Chapter 6 - Carbohydrate Metabolism Flashcards
After digestion of a piece of cake that contains flour, milk, and sucrose as its primary ingredients, the major carbohy- drate products entering the blood are which one of the following? Choose the one best answer. (A) Glucose (B) Fructose and galactose (C) Galactose and glucose (D) Fructose and glucose (E) Glucose, fructose, and galactose
The answer is E. The cake contains starch, lactose (milk sugar), and sucrose (table sugar). Digestion of starch produces glucose. Lactase cleaves lactose to galactose and glucose, and sucrase cleaves sucrose to fructose and glucose. Thus, the intestinal epithelial cells will absorb from the intestinal lumen, and then secrete into the blood, glucose, galactose, and fructose. The intestinal epithelial cells will not use these sugars as an energy source.
The immediate degradation of glycogen under normal conditions gives rise to which one of the following?
(A) More glucose than glucose-1-phosphate
(B) More glucose-1-phosphate than glucose (C) Equal amounts of glucose and
glucose-1-phosphate
(D) Neither glucose nor glucose-1-phosphate
(E) Only glucose-1-phosphate
The answer is B. Phosphorylase produces glucose-1-phosphate from glucose residues linked α-1,4. Free glucose is produced from α-1,6-linked residues at branch points by an α-1, 6-glucosidase activity of the debranching enzyme. Degradation of glycogen produces glucose- 1-phosphate and glucose in about a 10:1 ratio, which is the ratio of the α-1,4 linkages to α-1,6 linkages.
Phosphorylase kinase can be best described by which one of the following?
Activated in response to: / Target of enzyme activity: / Active in presence of caffeine? / Required substrate for enzymatic activity
(A) Insulin / Glycogen phosphorylase / Yes / ATP
(B) Insulin / Glycogen phosphorylase / Yes / GTP
(C) Insulin / Branching enzyme / No / ATP
(D) Epinephrine / Branching enzyme / No / GTP
(E) Epinephrine / Glycogen phosphorylase / Yes / ATP
(F) Epinephrine / Glycogen phosphorylase / Yes / GTP
The answer is E. Glucagon in the liver and epinephrine in both the liver and muscle
cause cAMP levels to rise, activating protein kinase A. Protein kinase A phosphorylates and activates phosphorylase kinase, which in turn phosphorylates and activates phosphorylase. These phosphorylation reactions require ATP. Branching enzyme is not a substrate for phosphorylase kinase. Phosphodiesterase inhibitors, such as caffeine, keep cAMP elevated, which allows protein kinase A to be active, which keeps phosphorylase kinase active, and in its phosphorylated form.
In an embryo with a complete deficiency of pyruvate kinase, how many net moles of ATP are generated in the conversion of 1 mole of glucose to 1 mole of pyruvate? (A) 0 (B) 1 (C) 2 (D) 3 (E) 4
The answer is A. Normally, 1 mole of ATP is used to convert 1 mole of glucose to 1 mole of glucose-6-phosphate and a second to convert 1 mole of fructose-6-phosphate to the bisphosphate. Two triose phosphates are produced by cleavage of fructose-1,6-bisphosphate. As the two triose phosphates are converted to pyruvate, four ATPs are generated; two by phosphoglycerate kinase and two by pyruvate kinase. Net, two ATPs are produced. If pyruvate kinase is completely deficient, two less ATPs will be produced, and thus the net ATP production will be zero. It is unlikely that the embryo would survive with a complete deficiency of this enzyme.
Which one of the following is a regulatory mechanism employed by muscle for glycolysis?
(A) Inhibition of PFK1 by AMP
(B) Inhibition of hexokinase by its product
(C) Activation of pyruvate kinase when glucagon levels are elevated
(D) Inhibition of aldolase by fructose 1,6-bisphosphate
(E) Inhibition of glucokinase by F-2,6-P
The answer is B. Hexokinase is inhibited by its product, glucose-6-phosphate. PFK1 is activated by AMP and F-2,6-P. F-2,6-P does not inhibit glucokinase, nor is glucokinase present in the muscle. Aldolase is not inhibited by its substrate, fructose 1,6-bisphosphate. Pyruvate kinase is inactivated by glucagon-mediated phosphorylation in the liver, but not in the muscle. The muscle isozyme of pyruvate kinase is not a substrate for protein kinase A. In addition, muscle cells do not respond to glucagon as they do not express glucagon receptors.
Caffeine, a methyl xanthine, has been added to a variety of cell types. Which one of the following would be expected in various cell types treated with caffeine and epinephrine?
