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Transcript
Chem 356 Structure and Function in Biochemistry Review Assignment #5 Answers 1. Soybeans and wheat contain starch, a polymer of glucose, which is broken down to glucose by microorganisms. The glucose is then broken down to pyruvate via glycolysis. Because the process is carried out in the absence of oxygen (i.e., it is fermentation), pyuvate is reduced to lactic acid and ethanol by the microorganisms. If oxygen were present, pyruvate would be oxidized to acetyl-CoA and then to CO2 and H2O. Some of the acetyl-CoA, however, would also be hydrolyzed to acetic acid (vinegar) in the presence of oxygen. Also, acetaldehyde could be oxidized to acetic acid in the presence of O2. 2. No; lactate dehydrogenase is required to recycle NAD+ from the NADH formed during the oxidation of glyceraldehyde 3-phosphate. 3. (a) There are two binding sites for ATP: a catalytic site and a regulatory site. Binding of ATP to an allosteric site inhibits PFK, either by reducing Vmax or increasing KM for ATP at the catalytic site. (b) The flux of intermediates through glycolysis is reduced when ATP is plentiful. (c) The graph indicates that the addition of ADP suppresses the inhibition of ATP. Because the adenine nucleotide pool is fairly constant, consumption of ATP leads to an increase in ADP levels. The data indicate that the activity of PFK is regulated by the [ATP]/[ADP] ratio, i.e., the energy charge of the cell. 4. (a) In muscle, glycogen is broken down to supply energy (ATP) via glycolysis. Glycogen phosphorylase catalyzes the conversion of stored glycogen to glucose-1phosphate, which is converted to glucose-6-phosphate. During strenuous muscle activity, skeletal muscle requires large quantities of glucose 6-phosphate. In the liver, the breakdown of glycogen is used to maintain a steady level of blood glucose between meals (glucose 6-phosphate is converted to free glucose). (b) In actively working muscle, ATP flux requirements are very high and glucose 1phosphate needs to be produced rapidly, requiring a high Vmax. 5. (a) ∆G’° = 3100 J/mol. The reaction: glycogenn + Pi ⇔ glycogenn-1 + glucose 1-P Keq’ = e-∆G’°/RT Keq’ = [glycogenn-1] [glucose 1-P]/[glycogenn] [Pi] K eq =e −( ( 3100 Jmol − 1 ) 8.315 Jmol −1 K −1 ( 298 K ) 0.286 = [glucose 1-P]/[Pi] [Pi]/[glucose 1-P] = 3.50 ) K ' eq = e −1.25107 Keq = 0.286 (b) The value of this ratio in the cell (>100:1) indicates that [glucose 1-P] is far below the equilibrium value. The rate at which glucose 1-P is removed (through entry into glycolysis) is greater than its rate of production (by the glycogen phosphorylase reaction). This indicates that metabolic flow is from glycogen to glucose 1phosphate, and that the glycogen phosphorylase reaction is probably the regulatory step in glycogen breakdown. 6. (a) Increases; (b) decreases; (c) increases. 7. Glucose is an allosteric inhibitor of phosphorylase a . Hence, crystals grown in the presence of glucose are in the T state. The addition of glucose 1-phosphate, a substrate, shifts the R ⇔ T equilibrium toward the R state. The conformational differences between these states are sufficiently large that the crystal shatters unless it is stabilized by chemical crosslinks. The shattering of a crystal caused by an allosteric transition was first observed by Haurowitz in the oxygenation of crystals of deoxyhemoglobin. 8. Water is excluded from the active site to prevent hydrolysis of the phosphate group from glucose 1-phosphate. The entry of H2O could lead to the formation of glucose rather than glucose 1-phosphate. A site-specific mutagenesis experiment is revealing in this regard. In glycogen phosphorylase, Tyr 573 is H-bonded to the 2’-OH of a glucose residue. The ratio of glucose 1-phosphate to glucose product is 9000:1 for the wild-type enzyme, and 500:1 for the Phe 573 mutant. Molecular modelling suggests that a H2O molecule occupies the site normally filled by the phenolic OH of Tyr 573 and occasionally attacks the oxocarbonium ion intermediate to form glucose. 9. (a) Muscle phosphorylase b will be inactive even when the AMP level is high. Hence, glycogen will not be degraded unless phosphorylase is converted into the a form by hormone-induced or Ca2+-induced phosphorylation. (b) Phosphorylase b cannot be converted into the much more active a form. Hence, the mobilization of liver glycogen will be markedly impaired. (c) The elevated level of the kinase will lead to the phosphorylation and activation of glycogen phosphorylase. Little glycogen will be present in the liver because it will be persistently degraded. (d) Protein phosphatase 1 will be continually active. Hence, the level of phosphorylase b will be higher than normal, and glycogen will be less readily degraded. (e) The absence of glycogenin will block the initiation of glycogen synthesis. Very little glycogen will be synthesized in its absence. 10. The deficiency is in the branching enzyme (Type IV, Andersen’s disease). The glycogen has very few branches and hence very long α(1→4)-linked branches. Normal glycogen has branches every 8 – 12 residues, yielding a G1P/glucose ratio of approximately 10 on glycogenolysis.