Chapter 7 - Lipid and Ethanol Metabolism

Question 1

A deficiency of pancreatic exocrine secretion can result in which one of the following?
(A) An increased pH in the intestinal lumen
(B) An increased absorption of fat-soluble vitamins
(C) A decreased formation of bile salt micelles
(D) Increased levels of blood chylomicrons
(E) Decreased amounts of fat in the stool

Answer 1

The answer is C. The pancreas produces bicarbonate (which neutralizes stomach acid) and digestive enzymes (including pancreatic lipase and colipase, enzymes that degrade dietary lipids). The decreased production of bicarbonate will lead to a decrease of intestinal pH. Lower levels of pancreatic lipase will result in the decreased digestion of dietary triacylglyc- erols, which will lead to the formation of fewer bile salt micelles. The reduced pH will also interfere with the ability of the bile salts to effectively form micelles. Intestinal cells will have less substrate for chylomicron formation, and less fat-soluble vitamins will be absorbed. More dietary fat will be excreted in the feces.

Question 2

Choose the one best answer that most accurately describes some properties of acetyl-CoA carboxylase.

Required cofactor / Intracellular location / Allosteric modifier / Enzyme that catalyzes a covalent modification

(A) Biotin / Mitochondrial / Citrate / PKA
(B) Biotin / Cytoplasmic / Citrate / AMP-activated protein kinase
(C) Thiamine / Mitochondrial / Acetyl-CoA / PKA
(D) Thiamine / Cytoplasmic / Acetyl-CoA / AMP-activated protein kinase
(E) None / Mitochondrial / Malonyl-CoA/ PKA
(F) None / Cytoplasmic / Malonyl-CoA / AMP-activated protein kinase

Answer 2

The answer is B. Biotin is required for the acetyl-CoA carboxylase reaction in which the sub- strate, acetyl-CoA, is carboxylated by the addition of CO2 to form malonyl-CoA. This reaction occurs in the cytosol. Malonyl-CoA provides the 2-carbon units that add to the growing fatty acid chain on the fatty acid synthase complex. As the growing chain is elongated, malonyl-CoA is decarboxylated. Citrate is an allosteric activator of the enzyme, and the enzyme is inhibited by phosphorylation by the AMP-activated protein kinase.

Question 3

The synthesis of fatty acids from glucose in the liver is best described by which one of the following?
(A) The pathway occurs solely in the mitochondria.
(B) It requires a covalently bound derivative of pantothenic acid.
(C) It requires NADPH derived solely from the pentose phosphate pathway.
(D) The pathway is primarily regulated by isocitrate.
(E) The pathway does not utilize a carboxylation reaction.

Answer 3

The answer is B. The synthesis of fatty acids from glucose occurs in the cytosol, except for the mitochondrial reactions in which pyruvate is converted to citrate (pyruvate to oxaloacetate, pyruvate to acetyl-CoA, and oxaloacetate and acetyl-CoA condense to form citrate). Biotin
is required for the conversion of pyruvate to oxaloacetate (a carboxylation reaction), which combines with acetyl-CoA to form citrate. Biotin is also required by acetyl-CoA carboxylase. Citrate, not isocitrate, is a key regulatory compound for acetyl-CoA carboxylase. Pantothenic acid is covalently bound to the fatty acid synthase complex as part of a phosphopantetheinyl residue. During the reduction reactions on the synthase complex, the growing fatty acid chain is attached to this residue. NADPH, produced by the malic enzyme as well as by the pentose phosphate pathway, provides the reducing equivalents.

Question 4

Which one of the following best describes the synthesis of triglyceride in adipose tissue?

Source of fatty acids / Source of backbone / Requires coenzyme A / Requires lipoprotein lipase / Requires 2-monoacylglycerol

(A) VLDL / Glycerol / Yes / No / Yes
(B) Chylomicrons / Glycerol / No / Yes / No
(C) VLDL and chylomicrons / DHAP / Yes / No / Yes
(D) VLDL and chylomicrons / DHAP / Yes / No / No
(E) Chylomicrons / DHAP / No / Yes / Yes
(F) VLDL / Glycerol / No / Yes / No

Answer 4

The answer is D. Fatty acids, cleaved from the triacylglycerols of chylomicrons and VLDL by the action of lipoprotein lipase, are taken up by adipose cells and react with coenzyme A to form fatty acyl-CoA. The lipoprotein lipase is not required to synthesize triglyceride within the adipocyte. Glucose is converted via DHAP to glycerol-3-phosphate, which reacts with fatty acyl-CoA to form phosphatidic acid. Adipose tissue lacks glycerol kinase and cannot use glycerol to directly form glycerol-3-phosphate. After inorganic phosphate is released from phosphatidic acid, the resultant diacylglycerol reacts with another fatty acyl-CoA
to form a triacylglycerol, which is stored in the adipose cells. (2-Monoacylglycerol is an intermediate for triglyceride synthesis only in intestinal cells, and is not produced in the adipocyte.)

