chromatin-modifying activity is most likely stimulated by the enzyme target of this drug? Select one: a. Methylation o f cytosine bases in DNA b. Deacetylation of core histone H4 c. Acetylation of core histones d. Deamination of cytosine bases in DNA e. Binding of histone Hl to nucleosomes

d. Fructose-6-phosphate and glyceraldehyde-3-phosphate In the absence of NADP+, the oxidative steps of the HMP shunt pathway are nonfunctional, so only the nonoxidative steps will occur. In addition, PFK-1 has been made nonfunctional, such that glyceraldehyde-3-phosphate (G3P) cannot be produced from either fructose-6-phosphate (F6P) or glucose-6-phosphate (G6P). In order to generate ribose-5-phosphate (R5P) under these conditions, both F6P and G3P need to be provided. These two substrates will react, using transketolase as a substrate, to generate erythrose-4-phosphate (E4P) and xylulose-5-phosphate (X5P). The X5P will be epimerized to ribulose-5-phosphate (Ru5P), and then isomerized to R5P. Glucose-6-phosphate cannot be used as a substrate because it cannot be converted to G3P (due to the block in PFK-1). Pyruvate cannot be used as a substrate in extracts of red blood cells because such cells do not have pyruvate carboxylase, so the pyruvate cannot be converted to either F6P or G3P.

A researcher is studying the HMP shunt pathway in extracts of red blood cells, in the absence of NADP+, and in which PFK-1 has been chemically inactivated. Which carbon substrates are required to generate ribose-5-phosphate in this system? Select one: a. Fructose-6-phosphate and pyruvate b. Glucose-6-phosphate and pyruvate c. Glucose-6-phosphate and sedoheptulose-7-phosphate d. Fructose-6-phosphate and glyceraldehyde-3-phosphate e. Glucose-6-phosphate and glyceraldehyde-3-phosphate

(E) Inhibition of the proton translocating ATPase Oligomycin blocks the F0 component of the proton-translocating ATPase, thereby blocking proton fl ow through the enzyme and ATP synthesis. Oligomy-cin does not affect any other complex of oxidative phos-phorylation.

A researcher was studying oxidative phosphorylation in a suspension of carefully washed and isolated mito-chondria. ATP, ADP, inorganic phosphate, lactate, lactate dehydrogenase, and oxygen were introduced to the sus-pension, and he was able to demonstrate ATP production within the mitochondria. The researcher then added oli-gomycin to the mixture, which stopped oxygen uptake. This occurred due to which of the following? (A) Inhibition of complex I (B) Inhibition of complex II (C) Inhibition of complex III (D) Inhibition of complex IV (E) Inhibition of the proton translocating ATPase

c. Western blot Western blotting is used to detect a target polypeptide or protein from within a mixed sample. First, a group of potential target proteins are separated by gel electrophoresis. The separated proteins are then transferred to a nitrocellulose membrane and probed with antibody specific for the protein of interest. The membrane is then washed and treated with an antibody bound to a marker, such as a radioactive isotope, that is specific for the antibody used in the second step described above. For instance, a serum sample from a patient with suspected human immunodeficiency virus (HIV) infection can be analyzed via Western blot to detect antibodies directed against specific viral proteins. Following separation of viral proteins by gel electrophoresis and protein transfer to a nitrocellulose membrane, the membrane is treated with patient serum. If the patient is HIV positive, they will have antibodies that react with viral p24, gp41, and gp120/160. If two of these three bands are positive, the test is considered positive.

A solution containing a mixture of several proteins is processed via gel electrophoresis and transfer to filter paper. Antibodies to a specific protein are then added to the filter. Next a radioactive protein that combines with the antibody-protein complex is used to determine whether the test is positive. This type of analysis is called a Select one: a. Southwestern blot b. Southern blot c. Western blot d. Microarray e. Northern blot

d. Fatty acid (beta) oxidation The defective organelle in MELAS syndrome (mitochondrial encephalomyopathy, lactic acidosis, and stroke-like episodes) is the mitochondria. Fatty acid oxidation (beta oxidation) occurs in the mitochondria and peroxisomes, and would be replicated if their project is successful. Beta oxidation is the process in which fatty acid molecules are metabolized to acetyl-CoA units to be used in the Krebs cycle to generate NADH and FADH2. The process occurs in four steps: 1. dehydrogenation of the bond between the alpha carbon (C2) and the beta carbon (C3) to form a trans double bond (C1 being the carboxylic acid of the fatty acid) 2.) Hydration of the trans double bond 3.) Oxidation of the hydroxyl bond to a ketone 4.) Finally, the bond is cleaved to release a two carbon acetyl-CoA group to be used in the Krebs cycle

A startup is working on a novel project in which they claim they can replicate the organelle that is defective in MELAS syndrome (mitochondrial encephalomyopathy, lactic acidosis, and stroke-like episodes). Which of the following metabolic processes must they be able to replicate if their project is to mimic the metabolic processes of this organelle? Select one: a. Cholesterol synthesis b. Glycolysis c. Hexose monophaste shunt d. Fatty acid (beta) oxidation e. Fatty acid synthesis

c. 5'-ATAGTACCG; 5'-GTGCTAGTC

A team of researchers is studying the mechanisms of genetic shift and genetic drift in HIV. To further the investigation, the scientists need to use a technique where they amplify the sequence of DNA shown below: They decide to use polymerase chain reaction (PCR) in order to do so. Which of the following would be the most appropriate DNA primer to use for this cycle of PCR? Select one: a. 5′-CGGTACTAT; 5'-CTGATCGTG b. 5'-ATAGTACCG; 5'-GACTAGCAC c. 5'-ATAGTACCG; 5'-GTGCTAGTC d. 5'-CGGTACTAT; 5'-CACGATCAG e. 5'-CGGTACTAT; 5'-GTGCTAGTA

a. malic enzyme

Acetyl-CoA is transported out of the mitochondria in order to serve as a substrate for fatty acid or cholesterol synthesis. Which of the following enzymes used in this transport process provides a cofactor required for these reductive biosynthesis reactions? Select one: a. malic enzyme b. malate dehydrogenase c. citrate synthase d. ATP-citrate lyase e. pyruvate carboxylase

d. UDP-glucose Blood glucose is rapidly converted to glucose-6-phosphate upon entering cells by hexokinase or, in the case of the liver, by glucokinase. Glucose-6-phosphate is in equilibrium with glucose-1-phosphate via the action of phosphoglucomutase. Glucose-1-phosphate is activated by UTP to form UDP-glucose, which is added to glycogen by an α-1,4 linkage in the presence of glycogen synthase. To increase the solubility of glycogen and to increase the number of terminal residues, glycogen-branching enzyme transfers a block of about 7 residues from a chain at least 11 residues long to a branch point at least 4 residues from the last branch point. The branch is attached by an α-1,6 linkage.

After a meal, blood glucose enters cells and is stored as glycogen, particularly in the liver. Which of the following is the donor of new glucose molecules in glycogen? Select one: a. glucose-6-phosphate b. UDP-glucose-1-phosphate c. UDP-glucose-6-phosphate d. UDP-glucose e. glucose-1-phosphate

d. Na+, K+ ATPase Na+, K+ ATPase. Most monosaccharides are transported with sodium from the intestinal lumen into the enterocyte. The energy for active transport of the carbohydrate is derived from the sodium gradient that is established by the Na+, K+ ATPase, which pumps sodium out of the cell (three atoms of sodium) in exchange for potassium (two atoms of potassium). This creates both a sodium gradient (outside concentration higher) and a charge gradient (outside positive as compared to inside the cell) across the plasma membrane. Due to these gradients, the entry of sodium into the cell is energetically favorable, and the monosaccharide piggybacks with the sodium for transport into the cell. The Na /H+ exchanger is not operative in intestinal epithelial cells, and none of the other enzymes (glucose-6-phosphate dehydrogenase, hexokinase, and chloride transporter) will create the necessary sodium gradient for monosaccharide transport.

After eating a meal containing carbohydrates, the monosaccharides must be absorbed from the intestinal lumen. This transport is dependent on which of the following enzymes? Select one: a. Na+/H+ antiporter b. Glucose-6-phosphate dehydrogenase c. Chloride transporter d. Na+, K+ ATPase e. Hexokinase

c. Muscle glycogen phosphorylase The answer is A: Muscle glycogen phosphorylase. The patient is lacking muscle glycogen phosphorylaseand cannot utilize muscle glycogen for energy. This is another glycogen storage disease, type V, McArdledisease. The lack of muscle glycogen phosphorylase is why lactate production during exercise is very low. Asshown in the fi gure below, there are many glycogen particles present in the muscle cells just below the sarcolemma,as the glycogen is not able to be degraded. Muscle damage also results from vigorous exercise, releasing myoglobin into the circulation, which is what leads to the reddish-brown urine after exercise. Alterations in liver enzymes (phosphorylase or PFK-1) would not affect exercise tolerance in the muscle. Muscle does not contain glucose-6-phosphatase, and this problem is not due to a lack of muscle GLUT4 transporters, as the muscle cannot utilize stored, internal glucose supplies.

An 18-year-old man visits the doctor due to exercise intolerance. His muscles become stiff or weak during exercise, and he sometimes cramps up. At times, his urine appears reddish-brown after exercise. An ischemic forearm exercise test indicates very low lactate production. A potential enzyme defect in this man is which of the following? Select one: a. Muscle GLUT4 transporters b. Liver glycogen phosphorylase c. Muscle glycogen phosphorylase d. Muscle glucose-6-phosphatase e. Liver PFK-1

a. Fabry disease Fabry disease is an X-linked disorder that results from a deficiency in alpha-galactosidase A. This leads to the deposition of neutral glycosphingolipids with terminal alpha-galactosyl moieties in most tissues and fluids. Most affected tissues are heart, kidneys, and eyes. The predominant glycosphingolipid accumulated is globotriaosylceramide [galactosyl-(14)- galactosyl-(14)-glucosyl-(11')-cer amide]. With increasing age, the major symptoms of the disease are due to increase in deposition of glycosphingolipid in the cardiovascular system. Indeed, cardiac disease occurs in most hemizygous males. Three types of Gaucher disease (choice B) have been characterized and are caused by defects in lysosomal acid beta-glucosidase (glucocerebrosidase). Defects in this enzyme lead to the accumulation of glucosylceramides (glucocerebrosides), which leads, primarily, to central nervous system (CNS) dysfunction and also hepatosplenomegaly and skeletal lesions. NPD (choice D) comprises three types of lipid storage disorder, two of which (type A and B NPD) result from a defect in acid sphingomyelinase. Type A is a disorder that leads to infantile mortality. Type B is variable in phenotype and is diagnosed by the presence of hepatosplenomegaly in childhood and progressive pulmonary infiltration. Pathologic characteristics of NPD are the accumulation of histiocytic cells that result from phingomyelin deposition in cells of the monocytemacrophage system. Tay-Sachs disease (choice E) results from a defect in hexosaminidase A leading to the accumulation of gangliosides, particularly in neuronal cells. This defect leads to severe mental retardation, progressive weakness, and hypotonia, which prevents normal motor development. Progression of the disease is rapid and death occurs within the second year. Krabbe disease (choice C), also called globoid-cell leukodystrophy, results from a deficiency in galactosylceramidase (galactocerebroside betagalactosidase). This disease progresses rapidly and invariably leads to infantile mortality

An adult man suffered from stable angina pectoris for 15 years, during which time there was progressive heart failure and repeated pulmonary thromboembolism. On his death at age 63, autopsy disclosed enormous cardiomyopathy (1100 g), cardiac storage of globotriaosylceramide (11 mg lipid/g wet weight), and restricted cardiocytes. Which of the following lipid storage diseases would result in these clinical findings? Select one: a. Fabry disease b. Krabbe disease c. Gaucher disease d. Tay-Sachs disease e. Niemann-Pick disease (NPD) type I A

a. Acyl-CoA dehydrogenase The acyl-CoA dehydrogenases catalyze the fi rst step of the fattyacid oxidation spiral in that these enzymes create a carbon-carbon double bond between carbons 2 and 3 of the fatty acyl-CoA, generating an FADH2 in the process. The FADH2 then donates its electrons to the electron transfer flavoprotein (ETF), which then transfers the electrons to coenzyme Q (via the ETF:CoQ oxidoreductase). A lack of the oxidoreductase activity will lead to an accumulation of mitochondrial FADH2, depleting FAD levels, and reducing the activity of the acyl-CoA dehydrogenases. The lack of FAD does not directly inhibit the β-ketothiolase or enoyl-CoA dehydrogenase steps, nor does it affect the activity of the carnitine acyltransferases. The figure below shows the normal transport of electrons from FADH2 to coenzyme Q when the FADH2 is generated by the acyl-CoA dehydrogenases.

