The answer is C: Apolipoprotein E. The patient has dysbetalipoproteinemia, a mutation in apolipoprotein E, such that the patient exhibits the rare E2 form instead of the normal E3 form. Apolipoprotein E has afﬁnity for the LDL receptor and the LDL receptor-related protein and, as such, is important for chylomicron remnant and IDL uptake from the circulation by the liver. With the homozygous E2 form, binding of the particles to their receptors is weak, and the particles circulate longer than normal, contributing to the high cholesterol and triglyceride levels seen in the circulation. Only about 10% of the individuals who are homozygous for E2 will develop this condition, and in those, obesity (BMI of 34) is a key factor which links the condition to the mutation. This disorder is not a problem with lipoprotein lipase (LPL) digesting triglycerides from particles, so neither LPL nor apo CII is defective. As both chylomicrons and VLDL are produced, it is not a defect in either apo B48 or B100 production or function. The answer is B. For a component to be in the oxidized state, it must have donated, or never received, electrons. Complex II will metabolize succinate to produce fumarate (generating FAD[2H]), but no succinate is available in this experiment. Thus, complex II never sees any electrons and is always in an oxidized state. The substrate malate is oxidized to oxaloacetate, generating NADH, which donates electrons to complex I of the electron-transport chain. These electrons are transferred to coenzyme Q, which donates electrons to complex III, to cytochrome c, and then to complex IV. Cyanide will block the transfer of electrons from complex IV to oxygen, so all previous complexes containing electrons will be backed up and the electrons will be "stuck" in the complexes, making these components reduced. Thus, answers A and C through E must be incorrect. Starting with glyceraldehyde 3-phosphate and synthesizing one molecule of pyruvate, the net yield of ATP and NADH would be which of the following?
A. 1 ATP, 1 NADH
B. 1 ATP, 2 NADH
C. 1 ATP, 4 NADH
D. 2 ATP, 1 NADH
E. 2 ATP, 2 NADH
F. 2 ATP, 4 NADH
G. 3 ATP, 1 NADH
H. 3 ATP, 2 NADH
I. 3 ATP, 4 NADH
When glycogen is degraded, glucose 1-phosphate is formed. Glucose 1-phosphate can then be isomerized to glucose 6-phosphate. Starting with glucose 1-phosphate and ending with two molecules of pyruvate, what is the net yield of glycolysis in terms of ATP and NADH formed? A. 1 ATP, 1 NADH
B. 1 ATP, 2 NADH
C. 1 ATP, 3 NADH
D. 2 ATP, 1 NADH
E. 2 ATP, 2 NADH
F. 2 ATP, 3 NADH
G. 3 ATP, 1 NADH
H. 3 ATP, 2 NADH
I. 3 ATP, 3 NADH
The answer is B. Fructose is converted to fructose 1-phosphate by fructokinase, and aldolase B in the liver splits the fructose 1-P into glyceraldehyde and dihydroxyacetone phosphate. Thus, the major regulated step of glycolysis, PFK-1, is bypassed and PEP is rapidly produced. As the [PEP] increases, pyruvate kinase produces pyruvate. As the glyceraldehyde-3-phosphate dehydrogenase reaction is proceeding rapidly (remember that fructokinase is a high Vmax enzyme, so there is a lot of substrate proceeding through the glycolytic pathway), the intracellular [NADH]/[NAD+] ratio is high, and the pyruvate produced is converted to lactate in order to regenerate NAD+. Thus, the pyruvate kinase step is not bypassed (thus, A is incorrect). Neither aldolase B nor lactate dehydrogenase is allosterically regulated (thus, C and D are incorrect), and even though the [ATP]:[ADP] ratio is high in the liver under these conditions, the ratio does not affect lactate formation (thus, E is incorrect). The answer is D. If, after fasting, the branches were shorter than normal, glycogen phosphorylase must be functional and capable of being activated by glucagon (thus, A and B are incorrect). The branching enzyme (amylo-4,6-transferase) is also normal because branch points are present within the glycogen (thus, E is incorrect). Because glycogen is also present, glycogenin is present in order to build the carbohydrate chains, indicating that C is incorrect. If the debranching activity is abnormal (the amylo-1,6-glucosidase), glycogen phosphorylase would break the glycogen down up to four residues from branch points and would then stop. With no debranching activity, the resultant glycogen would contain the normal number of branches, but the branched chains would be shorter than normal. The answer is C. The patient has McArdle disease, a glycogen storage disease caused by a deficiency of muscle glycogen phosphorylase. Because he or she cannot degrade glycogen to produce energy for muscle contraction, he or she becomes fatigued more readily than a normal person (thus, A is incorrect), the glycogen levels in her muscle will be higher than normal as a result of the inability to degrade them (thus, D is incorrect), and his or her blood lactate levels will be lower because of the lack of glucose for entry into glycolysis. He or she will, however, draw on the glucose in his or her circulation for energy, so his or her forearm blood glucose levels will be decreased (thus, B is incorrect), and because the liver is not affected, blood glucose levels can be maintained by liver glycogenolysis (thus, E is incorrect). The oxidation of fatty acids is best described by which of the following sets of reactions?
