Block 1

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2. Question The exclusive apolipoprotein of LDLs (low-density lipoproteins) that is critically important for interaction with LDL receptors is Answer Choices A Pancreatic lipase B Apo-B-100 C Apo-B-48 D Albumin E Apo-E

Correct Answer: Apo-B-100

97. Question A 40-year-old type II diabetic woman arrives for her regular check-up. Her fasting blood sugar is 200 mg/dL. The patient informs her physician that she has not been adhering to a strict diabetic diet. She is sent to a dietician who gives her a new diabetic diet chart. The patient is advised to cut down on her high carbohydrate intake. Your patient is curious about the chemistry behind her blood sugar reactions. You explain that the aerobic conversion of 1 mole of glucose to 2 moles of pyruvate is accompanied by the gross generation of how many moles of adenosine triphosphate (ATP)? Answer Choices A 2 B 4 C 8 D 10 E 12

Correct Answer: 4 Explanation The aerobic conversion of 1 mole of glucose to 2 moles of pyruvate is accompanied by the gross generation of 4 moles of ATP and 2 moles of nicotinamide adenine dinucleotide phosphate (NADH). This process of aerobic glucose glycolysis requires 2 equivalents of ATP to activate it and results in the production of 4 equivalents of ATP and 2 equivalents of NADH, resulting in a net generation of 2 moles of ATP. The reaction is as follows: Glucose + 2ATP +2 NAD+ → 2 Pyruvate + 4ATP + 2 NADH Reference: 1. Stavrianeas S, Silverstein T. Teaching glycolysis regulation to undergraduates using an electrical power generation analogy. Adv Physiol Educ. 2005 Jun; 29(2):128-30.

45. Question Human somatic cells have Answer Choices A 46 chromosomes B 23 chromosomes C 42 autosomes and 2 sex chromosomes D 36 chromosomes E 48 chromosomes

Correct Answer: 46 chromosomes Explanation Human somatic cells have 46 chromosomes (23 homologous pairs), 2 of which are sex chromosomes and 44 of which are autosomes. Each chromosome is made up of 2 chromatids, which are joined together by a centromere. The autosomes are designated by a number from 1 to 22, and a normal human has 2 chromosomes of each number. The sex chromosomes are designated either X or Y. Males have 1 X and 1 Y chromosome; females have 2 X chromosomes. Egg and sperm cells have only 23 chromosomes, the haploid chromosome number, because they have undergone meiosis. After the sperm fertilizes the egg, the fertilized egg regains the diploid (46) chromosome number.

63. Question The percentage of the end-diastolic ventricular volume ejected with each stroke (ejection fraction) is Answer Choices A 10% B 20% C 50% D 65% E 90%

Correct Answer: 65% Explanation The end-diastolic ventricular volume is about 130 mL. Thus, ejection fraction, the percentage of the ventricular volume ejected with each stroke, is about 65%, and about 50 ml of blood remains in each ventricle at the end of systole (end-systolic ventricular volume).

58. Question Blood samples from a healthy 25-year-old man show the results below. According to the results of sodium concentration in the blood, what answer corresponds to the correct sodium concentration across the cell? Ion distribution across cell membrane inside outside A 15 mmol/L 150 mmol/L B 150 mmol/L 5.5 mmol/L C 9.0 mmol/L 125 mmol/L D 1.0 mmol/L 120 mmol/L E 100 mmol/L 1200 mmol/L Answer Choices A A B B C C D D E E

Correct Answer: A Explanation A. In healthy individuals this difference (sodium in cytosol 145 mmol/L and sodium in extracellular space 12 mmol/L) provides the inward flow of sodium ions that increases the concentration of cations in the cell and causes depolarization, where the potential of the cell is higher than the cell's resting potential. This is called action potential. B. Sodium concentration inside the cell should be 10 or more times lower than in plasma in a healthy person. C. Sodium is the predominant extracellular cation, normally controlled physiologically in the range of 135 to 145 mEq/L (135-145 mmol/L) despite large variations in salt and water intake. Serum level below 125 mEq/L will usually have a cause, symptoms, and signs. D. 120 mEq/L (120 mmol/L) is the lower critical limit when patients usually have symptoms and signs of severe hyponatremia. E. The ratio of 1200:100 is neither possible nor is it compatible with life.

68. Question The classic definition of an endocrine hormone is Answer Choices A A hormone secreted from a gland into the circulatory system, which acts on a distant target cell (organ) B A hormone secreted from a cell, which acts upon neighboring cells C Hormones secreted from a cell, which act upon cells that are some distance away but do not have to go through the circulatory system D A hormone that acts upon itself

Correct Answer: A hormone secreted from a gland into the circulatory system, which acts on a distant target cell (organ) Explanation The classic definition of a hormone, such as TRH (thyroid releasing hormone), is a hormone that is secreted from the paraventricular nucleus (parvicellular or small neurons) of the hypothalamus. Once this neuropeptide is released, it acts upon the anterior pituitary thyrotropes to release TSH (thyroid stimulating hormone). This hormone is then released into the general circulation and travels to the thyroid gland to cause synthesis and release of the thyronines (T3, T4, FT4, FT3, and rT3). The thyronines are then released into the circulatory system, which act on distant cells to produce desired or undesirable responses.

36. Question The movement of Na+ and K+ across the plasma membrane against the gradient is mediated by Answer Choices A Simple diffusion B Facilitated diffusion (passive transport) C Ion channels D ATP-energized active transport E Ion gradient-energized active transport against the gradient

Correct Answer: ATP-energized active transport Explanation The enzyme Na+K+ ATPase uses the energy of ATP hydrolysis to move both Na+ and K+ through the plasma membrane against their concentration gradients.

118. Question The β-oxidation of fatty acids is an important process in the body, and deficiencies in this process can seriously affect the overall health of an individual. For example, a deficiency in the enzyme medium chain, acyl-CoA dehydrogenase, may be responsible for approximately 10% of sudden infant death syndrome cases. In the β-oxidation of fatty acids, carbon atoms are removed from the fatty acid as Answer Choices A Malonyl CoA B Acetoacetyl CoA C Acetyl acyl carrier protein D Acetyl CoA E Carbon dioxide

Correct Answer: Acetyl CoA 119. Question Acetyl CoA, required for fatty acid synthesis, is Answer Choices A Transported to the cytosol across the mitochondrial membrane B Condensed with oxaloacetate in mitochondria to form citrate, which is transported to the cytosol where it is cleaved to generate acetyl CoA C Carboxylated in mitochondria to form malonyl CoA, which is transported to the cytosol and decarboxylated to generate acetyl CoA D Produced in the cytosol in the course of the citric acid cycle E Synthesized in the cytosol from pyruvate by pyruvate dehydrogenase Correct Answer: Condensed with oxaloacetate in mitochondria to form citrate, which is transported to the cytosol where it is cleaved to generate acetyl CoA Explanation Acetyl CoA is produced in mitochondria by the oxidation of pyruvate by pyruvate dehydrogenase or in the course of the degradation of dietary fatty acids, amino acids, or ketone bodies. In mitochondria, acetyl CoA is condensed with oxaloacetate to form citrate, which is transported to the cytosol across the mitochondrial membrane. In cytosol, citrate is cleaved to generate acetyl CoA which is used for the de novo synthesis of fatty acids. Citric acid cycle occurs in mitochondria and consumes, rather than synthesizes, acetyl CoA; mitochondrial membrane is impermeable for acetyl CoA; carboxylation of acetyl CoA into malonyl CoA by acetyl CoA carboxylase occurs in the cytosol and is irreversible. See the attached image.

4. Question The β-oxidation of fatty acids is an important process in the body, and deficiencies in this process can seriously affect the overall health of an individual. For example, a deficiency in the enzyme medium chain, acyl-CoA dehydrogenase, may be responsible for approximately 10% of sudden infant death syndrome cases. In the β-oxidation of fatty acids, carbon atoms are removed from the fatty acid as Answer Choices A Malonyl CoA B Acetoacetyl CoA C Acetyl acyl carrier protein D Acetyl CoA E Carbon dioxide

Correct Answer: Acetyl CoA Explanation β-oxidation of fatty acids requires 4 reactions. The first reaction, catalyzed by an acyl-CoA dehydrogenase, is an oxidation forming a double bond between the α and β carbons; the second is a hydration; the third is another oxidation step; and the fourth step is the cleavage by thiolysis releasing acetyl CoA, leaving the fatty acyl chain shortened by 2 carbons. This sequence of reactions is then repeated until the fatty acid chain is completely degraded. 3 isozymes are required to completely oxidize a long chain fatty acid: long chain acyl-CoA dehydrogenase, medium chain acyl-CoA dehydrogenase, and short chain acyl-CoA dehydrogenase. Genetic deficiencies in all 3 isozymes have been reported. In the case of an individual with the medium chain acyl-CoA dehydrogenase (MCAD) deficiency, they appear normal at birth but begin to show developmental and behavioral disabilities, chronic muscle weakness, and failure to thrive. Under hypoglycemic conditions, such as an overnight fast, the child cannot properly utilize fatty acids as an energy source. The inability to break down these fatty acids leads to the abnormal accumulation of fatty acids in the liver and brain. MCAD is inherited as an autosomal recessive genetic trait and may be responsible for approximately 10% of deaths associated with sudden infant death syndrome (SIDS). β-oxidation, the degradation of fatty acids, removes 2-carbon units as acetyl CoA and can degrade almost any fatty acid. In a fatty acid with an even number of carbon atoms, such as the C-16 palmitate, the final result is 8 molecules of acetyl-CoA. In the case of a fatty acid with an odd number of carbon atoms, the final cleavage would yield 1 molecule of acetyl-CoA and 1 molecule of the 3-carbon propionyl CoA. Propionyl can be converted to succinyl CoA and enter the citric acid cycle. NADH and FADH2 are the major electron carriers in the oxidation of fuel molecules. In aerobic organisms, O2 is the ultimate acceptor of electrons derived from the oxidation of fuel molecules, such as glucose and fatty acids. Electrons are transferred via NADH and FADH2 to the electron transport chain found in the mitochondria matrix. This flow of electrons is used to synthesize ATP.

38. Question In the metabolic image attached, identify the metabolite marked by the "?" Answer Choices A Pyruvate B Acetoacetate C Lactate D Citrate E Acetyl-CoA F Acetoacetyl-CoA

Correct Answer: Acetyl-CoA Explanation Acetyl-CoA is one of the central metabolites. It is a product of the oxidative decarboxylation of pyruvate. It is also released at each cycle of the β-oxidation. At the same time, acetyl-CoA is a substrate for the citric acid cycle.

117. Question Which of the following compounds is formed as a result of the beta-oxidation of fatty acids? Answer Choices A Acetyl-CoA B Chylomicrons C Glucose D High-density lipoprotein E Low-density lipoprotein

Correct Answer: Acetyl-CoA Explanation The beta oxidation process is the mechanism by which fatty acids are degraded and oxidized to provide cellular energy. Triglycerides are hydrolyzed to glycerol and fatty acids. The glycerol is converted to glycerol-3-phosphate, which then enters the glycolysis pathway, and the fatty acids are transported into the mitochondria in an enzyme-catalyzed process with carnitine acting as a carrier. Once inside the mitochondria the enzymes of the beta-oxidation pathway cleave successive two-carbon groups (acetyl) off the fatty acid chains at the beta position, and combine them with coenzyme A to form Acetyl-CoA. The Acetyl-CoA is then available for use in the citric acid cycle pathway for energy production and other metabolic uses. After absorption across the intestinal mucosa, dietary fats aggregate as small (.03-.5 micron) droplets called chylomicrons. These are composed primarily of triglycerides but contain lesser amounts of cholesterol and phospholipids. Apoprotein B adsorbs to the outer surface of the chylomicrons, and helps to increase their suspension stability, and prevent adherence to vessel walls. Glucose is oxidized for energy production by the glycolysis and citric acid cycle pathways. Glycolysis may proceed anaerobically and results in pyruvic acid formation. After facilitated transport into the mitochondria, pyruvic acid is converted into acetyl-CoA, which enters the citric acid cycle. The acetyl portion is degraded to Co2 and hydrogen atoms that are used for energy production in oxidative phosphorylation. The lipoproteins are formed mainly in the liver and serve to transport lipids throughout the body. These lipid particles circulate in the blood. They are much smaller than chylomicrons but are similarly composed, containing mixtures of triglycerides, phospholipids, cholesterol, and proteins. The high-density lipoproteins are relatively high in protein content, about 50%. Low-density lipoproteins contain mostly cholesterol. Very low-density lipoproteins contain high concentrations of triglycerides with moderate amounts of cholesterol and phospholipids. The VLDL transports the triglycerides synthesized in the liver, mainly from carbohydrates, to the adipose for storage. After delivering their triglyceride, VLDL thus becomes LDL. The HDL transports cholesterol away from the peripheral tissues to the liver, which may explain its role in protecting against atherosclerosis. The fatty acids, which are the substrate of the beta-oxidation pathway, are transported in the blood as "free" fatty acids. These are fatty acids combined with the plasma protein albumin.

30. Question What are the two common chemical transmitters in the nerve cells? Answer Choices A Ca++ ions and norepinephrine B Ca++ ions and Na+ ions C Na+ ions and K+ ions D Acetylcholine and norepinephrine E K+ ions and norepinephrine F Na+ ions and acetylcholine G K+ ions and acetylcholine

Correct Answer: Acetylcholine and norepinephrine Explanation Nerve impulses are communicated across most synapses by chemical transmitters. The two common chemical transmitters in the nerve cells are acetylcholine and norepinephrine.

35. Question What happens to the postsynaptic receptor sites of neurons after the message has been transmitted? Answer Choices A Acetylcholine is hydrolyzed B Choline is hydrolyzed C Acetic acid is hydrolyzed D Ca++ is removed E Na+ is removed

Correct Answer: Acetylcholine is hydrolyzed Explanation If the acetylcholine molecule remains at the synaptic junctions after the message has been transmitted, no signal could be transmitted even if a new one was received. Therefore, the acetylcholine must be removed so that the neuron is reactivated. Enzyme acetylcholine esterase hydrolyses acetylcholine to acetic acid and choline.

19. Question Which of the following Krebs cycle enzymes requires iron? Answer Choices A Aconitase B Fumarase C Isocitrate dehydrogenase D Succinate dehydrogenase E Succinate thiokinase

Correct Answer: Aconitase Explanation The Krebs cycle progresses normally in a clockwise sequence, as in the attached image. However, it is thermodynamically possible to drive most of the reactions in the opposite direction if the required energy is applied. Cofactors are in green; the nucleoside diphosphate kinase [9] is not a Krebs cycle enzyme per se; the Krebs cycle enzymes are citrate synthase [1], aconitase[2], isocitrate dehydrogenase [3], α-ketoglutarate dehydrogenase[4], succinate thiokinase [5], succinate dehydrogenase[6], fumarase [7], and malate dehydrogenase [8]. See the attached image.

41. Question The Intestinal Disease Research Faculty of the Department of Pathology and Molecular Medicine, McMaster University, Canada, is participating in research on the sodium dependent glucose absorption in rat jejunal brush border membrane. They study the sodium glucose transporter 1 and glucose transporter (GLUT) expression by Western blot analysis. They find that glucose is transported by both active transport and facilitated diffusion across the cells. What is a major difference between an active and a facilitated diffusion transport system? Answer Choices A Active systems require carrier proteins to transport substances across a cell membrane and passive systems do not B Active systems can only carry a single molecule in one direction, while passive systems must move two molecules simultaneously C Active systems require ATP whose energy of hydrolysis they use for transport, while passive systems do not D Active systems transport only charged molecules and passive systems transport only neutral molecules E Active systems result in ATP synthesis, while passive systems do not

Correct Answer: Active systems require ATP whose energy of hydrolysis they use for transport, while passive systems do not Explanation Facilitated diffusion, or passive mediated transport, is characterized by the movement of molecules down a concentration gradient, the lack of a need for energy input, and the presence of specific carriers. These specific carriers exhibit saturation kinetics, show a high degree of specificity for the transported substance, and can be specifically inhibited. The chloride-bicarbonate transport system found in the erythrocyte membrane, and the transport of glucose into most cells, can be classified as facilitated diffusion. The chloride-bicarbonate system transport is carried out by the band 3 protein and both ions must move simultaneously. It is a reversible system that is driven by the concentration gradient. Active transport requires energy and involves transport against a concentration gradient. ATP is often used directly as a source of energy. The energy of Na+ going down a concentration gradient can also be used as a source of energy, but this is indirectly maintained by ATP hydrolysis. Active transport is unidirectional. Uniport involves the movement of a single molecule in one direction; Symport involves movement of 2 molecules, simultaneously, in the same direction; Antiport involves the movement of 2 molecules simultaneously in opposite directions. References 1. Grefner NM, Gromova LV, Gruzdkov AA et al; Structural-functional analysis of diffusion in glucose absorption by rat small intestine enterocytes; Tsitologiia, 2006; 48(4): 355-63. 2. Kaur Anand R, Kanwar U, Nath Sanyal S. Characteristics of glucose transport across the micro villous membranes of human term placenta. Nutr Hosp.,2006 Jan-Feb; 21(1): 38-46 3. Wright EM, Martin MG, Turk E, Intestinal absorption in health and disease—sugars, Best Pract Res Clin Gastroenterol. 2003 Dec; 17(6): 943-56

76. Question All proteins are composed of the same 20 amino acids arranged in different but specific sequences. The side chains of these amino acids are responsible for the various properties of the proteins. Which of the following amino acids would be most likely to be found in an alpha helix of a protein? Answer Choices A Proline B Glycine C Serine D Tyrosine E Alanine

Correct Answer: Alanine Explanation The basic structure of the 20 amino acids that make up mammalian proteins is composed of an amino group, a carboxyl group, a hydrogen atom, and a distinct side chain referred to as the "R" group, as shown in the image. The carbon to which all of these substituents are bonded is called the α-carbon. It is the R group that differs between the 20 amino acids and gives the amino acid its distinct properties. These "R" groups allow the protein chain to form various secondary structures. Protein alpha helices are stretches of amino acids in which there are hydrogen bonds between the carbonyl group of one amino acid and the amino group of another residue. Each helical turn has 3.6 residues and is a right-handed helix. Globin is an example of a protein comprised mostly of alpha helices. The amino acids alanine, glutamic acid, leucine, and methionine are the preferred amino acids in an alpha helix of a protein. The amino acids proline, glycine, tyrosine, and serine are almost never found in the protein alpha helix. Proline cannot fit into the helix and will actually introduce a kink in the structure. Proline is very common in structures called beta turns. The beta-turn is a short secondary structure containing only 4 amino acids and allows the peptide backbone to make a 180° turn.

12. Question All proteins are composed of the same 20 amino acids arranged in different but specific sequences. The side chains of these amino acids are responsible for the various properties of the proteins. Which of the following amino acids would be most likely to be found in an alpha helix of a protein? Answer Choices A Proline B Glycine C Serine D Tyrosine E Alanine

33. Question Hereditary fructose intolerance causes severe hypoglycemia, vomiting, jaundice, hepatomegaly, and convulsions due to a deficiency in Answer Choices A Fructokinase B Aldolase B C Aldolase A D Hexokinase E Phosphofructokinase-1

Correct Answer: Aldolase B Explanation Under normal conditions, fructose is phosphorylated by phosphofructokinase to fructose 1-phosphate which is then cleaved by aldolase B (1-phosphofructaldolase) to dihydroxyacetone phosphate and glyceraldehyde. Glycerol is formed from glyceraldehyde and later, by alcohol dehydrogenase. Fructose 1,6-bisphosphate is an intermediate of glycolysis produced by phosphofructokinase-1 from fructose 6-phosphate. Fructose is phosphorylated to form fructose 6-phosphate by hexokinase, an enzyme that has much higher affinity to glucose and under normal conditions (moderate supply of fructose and saturating concentrations of glucose) does not significantly participate in fructose metabolism. Deficiency in aldolase B leads to hereditary fructose intolerance which causes severe hypoglycemia, jaundice, hepatomegaly, and convulsions. Fructokinase deficiency usually results in a benign, asymptomatic condition and can be diagnosed by a moderate presence of fructose in the urine. See the attached image.

49. Question The DNA in a cell's nucleus contains Answer Choices A All the genetic information needed to generate any cell in any organism B All the genetic information needed to generate any cell in its particular organism C Only the genetic information needed to generate another cell in its particular organ system D Only the genetic information needed to generate another cell of its particular histologic type, e.g. endothelial cell, squamous cell, glandular cell, etc E Only the genetic information necessary to regenerate an exact copy of itself

Correct Answer: All the genetic information needed to generate any cell in its particular organism Explanation The most outstanding feature of the cell's nucleus is that it contains a copy of all the organism's DNA - all the structural information the organism and its ancestors have accumulated throughout evolution. In normal cells only the DNA needed to make one type of cell is "read", e.g. heart muscle cells only read the heart muscle DNA, kidney tubule cells only read kidney tubule DNA, but in diseased states, parts of the DNA not normally expressed may be read.

17. Question You are interested in learning more about the digestive process. You observe an experiment in which a cracker made from only flour, salt, and water is tested for the presence of starch after being ground up in water and after being chewed by one of your fellow students. The sample that was chewed by the student contains much less starch than the sample in water. What enzyme in saliva was most likely responsible for this change? Answer Choices A Lysozyme B Amylase C Sucrase D Lipase E Trypsin

Correct Answer: Amylase Explanation Starch is the storage form of glucose in plants. If the glucose residues are in unbranched α-1,4 linkages it is amylose. Amylopectin is the branched form with α-1,6 linkages occurring every 30 residues. Grinding of the cracker in water only would not break down the starch molecules. Chewing of the cracker by the student would add salivary enzymes to the mix. One of these enzymes (amylase) breaks down starch to yield maltose, maltotriose, and α-dextrin. Maltose contains 2 glucose residues.

62. Question Which of the following valves limits blood flow from the aorta to the left ventricle? Answer Choices A Mitral valve B Pulmonary valve C Tricuspid valve D Aortic valve E Coronary sinus valve

Correct Answer: Aortic valve Explanation Cardiac valves ensure that blood flows in only one direction. The tricuspid valve keeps blood from moving from the right ventricle back into the right atria. The pulmonary valve keeps blood from moving from the pulmonary artery to the right ventricle. The mitral valve keeps blood from moving from the left ventricle to the left atria and the aortic valve keeps blood from moving from the aorta to the left ventricle.

113. Question The exclusive apolipoprotein of LDLs (low-density lipoproteins) that is critically important for interaction with LDL receptors is Answer Choices A Pancreatic lipase B Apo-B-100 C Apo-B-48 D Albumin E Apo-E

Correct Answer: Apo-B-100 Explanation Important Properties of Lipoproteins: Chylomicrons- Transport of dietary cholesterol and triacyglycerols. 1-2% Protein. Apo-B-48, also Apo-A-1, Apo-A-II, -IV while in the lymph, Apo-C-II and Apo-E acquired from HDL in the blood. Assembled in the intestine. Apo-C-II activates endothelial lipoprotein lipase, which removes free fatty acides that become absorbed by tissues, Apo-B-48 comibes only with chylomicrons Chylomicron- Cholesterol, Apo-B-48, Apo-E. Appear as a result of the loss of Apo-C-II, removed by the liver through Apo-E. VLDL (very low density lipoproteins)- Transport of endogenous triacylglycerols from the liver. 7-10% Protein. Apo-B-100, Apo-E, Apo-C-I,-II,-III IDL (intermediate density lipoproteins)- Less triacyglycerols than VLDL 10-12% Protein. Like VLDL. Product of triacylglycerol removal from VLDL. Futher removal leads to LDL LDL (low density)- Primary carrier of cholesterol in the plasma. Delivers cholesterol from the liver to all tissues. 20% Protein. Apo-B-100 exclusively. LDL receptors on the cell membranes react with Apo-B-100 and thus mediate LDL HDL (high density lipoproteins)- Accumulates cholesterol esters. 55-66% Protein. Apo-A-I, -C-I, -C-II, - E. LDL receptors on the cell membranes react with Apo-B-100 and thus mediate LDL Albumin- Free fatty acids. Transports free fatty acids from adipose tissue.

114. Question The apolipoprotein combining exclusively with chylomicrons is Answer Choices A Pancreatic lipase B Apo-B-100 C Apo-B-48 D Albumin E Apo-E

Correct Answer: Apo-B-48 Explanation Important Properties of Lipoproteins: Chylomicrons- Transport of dietary cholesterol and triacyglycerols. 1-2% Protein. Apo-B-48, also Apo-A-1, Apo-A-II, -IV while in the lymph, Apo-C-II and Apo-E acquired from HDL in the blood. Assembled in the intestine. Apo-C-II activates endothelial lipoprotein lipase, which removes free fatty acides that become absorbed by tissues, Apo-B-48 comibes only with chylomicrons Chylomicron- Cholesterol, Apo-B-48, Apo-E. Appear as a result of the loss of Apo-C-II, removed by the liver through Apo-E. VLDL (very low density lipoproteins)- Transport of endogenous triacylglycerols from the liver. 7-10% Protein. Apo-B-100, Apo-E, Apo-C-I,-II,-III IDL (intermediate density lipoproteins)- Less triacyglycerols than VLDL 10-12% Protein. Like VLDL. Product of triacylglycerol removal from VLDL. Futher removal leads to LDL LDL (low density)- Primary carrier of cholesterol in the plasma. Delivers cholesterol from the liver to all tissues. 20% Protein. Apo-B-100 exclusively. LDL receptors on the cell membranes react with Apo-B-100 and thus mediate LDL HDL (high density lipoproteins)- Accumulates cholesterol esters. 55-66% Protein. Apo-A-I, -C-I, -C-II, - E. LDL receptors on the cell membranes react with Apo-B-100 and thus mediate LDL Albumin- Free fatty acids. Transports free fatty acids from adipose tissue.

18. Question The urea synthesis occurs almost exclusively in the liver because other tissues contain no or very little amounts of: Answer Choices A Carbamoyl phosphate synthetase I B Carbamoyl phosphate synthetase II C Argininosuccinate synthase D Ornithine transcarbamoylase E Arginase

Correct Answer: Arginase Explanation Urea is the major end product of nitrogen catabolism in humans. Urea is synthesized mostly in the liver in the course of the urea cycle. The first 2 reactions of the cycle, formation of carbamoyl phosphate and citrulline, occur in mitochondria. The newly formed citrulline is then transported into the cytosol, where the remaining cycle enzymes are located. One of the intermediates, ornithine, is regenerated in the course of the cycle and is transported back to mitochondria. The rate-limiting step of the urea cycle is the formation of carbamoyl phosphate by carbamoyl phosphate synthetase I (carbamoyl phosphate synthetase II is not involved in the urea cycle and is located in the cytosol). Carbamoyl phosphate synthetase I is activated by N-acetylglutamate and is driven by cleavage of 2 molecules of ATP. One more ATP molecule is consumed during the synthesis of argininosuccinate; therefore, the overall energy balance amounts to 3 ATP molecules consumed per 1 turn of the cycle. The urea synthesis occurs almost exclusively in the liver because other tissues contain little or no arginase, the enzyme that cleaves arginine releasing urea and regenerating ornithine. The nitrogens of the urea molecule are supplied by free ammonia and aspartate. See the attached image. •

21. Question Carbon skeletons of all amino acids are oxidized in the TCA cycle. However, carbon skeletons are fed into the cycle at several points. Some of the amino acids converted to -ketoglutarate are finally oxidized or utilized for gluconeogenesis. Among the amino acids listed below, which one belongs to this group? Answer Choices A Arginine B Valine C Isoleucine D Asparagine E Tyrosine

Correct Answer: Arginine Explanation Valine and isoleucine produce propionic acid, then to succinyl-CoA. Asparagine is converted to oxaloacetate. Tyrosine is converted to fumarate and acetoacetate. Only arginine among the above is converted to -ketoglutarate. The tricarboxylic acid cycle is the central feature of oxidative metabolism. It is the metabolic pathway in which acetyl-CoA is catalytically oxidized to carbon dioxide, with the concomitant reduction of NAD+ and FAD via a series of tricarboxylic (citric, cis-aconitic and isocitric) and dicarboxylic (succinic, fumaric, malic and oxaloacetic) acids. The actual carbon atoms that appear as CO2 after a single passage through the cycle, are not identical to those that entered as acetyl-CoA. TCA is also known as the citric acid cycle or the Krebs cycle.