Question 5

Which one of the following sequences places the lipoproteins in the order of most dense to
least dense?
(A) HDL/VLDL/chylomicrons/LDL (B) HDL/LDL/VLDL/chylomicrons (C) LDL/chylomicrons/HDL/VLDL (D) VLDL/chylomicrons/LDL/HDL (E) LDL/chylomicrons/VLDL/HDL

Answer 5

The answer is B. Because chylomicrons contain the most triacylglycerol, they are the least dense of the blood lipoproteins. Because VLDL contains more protein than chylomicrons, it is more dense than chylomicrons, but less dense than LDL. Because LDL is produced by the degradation of the triacylglycerols of VLDL, LDL is denser than VLDL. HDL is the most dense of the blood lipoproteins. It has the most protein and the least triacylglycerol (see Tables 7.2 and 7.3).

Question 6

Which one of the following best represents fasting conditions?

Activity of hormone-sensitive lipase / Fate of glycerol / VLDL production / Modification of acetyl-CoA carboxylase / Ketone body production

(A) Inactive / Glycolysis / High / Dephosphorylated / No
(B) Active / Glycolysis / High / Phosphorylated / Yes
(C) Inactive / Glycolysis / High / Dephosphorylated/ No
(D) Active / Gluconeogenesis / Low / Phosphorylated / No
(E) Inactive / Gluconeogenesis / Low / Dephosphorylated / Yes
(F) Active / Gluconeogenesis / Low / Phosphorylated / Yes

Answer 6

The answer is F. During fasting, the hormone-sensitive lipase of adipose tissue is activated
by a mechanism involving increased glucagon (and decreased insulin), cAMP, and protein kinase A. Phosphorylation of hormone-sensitive lipase activates the enzyme. Triacylglycerols are degraded, and fatty acids and glycerol are released into the blood. In the liver, glycerol is converted to glucose by gluconeogenesis and fatty acids are oxidized to produce ketone bodies. These fuels are released into the blood and supply energy to various tissues. During fasting, the liver does not produce significant quantities of VLDL. Fatty acid synthesis is reduced owing to the phosphorylation and inactivation of acetyl-CoA carboxylase by the AMP-activated protein kinase.

Question 7

A molecule of palmitic acid, attached to carbon 1 of the glycerol moiety of a triacylglycerol, is ingested and digested. It passes into the blood, is stored in a fat cell, and ultimately is oxidized to CO2 and H2O in a muscle cell. Choose the molecular complex in the blood in which the palmitate residue is carried from the first site to the second in each of the four questions that follow. An answer may be used once, more than once, or not at all.

From the lumen of the gut to the surface of the gut epithelial cell
(A) VLDL
(B) Chylomicron
(C) Fatty acid–albumin
complex
(D) Bile salt micelle
(E) LDL

Answer 7

The answer is D. A palmitate residue attached to carbon 1 of a dietary triacylglycerol is released by pancreatic lipase and carried from the intestinal lumen to the gut epithelial cell in a bile salt micelle, which will allow absorption of the fatty acid by the intestinal epithelial cell.

Question 8

A molecule of palmitic acid, attached to carbon 1 of the glycerol moiety of a triacylglycerol, is ingested and digested. It passes into the blood, is stored in a fat cell, and ultimately is oxidized to CO2 and H2O in a muscle cell. Choose the molecular complex in the blood in which the palmitate residue is carried from the first site to the second in each of the four questions that follow. An answer may be used once, more than once, or not at all.

From the gut epithelial cell to the blood
(A) VLDL
(B) Chylomicron
(C) Fatty acid–albumin
complex
(D) Bile salt micelle
(E) LDL

Answer 8

The answer is B. Palmitate is absorbed into the intestinal cell and utilized to synthesize a triacylglycerol, which is packaged in a nascent chylomicron and secreted via the lymph into the blood.

Question 9

A molecule of palmitic acid, attached to carbon 1 of the glycerol moiety of a triacylglycerol, is ingested and digested. It passes into the blood, is stored in a fat cell, and ultimately is oxidized to CO2 and H2O in a muscle cell. Choose the molecular complex in the blood in which the palmitate residue is carried from the first site to the second in each of the four questions that follow. An answer may be used once, more than once, or not at all.

From the intestine through the blood to a fat cell
(A) VLDL
(B) Chylomicron
(C) Fatty acid–albumin
complex
(D) Bile salt micelle
(E) LDL

Answer 9

The answer is B. The chylomicron, containing the palmitate, matures in the blood by accept- ing proteins from HDL. It travels to a fat cell. VLDL is the particle made in the liver with endogenous triglyceride.