A. Oxidation, hydration, oxidation, carbon-carbon bond breaking
B. Oxidation, dehydration, oxidation, carbon-carbon bond breaking
C. Oxidation, hydration, reduction, carbon-carbon bond breaking
D. Oxidation, dehydration, reduction, oxidation, carbon- carbon bond breaking
E. Reduction, hydration, oxidation, carbon-carbon bond breaking
The answer is D. A lack of carnitine would lead to an inability to transport fatty acyl-CoAs into the mitochondria. This would lead to a decrease in fatty acid oxidation (thus, A is incorrect), a decrease in ketone body production because fatty acids cannot be oxidized (thus, B is incorrect), a decrease in blood glucose levels because gluconeogenesis is impaired as a result of a lack of energy (thus, C is incorrect), and no increase in the levels of very long-chain fatty acids because these are initially oxidized in the peroxisomes and do not require carnitine for entry into that organelle (thus, E is incorrect). The beta-oxidation system, which creates dicarboxylic acids, is found in the endoplasmic reticulum, and as the concentration of fatty acyl-CoAs increase in tissues, they will be oxidized by this alternative pathway. The answer is d. (Murray, pp 102-110, 145-152. Scriver, pp 1521-1551.) Important carbohydrates include the disaccharides maltose (glucose-glucose), sucrose (glucose-fructose) and lactose (galactoseglucose), and the glucose polymers starch (cereals, potatoes, vegetables) and glycogen (animal tissues). Humans must convert dietary carbohydrates to simple sugars (mainly glucose) for fuel, employing intestinal enzymes and transport systems for enzymatic digestion and absorption. Simple sugars (galactose, fructose) are converted to glucose by liver enzymes, and the glucose is reversibly stored as glycogen. Enzymatic deficiencies in intestinal digestion (e.g., lactase deficiency in those with lactose intolerance), in sugar to glucose conversion (e.g., galactose to glucose conversion in galactosemia), or in glycogenesis/glycogenolysis (e.g., in those glycogen storage diseases) result in glucose deficiencies (low blood glucose or hypoglycemia) with potential accumulation and toxicity to hepatic tissues. The infant had been normal, excluding low glucose due to growth hormone deficiency, and could readily digest breast milk lactose with absorption and conversion to glucose. Low glucose during fasting and liver enlargement implies altered regulation of glycogen synthesis/release due to one of the enzyme deficiencies within the category of glycogen storage disease. The answer is e. (Scriver, pp 1521-1551. Murray, pp 145-152.) Under circumstances of intense muscular contraction, the rate of formation of NADH by glycolysis exceeds the capacity of mitochondria to reoxidize it. Consequently, pyruvate produced by glycolysis is reduced to lactate, thereby regenerating NAD+. Since erythrocytes have no mitochondria, accumulation of lactate occurs normally. Lactate goes to the liver via the blood, is formed into glucose by gluconeogenesis, and then reenters the bloodstream to be reutilized by erythrocytes or muscle. This recycling of lactate to glucose is called the Cori cycle. A somewhat similar phenomenon using alanine generated by muscles during starvation is called the glucosealanine cycle. All of the other substances listed—oxaloacetate, glycerol, and pyruvate—can be made into glucose by the liver. In muscle, glycogenolysis is synchronized with contraction by epinephrine (through cyclicAMP) and calcium activation of phosphorylase. In those with musclespecific phosphorylase defects (glycogen storage diseases V and VII), glucose is not mobilized as efficiently from glycogen, causing decreased contractile efficiency (cramping, fatigue), decreased yield of lactate from glycolysis, and maintenance of serum glucose by compensating liver metabolism. The answer is d. (Scriver, pp 4517-4554; Murray, pp 163-167.) Glucose-6-phosphate dehydrogenase (G6PD) is the first enzyme of the pentose phosphate pathway, a side pathway for glucose metabolism whose primary purpose is to produce ribose and NADPH. Its deficiency (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 high-energy 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 like 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. The answer is b. (Murray, pp 163-172. Scriver, pp 1553-1588.) Lactose in breast milk and infant formula is converted by intestinal lactase to glucose and galactose that are efficiently absorbed. In galactosemia (230400), deficiency of galactose-1-phosphate uridyl transferase prevents the conversion of galactose into glucose-6-phosphate by the liver or erythrocytes. Most other organs do not metabolize galactose. The severe symptoms of galactosemia are caused by the reduction of galactose to galactitol (dulcitol) in the presence of the enzyme aldose reductase. High levels of galactitol cause cataracts, the accumulation of galactose-1phosphate contributes to liver disease, and the accumulation of galactose metabolites in urine can be measured as reducing substances by the Clinitest method. Any carbohydrate, including glucose, with a C1 aldehyde registers as a reducing substance by Clinitest, so a Dextrostix (glucose only) test is often performed as a control. In normal children, galactose is first phosphorylated by ATP to produce galactose-1-phosphate in the presence of galactokinase. Next, galactose-1-phosphate uridyl transferase transfers UDP from UDP-glucose to form UDP-galactose and glucose-1-phosphate. Under the action of UDP-galactose-4-epimerase, UDP-galactose is epimerized to UDP-glucose. Finally, glucose-1-phosphate is isomerized to glucose6-phosphate by phosphoglucomutase. Infants with suspected galactosemia must be withdrawn from breast-feeding or lactose formulas and placed on nonlactose formulas such as Isomil. The answer is c. (Murray, pp 163-172. Scriver, pp 1489-1520.) Three enzymes for fructose metabolism, a specific fructokinase plus aldolase B and triokinase, are present at high levels in liver (also in kidney and small intestine). Fructokinase catalyzes the phosphorylation of fructose to fructose-1phosphate, which is then split to D-glyceraldehyde and dihydroxyacetone by aldolase B. Triokinase converts D-glyceraldehyde to glyceraldehyde3-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 as gluconeogenesis. 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. It provides a fuel for glycolysis, but also increases fatty acid, VLDL, and cholesterol-LDL production that is not desirable in diabetes mellitus. A normal female infant begins having jittery spells, vomiting, and falloff in growth when introduced to fruits and vegetables at age 6 months. Serum tests reveal low glucose and increased blood lactate, and her physician suspects hereditary fructose intolerance (229600), which is a deficiency of the enzyme aldolase B. The symptoms and serum abnormalities of this disease are due to which of the following?