66. Question The presence of a low pH in the duodenum will initiate the secretion of what pancreatic substance? Answer Choices A Trypsin B Amylase C Bicarbonate rich solution D Lipase E Glucose

Correct Answer: Bicarbonate rich solution Explanation Pancreatic fluid is comprised of an aqueous component with Na+ and K+ concentrations similar to plasma. Bicarbonate concentration is dependent on the rate of secretion of the aqueous component, which is primarily controlled by the hormone secretin. Acid conditions in the duodenum elicit the production of large volumes of pancreatic secretions which are low in enzyme content, and have high bicarbonate content. This is the result of stimulation of production of secretin by the duodenal mucosal cells in response to the acid environment. High levels of proteins, peptides, and amino acids in the duodenal chyme results in corresponding lower acid levels and increased enzyme content of the pancreatic secretions.

40. Question The largest difference in ion concentrations between the intracellular and extracellular fluid is found for Answer Choices A Calcium B Potassium C Chloride D Sodium E Magnesium

Correct Answer: Calcium Explanation A typical mammalian cell maintains a concentration gradient of many substances, including ions between the extracellular and intracellular fluid. The largest difference is for calcium which shows a 25,000-fold difference (2.5 mM extracellular compared to 0.1 μM). The difference for sodium is 14-fold (140 mM extracellular, 10 mM intracellular); potassium is 35-fold (4 mM extracellular, 140 mM intracellular); magnesium is 4-fold (2 mM extracellular, 0.5 mM intracellular); chloride is 25-fold (100 mM extracellular, 4 mM intracellular). These gradients are maintained across the cell membrane by preventing ion flux and by the active transport of ions from one side of the plasma membrane to the other. The specific transport of molecules across membranes is mediated by carriers. A carrier is not an enzyme and does not catalyze any chemical reactions. However, carriers have some properties in common with enzymes in that they are specific, have dissociation constants, exhibit saturation, and can be blocked by specific inhibitors. There are 2 types of transport, passive or facilitated diffusion and active transport. Active transport involves movement of molecules against a concentration gradient and therefore, requires energy usually as ATP. Examples include calcium transport, the sodium-potassium ATPase, and sodium-linked transport of glucose or amino acids. Passive transport involves movement of molecules down a concentration gradient and includes the movement of glucose in many cells.

5. Question Acyl-CoA derivatives of fatty acids are carried across the inner mitochondrial membrane by Answer Choices A Creatine B Creatinine C Carnitine D Keratin E Carotene

Correct Answer: Carnitine Explanation Creatine, creatinine and phosphocreatine undergo mutual transformation into each other (Figure 1). Creatine is a product of glycine metabolism in the liver, kidney, and pancreas. From there, it is transported to muscle and brain tissues. Phosphocreatine is a source of a high-energy phosphate group for ATP synthesis. Phosphocreatine undergoes a slow non-enzymatic transformation to creatinine, which has no function. This product of the phosphocreatine cyclization is eliminated mostly by renal glomerular filtration. The clearance of creatinine reflects the rate of glomerular filtration and is used as a kidney function test. Although dependent on age, sex, and skeletal muscle mass, the excretion of creatinine does not vary significantly from day to day in a healthy individual. Creatine clearance is calculated as a ratio: Urine volume per unit of time x creatinine in urine / creatinine in plasma. The level of serum creatinine in a healthy individual varies from 0.8 to 1.4 mg/dL. Under renal damage this value can increase up to 4 mg/dL. Carnitine transfers the acyl-CoA derivatives of fatty acids through the mitochondrial membrane for subsequent β-oxidation. (Figure 2) The most widely occurring and biologically active provitamin of vitamin A is β-carotene. (Figure 3) α-Keratin is a fibrous protein of mostly epithelial origin. It constitutes almost the entire material of wool, feathers, nails, hair, claws, etc., and much of the outer layer of skin. This structural protein gives all the structures strength and their protective properties.

10. Question The reaction 2H202--->2H20 +02 is catalyzed by: Answer Choices A Peroxidase B Monooxygenase C Dioxygenase D Catalase E Dehydrogenase

Correct Answer: Catalase [See Image 4] Oxygenases catalyze the direct incorporation of oxygen into a substrate molecule. True oxygenases (dioxygenases) incorporate 2 oxygen atoms into the substrate, while hydroxylases (monooxygenases) require additional electron donor (cosubstrate; B on the figure below) and transfer 1 oxygen atom to the substrate, reducing the other to water. [See Image 5] Many steroids, alcohols, and drugs, such as benzphetamine, aniline, and morphine, are hydroxylated by the microsomal cytochrome P-450 monooxygenase systems, which include cytochromes P-450 and b562. Mitochondrial cytochrome P-450 systems are abundant in steroidogenic tissues, where they are involved in the biosynthesis of steroid hormones from cholesterol, mediating its hydroxylation at C22 and C20 in side-chain cleavage and 11ß and 18 positions. In the liver, mitochondrial cytochrome P-450 systems are involved in the bile acid biosynthesis.

44. Question Oncocytes are Answer Choices A Malignant cells B Cells with large accumulations of glycogen C Cells which stain prominently with hematoxylin D Cells with large accumulations of mitochondria E Cells with large lipid vacuoles

Correct Answer: Cells with large accumulations of mitochondria Explanation Oncocytes are metaplastic or neoplastic cells which have accumulated large numbers of mitochondria. On light microscopy these cells have large amounts of granular, eosinophilic cytoplasm. Studies have shown that the mitochondria in oncocytes do not function properly; there is speculation that their large numbers may represent a compensatory hyperplasia.

116. Question The primary carrier(s) of dietary triacylglycerols and cholesterol in blood is (are) Answer Choices A Chylomicrons B VLDLs C LDLs D HDLs E Albumin

Correct Answer: Chylomicrons Explanation Important Properties of Lipoproteins: Chylomicrons- Transport of dietary cholesterol and triacyglycerols. 1-2% Protein. Apo-B-48, also Apo-A-1, Apo-A-II, -IV while in the lymph, Apo-C-II and Apo-E acquired from HDL in the blood. Assembled in the intestine. Apo-C-II activates endothelial lipoprotein lipase, which removes free fatty acides that become absorbed by tissues, Apo-B-48 comibes only with chylomicrons Chylomicron- Cholesterol, Apo-B-48, Apo-E. Appear as a result of the loss of Apo-C-II, removed by the liver through Apo-E. VLDL (very low density lipoproteins)- Transport of endogenous triacylglycerols from the liver. 7-10% Protein. Apo-B-100, Apo-E, Apo-C-I,-II,-III IDL (intermediate density lipoproteins)- Less triacyglycerols than VLDL 10-12% Protein. Like VLDL. Product of triacylglycerol removal from VLDL. Futher removal leads to LDL LDL (low density)- Primary carrier of cholesterol in the plasma. Delivers cholesterol from the liver to all tissues. 20% Protein. Apo-B-100 exclusively. LDL receptors on the cell membranes react with Apo-B-100 and thus mediate LDL HDL (high density lipoproteins)- Accumulates cholesterol esters. 55-66% Protein. Apo-A-I, -C-I, -C-II, - E. LDL receptors on the cell membranes react with Apo-B-100 and thus mediate LDL Albumin- Free fatty acids. Transports free fatty acids from adipose tissue.

111. Question Free fatty acids and monoacylglycerols yielded from dietary triacylglycerols by pancreatic lipases are transported by Answer Choices A LDLs B Albumin after complete hydrolysis of monoacylglycerols to fatty acids and glycerol C VLDL D HDL E Chylomicrons in the form of reassociated triacylglycerols

Correct Answer: Chylomicrons in the form of reassociated triacylglycerols Explanation Free fatty acids and monoacylglycerols yielded from dietary triacylglycerols by pancreatic lipases diffuse into intestinal epithelial cells, where they are reassembled into triacylglycerols. The triacylglycerols and dietary cholesterol are then included into chylomicrons and transported to the tissues.

94. Question The term "aerobic metabolism" refers to the fact that oxygen is available for use in mitochondria. The presence of oxygen allows which of the following biochemical reaction pathways to proceed? Answer Choices A Chylomicron hydrolysis B Citric acid cycle C Glycogenesis D Glycolysis E Pyruvic acid synthesis

Correct Answer: Citric acid cycle Explanation The citric acid cycle requires oxygen and does not function under anaerobic conditions. It is a series of reactions in which acetyl-CoA is metabolized to CO2 and H atoms. This process occurs in the mitochondria, where the enzymes of oxidative phosphorylation combine the hydrogen (produced by the CAC) with oxygen, to form H2O. This is accompanied by the generation of considerable quantities of ATP for use as an energy supply in other metabolic pathways. The hydrolysis of chylomicron triglycerides releases fatty acids and glycerol. This occurs primarily in the capillary endothelium of adipose tissues and in the liver where the requisite enzyme lipoprotein lipase is concentrated. It is not a part of aerobic metabolism per se. In glycolysis, glucose is broken down to pyruvic acid and/or lactic acid. This may be accomplished anaerobically. Pyruvic acid is then available for conversion to acetyl-CoA and entry into the citric acid cycle if oxygen is available. Glycogenesis is the process of glycogen formation from glucose. Glycogen is the major storage form of glucose and is present in most body tissues, especially liver and skeletal muscle. Glycogenesis is not directly oxygen dependent.

87. Question Endo- and exopeptidases perform respectively Answer Choices A Intracellular and extracellular proteolysis B Cleavage within a peptide chain and at a terminal amino acid residue C Lysosomal and cytoplasmic intracellular proteolysis D Enzymatic and nonenzymatic cleavage E Limited and complete proteolysis

Correct Answer: Cleavage within a peptide chain and at a terminal amino acid residue Explanation Several simple classifications of proteolysis are presented in the chart attached.

48. Question The term chromatin refers to Answer Choices A Nuclear DNA only B Nuclear DNA and histone proteins C All nuclear material D All nuclear material that stains with basic dyes E Complexes of DNA, histone proteins, and some non-histone proteins

Correct Answer: Complexes of DNA, histone proteins, and some non-histone proteins Explanation Chromatin refers to complexes formed between DNA strands, basic histone proteins, and certain acidic non-histone proteins. A small amount of RNA may also be included. There are five types of histone proteins - their function is to protect and package the DNA. There are many kinds of non-histone proteins with many different functions. Some of their functions include acting as initiators and controllers of DNA transcription.

12. Question Newborn infants can control their body temperature because of the presence of brown fat. The mitochondria in this tissue can generate heat instead of ATP normally produced by adult fat cell mitochondria. Brown fat mitochondria can generate heat because these mitochondria Answer Choices A Contain a protein that uncouples electron transport from ATP generation B Lack the enzyme ATP synthase C Lack the ATP translocase antiport system D Are not under respiratory control E Contain electron transport complexes that pump protons into the mitochondrial matrix

Correct Answer: Contain a protein that uncouples electron transport from ATP generation Explanation Brown fat contains a protein called uncoupling protein (UCP) or thermogenin that is a channel protein. UCP permeabilizes the inner mitochondrial membrane to protons, thus collapsing the proton gradient. Oxidative phosphorylation is the coupled process of generating a proton gradient from the passage of electrons derived from NADH and FADH2 through the respiratory chain complexes and using this proton gradient to generate ATP. The process occurs in the mitochondria of all cells including adult fat cells. If the process is uncoupled by breaking down the proton gradient then ATP cannot be synthesized. The uncoupling of oxidative phosphorylation generates heat and can be used to maintain body temperature in hibernating animals, newborns, and animals adapted to the cold. Electron transfer through the respiratory chain leads to pumping of protons from the matrix to the cytosolic side of the inner mitochondrial membrane in both brown fat and adult fat cell mitochondria. The protons then flow through the ATP synthase, resulting in the formation of ATP. The ATP translocase exchanges ADP from the intermembrane space into the matrix and ATP from the matrix into intermembrane space. This leads to acceptor control, meaning that ADP must be available for oxidative phosphorylation to proceed. However, acceptor control is not affected in brown fat tissue. In the presence of UCP, protons flow through the UCP uniporter and produce heat instead of ATP. The brown fat in human newborns is used to warm up the body during its adjustment to living outside the womb. The synthesis of UCP is regulated by fatty acids, through norepinephrine, and UCP is upregulated during cold adaptation. In addition to the protein uncouplers, chemical uncouplers such as dinitrophenol can collapse the proton gradient. These chemicals can pick up a proton in the intermembrane space and can then cross the inner mitochondrial membrane, resulting in the transport of a proton back into the matrix of the mitochondria. As with the proteins like UCP, this leads to uncoupling of electron transport from ATP synthesis. References: 1. Hagen T, Vidal-Puig A. Mitochondrial uncoupling proteins in human physiology and disease. Minerva Med. 2002 Feb;93(1):41-57. 2. Affourtit C, Crichton PG, Parker N, et al. Novel uncoupling proteins. Novartis Found Symp. 2007;287:70-80; discussion 80-91. 3. Berg, J.M., Tymoczko, J.L., Stryer, L. Biochemistry, 6th Edition, W. H. Freeman & Co., New York, 2007, p532-33.

22. Question Which product of glycine metabolism is eliminated by renal filtration and used as a kidney function test? Answer Choices A Creatine B Creatinine C Carnitine D Keratin E Carotene

Correct Answer: Creatinine Explanation Creatine, creatinine and phosphocreatine undergo mutual transformation into each other (Figure 1). Creatine is a product of glycine metabolism in the liver, kidney, and pancreas. From there, it is transported to muscle and brain tissues. Phosphocreatine is a source of a high-energy phosphate group for ATP synthesis. Phosphocreatine undergoes a slow non-enzymatic transformation to creatinine, which has no function. This product of the phosphocreatine cyclization is eliminated mostly by renal glomerular filtration. The clearance of creatinine reflects the rate of glomerular filtration and is used as a kidney function test. Although dependent on age, sex, and skeletal muscle mass, the excretion of creatinine does not vary significantly from day to day in a healthy individual. Creatine clearance is calculated as a ratio: Urine volume per unit of time x creatinine in urine / creatinine in plasma. The level of serum creatinine in a healthy individual varies from 0.8 to 1.4 mg/dL. Under renal damage this value can increase up to 4 mg/dL. Carnitine transfers the acyl-CoA derivatives of fatty acids through the mitochondrial membrane for subsequent β-oxidation. (Figure 2) The most widely occurring and biologically active provitamin of vitamin A is β-carotene. (Figure 3) α-Keratin is a fibrous protein of mostly epithelial origin. It constitutes almost the entire material of wool, feathers, nails, hair, claws, etc., and much of the outer layer of skin. This structural protein gives all the structures strength and their protective properties.

78. Question The side groups of several types of amino acid are capable of electrostatic interactions with other amino acids, and can contribute to a protein's native structure or to an enzyme's catalytic mechanism. Only one of the twenty types of amino acid has a side group that can covalently bond with another of its type. Which amino acid has a side group, which can covalently bond with another of its type to form an amino acid dimer within a protein molecule? Answer Choices A Glycine B Histidine C Serine D Cysteine E Phenylalanine

Correct Answer: Cysteine Explanation The amino acid, cysteine has a side group containing a thiol moiety (-CH2-SH). In oxidative conditions in the cell such as in mitochondria, the endoplasmic reticulum or lysosomes, two cysteines can form a disulfide bond (-CH2-S-S- CH2-), a covalent link between sulfur atoms. Disulfide bonds are necessary for the complete folding and native structure of many proteins. For example, disulfide bonds link the heavy and light chains of immunoglobulins and the enzyme ribonuclease, which cleaves RNA, has four disulfide bonds specifically between eight cysteine residues. The disulfide bond is the only covalent bond besides the peptide bond found in the native structure of a protein. Glycine is the simplest amino acid with the smallest side group, a hydrogen atom. The small size allows greater flexibility and glycine is often found in tight loops that connect two nearby regions of a protein molecule. Histidine has a five-member ring containing two nitrogen atoms in its side group. The nitrogen atoms are capable of electrostatic interactions but do not form covalent bonds. Serine is has an alcohol moiety (-CH2-OH) in its side group. Electrostatic hydrogen-bonding interactions can occur at serine residues but covalent bonds do not form between serine side groups. Phenylalanine is a large, hydrophobic amino acid. The hydrophobic quality means that electrostatic interactions (e.g. with H2O) are unfavorable. Phenylalanine is often found at the hydrophobic core of a protein or within the lipid bilayer-spanning region of a transmembrane protein.

38. Question A 23-year-old metal worker presents to your emergency room in a comatose state. Pulse oximetry indicates that his hemoglobin is 100%. When you draw blood for labs, you notice that the venous sample is bright red. You suspect cyanide poisoning. As cyanide blocks the cytochrome oxidase system, you understand that shortly after oxidative phosphorylation stops, glycolysis will stop too. Which of the following organelles is the major site for anaerobic metabolism? Answer Choices A Mitochondria B Cytoplasm C Golgi apparatus D Nucleolus E Centrioles

Correct Answer: Cytoplasm Explanation Oxidative phosphorylation occurs in the mitochondria. The TCA cycle occurs in the mitochondrial matrix. The electron transport chain occurs on the inner mitochondrial membrane. Anaerobic metabolism occurs in the cytoplasm of most cells. The mitochondria are the site of aerobic metabolism. The Golgi apparatus is responsible for packaging of material for intra- and extracellular use. The nucleolus is the site with in the nucleus of active transcription a gene to mRNA. Centrioles are cytoskeletal elements responsible for nuclear division during cell division.

70. Question Which of the following defines hyponatremia? Answer Choices A Increase in serum sodium concentration B Decrease in serum sodium concentration C Increase in serum potassium concentration D Decrease in serum potassium concentration E Equilibrium between sodium and potassium

Correct Answer: Decrease in serum sodium concentration Explanation Hyponatremia is defined as a serum sodium level less than 130 mmol/L. It is a common electrolyte disorder, occurring in about 15 percent of hospitalized patients. Most patients have elevated vasopressin concentrations.

27. Question Chronic ingestion of ethanol and hypoglycemia have a well-demonstrated relation. Ethanol decreases the rate of formation of glucose by gluconeogenesis. What is the mechanism behind this? Answer Choices A Decreased availability of biotin B Decreased availability of pyruvate C Decreased release of glucagons from pancreatic islets D Inactivation of fructose-1,6-bisphosphatase E Trapping of ATP by ethanol

Correct Answer: Decreased availability of pyruvate Explanation Metabolism of ethanol takes place in the liver, as shown in the included chart. NADH is generated in both reactions and so the cytosolic concentration of NADH rises. This favors the reduction of pyruvate to lactate and of oxaloacetate to malate. Both pyruvate and oxaloacetate are intermediates in the synthesis of glucose by gluconeogenesis. Since ethanol leads to the diversion of these into alternate metabolic pathways, synthesis of glucose by gluconeogenesis is decreased. This can precipitate hypoglycemia, especially in susceptible persons like malnourished or starved individuals.

82. Question Digitalis is a cardiac glycoside which acts on the Na+ K+ pump. Poisoning the Na+-K+ pump with digitalis causes which of the following changes in large axons? Answer Choices A Decreased intracellular Cl- concentration B Decreased intracellular K+ concentration C Decreased intracellular Na+ concentration D Immediate block in propagation of action potentials E Slow hyperpolarization of membrane potentials F Increased intracellular K+ concentration

Correct Answer: Decreased intracellular K+ concentration Explanation Digitalis is a cardiac glycoside that is a derivative of the Foxglove plant (Digitalis lanata). It has several direct and indirect actions on the cardiac musculature, both toxic and therapeutic. All cardiac glycosides act, at the molecular level, on the Nab K ATPase pump in the membrane. The effects of the drug on the cardiac musculature include: • Increased intracellular sodium • Decreased expulsion of calcium from the cell by the sodium calcium exchanger • Increased cardiac contractility Digitalis causes the following in a well-defined sequence: • Brief prolongation of the action potential Protracted shortening of the action potential, which is the result of increased potassium conduction, which is in turn caused by increased intracellular calcium. With more toxic concentrations the resting membrane potential is reduced as a result of inhibition of the sodium pump. This causes a decrease in the intracellular potassium (refer to the table). Effects of Digitalis> Increased IC cal & IC na, depolarization, conduction block over time

16. Question A 23-year-old Asian male visits his family doctor on experiencing abdominal pain, flatulence, and diarrhea. He informs the doctor that he had taken milk in the morning and felt terribly uncomfortable a few minutes later. On the doctor's advice, the patient drinks lactose, which results in a slight increase in the plasma glucose level. On drinking equal quantities of galactose and glucose, a significant increase in the plasma glucose is noticed. From this description it appears that the cause of the patient's problem is Answer Choices A Deficiency of maltase B Deficiency of sucrase C Deficiency of lactase D Diabetes mellitus E Deficiency of amylase

Correct Answer: Deficiency of lactase Explanation There are three major sources of carbohydrates in normal human diet - sucrose, lactose, and starch. The enzyme alpha-amylase, present in saliva, acts on cooked starch and converts into maltose, maltotriose, and alpha-limit dextrins. The pancreatic amylase in the small intestine acts on both cooked and uncooked starch and completely converts the starch into maltose and other glucose polymers. Hydrolysis of disaccharides and small glucose polymers into monosaccharides takes place in the small intestine by the enzymes sucrase, maltase, lactase, etc. Lactose present in the milk is a form of disaccharide. Deficiency of the enzyme lactase results in diarrhea, bloating, and excess of gas formation. A considerable variation in the prevalence of lactase deficiency among different races occurs in more than 65% of people of Asian and African origin.

56. Question The major function of the lysosomal compartment of the eukaryotic cell is to Answer Choices A Determine whether proteins are secreted B Degrade proteins, carbohydrates, and nucleic acids by hydrolysis C Utilize hydrogen peroxide (H2O2) D Supply the energy needs of the cell by synthesizing ATP E Synthesize proteins

Correct Answer: Degrade proteins, carbohydrates, and nucleic acids by hydrolysis Explanation The major function of the lysosome is cellular degradation. This cellular organelle is bounded by a single membrane and has a low interior pH. Within the lysosome are a variety of hydrolases capable of degrading proteins, nucleic acids, lipids, and polysaccharides. The low molecular weight molecules released by these enzymes can be cycled into the different metabolic pathways of the cell. No synthesis occurs in the lysosome. The ATP needs of the cell are met by the mitochondria. The peroxisome is another organelle whose function is to utilize or produce H2O2. They contain enzymes that oxidize a variety of compounds resulting in the formation of H2O2. The H2O2 is converted to water and oxygen by the enzyme catalase. A number of diseases are associated with lysosomal defects including Tay-Sachs disease (disorder of ganglioside breakdown), Hurler's disease (in which the enzyme necessary for the breakdown of glycosaminoglycans is missing), and I-cell disease (in which almost all of the hydrolytic enzymes are missing in the lysosomes of fibroblasts).

29. Question A connection between 2 adjacent epithelial cells responsible for tensile strength is the Answer Choices A Desmosome B Hemidesmosome C Extracellular collagen D Gap junction E Adherens junction

Correct Answer: Desmosome Explanation Hemidesmosomes are characteristic of all stratified squamous epithelia, such as the epidermis. They appear as electron-dense areas at the base of the innermost basal layer. The exterior of the hemidesmosome is attached to anchoring filaments composed of laminin V, while the interior attaches to keratin filaments. The 4 major components of the hemidesmosome are the α6β4 integrin, BPAG1-e, BPAG2, and DH1 (plectin). The α6β4 integrin attaches to the keratin filament network. BPAG2 is a transmembrane protein related to collagen, which together with the protein BPAG1-e, give the hemidesmosome its structure. The hemidesmosome is involved in the strengthening of substratum contacts since loss of this structure leads to a blistering disorder (junctional epidermolysis bullosa) and a drastic weakening of cell-substratum contacts. The keratin filament networks found in epithelial cells are connected to one another through the desmosomes and to the basal lamina through hemidesmosomes. Desmosomes are composed of cadherins, rather than integrins, and are present at all cell-cell borders of basal and suprabasal cells. The cell-cell contacts are mediated by transmembrane proteins related to cadherins. This provides the mechanical stability to the entire tissue. Cadherins are transmembrane proteins that mediate contacts at the adherens junctions, the region of cell-cell contact where the actin cytoskeleton is anchored. The inner surface interacts with keratin filaments.

26. Question Currently there is great expectation that the causes of many of the most devastating diseases will be known at the molecular level. Completion of the sequence of the human genome will give researchers the ability to Answer Choices A Identify all genes involved in a particular disease B Predict which proteins are expressed in which cell types C Cure any disease without any additional information D Determine mutations in specific genes involved in a given disease E Control the regulation of gene expression

Correct Answer: Determine mutations in specific genes involved in a given disease Explanation While all of the above give some information about the gene, only the 3-dimensional structure of the protein gives the complete information about gene function. Sequencing of the complete human genome, consisting of 3 x 109 base pairs of DNA, is currently underway in laboratories around the world. It is estimated that there are 100,000 human genes. Once they are all identified, it will provide new avenues for research and gene therapy for diseases by revealing the genetic basis for human disease.

34. Question Hemidesmosomes are responsible for connecting Answer Choices A Adjacent epithelial cells B Epithelial cells to the underlying basal lamina C Cytoskeletal components to the plasma membrane D Collagen to other extracellular matrix components E Epithelial cells to fibroblasts

Correct Answer: Epithelial cells to the underlying basal lamina Explanation Hemidesmosomes are characteristic of all stratified squamous epithelia, such as the epidermis. They appear as electron-dense areas at the base of the innermost basal layer. The exterior of the hemidesmosome is attached to anchoring filaments composed of laminin V, while the interior attaches to keratin filaments. The 4 major components of the hemidesmosome are the α6β4 integrin, BPAG1-e, BPAG2, and DH1 (plectin). The α6β4 integrin attaches to the keratin filament network. BPAG2 is a transmembrane protein related to collagen, which together with the protein BPAG1-e, give the hemidesmosome its structure. The hemidesmosome is involved in the strengthening of substratum contacts since loss of this structure leads to a blistering disorder (junctional epidermolysis bullosa) and a drastic weakening of cell-substratum contacts. The keratin filament networks found in epithelial cells are connected to one another through the desmosomes and to the basal lamina through hemidesmosomes. Desmosomes are composed of cadherins, rather than integrins, and are present at all cell-cell borders of basal and suprabasal cells. The cell-cell contacts are mediated by transmembrane proteins related to cadherins. This provides the mechanical stability to the entire tissue. Cadherins are transmembrane proteins that mediate contacts at the adherens junctions, the region of cell-cell contact where the actin cytoskeleton is anchored. The inner surface interacts with keratin filaments.

