Question 1.
- A – Wrong. Glucagon is not an enzyme, but a peptide hormone that increases blood glucose levels through glycogenolysis, stimulates gluconeogenesis and ketogenesis. The main target tissue is the liver.
- B – Wrong. Glucose-6-phosphate dehydrogenase is the enzyme that initiates the breakdown of glucose pentose phosphate cycle.
- C – Wrong. Glucokinase is a liver enzyme that phosphorylates glucose in the hepatocyte cytoplasm to glucose-6-phosphate. Its activity is greatly influenced by the level of glucose.
- D – Wrong. Hexokinase is an enzyme that catalyzes the phosphorylation of glucose to glucose-6-phosphate in extrahepatic tissues. Hexokinase activity for a very low Km value is not affected by blood glucose concentration.
- E - Correct answer. Glucose-6-phosphatase hydrolyzes glucose-6-phosphate to yield free glucose. It is present in the liver and kidney. It is missing (!) in muscle and fat tissue. Its presence enables the supply of glucose from the tissues to the blood circulation.
Question 2.
- A – Wrong. Under the catalysis of alanine aminotransferase, alanine can be metabolized to pyruvate, which then, with the participation of pyruvate carboxylase, Mg2+, ATP and CO2, gives rise to oxaloacetate and after reduction to malate (everything takes place in mitochondria). This crosses the mitochondrial membrane (it is impermeable to oxaloacetate), is oxidized again in the cytoplasm to oxaloacetate, and with the participation of phosphoenolpyruvate carboxykinase, GTP and CO2, it changes to phosphoenolpyruvate.
- B – Wrong. Pyruvate cannot be converted directly to phosphoenolpyruvate (pyruvate kinase can only work in one direction for thermodynamic reasons), but via pyruvate carboxylase and then phosphoenolpyruvate carboxykinase via oxaloacetate, malate and oxaloacetate again, phosphoenolpyruvate is formed in the gluconeogenetic pathway.
- C – Wrong. After dehydrogenation by lactate dehydrogenase, lactate gives pyruvate, which can be changed to phosphoenolpyruvate by a "detour", i.e. through gluconeogenesis enzymes (pyruvate carboxylase, phosphoenolpyruvate carboxykinase).
- D – Correct. Palmitate undergoes β-oxidation and gives rise to 8 acetylCoAs at the end, which can be further oxidized in the citrate cycle to CO2 and H2O already at the first stages of the cycle, so there are not too many carbon residues left converted into a sufficient amount of oxaloacetate needed for gluconeogenesis. This is because if oxaloacetate is removed from the cycle and not replaced, the cell's capacity to make much-needed ATP would be limited. Thus, acetyl-CoA is not a suitable substrate for gluconeogenesis; this is pyruvate arising from amino acids (alanine), which provides enough oxaloacetate through a reaction catalyzed by pyruvate carboxylase. In this way, gluconeogenesis does not interfere with metabolism in the citric acid cycle.
- E – Wrong. Oxaloacetate is an important substrate of gluconeogenesis. It is formed from pyruvate by the action of pyruvate carboxylase. However, it is impermeable across the mitochondrial membrane, so it must be converted to malate, which is permeable; in the cytoplasm it is again dehydrogenated to oxaloacetate, which gives phosphoenolpyruvate with the participation of phosphoenolpyruvate carboxykinase; the latter provides glucose in the gluconeogenetic pathway via triosaphosphates and hexosaphosphates.
Question 3.
- A – Wrong. Conversely, insulin stimulates hepatic glucokinase activity and phosphorylates intracellular glucose to glucose-6-phosphate. It thereby reduces the intracellular level of glucose and thus indirectly promotes the flow of glucose from the circulation to the hepatocyte. Insulin promotes hepatic glycolysis. On the contrary, it inhibits the activity of liver glucose-6-phosphatase, which releases glucose from the liver.