a. Accumulation of hexose phosphates, phosphate and ATP depletion, defective electron transport, and glycogen phosphorylase inhibition
b. Accumulation of triose phosphates, phosphate and ATP excess, defective glycolysis, and glycogen synthase inhibition
c. Accumulation of triose phosphates, phosphate and ATP depletion, defective electron transport, and glycogen synthase inhibition
d. Accumulation of hexose phosphates, phosphate and ATP depletion, defective electron transport, and glycogen phosphorylase stimulation
e. Accumulation of hexose phosphates, phosphate and ATP excess, defective electron transport, and glycogen phosphorylase stimulation
The answer is e. (Murray, pp 163-172. Scriver, pp 1489-1520.) Before urine test strips were designed with specific enzyme reagents like glucose oxidase, any sugar with a reducing aldehyde or ketone group would reduce the dye and produce a green color reaction. It was therefore important to differentiate glucosuria due to diabetes mellitus or renal tubular problems from other sugars in the urine, like galactose in galactosemia or fructose in essential fructosuria. 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 and then to xylitol by a xylulose reductase, the enzyme deficiency in essential pentosuria (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 are all 6-carbon hexoses. The answer is a. (Murray, pp 163-172. Scriver, pp 1521-1552.) Fructose is taken in by humans as sucrose, sucrose-containing syrups, and the free sugar. Fructose is mainly phosphorylated to fructose-1-phosphate by liver fructokinase. Aldol cleavage by fructose-1-phosphate-specific aldolase, not enolase, yields glyceraldehyde and dihydroxyacetone phosphate. The glyceraldehyde is phosphorylated to glyceraldehyde-3-phosphate by triose kinase, and both triose phosphates can enter glycolysis. Excess fructose from commercial foods can exercise adverse effects by raising blood lipids and uric acid. Fructose phosphorylation bypasses phosphofructokinase, a regulatory enzyme of glycolysis and provides excess glycerol metabolites and excess triglyceride/lipid biosynthesis. Fructose phosphorylation can also deplete liver cell ATP, lessening its inhibition of adenine nucleotide degradation and increasing production of uric acid. 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. The answer is b.(Murray, pp 136-144. Scriver, pp 1471-1488.)In the first steps of glycolysis, two ATPs are hydrolyzed, one in the phosphorylation of glucose to glucose-6-phosphate by glucokinase (hexokinase), and one in phosphorylation of fructose-6-phosphate to fructose 1,6-bisphosphate by phosphofructokinase. ATP is generated during conversion of 1, 3-bisphosphoglycerate to 3-phosphoglycerate by phosphoglycerate kinase and in the conversion of phosphoenolpyruvate to pyruvate by pyruvate kinase. Since two molecules of 1,3-bisphosphoglycerate are generated from one glucose molecule (and subsequently two molecules of phosphoenolpyruvate are generated), each of these steps results in generation of two ATPs. Thus, two ATPs are expended in the first step of glycolysis and four ATPs are subsequently generated in later stages, for a net total of two ATPs generated from glycolysis of one molecule of glucose. Correct answer = C. Clinitest is a nonspecific test that produces a change in color if urine is positive for reducing substances, including reducing sugars (glucose, fructose, galactose, xylulose, lactose), amino acids, ascorbic acid, and certain drugs and drug metabolites. Because sucrose is not a reducing sugar, it is not detected by Clinitest. Glucose oxidase method will not detect increased levels of galactose or other sugars in urine. It is therefore important that a copper reduction method be used as a screening test. In those instances when the copper method is positive and the glucose oxidase method is negative, glucosuria is ruled out. The answer is b. (Murray, pp 157-160; Scriver, pp 1521-1552.) Liver cells are permeable to glucose while extrahepatic tissues require insulin for glucose entry, reflecting different glucose transporters (GLUT) in different tissues. 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. 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. The answer is e. (Murray, pp 92-101. Scriver, pp 2261-2274. ) Under aerobic conditions, pyruvate is oxidized by pyruvate dehydrogenase to acetyl-CoA, which enters the citric acid cycle. The citric acid cycle generates reducing equivalents in the form of FADH and NADH that are converted to oxygen by the electron transport chain to yield abundant ATP. Under anaerobic conditions such as heavy exercise, pyruvate must be converted to lactate to recycle NADH to NAD+ to allow glycolysis to continue. In mitochondrial disorders resulting from mutations in cytochromes or pyruvate dehydrogenase, there is deficient NADH oxidation and ATP production. Lactate will accumulate as it does normally in tissues without mitochondria (erythrocytes) or in tissues with exercise stress (like muscle). The lactate can accumulate in serum, causing a decreased pyruvate to lactate ratio and lactic acidosis that are typical signs of mitochondrial disease. These abnormalities also occur with circulatory failure (shock) or hypoxemia, so they are suspect for inborn errors only when cardiorespiratory function is normal. Glycolysis produces only 2 ATP compared to the coupling of citric acid intermediates with electron transport that produces 12 ATP per cycle; tissues highly dependent on the respiratory chain (nerves, muscle, retina) are predominantly affected in mitochondrial disorders—for example, Leigh disease. Suggestive signs like the decreased pyruvate/lactate ratio must be followed by more specific tests like muscle biopsy (ragged red fibers), eye examination (retinal pigmentation), or mitochondrial DNA analysis (deletions, point mutations) to diagnose highly variable mitochondrial diseases. The answer is c. (Murray, pp 145-152. Scriver, pp 1521-1552.) 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. The answer is c. (Murray, pp 123-135. Scriver, pp 2367-2424.)In the citric acid cycle, the conversion of α-ketoglutarate to succinate results in decarboxylation, transfer of an H+/e− pair to NADH+, H+, and the substratelevel phosphorylation of GDP to GTP. 