9. Question Inherited deficiency in lysosomal ceramidase causes: Answer Choices A Farber's disease B Congenital abetalipoproteinemia C Niemann-Pick disease D Refsum's disease E Creutzfeldt-Jacob's disease

Correct Answer: Farber's disease Explanation Sphingolipidoses are caused by deficiencies in lysosomal enzymes responsible for the degradation of sphingolipids. Niemann-Pick disease is due to a deficiency in sphingomyelinase, an enzyme that hydrolyzes sphingomyelin to ceramide and phosphorylcholine. Deficiency in this enzyme causes an accumulation of sphingomyelin (and, occasionally, unesterified cholesterol) in reticuloendothelial cells of the liver, spleen (leading to the enlargement of the organs, hepatosplenomegaly), bone marrow, and central nervous system. An impaired degradation of ceramide, resulting from a deficiency in lysosomal enzyme ceramidase, causes Farber's disease (lipogranulomatosis) which is characterized by granulomatous lesions in the skin, joints, and larynx. Fabry's disease is due to a deficiency in α-galactosidase. Glycolipid, ceramide trihexoside (Gal-Gal-Glc-Cer), is accumulated in tissues causing kidney and heart failure, reddish-purple skin rash, and pain in low extremities. Tay-Sachs disease is caused by a deficiency in β-hexosaminidase A. Gangliosides are accumulated in tissues (in particular, the central nervous system), which results in mental retardation, blindness, muscular weakness, and seizures. A deficiency in β-glucosidase causes Gaucher's disease characterized by an accumulation of glucocerebrosides in reticuloendothelial cells causing hepatosplenomegaly, osteoporosis, and mental retardation. Gaucher's and Tay-Sachs diseases are comparatively frequent among Ashkenazi Jews (100 times more than in other populations). All sphingolipidoses are autosomal recessive disorders, except for Fabry's disease which is X-linked. A deficiency in apolipoprotein B-48 causes congenital abetalipoproteinemia. Refsum's disease is an inherited metabolic disease caused by an inability to metabolize phytanic acid. Creutzfeldt-Jacob's disease is a neurodegenerative disorder which is caused by infectious exogenous prion particles or inherited mutations in the prion protein gene.

11. Question The highest stores of energy (kcal/g) in the body are found in Answer Choices A Protein B Carbohydrate C Fat D Protein, carbohydrate, and fat equally E Protein and fat equally

Correct Answer: Fat Explanation Carbohydrate and protein each contain 4 kcal/g while fat contains 9 kcal/g. Therefore, the greatest energy reserve is found in fat.

39. Question Which of the following pathways takes place exclusively in the cytoplasm of the cell? Answer Choices A Urea cycle B Fatty acid synthesis C Fatty acid degradation D Oxidative phosphorylation E Gluconeogenesis

Correct Answer: Fatty acid synthesis Explanation Compartmentalization of parts of a metabolic pathway is an important regulatory mechanism. This is especially true for synthesis/degradation pathways such as for fatty acids where synthesis takes place in the cytoplasm while degradation takes place in the mitochondria. Parts of the urea cycle and gluconeogenesis take place in the mitochondria while other reactions take place in the cytoplasm. Oxidative phosphorylation takes place exclusively in the mitochondria.

24. Question A 32-year-old, vegetarian female in mid-pregnancy complains of lack of energy and says she becomes easily fatigued. Upon any strenuous movement, her heart pounds rapidly and she becomes short of breath. Nutritional supplement of what mineral may alleviate the female's symptoms? Answer Choices A Mg2+ B Ca2+ C Fe2+ D Cu2+ E Zn2+

Correct Answer: Fe2+ Explanation Pregnant females can experience iron deficiency anemia due to increased demands on their blood. Oxygen (O2) in the lungs binds to the iron ion, Fe2+, while complexed with the heme cofactor of hemoglobin in red blood cells. Iron supplements or foods with abundant iron, such as liver, lean meats, or vegetarian alternatives such as spinach, carrots, and raisins can alleviate anemic symptoms. The other minerals have biological roles but are not associated with anemic symptoms. Magnesium (Mg2+) coordinates with the negatively charged backbone of DNA and interacts with neurotransmitter receptors at excitatory synapses in the central nervous system. Mg2+ deficiency affects the nervous system, resulting in vasodilation, tremors, and depression. Calcium phosphate forms a hard material in bone and teeth. In addition, Ca2+ is a ubiquitous second messenger ion in cellular signaling coupled to G-protein signaling, hormone signaling, and ion channel activity. Ca2+ deficiency can give rise to muscle twitching or cramping and cardiac arrhythmias. Copper (Cu2+) participates in bone and blood formation and is an electron carrier in mitochondrial electron transfer proteins. Cu2+ deficiency is uncommon since the trace amounts needed are satisfied by most diets. Zinc (Zn2+) is a cofactor of many DNA and RNA binding proteins, including many transcription factors. Severe zinc deficiency can retard growth in children, can cause low sperm count in males, and can slow wound healing.

121. Question Macrocytic anemia and spina bifida are symptoms of which deficiency? Answer Choices A Folate (folic acid) B Vitamin B12 C Vitamin B2 (riboflavin) D Vitamin C E Pantothenic acid

Correct Answer: Folate (folic acid) Explanation Refer to table. Folate (folic acid)- Marcrocytic anemia and spina bifida B12- Pernicious anemia and nervous system disorders B2 (riboflavin)- Eye disorders and skin cracking, especially at corners of the mouth C- Scurvy-defective collagen formation and poor wound healing Pantothenic acid- Neuromuscular dysfunction and fatigue

52. Question Cells with large amounts of smooth endoplasmic reticulum (sER) Answer Choices A Appear basophilic B Have copious amounts of dense cytoplasm C Frequently manufacture steroids or lipids D Are only rarely identified in muscle E Modify proteins for export out of the cell

Correct Answer: Frequently manufacture steroids or lipids Explanation Cells with large amounts of sER have abundant clear or eosinophilic cytoplasm. These cells are usually involved in steroid synthesis, lipid synthesis, or drug detoxification. Adrenal cortical cells and testicular Leydig cells are good examples of steroid secreting cells. Liver cells are usually the site of drug detoxification. Modified sER is an important component of muscle cells, where it is called sarcoplasmic reticulum. The sarcoplasmic reticulum stores calcium that is required for muscle contraction. The Golgi apparatus is responsible for enzymatic modification of proteins.

25. Question Many different mechanisms are involved in the control on synthetic and degradative metabolic pathways. This is necessary to avoid futile cycles. In some pathways, some steps are reversible and others are not. The major regulatory step in the glycolytic pathway is the conversion of Answer Choices A Glucose 6-phosphate to fructose 6-phosphate B 3-phosphoglycerate to 2-phosphoglycerate C Glyceraldehyde 3-phosphate to 1,3 bisphosphoglycerate D Fructose 6-phosphate to fructose 1,6 bisphosphate E Fructose 6-phosphate to fructose 2,6 bisphosphate

Correct Answer: Fructose 6-phosphate to fructose 1,6 bisphosphate Explanation All but 3 of the reactions of glycolysis are readily reversible. The reactions involving 1) the conversion of glucose to glucose 6-phosphate, 2) the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate, and 3) the conversion of phosphoenolpyruvate to pyruvate, all involve the addition of a phosphate residue from ATP to a sugar. Since the reactions involve the hydrolysis of ATP, the ΔG° is negative for these reactions. Therefore, they are not readily reversible. The isomerization of glucose 6-phosphate to fructose 6-phosphate is a reversible reaction. The gluconeogenesis pathway (synthesis of glucose from pyruvate) uses 4 reactions referred to as "bypass" reactions to overcome the overall high negative ΔG° for glycolysis. The other reactions of the gluconeogenesis pathway are the reverse of those in glycolysis. The ΔG° for the conversion of glucose 6-phosphate to fructose 6-phosphate is +0.4 kcal/mol and therefore readily reversible. Gluconeogenesis is not a simple reversal of glycolysis because it would not be thermodynamically favorable. The conversion of fructose 6-phosphate to fructose 1,6 bisphosphate by the enzyme phosphofructokinase is the first committed step of glycolysis. This reaction commits the cell to metabolize glucose by the glycolytic pathway. Although in many metabolic pathways the first step in the pathway is the committed step, in the case of glycolysis it is the second step, the conversion of fructose 6-phosphate to fructose 1,6 bisphosphate, that is the committed step. The phosphoglucose isomerase reaction, the conversion of glucose 6-phosphate to fructose 6-phosphate is a reversible reaction. The production of fructose 2,6 bisphosphate from fructose 6-phosphate is not part of the main glycolytic pathway but is one of the regulatory mechanisms involved. It is produced in the liver by the dephosphorylated form of phosphofructokinase 2 in response to low glucose levels and slows glycolysis.

8. Question The enzyme deficiency associated with Gaucher's disease prevents the hydrolysis of Answer Choices A Glucocerebroside B Ceramide C Sphingolipids D Triacylglycerols E Sphingomyelin

Correct Answer: Glucocerebroside Explanation Gaucher's disease is an example of a lysosomal storage disease that is caused by the failure of the lysosome to break down a component that it would normally degrade. Accumulation of the undegraded compound leads to increased size and number of lysosomes. Eventually this disrupts cellular function. Gaucher's disease occurs in three types that vary in their severity. This disease is caused by a deficiency in the enzyme glucocerebrosidase, which is responsible for degrading glucocerebroside into ceramide and glucose. Macrophages, because of their phagocytic activity, are the cells most affected by this disease.

29. Question The transformation of oxaloacetate to phosphoenolpyruvate belongs to the process which is called Answer Choices A Glycolysis B Glycogenolysis C Glycogenesis D Gluconeogenesis E Glyoxylate cycle

Correct Answer: Gluconeogenesis Explanation Glycolysis is a breakdown of a molecule of glucose to yield two molecules of pyruvate. Glycogenolysis is a catabolic process of the glycogen breakdown. Glycogenesis is an anabolic process of the formation of glycogen. Gluconeogenesis is an anabolic process of the synthesis of glucose from non-carbohydrate precursors. Glyoxylate cycle is an anaplerotic pathway, a variation of the citric acid cycle. (See Image)

112. Question The initial event in the utilization of stored fat as an energy source is Answer Choices A Activation of fatty acids by acyl CoA synthetase B β-oxidation of fatty acids C Hydrolysis of triglycerides by lipase D Transport of fatty acids into mitochondrial matrix by carnitine

Correct Answer: Hydrolysis of triglycerides by lipase Explanation Fatty acids in the body are stored as triacylglycerols. These are highly concentrated stores of metabolic energy because they are reduced and anhydrous. The initial event in the utilization of this stored energy is the hydrolysis of triacylglycerol by lipases. The released fatty acid is then activated by linking it to coenzyme A. This reaction is carried out on the mitochondrial outer membrane by the enzyme acyl CoA synthetase. The conjugated fatty acid is then transported into the mitochondrial matrix by the carnitine shuttle system. Once inside the mitochondria, the fatty acids are metabolized by β-oxidation, which generates acetyl CoA, NADH, and FADH2.

100. Question During the majority of the time it takes a runner to complete a 1500 m race, he relies on the generation of ATP from the aerobic breakdown of glycogen. As part of the glycogen degradation pathway, the enzyme glycogen phosphorylase acts on glycogen to release Answer Choices A Free glucose B UDP-glucose C Glucose 6-phosphate D Fructose 6-phosphate E Glucose 1-phosphate

Correct Answer: Glucose 1-phosphate Explanation The enzyme glycogen phosphorylase carries out a phosphorolysis reaction. Inorganic phosphate is used to release glucose 1-phosphate from the α[1→4] glycosidic linkages of the glycogen molecule. The use of inorganic phosphate, instead of water (hydrolysis), conserves the energy of the glycosidic bond. Glucose 6-phosphate is formed from glucose 1-phosphate by the action of the enzyme phosphoglucomutase. The glucose 6-phosphate formed enters the glycolytic pathway where it is converted to fructose 6-phosphate. The branch points of glycogen (the α[1→6] linkages), are acted upon by the "debranching enzyme." This reaction is a hydrolysis and free glucose is released. The compound UDP-glucose is the activated form of glucose used in the synthesis of glycogen and is not a part of glycogen degradation.

102. Question A medical student is asked to name the enzyme deficiency, given only the following information: it is an inherited metabolic disorder of carbohydrate metabolism. It is characterized by an abnormally increased concentration of hepatic glycogen with normal structure and no detectable increase in serum glucose concentration after oral administration of fructose. What is the correct answer? Answer Choices A Fructokinase B Glucokinase C Glucose 6-phosphatase D Fructose-1, 6-diphosphotase deficiency E Phosphoglucomutase F UDPG-glycogen transglucosylase

Correct Answer: Glucose 6-phosphatase Explanation Normal structure of glycogen indicates that glycogen is being synthesized normally. An abnormally increased amount of hepatic glycogen indicates that it is not being broken down properly. There is no increase in serum glucose after administration of oral fructose, indicating that fructose has not been converted into glucose and exported. Normally in the liver, both glycogen and fructose are first transformed into glucose-6- phosphate, and then to glucose in order to leave the cell. This step is catalyzed by glucose 6-phosphatase. Refer to the image. A deficiency of glucose 6-phosphatase can explain all the given phenomena. Fructokinase deficiency results in fructosuria. It is a benign condition. Glucokinase deficiency would not allow the liver to remove excess glucose from the blood for storage. Fructose-1, 6-diphosphatase deficiency leads to failure of gluconeogenesis. A deficiency in phosphoglucomutase would not allow the interconversion of fructose and glucose. UDPG-glycogen transglucosylase deficiency would interfere with glycogen synthesis.

98. Question Many different mechanisms are involved in the control on synthetic and degradative metabolic pathways. This is necessary to avoid futile cycles. In some pathways, some steps are reversible and others are not. Which of the following steps of glycolysis are readily reversible? Answer Choices A Glucose to glucose 6-phosphate B Glucose 6-phosphate to fructose 6-phosphate C Fructose 6-phosphate to fructose 1,6-bisphosphate D Phosphoenolpyruvate to pyruvate E All reactions of glycolysis are readily reversible

Correct Answer: Glucose 6-phosphate to fructose 6-phosphate Explanation All but 3 of the reactions of glycolysis are readily reversible. The reactions involving 1) the conversion of glucose to glucose 6-phosphate, 2) the conversion of fructose 6-phosphate to fructose 1,6-bisphosphate, and 3) the conversion of phosphoenolpyruvate to pyruvate, all involve the addition of a phosphate residue from ATP to a sugar. Since the reactions involve the hydrolysis of ATP, the ΔG° is negative for these reactions. Therefore, they are not readily reversible. The isomerization of glucose 6-phosphate to fructose 6-phosphate is a reversible reaction. The gluconeogenesis pathway (synthesis of glucose from pyruvate) uses 4 reactions referred to as "bypass" reactions to overcome the overall high negative ΔG° for glycolysis. The other reactions of the gluconeogenesis pathway are the reverse of those in glycolysis. The ΔG° for the conversion of glucose 6-phosphate to fructose 6-phosphate is +0.4 kcal/mol and therefore readily reversible. Gluconeogenesis is not a simple reversal of glycolysis because it would not be thermodynamically favorable. The conversion of fructose 6-phosphate to fructose 1,6 bisphosphate by the enzyme phosphofructokinase is the first committed step of glycolysis. This reaction commits the cell to metabolize glucose by the glycolytic pathway. Although in many metabolic pathways the first step in the pathway is the committed step, in the case of glycolysis it is the second step, the conversion of fructose 6-phosphate to fructose 1,6 bisphosphate, that is the committed step. The phosphoglucose isomerase reaction, the conversion of glucose 6-phosphate to fructose 6-phosphate is a reversible reaction. The production of fructose 2,6 bisphosphate from fructose 6-phosphate is not part of the main glycolytic pathway but is one of the regulatory mechanisms involved. It is produced in the liver by the dephosphorylated form of phosphofructokinase 2 in response to low glucose levels and slows glycolysis.

32. Question The following enzyme helps protect erythrocytes against hemolysis: Answer Choices A 6-phosphogluconate dehydrogenase B Gluconolactone hydrolase C Glucose-6-phosphate dehydrogenase D Ribulose-5-phosphate-3 epimerase E Transaldolase F Transketolase

Correct Answer: Glucose-6-phosphate dehydrogenase Explanation The pentose phosphate pathway is one that interconnects within itself at many points. It is composed of two phases. The first is an irreversible, oxidative phase [dark cyan pathway] that transforms glucose-6-phosphate into ribulose-5-phosphate. The second is a reversible, non-oxidative phase [gray pathway] that converts three molecules of ribulose-5-phosphate into two molecules of fructose-6-phosphate and one molecule of glyceraldehyde-3-phosphate. The main enzyme deficiency resulting in a type of hemolytic anemia is glucose-6-phosphate dehydrogenase [1]. In the attached figure, cofactors are in BROWN. The other enzymes involved in this pathway are gluconolactone hydrolase [2], 6-phosphogluconate dehydrogenase [3], ribose-5-phosphate ketoisomerase [4], ribulose-5-phosphate 3-epimerase [5], transketolase [6], transaldolase [7], and 5-phosphoribosyl-1-pyrophosphate (PRPP) synthetase [8].

17. Question During conditions when the dietary supply of energy does not meet the energy requirement of the body, skeletal muscle proteins are degraded in the muscle tissue itself and carbon skeletons of amino acids are used for energy or transported to the liver for gluconeogenesis. This metabolic adjustment occurs during conditions like starvation, trauma, burns, and fever. In such situations, the NH3 group of amino acids is transported to the liver in what form? Answer Choices A Glutamine and alanine B Glutamate and serine C Glycine and serine D Ammonia E Aspartate

Correct Answer: Glutamine and alanine Explanation Glutamine is synthesized from glutamate and ammonia. Alanine is synthesized from pyruvate by transamination. The amino group from other amino acids is transferred to pyruvate through glutamate. Glutamine is 1 of the 20 amino acids commonly found in proteins. It is the amide at the gamma-carboxyl of the amino acid glutamate. Glutamine can participate in covalent cross-linking reactions between proteins by forming peptide-like bonds by a transamidation reaction with lysine residues. Alanine normally refers to L-alpha-alanine, the aliphatic amino acid found in proteins. The isomer beta-alanine is a component of the vitamin pantothenic acid and thus also of coenzyme A.

92. Question Glycogen is isolated from the muscle of a patient with McArdle's Disease. The patient complains of muscle cramps after exercise. A biopsy of the muscle reveals much more glycogen in this patient's muscle than in the muscle of a normal individual. The structure of the patient's glycogen is normal. Which of the following enzymes is most likely to be defective in this patient? Answer Choices A Glycogen synthase B Protein kinase A C Debranching enzyme D Glycogen phosphorylase E Phosphorylase kinase

Correct Answer: Glycogen phosphorylase Explanation Several different types of glycogen storage diseases have been identified that show defects in different enzymes involved in glycogen metabolism. Depending on the defect, the clinical features can vary widely. McArdle's disease is confined to the muscle and there is only an absence of the enzyme phosphorylase in the muscle. Glycogen phosphorylase is the primary enzyme responsible for glycogen breakdown. Strenuous exercise is limited in these patients because of severe muscle cramps. Otherwise, these patients are normal and well developed. The use of muscle glycogen is not essential for life. No increase in plasma lactate is observed because of the lower than normal rate of glycolysis in the muscle, since the muscle cannot metabolize glycogen. The cramps have been shown to correlate with high levels of muscle ADP.

31. Question The transformation of glucose-1-phosphate to UDP-glucose belongs to the process which is called Answer Choices A Glycolysis B Glycogenolysis C Glycogenesis D Gluconeogenesis E Glyoxylate cycle

Correct Answer: Glycogenesis Explanation Glycolysis is a breakdown of a molecule of glucose to yield two molecules of pyruvate. Glycogenolysis is a catabolic process of the glycogen breakdown. Glycogenesis is an anabolic process of the formation of glycogen. Gluconeogenesis is an anabolic process of the synthesis of glucose from non-carbohydrate precursors. Glyoxylate cycle is an anaplerotic pathway, a variation of the citric acid cycle. (See Image)

104. Question The reaction of glucose-1-phosphate production by glycogen phosphorylase belongs to the process which is called Answer Choices A Glycolysis B Glycogenolysis C Glycogenesis D Gluconeogenesis E Glyoxylate cycle

Correct Answer: Glycogenolysis Explanation Glycolysis is a breakdown of a molecule of glucose to yield two molecules of pyruvate. Glycogenolysis is a catabolic process of the glycogen breakdown. Glycogenesis is an anabolic process of the formation of glycogen. Gluconeogenesis is an anabolic process of the synthesis of glucose from non-carbohydrate precursors. Glyoxylate cycle is an anaplerotic pathway, a variation of the citric acid cycle. (See Image)

24. Question The transformation of 1,3-bisphosphoglycerate to 3-phosphoglycerate belongs to the process which is called Answer Choices A Glycolysis B Glycogenolysis C Glycogenesis D Gluconeogenesis E Glyoxylate cycle

Correct Answer: Glycolysis Explanation Glycolysis is a breakdown of a molecule of glucose to yield two molecules of pyruvate. Glycogenolysis is a catabolic process of the glycogen breakdown. Glycogenesis is an anabolic process of the formation of glycogen. Gluconeogenesis is an anabolic process of the synthesis of glucose from non-carbohydrate precursors. Glyoxylate cycle is an anaplerotic pathway, a variation of the citric acid cycle. (See Image)

22. Question A fermentation process which is the first step in the generation of large amounts of ATP in the body is called Answer Choices A Ketogenesis B Gluconeogenesis C Glycogenesis D Glycolysis E Glycogenolysis

Correct Answer: Glycolysis Explanation Glycolysis is a catabolic reaction in which glucose is degraded into 2 molecules of pyruvate. The energy released during this reaction is stored in the form of ATP. This reaction occurs without the need for oxygen and is beneficial in situations where a brief burst of energy is required for bodily functioning when oxygen levels are low. The breakdown products can then be funneled into the Kreb's cycle for further generation of large amounts of ATP. Ketogenesis results in the production of ketone bodies. AcetylCoA (produced from fatty acid or pyruvate oxidation) is then converted into free acetoacetate and D-B-hydroxybutyrate. These compounds are then further metabolized via the tricarboxylic acid cycle into CO2 and Hydrogen. In the case of diabetes, ketone bodies may accumulate faster than the peripheral tissues can metabolize them, and the state of ketosis occurs. Gluconeogenesis involves the synthesis of mono- and polysaccharides. It requires energy to build these more complex structures from lactate and pyruvate. In order for pyruvate to enter this metabolic pathway, it must be converted to glucose, and a phosphate group added to form glucose-6-phosphate, which is a high-energy molecule capable of entering this metabolic pathway. Glycogenesis is a synthesis reaction in which glycogen (a polysaccharide) is synthesized from glucose-1-phosphate. Glucose-6-phosphate is the typical reactant used, so the enzyme phosphoglucomutase is needed to convert glucose-6-phosphate to glucose-1-phosphate. Glycogenolysis is a catabolic reaction which results in the breakdown of glycogen into glucose molecules. This typically occurs in skeletal muscle and liver cells with the help of glycogen phosphorylase.

50. Question Where do proteins to be secreted move to after going from the lumen of the rough endoplasmic reticulum? Answer Choices A Extracellular space B Mitochondria C Golgi apparatus D Smooth endoplasmic reticulum E Cytoplasm

Correct Answer: Golgi apparatus Explanation The Golgi apparatus is continuous with the rough endoplasmic reticulum (rER) and, like the rER, is most prominent in cells that secrete large amounts of protein. Plasma cells have such large Golgi apparatuses that can be seen in the light microscope as a clear zone next to the nucleus - the perinuclear hof. The Golgi apparatus makes oligosaccharides, which can then be added to the newly synthesized protein to form glycoproteins. Sulfate groups are added to form proteoglycans. The products of these reactions are sorted, concentrated, and packaged into vesicles, and the vesicles are pinched off. These vesicles containing newly made complex proteins then migrate to their appropriate individual destinations.

91. Case A 6-year-old boy is mentally retarded and has a movement disorder. In addition, he compulsively gnaws on his fingers and his lips. He is seen by his pediatrician for follow-up for an underlying genetic disease. On physical examination, he is noted to be short for his age. There is quite a bit of maceration and damage to his fingers from his constant gnawing. His reflexes are found to be hyperactive. His sensory examination is within normal limits. His laboratory results are as follows: TEST RESULTS REFERENCE RANGE Hematocrit 38 (35-45 (age 6-12)) Hemoglobin 13.4 gm/dL (11.5-15.5 gm/dL (age 6-12)) Mean corpuscular volume 81 cu μ (77-95 cu μ (age 6-12)) WBC 6.4 x 103/mm3 (5.5-15.5 x 103/mm3) RBC 4.8 x 106/μl (3.7-5.2 x 106/μl (age 2-14)) Platelets 225,000/mm3 (130,000 - 400,000/mm3) Neutrophils 61% (40-75%) Lymphocytes 33% (15-45%) Monocytes 3% (1-10%) Eosinophils 2% (1-6%) Basophils 1% (0-2%) Uric acid 10.2 mg/dL (3.0-8.5 mg/dL (male) 2.5-7.0 mg/dL (female)) Question What can be a long-term complication of this condition? Answer Choices A Ankylosing spondylitis B Rheumatoid arthritis C Osteoarthritis D Gouty arthritis E Septic arthritis

Correct Answer: Gouty arthritis Explanation This male has Lesch-Nyhan syndrome. Mental retardation, self-mutilation, and movement disorders can be seen with Lesch-Nyhan syndrome, as the case here. The delay in growth, hyperreflexia, normal sensory examination, and tissue damage because of the self-mutilation are all also consistent with Lesch-Nyhan syndrome. Lesch-Nyhan syndrome is a disorder of purine metabolism. Lesch-Nyhan syndrome is due to a lack of an enzyme called hypoxanthine-guanine phosphoribosyltransferase (HPRT). Because of this, Lesch-Nyhan syndrome is sometimes also called HPRT deficiency. Due to the defects with purine metabolism, hyperuricemia can also be seen. The hyperuricemia sometimes leads to gout. Ankylosing spondylitis is an inflammatory spinal disease. It can lead to spinal fusion, which is sometimes referred to as "bamboo spine". It can also involve the sacroiliac joints, leading to a sacroiliitis. Ankylosing spondylitis occurs much more frequently in males. Ankylosing spondylitis is associated with HLA-B27. It is sometimes called Marie-Strümpell disease. Ankylosing spondylitis is not a long-term complication of Lesch-Nyhan syndrome. Rheumatoid arthritis is a chronic disease. It is an inflammatory disease. Subcutaneous nodules, morning stiffness, and symmetric joint swelling are some of the symptoms. Rheumatoid factor can be seen in approximately 70% of adults who have rheumatoid arthritis. This antibody is directed against the person's IgG. However, rheumatoid factor is not specific for rheumatoid arthritis. Rheumatoid arthritis is not a long-term complication of Lesch-Nyhan syndrome. Septic arthritis is associated with infections. Septic arthritis is not a long-term complication of Lesch-Nyhan syndrome.

