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Solution to Brainstorm
2) The answer is –a) Fatty acid break down provides Acetyl co A that serves as a precursor for ketone bodies. In Diabetes Mellitus glucose utilization is impaired due to absolute or relative insulin deficiency. Fatty acid breakdown occurs to provide energy and the resultant excessive Acetyl co A enters the pathway of ketogenesis. Protein breakdown provides amino acids, 6 amino acids are ketogenic, while 14 are glucogenic. Hence protein breakdown contributes only a little towards formation of Acetyl co A. The major contribution is through fatty acid breakdown. Glycogenolysis and Gluconeogenesis produce glucose only.
3) The answer is –a) Galactosemia. The clinical manifestations are typical of classical Galactosemia. Bilateral cataract rules out the possibility of Von Gierke’s disease and hereditary fructose intolerance, although other symptoms are there in both these diseases. In juvenile diabetes mellitus, jaundice and hepatomegaly are not observed.
4) The answer is-d)- Lactate, the end product of glycolysis in erythrocytes and during intense exercise in skeletal muscles ,is mobilized through Cori cycle to liver to provide glucose by the process of gluconeogenesis. (Erythrocytes lack mitochondria so the end product of glycolysis is always lactate. The mode of glycolysis during intense exercise is anaerobic; hence lactate is formed as a result of glycolysis.
Alanine is transported to liver through Glucose Alanine cycle. Glycerol is also similarly transported but not from the erythrocytes or skeletal muscles, rather from the adipose tissues. Glycerol is a waste product in adipose tissues since without phosphorylation it can not be utilized and the phosphorylating enzyme glycerol kinase is absent in adipose tissues.
5) The answer is-b) Branching enzyme. During the process of glycogen synthesis, branching enzyme creates branch points and further elongation is carried out by Glycogen synthase . In its deficiency stored glycogen is abnormal in chemistry, in the form of long polysaccharide chains with few branch points, resembling the structure of Amylopectin, thus this defect is also called Amylopectinosis. Alpha Amylase is an enzyme for digestion of starch and glycogen. Debranching enzyme deficiency results in the accumulation of abnormal glycogen, There is inability to remove the branch points, the resultant structure resembles Limit dextrin , thus it is also called Limit dextrinosis
6) The answer is-d) The hydrolysis of starch is catalyzed by salivary and pancreatic amylases, which catalyze random hydrolysis of alpha (1- 4) glycoside bonds, yielding dextrins, and further hydrolysis yields a mixture of glucose, maltose, isomaltose (from the branch points in amylopectin) and maltotriose.
7) The answer is- c) Palmitate, a fatty acid with 16 carbon atoms, is not a substrate for gluconeogenesis. Even chain fatty acids, predominantly present in our body, yield Acetyl co A upon oxidation, which can not contribute towards gluconeogenesis. The Pyruvate to Acetyl co A conversion is irreversible and moreover both of the carbon atoms of Acetyl co A are lost in the TCA cycle in the form of CO2.Oddchain fatty acid do act as substrates of gluconeogenesis, since propionyl co A the product of their oxidation can enter TCA cycle through formation of Succinyl co A, hence can contribute towards Glucose production.
8) The answer is-d) Hypoglycemic coma occurs as a result of insulin over dosage in Type 1diabetes Mellitus. It is not observed in Type 2 diabetes. Weight gain can occur in both types, it is the result of treatment with insulin or certain hypoglycemic drugs.
9) The answer is- a) Hexokinase is a non specific enzyme, it can phosphorylate fructose as well as other sugars but it has high km(low affinity) for fructose. Glucose is the true substrate for this enzyme.Fructose-6-phosphatethe end product of Hexokinase reaction can enter glycolytic pathway to be utilized further, so it does not accumulate to produce the toxic effects. Liver Aldolase (Aldolase B) cleaves Fructose-1-P only, the product of fructokinase catalyzed reaction. Aldolase A, present in all the cells of the body cleaves Fructose 1,6 bisphosphate, the product of PFK-1 catalyzed reaction of glycolysis.
10) The answer is- b) Free glucose is released by the action of α-1-6-amyloglucosidaseenzyme, a component of debranching enzyme. Debranching enzyme has two components. α-[1 4] -α-[1 4] Glucan transferase and α-1-6-amyloglucosidase.Glucan transferase shifts the trisaccharide on a branch bound by α-[1- 4] linkage to the straight chain and joins by α-[1 4] linkage. The exposed branch point is hydrolyzed by α-1-6-amyloglucosidase enzyme. Both components are present on the same polypeptide chain. Glucose-6- phosphatase does produce free glucose but it is absent in skeletal muscles.
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