Kaplan Biochem Flashcards

Question

``` 25. The majority of ATP generated during a 100 meter race is derived from which of the following? A. ATP stores B. Creatine phosphate C. Gluconeogenesis D. Glycolysis E. Lipolysis ```

Answer 1

The correct answer is D.The key to this question is understanding how and when the body utilizes fuel stores. The stores of ATP (choice A) will be used up in less than 1 second once the race has started. Creatine phosphate (choice B) will be the primary source of energy for the next 3 or 4 seconds. After the creatine phosphate stores are depleted, the majority of ATP needed to complete the race will be derived from glycolysis (anaerobic respiration). If the race were to last for an extended period of time, then the processes of gluconeogenesis (choice C) and lipolysis (choice E) might be utilized. Gluconeogenesis is the process of synthesizing glucose in the liver from non-carbohydrate sources, such as amino and fatty acids. Lipolysis is the splitting up or decomposition of fat in the body.

Answer 2

The correct answer is C. Circulating erythrocytes have a life span of about 60 days and are dependent on a functioning Na+/K+ ATPase in the plasma membrane. This pump provides the electrochemical force across the plasma membrane that helps to maintain the volume of the red cell 11 at a constant level by regulating salt, and consequently water, flow into and out of the cell. When the pump stops, the erythrocytes tend to lyse. It is therefore of extreme importance to the erythrocyte to supply the ATP necessary to keep the pump running. This is accomplished through the use of glycolysis, which converts glucose to pyruvate, with a net production of 2 ATP per glucose molecule. The Krebs, or citric acid, cycle (choice A) is the central metabolic degradative pathway of aerobic cells. However, it is located in mitochondria and requires the electron transport chain, making it unavailable for mature red cells to use. The electron transport chain (choice B) is located on the mitochondrial cristae and helps to convert the energy of NADH and FADH2 to ATP. The malate shuttle (choice D) is used to transport cytoplasmic NADH electrons into mitochondria and is not available to erythrocytes. The urea cycle (choice E) is used by liver cells for processing nitrogenous wastes, not generating energy.

Answer 3

The correct answer is E. Folic acid is a pteridine vitamin that exists as tetrahydrofolate (TH4) in its most reduced form. TH4 can accept methyl, methylene, or formyl carbons and transfer them as methyl groups. This function is vital in nucleotide and amino acid synthesis. Pantothenic acid is a key vitamin in acyl transfer reactions (choice A). It forms part of coenzyme A, which transfers acyl groups in thiol esters as acetyl-CoA, succinyl-CoA, and other acyl-CoA forms. Important vitamins in carboxylation reactions (choice B) include biotin and vitamin K. Biotin carries the carboxyl group in the pyruvate carboxylase and acetyl-CoA carboxylase reactions. Vitamin K is used in post-translational carboxylation of amino acid residues in blood clotting factors. Oxidative decarboxylation reactions (choice C) require thiamine (vitamin B1). Examples include the pyruvate dehydrogenase and alpha-ketoglutarate dehydrogenase complexes. Ascorbic acid (vitamin C) is a coenzyme in the hydroxylation (choice D) of lysyl and prolyl residues for collagen synthesis.

Answer 4

The correct answer is choice B. Prolonged treatment with broad-spectrum antibiotics eliminates intestinal bacteria that supply vitamin K. Vitamin K catalyses gamma carboxylation of glutamic acid residues on various proteins concerned with blood clotting. Deficiency is characterized by prolonged PT but normal bleeding time. 12 Choice A - Vitamin C deficiency leads to scurvy with increased bleeding time but normal PT. A diet deficient in vitamin C results in scurvy. Choice C - Vitamin B12 deficiency is usually caused by malabsorption and results in megaloblastic anemia and progressive peripheral neuropathy. Choice D - The most common cause of thiamine deficiency is alcoholism. Because alcohol interferes with absorption, deficiency results in dry and wet beri-beri.

Answer 5

The correct answer is E. The nucleolus is the site of manufacture of ribosomal RNA with its subsequent packaging into ribosomes. Consequently, very large nucleoli indicate an increased rate of ribsome production, which, in turn, suggests an increased rate of production of proteins. Large nucleoli are seen in many active cancers but can also be prominent in benign conditions characterized by high metabolic rate, such as tissue repair after trauma. Depending on the cells involved, the increased protein production associated with more ribosomes might produce more cell surface markers (choice A), but it is impossible to predict this simply from the presence of a large nucleolus. You should associate a large Golgi apparatus (choice B) with transport of material to the cell surface. Plasma cells that produce large amounts of immunoglobulins (choice C) may have large nucleoli, but more specifically have a very large endoplasmic reticulum and Golgi apparatus. The nucleolus is not involved with new DNA (choice D) synthesis; this is done elsewhere in the nucleus.

Answer 6

The correct answer is E. Folic acid deficiency results in the development of macrocytic anemia that yields macro-ovalocytes and hypersegmented neutrophils on the peripheral blood smear. Folic acid is a pteridine vitamin that exists as tetrahydrofolate (TH4) in its most reduced form. TH4 can accept methyl, methylene, or formyl carbons and transfer them as methyl groups. This function is vital in nucleotide and amino acid synthesis. By far the most common cause of folate deficiency is inadequate dietary intake. Alcoholics and those with poor diets are at the highest risk for developing this type of anemia. Pantothenic acid is a key vitamin in acyl transfer reactions (choice A). It forms part of coenzyme A, which transfers acyl groups in thiol esters as acetyl CoA, succinyl CoA, and other acyl CoA forms. 13 Important vitamins in carboxylation reactions (choice B) include biotin and vitamin K. Biotin carries the carboxyl group in the pyruvate carboxylase and acetyl CoA carboxylase reactions, and vitamin K is utilized in post-translational carboxylation of amino acid residues in blood clotting factors. Oxidative decarboxylation reactions (choice C) require thiamine (vitamin B1). Examples include the pyruvate dehydrogenase and alpha-ketoglutarate dehydrogenase complexes. Ascorbic acid (vitamin C) is a coenzyme in the hydroxylation (choice D) of lysyl and prolyl residues of collagen.

Answer 7

The correct answer is choice B. The key thing to remember here is that acetyl-CoA carboxylase catalyzes the first and rate-limiting step of fatty acid synthesis. If you got that far, you could have figured out which of the choices would inhibit the synthesis of fatty acids. Certainly glucagon, a catabolic hormone released in response to low blood glucose, would be a likely candidate to inhibit the synthesis of fatty acids. In fact, glucagon inhibits fatty acid synthesis by a cAMP-dependent phosphorylation of acetyl-CoA carboxylase. Conversely, glucagon stimulates fatty acid oxidation. Clinical correlate: Glycogen metabolism is profoundly affected by specific hormones. Insulin, a polypeptide hormone, increases the capacity of the liver to synthesize glycogen. When insulin levels are high, the production of glycogen is high. The action of insulin is opposed by both glucagon and epinephrine which will increase blood glucose levels. Citrate (choice A) is a key player in fatty acid synthesis (citrate shuttle). Therefore, the presence of citrate would stimulate, not inhibit, acetyl-CoA carboxylase. A high-carbohydrate, low-fat diet (choice C) would stimulate, not inhibit, the synthesis of fatty acids. In contrast to glucagon, insulin (choice D) is an anabolic hormone that promotes fatty acid synthesis and therefore would stimulate acetyl-CoA carboxylase. It does so by dephosphorylating the enzyme.

Answer 8

The correct answer is C. The hexose monophosphate shunt is an alternative route for the oxidation of glucose; it supplies the cell with NADPH and pentose sugars. The NADPH is used in many biosynthetic processes (e.g., fatty acid and cholesterol synthesis), whereas the pentoses are involved in the synthesis of nucleotides and some coenzymes. The pathway "shunts" from glucose-6- phosphate, which is oxidized in a series of NADPH-generating reactions, to ribulose-5-phosphate. The nonoxidative phase, which involves the transfer of C2 and C3 units from one sugar to another, follows. One resulting intermediate is fructose-6-phosphate (choice B), which can serve as a re-entry point to glycolysis, thereby closing the"shunt" loop. 14 Although all the other choices are glycolytic intermediates, they are not involved in the hexose monophosphate shunt.