7. Question A 52-year-old woman with a recent diagnosis of hypercholesterolemia is given a lipid-lowering agent. Her blood work is within normal limits. Vital signs are Temp 99.4 °F, BP 134/88 mm Hg, RR 12/min, P 74bpm. Physical examination reveals an obese woman sitting comfortably in no apparent acute distress. Cardiovascular examination reveals a systolic click and murmur. The patient takes the medication for a few weeks; however, she notes diffuse muscle aches after being placed on the medication. Which of the following enzymes is most likely inhibited? Answer Choices A HMG CoA synthesis B HMG CoA reductase C Geranyl - PP synthase D Farsenyl - PP synthase E Squalene synthase

Correct Answer: HMG CoA reductase • Explanation The correct answer choice is HMG CoA reductase, which acts at site B in the image. This patient was most likely given a statin. Statins are lipid lowering agents, which lower LDL by inhibiting the enzyme HMG CoA reductase. The major side effects of this class of medications are myalgias. Other effects to be aware of are liver toxicity. HMG CoA synthase catalyzes the reaction, which forms 3-hydroxy-3methyl glutaryl CoA. Farsenyl PP-synthase converts geranyl pyrophosphate to farsenyl pyrophosphate (reaction C). Farsenyl-pyrophosphate is converted to squalene by squalene synthase (reaction D). Squalene monooxygenase converts squalene to 2,3 oxidosqualene. This is an intermediate reaction at E in the image. References: 1. Lippincott's Illustrated Review: Biochemistry, 2nd edition. Copyright 1994 Lippincott Williams & Wilkins. 2. Wang W, Tong TJ.[The key enzyme of cholesterol synthesis pathway: HMG-CoA reductase and disease][Article in Chinese] Sheng Li Ke Xue Jin Zhan. 1999 Jan;30(1):5-9.

34. Question Two of the muscle types found in the human body are heart and skeletal muscle. A significant difference between the two types of muscle is that heart muscle Answer Choices A Stores large amounts of lipids while skeletal muscle stores very little lipids B Has a completely aerobic metabolism while skeletal muscle uses aerobic and anaerobic metabolism C Contains very few mitochondria compared to skeletal muscle D Cannot use ketone bodies as a fuel source while skeletal muscle uses ketone bodies exclusively E Has large glycogen stores while skeletal muscle has very little glycogen

Correct Answer: Has a completely aerobic metabolism while skeletal muscle uses aerobic and anaerobic metabolism Explanation Heart muscle is continually active while skeletal muscle has periods of activity and inactivity. The heart beats in a continual cycle of relaxation and contraction. Because heart muscle uses aerobic metabolism completely, this type of muscle contains a large number of mitochondria, almost half of the cell volume. The heart uses a mixture of glucose, free fatty acids, and ketone bodies as fuel sources. In contrast, skeletal muscle uses anaerobic and aerobic metabolism and contains fewer mitochondria. Neither heart nor skeletal muscle stores significant amounts of lipids or glycogen. Both phosphocreatine and muscle glycogen serve as fuel for ATP synthesis during heavy muscle activity. Resting skeletal muscle uses mostly free fatty acids derived from adipose tissue or ketone bodies from the liver as energy sources.

21. Question This enzyme catalyses 1 of the 3 irreversible steps in glycolysis: Answer Choices A Enolase B Glucokinase C Glyceraldehyde-3-phosphate dehydrogenase D Hexokinase E Lactate dehydrogenase F Pyruvate dehydrogenase

Correct Answer: Hexokinase Explanation Some toxic compounds are able to inhibit glycolysis. Iodoacetate inhibits glyceraldehyde-3-phosphate dehydrogenase, and fluoride inhibits enolase. In the attached figure, cofactors are in BLUE. The enzymes involved in glycolysis are as follows: glucokinase [1], hexokinase [2], phosphohexose isomerase [3], phosphofructokinase-1 [4], aldolase [5], phosphotriose isomerase [6], glyceraldehyde-3-phosphate dehydrogenase [7], phosphoglycerate kinase [8], phosphoglycerate mutase [9], enolase [10], pyruvate kinase [11], pyruvate dehydrogenase complex [12], and lactate dehydrogenase [13]. •

96. Question Suppose the enzyme glucokinase was isolated from mouse liver cells and shown to be inhibited by glucose 6-phosphate. These mice and control mice with normal glucokinase were fed and the serum was isolated 2 hours afterwards. Compared to control mice, what would you predict the serum from mutant mice to show? Answer Choices A Increased levels of glucokinase B High concentrations of glucose 1-phosphate C Low concentrations of glucose 6-phosphate D High concentrations of glucose E Normal levels of glucose

Correct Answer: High concentrations of glucose Explanation High concentrations of glucose would remain in the serum of the mutant mice after they were fed as a result of the mutation in glucokinase. Glucokinase is an enzyme found in the liver and is an isozyme of hexokinase. Both enzymes catalyze the reaction: Glucose + ATP Glucose 6-phosphate + ADP + Pi However, the Km for glucose is much higher for glucokinase than for hexokinase. In addition, hexokinase is inhibited by glucose 6-phosphate and wild type glucokinase is not. These properties are suited for the different functions of the enzymes. Hexokinase functions mainly as part of glycolysis, while the main function of glucokinase is to remove large amounts of glucose from the serum when blood glucose levels are high, such as after a meal. The fact that normal glucokinase is not inhibited by glucose 6-phosphate allows the enzyme to continue functioning under these conditions leading to high intracellular concentrations of glucose 6-phosphate, which would not occur in the mutant cells. In the mutant enzyme described here, glucokinase is inhibited by glucose 6-phosphate. Therefore, the enzyme would be inhibited before it could completely remove the excess glucose from the serum. The mutation affecting the inhibition properties of glucokiase would not affect levels of glucokinase which are not found in serum but in the cell. Glucose 1-phosphate and glucose 6-phosphate are not found in serum but only intracellularly. The presence of the phosphate in either position prevents it from leaving the cell.

108. Question A 47-year-old man comes into a free clinic. He appears to be intoxicated and malnourished. After obtaining additional information you make the assumption that he is a chronic alcoholic. Which of the following metabolic alterations are likely to be present in this man compared to a man who does not drink alcohol? Answer Choices A His levels of lactate/pyruvate are decreased compared to the non-drinker B His rate of gluconeogenesis is stimulated compared to the non-drinker C His levels of NADH/NAD+ are increased compared to the non-drinker D His rate of fatty acid oxidation is stimulated compared to the non-drinker E His level of ketone bodies is decreased compared to the non-drinker

Correct Answer: His levels of NADH/NAD+ are increased compared to the non-drinker Explanation In a chronic alcoholic, ethanol is metabolized by the enzyme alcohol dehydrogenase, which uses a hydrogen from ethanol to form NADH from NAD+ thereby significantly increasing the ratio of NADH to NAD+ in the chronic alcoholic. In a normal, non-drinking individual, the ratio of NADH/NAD+ is approximately 0.0014, that is nicotinamide dinucleotide is in the oxidized (NAD+) form by more than 700 times. Since many metabolic reactions are regulated by the levels of NADH, the increased amount of NADH in the chronic alcoholic affects his metabolism. Excess NADH inhibits gluconeogenesis, inhibits fatty acid oxidation, and favors the formation of lactate over pyruvate. This results in an increasing ratio of lactate/pyruvate. An increase in ketone body production also results because the high NADH/NAD+ ratio favors oxaloacetate conversion to malate. This reduces the oxaloacetate available to condense with acetyl CoA for the citric acid cycle. Therefore, the excess acetyl CoA is directed to ketone body production. References: 1. Lieber CS. Metabolism of alcohol. Clin Liver Dis. 2005 Feb;9(1):1-35. 2. Smith, C., Marks, A.D., Lieberman, M. 2005. Marks' Basic Medical Biochemistry; A Clinical Approach, 2nd Edition, Lippincott Williams, & Wilkins, Philadelphia, p458-469. 3. Hemat R. Principles Of Orthomolecularism. R.A.S. Hemat. 2004;118. 4. Frost G. Nutritional Management of Diabetes Mellitus. John Wiley and Sons. 2003;206.

14. Question There are many different types of bonding involved in the maintenance of protein conformation. Which of the following interactions are important for the stabilization of protein both secondary and tertiary structures? Answer Choices A Hydrophobic interactions B Ionic interactions C Peptide and disulphide bonds (covalent bonds) D Hydrogen bonds between side chains of amino acids E Hydrogen bonds between peptide groups

Correct Answer: Hydrogen bonds between peptide groups Explanation The term "secondary structure" reflects the regular folding of the polypetide chain. Particular examples are the α-helix and the β-plated sheet. This type of structure is stabilized by hydrogen bonding between petide groups. Although hydrogen bonds are individually weak, the cumulative effect of many bonds yields the stable polypeptide "backbone." Additional interactions between amino acid side chains can stabilize or destabilize this structure. The tertiary structure is the overall folding of the polypeptide chain. It is stabilized by hydrophobic and ionic interactions, disulphide bonds (or other covalent links in few proteins), and hydrogen bonding. The last type of interactions involve the peptide bonds and amino acid side chains. Thus, the hydrogen bonds between peptide groups participate in stabilization of protein both secondary and tertiary structures.

13. Question All proteins are composed of the same 20 amino acids arranged in different, but specific sequences. The side chains of these amino acids are responsible for the various properties of the proteins. The carboxyl group on a side chain of an amino acid in a polypeptide will be predominantly in the unprotonated form when it is Answer Choices A In a solution at pH 1 B Associated with other proteins C In a solution above pH 7 D Next to a proline residue E In a solution of 6N HCl

Correct Answer: In a solution above pH 7 Explanation The ionizable groups in a polypeptide or protein consist of the α-amino group (pKα = 7.6), the carboxyl group on the carboxy-terminal residue of the protein (pKα = 3.0), and the side chains of amino acids with ionizable R groups. These are lysine (pKα = 10), glutamic and aspartic acid (pKα = 4.6), histidine (pKα = 6-7), and arginine (pKα = 11.5-12.5). The exact pKα will depend on the environment of the side chain of the individual amino acid in the protein. The pKα is equal to the acid dissociation constant, therefore, when the pH is below the pKα the acid is protonated. The α-amino group (R-NH3+) will be protonated below pH = 7 since its pKa = 7.6. In solution above pH = 7, the carboxyl group of glutamic and aspartic acid will be in the basic, unprotonated (R-COO-) form. The Zwitterion form is the ionic form in which the number of positive charges equals the number of negative charges. Therefore, the net charge is zero. The pH at which an electrically neutral (zwitterionic form) form exists is known as the isoelectric point (pI). For the amino acid leucine, the pI = 6.

88. Question The major effect of an enzyme on a reaction is to Answer Choices A Alter the equilibria of the reaction B Allow it to proceed intracellularly C Increase the reaction rate D Prevent the substrate from achieving the transition state E Alter the products of the reaction

Correct Answer: Increase the reaction rate Explanation Enzymes are catalysts and act by increasing the reaction rate of a reaction by lowering its activation energy. The activation energy (ΔG‡) is defined as the energy required to convert all molecules from the ground state to the transition state. The equilibrium of the reaction is not affected by the enzyme. The energy required to decrease the activation energy is derived from weak, noncovalent interactions between enzyme and substrate. Some typical enzyme rate enhancements over uncatalyzed reactions range from 107 (carbonic anhydrase) to 1014 (urease). Enzymes exhibit high specificity due to several features of active sites. The active site of the enzyme occupies only a small part of the enzyme structure, and has a 3-dimensional configuration. It is structurally composed of clefts or crevices. Multiple weak interactions bind the substrate to the enzyme active site. The specificity depends on precise arrangements within the active site. The reaction transition state occurs when the optimal interactions between substrate and enzyme exist, and this only occurs in the transition state. To catalyze reactions, an enzyme must be complementary to transition state of the substrate. At this stage, the full complement of weak interactions occur. The best inhibitors of enzyme action are often transition state analogues.

35. Question A patient comes to your office and is diagnosed with Type I Glycogen Storage Disease. This disease is characterized by a lack of the enzyme glucose 6-phosphatase in the liver and kidney. This patient would be expected to have glycogen in the liver and kidney with what characteristics? Answer Choices A Increased in amount but of normal structure B Normal in amount but with increased branching C Increased in amount with increased branching D Normal in amount but with very short outer branches E Normal in amount and structure

Correct Answer: Increased in amount but of normal structure Explanation Glycogen is composed of glucose residues in α[1→4] glycosidic linkages. Branch points, composed of α[1→6] glycosidic linkages, occur approximately every 10 residues. This gives glycogen a highly branched structure. This increases its solubility greatly compared to a macromolecule containing only α[1→4] glycosidic linkages such as amylose, the unbranched form of starch found in plants, which is less soluble. In addition, the many "ends" that occur in glycogen can all be used to add or remove glucose residues as needed in a rapid manner. This allows glucose to be readily mobilized when needed for energy and also for excess glucose to be removed from circulation and stored as glycogen. All glycogen in the body is found in this highly branched structure. The enzyme glucose 6-phosphatase catalyzes the removal of phosphate from glucose 6-phosphate, resulting in the generation of free glucose. Glucose 6-phosphate cannot cross the cell membrane, and this enzyme is necessary to generate free glucose to be exported into the circulation. Without glucose 6-phosphatase, the glucose made as the result of gluconeogenesis cannot leave the cell. This lack of glucose 6-phosphatase is the cause of Type I Glycogen Storage Disease (von Gierke's disease) that is characterized by abnormally high amounts of glycogen in the liver and low blood glucose levels. This disease also affects the kidney. The structure of glycogen is not affected. Because of the large amounts of glucose 6-phosphate in the liver cells, there is an increase in the rate of glycolysis that leads to increased concentrations of lactate and pyruvate in the blood. The clinical features of Type I Glycogen Storage Disease are increased size of the liver, failure to thrive, severe hypoglycemia, ketosis, hyperuricemia, and hyperlipemia. Patients can now live longer lives with this disease. Several hours after eating a meal, blood glucose levels begin to fall. Under these conditions (overnight fast) there is a rapid mobilization of glycogen from the liver and muscle. Almost all of the liver glycogen and most of the muscle glycogen are used up. The muscle and adipose tissue also use less glucose under these conditions, allowing the level of glucose in the blood to be maintained. This occurs because of the lowered insulin levels present. Muscle and liver use fatty acids as fuel when glucose levels are low. The degradation of protein to be used in gluconeogenesis does not occur until prolonged starvation occurs.

120. Question How do aspirin and ibuprofen act as anti-inflammatory drugs? Answer Choices A Inhibit the synthesis of prostaglandins and thromboxanes B Inhibit the synthesis of leukotrienes C Are histamine receptor antagonists D Block protein biosynthesis E Inhibit the synthesis of serotonin (5-hydroxytryptamine)

Correct Answer: Inhibit the synthesis of prostaglandins and thromboxanes Explanation Aspirin and ibuprofen both inhibit prostaglandin endoperoxide synthase, an enzyme necessary for the synthesis of prostaglandins and thromboxanes. (See Image) Aspirin acetylates a Ser residue, which is critical for the catalytic action of the enzyme. Ibuprofen does not bind with the enzyme irreversibly; it is probably a structural analog of some intermediate in the reaction, catalyzed by the enzyme.

1. Which of the following statements about Lingual lipase is true? A Is active only at neutral and alkaline pH B Is responsible for the digestion of up to 30% of dietary triacylglycerol C Is secreted into gastric juice D Is essential for the lipid metabolism in adults E Specifically digests sphingolipids

Correct Answer: Is responsible for the digestion of up to 30% of dietary triacylglycerol Explanation Lingual lipase, an enzyme secreted by the glands at the back of the tongue, degrades triacylglycerol molecules. The enzyme is acid-stable and is therefore active in the adult stomach, where the pH is low. Although the rate of lipid hydrolysis by lingual lipase is slow (because in the stomach the lipid is yet not emulsified), the long retention time of up to 4 hours makes the enzyme responsible for the digestion of up to 30% of dietary triacylglycerol. Lingual lipase does not play a role in the digestion of sphingolipids.

37. Question Gene expression involves transcription and translation. Transcription Answer Choices A Is the assembly of polypeptide chains from mRNA transcript B Occurs at the ribosomes C Is the synthesis of single stranded RNA mediated by RNA polymerase using one strand of DNA as template D Occurs in 3'OH to 5'PO4 direction E Is the transfer of bacterial genes by a bacteriophage from one cell to another

Correct Answer: Is the synthesis of single stranded RNA mediated by RNA polymerase using one strand of DNA as template Explanation Transcription is the synthesis of single stranded RNA using one strand of DNA as template and is mediated by RNA polymerase and it occurs in template 5'PO4 to 3' OH direction. Translation is synthesis of specific protein from the mRNA code. The transfer of bacterial genes by a bacteriophage from one cell to another is known as transduction.

81. Question A 22-year-old man has a 1-week history of severe diarrhea. Now he complains that he feels dizzy when he stands up. The attending doctor examines him to find that his blood pressure (supine) is 114/76 mmHg with a pulse of 90/min and blood pressure (on standing) is 80/62 mmHg with a pulse of 118/min. He immediately starts an intravenous line. What is the most likely option to administer to the patient by this route? Answer Choices A Desmopressin B 5% dextrose in water C Fresh frozen plasma D Isotonic saline E Verapamil

Correct Answer: Isotonic saline Explanation The patient is dehydrated due to loss of body fluids. Dizziness and hypotension on standing confirm volume depletion. This is because on standing up the effect of gravity further reduces the return of blood to the right side of his heart, producing postural hypotension and dizziness. The increased pulse rate is a result of baroreceptor reflex. Adult diarrhea is isotonic, so there is no electrolyte imbalance between the extracellular and intracellular fluids and hence no net movement of fluid between the 2 compartments. Fluids can increase blood pressure in a volume-depleted patient only if they remain in the extracellular (ECF) compartment and have a tonicity similar to that of plasma. The least expensive method is the intravenous infusion of crystalloid solutions (0.9% normal saline or Ringer's lactate). Note: Since plasma volume accounts for 1/3 of the ECF (interstitial fluid accounts for the remaining two-thirds), it requires infusion of 3 liters of normal saline to add 1 liter to the plasma. Desmopressin, being a vasoconstrictor, can raise blood pressure; however, it is not a drug of first choice for treatment of volume depletion. 5% dextrose in water will not raise the blood pressure much. Of the total body water, the ICF compartment has 2/3 and the ECF compartment 1/3, so most of the water infused will shift into the ICF compartment. The ECF volume will not rise much. Theoretically, fresh frozen plasma meets all the requirements for an ideal fluid in such a case, but besides being very expensive, it has the potential for transmitting infectious diseases (e.g. hepatitis C), so it is rarely used. Verapamil is a calcium channel blocker and is most often used in treating hypertension. It has absolutely no role in this patient.

27. Question Why is the drug allopurinol effective in the treatment of gout? Answer Choices A It binds uric acid making it more soluble B It inhibits the conversion of purines into uric acid C It stimulates the secretion of uric acid D It regulates the enzyme hypoxanthine-guanine phosphoribosyl transferase E It binds phosphoribosyl pyrophosphate

Correct Answer: It inhibits the conversion of purines into uric acid Explanation Gout, Lesch-Nyhan Syndrome, and combined immunodeficiency syndrome are metabolic disorders of purine catabolism. Cystic fibrosis involves the defective regulation of a chloride channel, called the cystic fibrosis transmembrane regulator, found in the plasma membranes of epithelial cells. The enzyme phosphoribosyl pyrophosphate (PRPP) synthetase is defective in that it either has increased activity or is resistant to feedback inhibition. This is an X-linked recessive trait and leads to purine overproduction and overexcretion. A partial deficiency in the enzyme hypoxanthine-guanine phosphoribosyltransferase (HGPRTase) also leads to purine overproduction and overexcretion. However, a complete absence of this enzyme leads to the much more devastating disease, Lesch-Nyhan Syndrome. A salvage pathway exists for purine and pyrimidines, which result from the degradation of nucleic acids and nucleotides. In addition, purines are degraded to uric acid and pyrimidines mostly to urea. The salvage pathway for purines involves: Adenine + PRPP → AMP + PPi catalyzed by the enzyme adenine phosphoribosyltransferase. Hypoxanthine + PRPP → IMP + PPi and Guanine + PRPP → GMP + PPi Both of these reactions are catalyzed by the enzyme hypoxanthine-guanine phosphoribosyltransferase (HGPRT) PRPP=phosphoribosyl pyrophosphate. Urate is the final product of purine degradation and is excreted in the urine in humans. This reaction scheme is: AMP v IMP v Hypoxanthine v Xanthine v Urate <=> Guanine Allopurinol is an inhibitor of the enzyme xanthine oxidase, and it inhibits the conversion of hypoxanthine to xanthine and of xanthine to urate. As shown in the attached images, the structures of hypoxanthine and allopurinol are very similar and allopurinol was designed to be a competitive inhibitor of xanthine oxidase. Allopurinol is also metabolized by xanthine oxidase to oxypurinol, which also inhibits xanthine oxidase. Gout is caused by an elevated level of uric acid (urate) in the blood and tissues. The action of allopurinol causes the levels of xanthine and hypoxanthine in the blood to rise. These are more soluble than urate and are less likely to deposit as crystals in the joints. The joints become inflamed and painful due to the deposition of sodium urate crystals. Uric acid is also deposited in the kidney tubules. There is also an increase in PRPP in Lesch-Nyhan Syndrome, an increased rate of purine synthesis by the de novo pathway, as well as an overproduction of urate. The physiological effects of this disease are neurological defects, mental retardation, and self-mutilation. In Lesch-Nyhan Syndrome, allopurinol can lower urate levels, but it does not reverse the severe neurological damage. Combined immunodeficiency diseases have been described which involve adenosine deaminase and purine nucleoside phosphorylase. These 2 enzymes are involved in the degradation of purine nucleotides. 1 immunodeficiency syndrome is caused by an autosomal recessive mutation leading to a lack of the enzyme adenosine deaminase. This leads to a deficiency in the T and B cells. •

4. Question Which of the following statements about the compound fructose 2,6-bisphosphate is correct? Answer Choices A It is formed from glucose 2,6-bisphosphate B It is part of the glycolysis pathway C It is formed from fructose 1,6-bisphosphate D It is a regulator of the glycolysis pathway E It is split by the enzyme aldolase to form dihydroxyacetone phosphate and glyceraldehyde 3-phosphate

Correct Answer: It is a regulator of the glycolysis pathway Explanation The enzyme phosphofructokinase catalyzes the reaction: Fructose 6-phosphate + ATP ==> Fructose 1,6-bisphosphate + ADP + H+ as part of the glycolysis pathway. The enzyme is allosterically activated by fructose 6-phosphate, fructose 2,6 bisphosphate, and AMP. The enzyme is allosterically inhibited by citrate. The enzyme can also be phosphorylated, which decreases its affinity for fructose 2,6 bisphosphate. Fructose 2,6-bisphosphate is not part of the reaction pathway, but is a side reaction formed by a different enzyme phosphofructokinase 2 (PFK2). This enzyme is also referred to as the "tandem" or "bifunctional" enzyme because it contains 2 activities on the same polypeptide chain; PFK2 and a phosphatase activity that removes a phosphate generating fructose 6-phosphate. ATP interferes with the AMP activation of phosphofructokinase and therefore, inhibits the enzyme activity.

14. Question A patient has presented to the emergency clinic with a blood glucose level of 40 mg/dL. The physician on duty asked the intern to describe the storage polysaccharide in the liver that elevates the blood glucose level on breakdown. Which of the following best explains the structure of this storage polysaccharide in the human liver? Answer Choices A It is made up of 2 components, unbranched amylose and branched amylopectin B It is a branched polymer of glucose, which is linked by á1>4 glycosidic bonds at the branch point and á1>6 bonds in the main chain C It is a homopolysaccharide made up of glucose linked by â1>4 glycosidic bonds D It is made up of glucose units linked by á1>4 glycosidicbonds in the main chain and á1>6 bonds at the branch point E It is made up of glucose units linked by á1>4 glycosidic bonds in the main chain and â1>4 bonds at the branch point

Correct Answer: It is made up of glucose units linked by á1>4 glycosidicbonds in the main chain and á1>6 bonds at the branch point Explanation Glycogen is a branched storage homopolysaccharide present in human liver and muscle. It is made up of glucose units. The glucose units are linked by 1>4 glycosidic bonds in the main chain and 1>6 bonds at the branch point. It has many non-reducing ends and a single reducing end. Branching occurs at each 8-10 glucose residues. Refer to the image. Starch is a storage homopolysaccharide in plants. It is made up of 2 components: amylose and amylopectin. Amylose is not branched, and it exists in a helical conformation. It has glucose units linked by 1>4 glycosidic bonds. Amylopectin is the branched form. It has glucose units linked by 1>4 bonds in the straight chain and 1>6 bonds at the branch point. Cellulose, a structural homopolysaccharide in plants, is made up of glucose units linked by a 1>4 glycosidic bonds.

20. Question During the citric acid cycle, fumarate hydratase (fumarase) catalyses the reaction of fumarate with water. This process yields Answer Choices A Malate B Maleate C Malonate D Maltose E Mevalonate

Correct Answer: Malate Explanation L-malate is a product of reversible hydration of fumarate by fumarate hydratase (fumarase) in the citric acid cycle. (Figure 1) Maleate does not participate in the citric acid cycle. Its trans isomer, fumarate, is a regular substrate of fumarase, an enzyme of the citric acid cycle. (Figure 2) Malonate, -OOC-CH2-COO-, an analog of succinate, -OOC-CH2-CH2-COO-, is a strong competitive inhibitor of succinate dehydrogenase, an enzyme of the citric acid cycle. Therefore, it blocks the citric acid cycles at the step of oxidation of succinate to fumarate. Maltose is a carbohydrate, a disaccharide formed from two molecules of D-glucose. (Figure 3) The synthesis of mevalonate (Figure 4), is the first stage in cholesterol biosynthesis from acetate. (Figure 5) •

72. Question The kidneys contain approximately 2 million functional units, which are called nephrons. Which of the following anatomical structures performs the function of the control of blood pressure via Renin Angiotensin system? Answer Choices A Glomerulus. B Proximal tubules. C Collecting tubules and ducts. D Ascending part of the loop of Henle. E Juxtaglomerular apparatus.

Correct Answer: Juxtaglomerular apparatus. Explanation Each nephron is composed out of a glomerulus, which is surrounded by a capsule called Bowman's capsule. Bowman's capsule empties into a series of tubules leading to the pelvis of the kidney. Fluid filtration takes place in the glomerulus at a rate of about 125ml per minute for both kidneys together or roughly 180 liters per day. The glomerular filtrate has about the same composition as plasma, except that it contains almost no protein. The proximal tubules reabsorb glucose, amino acids, acetoacetic acid, protein, and about 65% of the sodium, potassium, and other electrolytes. They are relatively impermeable to waste products. Of the filtered 125ml of fluid only 1ml passes into the urine, 124ml are reabsorbed; 65% are absorbed in the proximal tubules, 15% in the loops of Henle, 10% in the distal tubules, and a little more than 9% in the collecting tubules and ducts. In the ascending part of the loop the Henle active transport of NaCl takes place, while the cells are only very little permeable for water or diffusion of NaCl, hence fluid leaving Henle's loop is hypotone. Several mechanisms regulate fluid osmolality, sodium and potassium concentration of plasma and urine. ADH increases the permeability of the collecting ducts for osmotic reabsorption of water. Aldosterone increases the secretion of potassium and the re-absorption of sodium in the renal tubules, primarily in the collecting tubules and ducts. The juxtaglomerular apparatus contains the cells for renin secretion. Renin reacts with plasma angiotensinogen to form angiotensin I that, in turn, is converted to angiotensinogen II by a lung enzyme called angiotensin converting enzyme. Angiotensin II causes vasoconstriction.