(A) Decreased activity of liver protein kinase A
(B) Decreased activity of muscle protein kinase A
(C) Increased activity of liver pyruvate kinase
(D) Decreased activity of liver glycogen synthase
(E) Decreased activity of liver glycogen phosphorylase
The answer is D. If the phosphodiesterase that degrades cAMP were inhibited (an effect
of caffeine) in the presence of epinephrine, cAMP levels would be elevated. Protein kinase A would become more active in the liver and muscle; pyruvate kinase would become less active in the liver; and glycogen synthase activity would be decreased in both muscle and liver. Phosphorylase activity would be increased in both muscle and liver owing to constant phosphorylation by phosphorylase kinase, which is activated by protein kinase A.
Which one of the following occurs during the conversion of pyruvate to glucose during gluconeogenesis?
(A) Biotin is required as a cofactor.
(B) The carbon of CO2, added in one reaction,
appears in the final product.
(C) Energy is utilized only in the form of GTP.
(D) All of the reactions occur in the cytosol.
(E) All of the reactions occur in the mitochondrion.
The answer is A. In the mitochondria, CO2 is added to pyruvate to form oxaloacetate. The enzyme is pyruvate carboxylase, which requires biotin and ATP. Oxaloacetate leaves the mitochondrion as malate or aspartate and is regenerated in the cytosol. Oxaloacetate is converted to phosphoenolpyruvate by a reaction that utilizes GTP and releases the same CO2 that was added in the mitochondrion. The remainder of the reactions occur in the cytosol.
A common intermediate in the conversion of glycerol and lactate to glucose is which one of the following?
(A) Pyruvate
(B) Oxaloacetate
(C) Malate
(D) Glucose-6-phosphate (E) Phosphoenolpyruvate
The answer is D. The only intermediate included on the list that the pathway of gluconeogenesis from glycerol has in common with the pathway of gluconeogenesis from lactate is glucose-6-phosphate. Glycerol enters gluconeogenesis as DHAP. Therefore, it bypasses the other compounds (pyruvate, oxaloacetate, malate and phosphoenolpyruvate) through which the carbons of lactate must pass on its pathway to glucose synthesis.
The pentose phosphate pathway generates which one of the following?
(A) NADH, which may be used for fatty acid synthesis.
(B) Ribose-5-phosphate, which may be used for the biosynthesis of ATP.
(C) Pyruvate and fructose 1,6-bisphosphate by the direct action of transaldolase and transketolase.
(D) Xylulose-5-phosphate by one of the oxidative reactions.
(E) Glucose from ribose-5-phosphate and CO2.
The answer is B. In the oxidative reactions of the pentose phosphate pathway, glucose is converted to ribulose-5-phosphate and CO2, with the production of NADPH. These reactions are not reversible. Ribose-5-phosphate and xylulose-5-phosphate are formed from ribulose- 5-phosphate by two of the nonoxidative reactions of the pathway. Ribose-5-phosphate is used for biosynthesis of nucleotides such as ATP. A series of reactions catalyzed by transketolase and transaldolase produce the glycolytic intermediates fructose-6-phosphate and glyceraldehyde-3-phosphate. Glucose is produced by gluconeogenesis in humans, and not directly by the hexose monophosphate shunt pathway.
Which one of the following metabolites is used by all cells for glycolysis, glycogen synthesis, and the hexose monophosphate shunt pathway? (A) Glucose-1-phosphate (B) Glucose-6-phosphate (C) UDP-glucose (D) Fructose-6-phosphate (E) Phosphoenolpyruvate
The answer is B. Glucose-6-phosphate is common to all pathways. It can be converted to glucose-1-phosphate for glycogen synthesis or go directly into the pentose phosphate pathway, or proceed through fructose-6-phosphate in glycolysis. UDP-glucose is formed from glucose- 1-phosphate and can be used to form glycogen, lactose, glycoproteins, and glycolipids.
A patient has a genetic defect that causes intestinal epithelial cells to produce disaccharidases of much lower activity than normal. Compared with a normal person, after eating a bowl of milk and oatmeal sweetened with table sugar, this patient will have higher levels of which one of the following? Choose the one best answer.
(A) Maltose, sucrose, and lactose in the stool
(B) Starch in the stool
(C) Galactose and fructose in the blood
(D) Glycogen in the muscles
(E) Insulin in the blood
The answer is A. In this patient, starch will be digested by salivary and pancreatic α-amylases to small oligosaccharides and maltose, but a lower than normal amount of glucose will be produced because of the deficiency of the brush border disaccharidases, which have maltase, isomaltase, sucrase, and lactase activity. Sucrose and lactose will not be cleaved. There will be more maltose, sucrose, and lactose in the stool and less monosaccharides in the blood and tissues. Insulin levels will be lower than normal, due to the reduced levels of glucose entering the blood. Muscle glycogen will not increase since there is less glucose in the circulation, and insulin, which is required for glucose entry into the muscle, may not be secreted under these conditions.