Answer 9

The correct answer is D.Major histocompatibility complex (MHC) class I proteins are found in the membranes of all nucleotide cells. If the cell is healthy and the peptides are normal, T cells will ignore them. If the cytoplasm contains abnormal peptides, they will appear in the cell membrane and the T cells will be activated which will lead to the destruction of the cell. After the MHC class I molecule has moved to the surface of the tumor cell, peptide fragments from the tumor are presented in a groove of the class I molecule. The peptide fragments are presented to cytotoxic CD8 T cells, which recognize the MHC class I molecules on the cell surface and kill the tumor cell. Loss of this molecule would therefore prevent the tumor cell from being killed. The CD3 molecule (choice A) is a marker on all T cells. It is involved in signal transduction, but not antigen recognition. This molecule would not be on the surface of tumor cells. The CD4 molecule (choice B) is not on the surface of a tumor cell, but it is on the surface of a CD4+ T helper lymphocyte. The CD8 molecule (choice C) is not on the surface of a tumor cell, but it is on the surface of a CD8+ cytotoxic T lymphocyte. MHC class II antigens (choice E) are not involved in killing of tumor cell targets. They present peptide fragments (derived from intracellular killing of extracellular organisms by macrophages) to CD4 T lymphocytes.

Answer 10

The correct answer is E. The porphyrin ring of heme is derived from the citric acid cycle intermediate succinyl-CoA and the amino acid glycine. The initial synthetic step, which is rate-limiting, is catalyzed by aminolevulinic acid synthase (ALA synthase).

Answer 11

The correct answer is D. Of the processes listed, only ketone body synthesis occurs exclusively in the mitochondria. Other mitochondrial processes include the production of acetyl-CoA, the TCA cycle, the electron transport chain, and fatty acid oxidation. The most common cause of ketone body formation is ketoacidosis secondary to diabetes. The essential diagnostic characteristics include: hyperglycemia, acidosis (with blood pH < 7.3), serum bicarbonate < 15 and serum positive for ketones. Processes that occur exclusively in the cytoplasm include cholesterol synthesis (choice A; in cytosol or in ER) and fatty acid synthesis (choice B). Note that gluconeogenesis (choice C) and the urea cycle (choice E) occur in both the mitochondria and the cytoplasm.

Answer 12

The correct answer is E. The human body is able to synthesize roughly half the amino acids necessary to build protein. These amino acids, termed nonessential, include alanine, arginine, asparagine, aspartate, cysteine, glutamate, glutamine, glycine, proline, serine, and tyrosine. The amino acids that must be supplied in the diet are termed essential; these include histidine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine, tryptophan, and valine. Phenylalanine undergoes hydroxylation to tyrosine, catalyzed by the enzyme phenylalanine hydroxylase. It is noteworthy that tyrosine becomes an essential amino acid in individuals lacking this enzyme. The hyperphenylalaninemias, which include phenylketonuria, result from impaired conversion of phenylalanine to tyrosine, and are also asociated with mental retardation. This condition is associated with increased phenylalanine in blood and increased phenylalanine and its by-products (e.g., phenylpyruvate, phenylacetate, and phenyllactate) in urine. Asparagine (choice A) is a member of the "oxaloacetate family"; its immediate precursor is aspartate. The immediate precursor of cysteine (choice B) is serine. Serine is also the precursor of the nonessential amino acid glycine. Glutamine (choice C), proline (choice D), and arginine are produced from glutamate. The synthesis of glutamate occurs by the reductive amination of alpha-ketoglutarate.

Answer 13

The correct answer is A. Vitamin B1, or thiamine, is the coenzyme required for the decarboxylation of alpha-ketoacids. An example of this reaction is pyruvate → acetyl CoA, catalyzed by the pyruvate dehydrogenase complex (although this particular reaction requires other cofactors, thiamine is required in all alpha-ketoacid decarboxylations). Thiamine is also required for the generation of pentose phosphates for nucleotide synthesis in the pentose phosphate pathway (hexose monophosphate shunt), serving as a cofactor for transketolase. 16 Vitamin B2 (choice B), or riboflavin, is a constituent of FMN (flavin mononucleotide) and FAD (flavin adenine dinucleotide). It functions in hydrogen and electron transport. Vitamin B3 (choice C), or niacin (nicotinic acid), is a coenzyme that is also involved in hydrogen and electron transport. Nicotinic acid functions in the form of NAD and NADP. Vitamin B5 (choice D), or pantothenic acid, is conjugated with coenzyme A to act as a carboxylic acid carrier. Vitamin B6 (choice E), or pyridoxine, is used as pyridoxal phosphate and pyridoxamine phosphate. Both of these cofactors are essential to protein metabolism and energy production.

Answer 14

The correct answer is A. The disease is albinism. The most common form of albinism is caused by a deficiency of copper-dependent tyrosinase (tyrosine hydroxylase), blocking the production of melanin from the aromatic amino acid tyrosine. Affected individuals lack melanin pigment in skin, hair, and eyes, and are prone to develop sun-induced skin cancers, including both squamous cell carcinomas and melanomas. Maple syrup urine disease is an example of a disorder of branched chain amino acids (choice B) causing motor abnormalities and seizures. Tay-Sachs disease is an example of a disorder of glycolipids (choice C). In this disorder, a deficiency of hexosaminidase A leads to accumulation of ganglioside GM2. Hunter's disease is an example of a disorder of glycoproteins (choice D). This mucopolysaccharidosis is inherited as an X-linked recessive trait. Homocystinuria disease is an example of a disorder of sulfur-containing amino acids (choice E).

Answer 15

The correct answer is E. To answer this question you need two separate pieces of information. First, riboflavin (one of the B vitamins) is used to make the flavin part of FAD (flavin adenine dinucleotide). Second, of the citric acid cycle enzymes listed, only succinate dehydrogenase (which catalyzes the conversion of succinate to fumerate) used FAD (which is converted to FADH2) as a cofactor. Aconitase (choice A) converts citrate to isocitrate and does not require a cofactor. 17 Citrate synthetase (choice B) combines acetyl-CoA and oxaloacetate to make citrate with release of coenzyme A (which requires pantothenic acid for synthesis). Isocitrate dehydrogenase (choice C) converts isocitrate to alpha-ketoglutarate and uses NAD+ (which is converted to NADH + H+). The NAD+ requires the vitamin niacin for synthesis. Malate dehydrogenase (choice D) converts malate to oxaloacetate and uses NAD+ (which converted to NADH + H+). The NAD+ requires the vitamin niacin for synthesis.

Answer 16

The correct answer is A. Gluconeogenesis is the process that results in the synthesis of glucose from non-carbohydrate precursors. This pathway is very important because the brain is highly dependent on glucose as the primary source of fuel. The three irreversible steps of glycolysis are catalyzed by hexokinase, phosphofructokinase-1 (choice B), and pyruvate kinase. In gluconeogenesis, other enzymes are needed to bypass these key steps. Pyruvate cannot be directly converted to phosphoenolpyruvate in gluconeogenesis. Therefore, pyruvate carboxylase (a mitochondrial enzyme; choice A) converts pyruvate to oxaloacetate, which can be converted to phosphoenolpyruvate by phosphoenolpyruvate carboxykinase (choice D), using two ATP equivalents per molecule of phosphoenolpyruvate. Fructose-1,6-bisphosphatase (choice A) is the enzyme that splits fructose-1,6-bisphosphate into fructose-6-phosphate and inorganic phosphate. It is also required for gluconeogenesis. Glucose-6-phosphatase (choices B and C) is a liver enzyme that hydrolyzes glucose-6-phosphate to glucose. A deficiency of this enzyme leads to von Gierke disease, also known as glycogen storage disease type I. Pyruvate dehydrogenase (choice C) is a mitochondrial enzyme that converts pyruvate to acetyl CoA. This enzyme requires thiamine pyrophosphate, lipoamide, and FAD as cofactors. Glucokinase (choice D) is a liver enzyme that converts glucose to glucose-6-phosphate. Unlike hexokinase, it is specific for glucose and is unresponsive to the level of glucose-6-phosphate. Its function is to store excess glucose, so it has a very high Km (ie, a low affinity) for glucose, becoming active only when the concentration of glucose is very high. Pyruvate kinase (choice E) catalyzes the conversion of phosphoenolpyruvate to pyruvate in the glycolytic pathway. It is activated by fructose-1,6-bisphosphate, the product of the committed step of glycolysis, and is allosterically inhibited by ATP, alanine, and acetyl CoA.

