Carbohydrate Metabolism and Regulation Practice Questions Flashcards
Precursors for DNA are especially important for actively dividing cells, such as cancer cells. Since cancer cells have greater requirements for DNA synthesis than other cells, the metabolic pathway needed for producing five carbon products is a sensible target to impair cancer cell growth. Based on this thinking, which enzyme would make a useful ‘metabolic target’ for inhibitor design to prevent five-carbon sugars from being produced?
A) galactose 4-epimerase
B) glucose-6-phosphate dehydrogenase
C) phosphoenolpyruvate carboxykinase (PEP-CK)
D) alcohol dehydrogenase
E) glucose-6-phosphatase
Correct answer is “B”. The metabolic pathway which produces five-carbon monosaccharides is the Pentose Phosphate Pathway. The rate-limiting enzyme, and the committed reaction step is the same for the pathway: Glucose-6- phosphate dehydrogenase. This enzyme is ‘product inhibited’ by NADPH, the other significant product of the pathway.
A patient who has been without food for four days has reduced glucose consumption in all tissues. This adaptation serves to conserve glucose for essential activities. The physician treating the patient after the fast provides an iv containing dextrose to rapidly replenish blood glucose. Which enzyme is allosterically inhibited in liver cells by a high intracellular free glucose concentration in this patient?
A. phosphoglucose isomerase
B. fructose-1,6-bisphosphatase
C.pyruvate carboxylase
D. glycogen phosphorylase
E. glucose-6-phosphatase
Correct answer is “D”. Glycogen metabolism is regulated by intracellular free glucose. This glucose concentration inside of cells directly reflects the plasma glucose concentration because the GLUT2 transporters act by facilitated diffusion transport across cell plasma membranes. The other enzymes are not regulated by free glucose. Phosphoglucose isomerease is not regulated and maintains equilibrium based on ‘mass action’ of substrate concentrations. Fructose-1,6-bisphosphatase is an enzyme which is regulated by Fuctose-2,6-bis-phosphate concentration. Pyruvate carboxylase is regulated allosterically by acetyl-CoA and ATP (activations). While Glucose-6- phosphatase is inhibited by free glucose at high intracellular concentration (~ 100 mM), this is too high to be physiologically relevant.
Why does it ‘makes sense’ that a high acetyl-CoA concentration serves to allosterically activate pyruvate carboxylase in liver to activate gluconeogenesis?
A. Acetyl-CoA can replace pyruvate and lactate as substrates for gluconeogenesis.
B. Acetyl-CoA is a precursor of glycerol which is a substrate for gluconeogenesis.
C.Acetyl-CoA is a substrate for the pyruvate carboxylase enzyme.
D.Acetyl-CoA cannot enter the Krebs (TCA) Cycle when it serves to activate pyruvate carboxylase.
E.Acetyl-CoA signals the availability of a substrate for ATP production needed for gluconeogenesis.
Correct answer “E”. The presence of acetyl-CoA in liver indicates that sufficient energetic reserves for glucose biosynthesis are being manufactured. Gluconeogenesis cannot operate without a ready supply of glucose.
Insulin has dramatic effects on the activity of enzymes required for the catabolism (breakdown) of glycogen. A key liver enzyme which responds to high extracellular insulin is glycogen synthase kinase (GSK). How does liver GSK activity change after a big meal, following insulin release from the pancreas?
A. GSK is activated producing the inhibitor, fructose 2,6 bisphosphate
B. GSK is activated to reduce the activity of glycogen phosphorylase ‘a’
C.GSK is activated to increase the concentration of glycogen synthase ‘a’
D.GSK is inhibited, preventing the conversion of glycogen synthase ‘a’ into glycogen synthase ‘b’
E.GSK is inhibited, resulting in increased glucose transport (GLUT4) activity in uptake of blood glucose
Correct answer “D”. Insulin acts on Akt protein (protein kinase B) which in turn modifies and inhibits basal activity of GSK. Therefore, when insulin is present the net effect is less conversion of glycogen synthase ‘a’ into glycogen synthase ‘b’, and more of the available glucose is stored as glycogen.
What is the role of PEP carboxykinase in gluconeogenesis?
A. It converts pyruvate to oxaloacetate
B. It is not involved in gluconeogenesis
C. It converts phosphoenol pyruvate to fructose 1,6 bisphosphate
D. It adds a carbon to phosphoenol pyruvate
E. It converts oxaloacetate to phosphoenol pyruvate
Correct Answer “E”. This is simply recollection of the second step of gluconeogenesis.
Which of the following is most likely used as a substrate to make glucose using the gluconeogenic pathway?
A. ATP
B. Fructose 1-phosphate
C. Acetyl CoA
D. Palmitate
E. Pyruvate
Correct answer “E”. ATP is required as an energy source, and thus is a substrate for some gluconeogenic enzymes, but is not really a substrate for the process. Fructose-1-phosphate can feed into glycolytic intermediates and could, in principle, be a gluconeogenic precursor, but this rarely happens in practice as fructose is typically present in the fed state and glycolysis active. Acetyl CoA cannot be a gluconeogenic precursor as it is effectively degraded early in the TCA cycle, and there is no enzymatic mechanism to convert it to another gluconeogenic precursor. Palmitate is a fatty acid that would be converted to acetyl CoA and therefore also cannot be converted to a gluconeogenic precursor. Pyruvate (answer E), is one of the principal gluconeogenic precursors.