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 complex, 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 coupled to the cleavage of the thioester bond of succinyl CoA. Thus, the production of succinate from α-ketoglutarate yields one substrate-level phosphorylation and the production of three ATP equivalents from NADH via oxidative phosphorylation. The answer is b. (Murray, pp 163-172. Scriver, pp 4517-4554.) NAD+/NADH is a favored redox couple used in pathways such as the citric acid cycle, glycolysis, and (with NADP/NADPH), the pentose phosphate pathway. Pyruvate/lactate is found as a redox couple in glycolysis, fumarate/succinate and oxaloacetate/malate in the citric acid cycle, and 6-phosphogluconate/ribulose-5-phosphate in the pentose phosphate pathway. The latter redox pair follows the glucose-6-phosphate to 6phosphogluconate pair, both steps generating NADPH that is essential for reducing oxidized glutathione. Since reduction involves the gain of electrons (loss of hydrogen protons) and oxidation the loss of electrons (gain of hydrogen protons), oxygen to water has the highest drive for oxidation in biological systems and thus the highest positive reduction (Redox) potential of +0.82. The NAD+/NADP+ to NADH/NADPH couple has the highest redox potential with a negative value of -.32, and other coupled reactions like those in the answer options will have intermediate redox potentials, allowing the oxygen to water potential to drive all metabolic reactions through the process of respiration. The answer is d. (Murray, pp 180-189. Scriver, pp 2327-2356.) Fats (triacylglycerols) are the most highly concentrated and efficient stores of metabolic energy in the body. This is because they are anhydrous and reduced. On a dry-weight basis, the yield from the complete oxidation of the fatty acids produced from triacylglycerols is approximately 9 kcal/g, compared with 4 kcal/g for glycogen and proteins. However, under physiologic conditions, glycogen and proteins become highly hydrated, whereas triacylglyceride stores remain relatively free of water. Therefore, although the energy yield from fat stores remains at approximately 9 kcal/g, the actual yields from the oxidation of glycogen and proteins are diluted considerably. Under physiologic conditions, fats yield three to four times the energy of glycogen stores. Patients with fatty acid oxidation disorders cannot switch to fat oxidation as efficiently when their glycogen is depleted by fasting, causing deficits in high energy requiring tissues like heart and skeletal muscle. The answer is b. (Murray, pp 136-144. Scriver, pp 2367-2424.) The conversion of glucose to glucose-6-phosphate is different in liver and muscle. In muscle and most other tissues, hexokinase regulates the conversion of glucose to glucose-6-phosphate. When the major regulatory enzyme of glycolysis, phosphofructose kinase, is turned off, the level of fructose-6phosphate 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. The answer is b. (Murray, pp 151-162. Scriver, pp 1327-1406.) Transamination of alanine to pyruvate, like that of other amino acids, provides citric acid cycle intermediates leading to oxaloacetate and gluconeogenesis. Urea is excreted (incorrect answer a) although fumarate from the urea cycle is a precursor to oxaloacetate, conversion of acetyl-CoA to malonyl-CoA initiates fatty acid synthesis (incorrect answer c), and diversion of oxaloacetate to citrate rather than pyruvate or glucose 6-P to glucose 1-P rather than glucose would decrease gluconeogenesis (incorrect answers d, e). Many amino acids are degraded and transaminated to yield citric acid cycle intermediates, including arginine, histidine, proline, and glutamine to glutamate; and then α-ketoglutarate, tyrosine, and phenylalanine to fumarate, asparagine to aspartate, and then to oxaloacetate. Others such as cysteine, glycine, serine, and threonine can be converted to alanine and transaminated to pyruvate that directly contributes to gluconeogenesis (through pyruvate carboxylase and phosphoenolpyruvate carboxykinase). During the early phases of starvation, the catabolism of proteins is at its highest level. Anabolic enzymes, which are not utilized during starvation, are targeted for degradation (with ubiquitin) and their synthesis repressed. The transamination of amino acids is a first step in amino acid degradation and also yields ketoacids for gluconeogenesis. The protein and amino acid degradation with ketogenesis results in a negative nitrogen balance, increasing ammonia and urea levels in the urine (fromnthe urea cycle). The glucose formed from gluconeogenic amino acids becomes the major source of blood glucose following depletion of liver glycogen stores. Completenoxidation of this glucose, as well as the ketone bodies formed from ketogenic amino acids, leads to a relative increase in the CO2 and H2O formed from amino acid
The answer is b. (Murray, pp 456-473. Scriver, pp 2367-2424.) Cholera toxin is an 87-kDa protein produced by Vibrio cholerae, a gramnegative bacterium. The toxin enters intestinal mucosal cells by binding to GM1 ganglioside. It interacts with Gs protein, which stimulates adenylate cyclase. By ADP-ribosylation of Gs, the toxin blocks its capacity to hydrolyze bound GTP to GDP. Thus, the G protein is locked in an active form and adenylate cyclase stays irreversibly activated. Under normal conditions, inactivated G protein contains GDP, which is produced by a phosphatase catalyzing the hydrolysis of GTP to GDP. When GDP is so bound to the G protein, the adenylate cyclase is inactive. Upon hormone binding to the receptor, GTP is exchanged for GDP and the G protein is in an active state, allowing adenylate cyclase to produce cyclic AMP. Because cholera toxin prevents the hydrolysis of GTP to GDP, the adenylate cyclase remains in an irreversibly active state, continuously producing cyclic AMP in the intestinal mucosal cells. This leads to a massive loss of body fluid into the intestine within a few hours. The answer is b. (Murray, pp 122-129. Scriver, pp 4637-4664.) Muscle cells are the only cells listed that are capable of utilizing all the energy sources available—glucose, fatty acids, and, during fasting, ketone bodies. Mitochondria are required for metabolism of fatty acids and ketone bodies. Since red blood cells (erythrocytes) do not contain mitochondria, no utilization of these energy sources is possible. Red cells are thus extremely dependent on glycolysis, accounting for anemias caused by deficiency of glycolytic enzymes like hexokinase (235700). Although the brain may utilize glucose and ketone bodies, fatty acids cannot cross the blood-brain barrier. Hepatocytes (liver cells) are the sites of ketone body production, but the mitochondrial enzyme necessary for utilization of ketone bodies is not present in hepatocytes. The answer is D: Providing substrate for glyceraldehyde3-phosphate dehydrogenase. Under anaerobic conditions, the NADH generated by the glyceraldehyde-3phosphate 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-3phosphate 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. The answer is 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, 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 answer is B: 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?