An inactivating mutation in the ETF : CoQ oxidoreductase will lead to an initial inhibition of which of the following enzymes in fatty acid oxidation? Select one: a. Acyl-CoA dehydrogenase b. Carnitine acyltransferase II c. Carnitine acyltransferase I d. Enoyl-CoA dehydrogenase e. β-ketothiolase

(C) Cytoplasmic malate dehydrogenase The cytoplasmic malate dehydrogenase is required in liver as part of the malate/aspartate shuttle in transferring reducing equivalents across the inner mitochondrial membrane. In the absence of such an activity, NADH levels will build up in the cytoplasm (since the electrons cannot be transferred to the mitochondrial matrix) and will lead to the reduc-tion of pyruvate to lactate to regenerate NAD+ for other cytoplasmic reactions. A defect in glucokinase will block glycolysis, with no pyruvate or lactate formation from glucose. The same is true for an inactivating mutation in PFK-1. If pyruvate kinase were defective, PEP would accumulate, which cannot be converted to lactate without forming pyruvate fi rst. A defect in glycerol-3-phosphate dehydrogenase will prevent the glycerol-3-phosphate shuttle from transferring electrons to the mitochondrial matrix, but the liver uses primarily the malate/aspartate shuttle for this activity. See the fi gure below for an over-view of the malate/aspartate shuttle system.

An inactivating mutation in which of the following enzymes would lead to lactic acid accumulation in the liver? (A) Glucokinase (B) Phosphofructokinase-1 (C) Cytoplasmic malate dehydrogenase (D) Pyruvate kinase (E) Glycerol-3-phosphate dehydrogenase

e. Adenylate cyclase Adenylate cyclase. If adenylate cyclase is defective, glucagon cannot initiate the activation of glycogenolysis and inhibition of glycolysis in the liver (cAMP levels will not increase, and PKA will stay inactive). Under such conditions, only the allosteric effectors in liver will be active, and there is no activator of glycogen phosphorylase b. When the hypoglycemia is severe enough, epinephrine release, working through its α-receptors, will activate phospholipase C, leading to calcium release. The increased calcium can activate phosphorylase kinase, which will activate phosphorylase, but fasting hypoglycemia will still occur. Defects in liver PFK-1 or glucokinase will not affect glycogenolysis or gluconeogenesis. Defects in liver galactokinase or fructokinase will not allow for metabolism of galactose or fructose, but do not affect the ability of the liver to degrade glycogen, or perform gluconeogenesis from other precursors.

An inactivating mutation in which of the following proteins can lead to fasting hypoglycemia? Select one: a. Fructokinase b. Liver glucokinase c. Galactokinase d. Liver PFK-1 e. Adenylate cyclase

b. glycogen phosphorylase activity will be increased

An increase in the state of phosphorylation, in hepatocytes in response to glucagon action, will be most accurately reflected by which of the following statements? Select one: a. phosphoprotein phosphatase-1 activity will be increased b. glycogen phosphorylase activity will be increased c. PFK-2 will exhibit an increased level of kinase activity d. phosphorylase kinase activity will be decreased e. PFK-1 activity will be unaffected

a. Acetyl-CoA carboxylase acetyl-CoA carboxylase . An inactivating mutation in acetyl-CoA carboxylase would lead to an inability to produce malonyl-CoA, which regulates fatty acid oxidation through an inhibition of carnitine acyl transferase 1. As malonyl-CoA levels increase, fatty acidoxidation is reduced, and as the levels decrease, fatty acid oxidation will increase. If malonyl-CoA decarboxylase were inactivated, malonyl-CoA levels would remain elevated, and fatty acid oxidation would be inhibited. Inactivating mutations in either carnitine acyltransferase 1 or 2 would lead to an inability to oxidize fatty acids, as they would not enter the mitochondria. A defect in medium-chain acyl-CoA dehydrogenase (MCAD) would also result in reduced fatty acid oxidation, as the initial step of the oxidation spiral would be inhibited once the fatty acid had been reduced to about 10 carbons in length. The reactions catalyzed by malonyl-CoA decarboxylase and acetyl-CoA carboxylase are shown below.

An individual contains an inactivating mutation in a particular muscle protein, which leads to weight loss due to unregulated muscle fatty acid oxidation. Such an inactivated protein could be which of the following? Select one: a. Acetyl-CoA carboxylase b. Malonyl-CoA decarboxylase c. Carnitine acyl transferase I d. Carnitine acyl transferase II e. Medium chain acyl-CoA dehydrogenase

d. á-ketoglutarate dehydrogenase The answer is A: a-ketoglutarate dehydrogenase. In order for citrate to be converted to glycogen, the citrate must fi rst be converted to oxaloacetate in the TCA cycle (which requires the participation of α-ketoglutarate dehydrogenase). From oxaloacetate, PEP carboxykinase will convert this to PEP, which will go through the gluconeogenic pathway up to glucose-6-phosphate. From there, G1P is produced, then UDPglucose, and finally incorporation of the glucose into glycogen. Pyruvate carboxylase, while being a gluconeogenic enzyme, converts pyruvate to OAA, which is not required in this series of reactions. PFK-1 and pyruvate kinase are irreversible enzymes of glycolysis and are not used in the gluconeogenic pathway. Glucose-6-phosphatase removes the phosphate from G6P, which is not required when glycogen is being synthesized.

An individual has been eating a large number of oranges during the winter months to protect against getting a cold. The excess carbons of citrate can be used to produce glycogen in the liver. Which one of the following liver enzymes is required for this conversion to occur? Select one: a. PFK-1 b. Pyruvate kinase c. Pyruvate carboxylase d. á-ketoglutarate dehydrogenase e. Glucose-6-phosphatase

e. Reduced ability to form malonyl-CoA Reduced ability to form malonyl-CoA. Biotinidase is required to remove covalentlybound biotin from proteins, which is how most of the biotin in our diet is received. In the absence of biotinidase, individuals can become functionally biotindeficient, due to the lack of free biotin in the body (as compared to being covalently bound to proteins). The formation of malonyl-CoA, via acetyl-CoA carboxylase, requires biotin as a required cofactor. Citrate lyase, malic enzyme, acetyl transacylase (an activity of fatty acid synthase) and acyl carrier protein (another component of fatty acid synthase) do not require biotin for their activity.

An individual with a biotinidase deficiency was shown to produce fatty acids at a greatly reduced rate (in the absence of supplements) as compared to someone who did not have the deficiency. This is due to which of the following? Select one: a. Defective acyl carrier protein b. Reduced activity of acetyl transacylase c. Reduced activity of malic enzyme d. Low activity of citrate lyase e. Reduced ability to form malonyl-CoA

A Ovary Ovarian, seminal vesicle, and liver cells express both aldose reductase, which reduces glucose to sorbitol, and sorbitol dehydrogenase, which converts sorbitol to fructose. Therefore, glucose can be converted into fructose in these tissues, and intracellular fructose concentrations are expected to be high in this experiment. In this experiment, fructose concentrations would be expected to be highest in liver cells, since they do not express hexokinase, which converts fructose into fructose-6-phosphate.

An investigator is conducting an experiment to study different pathways of glucose metabolism. He obtains cells cultured from various tissues to study the effect of increased extracellular glucose concentration. Following the incubation of these cells in 5% dextrose, he measures the intracellular fructose concentration. The concentration of fructose is expected to be highest in cells obtained from which of the following tissues? A Ovary B Kidney C Myelin sheath D Lens E Retina

D Glucose-6-phosphate to 6-phosphogluconolactone The formation of 6-phosphogluconolactone from glucose-6-phosphate by the enzyme glucose-6-phosphate dehydrogenase is the first step in the HMP shunt. HMP shunt reactions occur exclusively in the cytoplasm. Other metabolic processes that occur exclusively in the cytoplasm are glycolysis, nucleotide synthesis, nonsecretory protein synthesis, and fatty acid synthesis. The synthesis of secretory proteins occurs in the rough endoplasmic reticulum and the synthesis of cholesterol and steroids occurs in the smooth endoplasmic reticulum. Fatty acid beta-oxidation, Krebs cycle, oxidative phosphorylation, and ketogenesis are processes that occur exclusively in the mitochondria. Certain metabolic processes such as heme production and the urea cycle occur in both the mitochondria and cytoplasm. Gluconeogenesis is a process that occurs in the cytoplasm, mitochondria, and smooth endoplasmic reticulum.

An investigator is studying biomolecular mechanisms in human cells. A radioactive isotope that is unable to cross into organelles is introduced into a sample of cells. The cells are then fragmented via centrifugation and the isotope-containing components are isolated. Which of the following reactions is most likely to be present in this cell component? A Glucose-6-phosphate to glucose B Fatty acyl-CoA to acetyl-CoA C Carbamoyl phosphate to citrulline D Glucose-6-phosphate to 6-phosphogluconolactone E Isocitrate to α-ketoglutarate F HMG-CoA to acetoacetate G Pyruvate to oxaloacetate

C Uptake of fructose by small intestinal enterocytes Uptake of fructose by small intestine enterocytes is achieved via the GLUT-5 membrane transport protein. GLUT-5 is a uniporter that brings fructose from high concentrations in the intestinal lumen to lower concentrations in the intracellular compartment. This mechanism, like voltage-gated calcium channels, uses a specific transport protein that is not ATP-dependent and is thus also an example of facilitated diffusion. The uptake of glucose into cells via other types of transmembrane glucose transporters is also achieved via facilitated diffusion. Facilitated diffusion can take place either through carrier proteins or gated channels. While carrier proteins undergo a transformational change after binding of their substrate to enable transport across a membrane, gated channels form pores to allow free diffusion.

An investigator is studying membranous transport proteins in striated muscle fibers of an experimental animal. An electrode is inserted into the gluteus maximus muscle and a low voltage current is applied. In response to this, calcium is released from the sarcoplasmic reticulum of the muscle fibers and binds to troponin C, which results in a conformational change of tropomyosin and unblocking of the myosin-binding site. The membranous transport mechanism underlying the release of calcium into the cytosol most resembles which of the following processes? A Reabsorption of glucose by renal tubular cells B Secretion of doxorubicin from dysplastic colonic cells C Uptake of fructose by small intestinal enterocytes D Removal of calcium from cardiac myocytes E Absorption of LDL-cholesterol by hepatocytes

E Ribose-5-phosphate from fructose-6-phosphate Conversion of fructose-6-phosphate to ribose-5-phosphate is a nonoxidative (reversible) reaction of the pentose phosphate pathway (HMP shunt) that is increased when cells are unable to generate ribulose-5-phosphate (subsequently ribose-5-phosphate) and NADPH via the oxidative reaction. In the nonoxidative reaction, ribose-5-phosphate is produced by the enzymes transketolase and transaldolase using byproducts of glycolysis, fructose-6-phosphate, and glyceraldehyde-3-phosphate. Ribose-5-phosphate is used to synthesize nucleotides. There is no ATP produced in the pentose phosphate pathway. NADPH is required for fatty acid synthesis, cholesterol synthesis, and glutathione reduction.

An investigator is studying metabolic processes in cells from a mouse model. She identifies certain cells that are unable to generate enough reducing factor for respiratory burst. Increased production of which of the following substances is most likely to be present in these cells? A Reduced glutathione from glutathione disulfide B Ribulose-5-phosphate from glucose-6-phosphate C Palmitic acid from malonyl-CoA D Mevalonate from β-hydroxy-β-methylglutaryl-CoA E Ribose-5-phosphate from fructose-6-phosphate F 6-phosphogluconolactone from glucose-6-phosphate

D Tall villi with focal collections of goblet cells Tall villi lined by simple columnar epithelium and crypts with interspaced goblet cells (crypts of Lieberkühn) and Paneth cells describes normal jejunal architecture. Patients with lactose intolerance are deficient in lactase, permitting osmotically active lactose to pass undigested to the large bowel. The molecule subsequently binds water and acts as a substrate for colonic bacteria, leading to symptoms of flatulence, bloating, and abdominal pain. Biopsy shows normal histological findings, but samples will also show decreased activity of lactase.