Answer 17

The correct answer is D.Cyanide binds to the heme Fe3+ of cytochrome a3 in the electron transport system, blocking the transfer of electrons to oxygen and consequently the synthesis of ATP. In other 18 words, cyanide blocks the process of oxidative phosphorylation, which produces greater than 90% of the ATP used by the cells in our body. The major steps involved in this process occur within the "electron transport system (ETS)," or "respiratory chain," of the mitochondria. The steps at the end of the elctron transport system, where ATP is generated, are as follows: along the ETS, the repiratory enzymes continually pump hydrogen ions from the matrix of the mitochondria to the intermembrane space, which creates a large concentration gradient. At the end of the ETS, hydrogen ions pass through channels in the respiratory enzymes along the concentration gradient. As the hydrogen ions pass through these enzymes, the energy created is used to phophorylate ADP to ATP. Inhibition of mitochondrial respiration can lead to coma and death unless diagnosed and treated early. The breaking of covalent bonds in glucose molecules (choice A), the formation of carbon dioxide in the TCA cycle (choice B), and the splitting of water molecules into hydrogen and oxygen atoms (choice E) all require energy; these processes do not create energy. The formation of GTP from GDP (choice C) is a process that creates high energy phosphates in the tricarboxylic acid cycle.

Answer 18

The correct answer is D.This question is simple if you know that tyrosinase is an enzyme in the biosynthetic pathway for melanin formation from tyrosine. A lack of tyrosinase causes one form of albinism; a second form is caused by defective tyrosine uptake. Patients with albinism are vulnerable to developing cancers of the skin of all types, including basal cell carcinoma, squamous cell carcinoma, and melanoma. The melanomas are unusual in that they are non-pigmented (amelanotic) rather than black, since the patients cannot form melanin. Malignant melanoma is the leading cause of death from skin disease. Tumor thickness (Breslow's Classification) is the single most important prognostic factor. These lesions are often flat or raised. The borders are typically irregular. Remember many cases of melanoma occur with lesions on the head and neck.

Answer 19

The correct answer is C. Thiamine is a water-soluble vitamin that is converted to the coenzyme thiamine pyrophosphate. This coenzyme is used by pyruvate dehydrogenase to convert pyruvate to acetyl coenzyme A. In the absence of thiamine, pyruvate accumulates and can be converted by lactate dehydrogenase to lactate, which is spilled in the blood causing lactic acidosis. As a side note, thiamine deficiency is associated with the development of beri-beri - a neurological and cardiovascular disorder. The most common cause of beri-beri in the US today is seen in alcoholics. Damage to the nerves is expressed in terms of pain in the limbs and weakness of musculature. The heart can become enlarged and cardiac output can be decreased. Lactate dehydrogenase (choice A) produces lactate from pyruvate but does not use thiamine pyrophosphate. 19 Some lactic acidosis might be produced by decreased pyruvate carboxylase activity (choice B), but the enzyme requires biotin rather than thiamine pyrophosphate. Pyruvate kinase (choice D) makes pyruvate from phosphoenolpyruvate, but does not use thiamine pyrophosphate. Transketolase (choice E) requires thiamine pyrophosphate, but operates in another pathway (pentose phosphate pathway). Decreased transketolase activity is not associated with the development of lactic acidosis.

Answer 20

The correct answer is C. Cytochrome C oxidase is an electron transport chain component, in the mitochondria of muscle and brain. Cytochromes and the electron transport chain are involved in oxidative respiration. Defects can produce mental retardation, weakness, ataxia or seizures. The Golgi apparatus (choice A) is involved in packaging materials for secretion outside the cell. You should associate mucolipidosis I-cell disease with Golgi apparatus problems. Lysosomes (choice B) are the organelles that degrade many cellular products. Defects can produce a wide variety of lysosomal storage diseases, including Hurler and Hunter syndromes. Ribosomes (choice D) are the organelles that traslate mRNA into proteins. There are no known diseases of ribosomes, probably because their function is so crucial that any problems produce death in utero. Hypertrophy of the smooth endoplasmic reticulum (choice E) in the liver is associated with conditions that stimulate the cytochrome p450 detoxification systems (e.g., barbiturate use and alcoholism).

Answer 21

The correct answer is E. G proteins are involved in signal transduction, which allows macromolecules to affect a cell's biological functions without crossing the plasma membrane. The binding of an agonist, such as a hormone, to a receptor on the plasma membrane causes a conformational change in the receptor, which then interacts with a stimulatory G protein (Gs), accompanied by the binding of GTP to the alpha subunit of the G protein. The alpha subunit of the G protein (with attached GTP) dissociates from the beta and gamma subunits, and interacts with intracellular effector molecules such as adenylate cyclase, which it activates. Active transport (choice A) is the movement of molecules across a membrane from a region of low concentration to a region of high concentration; it requires both a carrier protein and the expenditure of energy (usually supplied by the hydrolysis of ATP by an ATPase). 20 Facilitated diffusion (choice B) is the movement of a molecule down its concentration gradient by means of a protein carrier. This type of diffusion is suited for molecules with low permeability resulting from inappropriate size or polarity. Pinocytosis (choice C) is also known as fluid-phase endocytosis and refers to the uptake of molecules that are in solution. Receptors are not involved. Receptor-mediated endocytosis (choice D) refers to the engulfment of a molecule that binds to a receptor on the surface of the cell, occurring commonly at the clathrin-coated pits on the plasma membrane.

Answer 22

The correct ansewer is choice B. Dietary fiber lowers blood cholesterol. Dietary fiber consists of nondigestable carbohydrate, including cellulose, lignin, and pectin. Other actions include the increasing of bowel motility, reduced exposure of the gut to carcinogens, softer stools, and interference with mineral absorption.

Answer 23

The correct answer is choice B. The key thing to remember here is that acetyl-CoA carboxylase catalyzes the first and rate-limiting step of fatty acid synthesis. If you got that far, you could have figured out which of the choices would inhibit the synthesis of fatty acids. Certainly glucagon, a catabolic hormone released in response to low blood glucose, would be a likely candidate to inhibit the synthesis of fatty acids. In fact, glucagon inhibits fatty acid synthesis by a cAMP-dependent phosphorylation of acetyl-CoA carboxylase. Conversely, glucagon stimulates fatty acid oxidation. Citrate (choice A) is a key player in fatty acid synthesis (citrate shuttle). Therefore, the presence of citrate would stimulate, not inhibit, acetyl-CoA carboxylase. A high-carbohydrate, low-fat diet (choice C) would stimulate, not inhibit, the synthesis of fatty acids. In contrast to glucagon, insulin (choice D) is an anabolic hormone that promotes fatty acid synthesis and therefore would stimulate acetyl-CoA carboxylase. It does so by dephosphorylating the enzyme.

Answer 24