- B – Correct. Insulin promotes glucose utilization in muscle by promoting the translocation of glucose via a transport system from the circulation into the cell, further stimulating the formation of glucose-6-phosphate, which can be isomerized to glucose-1-phosphate, a precursor for glycogen synthesis.
- C – Wrong. Insulin has the opposite effect, it stimulates lipogenesis and is a very effective inhibitor of lipolysis.
- D – Wrong. Insulin, on the other hand, has the effect of promoting the synthesis of glycogen from glucose.
- E – Wrong. Conversely, insulin stimulates the entry of amino acids into cells, stimulates protein synthesis (anabolic effect).
thumb|right|300px|Coriho cyklus
Question 4.
- A – Correct. The role of the Cori cycle (also the lactic acid cycle) is to transfer excess reducing equivalents from the muscles to the liver. In fast-working muscles (during contraction), the supply of O2 to the mitochondria is not fast enough to reoxidize the NADH formed during glycolysis. When the concentration of cytoplasmic NADH rises, lactate dehydrogenase catalyzes the transfer of reducing equivalents from NADH to pyruvate to form lactate. This penetrates from the muscles into the circulation and is taken up by the liver. In them, lactate dehydrogenase transfers electrons back to NAD to form pyruvate. Pyruvate is then converted into glucose in the gluconeogenetic pathway, which leaves the liver and enters the muscles from the circulation. This cycle allows skeletal muscle to work under anaerobic conditions for a short period of time.
- B – Wrong. Glucose-6-phosphate dehydrogenase is the initial enzyme that initiates the pentose phosphate pathway of glucose metabolism.
- C – Wrong. Pyruvate dehydrogenase converts pyruvate to acetyl CoA, which enters the citric acid cycle. Before that, the pyruvate of the cytoplasm must be transferred to the mitochondria. There, pyruvate undergoes oxidative decarboxylation catalyzed by a multienzyme complex collectively called the pyruvate dehydrogenase complex.
- D – Wrong. Glucokinase catalyzes the phosphorylation of glucose in the liver to glucose-6-phosphate, thus initiating the glycolytic pathway.
- E – Wrong. Hydroxymethylglutaryl-CoA reductase is a key enzyme in cholesterol synthesis.
Question 5.
- A – decreases
- B – rises
- C – decreases
- D – rises
- E – decreases
Question 6.
- C – Correct. By binding to the surface membrane receptor, glucagon activates the adenylate cyclase system and thus increases the content of cAMP in the cell (cAMP acts as a "second messenger"), thereby allosteric activation of cAMP-dependent protein kinase, which changes phosphorylase b into active phosphorylase a, which cleaves glycogen in the liver (glycogenolytic effect of glucagon in the liver; not in the muscles). At the same time, glycogen synthesis is inactivated (it is inactivated by protein phosphatase 1). After eating, glucose enters the vena portae, glucosemia rises; it reaches its maximum in 30-60 minutes. In a healthy individual, the value does not exceed 10 mmol/l (usually 7-8 mmol/l).
Question 7.
- A –Correctly. Absolute or relative lack of insulin prevents the transport of glucose across the cell membrane (except for the hepatocyte), reduces the activation of hexokinase and thus its phosphorylation to glucose-6-phosphate.
- B – Wrong. The site of gluconeogenesis is mainly the liver, not the muscles.
- C – Correct. Hepatic gluconeogenesis is significantly involved in hyperglycemia in diabetes. The main substrate of gluconeogenesis is amino acids (alanine), which is converted to pyruvate and to glucose by the gluconeogenetic pathway. The reaction requires energy, which is taken from the oxidation of fatty acids (lipolysis). Increased production of acetyl CoA (β-oxidation of fatty acids) allosterically activates pyruvate carboxylase (an enzyme that initiates the reverse conversion of pyruvate to phosphoenolpyruvate) and, conversely, inhibits pyruvate dehydrogenase, which is necessary for the entry of pyruvate (or acetyl CoA) into the citric acid cycle. The increased concentration of alanine and fatty acids inhibits the conversion of phosphoenolpyruvate to pyruvate and thereby inhibits glycolysis and promotes gluconeogenesis.