Question 10

A molecule of palmitic acid, attached to carbon 1 of the glycerol moiety of a triacylglycerol, is ingested and digested. It passes into the blood, is stored in a fat cell, and ultimately is oxidized to CO2 and H2O in a muscle cell. Choose the molecular complex in the blood in which the palmitate residue is carried from the first site to the second in each of the four questions that follow. An answer may be used once, more than once, or not at all.

From a fat cell to a muscle cell
(A) VLDL
(B) Chylomicron
(C) Fatty acid–albumin
complex
(D) Bile salt micelle
(E) LDL

Answer 10

The answer is C. The chylomicron triacylglycerol is digested by lipoprotein lipase, and the palmitate enters a fat cell and is stored as triacylglycerol. It is released as free palmitate and carried, complexed with albumin, to a muscle cell, where it is oxidized.

Question 11

A 6-month-old baby was doing well until he developed viral gastroenteritis and was unable to tolerate oral feeding for 2 days. He is admitted to the hospital with encephalopathy, cardiomegally and heart failure, poor muscle tone, and hypoketotic hypoglycemia. Blood work did not detect any medium-chain dicarboxylic acids.

Once this baby is diagnosed and treated, his diet will need to be very restricted. Theoretically, which one of the following fatty acids will he be able to consume and metabolize?
(A) An 8-carbon fatty acid
(B) A 14-carbon fatty acid
(C) A 20-carbon fatty acid
(D) Only unsaturated fatty acids, regardless of
chain length
(E) Only saturated fatty acids, regardless of
chain length

Answer 11

The answer is A. This baby has primary carnitine deficiency, an autosomal recessive disorder. The lack of medium-chain dicarboxylic acids in the blood rules out an MCAD deficiency. He is unable to transport blood-borne carnitine into the muscle and liver, thereby blocking fatty acid oxidation in those tissues. Carnitine is required to transfer most fatty acids from the cytoplasm to the matrix of the mitochondria. However, short- and medium-chain fatty acids (up to 10 or 12 carbons) are sufficiently water-soluble such that they can enter cells and be transferred into the mitochondria in the absence of carnitine. Once inside the mitochondria, an acyl-CoA synthetase will activate the fatty acid to an acyl-CoA such that β-oxidation can occur. The transfer is not affected whether the fatty acid is saturated or unsaturated; the chain length is the determining factor. Dietary restriction of long-chain fatty acids is essential to treat this disorder and alleviate the symptoms. The patient was doing well while feeding on a regular schedule because of the carbohydrate in the diet. Once the child had an extended fast, and needed to oxidize fatty acids for energy, the symptoms of carnitine deficiency became apparent. The hypoketotic hypoglycemia is a strong indication that the problem is in fatty acid oxidation.

Question 12

A 6-month-old baby was doing well until he developed viral gastroenteritis and was unable to tolerate oral feeding for 2 days. He is admitted to the hospital with encephalopathy, cardiomegally and heart failure, poor muscle tone, and hypoketotic hypoglycemia. Blood work did not detect any medium-chain dicarboxylic acids.

Which one of the following foods or supplements would be allowable on the above patient’s restricted diet?
(A) Coconut oil
(B) Tuna
(C) Walnuts
(D) Spinach
(E) Oleic acid supplements

Answer 12

The answer is A. The patient has a primary carnitine deficiency and can only metabolize medium-chain fatty acids. Coconut oil is high in medium-chain saturated fatty acids. Tuna and certain nuts are high in very long-chain fatty acids and omega-3 fatty acids. Spinach is a good source of ALA (alpha-linolenic acid), and omega-6 fatty acids. Oleic acid is a cis-Δ9 C18:1 fatty acid, and would not be metabolized in a child lacking carnitine in the cells.

Question 13

A 6-month-old baby was doing well until he developed viral gastroenteritis and was unable to tolerate oral feeding for 2 days. He is admitted to the hospital with encephalopathy, cardiomegally and heart failure, poor muscle tone, and hypoketotic hypoglycemia. Blood work did not detect any medium-chain dicarboxylic acids.

Dietary supplementation of which one of the following would be beneficial to this patient?
(A) Pantothenic acid
(B) Niacin
(C) Riboflavin
(D) Carnitine
(E) Thiamine

Answer 13

The answer is D. In many cases of primary carnitine deficiency, increasing the blood levels of carnitine is sufficient to allow some transport of carnitine into cells such that fatty acid oxida- tion can occur. While pantothenic acid (part of coenzyme A), niacin (the precursor for NAD1), and riboflavin (needed for FAD) are required for fatty acid oxidation, the rate-limiting step in these patients is the transport of the fatty acids from the cellular cytoplasm to the matrix of the mitochondria.

Question 14

A 50-year-old male patient has high cholesterol levels and is placed on lovastatin. He is counseled to stop drinking his usual glass of grapefruit juice every morning.