The correct answer is A. In the first step of pyrimidine synthesis, carbamoyl phosphate condenses with aspartate to form carbamoyl aspartate, in a reaction catalyzed by aspartate transcarbamoylase. In subsequent steps, ring closure occurs with the loss of water, followed by oxidation to yield orotic acid. Addition of ribose-5-phosphate produces orotidylic acid, which is decarboxylated by orotidylate decarboxylase to yield uridylic acid. The carbon dioxide that is evolved is derived from the alpha carboxyl group of aspartate. Carbamoyl phosphate (choice B) condenses with aspartate with the loss of inorganic phosphate to produce carbamoyl aspartate. The carbamoyl moiety of carbamoyl phosphate is retained. Glutamine (choice C), glycine (choice D) and N10-formyltetrahydrofolate (choice E) are all used in purine synthesis. Glutamine also donates an amino group to UTP to form CTP, but this step occurs after the synthesis of uridylic acid is complete.

Answer 25

The correct answer is E. The hormone in question is insulin, which helps the body utilize glucose in the bloodstream. Circulating insulin binds to a receptor on the plasma membrane of target cells. The insulin receptor is a transmembrane protein that stimulates cytosolic tyrosine kinase activity when activated. One of the important proteins activated by the tyrosine kinase is an insulin receptor substrate, which in turn activates the wide variety of enzymes and transport molecules that produce the actual metabolic changes in sugar, protein, and fat metabolism seen in the presence of insulin. The mitochondrial matrix (choice A) is not a significant receptor site for hormones. The nuclear matrix (choice B), sometimes together with cytoplasm, contains receptors for steroid hormones such as cortisol. The nuclear membrane (choice C) is not a significant receptor site for hormones.

Answer 26

The correct answer is E. Folates (pteroylglutamic acid and related compounds) are widely distributed in foodstuffs. Folic acid serves as an important mediator of one-carbon transfers Dietary deficiency is usually due to overcooked (folates are very labile) and old (folates rapidly decay with time) food. Folic acid deficiency is a macrocytic anemia that reveals macro-ovalocytes and hypersegmented neutrophils on the peripheral blood smear. Reduced folate levels can be seen in red blood cells or serum.

Answer 27

The correct answer is choice B. A key to excelling in biochemistry on the NBDE part 1 is to master the most clinically important elements of metabolic pathways, e.g., rate-limiting steps, irreversible steps, and steps involving enzymes affected by genetic diseases. In this case, glycogen phosphorylase is the enzyme involved in the rate-limiting step of glycogenolysis: Glycogen phosphorylase: (Glucose)n + Pi→ (Glucose)n–1 + Glucose-1-P Note that this enzyme is activated in response to the binding of glucagon to liver cell receptors or the binding of epinephrine to muscle receptors, via signal transduction. α±-1,4-Glucan transferase(choice A) is a debranching enzyme that removes three of four glucose units from a branch point and transfers them to the end of another chain. In this reaction, one α±-1,4 bond is cleaved and another is formed. The elongated chain then becomes a substrate for glycogen phosphorylase. Glycogen synthase (choice C) is the enzyme involved in the rate-limiting step of glycogen synthesis: Glycogen synthase: (Glucose)n + UDP-glucose → (Glucose) n+1 + UDP Note that this enzyme is inactivated in response to the binding of glucagon to liver cell receptors or the binding of epinephrine to muscle receptors. Phosphoglucomutase (choice D) is a key enzyme in glycogen synthesis that reversibly converts glucose-6-phosphate to glucose-1-phosphate. UDP-glucose pyrophophorylase (choice E) is another key enzyme in glycogen synthesis: UDP-glucose pyrophosphorylase: Glu-1-P + Uridine-P-P-P → UDP-Glu + PPi Note that the linkage formed between UDP and glucose is a high-energy bond that can provide energy to many biosynthetic reactions.

Answer 28

The correct answer is D.It is worth taking the time to learn how to read a Lineweaver-Burk plot. Lineweaver-Burk plots are used to determine the Vmax and Km of an enzyme; they are also used to differentiate between competitive and noncompetitive inhibition. Note that in a Lineweaver-Burk plot, the slope is Km/Vmax, the x-intercept is -1Km, and the yintercept is 1/Vmax. In the presence of a competitive inhibitor, the Km (choice A), and therefore the 23 slope (choice B), is increased. Similarly, if Km is increased, -1/Km will become less negative and the x-intercept (choice C) will shift to the right. Intuitively, this makes sense since a competitive inhibitor will increase the amount of substrate needed to reach half-maximal velocity (definition of Km). In contrast, the Vmax, and hence the y-intercept, is unchanged.

Answer 29

The correct answer is choice B. Sphingolipids are a class of lipids that are structural components of membranes. Specifically, sphingolipids are any lipid containing a long chain base like that of sphingosine (e.g., ceramides, cerebrosides, gangliosides). They are a common constituent of nerve tissue. Ceramide is a component of sphingolipids. Ceramide is composed of sphingosine, a longchain amino alcohol with a saturated fatty acid linked to the amino group. Sphingolipids can be differentiated on the basis of the "X" group that is esterified to the terminal hydroxyl group of ceramide. Carnitine (choice A) is involved in the oxidation of fatty acids. Carnitine is important in transferring fatty acids from the cytoplasm into the mitochondria (the carnitine shuttle). Diacylglycerol (choice C) is the alcohol common to all phospholipids. The second alcohol (e.g., choline, ethanolamine, serine) contributes the polar head that distinguishes the different classes of phospholipids. Like sphingolipids, phospholipids are found in membranes. Sphingomyelin (choice D) is a sphingolipid with phosphocholine as its "X" group. It is a component of the myelin sheath. Squalene (choice E) is a 30-carbon intermediate in the synthesis of cholesterol.

Answer 30

The correct answer is choice B. Sphingolipids are a class of lipids that are structural components of membranes. Ceramide is a component of sphingolipids. Ceramide is composed of sphingosine, a long-chain amino alcohol with a saturated fatty acid linked to the amino group. Sphingolipids can be differentiated on the basis of the "X" group that is esterified to the terminal hydroxyl group of ceramide. Carnitine (choice A) is involved in the oxidation of fatty acids. Carnitine is important in transferring fatty acids from the cytoplasm into the mitochondria (the carnitine shuttle). Diacylglycerol (choice C) is the alcohol common to all phospholipids. The second alcohol (e.g., choline, ethanolamine, serine) contributes the polar head that distinguishes the different classes of phospholipids. Like sphingolipids, phospholipids are found in membranes. 24 Sphingomyelin (choice D) is a sphingolipid with phosphocholine as its "X" group. It is a component of the myelin sheath. Squalene (choice E) is a 30-carbon intermediate in the synthesis of cholesterol.

Answer 31

The correct answer is A. Selective IgA deficiency (<5 mg/dL) is the most common of all the primary immunodeficiency diseases. The incidence reported in the US has ranged from 1:250 to 1:1000. IgA has two subclasses, IgA1 and IgA2. IgA1 predominates in the serum, while IgA2 predominates in mucosal secretions as a dimer bound together by a J chain with a secretory piece attached. Recurrent bacterial and viral infections of the respiratory tract can be attributed to a lack of secretory IgA (sIgA), the predominant immunoglobulin of the mucosal immune system. The most common types of infection include sinusitis, otitis and bronchitis. Atopic disease and autoimmune disorders can be associated with IgA deficiency. IgD (choice B) has not been given any particular function other than to act as a receptor on the B cell. It can be found in very low levels in serum. IgG (choice C) is the major immunoglobulin found in the humoral immune response. A patient with a low IgG will experience pyogenic infections. IgM (choice D) is found in the early response to an antigen. If the patient was deficient in IgM he would have also been characteristically low in IgG and would have experienced recurrent pyogenic infection, usually commencing by the age of 5-6 months.

Answer 32