An infant, who was nourished by a synthetic formula, had a sugar in the blood and urine. This compound gave a positive reducing-sugar test but was negative when measured with glucose oxidase. Treatment of blood and urine with acid (which cleaves glycosidic bonds) did not increase the amount of reducing sugar measured. Which of the following compounds is most likely to be present in this infant’s blood and urine? (A) Glucose (B) Fructose (C) Sorbitol (D) Maltose (E) Lactose
The answer is B. Fructose gives a positive result in a reducing-sugar test and a negative result in a glucose oxidase test. It is a monosaccharide, and, so, is not cleaved by acid. Glucose gives a positive test result with the enzyme glucose oxidase. Sorbitol has no aldehyde or ketone group, and, thus, cannot be oxidized in the reducing-sugar test. Maltose and lactose are disaccharides that undergo acid hydrolysis, which doubles the amount of reducing sugar. This infant probably has benign fructosuria or the more dangerous condition, HFI. A galactose oxidase test would rule out the possibility that the sugar was galactose.
A 3-year-old girl has been a fussy eater since being weaned, particularly when fruit is part of her diet. She would get cranky, sweat, and display dizziness, and lethargy, after eat- ing a meal with fruit. Her mother noticed this correlation, and as long as fruit was withdrawn from her diet the child did not display such symptoms. The problems the girl exhibits when eating fruit is most likely due to which one of the following?
(A) Decreased levels of fructose in the blood
(B) Elevated levels of glyceraldehyde in liver
cells
(C) High levels of sucrose in the stool
(D) Elevated levels of fructose-1-phosphate in
liver cells
(E) Decreased levels of fructose in the urine
The answer is D. The patient has HFI, which is due to a mutation in aldolase B. Sucrose would still be cleaved by sucrase, thus it would not increase in the stool. Fructose would not be metabolized normally, therefore it would be elevated in the blood and urine. Aldolase B would not cleave fructose 1-phosphate, thus its levels would be elevated and the product, glyceraldehyde, would not be produced.
A 1-year-old child, on a routine well child visit, was discovered to have cataract formation in both eyes. Blood work demonstrated elevated galactose and galactitol levels. In order to determine which enzyme might be defective in the child, which intracellular metabolite should be measured? (A) Galactose (B) Fructose (C) Glucose (D) Galactose-1-phosphate (E) Fructose-1-phosphate (F) Glucose-6-phosphate
The answer is D. The child has a form of galactosemia. The elevated galactitol enters the lens of the eye, and is trapped. The difference in osmotic pressure across the lens of the eye leads
to cataract formation. Galactose is phosphorylated by galactokinase to galactose 1-phosphate, which reacts with UDP-glucose in a reaction catalyzed by galactose-1-phosphate uridylyl trans- ferase to form UDP-galactose and glucose 1-phosphate. An epimerase converts UDP-galactose to UDP-glucose. Deficiencies in either galactokinase (nonclassical) or galactose-1-phosphate uridylyl transferase (classical) result in galactosemia, with elevated levels of galactose and galactitol (reduced galactose) in the blood. An intracellular measurement of galactose-1-phosphate can allow a definitive diagnosis to be obtained (such levels would be nonexistent
if the defect were in galactokinase, and the levels would be greatly elevated if the galactose- 1-phosphate uridylyl transferase enzyme were defective).
A pregnant woman who has a lactase deficiency and cannot tolerate milk in her diet is concerned that she will not be able to produce milk of sufficient caloric value to nourish her baby. The best advice to her is which one of the following?
(A) She must consume pure galactose in order to produce the galactose moiety of lactose.
(B) She will not be able to breastfeed her baby because she cannot produce lactose.
(C) The production of lactose by the mammary gland does not require the ingestion of milk or milk products.
(D) She can produce lactose directly by degrading α-lactalbumin.
(E) A diet rich in saturated fats will enable her to produce lactose.
The answer is C. The woman will be able to breastfeed her baby because she can produce lactose from amino acids and other carbohydrates. She will not have to eat pure galactose,
or even lactose, to do so. Glucose, which can be provided by gluconeogenesis or obtained from the diet, can be converted to UDP-galactose (glucose → glucose-6-phosphate → glucose- 1-phosphate → UDP-glucose → UDP-galactose). UDP-galactose reacts with free glucose to form lactose. α-Lactalbumin is a protein that serves as the modifier of galactosyl transferase, which catalyzes this reaction. The amino acids of α-lactalbumin can be used to produce glucose, but the immediate products of α-lactalbumin degradation are not lactose. Carbohydrates cannot be synthesized from fats.