Which of the following may occur when someone taking lovastatin chronically consumes grapefruit juice?
(A) Cholesterol levels increase
(B) Muscle pain and discomfort
(C) Steatorrhea
(D) Acid reflux
(E) A decrease in HDL levels

Answer 14

The answer is B. Grapefruit juice contains furanocoumarins, which inhibit the cytochrome P450 complex CYP3A4. This complex modifies various statins for rapid excretion from the body. Thus, in the presence of grapefruit juice, statin levels will be higher than expected. This will lead to prolonged inhibition of HMG-CoA reductase and a reduction of cholesterol levels (with minimal effect on HDL levels), but will also increase the probability of side effects from statin treatment, one of which is muscle pain and weakness. The grapefruit juice plus statin will not lead to gastric reflux or steatorrhea.

Question 15

A 50-year-old male patient has high cholesterol levels and is placed on lovastatin. He is counseled to stop drinking his usual glass of grapefruit juice every morning.

Given that grapefruit juice interferes with lovastatin action, which one of the following best explains this interaction?
(A) Grapefruit juice inhibits the cytochrome p450 enzyme which modifies lovastatin for excretion.
(B) Grapefruit juice stimulates the cytochrome p450 enzyme which modifies lovastatin for excretion.
(C) Grapefruit juice is a competitive inhibitor of lovastatin binding to cholesterol.
(D) Grapefruit juice is a competitive inhibitor of HMG-CoA reductase.
(E) Grapefruit juice reduces the maximal velocity of HMG-CoA reductase.

Answer 15

The answer is A. Grapefruit juice contains furanocoumarins, which inhibit the cytochrome P450 enzyme CYP3A4 that prepares statins for excretion. Grapefruit juice does not act as an inhibitor of HMG-CoA reductase. Lovastatin binds to HMG-CoA reductase, but not to
cholesterol.

Question 16

Aspirin is used in small daily doses to help prevent heart attacks and/or strokes. Aspirin can be used in this fashion because it inhibits which one of the following?
(A) Prostaglandin synthesis
(B) Thromboxane synthesis
(C) Arachidonic acid synthesis
(D) Leukotriene synthesis
(E) Linolenic acid synthesis

Answer 16

The answer is B. Thromboxanes promote platelet aggregation, and aspirin blocks this function through reducing the synthesis of thromboxanes. This decreases the chances of a clot forming in a coronary artery (MI) or in the artery that feeds the brain (CVA). Aspirin also inhibits prostaglandin synthesis, but this is an anti-inflammatory property. Leukotrienes are involved in allergies and asthma, and their synthesis requires lipoxygenase, which is not inhibited by aspirin. Arachidonic acid is derived from linoleic acid, and that synthesis (fatty acid elongation) is not inhibited by aspirin. Linolenic acid is an essential fatty acid, and cannot be synthesized by humans.

Question 17

A patient with a hyperlipoproteinemia would most likely benefit from a low- carbohydrate diet if the lipoproteins that are elevated in the blood belong to which class of lipoproteins? Choose the one best answer.
(A) Chylomicrons
(B) VLDL
(C) LDL
(D) HDL
(E) Chylomicrons and VLDL
(F) VLDL and LDL
(G) LDL and HDL

Answer 17

The answer is B. VLDL is produced mainly from dietary carbohydrate, LDL is produced from VLDL, and chylomicrons contain primarily dietary triacylglycerol. Elevated HDL levels are desirable and are not considered to be a lipid disorder. HDL also contains low levels of triglyceride. A low-carbohydrate diet would be expected to reduce the level of circulating VLDL due to reduced fatty acid and triglyceride synthesis in the liver.

Question 18

An individual has been determined to have hypertriglyceridemia, with a triglyceride level of 350 mg/dL (normal is <150 mg/dL). The patient decides to reduce this level by keeping his caloric intake the same, but switching to a low-fat, low-protein, high-carbohydrate diet. Three months later, after sticking faithfully to his diet, his triglyceride level was 375 mg/dL. This increase in lipid content is being caused by which component of his new diet?
(A) Phospholipids
(B) Triglycerides
(C) Amino acids
(D) Carbohydrates
(E) Cholesterol

Answer 18

The answer is D. Dietary glucose is the major source of carbon for synthesizing fatty acids in humans. In a high-carbohydrate diet, excess carbohydrates are converted to fat (fatty acids and glycerol) in the liver, packaged as VLDL, and sent into the circulation for storage in the fat cells. The new diet has reduced dietary lipids, which lower chylomicron levels, but the excess carbohydrate in the diet is leading to increased VLDL synthesis and elevated triglyceride levels. Dietary amino acids are usually incorporated into proteins, particularly in a low-protein diet.