The correct answer is choice B. In this question, you need to know that GTP is synthesized when CoASH is cleaved from succinyl-CoA to form succinate in the citric acid cycle. You also need to know that GTP, rather than ATP, is used as the energy source in protein synthesis, specifically in the formation of the activated elongation factor to which tRNA binds, and in the transfer of the elongating chain from the P to the A site in the ribosome. Cysteine degradation (choice A) requires O2 and produces sulfite and then sulfate, which is secreted in urine. Epinephrine synthesis from tyrosine (choice C) requires tetrahydrobiopterin, pyridoxal phosphate, oxygen, copper, and S-adenosylmethionine. 25 Isopentyl pyrophosphate synthesis from HMG CoA (choice D) requires NADPH and ATP, and releases CO2. Oxidative phosphorylation (choice E) used NADH, FADH2, coenzyme Q, oxygen, and a variety of cytochromes to produce ATP.

Answer 33

The correct answer is A. Bilirubin is a degradative product of hemoglobin metabolism. Bilirubin (pigment) stones are specifically associated with excessive bilirubin production in hemolytic anemias, including sickle cell anemia. Bilirubin stones can also be seen in hepatic cirrhosis and liver fluke infestation. Calcium oxalate stones (choice B) and cystine stones (choice E) are found in the kidney, rather than the gallbladder. Pure cholesterol stones (choice C) are less common than mixed gallstones, but have the same risk factors, including obesity and multiple pregnancies. Mixed stones (choice D) are the common "garden variety" gallstones, found especially in obese, middle aged patients, with a female predominance.

Answer 34

The correct answer is E. Pyruvate dehydrogenase (PDH) catalyzes the irreversible conversion of pyruvate to acetyl-CoA. Thiamine pyrophosphate (TPP), lipomide and FAD serve as catalytic cofactors in addition to CoA and NAD+, the stoichiometric cofactors in this reaction. If PDH is absent, pyruvate will be used in other pathways instead. Pyruvate will be converted to alanine via alanine aminotransferase (choice A) and to lactate via lactate dehydrogenase (choice C). Glutamate dehydrogenase (choice B) is involved in oxidative deamination, releasing ammonium ion for urea synthesis. Deficiency of this enzyme would not cause the symptoms described. Pyruvate carboxylase (choice D) is a gluconeogenic enzyme that catalyzes the conversion of pyruvate to oxaloacetate. Deficiency of this enzyme would not cause the symptoms described.

Answer 35

The correct answer is E. Blue hematoxylin binds to polyanions such as RNA and DNA. Nuclei contain large amounts of DNA and RNA, and they are consequently almost always blue. The nuclei of dysplastic and cancerous cells are often enlarged and hyperchromatic (e.g., darker blue) compared to normal cells of similar cell lines, because these altered cells often have extra DNA (are aneuploid) and/or RNA (are metabolically active). Pink eosin binds relatively nonselectively to cellular components, particularly proteins. Cytoplasm of different cell lines can be pink, purple, or blue, depending principally on the number of ribosomes in the cytoplasm. Consequently, blue-tinged cytoplasm tends to suggest high synthetic activity (e.g., abundant ribosomes).

Answer 36

The correct answer is A. Achondroplasia is an autosomal-dominant disease, typically seen in multiple generations of a pedigree. Men and women are affected in roughly equal frequencies. Autosomal recessive diseases (choice B) are clinically seen only in the homozygous individual. Skipped generations are typically seen. Both men and women are affected. An x-linked dominant disease phenotype (choice C) is seen in multiple generations. Both men and women are affected. X-linked recessive diseases (choice D) are seen much more commonly in men than women. Skipped generations are commonly seen. Male-to-male transmission is not seen in X-linked inheritance.

Answer 37

The correct answer is choice C. Vitamin E acts as antioxidant, and may prevent oxidation of LDL (low density lipoproteins). It protects membrane lipids from peroxidation. Oxidized LDL is thought to promote heart disease. 27 Choice A - Vitamin A is antioxidant, maintains vision and growth regulation. Consumption of betacarotene in the diet decreases the incidence of lung and skin cancer. Vitamin A deficiency results in night blindness and follicular hyperkeratosis. Choice B - In response to hypocalcemia, vitamin D helps normalize serum calcium level. Deficiency results in Rickets and osteomalacia. Choice D - Vitamin K exposes calcium binding sites on several calcium dependent proteins. Deficiency leads to prolonged bleeding and easy bruising with normal bleeding time.

Answer 38

The correct answer is A. In aerobic glycolysis, glucose is degraded to pyruvate, which is then converted to acetyl-CoA, the form in which it actually enters the citric acid cycle. Two acetyl-CoAs, each containing two of the glucose's carbons, are produced from each glucose molecule. In addition, a total of two carbons from glucose are released as CO2 when each of the two pyruvates is converted to acetyl-CoA. Citrate (choice B) is the product formed in the citric acid cycle when acetyl-CoA condenses with oxaloacetate. Oxaloacetate (choice C) is the citric acid cycle intermediate that condenses with acetyl-CoA to form citrate. Two pyruvates (choice D) are produced from degradation of glucose. These are then converted to the two acetyl-CoAs that enter the citric acid cycle. Succinate (choice E) is another citric acid cycle intermediate that forms when coenzyme A is removed from succinyl-CoA.

Answer 39

The correct answer is D.The administration of any product that contains phenylalanine, such as aspartame (an artificial sweetener), to an individual with any of the hyperphenylalaninemias could be detrimental to his or her general health. The hyperphenylalaninemias result from an impaired conversion of phenylalanine to tyrosine. The most common and clinically important of these is phenylketonuria, which is characterized by an increased concentration of phenylalanine in blood, increased concentration of phenylalanine and its by-products (such as phenylpyruvate, phenylacetate, and phenyllactate) in urine, and mental retardation. Phenylketonuria is caused by a deficiency of phenylalanine hydrolase. 28 Hyperornithinemia (choice A) is an inherited disorder of amino acid metabolism that results from a defect of the enzyme ornithine decarboxylase. This condition is associated with mental retardation, neuropsychiatric dysfunction, and protein intolerance. Hyperuricemia (choice B) is a condition associated with higher than normal blood levels of uric acid. Gout may be produced if the hyperuricemia persists. Hypervalinemia (choice C) is a genetic disorder of amino acid metabolism that results from a defect of the enzyme valine aminotransferase. This condition is associated with mental retardation, neuropsychiatric dysfunction, and protein intolerance. Wilson disease (choice E) is an autosomal recessive disorder associated with an abnormality of the hepatic excretion of copper, resulting in toxic accumulations of the metal in the brain, liver, and other organs.

Answer 40

The Correct Answer is choice B. Cortisol increases gluconeogenesis and is responsible for protein breakdown. It also has anti-inflammatory action. Aldosterone (choice A) stimulates renal reabsorption of sodium and excretion of potassium. Estrogen (choice C) controls the menstrual cycle and promotes the development of female secondary sex characteristics. Testosterone (choice D) stimulates spermatogenesis and promotes the development of male secondary sex characteristics.

Answer 41

The correct answer is D.The patient has a megaloblastic anemia, which can be due to deficiency of folate or B12. Pregnancy increases the need for folate and other nutrients used by both baby and mother, and may "unmask" a borderline dietary deficiency. For this reason, most obstetricians recommend vitamin supplements for pregnant women. Folic acid deficiency is a macrocyte anemia. Macro-ovalocytes and hypersegmented neutrophils are revealed on peripheral blood smear. Folic acid is the common term for pteroylmonoglutamic acid. Ascorbic acid (choice A) is vitamin C, and its deficiency predisposes for capillary fragility and oral lesions. Calcium deficiency (choice B) predisposes for osteoporosis/osteopenia. Copper deficiency (choice C) is rare; when it occurs, it may cause a hypochromic anemia, neutropenia, osteoporosis, or hypotonia. 29 Iron deficiency (choice E) causes a microcytic, hypochromic anemia, with reduced mental and physical performance.