50. Question A 65-year-old female presents to your service for a follow up on her insulin resistant diabetes mellitus. She is roughly 150 pounds overweight. She insists that she exercises regularly and eats only the prescribed diet six times daily. Her fasting blood glucose values run around 260 mg/dL. Her lab work indicates that she is losing renal function. You discuss the implications with her. She knows nothing about the kidneys so you start with the basics. Which of the following statements regarding nephrons is true? Answer Choices A Eighty-five percent of all nephrons in the kidneys are juxtamedullary nephrons B Only the cortical nephrons can form concentrated urine C Juxtamedullary nephrons have their loops of Henle in the medulla of the kidneys D Nephrons have an afferent arteriole which supplies them with blood, and an efferent vein which drains them E The vasa recta arises from the afferent arteriole and covers the loops of Henle in cortical nephrons only

Correct Answer: Juxtamedullary nephrons have their loops of Henle in the medulla of the kidneys Explanation Eighty-five percent of all nephrons are cortical nephrons. Reduction of volume of filtrate is accomplished by the loops of Henle of the juxtamedullary nephrons, which lie in the medulla of the kidney. The collecting tubules of both the cortical and juxtamedullary nephrons descend through the medullary gradient, concentrating the urine. Reduction of volume of filtrate is accomplished by the loops of Henle of the juxtamedullary nephrons, which lie in the medulla of the kidney. The afferent arteriole supplies the glomerulus and the efferent arteriole drains the glomerulus. The vas recta arise from the efferent arteriole and only in juxtamedullary nephrons.

60. Question Extracellular and intracellular fluids differ significantly in chemical composition. Which of the following components are found at much higher concentration in the intracellular fluid than in the extracellular fluid? Answer Choices A Ca2+ B K+ C Glucose D HCO3-

Correct Answer: K+ 61. Question Which one of the following intercellular chemical messengers act by inducing synthesis of enzymes in their target cells? Answer Choices A Catecholamines B Calmodulin C Steroid hormones D Peptide hormones ^Correct Answer: Steroid hormones Explanation Unlike any other chemical intercellular messenger, steroid hormones act by inducing the synthesis of enzymes in their target cells. The steroids enter cells and bind to specific receptor proteins in the cytoplasm. The binding changes the conformation of the receptor. The receptor protein-steroid complex then enters the nucleus, where it binds reversibly to DNA. This binding generally influences the gene to make more of a particular mRNA, and the result is increased formation of specific protein molecules with enzymatic activity. Catecholamines and peptide hormones act via receptors on the external surface of the membrane of their target cells to increase intracellular cyclic AMP, and cAMP mediates their physiological effects. Calmodulin is an acidic peptide with four Ca2+ binding domains. It contains 148 amino acid residues, has a molecular weight of 16,700 Æs, and is unique in that residue 115 is trimethylated lysine. It binds Ca2+, and the calmodulin-Ca2+ complex then binds to various enzymes, activating them. It is found in a wide variety of different cells in mammals and in the cells of simple invertebrates and plants.

3. Question The primary carrier (carriers) of cholesterol from the liver to all tissues is (are) Answer Choices A Chylomicrons B Albumin C HDLs D LDLs E Pancreatic lipase

Correct Answer: LDLs

115. Question The primary carrier (carriers) of cholesterol from the liver to all tissues is (are) Answer Choices A Chylomicrons B Albumin C HDLs D LDLs E Pancreatic lipase

Correct Answer: LDLs Explanation Important Properties of Lipoproteins: Chylomicrons- Transport of dietary cholesterol and triacyglycerols. 1-2% Protein. Apo-B-48, also Apo-A-1, Apo-A-II, -IV while in the lymph, Apo-C-II and Apo-E acquired from HDL in the blood. Assembled in the intestine. Apo-C-II activates endothelial lipoprotein lipase, which removes free fatty acides that become absorbed by tissues, Apo-B-48 comibes only with chylomicrons Chylomicron- Cholesterol, Apo-B-48, Apo-E. Appear as a result of the loss of Apo-C-II, removed by the liver through Apo-E. VLDL (very low density lipoproteins)- Transport of endogenous triacylglycerols from the liver. 7-10% Protein. Apo-B-100, Apo-E, Apo-C-I,-II,-III IDL (intermediate density lipoproteins)- Less triacyglycerols than VLDL 10-12% Protein. Like VLDL. Product of triacylglycerol removal from VLDL. Futher removal leads to LDL LDL (low density)- Primary carrier of cholesterol in the plasma. Delivers cholesterol from the liver to all tissues. 20% Protein. Apo-B-100 exclusively. LDL receptors on the cell membranes react with Apo-B-100 and thus mediate LDL HDL (high density lipoproteins)- Accumulates cholesterol esters. 55-66% Protein. Apo-A-I, -C-I, -C-II, - E. LDL receptors on the cell membranes react with Apo-B-100 and thus mediate LDL Albumin- Free fatty acids. Transports free fatty acids from adipose tissue.

28. Question A 14-year-old girl complains that she often feels light-headed and dizzy just before lunch at school. She tells you that she often does not eat breakfast. She almost always feels better after lunch. You explain to her that she is experiencing hypoglycemia. You tell her that the body can make glucose from non-carbohydrate precursors by the gluconeogenesis pathway. The enzyme glucose 6-phosphatase is part of this pathway and is: Answer Choices A Necessary for glucose to leave the cell B Found only in the brain C Acts as a key enzyme in glycolysis D Absent from the liver E Required for the phosphorylation of glucose

Correct Answer: Necessary for glucose to leave the cell Explanation Gluconeogenesis is the synthesis of glucose from non-carbohydrate precursors. It converts pyruvate to glucose. The gluconeogenesis pathway uses several of the same enzymes as glycolysis but is not a direct reversal of the glycolysis pathway. Glycolysis has a ΔG° of -20 kcal/mol. A direct reversal would have a ΔG° of +20 kcal/mol, which would be highly unfavorable. Therefore, the gluconeogenesis pathway utilizes "bypass reactions," which get around the glycolysis steps that when reversed would have a positive ΔG°. In addition, compartmentalization of part of the gluconeogenesis pathway serves to help regulate the pathway. While all of the reactions of glycolysis take place in the cytoplasm, the reaction pyruvate + CO2 + ATP → oxaloacetate + ADP + Pi takes place in the mitochondria. Oxaloacetate leaves the mitochondria as malate and is then re-oxidized to oxaloacetate in the cytoplasm. The remainder of the reactions, including the other "bypass reactions," occur in the cytoplasm. The enzyme glucose 6-phosphatase catalyzes the following reaction: glucose 6-phosphate + H2O → glucose + PI This is an important enzyme found in high levels in the liver, but it is not present in the brain. Glucose 6-phosphatase functions to produce free glucose as the last step in the gluconeogenesis and glycogen degradation pathway. The cell membrane is not permeable to glucose 6-phosphate. Therefore, without this enzyme, the other tissues of the body could not utilize glucose produced by the liver. Free glucose can diffuse out of the liver into the bloodstream to be carried to the other tissues of the body. The glucose 6-phosphatase reaction is essentially irreversible under cellular conditions. Therefore, this reaction cannot be used to form glucose 6-phosphate. The enzyme hexokinase is a key enzyme of glycolysis and is responsible for the formation of glucose 6-phosphate in an ATP requiring reaction. The enzyme that interconverts glucose 6-phosphate and glucose 1-phosphate is phosphoglucomutase. The hypoglycemia response involves an increase in glucagon, adrenal glucocorticoids, and epinephrine and a decrease in glucose uptake by cells. These conditions lead to an increase in lipolysis, gluconeogenesis, and protein degradation.

95. Question A 45-year-old man presents with nausea and rapid breathing and has a sweet-smelling odor on his breath. Laboratory tests indicate a blood glucose level of 260 mg/dl and a pH of 7.25. His serum and urine are positive for ketones, and the levels of β-hydroxybutyrate in serum are high. He appears to be in diabetic ketoacidosis. The high level of β-hydroxybutyrate results from Answer Choices A Lack of oxaloacetate to allow acetyl CoA to enter the citric acid cycle B Inability of the mitochondria to oxidize fatty acids as an energy source C High circulating levels of insulin D Inability of his liver to use β-hydroxybutyrate as an energy source E Inability to convert β-hydroxybutyrate to acetone

Correct Answer: Lack of oxaloacetate to allow acetyl CoA to enter the citric acid cycle Explanation This patient has elevated of levels of β-hydroxybutyrate, a ketone body, because of his body's lack of oxaloacetate, which is necessary for acetyl CoA to enter the citric acid cycle. In a patient with untreated diabetes mellitus, glucose does not enter the cells to be metabolized to oxaloacetate. At the same time, fatty acids are being oxidized in the mitochondria to produce acetyl CoA. Normally, there is sufficient oxaloacetate in the cell to condense with the acetyl CoA to form citrate as part of the citric acid cycle. The excess acetyl CoA is converted to the ketone bodies acetoacetate, β-hydroxybutyrate, and acetone. β-Hydroxybutyrate is the predominate form. β-Hydroxybutyrate and acetoacetate are secreted into the bloodstream and can be used by various tissues to derive energy. Once taken up by these tissues, the ketone bodies are converted back to acetyl CoA through a series of reactions involving conversion of β-hydroxybutyrate to acetoacetate, then to acetoacetyl CoA, and finally 2 molecules of acetyl CoA. The acetyl CoA can then be used by the citric acid cycle to generate ATP. The liver lacks the enzyme acetoacetate: succinyl-CoA transferase and therefore cannot use ketone bodies as an energy source under any condition. References: 1. Devlin, T.M. Textbook of Biochemistry with Clinical Correlations, 6th Edition, Wiley-Liss, New Jersey, 2006, p.107; p687-689. 2. Insel P. Turner E. Ross D. Discovering Nutrition. Jones & Bartlett Publishers. 2005:262 3. Smith C. Marks A. Lieberman M. Marks' Basic Medical Biochemistry: A Clinical Approach. Lippincott Williams& Wilkins. 2005:435

5. Question A patient comes to your office complaining of gastrointestinal discomfort. She tells you that it is worst at mid-morning and that she has cereal with milk and a large glass of milk for breakfast each day. This patient likely has a defect in which of the following enzymes? Answer Choices A Lactase B Sucrase C Glucokinase D α-amylase E Phosphoglucomutase

Correct Answer: Lactase Explanation The surface of the small intestine has enzymes, intestinal disaccharidases, which hydrolyze disaccharides to monosaccharides that can then be absorbed. Lactase is one of these enzymes and hydrolyzes the disaccharide lactose to glucose and galactose. Lactose is the sugar found in milk products. In the absence of the enzyme lactase, lactose can not be hydrolyzed and absorbed in the upper small intestine. The bacteria in the lower small intestine ferment lactose resulting in gas production. In addition, water is drawn into the intestinal lumen because of the osmotically active solutes. This leads to diarrhea. Commercial sources of lactase are now available which can be added to milk to pre-digest the lactose. During the process of making yogurt, the lactose present is partially hydrolyzed and therefore, patients lacking lactase often do not have problems eating yogurt.

23. Question This enzyme catalyses the final glycolytic step in erythrocytes Answer Choices A Enolase B Glucokinase C Glyceraldehyde-3-phosphate dehydrogenase D Hexokinase E Lactate dehydrogenase F Pyruvate dehydrogenase

Correct Answer: Lactate dehydrogenase Explanation Some toxic compounds are able to inhibit glycolysis. Iodoacetate inhibits glyceraldehyde-3-phosphate dehydrogenase, and fluoride inhibits enolase. In the attached figure, cofactors are in BLUE. The enzymes involved in glycolysis are as follows: glucokinase [1], hexokinase [2], phosphohexose isomerase [3], phosphofructokinase-1 [4], aldolase [5], phosphotriose isomerase [6], glyceraldehyde-3-phosphate dehydrogenase [7], phosphoglycerate kinase [8], phosphoglycerate mutase [9], enolase [10], pyruvate kinase [11], pyruvate dehydrogenase complex [12], and lactate dehydrogenase [13].

8. Question An 8-year-old boy presents with acute abdominal pain. On examination, there is mild hepatosplenomegaly. Biochemical investigations reveal elevated serum amylase and lipase levels. The plasma sample of the patient shows a thick creamy layer when kept overnight at 4 C. Fasting serum triglycerides level are 2200mg/dl. There have been similar complaints of abdominal pain in the past 4-5 years. The Apo CII levels were normal. What is the most likely diagnosis? Answer Choices A Abetalipoproteinemia B Apo CII deficiency C Familial combined hyperlipidemia D Familial hypercholesterolemia E Familial type III hyperlipoproteinemia F Lipoprotein lipase deficiency

Correct Answer: Lipoprotein lipase deficiency Explanation Lipoprotein lipase deficiency is characterized by a defect in the degradation of chylomicrons. Lipoprotein lipase is required for the catabolism of triacylglycerol present in chylomicron to free fatty acids and glycerol. Lipoprotein lipase is present in the endothelium of blood vessels of the adipose tissue and muscle. A high serum triglyceride level characterizes this disorder. The serum triglyceride level is usually >1500mg/dl. The serum cholesterol is marginally raised. Lipoprotein lipase may be released from the tissues by administration of intravenous heparin. Diagnosis is confirmed by measuring lipoprotein lipase using specific immunologic techniques. Serum triglycerides levels are markedly elevated (>1500mg/dl). The most frequent presentation is as recurrent abdominal pain accompanied by hepatosplenomegaly. In many cases, patients present with acute pancreatitis. Observation of plasma, after keeping it overnight at 4 C, shows the presence of a creamy layer on the plasma. Management involves the administration of medium-chain triglycerides in the diet that enter the portal circulation directly without formation of chylomicrons. Familial hypercholesterolemia (FH) is an autosomal dominant genetic defect in which there is deficiency of functional LDL receptors. These LDL receptors recognize the Apo B100 present in IDL and LDL. LDL/IDL binds to the receptor and is taken up into the cell by receptor-mediated endocytosis. In FH, there is decreased peripheral uptake of LDL and IDL, which causes a marked increase in the serum cholesterol level. Serum triglycerides are within normal limits. There is a large accumulation of cholesterol in the macrophages of the tendons that causes xanthomas, indicating that the scavenger cells are taking up and degrading large amounts of LDL. Patients manifest with an increased risk of atherosclerosis and coronary artery disease at an early age. There is cholesterol accumulation in the arteries, especially at the root of the aorta, causing aortic stenosis. Apo C II is the apolipoprotein present in chylomicrons that is required for the activation of lipoprotein lipase. Absence of Apo CII is inherited as a recessive disorder. It is characterized by high levels of chylomicrons and serum triglycerides. It is distinguished from lipoprotein lipase deficiency by the assay of lipoprotein lipase after the administration of intravenous heparin. In this disorder, lipoprotein lipase levels are normal; however, Apo CII levels are low. Symptoms of presentation and management of the condition are similar to lipoprotein lipase deficiency. Familial combined hyperlipidemia (Frederick son's type IIB hyperlipoproteinemia) is characterized by an increase in LDL and VLDL. In these patients, there is an increased fasting serum cholesterol and serum triglycerides level. There is an increased risk of coronary artery disease in these patients. Serum cholesterol level is usually in the range of 200-300 mg/dl and serum triglyceride is in the range of 200-500mg/dl. Abetalipoproteinemia is characterized by an absence of lipoprotein species containing apoprotein B, which are LDL, chylomicrons, and VLDL. The levels of triglycerides and total cholesterol are very low (<50mg/dl). The predominant features include malabsorption of dietary fat and steatorrhea. There is also malabsorption of the fat-soluble vitamins. Lack of vitamin E causes progressive degeneration of the CNS, which manifests as neurological abnormalities by the first decade. There is decreased visual acuity and night blindness as a result of vitamin A deficiency. Familial type III hyperlipoproteinemia (dysbetalipoproteinemia) is characterized by increased levels of IDL in circulation. There is also an increase in the chylomicron remnant particles. There is a defect in apolipoprotein E, which is required for the uptake of chylomicron remnants and IDL by the liver. In these individuals, the fasting serum cholesterol and triglycerides are raised. Patients present with an increased risk of premature atherosclerosis and palmar xanthomas.

93. Question A 6-month-old boy with developmental delay is found to have hepatosplenomegaly. Analysis of tissue obtained on biopsy of the liver shows unusually large amounts of glucocerebroside. Which of the following enzymes is most likely to be deficient? Answer Choices A Glucose-6-phosphatase B Lipoprotein lipase C Liver phosphorylase D Lysosomal hydrolase E Sphingolipid synthase F Tissue phospholipase

Correct Answer: Lysosomal hydrolase Explanation This child has Gaucher's disease. It is one of the most prevalent lysosomal storage diseases and is due to a deficiency of a lysosomal hydrolase, more specifically a glucocerebroside - beta-glucosidase. This results in accumulation of glucocerebroside in live bone marrow and spleen. Gaucher's cells are seen in the brain. A presentation, as in this case, is seen with acute Gaucher's disease. The more common type presents in adulthood with hepatosplenomegaly and pigmentation of exposed parts. A deficiency of glucose-6-phosphatase is seen in Von Gierke disease. It is a type I glycogen storage disease. Deficiency of liver phosphorylase is seen in another glycogen storage disease, type VI - Hers. Lipases are not necessary for degradation of glucocerebrosides. Sphingolipid synthetase would affect synthesis of sphingolipids, not their degradation.

43. Question The mucopolysaccharidoses are a group of genetically independent disorders caused by a deficiency of enzymes catalyzing the stepwise degradation of glycoaminoglycans. The clinical features, while variable, are progressive and involve multiple organs. The cellular manifestations of these different genetic disorders include many changes that are secondary to the primary cellular change, which is characterized by Answer Choices A Decreased protein synthesis B Abnormal lipid synthesis C Loss of membrane integrity D Lysosomal storage E Increased carbohydrate synthesis

Correct Answer: Lysosomal storage Explanation Lysosomal storage of undegraded substrates, primarily glycoaminoglycans, is the most characteristic feature of the mucopolysaccharidoses. The accumulated substrates are due to a deficiency in an enzyme required in the stepwise catabolism of the glycoaminoglycans. Decreased protein synthesis is not the correct choice because it is variable and is a secondary manifestation of the mucopolysaccharidosis. Lipid synthesis is not impaired in the mucopolysaccharidoses. The cellular and subcellular membranes in the mucopolysaccharidoses remain intact. There is not an increase in carbohydrate synthesis in the mucopolysaccharidoses.

46. Question Which of the following cell structures is responsible for energy release? Answer Choices A Nucleus B Ribosome C Endoplasmic reticulum D Mitochondria E Lysosome

Correct Answer: Mitochondria Explanation The cell is composed of two major microscopically visible parts: the nucleus and the cytoplasm. The cytoplasm is encircled by the cell membrane, a permeable membrane separating the cell from surrounding structures. The cell contains the nucleus, which is enclosed by the highly permeable nuclear membrane. It contains the 23 pairs of chromosomes, the genetic code that regulates cell function and reproduction. This is achieved by transcribing and translating the genetic code from DNA via RNA into protein molecules. Simplified, the ribosome 'reads' the code and causes the proper succession of amino acids to form a protein. A vast network of tubes and vesicles, found in most areas of the cytoplasm, is called endoplasmic reticulum. Its membrane functions as a manufacturing plant for multiple substances such as proteins, lipids, carbohydrates, and structures like lysosomes and secretory granules. Lysosomes are organelles found in great numbers in all cells. They are small, spherical vesicles containing digestive enzymes, which are enclosed by a membrane. They perform various digestive functions like digesting phagocytized particles, damaged portions of the cell, and in case of atrophy, portions of the cellular mass itself. Mitochondria are responsible for energy release in the cell. They are the power plants of the cell, so to speak. They contain the enzymes of the citric acid cycle and the oxidative enzyme system, which finally produces adenosine triphosphate (ATP) and other substances that provide the cell with energy.

55. Question Which of the following cell structures is responsible for energy release? Answer Choices A Nucleus B Ribosome C Endoplasmic reticulum D Mitochondria E Lysosome

28. Question Your lab is interested in inherited mitochondrial myopathies, and you have developed a murine model for the various diseases found under this rubric. One of your rotating surgical residents in the lab wants to irradiate mouse sperm to develop mutant lines of these mice. His efforts are most likely to fail because Answer Choices A Mitochondrial genes are only found in ova B Mouse sperm is radiation resistant C Mitochondrial membranes protect its genome from radiation D Mutations produced by radiation are randomly distributed among the genome E Radiation damage only occurs to nuclear genes

Correct Answer: Mitochondrial genes are only found in ova Explanation Mitochondria have their own circular genomes of approximately 16,000 to 17,000 base pairs. Although many of the proteins found in the mitochondria are encoded in the nucleus and imported from the cytoplasmic site of synthesis, some mitochondrial genes encode for proteins, several tRNA molecules, and mitochondrion-specific rRNA. In addition, mitochondria have a specific protein-synthesizing complex with RNA polymerase, tRNA-syntheses, and ribosomes. The overall scheme of protein synthesis in mitochondria is similar to that of cytoplasmic protein synthesis, but the components used in the process are slightly different in size and in number from those of the cytoplasmic situation. As an example, in the case of eukaryotic cells, the cytoplasmic ribosomal monomer size is 80S; the mitochondrial monomer size ranges between 55S-60S. The eukaryotic cytoplasmic ribosomal small subunit is 40S in size; the mitochondrial counterpart ranges between 30S-35S. Whereas the eukaryotic large ribosomal subunit is 60S, the mitochondrial large ribosomal subunit ranges in size from 40S to 45S. Because many of the characteristics of the mitochondria resemble those of prokaryotes, a prevailing theory considers that the mitochondrion developed from a symbioant prokaryote that invaded the eukaryotic cell in the very distant evolutionary past. Still unresolved, however, is the mechanism coordinating protein synthesis in the mitochondrion and cytoplasm. Mitochondria have recently come to the clinical forefront because of the delineation of mitochondrial myopathies. Typically, patients with these syndromes have complaints of weakness and muscular cramping upon minimal exertion; infants tend to exhibit poor feeding and crawling activities. Some of these mitochondrial myopathies involve defects in the proteins encoded by mitochondrial genes; as such, they are inherited exclusively from the mother, since mitochondria are only derived from the ovum.

49. Question When the left ventricular pressure is less than the left atrial pressure, and less than the aortic pressure, blood is Answer Choices A Stagnant, not moving anywhere B Pooling in the left atrium which will lead to pulmonary edema C Moving normally through the heart D In danger of clotting due to stasis

Correct Answer: Moving normally through the heart Explanation If the pressure in the left ventricle is less than the left atria, then blood is moving into the ventricle. As the pressure is also less that the aortic pressure, the blood is not moving into the aorta, yet. The length of diastole is sufficiently short to prevent any pooling leading to stasis.

105. Question An African-American man comes to your office exhibiting signs of hemolytic anemia. He says that he has been taking the drug primaquine for 10 days since his return to the United States from Zaire. You explain that he probably has a genetic mutation affecting an enzyme found in the pentose phosphate pathway. The flow of glucose 6-phosphate through the pentose phosphate pathway is regulated by the cell's need for what? Answer Choices A NADPH B NADH C Glucose D Fructose 6-phosphate E Glyceraldehyde 3-phosphate

Correct Answer: NADPH Explanation 2 molecules of NADPH are produced for each glucose molecule. Both NAD (nicotinamide adenine dinucleotide) and NADP (nicotinamide adenine dinucleotide phosphate) serve as coenzymes and are derived from the vitamin niacin. NAD is composed of adenosine and N-ribosyl-nicotinamide linked through pyrophosphate linkage between the 5' OH groups of the 2-ribosyl moieties. The only difference in the structure of NADP is the presence of a phosphate on the 2' OH of the ribose moiety attached to adenine. This additional phosphate is enough to provide specificity such that in general a reaction uses only NAD and not NADP, or NADP but not NAD. In general, NADP is used in biosynthetic reactions such as fatty acid biosynthesis while NAD is used in degradative reactions This patient has a deficiency in the first enzyme in the pentose phosphate pathway, glucose-6-phosphate dehydrogenase (G6PDH), catalyzes the following reaction: Glucose 6-phosphate + NADP+ → 6-phosphoglucono-δ-lactone + NADPH + H+ The pentose phosphate pathway is composed of an oxidative and a non-oxidative branch. The main function of the oxidative branch is to produce NADPH by the oxidation of glucose 6-phosphate, while the main function of the non-oxidative branch is to rearrange 5-carbon sugars produced in the oxidative branch. These sugars are used as precursors for the biosynthesis of DNA, RNA, or nucleotide coenzymes. 2 NADPH molecules and 1 ribose 5-phosphate are produced for each molecule of glucose 6-phosphate. The NADPH produced by this pathway can be utilized for reductive biosynthesis such as the synthesis of fatty acids. The cell's need for NADPH or ribose phosphates determines whether the oxidative or non-oxidative pathway is more active in the cell. If the cell needs NADPH, then glucose 6-phosphate is oxidized to ribulose 5-phosphate producing 2 NADPH molecules. A deficiency in G6PDH results in a decreased capacity to detoxify peroxides and leads to oxidative damage of red blood cells and hemolysis. Depending on the degree to which the level of the enzyme is reduced the hemolysis will vary. This problem can be amplified in patients with the deficiency who are taking certain drugs such as the antimalarial drug primaquine. In the absence of the drug the enzyme defect does not affect the red blood cells. This form of the disease occurs in 10-14% of African-Americans and is found on the X chromosome; therefore, it is sex-linked. Accumulation of peroxides within the red blood cell leads to damage and hemolysis of these cells. The consequence of this can be hemolytic anemia. Normally, these peroxides are reduced to water by the enzyme glutathione peroxidase. This enzyme uses glutathione and the product of the reaction, oxidized glutathione, and must be regenerated by being reduced by NADPH. The pentose phosphate pathway is the sole source of NADPH in the red blood cell. Therefore, a deficiency in G6PDH reduces the amount of NADPH available. As a consequence of this, the peroxides build up in the cell and cause lysis of the red blood cell. Hemolytic anemia exhibits symptoms similar to other types of anemia such as darkening of the urine and a marked fall in hemoglobin or erythrocyte count.

106. Question An African-American man comes to your office exhibiting signs of hemolytic anemia. He says that he has been taking the drug primaquine for 10 days since his return to the United States from Zaire. You explain that he probably has a genetic mutation affecting an enzyme required for the production of which of the following compounds? Answer Choices A NADH B FADH2 C AMP D NADPH E ATP

11. Question What part of the NAD+ molecule directly participates in oxidation-reduction reactions? Answer Choices A Adenine B Nicotinamide C Adenine-attached ribose D Nicotinamide-attached ribose E Phosphate group

Correct Answer: Nicotinamide Explanation (See Image) •

36. Question Mutations are changes that can occur in the DNA code, which changes the coded protein or prevents its synthesis. A mutation that changes a codon encoding an amino acid to a stop codon is called Answer Choices A Silent mutation B Missense mutation C Nonsense mutation D Frameshift mutation E Null mutation

Correct Answer: Nonsense mutation Explanation Silent mutation is a change at the DNA level that does not result in any change of amino acid in the encoded protein. Missense mutation results in a different amino acid being inserted in the protein. Nonsense mutation changes a codon encoding an amino acid to a stop codon. Frameshift mutation is a result of insertion or deletion in the genome, leading to disruption of the gene and production of incomplete or inactive proteins. Null mutation occurs when there is extensive insertion or deletion or gross rearrangement of the chromosome structure, which results in complete destruction of gene function.