Question 19

An alcoholic who went on a weekend binge without eating any food was found to have severe hypoglycemia. Hypoglycemia occurred because the metabolism of ethanol prevented the production of blood glucose from which one of the following? Choose the one best answer.
(A) Glycogen
(B) Lactate
(C) Glycerol
(D) Oxaloacetate
(E) Lactate, glycerol, and oxaloacetate

Answer 19

The answer is E. Ethanol metabolism (which produces high NADH levels) does not prevent glycogen degradation. In fact, glycogen stores would be rapidly depleted under these conditions because of decreased gluconeogenesis. Lactate is converted to pyruvate during gluconeogenesis. The pyruvate–lactate equilibrium greatly favors lactate when NADH is high. Thus, alanine and lactate are prevented from producing glucose. Lactate levels are elevated, and
a lactic acidosis can result. Glycerol normally enters gluconeogenesis by forming glycerol- 3-phosphate, which is oxidized to DHAP. High NADH levels prevent this oxidation. Aspartic acid is converted to oxaloacetate (via transamination), as do other amino acid degradation products that enter the TCA cycle (α-ketoglutarate, succinyl-CoA, fumarate). However, the high NADH levels favor malate formation from oxaloacetate, reducing the amount of oxaloacetate available for gluconeogenesis (through the phosphoenolpyruvate carboxykinase reaction). Thus, the three major gluconeogenic precursors (alanine, glycerol, and lactate) do not form glucose because of the high NADH levels, and as glycogen stores are depleted, hypoglycemia results.

Question 20

A man has just received his fourth DUI citation. The judge orders an alcohol dependency program complete with a medication that makes him have nausea and vomiting if he drinks alcohol while taking the medication. The drug-induced illness is due to the buildup of which one of the following?
(A) Ethanol
(B) Acetaldehyde
(C) Acetate
(D) Acetyl-CoA
(E) Acetyl phosphate

Answer 20

The answer is B. The court-ordered medication is disulfiram. Disulfiram inhibits aldehyde dehydrogenase, which greatly reduces the amount of acetaldehyde that is converted to acetate. This causes an accumulation of acetaldehyde, which is the substance responsible for the symptoms of a “hangover,” including nausea and vomiting. Alcohol dehydrogenase reduces ethanol to acetaldehyde. Acetyl-CoA synthetase converts acetate to acetyl-CoA.

Question 21

A 52-year-old man, after suffering a
heart attack, was put on 81 mg of aspirin daily by his cardiologist. The purpose of this treatment is to reduce the levels of which one of the following?
(A) Cytokines
(B) Leukotrienes
(C) Thromboxanes
(D) Cholesterol
(E) Triglycerides

Answer 21

The answer is C. Platelet aggregation is often a determining factor in heart attacks. Thromboxane A2, produced by platelets, promotes platelet aggregation when clotting is required, and inhibition of thromboxane A2 synthesis by aspirin reduces the potential for inappropriate clot formation, and further heart attacks. Thromboxane A2 is produced from arachidonic acid by the action of cyclooxygenase, the enzyme covalently modified and irreversibly inhibited by aspirin. Leukotrienes are also synthesized from arachidonic acid, but utilize lipoxygenase in their synthesis, which is not inhibited by aspirin. Cholesterol, triglyceride, and cytokine synthesis do not require cyclooxygenase activity.

Question 22

Remembering the distribution and solubility of ethanol, after several drinks with an
evening meal, in which of the following tissues would you find the LEAST amount of alcohol?
(A) Brain
(B) Liver
(C) Fatty tissue
(D) Central cornea

Answer 22

The answer is D. Ethanol is both water- and lipid-soluble. It is easily absorbed from the gastrointestinal tract and is distributed throughout the body via the blood stream. It is mostly metabolized in the liver, so the level would be high in this tissue. It is lipid-soluble, so it would be found in fatty tissue. It has many central effects in the brain, so it easily passes the blood– brain barrier. The central cornea has no arterial supply. The only way alcohol could accumulate in the central cornea would be through diffusion into the aqueous humor and then into the central cornea, a slower and less-efficient system. Therefore, the tissue with the lowest level of alcohol would be the central cornea.

Question 23

A 14-year-old girl with Type 1 diabetes has had viral gastroenteritis for 5 days, and she has been vomiting, been nauseous, and had trouble taking fluids by mouth. Because she was not eating, she did not take any insulin during her illness. She becomes weak and confused and is taken to the emergency room (ER) by her parents. The ER doctor notices a fruity odor to her breath, hyperventilation, and a blood glucose level of 600 mg/dL.

A blood pH measurement is taken. You would expect this value to be which one of the following?
(A) 6.75
(B) 7.15
(C) 7.40
(D) 7.65
(E) 8.00

Answer 23

The answer is B. The patient is exhibiting the symptoms of diabetic ketoacidosis. Normal blood pH is in the range of 7.40. Diabetic ketoacidosis reduces the blood pH since ketone bod- ies accumulate and produce acid, which the blood has trouble buffering. A mild ketoacidosis would reduce the pH to about 7.25; one in which the patient exhibits neurological changes (weak and confused) would lower the pH even further to 7.15. Life-threatening diabetic ketoacidosis would be a pH of 7.0. Answers D and E are incorrect because they represent an alkalization of the blood, which does not occur when acids accumulate.