Answer 42

The Correct Answer is choice B. Collagen formation begins with transcription of mRNA from appropriate DNA genes in the nucleus. While still within the nucleus, the mRNA is spliced. It is then transported through the cytoplasm to the ribosomes on the rough endoplasmic reticulum. Individual chains are translated on the ribosomes, with the ends feeding into the endoplasmic reticulum lumen. Within the lumen, glycosylation of the individual chains occurs. The material then moves toward the Golgi bodies (whose lumens are connected to the endoplasmic reticulum) where the triple helices of procollagen form. The procollagen is then secreted into the extracellular space, where cleavage of pro-peptides and cross- linking of different triple helices occurs, maturing the collagen. The extracellular space (choice A) is the site of procollagen cleavage and cross- linking. The nucleus (choice C) is the site of mRNA transcription and splicing. The rough endoplasmic reticulum (choice D) is the site of chain translation and glycosylation. The smooth endoplasmic reticulum (choice E) does not participate in collagen synthes

Answer 43

The Correct answer is choice B. The disease is I cell disease, which is a rare genetic disease that may receive disproportionate attention both because it is a mucolipidosis (a form of generalized lysosomal disorder) with accumulation of abnormal chemical material in lysosomes, and because the biochemical basis illustrates an interesting mechanism. Specifically, the Golgi apparatus in cells of these patients has an abnormal N-acetyl-glucosaminotransferase (N- acetylglucosamine-1- phosphotransferase), and is not able to add the necessary recognition marker mannose phosphate to enzymes usually destined to enter lysosomes. Physiologically, the Golgi apparatus is a cellular organelle consisting of a series of membranous plates that give rise to lysosomes and secretory vesicles. A complete deficiency of this enzyme (type I form of I cell disease) causes death early in life. Endoplasmic reticulum (choice A) receives the growing peptide chain of the enzymes, but is not the site of addition of recognition markers. Lysosomes (choice C) are the normal destination for the enzymes with the mannose phosphate marker, but do not receive them if the marker is not present. 30 Mitochondria (choice D) supply the ATP to drive these chemical reactions. Ribosomes (choice E) are the site of synthesis of the amino acid chains of proteins, but not the site of addition of the destination marker.

Answer 44

The correct answer is A. Long-term starvation induces many biochemical changes. Much of the body's energy requirements are normally supplied by serum glucose, but in starvation are supplied by both glucose and lipid-derived ketone bodies, including acetoacetic acid and beta-hydroxybutyric acid. Glucose cannot be synthesized from lipids, and is instead made from amino acids such as alanine in the process of gluconeogenesis. Ketosis as seen in diabetic patients would also be expected. Serum alanine (choice B) drops dramatically in starvation, due to its conversion to glucose. Bicarbonate (choice C) levels drop as the bicarbonate buffers the hydrogen ions produced by the ketone bodies. Chylomicrons (choice D) are the lipid form seen after absorption of dietary fat, and would drop because the person is not feeding. Glucose (choice E) is maintained in the blood at a much lower than normal level during starvation.

Answer 45

The correct answer is choice B. Endemic goiter, such as in this patient, is due to dietary iodine deficiency. This disorder is common world-wide in mountainous areas (where fish are not available), although the use of iodized salt in the United States has limited its prevalence here. Frank symptoms of hypothyroidism may or may not be present, possibly because of the increased synthesis of the more potent triiodothyronine (T3) at the expense of thyroxine (T4). Goiters may become multinodular and grow to great size. Impaired cognition and hearing may be severe in congenital hypothyroidism. Copper deficiency (choice A) can cause anemia, neutropenia, hypotonia, psychomotor retardation, osteoporosis, depigmentation of hair, and glucose intolerance. Iron deficiency (choice C) can cause anemia, cognitive dysfunction, impaired immunity, impaired thermoregulation, and reduced levels of physical activity. Iron deficiency most commonly occurs during pathological bleeding, such as an ulcer. 31 Selenium deficiency (choice D) can cause congestive cardiomyopathy and skeletal muscle degeneration. Zinc deficiency (choice E) causes rash, growth retardation, and impairments of immunity, wound healing, mentation, sexual function, and night vision. Zinc deficiency most commonly occurs in the elderly.

Answer 46

The correct answer is C. Histones are enriched in the amino acids lysine and arginine. They are positively charged and bind tightly to DNA, which is negatively charged from phosphate groups. Proline (choice D) and glycine (choice B) are neutral, aliphatic amino acids. Aspartate (choice A) is negatively charged. Tyrosine (choice E) is a neutral, aromatic amino acid.

Answer 47

The correct answer is C. Hydroxylation of proline in fibroblasts generates the modified amino acid hydroxyproline. This is an example of post-translational modification. Hydroxyproline is involved in stabilizing the three-dimensional triple helix structure of collagen. Proline also has an alphatic side chain but differs from other amino acids in that its side chain is bonded to both the nitrogen and carbon atoms. Note: Proline contains a secondary rather than a primary amino group, which specifically makes it an imino acid. Cysteine (choice A) is unique in its ability to form a covalent disulfide bond with another cysteine residue elsewhere in the protein molecule, thereby forming a cysteine bond. Such strong disulfide bonds stabilize the three-dimensional structure of the protein. Glycine (choice B) is abundant in fibroblasts since it constitutes every third amino acid in the primary sequence of collagen. However, glycine is not hydroxylated. Serine (choice D), tyrosine (choice E), and threonine can all be phosphorylated post-translationally to form phosphoserine, phosphotyrosine, and phosphothreonine, respectively. These phosphorylated amino acids are believed to play a role in signal transduction.

Answer 48

The correct answer is D.It is worth taking the time to learn how to read a Lineweaver-Burk plot. Lineweaver-Burk plots are used to determine the Vmax and Km of an enzyme; they are also used to differentiate between competitive and noncompetitive inhibition. Note that in a Lineweaver-Burk plot, the slope is Km/Vmax, the x-intercept is -1/Km, and the y-intercept is 1/Vmax. In the presence of a competitive inhibitor, the Km (choice A) and therefore the slope (choice B) are both increased. Similarly, if Km is increased, -1/Km will become less negative and the xintercept will shift to the right. Intuitively, this makes sense since a competitive inhibitor will increase the amount of substrate needed to reach half-maximal velocity (definition of Km). In contrast, the Vmax, and hence the y-intercept, is unchanged (choice D). It is important to note that it is convenient to transform the Michaelis-Menton equation into one that gives a straight line plot, which can be done by taking the reciprocal of both sides to 1/V = 1/Vmax + Km/Vmax + 1/[s] A plot of 1/V versus 1/[s] is called the Lineweaver-Burk plot.

Answer 49

The correct answer is C. The patient has a classic case of Lesch-Nyhan syndrome, an X-linked recessive disorder caused by severe deficiency of the purine salvage enzyme hypoxanthine-guanine phosphoribosyl transferase (HPRT). This defect is associated with excessive de novo purine synthesis, hyperuricemia. The clinical and symptoms include spasticity, writhing movements, compulsive biting of fingers and lips, and head-banging. At puberty, the child often develops arthritis and death from renal failure occurs at age 25 years. The biochemical basis of the often striking selfmutilatory behavior (which may require restraints and even tooth extraction) has never been established. Treatment with allopurinol inhibits xanthine oxidase and reduces gouty arthritis, urate stone formation, and urate nephropathy. It does not, however, modify the neurologic/psychiatric presentation. Mental deficiency, spasticity, and self-mutilation are the most common characteristics of this condition.

Answer 50