An otherwise healthy 45-year-old woman comes to the physician because of a 1-year history of episodic abdominal cramps, bloating, and flatulence. The symptoms worsen when she has pizza or ice cream and have become more frequent over the past 4 months. Lactose intolerance is suspected. Which of the following findings would most strongly support the diagnosis of lactose intolerance? A Partial villous atrophy with eosinophilic infiltrates B Periodic acid-Schiff-positive foamy macrophages C Crypt abscesses and colonic ulcerations D Tall villi with focal collections of goblet cells E Duodenal epithelium with dense staining for chromogranin A F Villous atrophy and crypt hyperplasia G Noncaseating granulomas with lymphoid aggregates

d. Acyl-carnitine Primary carnitine deficiency is a lack of carnitine within the cell (such as a mutation in the carnitine transporter); secondary carnitine deficiency occurs when the carnitine is sequestered in the form of acyl-carnitine (the carnitine cannot be removed from the acyl group, such as a defect in carnitine acyl transferase 2). Thus, elevated levels of acylcarnitine would be expected in a secondary carnitine deficiency, but not in a primary carnitine deficiency. In both types of carnitine deficiencies, fatty acid oxidation is significantly reduced, so the levels of ketone bodies, glucose, lactate, and fatty acids would be similar under both conditions.

Carnitine deficiency can occur in a number of ways. Secondary carnitine deficiency can be distinguished fromprimary carnitine deficiency by measuring which of the following in the blood? Select one: a. Glucose b. Ketone bodies c. Lactic acid d. Acyl-carnitine e. Fatty acids

b. Sphingomyelin The most common sphingolipid in mammals is sphingomyelin. Ceramide, the basic structure from which all sphingolipids are derived, is composed of the 18-carbon sphingosine connected via itsa mino group to a fatty acid by an amide linkage. The fatty acid is usually long-chain (18 to 26 carbons) and is saturated or monosaturated. Except for the lack of the glycerol backbone, sphingolipids are quite similar in structure and physical properties to the phospholipids phosphatidyl choline and phosphatidyl ethanolamine. Either phosphoryl choline or phosphoryl ethanolamine is the head group attached to ceramide. If a neutral sugar residue is the polar head group attached to ceramide, a cerebroside is formed. If oligosaccharide head groups containing sialic acid are used, gangliosides are formed. All sphingolipids are important membrane constituents.

Ceramide is a precursor to which of the following compounds? Select one: a. Phosphatidyl glycerol b. Sphingomyelin c. Phosphatidyl serine d. Phosphatidyl choline e. Phosphatidyl ethanolamine

c. Acetyl CoA carboxylase Under conditions in which the entry charge of liver cells is high, intermediates of the citric acid cycle are abundant. Citrate, an early intermediate in the cycle, readily diffuses across the inner membrane of mitochondria and out into the cytosol. Citrate allosterically inhibits phosphofructokinase and, conversely, stimulates fructose-1,6-diphosphatase. Thus, when the energy level of hepatocytes is low and biosynthetic precursors are not abundant, phosphofructokinase is stimulated and glycolysis is favored. When the energy level is high, citrate inhibits phosphofructokinase, stimulates the diphosphatase, and thereby promotes gluconeogenesis. The diffusion of high levels of citrate into the cytosol also stimulates synthesis of fatty acids. Citrate activates acetyl CoA carboxylase, the first step in the synthesis of fatty acids, as well as provides its substrates, acetyl CoA and NADPH. However, citrate does not allosterically activate fatty acid synthetase. Enolase, an enzyme of the glycolytic pathway, is not regulated.

Citrate has a positive allosteric effect on which one of the following enzymes? Select one: a. Phosphofructokinase b. Enolase c. Acetyl CoA carboxylase d. Pyruvate kinase e. Fatty acid synthetase

c. E The answer is E. Under the conditions described, DNA synthesis is occurring without any requirement for NADPH (such as fatty acid synthesis). Under these conditions, NADPH levels are high and glucose-6-phosphate dehydrogenase is inactive. The cell requires ribose-5-phosphate, however, for nucleotide biosynthesis, and this is synthesized from fructose-6-phosphate and glyceraldehyde-3-phosphate using the nonoxidative reactions of the pathway. Thus, both transketolase and transaldolase will be active under these conditions. PFK-1 is active as well, as the only way to generate glyceraldehyde-3-phosphate from a sugar precursor is via enzymes of the glycolytic pathway.

Consider an intestinal epithelial cell in S phase, and for which the major, active biosynthetic pathway is nucleotide synthesis. Which one of the following best represents the activity state of a series of key enzymes under these conditions? Select one: a. D b. C c. E d. B e. A

d. Inhibition of cytochrome c oxidase

Cyanide is poisonous as it stops respiration because of its: Select one: a. Combination with RB cell membrane b. Formation of complex with Hb. c. Inhibition of myoglobin d. Inhibition of cytochrome c oxidase e. Inhibition of TCA cycle

a. Glycogen Phosphorylase Glycogen is the main source of glucose during the first 24 hours of a prolonged fast. Lack of glycogen phosphorylase, the major enzyme responsible for hydrolysis of glycogen (glycogenolysis), would severely impair the ability of the liver to make glucose from glycogen. The only other enzyme listed that would have any potential effect would be debranching enzyme, which helps remove the α-1,6-linked branches from glycogen and is required for complete degradation of glycogen. The other enzymes are involved either in glycogen synthesis or gluconeogenesis and would not have any effect on glucose production from glycogen.

Deficiency of which of the following enzymes would impair the body's ability to maintain blood glucose concentration during the first 24 hours of a prolonged fast? Select one: a. Glycogen Phosphorylase b. PEP carboxykinase c. Glycogen synthase d. Fructose 1,6-bisphosphatase e. Debranching enzyme

c. Fructose-2,6-bisphosphate

During the first week of a diet of 1500 calories per day (fasting state), the oxidation of glucose via glycolysis in the liver of a normal 59-kg (130-lb) woman is inhibited by the lowering of which one of the following? Select one: a. Citrate b. Fatty acyl-CoA c. Fructose-2,6-bisphosphate d. Ketone bodies e. ATP

e. There is one mRNA coding for both enzymes Lac operon Includes multiple genes, encoding proteins - enzymes that metabolize lactose (one of possible sources of energy production for E. coli)

E. coli colonies grown on a lactose-containing medium up-regulate the production of the enzymes I-galactosidase and galactoside permease. Which of the following best explains the synchronous production of both enzymes in response to lactose? Select one: a. There are two activator binding sites for one activator protein b. There are two operators for one repressor protein c. There are two repressors for one inducer d. There are two promoters in close proximity to each other e. There is one mRNA coding for both enzymes

e. There is one mRNA coding for both enzymes Lac operon Includes multiple genes, encoding proteins - enzymes that metabolize lactose (one of possible sources of energy production for E. coli)

E. coli colonies grown on a lactose-containing medium up-regulate the production of the enzymes I-galactosidase and galactoside permease. Which of the following best explains the synchronous production of both enzymes in response to lactose? Select one: a. There are two promoters in close proximity to each other b. There are two activator binding sites for one activator protein c. There are two repressors for one inducer d. There are two operators for one repressor protein e. There is one mRNA coding for both enzymes

a. 1 mol of glucose can be synthesized in gluconeogenesis During lipolysis, triglycerides are split into three free fatty acids and glycerol. The free fatty acids, as well as the free glycerol, diffuse into the bloodstream, where they are circulated throughout the body. The free fatty acids are used as an energy source for many tissues, primarily muscle. The free glycerol that is released cannot be phosphorylated back to glycerol-3-phosphate in the adipose tissue because it lacks glycerol kinase. However, the free glycerol released in lipolysis is taken up by the liver, where it can be phosphorylated to glycerol-3-phosphate. The phosphorylated glycerol can enter glycolysis or gluconeogensesis at the level of triose phosphates. If gluconeogenesis occurs, for every 2 mol of glycerol-3-phosphate, 1 mol of glucose can be synthesized.

For every 2 mol of free glycerol released by lipolysis of triacylglycerides in adipose tissue Select one: a. 1 mol of glucose can be synthesized in gluconeogenesis b. 1 mol of triacylglyceride is released c. 2 mol of triacylglycerides is released d. 3 mol of acyl CoA is produced e. 2 mol of free fatty acids is released

a. The ability of nucleases to act on mRNA 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).

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? Select one: a. The ability of nucleases to act on mRNA b. The rate of binding of ribosomes to mRNA c. The activity of RNA polymerase III d. The activity of RNA polymerase II e. The rate of mRNA translation

a. The ability of nucleases to act on mRNA 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).

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? Select one: a. The ability of nucleases to act on mRNA b. The rate of binding of ribosomes to mRNA c. The activity of RNA polymerase III d. The rate of mRNA translation e. The activity of RNA polymerase II

a. Be more specifically defined as UDP-glucose-glycogen glucosyl transferase Glycogen synthetase is an enzyme that transfers glucosyl moieties from UDP-glucose to a glycogen polymer primer. In plants, ADP-glucose plays a part similar to that of UDP-glucose in animals. The enzyme exists in two forms: an active, dephosphorylated form and an inactive, phosphorylated form. It is inactivated by phosphorylation of a specific serine residue. Glycogen breakdown, not synthesis, is positively affected by increased calcium levels.

Glycogen synthetase, the enzyme involved in the biosynthesis of glycogen, may . Select one: a. Be more specifically defined as UDP-glucose-glycogen glucosyl transferase b. Employ UDP-D-glucose as a glucosyl donor in both plants and animals c. Be activated by increased calcium levels d. Be activated by the phosphorylation of a specific serine residue e. Synthesize glycogen without a polymer primer

c. Increased melting point

Hydrogenation of an unsaturated fatty acid usually results in a compound having Select one: a. Decreased solubility in acid b. Decreased melting point c. Increased melting point d. Increased solubility in base

c. C-18 cis-Δ9,Δ12 In mammals, a variety of fatty acids are considered essential and cannot be synthesized. These include linoleate (C-18 cis-Δ9, Δ12) and linolenate (C-18 cis-Δ9, Δ12, Δ15). Either these fatty acids or fatty acids for which they are precursors must be supplied in the diet as starting points for synthesis of a variety of other unsaturated fatty acids that lead to the synthesis of prostaglandins, thromboxanes, and leukotrienes. For example, arachidonate, a 20-carbon fatty acid with four double bonds, is derived from linolenate. Arachidonate gives rise to some prostaglandins, thromboxanes, and leukotrienes. Some fatty acids must be obtained in the diet because of the limitations governing enzymes of fatty acid synthesis in humans; that is, double bonds cannot be introduced beyond the 9-10 bond position of carbons in the fatty acid chain, and subsequent double bonds after the first must be separated by two single bonds. Thus, linolenate and linoleate cannot be synthesized in humans.

In humans, the formation of the fatty acid C-18-Δ9,Δ12 can be derived from which of the following? Select one: a. C-18 cis-Δ6 b. C-16 cis-Δ6,Δ9 c. C-18 cis-Δ9,Δ12 d. C-18 cis-Δ9 e. C-18

c. skeletal muscle

In which of the following tissues would pentose phosphate pathway be expected to be the least active? Select one: a. erythrocytes b. lactating mammary tissue c. skeletal muscle d. adrenal gland e. adipose tissue

a. COX-2 is specifically induced during inflammation COX-2 is specifically induced during inflammation. COX-2 is induced during inflammatory conditions, while COX-1 is constitutively expressed. Thus, when an injury occurs, and an immune response is mounted at the site of injury, COX-2 is induced in those cells to produce second messengers that play a role in mediating the pain response. Specifically inhibiting the COX-2 isozyme will block the production of those second messengers, without affecting the normal function of COX-1. Inhibiting COX-1 may reduce the frequency of heart attacks, and inhibiting COX-2 will block prostaglandin production via the cylco-oxygenase. Recent data suggests that certain drugs that specifically block COX-2 have unwanted side effects, such as an increase in heart attacks.