Question 24

A 14-year-old girl with Type 1 diabetes has had viral gastroenteritis for 5 days, and she has been vomiting, been nauseous, and had trouble taking fluids by mouth. Because she was not eating, she did not take any insulin during her illness. She becomes weak and confused and is taken to the emergency room (ER) by her parents. The ER doctor notices a fruity odor to her breath, hyperventilation, and a blood glucose level of 600 mg/dL.

The patient is hyperventilating because of which one of the following?
(A) The low pH of the blood
(B) The elevated pH of the blood
(C) The increased glucagon/insulin ratio in
the blood
(D) Lack of fluids in the body
(E) Difficulty in breathing due to the lack of food

Answer 24

The answer is A. Hyperventilation is the body’s way to try and raise the lowered blood pH by exhaling carbon dioxide rapidly. Carbon dioxide will form carbonic acid and a proton in the blood; as the carbon dioxide is exhaled, the acid and proton will associate so that the carbonic acid can form carbon dioxide and water. This will decrease the proton concentration in the blood, and raise the pH. The hyperventilation is not due to the altered hormonal ratios in the blood, the lack of fluids, or the lack of food.

Question 25

A 14-year-old girl with Type 1 diabetes has had viral gastroenteritis for 5 days, and she has been vomiting, been nauseous, and had trouble taking fluids by mouth. Because she was not eating, she did not take any insulin during her illness. She becomes weak and confused and is taken to the emergency room (ER) by her parents. The ER doctor notices a fruity odor to her breath, hyperventilation, and a blood glucose level of 600 mg/dL.

The fruity odor noticed by the ER physician is due to which one of the following?
(A) Oxidation of acetoacetate
(B) Reduction of acetoacetate
(C) Conversion of acetoacetate to acetoacetyl-CoA
(D) Decarboxylation of acetoacetate
(E) Carboxylation of acetoacetate

Answer 25

The answer is D. The fruity odor is due to acetone, which is being exhaled. The acetone is derived from the spontaneous decarboxylation of acetoacetate (one of the ketone bodies) to acetone within the blood and tissues.

Question 26

A 2-day-old infant born at 32 weeks gestation has had breathing difficulties since birth and is currently on a respirator and 100% oxygen. These difficulties occur due to which one of the following?
(A) An inability of the lung to contract to exhale
(B) An inability of the lung to expand when
taking in air
(C) An inability of the lung to respond to insulin
(D) An inability of the lung to respond to glucagon
(E) An inability of the lung to produce energy

Answer 26

The answer is B. The baby has respiratory distress syndrome, due to an inability to produce surfactant, a hydrophobic molecule that is secreted by the type II cells in the lung and coats
the airways, reducing surface tension during contraction, and allowing relatively easy expan- sion of the lung during inhalation. This is due to the lungs not yet producing surfactant, which contains a few proteins and a large amount of dipalmitoylphosphatidyl choline. Respiratory distress syndrome is not related to insulin or glucagon response by the lung, or the ability of the lung cells to generate energy.

Question 27

An 8-month-old baby girl had normal growth and development for the first few months, but then progressively deteriorated with deafness, blindness, atrophied muscle, inability to swallow, and seizures. Early on in the diagnosis of the child, it was noticed that a cherry red macula was present in both eyes. Considering the child in the above case, measurement of which one of the following would enable one to determine whether the mutation were in the hex A or hex B gene?
(A) GM1
(B) GM2
(C) Globoside
(D) Glucocerebroside
(E) Ceramide

Answer 27

The answer is C. The child is exhibiting the symptoms of either Tay–Sachs or Sandhoff’s disease, both of which are sphingolipidoses. The hex A gene codes for hexosaminidase A, whereas the hex B gene codes for hexosaminidase B. The hex A protein consists of two A and two B subunits, and cleaves only GM2. The hex B protein is a B tetramer, and cleaves both GM2 and globoside. In Tay–Sachs disease, a loss of hex A activity, globoside degradation is normal as the hex B protein is normal. The loss of hex B activity affects both hex A (since two subunits
are of the B variant) and hex B (tetramer) activity, and globoside will accumulate in Sandhoff’s disease, but not in Tay–Sachs disease.