``` The correct answer is D.The patient has gout, which is due to precipitation of monosodium urate crystals in joint spaces (notably the great toe) and soft tissues (causing tophi, which are often found on the external ears). Gout is often an acute condition that is typically nocturnal and usually monarticular, often involving the first metatarsophalangeal joint. Hyperuricemia occurs in most cases. Identification of urate crystals in joint fluid or tophi is diagnostic. Bile pigments (choice A) are found in some gallstones. Calcium pyrophosphate (choice B) crystals are deposited in pseudogout, which classically affects the knee or other large joints. Cystine (choice C) and struvite (choice E) can form kidney stones ```

Answer 51

The correct answer is choice B. You may be asked to make this type of calculation on the NBDE part 1. You should know that 1 g of either protein or carbohyrate produces about 4 kcal = 4 Calories (kcal) of energy, and 1 g of fat produces 9 kcal = 9 Calories of energy. The calculation is then straightforward. The Calories from carbohydrates are 125 X 4 = 500; from protein are 15 X 4 = 60; and from fat are 10 X 9 = 90. The total is 500 + 60 + 90 = 650 kcal/day.

Answer 52

The correct answer is C. Citrate is produced by citrate synthase from acetyl CoA and oxaloacetate. Note that this is the first step in the citric acid cycle. This reaction takes place in the mitochondria, but citrate can move freely from the mitochondria into the cytosol. When the citric acid cycle slows down, citrate accumulates. In the cytosol, it acts as a negative allosteric regulator of phosphofructokinase I, the enzyme that catalyzes the committed step of glycolysis. Fructose-2,6-bisphosphatase (choice A) breaks down fructose-2,6-bisphosphate, a potent allosteric activator of phosphofructokinase I. Fructose-2,6-bisphosphatase is activated by cyclic AMPdependent protein kinase. Isocitrate dehydrogenase (choice B) converts isocitrate to alpha-ketoglutarate in the citric acid cycle. It is allosterically stimulated by ADP and inhibited by ATP and NADH. This reaction produces NADH and CO2. Pyruvate carboxylase (choice D) is a mitochondrial enzyme that converts pyruvate to oxaloacetate. It is important in gluconeogenesis and replenishes the oxaloacetate in the citric acid cycle. 6-phosphogluconate dehydrogenase (choice E) converts 6-phosphogluconate to ribulose 5- phosphate in the pentose phosphate shunt pathway.

Answer 53

The correct answer is C. Glycogen is the storage form of glucose and a readily mobilizable fuel store. When individuals do not consume adequate quantities of carbohydrates, the body responds by breaking down glycogen stores in the liver to maintain normal blood glucose levels. The two enzymes involved in the degradation and synthesis of glycogen are glycogen phosphorylase (choice C) and glycogen synthase (choice D), respectively. The processes of glycogen synthesis and degradation are coordinated by a hormone-triggered cascade that ensures that when one enzyme is active, the other enzyme is inactive. Acetyl-CoA carboxylase (choice A) is the key enzyme involved in fatty acid synthesis. Glucose 6-phosphate dehydrogenase (choice B) is involved in the pentose phosphate pathway. Thiolase (choice E) converts acetoacetyl-CoA into acetyl-CoA.

Answer 54

The correct answer is E. This question is asking you what forces are responsible for retaining integral membrane proteins within the lipid bilayer, i.e., hydrophobic interactions. Amphipathic agents, such as detergents, are used to solubilize integral membrane proteins; they do so by disrupting hydrophobic interactions between integral membrane proteins and other membrane constituents, such as phospholipids. Choices A, B, C, and D would destabilize the association of peripheral membrane proteins with the membrane lipid bilayer.

Answer 55

The correct answer is choice B. Most of the calcium in the body is found as calcium phosphate crystals in the bones and teeth that contribute to the physical strength of these structures. Vitamin D (cholecalciferol) increases intestinal absorption of calcium and phosphate. 35 Choice A - Calcitonin is secreted by specialized parafollicular cells of the thyroid gland. It decreases the resorption of bone, reduces serum calcium, and improves bone architecture. Choice C - Vitamin A (carotene) is converted to several active forms in the body and is associated with important functions, including maintenance of healthy epithelium and vision. Choice D - Vitamin E (alpha tocopherol) prevents peroxidation of fatty acids in cell membranes, helping to maintain their normal fluidity.

Answer 56

The correct answer is D.Thiamine deficiency is most frequently encountered in alcoholics and in developing countries. Deficiency of this vitamin can take several forms: dilated cardiomyopathy (wet beriberi), polyneuropathy (dry beriberi), and mamillary body degeneration (Wernicke-Korsakoff syndrome). Ascorbic acid (choice A, Vitamin C) deficiency causes scurvy, associated with capillary fragility, bony abnormalities, and poor wound healing. Retinol (choice B, Vitamin A) deficiency causes blindness and impaired immune responses. Riboflavin (choice C) deficiency causes cheilosis, glossitis, and dermatitis. Vitamin K (choice E) deficiency causes impaired blood clotting because of decreased production of factors II, VII, IX, and X.

Answer 57

The correct answer is C. Regardless of how long the untranslated regions are, the number of nucleotides in the coding region of an mRNA is three times the number of amino acids, since three nucleotides are required to code for each amino acid, and 3 X 64 = 192. In reality, three nucleotides code for the first amino acid (formyl-methionine in prokaryotes, methionine in eukaryotes), which may be removed in posttranslational steps, and three nucleotides at the 3' end (are needed to terminate the process (i.e., a STOP codon), so the actual number would likely be slightly higher.