Inhibitors specific for cyclooxygenase 2 (COX-2) were deemed more efficacious for certain conditions than inhibitors which blocked both COX-1 and COX-2 activities. This is due to which of the following? Select one: a. COX-2 is specifically induced during inflammation b. Inhibiting COX-2 did not alter prostaglandin production c. Specifically inhibiting COX-2 reduces the rate of heart attacks d. Inhibiting COX-1 increased the frequency of heart attacks e. COX-1 is inducible and only expressed during wound repair, while COX-2 is expressed constitutively

d. Deficiency of fatty acid desaturase greater than Δ9 Infants placed on chronic low-fat formula diets often develop skin problems, impaired lipid transport, and eventually poor growth. This can be overcome by including linoleic acid to make up 1 to 2% of the total caloric requirement. Essential fatty acids are required because humans have only Δ4, Δ5, Δ6, and Δ9 fatty acid desaturase. Only plants have desaturase greater than Δ9. Consequently, certain fatty acids such as arachidonic acid cannot be made "from scratch" (de novo) in humans and other mammals. However, linoleic acid, which plants make, can be converted to arachidonic acid. Arachidonate and eicosapentaenoate are 20-carbon prostanoic acids that are the starting point of the synthesis of prostaglandins, thromboxanes, and leukotrienes.

It has been noted that infants placed on extremely low-fat diets for a variety of reasons often develop skin problems and other symptoms. This is most often due to Select one: a. Glycogen storage diseases b. Deficiency of chylomicron and VLDL production c. Lactose intolerance d. Deficiency of fatty acid desaturase greater than Δ9 e. Antibody abnormalities

a. homeotic gene HOX gene Multiple developmental abnormalities due to mutation in a single gene.

Klein-Waardenburg syndrome is a single-gene disorder that includes dystopia canthorum (lateral displacement of the inner corner of the eye), impaired hearing, and pigmentary abnormalities. The gene involved is most likely to be a Select one: a. homeotic gene b. pseudogene c. tumor suppressor gene d. transgene e. proto-oncogene

e. Is homozygous for Tay-Sachs disease Gangliosides are continually synthesized and broken down. The specific hydrolases that degrade gangliosides by sequentially removing terminal sugars are found in lysosomes. In the lipid storage disease known as Tay-Sachs disease, ganglioside GM2 accumulates because of a deficiency of β-N-acetylhexosaminidase, a lysosomal enzyme that removes the terminal N-acetylgalactosamine residue. Homozygotes produce virtually no functional enzyme and suffer weakness, retardation, and blindness. Death usually occurs before infants are 3 years old. Carriers (heterozygotes) of the autosomal recessive disease produce approximately 50% of the normal levels of enzyme but show no ill effects. In high-risk populations, such as Ashkenazi Jews, screening for carrier status may be performed.

Leukocyte samples isolated from the blood of a newborn infant are homogenized and incubated with ganglioside GM2. Approximately 47% of the expected normal amount of N-acetylgalactosamine is liberated during the incubation period. These results indicate that the infant Select one: a. Will most likely have mental deficiency b. Is a heterozygote (carrier) for Tay-Sachs disease c. Has relatively normal β-N-acetylhexosaminidase activity d. Has Tay-Sachs syndrome e. Is homozygous for Tay-Sachs disease

c. Activation of pyruvate carboxylase Activation of pyruvate carboxylase. Fatty acid oxidation increases the levels of acetyl-CoA within the mitochondrial matrix, and acetyl-CoA is a potent activator of pyruvate carboxylase, a key gluconeogenic enzyme (it will convert pyruvate to oxaloacetate, a necessary fi rst step to bypass the irreversible pyruvate kinase reaction). Acetyl-CoA cannot be used to synthesize net glucose, so it is not an effective precursor of glucose production. Acetyl-CoA does not activate PEP carboxykinase (that enzyme is transcriptionally controlled), nor does it affect pyruvate kinase (a cytoplasmic enzyme). PFK-2 is not regulated by acetyl-CoA (phosphorylation by protein kinase A is the key regulator effect for PFK-2 in the liver).

Liver fatty acid oxidation leads to an enhancement of gluconeogenesis via which of the following? Select one: a. Inhibition of PFK-2 b. Inhibition of pyruvate kinase c. Activation of pyruvate carboxylase d. Generation of precursors for glucose synthesis e. Activation of phosphoenolpyruvate carboxykinase

e. 5'-TATTTATATTATTATTA-3'

Many different mechanisms are involved in regulation of gene expression during the differentiation of tissues and biologic processes in the multicellular organisms. One important mechanism regulates expression of different genes at the level of transcription. A DNA sequence in the human genome which is responsible for binding of multiple proteins, activating transcription is which of the following? Select one: a. 5'-CACACACACACACA-3' b. 5'-CGCGCGCGCGCGC-3 c. 5'-GGGTTTTGGGTGTGT-3 d. 5'-AAAGAGGAGAGGAG-3' e. 5'-TATTTATATTATTATTA-3'

a. Muscle phosphorylase Muscle phosphorylase deficiency leads to a glycogen storage disease [McArdle's disease] and, in young adults, an inability to do strenuous physical work because of muscular cramps resulting from ischemia. The compromised phosphorylation of muscle glycogen characteristic of McArdle's disease compels the muscles to rely on auxiliary energy sources such as free fatty acids and ambient glucose.

McArdle's disease causes muscle cramps and muscle fatigue with increased muscle glycogen. Which of the following enzymes is deficient? Select one: a. Muscle phosphorylase b. Muscle debranching enzyme c. Muscle glycogen synthetase d. Hepatic hexokinase e. Muscle hexokinase

e. The pentose phosphate pathway The sources of NADPH for synthesis of fatty acids are the pentose phosphate pathway and cytosolic malate formed during the transfer of acetyl groups to the cytosol as citrate. The enzyme citrate lyase splits citrate into acetyl CoA and oxaloacetate. The oxaloacetate is reduced to malate by NADH. NADP-linked malate enzyme catalyzes the oxidative decarboxylation of malate to pyruvate and carbon dioxide. Thus, the diffusion of excess citrate from the mitochondria to the cytoplasm of cells not only provides acetyl CoA for synthesis of fatty acids but NADPH as well. One NADPH is produced for each acetyl CoA produced. However, most of the NADPHs needed for synthesis of fatty acids are derived from the pentose phosphate pathway. For this reason, adipose tissue has an extremely active pentose phosphate pathway.

Most of the reducing equivalents utilized for synthesis of fatty acids can be generated from Select one: a. Glycolysis b. Citrate lyase c. Mitochondrial malate dehydrogenase d. The citric acid cycle e. The pentose phosphate pathway

c. Membrane lipids Multiple sclerosis is a demyelination disease characterized by chronic inflammation. The primary losses are of the phospholipids and sphingolipids composing myelin membrane sheets of nerves in white matter. Consequently, brain white matter eventually looks like gray matter. High levels of sphingolipids and phospholipids.

Multiple sclerosis is a disease characterized by chronic inflammation. There are significant data to indicate that susceptibility to multiple sclerosis is inherited and causes a primary change in Select one: a. Nucleotide metabolism b. Stored carbohydrates c. Membrane lipids d. Blood proteins e. Anti-inflammatory steroids

e. Pribnow boxes

Nucleotide sequences located a short distance upstream of the coding region of DNA are targeted by DNA gyrase to unwind and separate the double helix for transcription. These sequences are known as: Select one: a. Enhancer regions b. Shine-Delgarno sequences c. Palindromic sequences d. Termination sequences e. Pribnow boxes

d. CAP protein (cAMP-binding protein) will be bound to the lac promoter 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.

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? Select one: a. Attenuation of transcription of the trp operon will increase b. cAMP levels will be lower than cells growing in the presence of glucose c. RNA polymerase will not bind to the trp promoter d. CAP protein (cAMP-binding protein) will be bound to the lac promoter e. The lac repressor will be bound to the lac operator

e. CAP protein (cAMP-binding protein) will be bound to the lac promoter 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.

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? Select one: a. cAMP levels will be lower than cells growing in the presence of glucose b. The lac repressor will be bound to the lac operator c. Attenuation of transcription of the trp operon will increase d. RNA polymerase will not bind to the trp promoter e. CAP protein (cAMP-binding protein) will be bound to the lac promoter

a. Catalase

Periodontitis induces development of lipid peroxidation in the periodontal tissues, as well as increase in malondialdehyde and hydrogen peroxide concentration in the oral cavity. Which of the following enzymes provides antioxidant protection? Select one: a. Catalase b. ATP synthase c. Lactase d. Amylase e. Maltase

a. Phosphatidylinositol acts as a substrate for intracellular processes Phosphatidylinositol acts as a substrate for intracellular processes. Phosphatidylinositol is used as the substrate to provide signaling molecules in response to the appropriate stimuli (the phosphatidylinositol cycle). As such, it must face the cytoplasm of the cell such that when the inositol phosphate derivatives are produced, such as IP3, they can move to their target receptors to elicit a cellular response. Inositol contains six hydroxyl groups and is a very hydrophilic molecule. Inositol's structure is quite different from glucose (there is no carbonyl group in inositol), so it is unlikely that glucose and inositol would compete for binding to the same receptors. Phosphatidylinositol does not bind to phosphatidylserine in the inner leaflet of membranes. Inositol also does not interact with the actin cytoskeleton.

Phosphatidylinositol contributes to phospholipid bilayer asymmetry by being in the inner leaflet of membranes, facing the cytoplasm of the cell. This is most likely due to which of the following? Select one: a. Phosphatidylinositol acts as a substrate for intracellular processes b. Inositol is very similar in structure to glucose and could compete with glucose for binding of ligands to the extracellular surface c. The hydrophobic nature of inositol is unstable facing the cellular exterior d. Phosphatidylinositol binds to phosphatidylserine, another inner leaflet specific phospholipid e. Inositol interacts with intracellular actin, linking the inner leaflet to a cell's cytoskeleton

c. glucose 6-phosphate

Regulation of glycogen metabolism is tightly controlled at the level of the activity of glycogen phosphorylase. Which of the following is known to act as a negative effector of the phosphorylated (most active) form of glycogen phosphorylase? Select one: a. Ca2+ b. cAMP c. glucose 6-phosphate d. AMP e. protein kinase A

c. 2-phosphoglycerate Fluoride inhibits the glycolytic enzyme enolase, which catalyzes the dehydration of 2-phosphoglycerate to phosphoenolpyruvate. Thus, 2-phosphoglycerate accumulates under these conditions.

Streptococcus mutans, found in dental plaque, produces acids from the metabolism of carbohydrates. Topical fluoride treatment in the dental office can slow the production of acids, resulting in the accumulation of which metabolite? Select one: a. Fructose-1,6-bisphosphate b. Glucose-6-phosphate c. 2-phosphoglycerate d. Phosphoenolpyruvate e. Glyceraldehyde-3-phosphate

d. E The answer is E. Under fasting conditions, the liver is exporting glucose, so the pathways of glycogenolysisand gluconeogenesis will be active, while glycolysis will be inhibited (all due to the effects of glucagon andactivation of PKA). In glycolysis, PFK-2 is phosphorylated, activating its phosphatase activity, which leads toa reduction in fructose-2,6-bisphosphate levels. This results in a reduction of PFK-1 activity (thus, PFK-1 isnot active, but is not phosphorylated). Glycogen degradation has been activated, and synthesis inhibited, viathe phosphorylation of glycogen synthase, inactivating the enzyme (thus, glycogen synthase is not active,but is phosphorylated). Phosphorylase kinase has been activated, and phosphorylated, by PKA (so phosphorylase kinase is active, and phosphorylated). Pyruvate dehydrogenase is inactive under these conditions (due to fatty acid oxidation in the mitochondria acetyl-CoA levels and NADH levels are high, which slows down the TCA cycle and inhibits pyruvate dehydrogenase), and it is also phosphorylated by the PDH- kinase, which is activated by NADH.

Ten hours into a fast, in a normal individual, which of the following best represents the activity and phosphorylation state of a number of key enzymes within the liver? Select one: a. A b. C c. B d. E e. D

d. Acetyl CoA The enzyme controlling the first step in gluconeogenesis is pyruvate carboxylase. It catalyzes the conversion of pyruvate to oxaloacetate. Pyruvate is absolutely dependent upon the presence of the allosteric effector acetyl CoA or a closely related acyl CoA for its function. Under conditions of high-energy charge and high levels of acetyl CoA, oxaloacetate is utilized for gluconeogenesis. If low amounts of ATP are present, oxaloacetate is consumed in the citric acid cycle.