Which of the following steps in glycolysis is bypassed by fructose 1-Phosphate?
A. All steps
B. None
C. PFK-1
D. Aldolase B
E. PFK-2
Correct answer “C”, phosphofructokinase 1. Aldolase is still used, and PFK-2 is not a glycolytic enzyme. Since some glycolytic enzymes are used, neither A nor B are correct.
High levels of sorbitol can lead to the formation of Advanced Glycation End Products and affect intracellular osmolarity, both of which can damage tissues such as the retina. Which of the following pathways produces sorbitol?
A. Polymerization of tyrosine
B. Pentose phosphate pathway
C. Gluconeogenesis
D. Polyol pathway
E. Glycolysis
Correct answer “D”, the polyol pathway. This is the pathway that converts glucose to sorbitol to fructose.
Which of the following results in activation of glycogen synthase?
A. cGMP
B. cAMP
C. Phosphorylation
D. Inhibition of AKt/PKB
E. Dephosphorylation
Correct answer “E”, dephosphorylation. This is the most direct mechanism, however, D is also correct as described in question 4, but E is the better answer.
The deficiency of which enzyme results in von Gierke glycogen storage disorder?
A. Glycogen phosphorylase
B. Glucosidase
C. Glucose 6-phosphatase
D. Glucokinase
E. Glycogen synthase
Correct answer “C”, glucose-6-phosphatase. This is one of the glycogen storage disorders.
Fructose 1,6 bisphosphatase is one of three unique enzymes in gluconeogenesis. Which small molecule can inhibit this enzyme?
A. Fructose 6-phosphate
B. Galactose
C. Glucose 6-phosphate
D. Calcium
E. Fructose 2,6 bisphosphate
Correct answer “E”, fructose 2,6 bisphosphate. This small regulatory molecule directly inhibits fructose 1,6 bisphosphatase, competitively. This is part of the insulin/glucagon regulation of the PFK-1/F16BPase regulatory bubble. The other choices do not inhibit this enzyme.
Patients who are deficient in the enzyme galactose-phosphate uridyl transferase will not be able to carry out which of the following conversions?
A. Lactose to lactulose
B. Lactose to glucose
C. Lactose to galactose
D. Galactose to glucose
E. Galactose to galactitol
Correct answer “D”, galatose to glucose. Galactose-phosphate uridyl tranferase makes UDP-galactose as part of the pathway that converts the galactose to glucose. It is not found in any of the other conversions listed.
The rate limiting step for glucose entry to cells differs depending on the tissue. What are the rate limiting steps in muscle versus liver and why are they important?
In muscle, hexokinase is lower Km, below the circulating concentration, so it could be limiting if it was saturated constantly. However, the rate of glucose entry is limited by by the Glut1 transporters and can be increased by recruitment of the Glut4 transporters. These essentially determine the rate of uptake into muscle. Most glucose entering becomes immediately trapped by hexokinase mediated phosphorylation.
In liver, both glucokinase and Glut 2 transporters are high Km to reflect increased activity in high glucose concentrations. In addition, glucokinase itself is regulated by glucose, F6P, F1P, and AMP. This is the key regulated step for glucose entry into cellular metabolism.
A newborn two weeks of age was brought to the pediatrician due to frequent vomiting, lethargy, and diarrhea. Family history revealed that the child never ate well and had only been breast fed. The physical examination revealed an enlarged liver and jaundice. Suspecting a congenital enzyme defect, the patient was referred to an ophthalmologist for a slitlamp exam. The result is shown in the image.
Which of the following enzymes is most likely defective?
A. Fructose-1,6-bisphosphate
B. Galactose-1-phosphate uridyltransferase
C. Galactokinase
D. Glycogen Synthase
E. Fructokinase
Correct answer “B”, galactose-1-phosphate uridyl transferase. A deficiency in galactose-1-phosphate uridyltransferase leads to accumulation of galactose in the blood. In the eye this leads to formation of galactitol with an accompanying osmotic imbalance across the lens, and formation of cateracts. The high galactose levels block phosphoglucomutase in the liver, and leads to insufficient glycogen degradation. This can be distinguished from the non-classical galactosemia which is a defect in galactokinase. That has the issues with cataract formation but no accompanying feeding and liver problems. The other choices will not produce the symptoms described, including cataract formation.
An infant about three months of age is switched from breast feeding to formula and fruit juices. After the change in diet the infant vomits and has severe hypoglycemia after eating. Removing the juices from the diet reduced the severity of the symptoms. A congenital defect in metabolism was considered, which could explain the symptoms of hypoglycemia. What is the result of this defect?
A. Fructose inhibition of debranching enzyme.
B. Galactose-1-phosphate inhibition of glycogen phosphorylase.
C. Fructose-1-phosphate accumulation, causing phosphate depletion.
D. Fructose-6-phosphate inhibition of glycogen phosphorylase.
E. Galactose inhibition of aldolase B.
Correct answer “C”. The symptoms are consistent with hereditary fructose intolerance which leads to an accumulation F1P. This accumulation effectively traps the phosphate pool so it is unavailable for ATP synthesis and unavailable for glycogen phosphorylase as a substrate. Thus, there is little glycogen degradation resulting in fasting hypoglycemia. The key to answering the question is noting that the infant thrived on breast milk. Thus, galactose metabolism is not implicated (ruling out B and E). The switch to juices introduced fructose. A and D are also incorrect answers; although G6P is an important inhibitor of glycogen phosphorylase, F6P is not.