Question 28

A patient with high blood cholesterol levels was treated with lovastatin. This drug lowers blood cholesterol levels due primarily to which one of the following?
(A) Inhibition of absorption of dietary cholesterol
(B) Inhibition of lipoprotein lipase in adipose tissue
(C) Inhibition of citrate lyase in the liver
(D) Induction of LDL receptors in the liver and
peripheral tissues
(E) Inhibition of HMG-CoA reductase in the
liver and peripheral tissues

Answer 28

The answer is D. The class of drugs known as the statins (e.g., lovastatin) lower blood choles- terol levels through the induction of LDL receptor expression on the liver and peripheral tissue cell surface. Statins directly inhibt HMG-CoA reductase, a key regulatory enzyme in cholesterol biosynthesis, which reduces intracellular cholesterol levels. The reduction of intracellular cholesterol leads to the induction of LDL receptors, as the cells now need to obtain their cholesterol from the circulation. Ezetimibe inhibits the intestinal absorption of cholesterol. Statins do not inhibit lipoprotein lipase or citrate lyase.

Question 29

A patient is taking a statin primarily to reduce the risk of atherosclerosis. A potential problem of statin treatment is which one of the following?
(A) Reduced intracellular cholesterol
(B) Reduced synthesis of coenzyme Q
(C) Reduced synthesis of intracellular
phospholipids
(D) Reduced intestinal absorption of
cholesterol
(E) Reduced levels of chylomicrons

Answer 29

The answer is B. Statins inhibit HMG-CoA reductase, thereby reducing intracellular choles- terol levels. The reduced intracellular cholesterol induces the expression of the LDL receptor, which binds LDL and internalizes it within the cell, thereby reducing circulating cholesterol levels, and reducing the risk of developing atherosclerosis. The inhibition of HMG-CoA reductase reduces mevalonate production, which is the precursor for isoprene biosynthesis. Isoprenes are required for the synthesis of coenzyme Q, as well as dolichol phosphate. Thus, taking statins may reduce endogenous coenzyme Q levels. HMG-CoA reductase does not play a role in intestinal cholesterol absorption, nor will it affect the production of chylomicrons,
which carry dietary (exogenous) cholesterol throughout the body.

Question 30

A person with an intestinal infection caused by a proliferation of bacteria in the gut would most likely have an increase in which one of the following?
(A) The synthesis of bile salts in the liver
(B) The amount of conjugated bile salts in the (C) The absorption of dietary lipid by intesti- nal cells
(D) Body stores of fat-soluble vitamins
(E) Body stores of water-soluble vitamins

Answer 30

The answer is A. Bacteria in the intestine deconjugate and dehydroxylate bile salts, convert- ing them to secondary bile salts. Therefore, the bile salts become less water-soluble and less effective as detergents, less readily absorbed, and more likely to be excreted in the feces than recycled by the liver. Fewer micelles would be produced, so less dietary lipid (including the fat-soluble vitamins) would be absorbed. Because fewer bile salts would return to the liver, more bile salts would be synthesized. Bile salts inhibit the 7α-hydroxylase that is involved in their synthesis. In addition, the person’s food intake might decrease, which would augment some of the effects noted above. Charges on bile salts will not affect the uptake of water-soluble vitamins. Since the bacteria in the gut deconjugate and dehydroxylate the bile salts, the amount of conjugated bile salts in the intestine will decrease.

Question 31

A 28-year-old man was found to have elevated cholesterol levels of 325 mg/dL on a routine checkup. His father died of a heart attack at the age of 42, and also had greatly elevated cholesterol levels throughout his life. The man’s physician placed him on lovastatin, and his cholesterol levels dropped to 170 mg/dL. The nature of the elevated cholesterol in this patient is most likely due to a mutation in which one of the following proteins?
(A) HMG-CoA reductase
(B) Protein kinase A
(C) ACAT
(D) LDL receptor
(E) Lipoprotein lipase

Answer 31

The answer is D. The patient most likely has familial hypercholesterolemia due to a mutated LDL receptor. Lovastatin is an inhibitor of HMG-CoA reductase. HMG-CoA reductase inhibitors cause cells to decrease the rate of cholesterol synthesis. Lower cellular levels of cholesterol cause a decreased conversion of cholesterol to cholesterol esters (by the ACAT reaction) for storage and an increased production of LDL receptors. An increased number of receptors will cause more LDL to be taken up by cells from the circulation and degraded by lysosomes. Thus, blood cholesterol levels will decrease. Blood triacylglycerol levels will also decrease but not to a great extent because LDL contains only small amounts of triacylglycerol. The patient most likely is heterozygous for a mutation in the LDL receptor. If the patient were homozygous for such a mutation, then lovastatin would not reduce circulating cholesterol levels, as there would be no functional LDL receptors to upregulate. The patient does not have a mutation in HMG-CoA reductase, as that is the target of lovastatin, and if an HMG- CoA reductase mutation were to reduce HMG-CoA reductase activity, then lovastatin would have little to no effect. A mutation in protein kinase A, rendering it inactive, would not affect intracellular cholesterol synthesis (as it is the AMP-activated protein kinase that regulates the HMG-CoA reductase activity). An inactivating mutation in ACAT would not lead to a reduction of circulating cholesterol in response to lovastatin. Inactivating mutations in lipoprotein lipase lead to hypertriglyceridemia, and do not respond to lovastatin.