The activity of pyruvate carboxylase is dependent upon the positive allosteric effector Select one: a. Succinate b. AMP c. Citrate d. Acetyl CoA e. Isocitrate

e. Acetyl-CoA Acetyl-CoA is an allosteric regulator of pyruvate carboxylase which acts to increase gluconeogenesis. The purpose of gluconeogenesis is to create de novo glucose to maintain blood glucose levels when glycogen stores have been exhausted. Gluconeogenesis mainly occurs in the liver where the newly manufactured glucose is used to supply red blood cells and the brain with the glucose it needs to function; whereas, the liver does not use the glucose but rather uses beta oxidation to supply itself with energy. Regulation of gluconeogenesis can be favored in one direction over the other with stimulation occurring with glucagon, acetyl CoA, and citrate; inhibition can occur when there is a high NADH/NAD+ ratio (intoxication with ethanol). Incorrect Answers: AMP and fructose-2,6-biphosphate increases glycolysis via positive feedback to phosphofructokinase 1. ADP stimulates the TCA cycle via positive feedback to isocitrate dehydrogenase. Insulin acts to stimulate glycogen synthesis via activation of glycogen synthase.

The balance between glycolysis and gluconeogenesis is regulated at several steps, and accumulation of one or more products/chemicals can either promote or inhibit one or more enzymes in either pathway. Which of the following molecules if increased in concentration can promote gluconeogenesis? Select one: a. ADP b. Insulin c. Fructose-2,6-biphosphate d. AMP e. Acetyl-CoA

a. Structure C Fatty acid synthesis in the cytosol terminates at palmitate, a C16 saturated fatty acid. Elongation of acyl groups can occur from palmitate as well as from other dietary saturated and unsaturated fatty acids of lengths C10 and greater. Longer-chain elongation of fatty acids occurs in the endoplasmic reticulum (structure C) using malonyl CoA as the acetyl donor and NADPH as the reductant. Very-long-chain and long-chain fatty acids are preferentially catabolized in peroxisomes (structure J). Other structures diagrammed in the figure are the plasma membrane (A), mitochondrion (B), nucleoplasm (D), nucleolus (E), Golgi apparatus (F), secretory vesicles (G), caveolae such as those taking up low-density lipoprotein from its receptor (H), and lysosomes (I).

The elongation of fatty acids occurs in which of the diagrammatic structures shown below? Select one: a. Structure C b. Structure I c. Structure J d. Structure E e. Structure B f. Structure A g. Structure G h. Structure D i. Structure H j. Structure F

d. Palmitic acid In humans, the end product of fatty acid synthesis in the cytosol is palmitic acid. The specificity of cytosolic multienzyme, single-protein fatty acid synthetase is such that once the C16 chain length is reached, a thioesterase clips off the fatty acid. Elongation as well as desaturation of de novo palmitate and fatty acids obtained from the diet occur by the action of enzymes in the membranes of the endoplasmic reticulum.

The end product of cytosol fatty acid synthase in humans is Select one: a. Linoleic acid b. Oleic acid c. Arachidonic acid d. Palmitic acid e. Palmitoleic acid

c. Is composed of covalently linked enzymes The fatty acid synthase complex of mammals is composed of two identical subunits. Each of the subunits is a multienzyme complex of seven enzymes and the acyl carrier protein component. All the components are covalently linked together; thus, all the components are on a single polypeptide chain, which functions in the presence of another identical polypeptide chain. Each cycle of fatty acid synthesis employs the acyl carrier protein and six enzymes: acetyl transferase, malonyl transferase, β-ketoacyl synthase, β-ketoacyl reductase, dehydratase, and enoyl reductase. When the final fatty acid length is reached (usually C16), thioesterase hydrolyzes the fatty acid off of the synthase complex.

The fatty acid synthase complex of mammals Select one: a. Dissociates into eight different proteins b. Is composed of seven different proteins c. Is composed of covalently linked enzymes d. Is a dimer of unsimilar subunits e. Catalyzes eight different enzymatic steps

d. Decreased concentrations of glucose, increased concentration of cAMP The lactose (lac) operon is a classic model for understanding gene regulation. It is negatively controlled through two regulatory genes—the lac I gene that constitutively (always) expresses a repressor protein and the operator (o) region to which the repressor binds. The lac operon is inducible by lactose and lactose analogues, inactivating the repressor and uncovering the operon and its neighboring promoter (p) sequence. RNA polymerase then transcribes the inducible, structural genes β-galactosidase (z), permease (y), and transacetylase (a). The RNA transcript is polycistronic, so that one regulatory site allows transcription of all three genes needed for the metabolism of lactose. The bacterial ribo- somes immediately attach to the nascent RNA transcript, allowing for simultaneous transcription and translation. When all the lactose is metab- olized, the repressor returns to its native conformation, binds to the oper- ator, and shuts down lac operon transcription

The lactose operon is negatively controlled by the lactose repressor and positively controlled by which of the following? Select one: a. Decreased concentrations of glucose and cAMP b. Increased concentrations of glucose and adenosine triphosphate (ATP) c. Increased concentrations of glucose, decreased concentration of cAMP d. Decreased concentrations of glucose, increased concentration of cAMP e. Increased concentrations of glucose and cyclic AMP (cAMP)

b. Decreased concentrations of glucose, increased concentration of cAMP The lactose (lac) operon is a classic model for understanding gene regulation. It is negatively controlled through two regulatory genes—the lac I gene that constitutively (always) expresses a repressor protein and the operator (o) region to which the repressor binds. The lac operon is inducible by lactose and lactose analogues, inactivating the repressor and uncovering the operon and its neighboring promoter (p) sequence. RNA polymerase then transcribes the inducible, structural genes β-galactosidase (z), permease (y), and transacetylase (a). The RNA transcript is polycistronic, so that one regulatory site allows transcription of all three genes needed for the metabolism of lactose. The bacterial ribo- somes immediately attach to the nascent RNA transcript, allowing for simultaneous transcription and translation. When all the lactose is metab- olized, the repressor returns to its native conformation, binds to the oper- ator, and shuts down lac operon transcription

The lactose operon is negatively controlled by the lactose repressor and positively controlled by which of the following? Select one: a. Increased concentrations of glucose and cyclic AMP (cAMP) b. Decreased concentrations of glucose, increased concentration of cAMP c. Increased concentrations of glucose and adenosine triphosphate (ATP) d. Decreased concentrations of glucose and cAMP e. Increased concentrations of glucose, decreased concentration of cAMP

a. HMP-shunt

The most important source of reducing equivalents for FA synthesis in the Liver is: Select one: a. HMP-shunt b. TCA cycle c. Gluconeogenesis d. Glycolysis e. Uronic acid pathway

d. 19 Acetoacetate will react with succinyl-CoA to produce acetoacetyl-CoA and succinate (this costs 1 GTP, as the succinate thiokinase step is skipped). The acetoacetyl-CoA is converted to two acetyl-CoA, each of which can generate 10 ATP when completely oxidized (each acetyl-CoA generates 1 GTP, 3 NADH, and 1 FADH2). The sum, then, is 20 minus the 1 lost in the CoA transferase step, for a net yield of 19 ATP.

The net energy yield obtained (moles of ATP per mole of substrate oxidized) when acetoacetate is utilized by the nervous system as an alternative energy source is which of the following? Consider that acetoacetate must be oxidized to four molecules of carbon dioxide during the reaction sequence. Select one: a. 20 b. 18 c. 17 d. 19 e. 21

c. DNA methylation DNA methylation occurs mainly at CpG dinucleotides that often cluster in at the upstream promoter regions of genes (CpG islands). While these are generally correlated with gene inactivation, there are many exceptions. Double crossovers at meiosis can substitute a normal allele for a mutant allele (conversion), and reverse transcriptases can copy intronless mRNA into complementary DNAs (cDNAs) that integrate into the genome as pseudogenes. Immunoglobulin genes undergo gene rearrangement to unite variable, joining, and constant regions for expres- sion of a unique antibody. Unequal crossing over between sister chromatids is thought to be an important mechanism for variation in copy number within gene clusters.

The process that occurs at the 5 position of cytidine and often correlates with gene inactivation is Select one: a. Pseudogene b. Gene rearrangement c. DNA methylation d. Sister chromatid exchange e. Gene conversion

d. DNA methylation DNA methylation occurs mainly at CpG dinucleotides that often cluster in at the upstream promoter regions of genes (CpG islands). While these are generally correlated with gene inactivation, there are many exceptions. Double crossovers at meiosis can substitute a normal allele for a mutant allele (conversion), and reverse transcriptases can copy intronless mRNA into complementary DNAs (cDNAs) that integrate into the genome as pseudogenes. Immunoglobulin genes undergo gene rearrangement to unite variable, joining, and constant regions for expres- sion of a unique antibody. Unequal crossing over between sister chromatids is thought to be an important mechanism for variation in copy number within gene clusters.

The process that occurs at the 5 position of cytidine and often correlates with gene inactivation is Select one: a. Sister chromatid exchange b. Gene rearrangement c. Pseudogene d. DNA methylation e. Gene conversion

e. dephosphorylation The level of active PDHc is controlled by its state of phosphorylation. A family of enzymes called the PDH kinases phosphorylates the PDHc. In order to return the PDHc to full activity following PDH kinase-mediated phosphorylation, the phosphates are removed by PDH phosphatase.

The pyruvate dehydrogenase complex (PDHc) is a multi-subunit enzyme whose activity is regulated by both allosteric effectors and covalent modification. Which of the following exerts a positive influence on the activity of PDHc toward pyruvate? Select one: a. phosphorylation b. acetyl-CoA c. NADH d. ATP e. dephosphorylation

a. The size of the DNA fragments

The rate of migration of DNA within an agarose gel in the gel electrophoresis technique is primarily based on what factor? Select one: a. The size of the DNA fragments b. The number of DNA fragments c. The negative charge of the DNA d. The size of the wells of the gel e. The volume of the DNA sample loaded

c. I-cell disease I-cell disease. The sphingolipidoses and I-cell disease are both lysosomal storage diseases, whereas the other disorders listed do not involve lysosomal dysfunction. Mitochondrial myopathy, encephalopathy, lactic acidosis, and stroke (MELAS) is a mitochondrial disorder, and Zellweger's is a disorder of peroxisomal biogenesis. G6PDH (glucose-6-phosphate dehydrogenase) deficiency and von Gierke disease are single gene mutations which do not alter lysosomal function (although type II glycogen storage disease, Pompe disease, is a lysosomal storage disease).

The sphingolipidoses, as a class, are most similar to which one of the following disorders? Select one: a. MELAS b. Glucose-6-phosphate dehydrogenase deficiency c. I-cell disease d. Zellweger syndrome e. von Gierke disease

d. In both the cytosol and mitochondria The synthesis of 3-hydroxy-3-methylglutaryl CoA requires the condensation of three acetyl CoA groups. The two enzymatic steps involved are the first two steps of cholesterol synthesis and ketone body synthesis. While cholesterol synthesis occurs in the cytosol of most mammalian tissues, ketone body synthesis can only occur in the mitochondria of liver cells. Not only are cholesterol synthesis and ketone body synthesis separated by compartmentalization, they are separated by metabolic needs. Cholesterol synthesis is an anabolic pathway that takes place when acetyl CoA production from excess dietary precursors is possible. In contrast, ketone body production by the liver occurs when acetyl CoA levels from β oxidation are high. This catabolic situation exists during fasting, starvation, and uncontrolled diabetes.

The synthesis of 3-hydroxy-3-methylglutaryl CoA can occur Select one: a. Only in the endoplasmic reticulum of all mammalian tissues b. Only in the cytosol of all mammalian tissues c. In lysosomes d. In both the cytosol and mitochondria e. Only in mitochondria of all mammalian tissues

b. Phosphoglycerate kinase Phosphoglycerate kinase. In gluconeogenesis, phosphoglycerate kinase catalyzes thephosphorylation of 3-phosphoglycerate to 1,3-bisphosphoglycerate, a step which requires ATP. The other twosteps requiring a high-energy phosphate bond in the conversion of pyruvate to glucose are pyruvate carboxylaseand phosphoenolpyruvate carboxykinase. Fructose-1,6-bisphosphatase and glucose-6-phosphatase are enzymes that remove phosphates from substrates, releasing the phosphates as inorganic phosphate. They do not require, nor generate, ATP. Pyruvate kinase is not utilized for gluconeogenesis, and triose phosphate isomerase catalyzes the conversion of dihydroxyacetone phosphate and glyceraldehyde-3-phosphate, without the involvement of a high-energy phosphate bond.