(A) Reduced levels of electron acceptors in her system, leading to 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 effectiveness of protein hydroxylation, leading to reduced enzymatic activity and reduced glucose production
(E) Reduced levels of electron donors in her system
The answer is E: Mechanical disruption of the intestinal epithelial cells, leading to loss of lactase from their surface. Mechanical disruption of the intestinal epithelial cells (as brought about by acute viral gastroenteritis) leads to a loss of cell surface enzymes, but lactase is the most severely affected, as it is present at the lowest levels on these cells. While glucoamylase is also lost, its activity is in vast excess of what is required and its partial loss does not affect its ability to hydrolyze glucose-glucose linkages (it does not hydrolyze lactose). A lack of lactase means that the lactose in the diet passes undigested through the small intestine to the large intestine where the bacterial ﬂ ora metabolize the lactose, producing gases and acids that disrupt the osmotic balance between the lumen of the bowel and the cells lining it. This leads to water secretion by the cells into the lumen of the bowel, resulting in diarrhea. Lactose is not directly transported by intestinal epithelial cells (its components, glucose and galactose, are, after hydrolysis of β-1,4 linkage between the two sugars), and a mechanical disruption of intestinal cells does not alter transcription of galactokinase and fructokinase. Mr Smith recently had a bout of ﬁve days of severe nausea, vomiting, low-grade fever, and diarrhea. This condition had afﬂicted a number of people in Mr Smith's ofﬁce. After recovering from this disorder, Mr Smith found that he could no longer drink milk before going to bed as he became ﬂatulent, his stomach hurt, and he would develop diarrhea. If he did not drink milk, these conditions did not occur. He had never experienced these problems before the afﬂiction. A possible explanation for Mr Smith's problem is which one of the following?
(A) Mechanical disruption of the intestinal epithelial cells, leading to reduced transcription of the galactokinase gene
(B) Mechanical disruption of the intestinal epithelial cells, leading to reduced transcription of the fructokinase gene
(C) Mechanical disruption of the intestinal epithelial cells, leading to loss of lactose transport into the cells
(D) Mechanical disruption of the intestinal epithelial cells, leading to loss of the glucoamylase complex from their surface
(E) Mechanical disruption of the intestinal epithelial cells, leading to loss of lactase from their surface
The answer is D: The high NADH/NAD+ ratio impaired gluconeogenesis. Ethanol oxidation to acetic acid (via acetaldehyde) generates large amounts of NADH. As liver glycogen stores have been depleted within 36 h of the fast, gluconeogenesis is required to maintain blood glucose levels. The major precursors for gluconeogenesis are glycerol, lactate, and amino acids (which give rise to pyruvate or TCA cycle precursors, which generate oxaloacetate). Because of the high NADH/NAD+ ratio (due to the ethanol metabolism), pyruvate destined for gluconeogenesis is shunted to lactate in order to regenerate NAD+ to allow alcohol metabolism to continue. Similarly, oxaloacetate is shunted to malate, also to regenerate NAD+ for ethanol metabolism. Glycerol, which is converted to glycerol-3-phosphate, cannot go to dihydroxyacetone phosphate due to the high NADH levels in the liver. Thus, the high NADH/NAD+ ratio diverts gluconeogenic precursors from entering gluconeogenesis, and the liver has trouble maintaining adequate blood glucose levels. Liver glycogen stores have been depleted within the ﬁ rst 36 h of the fast, but glycogen regulation is not affected by the NADH/NAD+ ratio. Under conditions in which the liver is exporting glucose (glucagon administration, for example), liver glycolysis is inhibited by covalent modiﬁcation of key regulatory enzymes, not the NADH/NAD+ ratio. The answer is C: Aldolase. 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), which enables glucose metabolism to be very close to normal. A deﬁ ciency 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. The fructose pathway (indicating the reaction catalyzed by aldolase B), and its relationship to glycolysis, is shown below.
The answer is C: Glycerol, lactate, and glutamine. The woman has a BMI in the unhealthy range (15.7), indicating inadequate nutrient uptake. Since her nutrient uptake is poor, most of the glucose present in her blood is derived from gluconeogenesis, as her glycogen stores are most likely depleted. The substrates for gluconeogenesis are lactate (derived from red blood cell metabolism), glycerol (from triglyceride degradation), and amino acids derived from protein muscle degradation. Glutamine is a glucogenic amino acid, and is also used to transport nitrogen groups from the muscle to the liver for safe excretion via the urea cycle. Leucine is a strictly ketogenic amino acid (giving rise only to acetyl-CoA), and leucine carbons cannot be used to make glucose via gluconeogenesis. Fatty acids are also strictly ketogenic, and cannot be used for glucose production (fatty acids give rise to acetyl-CoA, which cannot be used to produce net glucose). Lysine is also a strictly ketogenic amino acid, and cannot be used for glucose production. Heme degradation gives rise to bilirubin, which cannot be further degraded, and none of the carbons of heme can be used for glucose production. A thin, anxious woman, who is 5′ 6″ tall, weighs 92 lb. Blood work indicates a glucose level of 70 mg/dL under fasting conditions. Her liver is using which of the following as precursors for glucose production under these conditions?
(A) Glycerol, lactate, and leucine
(B) Fatty acids, alanine, and glutamine
(C) Glycerol, lactate, and glutamine
(D) Glycerol, fatty acids, and glutamine
(E) Lactate, heme, and lysine
The answer is A: Pancreatic glucokinase. The boy has developed MODY (maturity onset diabetes of the young), and one variant of MODY is a mutated glucokinase (an inheritable disorder) such that the Km for glucose 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 hyperglycemia is not sufﬁ cient to induce insulin release. As insulin release occurs from the pancreas, liver, muscle, or intestinal 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 t ranscription 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 differentiation factor 1. MODY is not insulin resistance. Therefore, all the other aspects of insulin resistance syndrome are not present (obesity, hypertension, and hypertriglyceridemia). 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. The answer is C: Galactokinase. The child has nonclassical 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 galactose1-phosphate is not accumulating (as occurs in classical galactosemia, a defect in galactose-1-phosphate uridylyl transferase), the other effects seen with classical galactosemia (hypoglycemia and neurological deﬁcit) do not occur. The sugar that is accumulating in the urine is galactose, which contains an aldehyde, which generates 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 nonclassical galactosemia. The answer is 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 fructose2,6-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 glucokinase (which has a high Km) to phosphorylate glucose. Glucokinase is not regulated by phosphorylation. The answer is A. Upon insulin release, the cAMP phosphodiesterase 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 phosphorylated, which leads to an active kinase activity and an
enzymes 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).