51. Question The enzymes of the electron transport system are located Answer Choices A In the cell's cytoplasm B In the smooth endoplasmic reticulum C On the mitochondrial cristae D In the mitochondrial matrix E On the outer mitochondrial membrane

Correct Answer: On the mitochondrial cristae Explanation The enzymes of the electron transport system and an ATPase are located in elementary particles, which are racquet shaped particles attached to the mitochondrial cristae, or inner mitochondrial membrane. These particles protrude into the mitochondrial matrix, where the enzymes for the Krebs cycle are. This allows substrates of these cycles to pass easily down the enzyme chains.

42. Question When uniport transport of a molecule occurs by a carrier Answer Choices A Two molecules are moved simultaneously in the same direction B Energy is never required C Only nonionic compounds can be transported D One molecule is moved in one direction E Two molecules are moved simultaneously in opposite directions

Correct Answer: One molecule is moved in one direction Explanation Facilitated diffusion, or passive mediated transport, is characterized by the movement of molecules down a concentration gradient, the lack of a need for energy input, and the presence of specific carriers. These specific carriers exhibit saturation kinetics, show a high degree of specificity for the transported substance, and can be specifically inhibited. The chloride-bicarbonate transport system found in the erythrocyte membrane and the transport of glucose into most cells can be classified as facilitated diffusion. The chloride-bicarbonate system transport is carried out by the band 3 protein, and both ions must move simultaneously. It is a reversible system that is driven by the concentration gradient. Active transport requires energy and involves transport against a concentration gradient. ATP is often used directly as a source of energy. The energy of Na+ going down a concentration gradient can also be used as a source of energy, but this is indirectly maintained by ATP hydrolysis. Active transport is unidirectional. Uniport involves the movement of a single molecule in one direction; Symport involves movement of two molecules simultaneously in the same direction; Antiport involves the movement of two molecules simultaneously in opposite directions.

19. Question Which of the following is a mitochondrial enzyme of the urea cycle? Answer Choices A Arginase B Argininosuccinase C Argininosuccinic acid synthetase D N-acetyl-glutamate synthetase E Ornithine transcarbamoylase

Correct Answer: Ornithine transcarbamoylase Explanation Ornithine transcarbamoylase [2] converts, within liver mitochondria, ornithine into citrulline by addition of a carbamoyl radical. A deficiency of this enzyme will cause hyperammonemia in both the infant and the mother. This disorder is an X-linked known as Hyperammonemia Type 2. The other enzymes of the urea cycle are found in the cytosol of hepatocytes. They include, argininosuccinic acid synthetase [3], which deficiency causes the rare disorder citrullinemia; argininosuccinase [4], which deficiency causes the rare disorder Argininosuccinicaciduria; and arginase [5], which deficiency causes hyperargininemia. Another enzyme important to the urea cycle, although not a part of it proper, is carbamoyl phosphate synthetase [1], which is found in liver mitochondria. Its deficiency causes the very rare Hyperammonemia Type 1. •

45. Question Which of the following words is best defined as the diffusion of water molecules through semipermeable membranes? Answer Choices A Osmosis B Osmotic pressure C Capillary pressure D Diffusion E Oncotic pressure

Correct Answer: Osmosis Explanation If fluids are separated by a membrane that is semipermeable, allowing a solvent but not the solute to pass, such as a cell membrane, diffusion will lead to equal concentration of the solvent on both sides of the membrane, while the absolute amounts of the solvent on either side may differ considerably. This process is called osmosis. The pressure applied in the opposite direction of the osmotic flow, which is necessary to stop osmosis, is called the osmotic pressure. The blood pressure inside capillaries is called capillary pressure. Capillaries are porous. The pores are big enough to allow large quantities of dissolved substances to mix continuously with interstitial fluid but small enough to prevent free passage for proteins. Hence, those proteins develop a certain osmotic force, the capillary osmotic pressure, which is called colloid osmotic pressure or oncotic pressure. Diffusion is the physical process of molecules randomly moving around in any given space to reach a state of equal distribution. Diffusion, although the basis of osmosis, does not fit any of the given definitions.

47. Question Apoptosis is a term used to describe Answer Choices A Programmed cell death B Cell death by necrosis C Protease activation in the intestine D An early phase in cell division E Differentiation of a stem cell population

Correct Answer: Programmed cell death Explanation Apoptosis, or programmed cell death, plays a role in the maintenance of adult tissues and in embryonic development. Apoptosis is very different from cell death due to injury. There are specific genes responsible for the regulation and execution of this process, and these genes are conserved from nematodes to humans. The process involves DNA fragmentation and chromatin condensation. This is followed by fragmentation of the nucleus and then fragmentation of the cell. The regulators of this process include the bcl-2 family of proteins believed to be involved in regulating the ICE protease family. Bcl-2 is an integral membrane protein mostly found in the outer membrane of the mitochondria. Overexpression of bcl-2 prevents apoptosis induced by a wide variety of agents. The ICE proteases are a family of cysteine proteases that play a key role in apoptosis, indicating that induction of apoptosis involves proteolytic cleavage of one or more target proteins. The bcl-2 protein can act as an oncogene if it is overexpressed and appears to contribute to lymphomas by protecting against apoptosis rather than stimulating cell division. Abnormal expression of bcl-2 blocks apoptosis and maintains cell survival under conditions that would normally induce cell death.

25. Question A 22-year-old female has a family history of phenylketonuria but does not have the metabolic disorder herself. The female is concerned that her unborn child may have phenylketonuria and asks her physician if the child's development could be affected. The physician reassures the female that her normal metabolism would supplement the needs of the developing fetus and that phenylketonuric symptoms only develop after birth. In an infant with phenylketonuria, treatment is possible by strictly monitoring the dietary intake of which amino acid? Answer Choices A Tyrosine B Tryptophan C Phenylpyruvate D L-dopa E Phenylalanine

Correct Answer: Phenylalanine Explanation Strict monitoring of the dietary intake of phenylalanine is a successful method of treating phenylketonuria. Phenylketonuria is a recessively transmitted genetic disorder of phenylalanine metabolism. In normal metabolism, phenylalanine is converted by the enzyme phenylalanine hydroxylase to tyrosine. In phenylketonuria, phenylalanine hydroxylase is inactive. Phenylketonuria is characterized by abnormally high levels of phenylalanine in the blood and urine with normal tyrosine levels. Without dietary treatment, mental retardation may result in phenylketonuric infants. Foods such as breads, meat, poultry, fish, and nuts have high phenylalanine content and should be avoided whenever possible. Fats such as butter and mayonnaise have little phenylalanine content and may be used conservatively. Tyrosine is converted to L-dopa by the enzyme tyrosine hydroxylase. L-dopa levels are not affected in phenylketonurics. Phenylpyruvate is formed by deamination of excess phenylalanine. Phenylpyruvate is also detected in the blood and urine of phenylketonurics. Tryptophan, like phenylalanine and tyrosine, is an aromatic amino acid but is not directly related to phenylalanine metabolism.

1. Question A deficiency in dihydrobiopterin reductase causes Answer Choices A Phenylketonuria B Alcaptonuria C Maple syrup urine disease D Hunter's syndrome E Fabry's disease

Correct Answer: Phenylketonuria Explanation Phenylketonuria (PKU) is typically caused by a mutation in the gene that codes for phenylalanine hydroxylase, the enzyme that converts phenylalanine to tyrosine. In patients with PKU, tyrosine becomes essential and must be supplied in the diet. The disease is the most common inborn deficit of amino acid metabolism (prevalence in the U.S. approximately 1:14,000). The patients with untreated PKU show symptoms of mental retardation typically by the age of one year. Female patients with PKU must go on a low phenylalanine diet prior to conception in order to avoid neurologic damage to the fetus. High blood phenylalanine levels in the mother can lead to microcephaly, mental retardation, and congenital heart abnormalities in the fetus. A deficiency in dihydrobiopterin reductase causes a rare form of PKU (accounted for about 2 % of PKU). It prevents synthesis of tetrahydrobiopterin, the coenzyme of phenylalanine hydroxylase. Alcaptonuria is due to a deficiency in homogentisate oxidase, one of the enzymes of tyrosine catabolism. Maple syrup urine disease is caused by a deficiency in branched-chain α-ketoacid dehydrogenase which catalyzes the first step in the metabolism of leucine, valine, and isoleucine. Fabry's disease is due to deficits in a-galactosidase; Hunter's syndrome is associated with iduronate sulfatase deficiency which blocks the lysosomal degradation of heparan sulfate.

107. Case A 16-year-old girl has not been eating much because she has been attempting to lose weight for the past 6 months. She currently weighs 90 pounds; she is 5'2'', and she has not had her period in 6 months. You explain that, as a consequence of her condition, levels of several metabolic enzymes are affected. Question What enzyme will be induced in the liver of this patient? Answer Choices A Glucokinase B Phosphoenolpyruvate carboxykinase C Pyruvate kinase D Phosphofructokinase E Triose phosphate isomerase

Correct Answer: Phosphoenolpyruvate carboxykinase Explanation Under fasting conditions, the liver synthesizes glucose by the process of gluconeogenesis in order to provide glucose for the organs and cells, such as the brain and red blood cells, which rely exclusively on glucose for energy under normal conditions. The gluconeogenic enzyme phosphoenolpyruvate (PEP) carboxykinase can be induced under these conditions. In addition, 2 other gluconeogenic enzymes, fructose 1,6-bisphosphatase and glucose 6-phosphatase, can be induced during fasting/starvation conditions. At the same time, the liver glycolytic enzymes (glucokinase, phosphofructokinase, and pyruvate kinase) are repressed during fasting/starvation conditions in order to preserve glucose, which is released into the bloodstream from the liver. While gluconeogenesis uses several of the same enzymes as glycolysis, it also uses 4 unique enzymes to bypass the thermodynamically unfavorable reactions of glycolysis. In gluconeogenesis, pyruvate is converted to oxaloacetate by pyruvate carboxylase and then to phosphoenolpyruvate by phosphoenolpyruvate carboxykinase. The enzymes, fructose 1,6-bisphosphatase and glucose 6-phosphatase, remove a phosphate forming fructose 6-phosphate and glucose, respectively, as part of gluconeogenesis. The mechanism of the induction of these gluconeogenesis enzymes involves the increase in cAMP levels in response to glucagon, the hormone released in response to low blood glucose levels. This leads to kinase activation, and that in turn activates coactivators of transcription factors required to upregulate the transcription of phosphoenolpyruvate carboxykinase, fructose 1,6-bisphosphatase, and glucose 6-phosphatase. Triose phosphate isomerase is an enzyme used by both gluconeogenesis and glycolysis to interconvert glyceraldehyde 3-phosphate and dihydroxyacetone phosphate. References 1. Desvergne, B, Michalik, L., Wahli, W. 2006. Transcriptional regulation of metabolism. Physiol Rev 86:465-514. 2. Lieberman, M. Marks, AD. 2008. Marks' Basic Medical Biochemistry, A Clinical Approach, 3rd Ed., Wolters Kluwer, Philadelphia. p569-578. 3. Kirchner S, Panserat S, Lim PL, et al. The role of hepatic, renal and intestinal gluconeogenic enzymes in glucose homeostasis of juvenile rainbow trout. J Comp Physiol [B]. 2008 Mar;178(3):429-38. Epub 2008 Jan 8. 4. Storey KB. Functional Metabolism: Regulation and Adaptation. Wiley-IEEE. 2004:249.

30. Question The intracellular glycogen is mobilized by the enzyme yielding glucose-1-phosphate and a shorter chain of glycogen. What is the type of this reaction? Answer Choices A Hydrolysis B Phosphorolysis C Thiolysis D Proteolysis E Glycolysis

Correct Answer: Phosphorolysis Reactions of this type are widely spread in metabolism. A typical example of this type reaction is protein and carbohydrate digestion. Phosphorolysis is the cleavage of a bond by the addition of phosphate: R1-R2+H3PO4--------> R1-H + H2PO3-O-R2 A typical example of this reaction is glycogen phosphorolysis yielding glucose-1-phosphate. See Figure 1 in the attached image. Thiolysis is the cleavage of a bond by the addition of thiol: R1-R2+HS-R3--------> R1-H + R3-S-R2 At the last step of the fatty acid oxidation cycle, acetoacetyl-CoA reacts with free coenzyme A, which is a thiol. See Figure 2 in the attached image. •

41. Question Which of the following ions has a higher concentration inside the cell than in the extracellular fluid? Answer Choices A Sodium B Chloride C Calcium D Potassium E Magnesium

Correct Answer: Potassium Explanation A typical mammalian cell maintains a concentration gradient of many substances, including ions between the extracellular and intracellular fluid. The largest difference is for calcium which shows a 25,000-fold difference (2.5 mM extracellular compared to 0.1 μM). The difference for sodium is 14-fold (140 mM extracellular, 10 mM intracellular); potassium is 35-fold (4 mM extracellular, 140 mM intracellular); magnesium is 4-fold (2 mM extracellular, 0.5 mM intracellular); chloride is 25-fold (100 mM extracellular, 4 mM intracellular). These gradients are maintained across the cell membrane by preventing ion flux and by the active transport of ions from one side of the plasma membrane to the other. The specific transport of molecules across membranes is mediated by carriers. A carrier is not an enzyme and does not catalyze any chemical reactions. However, carriers have some properties in common with enzymes in that they are specific, have dissociation constants, exhibit saturation, and can be blocked by specific inhibitors. There are 2 types of transport, passive or facilitated diffusion and active transport. Active transport involves movement of molecules against a concentration gradient and therefore, requires energy usually as ATP. Examples include calcium transport, the sodium-potassium ATPase, and sodium-linked transport of glucose or amino acids. Passive transport involves movement of molecules down a concentration gradient and includes the movement of glucose in many cells.

15. Question Cellulose can not be digested by humans because it contains Answer Choices A β-(1→4) glycosidic bonds B β-(1→6) glycosidic bonds C α-(1→4) glycosidic bonds D α-(1→6) glycosidic bonds E α-(2→6) glycosidic bonds

Correct Answer: β-(1→4) glycosidic bonds Explanation Cellulose is the main structural polysaccharide of plants. It consists of β-D-glucopyranose units linked by β-(1→4) glycosidic bonds. Humans and many other mammals do not have a glycosidase that attacks the β linkage and, therefore, cannot digest cellulose.

59. Question Apoptosis is a term used to describe Answer Choices A Programmed cell death B Cell death by necrosis C Protease activation in the intestine D An early phase in cell division E Differentiation of a stem cell population

67. Question Hormones can exist in many forms before becoming mature hormones. The hormone ACTH is a derived hormone that is responsible for release of cortisol from the adrenal gland. What is the term used to describe a hormone that has a precursor form? Answer Choices A Prohormone B Preprohormone C Hormone D Prior hormone E Fake hormone

Correct Answer: Prohormone Explanation A prohormone is a hormone that has to be cleaved at certain sites in order to become functional. Some prohormones occur as preprohormones. ACTH or adrenocorticotropin hormone is a hormone that exists in a prohormone (proopiomelanocortin). The length of the prohormone is around 265 base pairs. From this peptide there are several other hormones created besides ACTH. Beta-endorphin, melanocyte stimulating hormone, met-enkephalin, and lipocortin.

79. Question All proteins are composed of the same 20 amino acids arranged in different, but specific sequences. The side chains of these amino acids are responsible for the various properties of the proteins. Which of the following amino acids would be most likely to be found in a beta turn of a protein? Answer Choices A Proline B Glycine C Serine D Tyrosine E Alanine

Correct Answer: Proline Explanation The basic structure of the 20 amino acids that make up mammalian proteins is composed of an amino group, a carboxyl group, a hydrogen atom, and a distinct side chain referred to as the "R" group, as shown in the image. The carbon to which all of these substituents are bonded is called the α-carbon. It is the R group that differs between the 20 amino acids and gives the amino acid its distinct properties. These "R" groups allow the protein chain to form various secondary structures. Protein alpha helices are stretches of amino acids in which there are hydrogen bonds between the carbonyl group of one amino acid and the amino group of another residue. Each helical turn has 3.6 residues and is a right-handed helix. Globin is an example of a protein comprised mostly of alpha helices. The amino acids alanine, glutamic acid, leucine, and methionine are the preferred amino acids in an alpha helix of a protein. The amino acids proline, glycine, tyrosine, and serine are almost never found in the protein alpha helix. Proline cannot fit into the helix and will actually introduce a kink in the structure. Proline is very common in structures called beta turns. The beta-turn is a short secondary structure containing only 4 amino acids and allows the peptide backbone to make a 180° turn.

74. Question All proteins are composed of the same 20 amino acids arranged in different, but specific sequences. The side chains of these amino acids are responsible for the various properties of the proteins. In which of the following amino acids is the side chain bonded to the backbone nitrogen as well as the α-carbon? Answer Choices A Glycine B Proline C Phenylalanine D Alanine E Lysine

65. Question Peristalsis is a form of gastrointestinal motility that primarily functions to: Answer Choices A Mix chyme B Prevent chyme reflux C Separate chyme D Propel chyme aborad E Stimulate chyme synthesis

Correct Answer: Propel chyme aborad Explanation There are 3 basic types of motility in the gastrointestinal tract: peristalsis, rhythmic segmentation, and tonic segmentation. Peristalsis is coordinated by nerves and involves successive contractions of the gut wall. Peristalsis can be stimulated by neural or humoral agents or when the gut wall is distended by intestinal contents (chyme) and results in muscle contraction proximal (orad) and relaxation distal (aborad) to the chyme bolus. This process continues over a short distance and the bolus is propelled in the aboral direction. Esophageal peristalsis is an important component of the swallowing reflex. Isolated contraction and relaxation of rings of circular smooth muscle (rhythmic segmentation) serve to liquefy food to form chyme and mix it, rather than propel it. Other areas of the gastrointestinal tube have high neurogenic and myogenic tone that produce a closure or narrowing of the gut lumen (tonic segmentation). These "sphincter" regions act to separate chyme and prevent its reflux between compartments.

6. Question A 30-year-old man has a medical history of well-controlled asthma. He began taking ibuprofen daily 2 months ago after he injured his knee and has been noticing a worsening of his asthma. He says he has been using his inhalers more often than prior to his ibuprofen use. He is counseled to stop taking ibuprofen because it may be contributing to the loss of control of his asthma. Nonsteroidal anti-inflammatory drugs, such as ibuprofen, inhibit the synthesis of what substance, which can exacerbate asthma? Answer Choices A Arachidonic acid B Hydroxyeicosatetraenoic acid C Leukotrienes D Prostaglandins E Histamine

Correct Answer: Prostaglandins Explanation Nonsteroidal anti-inflammatory drugs (NSAIDs) like ibuprofen inhibit the production of prostaglandins but not leukotrienes by inhibiting the cyclooxygenase enzyme that converts arachidonic acid to the prostaglandins. Aspirin covalently modifies the cyclooxygenase, leading to its inhibition, while the other NSAIDs such as ibuprofen noncovalently bind and inhibit the cyclooxygenase. A second set of reactions catalyzed by 5-lipoxygenase converts arachidonic acid to 5-hydroperoxyeicosatetraenoic acid (5-HPETE), which is further processed to the leukotrienes. Thus, by blocking the use of arachidonic acid to make the prostaglandins, the arachidonic acid produced is used to make increased amounts of the leukotrienes. These leukotrienes are responsible for the bronchoconstriction and vasoconstriction characteristic of asthma. Anti-asthma drugs target 5-lipoxygenase, thereby reducing the amount of leukotrienes present and reducing the asthma symptoms caused by bronchoconstriction and vasoconstriction. 5-Lipoxygenase is not inhibited by the NSAIDs. Arachidonic acid is formed from the action of phospholipase A2 on membrane phospholipids. Formation of arachidonic acid is inhibited by corticosteroids but not NSAIDs. Histamine is synthesized from the amino acid histidine by a decarboxylation reaction. It is found in practically all tissues, with large amounts in the skin, lungs, and gastrointestinal tract. It is stored in high concentrations in the granules of mast cells and basophils. It is metabolized by monoamine oxidase enzymes and by methylation. References: 1. American Lung Association Asthma Tip Sheet. http://www.lungusa.org/site/pp.asp?c=dvLUK9O0E&b=22592. 2. Devlin, T.M. Textbook of Biochemistry with Clinical Correlations, 6th Edition, Wiley-Liss, New Jersey, 2006, 730-738. 3. Sawdy R, Pan H, Sullivan M, et al. Effect of selective vs. non-selective cyclo-oxygenase inhibitors on fetal membrane prostaglandin synthesis. J Obstet Gynaecol. 2003 May;23(3):239-43. 4. Weber A, Yildirim H, Schrör K. Cyclooxygenase-independent inhibition of smooth muscle cell mitogenesis by ibuprofen. Eur J Pharmacol. 2000 Feb 11;389(1):67-9.

103. Question Glycogenin is a Answer Choices A Protein that serves as an acceptor of glucose residues for the initiation of glycogen synthesis B Polysaccharide that is used as a metabolic reserve when glycogen stores are totally depleted C Component of glycogen D Glycogen-like polysaccharide found in plant tissues

Correct Answer: Protein that serves as an acceptor of glucose residues for the initiation of glycogen synthesis Explanation The only direct source of all the glycosyl residues that are added to the growing glycogen chain during its elongation is UDP-glucose that is synthesized from glucose 1-phosphate and UTP by UDP-glucose pyrophosphorylase. Elongation of glycogen molecules is carried out by glycogen synthase, an enzyme responsible for making α-1,4-linkages. This enzyme cannot initiate the synthesis using free glucose as an acceptor of a molecule of glucose from UDP-glucose and needs a primer that can be either a fragment of glycogen or (if glycogen stores are depleted) a specific protein, glycogenin. Transfer of the first molecule of glucose from UDP-glucose to glycogenin is catalyzed by glycogen initiator synthase. See the attached image.

81. Question The kidneys contain approximately 2 million functional units, which are called nephrons. Which of the following anatomical structures performs the function of the absorption of glucose, amino acids, acetoacetic acid, and protein? Answer Choices A Glomerulus B Proximal tubules C Collecting tubules and ducts D Ascending part of the loop of Henle E Juxtaglomerular apparatus

Correct Answer: Proximal tubules Explanation Each nephron is composed out of a glomerulus, which is surrounded by a capsule called Bowman's capsule. Bowman's capsule empties into a series of tubules leading to the pelvis of the kidney. Fluid filtration takes place in the glomerulus at a rate of about 125ml per minute for both kidneys together or roughly 180 liters per day. The glomerular filtrate has about the same composition as plasma, except that it contains almost no protein. The proximal tubules reabsorb glucose, amino acids, acetoacetic acid, protein, and about 65% of the sodium, potassium, and other electrolytes. They are relatively impermeable to waste products. Of the filtered 125ml of fluid only 1ml passes into the urine, 124ml are reabsorbed; 65% are absorbed in the proximal tubules, 15% in the loops of Henle, 10% in the distal tubules, and a little more than 9% in the collecting tubules and ducts. In the ascending part of the loop the Henle active transport of NaCl takes place, while the cells are only very little permeable for water or diffusion of NaCl, hence fluid leaving Henle's loop is hypotone. Several mechanisms regulate fluid osmolality, sodium and potassium concentration of plasma and urine. ADH increases the permeability of the collecting ducts for osmotic reabsorption of water. Aldosterone increases the secretion of potassium and the re-absorption of sodium in the renal tubules, primarily in the collecting tubules and ducts. The juxtaglomerular apparatus contains the cells for renin secretion. Renin reacts with plasma angiotensinogen to form angiotensin I that, in turn, is converted to angiotensinogen II by a lung enzyme called angiotensin converting enzyme. Angiotensin II causes vasoconstriction.

16. Question Which vitamin is the essential factor in transamination reactions? Answer Choices A Cobalamine B Niacin C Pyridoxine D Riboflavin E Thiamine

Correct Answer: Pyridoxine Explanation As shown in the attached figure, pyridoxal phosphate will serve as the cofactor for, in this case, aspartate transaminase (1), allowing the enzyme to form bound intermediate metabolites, shown in brackets [ ]. The resulting glutamate may be used to generate the NH4+ needed to enter the Urea cycle by the action of the glutamate dehydrogenase (2). •

3. Question Which vitamin is the essential factor in transamination reactions? Answer Choices A Cobalamine B Niacin C Pyridoxine D Riboflavin E Thiamine

86. Question Enzymes do not affect reaction equilibria. Why are they so important biologically? Answer Choices A They increase the rate of direct reactions only B Reactions in vivo are never in an equilibrium state C They decrease the rate of reverse reaction only D Enzymes are proteins E Many enzymes are regulated

Correct Answer: Reactions in vivo are never in an equilibrium state Explanation Enzymes do not affect reaction equilibria. In a reaction illustrated in the attached image, both direct and reverse processes go with increased rates in the presence of enzyme. Enzymes decrease the period of time necessary to approach chemical equilibrium. If biological reactions were at their equilibria, the enzymatic catalysis would not make any biological sense. Nevertheless, biological systems are open and they exchange energy and matter with their environment. The final products of many biological reactions are eliminated by the subsequent processes, and finally from the system, and thus do not enter reverse reactions. Enzymes accelerate approaching to the equilibrium states which nevertheless never happen. The final effect of the enzymatic action looks like acceleration of direct reactions only.

69. Question Target cell membranes contain _____ for interaction with the specific hormones in order to elicit a response. Answer Choices A Acceptors B Introns C DNA binding site D Receptors

Correct Answer: Receptors Explanation Most hormones, other than the steroid and thyroid hormones, act through membrane receptors in order to elicit responses. The cascade of events that happens post-receptor hormone interaction are complicated. These interactions involve various second messenger systems and some even "third" messenger systems to elicit responses.

26. Case A novice runner is training for a charity marathon race and tries to keep pace with the experienced runners. He runs to the point where ATP utilization is faster than ATP can be regenerated by oxidative phosphorylation. Question What process is required for ATP production to continue in the skeletal muscle of this novice runner? Answer Choices A Conversion of acetyl CoA to acetoacetate B Transamination of pyruvate to alanine C Conversion of lactate to pyruvate by lactate dehydrogenase D Regeneration of NAD+ from NADH by lactate dehydrogenase E Transfer of electrons from cytoplasmic NADH to the mitochondria

Correct Answer: Regeneration of NAD+ from NADH by lactate dehydrogenase Explanation Under anaerobic conditions, such as the one occurring in this runner, the muscle must rely on glycolysis to produce ATP. Glycolysis requires NAD+ that must be regenerated from NADH by the lactate dehydrogenase reaction. After muscles use up available ATP and creatine phosphate during exercise, additional ATP must be generated either by aerobic or anaerobic means. Aerobic metabolism relies on oxygen to utilize the mitochondrial oxidative phosphorylation system, which couples electron transport from NADH and FADH2 to production of ATP by ATP synthase. In the absence of oxygen, the muscle relies on anaerobic glycolysis to generate ATP from glucose. Glycolysis requires an adequate supply of NAD+, which is converted to NADH by the enzyme glyceraldehyde 3-phosphate dehydrogenase. A specific isozyme of lactate dehydrogenase converts another product of glycolysis, pyruvate, to lactate, and at the same time it converts NADH to NAD+. This allows glycolysis to continue in the muscle and the production ATP through the metabolism of glucose. The lactate generated in the muscle is transported to the liver where a different isozyme of lactate dehydrogenase converts the lactate to pyruvate that can be used by the gluconeogenesis pathway to generate more glucose. This glucose can then be transported to the muscle. The conversion of acetyl CoA to acetoacetate, a ketone body, occurs in liver mitochondria. The electrons of NADH generated in the cytoplasm can be transferred to mitochondrial NADH via the malate-aspartate shuttle. Once in the mitochondria, these electrons can be converted to ATP by oxidative phosphorylation, but only under aerobic conditions. The transamination of pyruvate to form alanine occurs in the liver and involves the enzyme alanine transaminase.

