Block 1 culled from slpanday8

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.

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

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.

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.

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. •

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

3. 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).

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.

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.

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.

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.

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.

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

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

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. •

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.

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.

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.

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.

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.

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.

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.

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.

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.

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)

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.

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)

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.

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]. •

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.

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.

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.

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.

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.

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.

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.

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) •

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.

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.

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.

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.

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.

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.

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.

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.

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.

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). •

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.

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.

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.

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.

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.

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.

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.

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).

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.