Correct answer = E. The patient appears to have a thiamine-responsive PDH deficiency. The enzyme fails to bind thiamine pyrophosphate at low concentration, but shows significant activity at a high concentration of the coenzyme. This mutation, which affects the Km of the enzyme for the coenzyme, is present in some, but not all, cases of PDH deficiency. All inborn errors of PDH are asso ciated with elevated levels of lactate, pyruvate, and alanine (the transamination product of pyruvate). Patients routinely show neuroanatomic defects, developmental delay, and often early death. Elevated lactate and pyruvate are also observed in pyruvate carboxylase deficiency, another rare defect in pyruvate metabolism. Because PDH is an integral part of carbohydrate metabol ism, a diet low in carbohydrates would be expected to blunt the effects of the enzyme deficiency. By contrast, fatty acid degradation occurs via conversion to acetyl CoA by β-oxidation, a process that does not involve pyruvate as an intermediate. Thus, fatty acid metabolism is not disturbed in this enzyme deficiency. The answer is C. Without oxygen, aerobic metabolism cannot function, so the electron transfer chain will stop (without oxygen, cytochrome oxidase cannot remove electrons from the chain to reduce oxygen to water). ATP synthesis in the mitochondria will stop because of the coupling of oxidation and phosphorylation. The TCA cycle will stop because of the accumulation of NADH in the mitochondria (as the electron transfer chain is fully reduced, owing to the lack of oxygen, NADH cannot donate electrons to a reduced complex I), and NADH inhibits key enzymes of the TCA cycle. However, the tissues still need energy, and so they use anaerobic glycolysis to generate ATP. The end product, pyruvate, is converted to lactate to regenerate NAD+ such that glycolysis can continue. An accumulation of lactate (which is a metabolic dead end) can lead to lactic acidosis in the patient. The answer is C. Metformin can increase glucose uptake by tissues and lead to increased lactate formation. In addition, metformin, through unknown mechanisms, appears to block lactate uptake by the liver (this could occur because of the reduced gluconeogenesis occurring in the liver in the presence of metformin, as lactate is a key substrate for gluconeogenesis). The cardiac muscles, with their massive amount of mitochondria, will utilize lactate as fuel and can overcome a lactate buildup from therapeutic doses of metformin. It is only in the rare case (about 1 in 10,000) that metformin treatment will lead to lactic acidosis. Lactate does not enter the TCA cycle to be oxidized, as it first has to be converted to pyruvate, then to acetyl-CoA before it can enter the cycle. Red blood cells generate lactate, but do not utilize it as a fuel. Kidney cells, and other muscles, do not utilize lactate to an appreciable extent as a fuel, as compared to the heart muscle. The answer is C. Metformin tends to increase circulating blood lactate levels owing to reduced use of lactate by the liver for gluconeogenesis, which is inhibited in the presence of metformin. The excess lactate, however, can be used by the heart as an energy source, which reduces circulating lactate levels. If the loss of cardiac muscle cells (and their mitochondria) is significant due to a myocardial infarction, the elevated lactate due to metformin use will accumulate since it is no longer being metabolized by the heart. This could lead to lactic acidosis, which may be fatal. Lactate can also accumulate in renal failure; however, none of the other conditions listed will lead to renal failure. Pyelonephritis usually does not lead to renal failure. Torn muscles may cause myoglobinuria, and this may lead to renal failure, but it would be an uncommon outcome. Red blood cells produce lactate, and so anemia would cause reduced lactate formation. A significant weight gain would not be the cause to stop taking metformin. The answer is C. The elevation of lactate, as well as α-ketoglutarate in the TCA cycle, suggests a defect in reactions that catalyze oxidative decarboxylations. Pyruvate would accumulate if pyruvate dehydrogenase was defective, and the increase of concentration of pyruvate would increase the production of lactate. Five cofactors are needed for oxidative decarboxylation reactions, and they are NAD+, FAD, lipoic acid, thiamine pyrophosphate, and coenzyme A (which is derived from pantothenic acid). The drug appears to be acting by blocking the conversion of pantothenic acid to coenzyme A. Biotin is used for carboxylation reactions, as is vitamin K. Ascorbate is utilized for proline hydroxylation, and vitamin B6 helps to catalyze a variety of reactions involving amino acids. The answer is B. The patient is experiencing ischemic-reperfusion injury. The patient has suffered a heart attack, in which a region of the heart muscle has become anaerobic due to a lack of oxygen delivery. In these cells, the mitochondrial electron transfer chain is reduced, since the terminal electron acceptor (oxygen) is missing. When tPA is administered to dissolve the clot that is causing the blockage of oxygen delivery to the damaged heart muscle, it allows oxygen to rapidly reenter the damaged cells. In these cells, due to the hypoxia, coenzyme Q is fully reduced in the mitochondrial membrane, and the sudden influx of oxygen leads to some accidental electron transfers to oxygen, generating the superoxide radical. This leads to increased radical damage to the already damaged heart muscle, and further damage to the tissue. Cytochrome a is not released from mitochondria (there may be release of cytochrome c if the damage is sufficient, which would initiate apoptosis in the heart cells). Lactic acid levels are elevated owing to anaerobic glycolysis being used to generate energy, but the high lactate levels are not contributing to further damage to the heart muscle. tPA does not have a direct effect on the activity of the TCA cycle enzymes, and is not an uncoupler. The answer is E. During aerobic exercise, the muscle needs to generate ATP via oxidative phosphorylation. For this to occur, NADH levels will decrease (an increase in NADH levels would inhibit the TCA cycle enzymes), leading to a decrease in the NADH/NAD+ ratio. While citrate is an inhibitor of citrate synthase, if this were to occur during exercise, the TCA cycle would stop, not push forward. The activity of isocitrate dehydrogenase is inhibited (not activated) by NADH, and fumarase is not a regulated enzyme. If TCA cycle intermediates were to decrease in concentration, then the cycle would slow down as well, the opposite of what is needed under conditions in which energy needs to be generated, such as during exercise. The answer is 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 pyruvate dehydrogenase complex consists of three major catalytic subunits, designated E1, E2, and E3. The E1 subunit is the one that binds thiamine pyrophosphate 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 metabolism, 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. The answer is C: The E3 subunit of pyruvate dehydrogenase. The child is defective in a variety of oxidative decarboxylation reactions (pyruvate dehydrogenase, 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 subunits, 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. The answer is E: 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 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 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 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 synthesis, and, due to coupling, electron transfer through the chain. Rotenone blocks complex I transfer to coenzyme Q, which signiﬁ cantly reduces electron ﬂ ow, and will not lead to