Question 32

A person with a known hyperlipoproteinemia went in for a scheduled checkup. The lab values revealed a blood cholesterol level of 360 mg/dL (the recommended level is below 200 mg/dL) and a blood triglyceride (triacylglycerol) level of 140 mg/dL (the recommended level is below 150 mg/dL). The hyperlipoproteinemia is most likely due to which one of the following?
(A) A decreased ability for receptor-mediated endocytosis of LDL
(B) A decreased ability to degrade the triacylg- lycerols of chylomicrons
(C) An increased ability to produce VLDL
(D) An inactive lipoprotein lipase
(E) A decreased ability to convert VLDL to IDL

Answer 32

The answer is A. Of the blood lipoproteins, LDL contains the highest concentration of cho- lesterol and lowest concentration of triacylglycerols. Elevation of blood LDL levels (the result
of decreased endocytosis of LDL) would result in high blood cholesterol levels and relatively normal triacylglycerol levels. A decreased ability to degrade the triacylglycerols of chylomicrons or to convert VLDL to IDL, as well as an increased ability to produce VLDL, would all result
in elevated triacylglycerol levels. An inactive lipoprotein lipase would keep chylomicron and VLDL levels elevated, so an increased triglyceride level would have been seen if this were the defect.

Question 33

A 4-month-old child exhibited extreme tiredness, irritable moods, poor appetite, and fasting hypoglycemia associated with vomiting
and muscle weakness. Blood work showed elevated levels of free fatty acids, but low levels of acylcarnitine. A muscle biopsy demonstrated a significant level of fatty acid infiltration in the cytoplasm. The most likely molecular defect
in this child is in which one of the following enzymes?
(A) Medium-chain acyl-CoA dehydrogenase
(B) Carnitine transporter
(C) Acetyl-CoA carboxylase
(D) Carnitine acyltransferase II
(E) HMG-CoA synthase

Answer 33

The answer is B. The child has the symptoms of primary carnitine deficiency. Carnitine can- not be transported from the blood into the liver and muscle, and fatty acid oxidation in those tissues is severely impaired. The inability to utilize fatty acids for energy give rise to muscle weakness, and an accumulation of fatty acids can occur within the muscle tissue. The inability of the liver to oxidize fatty acids will lead to fasting hypoglycemia as there is insufficient energy for gluconeogenesis. An MCAD deficiency would not show fatty infiltration in the muscle, nor elevated levels of free fatty acids (the presence of medium-chain dicarboxylic acids, or acylglycines, would be observed instead). A defect in carnitine acyltransferase II would result in acylcarnitines in the circulation. An HMG-CoA synthase deficiency would not allow ketone body formation, and would not present with these symptoms. A lack of acetyl-CoA carboxylase would greatly reduce the fatty acid content within the fat cell, as endogenous fatty acids would not be able to be synthesized from acetyl-CoA.

Question 34

Type 1 diabetes mellitus is caused by a decreased ability of the β cells of the pancreas to produce insulin. A person with Type 1 diabetes mellitus who has neglected to take insulin injections will exhibit which one of the following?
(A) Increased fatty acid synthesis from glucose in liver
(B) Decreased conversion of fatty acids to ketone bodies
(C) Increased stores of triacylglycerol in adipose tissue
(D) Increased production of acetone
(E) Increased glucose transport into
muscle cells

Answer 34

The answer is D. Decreased insulin levels cause fatty acid synthesis to decrease and glucagon levels to increase. Adipose triacylglycerols are degraded, and fatty acids are released. They are converted to ketone bodies in the liver, and a ketoacidosis can occur. Nonenzymatic decarboxylation of acetoacetate forms acetone, which causes the odor associated with diabetic ketoacidosis. Insulin is required for efficient glucose transport into muscle cells.

Question 35

A premature infant, when born, had low Apgar scores and was having difficulty breathing. The NICU physician injected a small amount of a lipid mixture into the child’s lungs, which greatly reduced the respiratory distress the child was experiencing. In addition to proteins, a key component of the mixture was which one of the following?
(A) Sphingomyelin
(B) A mixture of gangliosides
(C) Triacylglycerol
(D) Phosphatidylcholine
(E) Prostaglandins E and F

Answer 35

The answer is D. The premature infant is experiencing respiratory distress syndrome, which is caused by a deficiency of lung surfactant. The lung cells do not begin to produce surfactant until near birth, and premature infants frequently are not producing sufficient surfactant to allow the lungs to expand and contract as needed. The surfactant is composed of a number of hydrophobic proteins and dipalmitoylphosphatidylcholine. Sphingomyelin, gangliosides, triglyceride, and prostaglandins are not components of the surfactant. The phosphatidylcholine content of the surfactant is 85% of the total lipids associated with the complex.