101. Question A 50-year-old woman has type II diabetes mellitus and has been taking 500mg metformin daily for approximately 1 year. She has a hemoglobin A1c of 6.5% and tells you that she carefully watches her diet, walks 1 mile almost every day, and monitors her blood sugar levels in the evening and before eating breakfast. She says that her blood sugar levels are often higher in the morning than they were the previous evening even though she does not eat anything. You explain to her that this is due to what process? Answer Choices A Release of glucose to the bloodstream from glycogen in muscle B Release of glucose to the bloodstream from glycogen in the liver C Release of glucose to the bloodstream formed by gluconeogenesis from amino acids occurring in the liver D Continuous absorption of glucose into the bloodstream formed by the digestive system E Release of glucose to the bloodstream from gluconeogenesis from lactate occurring in the muscle

Correct Answer: Release of glucose to the bloodstream from glycogen in the liver Explanation The major source of glucose for the body during an overnight fast is the glucose released from glycogen stored in the liver. As blood glucose levels decrease, glucagon levels increase. The binding of glucagon to receptors in the liver triggers the activation of a cascade, resulting in the activation of the enzyme glycogen phosphorylase. This enzyme catalyzes the release of glucose 1-phosphate from glycogen. The glucose 1-phosphate is converted to glucose 6-phosphate by the action of phosphoglucomutase, and the resulting glucose 6-phosphate is converted to glucose by glucose 6-phosphatase. The glucose formed is then released to the bloodstream. Therefore, even though the patient did not eat overnight, the action of glucagon on liver generates glucose which could increase the patient's serum blood glucose. Although muscle contains large amounts of glycogen which can be broken down, the muscle lacks the enzyme glucose 6-phosphatase. Therefore, glucose 6-phosphate remains in the muscle to be used to generate ATP for muscle contraction. Most individuals have adequate liver glycogen stores to provide glucose for an overnight fast so they will not be breaking down protein to generate amino acids, which could be used for gluconeogenesis, the generation of glucose from non-carbohydrate sources. References: 1. Devlin, T.M. Textbook of biochemistry with Clinical Correlations, 6thEdition, Wiley-Liss, New Jersey, 2006, p617-25, 1079-82. 2. Smith C. Marks' Basic Medical Biochemistry: A Clinical Approach. Lippincott Williams & Wilkins. 2005;35. 3. McPhee S. Pathophysiology of Disease: An Introduction to Clinical Medicine. McGraw-Hill Professional. 2006;520.

54. Question Which of the following cell structures is responsible for protein synthesis? Answer Choices A Nucleus B Ribosome C Endoplasmic reticulum D Mitochondria E Lysosome

Correct Answer: Ribosome

53. Question The particles seen lining the rough endoplasmic reticulum on transmission electron microscopy are Answer Choices A Clusters of glycogen molecules B Strands of tubulin cut in cross-section C Exocytotic vesicles D Ribosomes E Centrioles

Correct Answer: Ribosomes Explanation Ribosomes lining endoplasmic reticulum create the cytoplasmic element called rough endoplasmic reticulum (rER). rER is a major site of protein synthesis in the cell. The ribosomes along the ER make strands of protein by transcribing messenger RNA; the strands of protein are either released into the cytoplasm or, if they are to be modified or exported out of the cell, directed into the interior of the endoplasmic reticulum.

123. Question A 72-year-old man presents with swollen and bleeding gums. The examining doctor finds anemia, perifollicular hemorrhages, and loose teeth. A careful history reveals that the patient has been living alone since the death of his wife 18 months ago. His diet consists largely of cola and hot dogs or cheese sandwiches from a nearby deli. What is the most likely diagnosis? Answer Choices A Beriberi B Iron deficiency C Osteomalacia D Pellagra E Scurvy

Correct Answer: Scurvy Explanation This is a classic case of vitamin C deficiency. Vitamin C (ascorbic acid) is a water-soluble antioxidant. It is necessary for hydroxylation of proline in collagen synthesis, degradation of tyrosine, epinephrine synthesis, and bile acid formation. Defective collagen in vessel walls leads to bleeding, producing ecchymoses and petechiae. Vitamin C is found in almost all fruits and vegetables, especially in citrus fruits. Deficiency is seen mainly in infants fed boiled milk, the elderly, and single people. Treatment involves administration of 250 mg of ascorbic acid daily, followed by 40 mg daily maintenance doses. The patient should be encouraged to eat fresh fruit and vegetables. Clinical features of vitamin C deficiency- Swollen, spongy gums with bleeding and superadded infection, loosening of teeth; Perifollicular hemorrhages; Spontaneous bleeding and bruising; Corkscrew hair; Anemia; Failure of wound healing

32. Question Secondary active transport mechanisms occur throughout nature. One example of this type of transport is found in mitochondria and bacteria that use energy stored in H + gradients set up during metabolism to make ATP. Another example of secondary active transport involves the symport or cotransport of amino acids or glucose with Na + in animal cells. Pick the choice that best matches the process of secondary active transport with its dominant characteristic Answer Choices A Secondary active transport: another process makes the energy needed to drive the transport process B Secondary active transport: molecule being transported interacts with a molecule in the membrane, but no energy is involved in the process C Secondary active transport: energy is used directly by the carrier involved in the transport D Secondary active transport: molecules move from an area of high concentration to an area of low concentration E Secondary active transport: the permease or carrier molecule that performs the translocation chemically modifies the molecule being transported

Correct Answer: Secondary active transport: another process makes the energy needed to drive the transport process Explanation Transport mechanisms can be classified in many ways. One way is to determine if energy is required in the transport process. By this method, the processes may be divided into those that do not require energy and those that do so. The processes that do not require energy are simple diffusion and facilitated diffusion. Those processes requiring energy are primary active transport, secondary active transport, and group translocation. Simple diffusion requires that the molecule being transported be soluble in the membrane. The molecules travel from an area of high concentration to an area of low concentration for that particular molecule. No molecule in the membrane interacts with the molecule that crosses the membrane in simple diffusion; in contrast, in facilitated diffusion a molecule in the membrane does interact with the molecule coming across the membrane. This interacting molecule in the membrane is usually a protein that acts as a carrier or as a pore through the membrane, but, just as in simple diffusion, no energy is involved in the process. When energy is involved in the translocation process, and the energy is used directly by the carrier involved in the transport, then the process is called primary active transport. The energy can take many forms in this type of transport. Some of these forms are ATP, light, the high-energy bond of a molecule, and electron flow. On the other hand, if another process makes the energy needed to drive the transport process, then the process is called secondary active transport. Secondary active transport may use ion gradients created by other active transport systems in order to generate energy. When the gradients collapse, the potential energy is transformed into energy to transport the target molecule. Mitochondria and bacteria use energy stored in H + gradients set up during metabolism to make ATP. Another example of secondary active transport involves the symport or cotransport of amino acids or glucose with Na + in animal cells. In order to maintain the Na + concentration constant, the cell must drive out the Na + in exchange for K + through a Na + - K + - ATPase pump. Many other organisms use a symport system similar to this one. Some of the proteins involved in these systems show sequence similarity in some cases but not in other cases.

43. Question Proteins synthesized on endoplasmic reticulum membrane-bound ribosomes would be expected to end up in Answer Choices A The cytosol B Peroxisomes C The nucleus D Secretory vesicles E The mitochondria

Correct Answer: Secretory vesicles Explanation Proteins which are synthesized of free ribosomes either remain in the cytosol or are transported to the nucleus, peroxisome, mitochondria, or chloroplast. Proteins synthesized on membrane-bound ribosomes are translated into the endoplasmic reticulum (ER) while translation is in progress. These proteins can be retained within the ER or transported to the plasma membrane, lysosomes, or secretory vesicles. The targeting of secretory proteins to the ER is due to a hydrophobic signal sequence which is normally found at the amino terminal end of the secretory protein. Translation begins on soluble ribosomes. Following translation of the signal sequence and its emergence from the ribosome the ribosome is bound to the signal recognition particle in the ER membrane. The signal sequence is inserted into a membrane channel and translation continues on the bound ribosome. The signal sequence is cleaved by the signal peptidase and the growing polypeptide is released into the lumen of the ER.

42. Question One of the ways that cells communicate with each other is by secretion of various molecules. The molecules which are secreted for this purpose are known as Answer Choices A Receptor molecules B Signaling molecules C Spectrin molecules D Integrin tetramers E Anticodons

Correct Answer: Signaling molecules Explanation Cells can communicate with each other by releasing signaling molecules that attach to receptor molecules on target cells. Cells communicate with each other via chemical messengers. Within a given tissue, some chemicals move from cell to cell via gap junctions without entering the ECF. There are three general types of intercellular communication mediated in this fashion: (i) neural communication, in which neurotransmitters are released at synaptic junctions from nerve cells and act across a narrow synaptic cleft on a postsynaptic cell; (ii) endocrine communication, in which hormones reach cells via the circulating blood; and (iii) paracrine communication, in which the products of cells diffuse in the ECF to affect neighboring cells that may be some distance away. It should be noted that in various parts of the body, the same chemical messenger can function as neurotransmitter, a paracrine mediator, a hormone secreted by neurons into the blood (neural hormone), and a hormone secreted by gland cells into the blood.

31. Question No transport process can violate the laws of thermodynamics. The free energy of the transport process must be less than 0 for the transport process to operate in the intended direction. Pick the choice that correctly associates the transport process with the correct thermodynamic concept. Answer Choices A Simple diffusion: the system has an increase in entropy after the molecules move from an area of high concentration to an area of low concentration B Facilitated diffusion: molecule being transported interacts with a molecule on the exterior surface of the membrane, but no energy is involved in the process C Primary active transport: energy is used indirectly by the carrier involved in the transport D Secondary active transport: another process transports a chemically modified form of the primary molecule being transported by the primary transport process E Group translocation: the permease or carrier molecule that performs the translocation only transfers part or a group of the original molecule being transported by the primary transport process

Correct Answer: Simple diffusion: the system has an increase in entropy after the molecules move from an area of high concentration to an area of low concentration Explanation Transport mechanisms can be classified in many ways. One way is to determine if energy is required in the transport process. By this method, the processes may be divided into those that do not require energy and those that do so. The processes that do not require energy are simple diffusion and facilitated diffusion. Those processes requiring energy are primary active transport, secondary active transport, and group translocation. Simple diffusion requires that the molecule being transported be soluble in the membrane. The molecules travel from an area of high concentration to an area of low concentration for that particular molecule. No molecule in the membrane interacts with the molecule that crosses the membrane in simple diffusion; the process is driven by an increase in entropy of the system. An equal concentration of molecules across the membrane is a more disorganized arrangement of molecules than an unequal concentration across the membrane of the same molecules. In thermodynamics terms, this means that for simple diffusion to take place, the DG of the process has to be less than 0. If DS, the change in entropy, is large this will be the case, since DG = DH -T DS. The actual value of DG can be predicted using the following relation: DG = 2.3 RT log [(initial concentration of molecule, side 1)/(initial concentration of molecule, side 2)], where side 1 and side 2 refer to each side of the membrane. R is the gas constant, 8.32 Joules deg -1 mole -1, and T is in degrees Kelvin, where ° K = ° C +273. The direction of movement of the molecules will be such as to make the value of DG less than 0, and at equilibrium, DG = 0, so that the concentration of the molecule on each side of the membrane will be the same.

39. Question The cell is highly organized with numerous vesicles and compartments. The organization of subcellular vesicles allows for increased specialization of function. The site within the cell responsible for lipid synthesis is the Answer Choices A Nucleus B Lysosome C Rough endoplasmic reticulum D Golgi apparatus E Smooth endoplasmic

Correct Answer: Smooth endoplasmic Explanation The smooth endoplasmic reticulum is free of ribosomes and is the site of lipid synthesis. The nucleus is the site of DNA compartmentalization. The lysosome functions to break down glycoproteins and glycoaminoglycans under acidic conditions, and the rough endoplasmic reticulum is the site of protein synthesis with the presence of ribosomes interacting with the endoplasmic reticulum membrane. The Golgi apparatus is the membranous region within the cell that fatty acids, sugars, phosphate, and sulfate are added to newly synthesized proteins. These posttranslational processes are often markers for subsequent targeting of proteins to different locations within the cell.

33. Question The RBCs are very pliable cells as they need to squeeze through narrow capillaries and the cords of Billroth in the spleen, without disruption of their membranes. Which of the following enables the erythrocyte to withstand the stress on its plasma membrane as it presses through narrow capillaries? Answer Choices A Desmosomes B Hemoglobin C Integrin D Myosin E Spectrin

Correct Answer: Spectrin Explanation Spectrin is the major protein of the RBC cell membrane cytoskeleton. Along with other proteins, it forms a cable meshwork that is responsible for the shape, strength and flexibility of the red cell membrane. A deficiency of spectrin results in spontaneous loss of the red cell membrane, forcing the cell to take up a spherical configuration. This is the molecular basis of hereditary spherocytosis. Desmosomes are structures that anchor cells together and do not contribute to membrane pliability. Hemoglobin does not contribute to flexibility of the membrane. Integrins are a large class of receptor proteins that bind the cell to the extracellular matrix and mediate transmembrane signaling. Myosin is a contractile protein found in muscle fibers.

44. Question Which of the following statements is a characteristic of a carrier operating by active transport Answer Choices A No energy is required B The carrier operates in both directions C Substances can be moved against a concentration gradient D There is no specificity for the substance to be carried E The process cannot be specifically inhibited

Correct Answer: Substances can be moved against a concentration gradient Explanation A typical mammalian cell maintains a concentration gradient of many substances, including ions between the extracellular and intracellular fluid. The largest difference is for calcium which shows a 25,000-fold difference (2.5 mM extracellular compared to 0.1 μM). The difference for sodium is 14-fold (140 mM extracellular, 10 mM intracellular); potassium is 35-fold (4 mM extracellular, 140 mM intracellular); magnesium is 4-fold (2 mM extracellular, 0.5 mM intracellular); chloride is 25-fold (100 mM extracellular, 4 mM intracellular). These gradients are maintained across the cell membrane by preventing ion flux and by the active transport of ions from one side of the plasma membrane to the other. The specific transport of molecules across membranes is mediated by carriers. A carrier is not an enzyme and does not catalyze any chemical reactions. However, carriers have some properties in common with enzymes in that they are specific, have dissociation constants, exhibit saturation, and can be blocked by specific inhibitors. There are 2 types of transport, passive or facilitated diffusion and active transport. Active transport involves movement of molecules against a concentration gradient and therefore, requires energy usually as ATP. Examples include calcium transport, the sodium-potassium ATPase, and sodium-linked transport of glucose or amino acids. Passive transport involves movement of molecules down a concentration gradient and includes the movement of glucose in many cells.

20. Question Isoleucine, valine, methionine and threonine enter a common metabolic pathway at the point of Answer Choices A Pyruvate B Acetyl-CoA C Succinyl-CoA D α-Ketoglutarate E Oxaloacetate

Correct Answer: Succinyl-CoA Explanation The metabolism of amino acids isoleucine, valine, methionine and threonine, and fatty acids with an odd number of carbon atoms yields propionyl-CoA. The reactions in the attached image turn it to succinyl-CoA.

47. Question The purpose of the nucleolus is to Answer Choices A Synthesize ribosomal subunits B Synthesize protein C Synthesize DNA D Synthesize messenger RNA E Replicate viral DNA

Correct Answer: Synthesize ribosomal subunits Explanation The nucleolus is the site of synthesis of ribosomal subunits. The subunits are then transported through the nuclear pores and assembled in the cytoplasm. Because the purpose of ribosomes is to assemble proteins guided by a strand of messenger RNA, the nucleolus is very large in cells that are producing large amounts of protein and in many malignant cells.

109. Question The action of leptin on glucose metabolism is mediated by Answer Choices A The adipose tissue B The hypothalamus C The gonads D The liver E The thyroid

Correct Answer: The hypothalamus Explanation The regulation of feeding, and in the case of obesity, overfeeding, is complex. Under "coupled" conditions, the amount of ADP determines mitochondrial energetic activity. Thus, if work is being done, ATP is being produced and H+ is being transported. Under "uncoupled" conditions, either thermogenin (UCP; uncoupling protein) will dissipate H+, driving mitochondrial metabolism to produce heat as its product. They are also involved in the mitochondrial transport of free fatty acids (FFA). In the image below, the GREEN ARROWS indicate stimulation or induction; the RED ARROWS indicate inhibition; and the BLACK ARROWS indicate unknown type of regulation. The GRAY ARROWS indicate metabolic steps; the CYAN ARROWS indicate membrane transit or transport; and the BROWN ARROWS or BOXES indicate mitochondrial membrane or its components. The YELLOW BOX represents the hypothalamic nuclei circuitry involved that are explained in the second image. T3 and T4 represent triiodothyronine and thyroxine, respectively. [See Image 1] Leptin is produced by white adipocytes in response to nutritional intake in order to induce high efficiency in the utilization of carbohydrates and lipids. Its action on the central nervous system is directed towards inhibiting overfeeding. Thus, leptin binds to its receptor (Lep-R or OBR), and using the Janus kinase 2 (Jak2) second messenger system, inhibits the release of neuropeptide Y (NPY) from the arcuate nucleus (ARC). This ensures the release of POMC-derived hormones that will, in turn, inhibit the release of NPY from the dorsal medial hypothalamic nucleus (DMH), as well as the stimulation of feeding by the paraventricular hypothalamic nucleus (PVN). This system can be disrupted, and excessive release of NPY may lead to overfeeding. This can occur due to a number of alterations. The most severe case is associated with mutation of the gene coding for the Lep-R, a slightly less severe presentation is associate with mutation of the gene coding for leptin. Both types of alterations lead to severe cases of obesity due primarily to the excessive secretion of NPY from ARC. Abnormal ubiquitous synthesis of the agouti signaling peptide (ASP) leads to excessive secretion of NPY from DMH, which in turn results in a medium severity type of obesity. ASP is a highly competitive antagonist for melanocortin receptors (MC-R). Finally, a light to medium severity obesity can result from alterations in the function of thermogenin 2. In the image below, receptors are represented in GRAY BOXES. Ligands are represented in LIME BOXES if they have a stimulatory action or in RED BOXES if they have an inhibitory action. A similar color code applies to arrows, with GRAY ARROWS indicating metabolic steps. [See Image 2] The following table summarizes the obesity and diabetic phenotypes associated with those disorders that are not related to alterations in the insulin regulatory axis. Please note that when diabetes is mentioned as presenting as a "mild" syndrome, it refers to a pathological condition requiring medical treatment, but that is relatively mild compared to the other presentation. Because humans have relatively little brown adipose tissue, if thermogenin 1 is affected, obesity will not result, as is the case when thermogenin 2 is abnormal. Thermogenin 2 is found in other tissue throughout the body, in addition to the white adipose tissue. Affected Molecule Degree of Obesity Degree of Diabetes Leptin Receptor Severe Very severe Leptin Very Severe Severe Agouti signaling peptide Medium Mild Thermogenin 2 Medium to Mild Mild Thermogenin 1 None None Although leptins have been found to have important direct effects on the reproductive and hemopoietic systems, their effects on carbohydrate and lipid metabolisms are mediated through the hypothalamus. Through its central effects, they then regulate hepatic, thyroid, and adipose tissue function. The pancreas, through its secretion of insulin, glucagon, and C-peptide, obviously plays an important role in these metabolisms, but it is an independent role from that played by leptins.

57. Question If the concentration of potassium outside a neuron were to increase from 4 mEq/L to 8 mEq/L, what would you expect to happen to the minimal stimulus required for initiation of an action potential? Answer Choices A The minimal stimulus required for initiation of an action potential would remain the same B The minimal stimulus required for initiation of an action potential would increase C The minimal stimulus required for initiation of an action potential would decrease D The minimal stimulus required for initiation of an action potential would stay the same, but the peak of the action potential would increase E It is impossible to predict what would occur

Correct Answer: The minimal stimulus required for initiation of an action potential would decrease Explanation As the resting membrane potential increases from - 91 mV to - 76 mV, the amount of energy required to cross the "threshold" and initiate an action potential decreases from 30 mV to 15 mV.

85. Question Apoenzymes are Answer Choices A The proteins that catalyze the apoptosis, programmed cell death B The enzymes that perform metabolic conversions of apolipoproteins C The protein portion of an enzyme that requires a prosthetic group, coenzyme, or cofactor to develop the full enzymatic activity D Specific substrates of the enzymes E Competitive inhibitors of the enzymes

Correct Answer: The protein portion of an enzyme that requires a prosthetic group, coenzyme, or cofactor to develop the full enzymatic activity Explanation Enzymes are biological catalysts. Almost all enzymes are proteins, with the exception of a small group of ribozymes, which are catalytically active RNA molecules. Many protein enzymes require additional nonprotein components to develop their catalytic activity. The term cofactor is used for metal ions, such as Fe2+, Mg2+, Mn2+, Zn2+, or for organic molecules, which participate in the enzymatic catalysis. Nonprotein components that are tightly bound to the enzyme are called prosthetic groups. Examples of prosthetic groups are FAD, FMN, and heme. Less tightly bound complex organic or metalloorganic compounds are called coenzymes. Coenzymes can dissociate from the corresponding protein molecule and be bound with another similar or different enzyme. Examples of coenzymes are NAD+, CoA, thiamine pyrophosphate, and pyridoxal phosphate. The protein part of the complex enzyme is called apoenzyme (apoprotein in a general case), and a complete fully functional enzyme together with its nonprotein part is called holoenzyme. Ribozymes are known to catalyze some steps of RNA processing. Examples are self-splicing introns and self-cleaving RNAs.

46. Question A nucleosome is Answer Choices A A DNA packaging unit found only in chromosomes during mitosis B A DNA packaging unit found only in heterochromatin C A DNA packaging unit found only in euchromatin D The smallest unit of chromatin E A complex of DNA and transcribed RNA

Correct Answer: The smallest unit of chromatin Explanation A nucleosome is the smallest unit of chromatin. It consists of a strand of DNA wrapped twice around a complex made up of eight histone molecules. Nucleosomes are present along DNA strands at approximately 1.5 nm intervals and give the appearance of "beads on a string". The strands of DNA containing nucleosomes are then coiled to form fibrils, which are the packing units for both chromosomes and interphase chromatin. In euchromatin the coiling is loose and in heterochromatin the coiling is tight.

37. Question The committed step of a metabolic pathway is Answer Choices A The step of a pathway which controls the overall efficiency of the pathway B The step which is common between several pathways C Almost always a readily reversible reaction D Often the last step of the pathway E Always the first step in the pathway

Correct Answer: The step of a pathway which controls the overall efficiency of the pathway Explanation The regulation of any metabolic pathway depends on the activities of all the enzymes in that pathway. However, most pathways have one step, the committed step, which is unique to that pathway and this step is highly regulated. The enzyme associated with this step can be regulated by several mechanisms including modification by activators, inhibitors, covalent modifications, or compartmentalization. Long term levels of an enzyme can be altered by regulating its synthesis via hormones or substrate levels. The committed step is usually not a readily reversible step and although it does not have to be the first step in the pathway it is usually one of the first unique steps of a pathway. If the step was common to several pathways it could not be uniquely regulated. For example, inhibiting a step common to pathways A→B→C→D and A→B→Y→Z, namely A→B, would inhibit the synthesis of D and Z under conditions where it might be necessary to continue to synthesize high levels of Z. Regulation of the unique steps B→C or B→Y would inhibit the synthesis of D while allowing Z to be produced, or vice versa.

48. Question The major property of biological membranes which makes them impermeable to most ions and polar molecules is Answer Choices A The presence of cholesterol B The absence of all proteins from the membrane C The structure of the lipid bilayer D The absence of charged groups on the membrane surface E The hydrophilic core of the membrane

Correct Answer: The structure of the lipid bilayer Explanation A typical biological membrane is composed of lipids and protein. The model used to describe membrane structure is the "fluid mosaic" model. In this model, the lipids form a bilayer with the hydrophobic "tails" pointed towards the center and the hydrophilic "heads" pointed to the outside of the bilayer. The lipids can diffuse laterally in this bilayer and the proteins "float" within the sea of lipids. Some proteins are anchored to the cytoskeleton and do not freely diffuse within the bilayer. The membrane is selectively permeable and is impenetrable to most charged species. However, water and hydrophobic molecules can readily diffuse across it. Specific transport, channel, or pore proteins function to transport these compounds across membranes. The inner hydrophobic core of the lipid bilayer does not allow ions or polar molecules to freely diffuse across it. Cholesterol can affect the fluidity of the membrane, but does not affect transport.

15. Question The protein folding to the tertiary structure is determined mainly by Answer Choices A pH of their environment B Temperature C Their post-translational modification D Their amino acids sequences E Presence of the denaturative agents

Correct Answer: Their amino acids sequences Explanation The tertiary structure is unique for the folding of each protein, though some common patterns can be found. This type of protein organization is stabilized by hydrophobic and ionic interactions, hydrogen and covalent bonds. Temperature, pH, and presence of denaturative agents can influence the tertiary structure by affecting the mentioned interactions. Nevertheless, these influences do not determine the protein folding. The importance of the amino acid sequence for protein folding has been shown in experiments with protein renaturation. Many proteins denaturated by heat, denaturating agents or extremes of pH, can be returned to their native structure and functional activity. In the state of complete denaturation, proteins retain only their primary structure and do not have any biological activity. After a renaturation procedure, when the denaturative agent has been removed, they restore their functional properties and all higher levels of their organization. The comparison of homologous proteins with invariant amino acids in some important positions also shows the similarity of their tertiary structures. The post-translational modification, such as glycosylation or limited proteolysis, can influence the tertiary structure and may be necessary to achieve the biological activity.

64. Question At rest, normal expiration is the result of the Answer Choices A Diaphragm B Scalenes C Serratus posterior-superior muscle D Thoracic wall and lung elastic tissue

Correct Answer: Thoracic wall and lung elastic tissue Explanation All of the muscles listed assist with the active process of inspiration. Expiration is a passive process that is assisted by the contraction of elastic tissues of the lung.