an increase in temperature. The answer is D: 22.5 moles of ATP per mole of citrate. The following steps (see the ﬁgure on page 95) are required for the complete oxidation of citrate to carbon 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 succinyl-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 oxaloacetate, 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 p yruvate reenters 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. The answer is B: Metformin blocks hepatic gluconeogenesis. Metformin leads to a reduction of hepatic gluconeogenesis. This is accomplished through the activation of the AMP-activated protein kinase, which phosphorylates and sequesters within the cytoplasm TORC2, which is a coactivator of CREB activity (a transcription factor needed for expression of two gluconeogenic enzymes, PEP carboxykinase and glucose6-phosphatase). Thus, when TORC2 is absent from the nucleus, gluconeogenesis is impaired as the synthesis of two key enzymes is greatly reduced. One of the major gluconeogenic precursors is lactate, generated from the red blood cells and exercising muscle. In the Cori cycle, two lactates are converted to one glucose, which is then exported. If gluconeogenesis is blocked, lactate is not utilized and its levels can increase, and potentially lead to lactic acidosis. However, in the absence of congestive heart failure or renal insufﬁ ciency, this does not occur. The heart, with its massive amount of muscle and mitochondria, can utilize the lactate for energy unless the heart is dysfunctional or has lost muscle mass. Good, functional kidneys can also overcome the lactate imbalance caused by metformin treatment.
Metformin does decrease the insulin resistance, but this does not increase lactate in the aerobic state. Metformin does not inhibit the TCA cycle, glycolysis, or dietary protein absorption. These interactions are outlined in the ﬁgure below.
The answer is C: Muscle PFK-1. The child has a form of glycogen storage disease known as type VII, Tarui disease, 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 activities, 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. The answer is C: Fructose-1-phosphate inhibition of glycogen phosphorylase. The child has hereditary fructose intolerance, a defect in aldolase B activity in the liver. This leads to an accumulation of fructose-1phosphate in the liver (and, as fructokinase has a high Vmax, a large amount of fructose-1-phosphate accumulates). At high levels, fructose-1-phosphate, through similarity in structure to glucose-1-phosphate, inhibits glycogen phosphorylase activity, leading to hypoglycemia (glycogen degradation is inhibited when blood glucose 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). Galactose is found in lactose, which, while present in milk, is not found in fruit juice. The answer is 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 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. The answer is E: 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. The answer is 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 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. The answer is E: 1,000,000. One active PKA can activate 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. The answer is D: An altered glycogenin with an increased K m for UDPglucose. A reduction in overall 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 carbohydrate 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 glycogen primers will be decreased, as the levels of UDPglucose 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 phosphorylase, 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 phosphorylase 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, leading to enhanced glycogen synthesis. The answer is B: Phosphoglucomutase. For this woman to synthesize lactose, she needs to synthesize the precursors 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 glycolytic enzyme, which is deﬁ cient in hereditary fructose intolerance. Phosphohexose isomerase coverts glucose6-phosphate to fructose-6-phosphate, which is not required for lactose synthesis. Classical galactosemia (severe, type 1) is a deﬁcit 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 deﬁcit in galactokinase, such that galactose cannot be phosphorylated. The complications in type 1 galactosemia due to the accumulation of galactose-1phosphate are not seen in type 2 galactosemia. In either case, the missing enzymes are not required for the synthesis of lactose. The answer is A: a-ketoglutarate dehydrogenase. In order for citrate to be converted to glycogen, the citrate must ﬁrst 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. The answer is 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, ﬂax, 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.
The answer is E: 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. The answer is C: cisD9 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. The answer is 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 m itochondrial 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. The answer is 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. The answer is D: 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. The answer is 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 ﬁrst begins taking the drug, a low concentration of drug is sufﬁcient to exert a physiological effect, as the drug detoxifying system has not yet been induced. As the detoxifying system is induced, however, higher concentrations 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 expression, leading to enhanced drug action, nor are opposing neurotransmitters expressed. The tolerance comes about due to enhanced inactivation of the drug due to the induction of drug-metabolizing enzymes. The answer is C: 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
3 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 g lucose-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 xylulose5-phosphate (X5P, step 1 in the ﬁgure 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 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.
The answer is 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 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. The answer is 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 reaction catalyzed by superoxide dismutase.
The answer is 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 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 answer is A: Steatorrhea. The primary reason for synthesizing conjugated bile acids is to lower the pKa of the acid, so that a higher percentage of the acid will be ionized in the intestine. The greater a bile acid is ionized, the more efﬁcient the emulsiﬁcation is for the digestion of the triglyceride. Without conjugation with glycine or taurine, the pKa of the bile salts is about 6.0; at a pH of 6.0, only 50% of the bile salts will be ionized in the intestinal lumen, which would produce inefﬁcient triglyceride digestion, and the triglyceride content of the stool would increase. By reducing the pKa to 4.0 (conjugated with glycine) or 2.0 (conjugated with taurine), greater than 99% of the bile acids will be ionized, and triglyceride digestion will be maximal. If an inability to conjugate the bile acids leads to inefﬁcient triglyceride digestion, then intestinal chylomicron formation will be reduced, not
elevated (due to reduction of lipid uptake into the enterocyte). Transport of the water soluble B vitamins into the intestinal cells is not dependent on lipid digestion, as is fat-soluble vitamin absorption. The conjugation of bile acids will not affect the pH of the intestinal lumen, nor will it affect the secretion of zymogens from the pancreas to the intestine.