The synthesis of one mole of glucose from two moles of lactate requires six moles of ATP. Which one of the following steps requires ATP in the gluconeogenic pathway? Select one: a. Fructose-1,6-bisphosphatase b. Phosphoglycerate kinase c. Glucose-6-phosphatase d. Triosephosphate isomerase e. Pyruvate kinase

d. Southern blot

The technique that utilizes probes to detect specific DNA sequences is known as what? Select one: a. Eastern blot b. Western blot c. Northern blot d. Southern blot e. Northwestern blot

b. TPP

The transketolase enzyme of the pentose phosphate pathway requires which of the the following for maximal activity? Select one: a. coenzyme A b. TPP c. tetrahydrofolate d. biotin e. calcium ions

a. Catabolite activator protein

To begin transcription of the lac operon, RNA polymerase binds to a region of the DNA molecule called the promoter. A small homodimer aids in binding of the RNA polymerase. Name the homodimer protein. Select one: a. Catabolite activator protein b. Instigator c. Operator d. Terminator e. Stop codon

e. Catabolite activator protein

To begin transcription of the lac operon, RNA polymerase binds to a region of the DNA molecule called the promoter. A small homodimer aids in binding of the RNA polymerase. Name the homodimer protein. Select one: a. Stop codon b. Operator c. Instigator d. Terminator e. Catabolite activator protein

(C) Rotenone At point 1, an oxidizable substrate was added to the mixture as indicated in the fi gure (pyruvate), which is oxidized to form NADH. The NADH can add electrons to complex I to initiate elec-tron fl ow across the chain. Since at point 2 the addi-tion of succinate allows electron fl ow to reoccur, after being inhibited, it suggests that the inhibitor added at point A blocks electron fl ow from complex I to complex III (recall, succinate will add electrons at complex II, bypassing complex I). The only inhibitor in the list that does this is rotenone. Antimycin A blocks electron fl ow from complex III to complex IV. Atractyloside blocks ATP/ADP exchange across the inner mitochondrial mem-brane and will stop electron fl ow due to an inhibition of phosphorylation. The addition of succinate would not be able to overcome an inhibition of ATP synthesis due to lack of substrate (ADP). Dinitrophenol is an uncoupler, but would not allow electron fl ow from complex 1 in the presence of rotenone. Lactate is another oxidizable sub-strate, which would not overcome the block of electron transfer from complex I as lactate oxidation will gener-ate NADH, which adds electrons to complex I.

What compound was added at the point indicated as A? (A) Antimycin A (B) Atractyloside (C) Rotenone (D) Dinitrophenol (E) Lactate

(D) Dinitrophenol The increase in oxygen uptake stimulated by succinate (which is allowing electron fl ow from complex II to oxygen) is being blocked by oli-gomycin, which inhibits ATP synthesis. The block in ATP synthesis leads to the cessation of oxygen consumption due to the coupling of oxidation and phosphorylation. The only drug that can allow electron fl ow, in the absence of ATP synthesis, is an uncoupler, which uncouples the link between oxygen consumption and ATP production. Dini-trophenol is the only uncoupler on the list of answers. Note also that the rate of oxygen consumption has increased as compared to that when either NADH or succinate was donating electrons. This is due to the lack of a proton gra-dient in the presence of an uncoupler, so there is no "back pressure" to oxygen consumption, and the electron fl ow is faster than in the absence of the uncoupler.

What compound was added at the point indicated as B? (A) Antimycin A (B) Atractyloside (C) Rotenone (D) Dinitrophenol (E) Lactate

a. b 1→4 linkage

What is the kind of glycosidic linkage seen here in lactose? Select one: a. b 1→4 linkage b. a 1→4 linkage c. b 1→2 linkage d. b 1→6 linkage e. a 1→2 linkage

a. 5' AGGCCT 3'

What sequence is a palindrome? Select one: a. 5' AGGCCT 3' b. 5' CCAGG 3' c. 5' ACGGATTCGC 3' d. 5' ATG 3' e. 5' CCATT 3'

c. Inhibition of a translocation between cellular compartments Under conditions of active synthesis of fatty acids in the cytosol of hepatocytes, levels of malonyl CoA are high. Malonyl CoA is the activated source of two carbon units for fatty acid synthesis. Malonyl CoA inhibits carnitine acyltransferase I, which is located on the cytosolic face of the inner mitochondrial membrane. Thus, long-chain fatty acyl CoA units cannot be transported into mitochondria where β oxidation occurs, and translocation from cytosol to mitochondrial matrix is prevented. In this situation compartmentalization of membranes as well as inhibition of enzymes comes into play.

When the liver is actively synthesizing fatty acids, a concomitant decrease in β oxidation of fatty acids is due to Select one: a. Inhibition by an end product b. Activation of an enzyme c. Inhibition of a translocation between cellular compartments d. Decreases in adipocyte lipolysis e. Detergent effects

b. Cellulose Cellulose, the most abundant compound known, is the structural fiber of plants and bacterial walls. It is a polysaccharide consisting of chains of glucose residues linked by β 1→ 4 bonds. Since humans do not have intestinal hydrolases that attack β 1→ 4 linkages, cellulose cannot be digested but forms an important source of "bulk" in the diet. Lactose is a disaccharide of glucose and galactose found in milk. Amylose is an unbranched polymer of glucose residues in α-1,4 linkages. Glycogen is a branched polymer of glucose with both α-1,4 and α-1,6 linkages. Maltose is a disaccharide of glucose, which is usually the breakdown product of amylose.

Which of the following carbohydrates would be most abundant in the diet of strict vegetarians? Select one: a. Maltose b. Cellulose c. Glycogen d. Lactose e. Amylose

b. Cellulose Cellulose, the most abundant compound known, is the structural fiber of plants and bacterial walls. It is a polysaccha- ride consisting of chains of glucose residues linked by β 1→ 4 bonds. Since humans do not have intestinal hydrolases that attack β 1→ 4 linkages, cellu- lose cannot be digested but forms an important source of "bulk" in the diet. Lactose is a disaccharide of glucose and galactose found in milk. Amylose is an unbranched polymer of glucose residues in α-1,4 linkages. Glycogen is a branched polymer of glucose with both α-1,4 and α-1,6 linkages. Maltose is a disaccharide of glucose, which is usually the breakdown product of amylose.

Which of the following carbohydrates would be most abundant in the diet of strict vegetarians? Select one: a. Lactose b. Cellulose c. Maltose d. Amylose e. Glycogen

a. Acetyl-CoA carboxylase

Which of the following enzymes is most responsive to changes in the level of circulating insulin such that as insulin levels rise fatty acid synthesis is increased? Select one: a. Acetyl-CoA carboxylase b. Fatty acid synthase c. Malonyl-CoA decarboxylase d. Stearoyl-CoA desaturase e. ATP-citrate lyase

a. citrate

Which of the following exerts a positive allosteric effect on de novo fatty acid synthesis at the level of acetyl-CoA carboxylase? Select one: a. citrate b. AMPK c. acetyl-CoA d. ATP e. biotin

g. Glucagon The balance and integration of the metabolism of fats and carbohydrates are mediated by the hormones insulin, glucagon, epinephrine, and norepinephrine. All of these hormones exercise acute effects upon metabolism. Glucagon stimulates gluconeogenesis and blocks glycolysis. When blood sugar levels get low, the α cells of the pancreas release glucagon. The main targets of glucagon are the liver and adipose tissue. In the liver, glucagon stimulates the cyclic AMP-mediated cascade that causes phosphorylation of phosphorylase and glycogen synthesis. This effectively turns off glycogen synthase and turns on glycogen phosphorylase, thereby causing a breakdown of glycogen and a production of glucose in liver, which ultimately raises blood glucose levels. Insulin and glucagon are two antagonistic hormones that maintain the balance of sugar and fatty acids in blood. Insulin is produced by the β cells of the pancreas and its release is stimulated by high levels of glucose in the blood. It has a number of effects, but its major effect is to allow the entry of glucose into cells. Insulin also allows the dephosphorylation of key regulatory enzymes. The consequence of these actions is to allow glycogen synthesis and storage in both muscle and liver, suppression of gluconeogenesis, acceleration of glycolysis, promotion of the synthesis of fatty acids, and promotion of the uptake and synthesis of amino acids into protein. All in all, insulin acts to promote anabolism.

Which of the following hormones stimulates gluconeogenesis? . Select one: a. Growth hormone b. Insulin c. Thyroxine d. Progesterone e. Glucocorticoids f. Epinephrine g. Glucagon h. Aldosterone

c. glucose-6-phosphate dehydrogenase

Which of the following is a congenital defect in which enzyme may cause hemolytic anemia following administration of the antimalarial drug primaquine? Select one: a. hexokinase b. pyruvate kinase c. glucose-6-phosphate dehydrogenase d. phosphofructokinase e. glyceraldehyde-3-phosphate dehydrogenase

b. Adenosine monophosphate (AMP) The glycolytic pathway has three key irreversible enzymes: hexokinase, phosphofructose kinase, and pyruvate kinase. Under conditions of limiting cellular energy (low-energy charge), ADP and AMP accumulate and positively regulate phosphofructokinase. Under conditions of cellular "plenty," ATP and citrate, both negative effectors of phosphofructokinase, accumulate. When phosphofructokinase is inhibited, glucose-6-phosphate accumulates and shuts off hexokinase ATP, inhibiting the regulatory enzymes of glycolysis, while a lower energy charge actually stimulates glycolysis.

Which of the following is an allosteric effector that enhances activity of phosphofructokinase of the glycolytic pathway? Select one: a. Adenosine triphosphate (ATP) b. Adenosine monophosphate (AMP) c. Citric acid d. Glucose-6-phosphate e. Glucose

a. Accumulation of ceramide-containing lipids Sphingolipidoses are lipid storage diseases that exhibit Mendelian (autosomal or X-linked recessive) inheritance. To date, all lipidoses studied demonstrate accumulation of a ceramidecontaining sphingolipid due to the genetic deficiency of a specific hydrolytic enzyme involved in the breakdown of the sphingolipid in question. This leads to the accumulation of the sphingolipid because its synthetic rate is normal. Since the decrease in activity of the abnormal hydrolytic enzyme is similar in all tissues, diagnostic tests measuring the enzyme can easily be set up using skin biopsies or blood cell measurements. Heterozygous carriers can be screened and receive genetic counseling.

Which of the following is most characteristic of a sphingolipidosis? Select one: a. Accumulation of ceramide-containing lipids b. Multifactorial inheritance c. Deficiency of a hydrolytic enzyme d. Variable activities of abnormal enzyme in different patient tissues e. Abnormalities of sphingolipid synthesis

c. FADH2 Two major enzyme complexes are involved in the synthesis of fatty acids. The first is acetyl CoA carboxylase, which synthesizes malonyl CoA by the steps shown below for the synthesis of palmitate: 7 acetyl CoA + 7 HCO3− + 7 ATP → 7 malonyl CoA + 7 ADP + 7 Pi Using the malonyl CoA, palmitate is then synthesized by seven cycles of the fatty acid synthetase complex, whose stoichiometry is summarized below: acetyl CoA + 7 malonyl CoA + 14 NADPH → palmitate + 7 CO2 + 14 NAD+ + 8 CoA + 6 H2O As can be seen from the equations above, the necessary amount of malonyl CoA is synthesized. Palmitate is subsequently synthesized from malonyl CoA and one initial acetyl CoA. Thus, acetyl CoA, NADPH, ATP, and HCO3− are all necessary in this process. In contrast, FADH2 is not utilized in fatty acid synthesis, but is one of the products of fatty acid oxidation. Vitamin B12 is required for conversion of propionic acid to methylmalonic acid, a step in the β oxidation of odd-numbered fatty acid

Which of the following is not used in the synthesis of fatty acids? Select one: a. Cobalamin (vitamin B12) b. HCO3 c. FADH2 d. AMP e. NADPH

b. Liver Only the liver and kidneys can synthesize glucose by gluconeogenesis.All the other organs listed are dependent on provision of glucose from blood, either supplied by the diet or by gluconeogenesis in liver and the kidneys.