- D – Wrong. Glucose transfer to the hepatocyte is not dependent on insulin. In other cells, this protein glucose carrier facilitates the translocation of glucose across the membrane and requires insulin. Indirectly, however, the lack of insulin affects the entry of glucose from the circulation into the hepatocyte in diabetes by insufficient activation of glucokinase, which under normal circumstances increases the gradient between "free" glucose outside and inside the liver cell.
- E – Wrong. An increase in the renal threshold for glucose occurs later in the course of diabetes and is a consequence of hyperglycemia rather than its primary cause. However, the increased renal threshold in advanced diabetes may contribute to the discrepant finding of high hyperglycemia with relatively little glycosuria.
- F – Correct. In the absence of insulin, the effect of glucagon prevails, which promotes hyperglycemia through increased hepatic glycogenolysis and stimulation of gluconeogenesis.
- G – Wrong. Conversely, a deficit of insulin (due to a relative excess of glucagon) promotes lipolysis, which is needed as an energy source for gluconeogenesis.
Question 8.
- A – Wrong. It is not about reduced breakdown, but about increased production of ketone bodies in the liver.
- B – Correct. The combination of a lack of insulin (or its effect) with an increased effect of glucagon leads to insufficient utilization of glucose, gluconeogenesis and stimulation of lipolysis (increasing β-oxidation of fatty acids and overproduction of acetyl CoA in the liver).
- C – Wrong. The non-enzymatic conversion of acetoacetate to acetone, on the other hand, leads to the possibility of excretion of ketone bodies also through the lungs (a fruity odor can be felt in the breath).
- D – Correct. Increased catabolism of fatty acids in the liver, which get there from lipolysis from fat depots, leads to an overproduction of acetyl CoA, which cannot be broken down in the citric acid cycle (lack of oxaloacetate formed during glycolysis); from the acetyl-CoA molecule, it condenses to acetoacetyl-CoA, from which acetoacetate is formed; in the absence of O2 it can be reduced to hydroxybutyrate. Both carboxylic acids are moderately strong (pK around 4) and significantly affect blood pH in the sense of acidemia.
- E – Correct. Increased production of acetyl-CoA in the liver leads to accumulation of the condensation product - acetoacetate. It is therefore an excessive production of ketone bodies in the liver, not a reduced breakdown.
- F – Correct. The formation of hydroxymethylglutaryl-CoA (HMG-CoA) is one of the pathways from which acetoacetate is formed. Acetoacetyl-CoA takes on another molecule of acetyl CoA; this creates HMG-CoA, which cleaves acetyl-CoA through the action of HMG-CoA-lyase, leaving "free" acetoacetate. This mode of formation of acetoacetate from aceto-acetyl-CoA appears to be more significant than simple deacylation.
Question 9.
- A – Right. Prolonged significant hyperglycemia in noninsulin dependent diabetes induces osmotic diuresis. In a state of insufficient water supply, especially in the elderly after a stroke or infection, ECT hyperosmolarity occurs. Residual insulin secretion is enough to prevent excessive ketogenesis, but not enough to affect hyperglycemia. Therefore, hyperosmolar coma occurs more often in noninsulin-dependent diabetes; has a high mortality rate (up to 50%).
- B – Wrong. Lack of insulin combined with excess glucagon is the cause of ketoacidotic coma in IDDM. In NIDDM, residual insulin secretion will prevent increased ketone body production but not hyperglycemia.
- C – Wrong. Insulin does not affect glucose utilization by the brain.
- D – Wrong. Glycation of collagen in the basement membrane of the glomeruli does indeed cause increased permeability, which first leads to an increase in albuminuria as the first laboratory sign starting diabetic nephropathy; however, it is not the cause of osmotic diuresis.
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