The answer is B: LCAT. HDL is protective, in part, due to its ability to remove excess cholesterol from cell membranes and return it to the liver. In order to accomplish this, the cholesterol, after transport to the HDL particle via the participation of ABC1, needs to be trapped within the core of the HDL particle, and this is accomplished by esteriﬁcation and converting the cholesterol to a cholesterol ester. LCAT (lecithin cholesterol acyl transferase) is the enzyme that creates a cholesterol ester. The reaction, on page 153, is the transfer of a fatty acid from phosphatidyl choline (lecithin) to cholesterol, creating the cholesterol ester and lysophosphatidyl choline. ACAT (acyl-CoA cholesterol acyl transferase) c reates cholesterol esters in cells, but not in the HDL particles. CETP exchanges HDL cholesterol esters for VLDL triglyceride. The answer is C: LDL receptors are nonfunctional. Statins are effective in lowering circulating cholesterol levels due to a series of events. First, the statins inhibit HMG-CoA reductase, reducing intracellular synthesis of cholesterol. The reduced cholesterol levels in the cell then upregulate the synthesis of LDL receptors, which remove LDL from circulation, thereby reducing circulating cholesterol levels. Familial hypercholesterolemia (FH) is a mutation in the LDL receptor, making the receptor unable to bind LDL. In homozygous familial hypercholesterolemia, both LDL receptor genes are mutated, and the LDL receptors are nonfunctional. Upregulating nonfunctional LDL receptors will not lead to a reduction of LDL in the circulation, so such individuals are resistant to statin action. FH is not due to a resistant HMG-CoA reductase, nor an inability of statins to reach their target. FH is not related to reverse cholesterol transport, nor to LCAT. The answer is C: MTTP. The patient has abetalipoproteinemia, an absence of apo B-containing proteins in the circulation. This leads to low chylomicron and VLDL levels. The problem is the synthesis of the chylomicrons and VLDL, both of which require the activity of the microsomal triglyceride transfer protein (MTTP). In the absence of MTTP activity, triglycerides cannot be transferred to the core particle as it is being synthesized, leading to little, if any, synthesis of these particles. The intestinal cells become laden with lipids obtained from the diet and those which cannot be exported due to the inability to produce chylomicrons. Mutations in LPL or apolipoprotein CII will not interfere with chylomicron or VLDL synthesis; mutations in those proteins would lead to an inability to remove triglyceride from those circulating particles. Deﬁciencies in LCAT or ABC1, which are related to HDL metabolism, would not affect the synthesis of chylomicrons or VLDL. MTTP is required to transfer lipid to apo B48 as it is synthesized, and to transfer lipid from the cytoplasm to the lumen of the endoplasmic reticulum as the particle (chylomicrons in the intestine, and VLDL in the liver) is being synthesized. The answer is D: Reduced secretion of LPL. Insulin release stimulates the secretion of lipoprotein lipase (LPL) from fat and muscle cells such that the capillaries inﬁltrating these tissues have the lipase bound to extracellular matrix material. Then, as the triglyceride-rich particles move through the tissues, they bind to LPL via apolipoprotein CII, and the triglyceride is digested and the fatty acids used by the tissues. In the absence of insulin, LPL levels are low, and the particles have a longer half-life in circulation due to the reduced rate of digestion, which contributes to hypertriglyceridemia. If there were reduced synthesis of VLDL, triglycerides in the circulation would be reduced, not increased. Insulin does not alter the rate of apolipoprotein CII production. The release of insulin decreases fatty acid oxidation (promoting fatty acid synthesis), but if increased fatty acid oxidation did occur, then triglycerides would not accumulate in the circulation. Insulin also does not alter the synthesis of apolipoprotein B100 in the liver, which is required for VLDL synthesis. The answer is D: Secretin. Secretin is released from the intestine when food enters, and it signals the pancreas to release a watery mixture of bicarbonate into the intestine, in order to help neutralize the acid present from the digestion that occurred in the stomach. If the pH of the intestinal lumen is too low, the bile salts will not be ionized, and emulsiﬁcation of the dietary fats will be inefﬁcient, as will be the formation of mixed micelles to allow intestinal absorption of fat components. Digestion of carbohydrates and protein is not dependant on bile salt ionization. A loss of cholecystokinin would result in no pancreatic zymogens being secreted, and there would be no digestion of carbohydrates, proteins, or lipids within the intestine. A lack of insulin secretion, or glucagon secretion, does not affect digestion in the intestinal lumen. Cortisol secretion also does not alter intestinal digestion of nutrients. The answer is D: ABCG5. The patient has sitosterolemia, an accumulation of plant sterols (phytosterols) in cells and tissues. Under normal conditions, phytosterols can diffuse into the epithelial cells, but they are actively transported back into the intestinal lumen by an ABC-cassette (ATP-binding) containing protein, ABCG5 (the other protein responsible for phytosterol efﬂ ux is ABCG8).Those sterols which make it to the liver are exported by the same proteins in the liver to
the bile duct, where they will be released along with the bile during fat digestion. In the absence of activity of either ABCG5 or ABCG8, the phytosterols are packaged into chylomicrons and are eventually delivered to the liver, where they are packaged into VLDL. While human cells cannot utilize phytosterols, their increased presence interferes with the synthesis of cholesterol and the normal cholesterol recycling within the affected patient. Patients with this disorder develop premature coronary artery disease. It has been hypothesized that the high levels of plant sterols in the circulating lipoprotein particles accelerate the deposition of these sterols in the walls of the arteries, promoting atherosclerosis. This disorder is not due to mutations in either apo B100 or apo B48, as both VLDL and chylomicrons are synthesized normally in the patient. The defect is not in ABC1, as the patient does not display the symptoms of Tangier disease. The defect is also not in MTTP, as a defect in that protein leads to abetalipoproteinemia.