Which of the following organs or tissues does NOT need to be supplied with glucosefor energy production during a prolonged fast? Select one: a. Brain b. Liver c. Lens d. Cornea e. RBCs

e. fructokinase

Which of the following represents the enzyme deficiency that leads to "essential fructosuria"? Select one: a. 6-Phosphofructo-l kinase, PFK1 b. hexokinase c. fructose-1-phosphate aldolase (aldolase B) d. fructose-1,6-bisphosphate aldolase (aldolase A) e. fructokinase

c. Phosphorylase kinase is activated by phosphorylation The sequential cleavage of the α-1,4-glycosidic bonds of glycogen to release successive glucose-1-phosphate residues is known as glycogenolysis. The enzyme catalyzing this reaction is glycogen phosphorylase a, an active, phosphorylated tetramer formed by covalent modification of phosphorylase b, an inactive dimer. In glycogenesis or glycogen synthesis, activated glycogen synthase adds the glucose of uridine diphosphate (UDP)-glucose units to a growing glycogen polymer by forming α-1,4 linkages. In contrast to phosphorylase, glycogen synthase is inactivated by covalent phosphate binding. The same enzyme that inactivates glycogen synthase by catalyzing its phosphorylation activates another enzyme, phosphorylase kinase, which activates glycogen phosphorylase by phosphorylation.

Which of the following statements about glycogen metabolism is true? Select one: a. Cyclic AMP levels are lowered by epinephrine and glucagon stimulation ofadenylate cyclase b. Cyclic AMP-activated protein kinase stimulates glycogen synthase c. Phosphorylase kinase is activated by phosphorylation d. Phosphorylase b is inactivated by phosphorylation e. Glycogen synthesis is stimulated by glucago

b. There are more branch residues than residues in straight chains Glycogen is a highly branched polymer of α-D-glucose residues joined by α-1,4-glycosidic linkage. Under the influence of glycogen synthase, the C4 alcohol of a new glucose is added to the C1 aldehyde group of the chain terminus. The branched chains occur about every 10 residues and are joined in α-1,6-glycosidic linkages. Large amounts of glycogen are stored as 100- to 400-Å granules in the cytoplasm of liver and muscle cells. The enzymes responsible for making or breaking the α-1,4-glycosidic bonds are contained within the granules. Thus glycogen is a readily mobilized form of glucose.

Which of the following statements about the structure of glycogen is true? Select one: a. Branch points contain α-1,4 glycosidic linkages b. There are more branch residues than residues in straight chains c. The monosaccharide residues alternate between D- and L-glucose d. Glycogen is a copolymer of glucose and galactose e. New glucose molecules are added to the C1 aldehyde group of chain termini, forming a hemiacetal

c. 3-hydroxy-3-methylglutaryl CoA → mevalonic acid Regulation of cholesterol metabolism is by definition exerted at the "committed" and rate-controlling step. This is the reaction catalyzed by 3-hydroxy-3-methylglutaryl CoA reductase. Reductase activity is reduced by fasting and by cholesterol feeding and thus provides effective feedback control of cholesterol metabolism. The statin class of drugs act at this site.

Which of the following steps in the biosynthesis of cholesterol is thought to be rate-controlling and the locus of metabolic regulation? Select one: a. Lanosterol → cholesterol b. Mevalonic acid → geranyl pyrophosphate c. 3-hydroxy-3-methylglutaryl CoA → mevalonic acid d. Geranyl pyrophosphate → farnesyl pyrophosphate e. Squalene → lanosterol

d. Glycerol from lipolysis is phosphorylated, converted to fructose-1,6-bisphosphate, and eventually converted to glucose The answer is Glycerol from lipolysis is phosphorylated, converted to fructose-1,6-bisphosphate, and eventually converted to glucose Lipolysis in adipose tissue leads to increased blood levels of fatty acids and glycerol. Since most tissues have little glycerol kinase, the liver takes up most of the free glycerol, phosphorylates it using glycerol kinase, and oxidizes it to dihydroxyacetone phosphate (DHAP) using a dehydrogenase: glycerol + ATP → glycerol-3-P + ADP glycerol-3-P + NAD+ → DHAP + NADH + H+ Aldolase allows both of the triose phosphates DHAP and glyceraldehyde-3- phosphate to condense and form fructose-1,6-bisphosphate. In this manner, aldolase allows adipocyte glycerol to hepatic gluconeogenesis.

Which of the following steps is involved in the generation of glucose from lipolysis? Select one: a. Glycerol from lipolysis is converted to triglycerides b. Fatty acids from lipolysis stimulate the citric acid cycle c. Glycerol from lipolysis is taken up by liver cells and dimerized to fructose d. Glycerol from lipolysis is phosphorylated, converted to fructose-1,6-bisphosphate, and eventually converted to glucose e. Fatty acids from lipolysis are oxidized, producing NADH and stimulating gluconeogenesis

c. Cis Δ5, 8, 13 C20:3 Humans can synthesize fatty acids of the ω-7 series or higher, but not of ω-6 or lesser. This is due to the limitation of the desaturase system (see the fi gure below), which can only introduce double bonds at positions 4, 5, 6 and 9, in a substrate that contains at least 16 carbons. Answer Cis Δ9, 12 C18:2 is an ω-6 fatty acid, as are answers Cis Δ5, 8, 11, 14 C20:4 and Cis Δ10 C16:1. Answer Cis Δ9, 12, 15 C18:3 is an ω-3 fatty acid. Answer Cis Δ5, 8, 13 C20:3 is an ω-7 fatty acid, and would be synthesized as follows. Start with C16:0 (palmitic acid), and add a double bond at position 9, creating a cis Δ9 C16:1. Elongate that fatty acid by two carbons, creating a cis Δ11 C18:1. Desaturate this 18-carbon fatty acid at position 6, creating a cis Δ6Δ11 C18:2, which is elongated by two carbons, producing a cis Δ8Δ13 C20:2. Desaturate this fatty acid at position 5 and the final product is obtained.

Which one of the fatty acids listed below can be synthesized by humans? Select one: a. Cis Δ9, 12, 15 C18:3 b. Cis Δ9, 12 C18:2 c. Cis Δ5, 8, 13 C20:3 d. Cis Δ5, 8, 11, 14 C20:4 e. Cis Δ10 C16:1

b. Glycolysis Epinephrine stimulates both muscle and liver adenylate cyclase to produce cyclic AMP. In the liver, the increased cyclic AMP levels activate a phosphatase that dephosphorylates fructose- 2,6-bisphosphate (F-2,6-BP) while deactivating a kinase that produces F-2,6-BP. Thus, F-2,6-BP levels are decreased and phosphofructokinase activity is decreased. In liver and muscle, F-2,6-BP is the major allosteric activator of phosphofructokinase. In skeletal muscle, however, the kinase responsible for the synthesis of F-2,6-BP is activated, not inhibited, by cyclic AMP. Thus, muscle sees an increase in glycolysis following epinephrine stimulation, while the liver experiences a decrease in glycolytic activity. In both tissues, glycogen phosphorylase is activated and glycogenolysis occurs. Under these conditions, glucose is utilized in muscle for ATP production elative to contractile activity, while the liver produces glucose for export to the blood.

Which one of the following activities is simultaneously stimulated by epinephrine in muscle and inhibited by epinephrine in the liver? Select one: a. Fatty acid oxidation b. Glycolysis c. Glycogenolysis d. Cyclic AMP synthesis e. Activation of phosphorylase

c. Phosphatidate Diacylglycerol-3-phosphate, more commonly known as phosphatidate, is an intermediate common to the synthesis of both triacylglycerol and phospholipids. In a two-step process, glycerol phosphate is successively acylated by two acyl CoAs to lysophosphatidate, which contains a fatty acid group in the 1′ position, and then phosphatidate, which contains fatty acid groups in the 1′ and 2′ positions with a phosphate group in the 3′ position. From that point, pathways for synthesis of phospholipids and triacylglycerol diverge. If storage lipid is to be produced, phosphatidate is dephosphorylated by a phosphatase and then acylated by acyl CoA to form triacylglycerol. In contrast, if phospholipids are to be produced, phosphatidate is activated by CTP in a reaction that produces CDP-diacylglycerol and pyrophosphate. Phosphatidylserine, phosphatidylinositol, phosphatidylethanolamine, and phosphatidylcholine can all be derived from CDP-diacylglycerol.

Which one of the following compounds is a key intermediate in the synthesis of both triacylglycerols and phospholipids? Select one: a. CDP-diacylglycerol b. CDP-choline c. Phosphatidate d. Triacylglyceride e. Phosphatidylserine

e. Glucose-6-phosphatase deficiency The HMP shunt can have increased activity under two conditions, one being an increase in the cofactor NADP+ levels and the other being an increase in the substrate levels (glucose-6-phosphate). The only enzyme listed, which when defective would lead to an increase in either glucose-6-phosphate or NADPH, is glucose-6-phosphatase. A defi ciency in glycogen phosphorylase would not produce glucose-1-phosphate; thus, there would not be an increase in the HMP shunt under these conditions. A deficiency in fructose-1,6-bisphosphatase deficiency would impair gluconeogenesis and would not lead to the synthesis of glucose-6-phosphate. Deficiencies in either pyruvate carboxylase or pyruvate dehydrogenase would lead to pyruvate accumulation and NAD+ accumulation, but not NADP+ or glucose-6-phosphate accumulation.

Which one of the following disorders would lead to increased activity of the HMP shunt pathway? Select one: a. Fructose-1,6-bisphosphatase deficiency b. Pyruvate kinase deficiency c. Glycogen phosphorylase deficiency d. Pyruvate dehydrogenase deficiency e. Glucose-6-phosphatase deficiency

d. Glyceraldehyde-3-phosphate dehydrogenase All the enzymes named are glycolytic enzymes that carry out phosphorylation of glucose-derived substrates or of ADP to form ATP. However, only the reaction catalyzed by glyceraldehyde-3-phosphate dehydrogenase is a phosphorylation reaction coupled to oxidation that uses inorganic phosphate. In this reaction, glyceraldehyde-3-phosphate is converted to 1,3 bisphosphoglycerate by the addition of inorganic phosphate and the oxidation of glyceraldehyde- 3-phosphate with the concomitant reduction of NAD+ to NADH + H+. This reaction is an example of a high-energy phosphate compound being produced by an oxidation-reduction reaction. The oxidation of the aldehyde group at C1 of glyceraldehyde-3-phosphate provides the energy for the reaction. The 1,3-bisphosphoglycerate can then be utilized to phosphorylate ADP to ATP through the action of phosphoglycerate kinase, which is the next step in the glycolytic pathway.

Which one of the following enzymes catalyzes phosphorylation with the use of inorganic phosphate? Select one: a. Phosphoglycerate kinase b. Pyruvate kinase c. Phosphofructokinase d. Glyceraldehyde-3-phosphate dehydrogenase e. Hexokinase

e. Phosphoglycerate kinase All the enzymes listed are specific to either glycolysis or gluconeogenesis, except for phosphoglycerate kinase. It is one of seven enzymes common to both glycolysis and gluconeogenesis. The enzymes hexokinase, phosphofructokinase, and pyruvate kinase catalyze irreversible reactions unique to glycolysis. In order for gluconeogenesis to occur, the three irreversible reactions must be replaced. Pyruvate is synthesized into phosphoenolpyruvate by a two-step reaction. First, oxaloacetate is formed by carboxylation in the presence of pyruvate carboxylase. Then, phosphoenolpyruvate carboxykinase decarboxylates and phosphorylates oxaloacetate in the presence of GTP. The next irreversible step to be bypassed in gluconeogenesis requires fructose-6-phosphate to be produced by the action of fructose-1,6-phosphatase on fructose-1,6-phosphate. When glucose-6-phosphate is finally produced during gluconeogenesis, it is converted to glucose by glucose-6-phosphatase, an enzyme unique to the endoplasmic reticulum. The free glucose may then diffuse from the liver into the bloodstream. Of the enzymes given as possible answers, only phosphoglycerate kinase catalyzes a reversible reaction common to both glycolysis and gluconeogenesis.

Which one of the following enzymes is common to both glycolysis and gluconeogenesis? Select one: a. Hexokinase b. Pyruvate kinase c. Pyruvate carboxylase d. Fructose-1,6-bisphosphatase e. Phosphoglycerate kinase

a. Step C Ceramide is the basic unit composing all sphingolipids, which include sphingomyelin and gangliosides. Sphingomyelin, which usually contains phosphocholine as a polar head group, is the only phospholipid that does not have a glycerol backbone. In contrast, gangliosides have complex oligosaccharide head groups.

Which one of the following steps results in the formation of a phospholipid? Select one: a. Step C b. Step E c. Step D d. Step B e. Step A

a. Acyl-carnitine With a carnitine deficiency, fatty acids cannot be added to carnitine, and acyl-carnitine would not be synthesized. With a carnitine acyl-transferase 2 deficiency, the fatty acids are added to carnitine, but the acyl-carnitine cannot release the acyl group within the mitochondria. This will lead to an accumulation of acylcarnitine, which will lead to an accumulation in the circulation. The end result of either deficiency is a lack of fatty acid oxidation, such that ketone body levels would be minimal under both conditions, and blood glucose levels would also be similar in either condition. Insulin release is not affected by either deficiency, and carnitine levels, normally low, would not be significantly modified in either deficiency.