Answer 58

The correct answer is C. The first step in killing bacteria is the production of superoxide ion, O2- , by the action of NADPH oxidase on NADPH and O2. The superoxide is then converted to hydrogen peroxide, either spontaneously or through the action of superoxide dismutase. The hydrogen peroxide can also be converted to the toxic HOCl. radical by the action of myeloperoxidase. Hydrogen peroxide(choice A) is not an enzyme but is used as an antiseptic for dental procedures and for the treatment of apthous ulcers in the mouth. Myeloperoxidase (choice B) is part of the killing pathway, but is active later in the sequence, when H2O2 Peroxidase (choice D) is also a heme enzyme. It catalyzes the reaction in which hydrogen peroxide is reduced to water. Superoxide dismutase (choice E) is part of the killing pathway, but is active later in the sequence, when superoxide is converted to hydrogen peroxide. This enzyme is present in all aerobic organisms.

Answer 59

The correct answer is C. Hydroxylation of proline in fibroblasts generates the modified amino acid hydroxyproline. This is an example of post-translational modification. Hydroxyproline is involved in stabilizing the three-dimensional triple helix structure of collagen. Cysteine (choice A) is unique in its ability to form a covalent disulfide bond with another cysteine residue elsewhere in the protein molecule, thereby forming a cystine residue. Such strong disulfide bonds stabilize the three-dimensional structure of the protein. Glycine (choice B) is abundant in fibroblasts since it constitutes every third amino acid in the primary sequence of collagen. However, glycine is not hydroxylated. Serine (choice D), tyrosine (choice E), and threonine can all be phosphorylated post-translationally to form phosphoserine, phosphotyrosine, and phosphothreonine, respectively. These phosphorylated amino acids are believed to play a role in signal transduction.

Answer 60

The correct answer is D.The primer molecule required by DNA polymerase is a short strand of RNA (4-10 bases) complementary to the template strand of the DNA molecule. The primer is synthesized by a specific RNA polymerase known as primase. The growing end of the RNA primer is a free 3'-OH group. The primer RNA is excised at a later stage of replication. The primase does not itself require a primer for initiation of nucleotide synthesis. DNA ligase (choice A) catalyzes the formation of a phosphodiester bond between the 3'-OH of one fragment of DNA and the 5'-monophosphate group of an adjacent DNA fragment. Polymerase I (choice B) catalyzes the polymerization of nucleotides and also functions in processing and repair mechanisms. Polymerase III (choice C) is a part of a multiprotein complex and is the major replicating enzyme in E. coli. Topoisomerases (choice E) produce swivel points in the DNA molecule that relieve the strain induced by the replication fork. These enzymes cut and reseal the DNA.

Answer 61

The correct answer is C. Histones are enriched with the amino acids lysine and arginine. They are positively charged and bind tightly to DNA, which is negatively charged because of its phosphate groups. Proline (choice D) and glycine (choice B) are neutral, aliphatic amino acids. Aspartate (choice A) is negatively charged. Tyrosine (choice E) is a neutral, aromatic amino acid.

Answer 62

The correct answer is choice B. This question requires two logical steps: first, you need to appreciate that the hydrophilic amino acids are more likely to appear on the surface of a protein molecule, while hydrophobic amino acids are most likely be found in its interior. Next, you need to figure out which of the amino acids listed is hydrophilic. If you recall that arginine is a basic amino acid that is positively charged at physiologic pH, you should be able to answer this question right away. In addition to arginine, lysine and histidine also have basic side chains. Structurally the planar outer part of arginine's side chain consists of 3 nitrogens bonded to a carbon atom, which is called a guanidium group. All of the other choices have neutral side chains and are uncharged at physiologic pH. They would most likely be found in the hydrophobic core of the protein structure. Alanine (choice A), isoleucine (choice C), and leucine (choice D) all have aliphatic side chains; phenylalanine (choice E) has aromatic side chains.

Answer 63

The correct answer is D.Western blot is used as the confirmatory test for HIV exposure and is highly specific. The combination of ELISA and Western blot has a positive predictive value greater than 99%. The Western blot primarily eliminates a small percentage of false positives found by the ELISA test alone. Choice A - ELISA is used as the primary screening assay because it is very sensitive. Choice B - Southern blotting is a technique that can be used to detect mutations in DNA. Choice C - Northern blots analyze RNA extracted from a tissue and are typically used to determine which genes are being expressed.

Answer 64

The correct answer is C. Ketone bodies, which include acetoacetate, beta-hydroxybutyrate, and acetone, are produced by the liver in the fasting state by beta-oxidation of fatty acids. They are then released into the blood stream, where they can be used as alternative energy sources for other organs, such as muscle, kidney, and brain. The brain specifically still requires a small amount of circulating glucose to function, but the amount required is reduced when ketone bodies are available. Apoprotein B (choice A) is one of the proteins that hold lipoproteins together. Beta-carotene (choice B) is a vitamin with antioxidant properties. C-reactive protein (choice D) is a serum protein produced by the liver that rises during infections and in inflammatory states. 39 Coenzyme A (choice E) is found in mitochondria and carries acetyl groups into the citric acid, or tricarboxylic acid, cycle.

Answer 65

The correct answer is D.Peroxisomes are cell organelles that are present in only small numbers in most mammalian cells. In the liver, however, these single membrane-bound organelles are present in large numbers and are import in detoxification and long chain fatty acid metabolism. The clinically important degradation of ethanol to (potentially toxic) acetaldehyde occurs in humans in both peroxisomes and the smooth endoplasmic reticulum (P450 system). Acetaldehyde is then oxidized to (non-toxic) acetate. This reaction, catalyzed by aldehyde dehydrogenase, occurs in the mitochondria. Golgi bodies (choice A) are involved with packaging substances that are transported out of the cell. Lysosomes (choice B) are sites of degradation of intracellular waste products. The mitochondria (choice C) are involved with ATP production and contain both the electron transport chain and the citric acid (Krebs) cycle. Ribosomes (choice E) are the sites of protein synthesis.

Answer 66

The correct answer is D.Levodopa is used in the treatment of Parkinson disease, a condition caused by dopamine deficiency in the CNS. When levodopa enters the CNS, it is metabolized to dopamine, some of which is subsequently metabolized to norepinephrine (NE) in noradrenergic neurons. In other words, dopamine is the immediate precursor to NE. The synthesis of NE begins in the axoplasm of the terminal nerve endings of adrenergic fibers; however, its synthesis is completed inside the vesicles of these fibers. With respect to the endogenous syntheses of NE, the basic steps are as follows: tyrosine (choice E) is converted to dihydroxyphenylalanine (DOPA) through the process of hydroxylation. Then, DOPA (choice B) undergoes decarboxylation to the neurotransmitter dopamine. Dopamine is then transported into the vesicles of the adrenergic fibers, where it undergoes hydroxylation to NE (choice D). In the adrenal medulla, NE is transofrmed into epinephrine (choice C) through the process of methylation. Choline (choice A) is combined with acetyl-CoA to become acetylcholine.

Answer 67

The correct answer is choice B. The process of glycolysis is defined as the sequence of reactions that converts glucose into pyruvate with the concomitant production of ATP. Glycolysis begins when glucose is converted by hexokinase to glucose-6-phosphate. When this compound interacts with the enzyme phosphoglucose isomerase (choice D), fructose-6-phosphate is formed. Fructose-6-phosphate is then converted by phosphofructokinase (choice C) to form fructose 1,6-biphosphate, which is subsequently converted to glyceraldehyde 3-phosphate by aldolase (choice A). After a number of enzymatic reactions, phosphoenolpyruvate is formed. Phosphoenolpyruvate is converted to pyruvate by pyruvate kinase (choice E), and the glycolytic pathway is then completed.

Kaplan Biochem Flashcards

(91 cards)