NADH can act as an inhibitor of isocitrate dehydrogenase and a-ketoglutarate dehydrogenase. This is an example of:
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A:
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- A genetic defect in coagulation factor IX causes hemophilia b, a disease characterized by a tendency to bleed profusely after very minor trauma. However, a genetic defect in coagulation factor XI has only mild clinical symptoms. Explain this discrepancy in terms of the mechanism for activation of coagulation proteases shown in Box.Intramitochondrial ATP concentrations are about 5 mM, and phosphate concentration is about 10 mM. If ADP is five times more abundant than AMP, calculate the molar concentrations of ADP and AMP at an energy charge of 0.85. Calculate AG for ATP hydrolysis at 37 °C under these condi- tions. The energy charge is the concentration of ATP plus half the concen- tration of ADP divided by the total adenine nucleotide concentration: [ATP] + 1/2[ADP] [ATP] + [ADP] + [AMP]A 65-year-old man with severe atherosclerotic coronary artery disease comes to the emergency department because of a 12-hour history of chest pain. Plasma activity of the MB isozyme of creatine kinase (MB-CK) is markedly increased. Which of the following processes is the most likely explanation for the increased plasma MB-CK? (A) Cell membrane damage (B) Endoplasmic reticulum dilation (C) Mitochondrial swelling (D) Polysome dissociation (E) Sodium pump dysfunction
- Kinases are enzymes that transfer a phosphoryl group from a nucleosidetriphosphate. Which of the following are valid kinase-catalyzed reactions? (a) ADP + CMP → AMP + CDP. (b) AMP + ATP → 2 ADP.Many phospholipase C enzymes contain pleckstrin homology (PH) domains. What is the function of these domains? (a) Bind DNA, (b) Phosphorylate proteins, (c) Phosphatase proteins, (d) Bind proteins, (e) Bind phosphatidylinositol. I note that PH domain interacts with 3’phosphoinositides, contributing to recruitment of AKT to the plasma membrane, thus thinking of the possible answer (e) which is phosphatidylinositol (ie. PI) of a family of lipids consisting PIP3 or PIP2 etc. However, I wonder if “3’phosphoinositides” is the same as “PIP3”.As an alternative, I note that PH domain is a protein domain that occurs in a wide range of proteins involved in intracellular signaling, I tend to choose (d) as an answer also.I should be grateful for receiving your expert advice as to which (a) to (e) is the best answer.Intramitochondrial ATP concentrations are about 5 mM, and phos- phate concentration is about 10 mM. If ADP is five times more abundant than AMP, calculate the molar concentrations of ADP and AMP at an energy charge of 0.85. Calculate AG for ATP hydrolysis at 37 °C under these conditions. The energy charge is the concentra- tion of ATP plus half the concentration of ADP divided by the total adenine nucleotide concentration: [ATP] + 1/2[ADP] [ATP] + [ADP] + [AMP]
- If intracellular [ATP] = 5 mM, [ADP] = 0.5 mM, and [Pi] = 1.0 mM, calculate the concentration of AMP at pH 7 and 25°C under the condition that the adenylate kinase reaction is at equilibrium.TPCK and TLCK are irreversible inhibitors of serine proteases. One ofthese inhibits trypsin and the other chymotrypsin. Which is which? Explainyour reasoning. Suggest the effects of each of the following mutations on the physiologicalrole of chymotrypsinogen:(a) R15S(b) C1S(c) T147SSketch curves for reaction velocity versus [fructose-6-phosphate] for the phosphorylated and nonphosphorylated forms of PFK-2 in liver.
- The text states that ATP is synthesized primarily by energy metabolism,whereas other nucleoside triphosphates are formed from the action of nucleoside diphosphate kinase. What additional pathway exists for GTP synthesis?The enzyme lysozyme kills certain bacteria by attacking a sugar called N-acetylglucosamine (NAG) in their cell walls. At an enzyme concentration of 2 × 10-6 M, the maximum rate for substrate (NAG) reaction, found at high substrate concentration, is 1 × 10-6 mol L¯'s1. The rate is reduced by a factor of 2 when the substrate concentration is reduced to 6 x 10-6 M. Determine the Michaelis-Menten constants Km and k, for lysozyme.Residue Asn 204 in the glucose binding site of hexokinase IV was mutated, in two separate experiments, to either Ala or Asp. The Asn → Ala mutant had a KM nearly 50-fold greater than the wild-type enzyme, and the Asn → Asp mutant had a 140-fold greater KM value than the wild-type enzyme. These mutations impact the intermolecular interactions between the enzyme and the glucose substrate.The amide functional group of the Asn side chain can form with the hydroxyl groups of the glucose substrate and can potentially function as either a . The methyl group of Ala cannot participate in hydrogen bond formation, which explains the in glucose affinity as indicated by the higher KM for the mutant enzyme. The side chain of Asp could potentially serve as a , but…