You are examining a patient who exhibits fasting hypoglycemia and need to decide between a carnitine deficiency and a carnitine acyltransferase-2 deficiency as the possible cause. You order a blood test to specifically examine the levels of which one of the following? Select one: a. Acyl-carnitine b. Glucose c. Ketone bodies d. Carnitine e. Insulin

a. Linoleic

You are participating in the studying of human nutritional requirements. For the normal production of all important prostaglandins, a person should daily ingest about 10 grams of which of the following fatty acids? Select one: a. Linoleic b. Oleic c. Suberic d. Palmitic e. Stearic

e. Glyceraldehyde-3-phosphate dehydrogenase Glucose-6-phosphate dehydrogenase. Given the demographics of the patient's ancestry (andthe need for obtaining an accurate history), and the fact that the patient is a male, the patient may have glucose-6-phosphate dehydrogenase deficiency (an X-linked disorder). If a person with this enzyme deficiency is given primaquine, which is a strong oxidizing agent, hemolytic anemia is likely to develop. If a physician suspects that a patient may have such an enzymatic deficiency, it is imperative to check before prescribing strong oxidizing agents to the patient,or prescribe another antimalarial prophylaxis that is not a strong oxidizing agent (such as tetracycline). If individuals were deficient in transketolase, pyruvate dehydrogenase, α-ketoglutarate dehydrogenase, or glyceraldehyde-3-phosphate dehydrogenase, red cell lysis would not occur. One should also recall that the red cells lack mitochondria, so these cells do not contain pyruvate dehydrogenase or α-ketoglutarate dehydrogenase.

You are seeing a male patient of African American descent, whose grandparents live in a chloroquine resistant malaria belt in Africa. He wants to visit his grandparents, and you want to give him primaquine as a malaria prophylaxis, but before you do so, you should test the patient for which of the following nonsymptomatic enzymatic deficiencies? Select one: a. Pyruvate dehydrogenase b. α-Ketoglutarate dehydrogenase c. Glucose-6-phosphate dehydrogenase d. Transketolase e. Glyceraldehyde-3-phosphate dehydrogenase

e. decreased binding of a transcription factor to a DNA sequence The zinc finger domain is a DNA-binding motif consisting of specific spacings of cysteine and histidine residues that allow the protein to bind zinc atoms. The metal atom coordinates the sequences around the cysteine and histidine residues into a fingerlike domain. The finger domains can interdigitate into the major groove of the DNA helix. The spacing of the zinc finger domain in this class of transcription factor coincides with a half turn of the double helix

You are studying the changes that have occurred in the genomes of members of a family susceptible to an inherited form of liver cancer. You find that afflicted family members harbor a mutation in the zinc-finger motif of a cell-cycle regulatory gene. Which of the following is the most likely result of this mutation? Select one: a. stimulation of mRNA synthesis b. enhanced transport of a hormone receptor complex into the nucleus c. "unzipping" of leucine-rich helices d. enhanced binding of hormones to receptors e. decreased binding of a transcription factor to a DNA sequence

e. decreased binding of a transcription factor to a DNA sequence The zinc finger domain is a DNA-binding motif consisting of specific spacings of cysteine and histidine residues that allow the protein to bind zinc atoms. The metal atom coordinates the sequences around the cysteine and histidine residues into a fingerlike domain. The finger domains can interdigitate into the major groove of the DNA helix. The spacing of the zinc finger domain in this class of transcription factor coincides with a half turn of the double helix

You are studying the changes that have occurred in the genomes of members of a family susceptible to an inherited form of liver cancer. You find that afflicted family members harbor a mutation in the zinc-finger motif of a cell-cycle regulatory gene. Which of the following is the most likely result of this mutation? Select one: a. "unzipping" of leucine-rich helices b. enhanced transport of a hormone receptor complex into the nucleus c. stimulation of mRNA synthesis d. enhanced binding of hormones to receptors e. decreased binding of a transcription factor to a DNA sequence

a. cysteine will be incorporated into growing protein in the position where alanine should be normally present

You are studying the effects of chemical treatment of components of the translational machinery. You discover that one chemical in particular results in the inhibition of aminoacyl tRNA synthetase proofreading activity. In result of this inhibition the cysteine is attached to a tRNA specific for alanine by affected aminoacyl tRNA synthetase. Which of the following would best describe the results of this chemical-induced process? Select one: a. cysteine will be incorporated into growing protein in the position where alanine should be normally present b. cysteine will be incorporated randomly into growing protein c. another amino acid (neither alanine nor cysteine) will be incorporated into growing protein in the position where alanine should be normally present d. protein synthesis will stop immediately and newly synthesized protein will be degraded e. cysteine will be incorporated into growing protein in the position where cysteine is normally present

d. Eukaryote RNA polymerases require primer RNA molecules.

You are studying the process of RNA synthesis in a Molecular Biology laboratory for your medical school research project. Which of the following statements concerning transcription is FALSE? Select one: a. Rho factor is sometimes required for termination of transcription in prokaryotes. b. The drug rifamycin inhibits initiation of transcription in prokaryotes. c. In eukaryotes transcription occurs in the nucleus. d. Eukaryote RNA polymerases require primer RNA molecules. e. Prokaryote promoters contain a consensus sequence called the Pribnow Box.

d. absence of a Shine-Dalgarno sequence

You are studying the translation of mRNAs derived from human erythrocytes in cell-free system derived from the E coli bacterium. You find the human globin mRNA is translated with extremely low efficiency in this prokaryotic system. Which of the following is the best explanation for these findings? Select one: a. presence of introns b. presence of a poly(a) tail c. presence of enhancers d. absence of a Shine-Dalgarno sequence e. inadequate amount of amino acids

d. α-amylase The digestion of starch (amylopectin and amylose) begins in the mouth, where chewing mixes the food with saliva. The salivary glands secrete approximately 1 L of liquid per day into the mouth, containing salivary !-amylase and other components.a-Amylase is an endoglucosidase, which means that it hydrolyzes internal a-1,4-bonds between glucosyl residues at random intervals in the polysaccharide chains. The shortened polysaccharide chains that are formed are calleda-dextrins. Salivary !-amylase is largely inactivated by the acidity of the stomach contents, which contain HCl secreted by the parietal cells

You are tending to a 27-year-old woman whose chief complaint is that she can no longer taste sweetness in food, particularly fruits. You suspect that she may have a defect or disease of her tongue but physical examination as well as a small tissue biopsy show no anomalies. You order an assay of her saliva because you suspect she may have a defect in the secretion of which of the following enzymes? Select one: a. lingual lipase b. haptocorrin c. lysozyme d. α-amylase e. mucin

b. Enterohepatic circulation reabsorption of bile salts Enterohepatic circulation reabsorption of bile salts. Because of the elevated liver enzymes (suggestive of liver damage), a statin would be relatively contraindicated in this patient, as a potential side effect of statins is liver damage. Cholestyramine would be a reasonable alternative to statins. Cholestyramine is one of the "bile acid binders" and prevents the reabsorption of bile salts. Since cholesterol is the precursor of bile salts, and 95% of bile salts are usually reabsorbed back into the enterohepatic circulation, losing bile salts in the feces would require increased synthesis of bile salts, thereby reducing the levels of free cholesterol in the body. Statins work by inhibiting HMG-CoA reductase. Cholestyramine does not reduce hepatic cholesterol synthesis, inhibit the release of bile salts, or interfere with the production of chylomicrons. Its sole action is in the lumen of the intestine, where it binds the bile salts so that they cannot be resorbed and sent back to the liver.

You have a patient whose blood work indicates high total cholesterol and elevated liver enzymes. You place him on cholestyramine to lower his cholesterol. Cholestyramine acts to lower cholesterol by inhibiting which of the following enzymes/pathways? Select one: a. The production of chylomicrons b. Enterohepatic circulation reabsorption of bile salts c. Release of bile salts from the gall bladder d. Hepatic cholesterol synthesis e. HMG-CoA reductase

c. HMG-CoA reductase The first stage of cholesterol synthesis leads to the production of the intermediate mevalonate. Two molecules of acetyl-CoA condense to form acetoacetyl-CoA which condenses with another acetyl-CoA to form β-hydroxymethylglutaryl-CoA (HMG-CoA). HMGCoA synthase catalyses this step. Next, HMG-CoA reductase catalyzes the reduction of HMG-CoA to mevalonate. Statins (the class of drugs to which pravastatin belongs) directly inhibit HMG-CoA reductase, so mevalonate cannot be formed and cholesterol synthesis cannot continue. Statins do not inhibit the enzymes MCAD (required for fatty acid oxidation), CAT-1 (required for acyl-CoA transportinto the mitochondria), or citrate lyase (needed to provide acetyl-CoA in the cytoplasm).

You have placed a patient on Pravachol pravastatin to reduce her cholesterol. This class of drugs is effective due to a direct inhibition of which of the following? Select one: a. Medium chain acyl-CoA dehydrogenase (MCAD) b. Carnitine acyltransferase 1 (CAT-1) c. HMG-CoA reductase d. Citrate lyase e. HMG-CoA synthase

a. Prostaglandin synthesis Prostaglandin synthesis. Eicosanoids are potent regulators of cellular function. They are derived from arachidonic acid and are metabolized by three pathways: the cyclooxygenase pathway (prostaglandins and thromboxanes), lipoxygenase pathway (leukotrienes), and the cytochrome P450 pathway (epoxides). Nonsteroidal anti-inflammatory drugs (NSAIDs) do not block arachidonic acid release from the membrane (which would block all eicosanoid synthesis); however, they do interfere with the cyclooxygenase pathway. Prostaglandins affect inflammation, thromboxanes affect formation of blood clots, and leukotrienes affect bronchoconstriction and bronchodilatation. NSAIDs block prostaglandins as one of their antiinflammatory mechanisms. Thus, while NSAIDS will block both prostaglandin and thromboxane synthesis, it is the blockage of prostaglandin synthesis which will block the inflammatory symptoms.

You prescribe ibuprofen to help reduce your patient's inflammation. Which of the following pathways is blocked as an antiinflammatory mechanism of action of nonsteroidal antiinflammatory drugs? Select one: a. Prostaglandin synthesis b. All eicosanoid synthesis c. Arachidonic acid release from the membrane d. Leukotriene synthesis e. Thromboxane synthesis

a. -1183 to -790

Your lab has discovered that interferon-γ (IFNγ) suppresses the expression of the putative anion transporter (PAT-1) gene in the small intestine. You used a reporter assay to perform a functional analysis of the PAT-1 promoter to identify the IFNγ response element. A series of deletions of the promoter ranging from -179 to -1183, as shown in the figure below, were linked upstream of the luciferase (Luc) gene. You then tested the activity of these constructs in intestinal cells. Based on the results shown below, the IFNγ responsive element is located in which region of the promoter? Select one: a. -1183 to -790 b. -790 to +114 c. -398 to -179 d. -790 to -398 e. -179 to +114

e. -1183 to -790

Your lab has discovered that interferon-γ (IFNγ) suppresses the expression of the putative anion transporter (PAT-1) gene in the small intestine. You used a reporter assay to perform a functional analysis of the PAT-1 promoter to identify the IFNγ response element. A series of deletions of the promoter ranging from -179 to -1183, as shown in the figure below, were linked upstream of the luciferase (Luc) gene. You then tested the activity of these constructs in intestinal cells. Based on the results shown below, the IFNγ responsive element is located in which region of the promoter? Select one: a. -179 to +114 b. -398 to -179 c. -790 to +114 d. -790 to -398 e. -1183 to -790

(B) Metformin blocks hepatic gluconeogenesis

Your obese patient has type 2 diabetes mellitus and you have started him on metformin. One of the possible complications of metformin therapy is lactic acidosis. Why is this a concern with metformin therapy? (A) Metformin reduces insulin resistance (B) Metformin blocks hepatic gluconeogenesis (C) Metformin blocks the TCA cycle (D) Metformin inhibits glycolysis (E) Metformin inhibits dietary protein absorption

a. They are epimers

ow are the following carbohydrates related to each other? Select one: a. They are epimers b. One is an aldose while the other is a ketose sugar c. They are D and L isomers d. They are anomers