110. Question A 12-year-old boy comes to your office and tells you that he has been trying out for his school's track team. After running many sprints for the coach he experienced muscle pain. He says he never had this pain before. You suspect a defect in fatty acid metabolism. A muscle biopsy reveals altered levels of the enzyme carnitine palmitoyl transferase compared to normal muscle. The function of carnitine in fatty acid breakdown is to Answer Choices A Serve to activate lipases B Transport long chain fatty acids into the mitochondria C Inhibit the action of a translocase D Supply reducing power for fatty acid breakdown E Activate coenzyme A

Correct Answer: Transport long chain fatty acids into the mitochondria Explanation Carnitine is a zwitterionic compound derived from lysine which serves as the carrier of acyl groups across the mitochondrial membrane. Long chain fatty acids must enter the mitochondria to undergo b-oxidation. However, the mitochondrial membrane is impermeable to these long chain fatty acids. The activated fatty acyl CoA is transferred to carnitine by the enzyme carnitine acyltransferase I forming fatty acyl-carnitine on the outer mitochondrial surface. The fatty acyl-carnitine then exchanges with free carnitine via a translocase putting the fatty acyl-carnitine on the inner surface of the mitochondria. The fatty acid is then transferred back to CoA by carnitine acyltransferase II. The fatty acyl-CoA is now available for b-oxidation. Short chain fatty acids can enter the mitochondria without carnitine so their metabolism will not be affected by the lack of the enzyme carnitine palmitoyl transferase. The structure of carnitine is as shown in the image.

99. Question A runner has just completed a competitive 5K run. During the race, he experienced muscle pain and cramping due to the buildup of lactic acid. The Cori Cycle exists in the body to relieve the stress of lactic acid buildup. What is a key component of the Cori Cycle? Answer Choices A Transport of lactic acid from muscle to liver B Regeneration of NAD+ in the liver C Generation of a proton gradient D Conversion of pyruvate to acetyl CoA E Stimulation of glucose 6-phosphatase in muscle

Correct Answer: Transport of lactic acid from muscle to liver Explanation Transport of lactic acid from the muscle to the liver is a key component of the Cori Cycle and relieves the metabolic stress in muscle resulting from the anaerobic conditions. After extended periods of strenuous exercise, pyruvate and NADH build up in the muscle from the glycolytic pathway. Under these conditions, muscle metabolism is anaerobic and glycolysis is the major source of ATP. In order to relieve this burden on the muscle and regenerate the NAD+ required for glycolysis to continue, the Cori Cycle operates. The Cori Cycle involves the liver and muscle. Under anaerobic conditions, the pyruvate produced by glycolysis is converted to lactate by the action of the enzyme lactate dehydrogenase (LDH), which catalyzes the following reaction: pyruvate + NADH → lactate + NAD+ This reaction regenerates the NAD+ in muscle that can be used to oxidize glyceraldehyde 3-phosphate to 1,3 bisphosphoglycerate and allow the reactions of glycolysis to continue. The lactate in the muscle enters the circulation and is picked up by the liver. In the liver, a different isozyme of LDH converts this lactate to pyruvate by the following reaction: lactate + NAD+ → pyruvate + NADH The pyruvate generated in the liver enters the gluconeogenesis pathway, resulting in the synthesis of glucose. This glucose diffuses into the bloodstream, where it can be picked up by the muscle and used to generate energy via glycolysis. Therefore, this provides a functional cycle between the liver and tissues such as the muscle that cannot completely oxidize glucose to CO2 due to lack of oxygen. The pyruvate dehydrogenase reaction that converts pyruvate to acetyl CoA does not occur under anaerobic conditions. Glucose 6-phosphatase is not expressed in muscle but is expressed in liver and allows the liver to generate glucose to be released into circulation. References: 1. Mougios V. Exercise Biochemistry. Human Kinetics. 2006;170. 2. Wildman R. Advanced Human Nutrition. CRC. 2000;350. 3. Meisenberg, G. and Simmons, W.H. Principles of Medical Biochemistry Mosby, Elsevier, Philadelphia, 2006, p566-8.

2. Question Proteases of the digestive tract are secreted as inactive proenzyme precursors to avoid proteolytic damage of the secreting cells and organs. Later, they are activated by partial proteolysis. In particular, chymotrypsinogen is activated to chymotrypsin by Answer Choices A Pepsin B Trypsin C Chymotrypsin D Enteropeptidase E Both enteropeptidase and trypsin

Correct Answer: Trypsin Explanation The chart below explains the sequence of proteolytic activation of digestive enzymes. As it can be seen from the chart, conversion of pepsinogen and trypsinogen into their active enzymatic form is an autocatalytic process. Trypsin is also activated by enteropeptidase, but in turn activates elastase, chymotrypsin, and carboxypeptidase.

13. Question One of the ways that a body of water such as a lake can be reclaimed for sport fishing is to treat the lake with rotenone. At the dose used, rotenone kills the fish but allows other organisms to survive. Once the rotenone has degraded, which occurs quickly, a new sport fish can be reintroduced into the lake. Rotenone inhibits the transfer of electrons from NADH-Q reductase to Answer Choices A NADH-Q reductase B Molecular oxygen C Ubiquinone D Cytochrome c E Cytochrome oxidase

Correct Answer: Ubiquinone Explanation The respiratory chain of electron transport is found on the inner mitochondrial membrane. The purpose of the respiratory chain is to generate the energy necessary for the cell to synthesize ATP. Electrons are passed from NADH and FADH2 into the respiratory complexes. Energy transfer through the respiratory complexes leads to the pumping of protons from the matrix side to the cytosolic side of the inner mitochondrial membrane. This is the main energy-conserving event. The proton gradient is used by the mitochondrial ATP synthase (ATPase) to form ATP. The respiratory chain is composed of 3 complexes (NADH-Q reductase, cytochrome reductase, and cytochrome oxidase), which pump protons and 2 mobile carriers (ubiquinone and cytochrome c). The electrons from NADH enter the chain at the NADH-Q reductase complex, while electrons from FADH2 are passed to ubiquinone. The discovery of compounds that inhibit the respiratory chain at discrete sites helped elucidate the sequence of events involved. Rotenone and Amytal are 2 compounds that inhibit the transfer of electrons from NADH-Q reductase to ubiquinone. This results in the failure of respiratory functions. Electron transfer from ubiquinone to cytochrome c1 is blocked by antimycin A. The final step in the respiratory chain is the passage of electrons from cytochrome oxidase to molecular oxygen (O2). Cyanide, azide, and carbon monoxide block this step.

75. Question A woman with metastatic breast cancer is seen by an oncologist. He performs tests to determine whether the HER2 protein is overexpressed in this patient's tumors. The results indicate that HER2 is overexpressed and the oncologist suggests trying trastuzumab. This is a monoclonal antibody directed against the HER2 protein, one of the EGF receptors. Since the EGF receptor is a transmembrane protein, which of the following amino acids would be among those found in its transmembrane domain? Answer Choices AArginine and lysine B Glutamate and aspartate C Valine and leucine D Glutamine and asparagine E Glycine and histidine

Correct Answer: Valine and leucine Explanation The drug trastuzumab (Herceptin®) is recommended for treatment of patients with metastatic breast cancer overexpressing the epidermal growth factor receptor (HER2). This drug is a humanized monoclonal antibody, which binds to HER2 on the surface of tumor cells. This blocks the ability of EGF-like molecules to bind to the receptor. In the absence of ligand binding, the tyrosine kinase activity of the receptor is not activated, and cell growth cannot be stimulated. The growth factor (GF) receptor is composed of 3 domains: the extracellular domain binds the ligand; the transmembrane domain spans the membrane 1 time; and the cytoplasmic domain contains the tyrosine kinase enzymatic domain. In the absence of ligand binding, the tyrosine kinase is inactive. After binding of the GF to the receptor, receptor dimerization occurs with a change in conformation. This conformational change is responsible for the activation of the tyrosine kinase activity. In addition to autophosphorylation of tyrosine residues on the receptor molecule itself, other cellular proteins are phosphorylated on tyrosine by the receptor kinase. These phosphorylated molecules are part of a cascade of events that lead to a biological response to the GF. GF receptor binding does not directly lead to activation of adenylate cyclase. This occurs in response to hormones such as epinephrine. The GF-receptor complex is internalized but this is not directly part of the mechanism of GF action. A transmembrane protein is a protein that extends through the cell's plasma membrane. The transmembrane domain of such a protein is the part of the protein (12-20 amino acids) that extends through the membrane. The plasma membrane is composed of a lipid bilayer with the hydrophobic chains of the lipid molecules on the interior. Therefore, it is expected that the transmembrane domain, the part of the protein in contact with the lipid bilayer, would contain mostly hydrophobic amino acids. Of the choices given, only valine and leucine are considered hydrophobic amino acids. Other hydrophobic amino acids include isoleucine, alanine, phenylalanine, tyrosine, and tryptophan. The charged amino acids arginine, lysine, glutamate, aspartate, glutamine, asparagine, and histidine would not be expected to be found in transmembrane domains of proteins.

73. Question The largest fractional volume of blood in the circulation is normally contained in the Answer Choices A Veins B Vena cavae C Arterioles D Capillaries E Aorta

Correct Answer: Veins Explanation The veins, including the venules, small veins, large veins, and venous sinuses, and the vena cavae normally contain 60% to 70% of the total blood volume.

9. Question A patient lacking the enzyme glucose 6-phosphatase in the liver would be expected to have Answer Choices A A decreased rate of glycolysis in the liver B High blood glucose levels C Very high glycogen levels in the liver D Glycogen with an abnormal number of branch points E No severe clinical manifestations

Correct Answer: Very high glycogen levels in the liver Explanation The enzyme glucose 6-phosphatase catalyzes the removal of phosphate from glucose 6-phosphate resulting in the generation of free glucose. Glucose 6-phosphate cannot cross the cell membrane, and this enzyme is necessary to generate free glucose to be exported into the circulation. Without glucose 6-phosphatase, the glucose made as the result of gluconeogenesis cannot leave the cell. This lack of glucose 6-phosphatase is the cause of von Gierke's disease which is characterized by abnormally high amounts of glycogen in the liver and low blood glucose levels. This disease also affects the kidney. The structure of the glycogen is not affected. Because of the large amounts of glucose 6-phosphate in the liver cells, there is an increase in the rate of glycolysis which leads to increased concentrations of lactate and pyruvate in the blood. The clinical features of von Gierke's disease are increased size of the liver, failure to thrive, severe hypoglycemia, ketosis, hyperuricemia, and hyperlipemia.

122. Question Rhodopsin deficiency, night blindness, retarded growth, skin disorders, and increased infection risk are symptoms of which deficiency? Answer Choices A Vitamin A B Vitamin B1 (thiamine) C Vitamin B2 (riboflavin) D Vitamin B3 (niacin) E Pantothenic acid

Correct Answer: Vitamin A Explanation A (retinoids, carotenoids)- Rhodopsin deficiency, night blindness, retarded growth, skin disorders, and increased risk of infections B1 (thiamine)- Beriberi-muscle weakness (including cardiac muscle), neuritis, and paralysis B2 (riboflavin)- Eye disorders and skin cracking, especially at corners of the mouth B3 (niacin)- Pellagra-diarrhea, dermatitis, and mental disturbances Pantothenic acid- Neuromuscular dysfunction and fatigue

83. Question Eye disorders and skin cracking, especially at corners of the mouth, are the symptoms of which deficiency? Answer Choices A Vitamin A B Vitamin B1 (thiamine) C Vitamin B2 (riboflavin) D Vitamin B3 (niacin) E Pantothenic acid

Correct Answer: Vitamin B2 (riboflavin) Explanation A (retinoids, carotenoids)- Rhodopsin deficiency, night blindness, retarded growth, skin disorders, and increased risk of infections B1 (thiamine)- Beriberi-muscle weakness (including cardiac muscle), neuritis, and paralysis B2 (riboflavin)- Eye disorders and skin cracking, especially at corners of the mouth B3 (niacin)- Pellagra-diarrhea, dermatitis, and mental disturbances Pantothenic acid- Neuromuscular dysfunction and fatigue

18. Question Which of the following is a component of nicotinamide adenine dinucleotide and is involved in glycolysis and the citric acid cycle? Answer Choices A Vitamin B1 (thiamine) B Vitamin B2 (riboflavin) C Vitamin B3 (niacin) D Pantothenic acid E Biotin

Correct Answer: Vitamin B3 (niacin) Explanation o B1 (thiamine) is involved in carbohydrate and amino acid metabolism, necessary for growth. o B2 (riboflavin) is a component of flavin adenine dinucleotide (FAD) and is involved in the citric acid cycle. o B3 (niacin) is a component of nicotinamide adenine dinucleotide and is involved in glycolysis and the citric acid cycle. o Pantothenic acid is a constituent of coenzyme A, glucose production from lipids and amino acids, and steroid hormone synthesis. o Biotin is involved in fatty acid and purine synthesis and the movement of pyruvic acid into the citric acid cycle.

80. Question The addition of specific monomers to each of the three biological macromolecules (proteins, polysaccharides, and nucleic acids) involves the loss of a molecule of Answer Choices A Carbon dioxide B Pyrophosphate C Glucose D Ammonia E Water

Correct Answer: Water Explanation The major macromolecules in the cell are the polynucleotides (DNA, RNA) synthesized from nucleotides, polysaccharides synthesized from various sugar monomers, and proteins synthesized from amino acids. The polymerization reactions that take place to form these macromolecules all involve the loss of water, called a dehydration reaction. The breakdown of all three macromolecules releasing monomers involves a hydrolysis reaction. Synthesis of all three macromolecules requires energy input. Amino acids are added as amino acyl-tRNA, polysaccharides as UDP-sugars, and nucleotides as the triphosphate.

77. Question All proteins are composed of the same 20 amino acids arranged in different but specific sequences. The side chains of these amino acids are responsible for the various properties of the proteins. At neutral pH, the amino acid leucine will be predominantly in the Answer Choices A Completely un-ionized form B Fully protonated form C Hydrogen bonded form D Partially ionized form E Zwitterionic form

Correct Answer: Zwitterionic form Explanation The ionizable groups in a polypeptide or protein consist of the α-amino group (pKα = 7.6), the carboxyl group on the carboxy-terminal residue of the protein (pKα = 3.0), and the side chains of amino acids with ionizable R groups. These are lysine (pKα = 10), glutamic and aspartic acid (pKα = 4.6), histidine (pKα = 6-7), and arginine (pKα = 11.5-12.5). The exact pKα will depend on the environment of the side chain of the individual amino acid in the protein. The pKα is equal to the acid dissociation constant, therefore, when the pH is below the pKα the acid is protonated. The α-amino group (R-NH3+) will be protonated below pH = 7 since its pKa = 7.6. In solution above pH = 7, the carboxyl group of glutamic and aspartic acid will be in the basic, unprotonated (R-COO-) form. The Zwitterion form is the ionic form in which the number of positive charges equals the number of negative charges. Therefore, the net charge is zero. The pH at which an electrically neutral (zwitterionic form) form exists is known as the isoelectric point (pI). For the amino acid leucine, the pI = 6.

89. Question Almost all reactions that occur inside the cell are catalyzed by enzymes. This allows reactions that might not normally occur to occur in the cell because the enzyme acts as a catalyst increasing the reaction rate. Enzymes increase the rates of reactions by Answer Choices A altering the reaction equilibrium B altering the free energy of the ground state C being consumed by the reaction D lowering the activation energy E altering the desired reaction products

Correct Answer: lowering the activation energy Explanation Enzymes function as catalysts and as such obey all of the thermodynamic rules associated with catalysts. Catalysts and enzymes do not affect the reaction equilibria but increase the reaction rate. They enhance the reaction rates by lowering the activation energy (ΔG‡). The activation energy is defined as the difference between the energy level of the ground state and the transition state. The transition state is the point of highest free energy between the substrates and products. The higher the activation energy, the slower the reaction. The interconversion of substrate and product is sped up but the equilibrium point is not changed. The amount of substrate and product present at equilibrium is the same with or without the enzyme being present. Only the rate at which equilibrium is achieved is changed. The ground state, the energy of the substrate or products under normal conditions, is not changed. The enzyme or catalyst is not used up in the reaction. Often there are different intermediates in an enzymatic reaction that occur without the enzyme being present but the final reaction products are not changed. Some typical enzyme rate enhancements over uncatalyzed reactions range from 107 (carbonic anhydrase) to 1014 (urease). Enzymes exhibit high specificity due to several features of active sites. The active site of the enzyme occupies only a small part of the enzyme structure and has a 3-dimensional configuration. It is structurally composed of clefts or crevices. Multiple weak interactions bind the substrate to the enzyme active site. The specificity depends on precise arrangements within the active site. The reaction transition state occurs when the optimal interactions between substrate and enzyme exist and this only occurs in the transition state. To catalyze reactions, an enzyme must be complementary to transition state of the substrate. At this stage, the full complements of weak interactions occur.

84. Question The attached image shows dependence of the rate of an enzyme-catalyzed reaction versus Answer Choices A pH or temperature B Concentration of an enzyme C Concentration of a substrate D Concentration of an irreversible inhibitor E Concentration of a substrate for allosteric enzyme

Correct Answer: pH or temperature Explanation Enzymatic activity of most enzymes depends on pH according to a bell-shaped curve with the point of optimal pH where enzyme demonstrates highest activity. The enzyme activity declines on either side of optimal interval even under small changes of pH. Slight changes of pH causes changes of ionization of substrate and some amino acids in the enzyme molecule (glutamic and aspartic acids, lysis, arginine, histidine). This creates suboptimal conditions for the enzyme - substrate electrostatic interactions and causes slight conformational changes in the enzyme at the levels of tertiary and quaternary structure. The structure of the active catalytic site becomes affected by these changes. As a result, the enzyme declines its activity. The temperature influences the rate of an enzyme-catalyzed reaction according to a similar bell-shaped curve, though due to other reasons. At low temperatures, the rate of reaction increases with temperature, like for any chemical reaction (approximately doubles for every rise of 10° C). Later, at higher temperatures, enzymes undergo denaturation and become inactivated. As a result, every enzyme has an optimal temperature, where catalysis occurs at the maximum rate. See Explanation Image.

10. Question Many types of covalent and non-covalent bonds are involved in maintaining biological systems. Each bond has a distinct bond energy, and these energies are important for the interactions within and between biological molecules. The driving force for the formation of "hydrophobic bonds" is the Answer Choices A attraction between hydrophobic molecules B repulsion of hydrophobic molecules from water C formation of covalent bonds D basis of stabilization of the DNA helix E interaction with membrane phospholipids

Correct Answer: repulsion of hydrophobic molecules from water Explanation The bond energy can be thought of as the amount of energy required to break a bond; the more energy required, the stronger the bond. The hydrogen bond, when a hydrogen atom is shared between 2 other electronegative atoms such as oxygen and nitrogen, has a bond strength of 1 kcal/mol. Covalent bonds, of which a peptide bond is an example, are the strongest bonds with bond energies of approximately 90 kcal/mol. The individual bond energy will vary a great deal depending on the atoms involved and the environment. Ionic bond strengths are approximately 3 kcal/mol. Therefore, the hydrogen bond is the weakest of these bonds. Hydrophobic forces often called "hydrophobic bonds" are formed when water forces hydrophobic groups together in order to minimize the disruption of hydrogen bonds between water molecules. The real attraction between hydrophobic groups is actually caused by repulsion from water. These interactions are noncovalent. Hydrophobic interactions are very important in maintaining protein structure and in interactions between proteins and membrane phospholipids. In the aqueous cellular environment, the hydrophobic amino acid residues of a protein are found at the center away from the surface of the protein in order to minimize interactions with surrounding water. Regions of a protein, for example the transmembrane domain of receptor molecules, contain stretches of hydrophobic molecules that can interact with the lipid components of membrane phospholipids. While this is important, it is not the driving force for their formation. This stabilizes the protein within the membrane. The cytoplasmic and extracellular domains of such a protein will have their hydrophobic residues towards the interior as would any other cytoplasmic protein. The two chains of the DNA double helix are arranged in an antiparallel orientation. Hydrogen bonding between the bases on opposite strands stabilizes the helix structure. Specific hydrogen bonds occur between guanine and cytosine and between adenine and thymine. The hydrogen bonds are relatively weak individually, but the sum of all of these bonds in the DNA molecule makes the double helix a stable molecule. These non-covalent hydrogen bonds can easily be broken to separate the two strands as is required for DNA replication or transcription.

90. Question Almost all reactions that occur inside the cell are catalyzed by enzymes. This allows reactions that might not normally occur to occur in the cell because the enzyme acts as a catalyst. Which of the following is an important characteristic of the active site of all enzymes? Answer Choices A a serine residue is usually present B all residues in the active site must be co-linear C the substrate binds due to multiple weak interactions D the product is released very slowly E it is located at the N-terminal end of the protein

Correct Answer: the substrate binds due to multiple weak interactions Explanation Enzymes function as catalysts and as such obey all of the thermodynamic rules associated with catalysts. Catalysts and enzymes do not affect the reaction equilibria but increase the reaction rate. They enhance the reaction rates by lowering the activation energy (ΔG‡). The activation energy is defined as the difference between the energy level of the ground state and the transition state. The transition state is the point of highest free energy between the substrates and products. The higher the activation energy, the slower the reaction. The interconversion of substrate and product is sped up but the equilibrium point is not changed. The amount of substrate and product present at equilibrium is the same with or without the enzyme being present. Only the rate at which equilibrium is achieved is changed. The ground state, the energy of the substrate or products under normal conditions, is not changed. The enzyme or catalyst is not used up in the reaction. Often there are different intermediates in an enzymatic reaction that occur without the enzyme being present but the final reaction products are not changed. Some typical enzyme rate enhancements over uncatalyzed reactions range from 107 (carbonic anhydrase) to 1014 (urease). Enzymes exhibit high specificity due to several features of active sites. The active site of the enzyme occupies only a small part of the enzyme structure and has a 3-dimensional configuration. It is structurally composed of clefts or crevices. Multiple weak interactions bind the substrate to the enzyme active site. The specificity depends on precise arrangements within the active site. The reaction transition state occurs when the optimal interactions between substrate and enzyme exist and this only occurs in the transition state. To catalyze reactions, an enzyme must be complementary to transition state of the substrate. At this stage, the full complements of weak interactions occur.

40. Question The accompanying image represents the levels of three metabolic components in the blood plasma in a patient that has not had anything to eat or drink for the last six days. The line labeled "C" in the image represents the concentration of Answer Choices A glucose B triacylglycerides C ketone bodies D amino acids E bicarbonate

Correct Answer: triacylglycerides Explanation Under starvation conditions, ketone bodies (represented by line A in the image) in the circulation rise dramatically. This occurs because of the lack of glucose present. Without glucose degradation, there is a reduction in the level of oxaloacetate. Oxaloacetate is necessary for acetyl CoA to enter the citric acid cycle. The absence of oxaloacetate forces acetyl CoA, derived from fatty acid breakdown, into the ketone body biosynthetic pathway. Without oxaloacetate, the acetyl CoA cannot enter the citric acid cycle. Therefore, the amount of ketone bodies under starvation conditions increases. These ketone bodies become the main source of fuel for the brain. The glucose (represented by line B in the image) output of the liver decreases approximately 2-fold while the output of fatty acids (represented by line C in the image) increases somewhat in the first 2-4 days and then remains constant. The degradation of fatty acids provides the ketone bodies needed for the brain after several days of starvation. The blood levels of the different metabolites reflect the shift in fuel usage result by the different organs. This switch away from glucose accomplishes 2 things. First, fuel other than glucose is provided so the brain can continue to function. Second, the decreased need for glucose for the brain means that muscle protein breakdown (and therefore, level of amino acids) is decreased. Muscle protein can be broken down to give pyruvate and alanine, which are used by gluconeogenesis in the liver. Therefore, the decreased need for glucose under prolonged starvation conditions leads to a decreased need for protein breakdown, thus sparing the muscle. This is important for survival. Bicarbonate levels are not directly involved in fuel metabolism.

7. Question A patient lacks pigment in his skin, hair, and eyes and is very sensitive to sunlight. This individual lacks the enzyme necessary to convert what amino acid into melanin? Answer Choices A leucine B tyrosine C lysine D tryptophan E phenylalanine

Correct Answer: tyrosine Explanation Albinism is a disease resulting from the inability to form melanin from tyrosine and results from defects in the enzyme tyrosinase. In addition to being incorporated into proteins, several amino acids serve as precursors of other biologically important molecules. The amino acid tyrosineis one of these amino acids and serves as precursor to a wide variety of molecules that serve as skin pigment (melanin), hormones and neurotransmitters (dopamine, noradrenaline, adrenaline), and thyroid hormone (thyroxin). The biosynthesis of adrenaline (norepinephrine) occurs in the chromaffin cells of the adrenal medulla. The synthesis of thyroxin in the thyroid results from the iodination of thyroglobulin. Melanin is a family of high molecular weight polymers. These polymers are insoluble, very dark in color, and are found in the melanocytes of the skin, retina and ciliary body of the eye, and the substantia nigra and choroid plexus of the brain. Tryptophan is a precursor of the neurotransmitter serotonin. Phenylalanine is converted to tyrosine as part of the amino acid degradation pathway and an absence of the required enzyme, phenylalanine hydroxylase, is the cause of the disease phenylketonuria (PKU). Leucine and lysine are not precursors to other biologically active compounds in humans.

23. Question In addition to being incorporated into proteins, amino acids serve as precursors for many different types of biologically active molecules. Among these are the catecholamines dopamine, epinephrine, and norepinephrine. The amino acid that serves as a precursor for these three neurotransmitters is Answer Choices A glycine B histidine C tryptophan D tyrosine E phenylalanine

Correct Answer: tyrosine Explanation Tyrosine is converted to a family of catecholamine compounds. Initially tyrosine is converted to dopa by the action of the enzyme tyrosine hydroxylase. Dopa is then decarboxylated by dopamine decarboxylase, a pyridoxal phosphate-containing enzyme, to form dopamine. Norepinephrine is formed from dopamine by the action of dopamine β-hydroxylase in a reaction requiring molecular oxygen and ascorbate. Methylation of norepinephrine results in the formation of epinephrine. The amino acid histidine is decarboxylated by the PLP-dependent enzyme histidine decarboxylase to give histamine. Histamine is released in large amounts as part of the allergic reaction. In addition, it is responsible for the stimulation of acid secretion by the stomach. The drug cimetidine (Tagamet®) is a structural analog of histamine that inhibits the secretion of gastric acid by acting as a histamine receptor antagonist. Glutamate is decarboxylated to form γ-aminobutyrate (GABA) while tryptophan serves as a precursor to serotonin.

6. Question Some cases of emphysema are associated with a deficiency in which protein? Answer Choices A α1-antitrypsin B α2-macroglobulin C Prothrombin D Elastase E Immunoglobulin G

Correct Answer: α1-antitrypsin Explanation α1-antitrypsin also known as α1-antiprotease is a 52,000 dalton glycoprotein that is the major component of the α1 fraction of human plasma. It is synthesized by hepatocytes and macrophages and is a protease inhibitor of general specificity. α1-antitrypsin inhibits trypsin, elastase, and certain other proteases by complex formation. A deficiency of the protein plays a role in emphysema. These patients synthesize a less stable form of α1-antitrypsin and much less of the protein is secreted. The interaction of α1-antitrypsin with proteases involves a methionine residue. This methionine is oxidized to methionine sulfoxide by cigarette smoke, thereby preventing α1-antitrypsin from inactivating proteases in the lung. This leads to proteolytic destruction of the lung and accelerates emphysema development. This is especially true in patients who genetically produce low levels of α1-antitrypsin. There is also a liver disease associated with a deficiency of α1-antitrypsin (α1-antitrypsin deficiency liver disease).

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