The inhibition of PFK-1 by ATP diminishes when the ADP concentration is high, as shown in the graph What explains this observation? ADP levels regulate the association of the individual PFK-1 monomers. The [ATP]/[ADP] ratio regulates PFK-1 activity. ADP outcompetes ATP as a phosphoryl donor for PFK-1 at high concentrations. O Binding of ADP to the catalytic site increases PFK-1 activity.
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- Using the ActiveModel for enoyl-CoA dehydratase, give an example of a case in which conserved residues in slightly different positions can change the catalytic rate of reaction.Distinguishing the Mechanisms of Class I and Class I Aldolases Fructose bisphosphate aldolase in animal muscle is a class 1 aldolase, which forms a Schiff base intermediate between substrate (for example. fructose-1, 6-bisphosphate or dihydroxyacetone phosphate) and a lysine at the active site (see Figure I8.12). The chemical evidence for this intermediate conies from studies with aldolase and the reducing agent sodium borohydride, NaBH4. Incubation of the enzyme with dihydroxyacetone phosphate and NaBH4 inactivates the enzyme. Interestingly, no inactivation is observed if NabH4 is added to the enzyme in the absence of substrate. Write a mechanism that explains these observations and provides evidence for the formation of a Schiff base intermediate in the aldolase reaction.Fructose 2,6-bisphosphate is a potent stimulator of phosphofructokinase. Explain how fructose 2,6-bisphosphate might function in the concerted model for allosteric enzymes.
- Phosphofructokinase-1 (PFK-1) catalyzes the transfer of a phosphoryl group from ATP to fructose 6-phosphate, Shown is a graph of the rate of PFK-1 vs. [ATP], Concisely explain the shape of the graph. 2 (ATP) (mmol l) PFK-1 reaction rateIn class, I mentioned that fructose is metabolized differently in the liver compared to glucose. Refer to the figure shown below to calculate the number ofATPs you would expect from the metabolism of fructose in the liver. Show your work! Fructokinase Fructose Fructose-1-P АТР ADP Aldolase B Dihydroxy- acetone phosphate Glyceraldehyde АТР Triose kinase Triose phosphate isomerase ADP 4 - Glyceraldehyde-3-P Glycolysis Руruvate Acetyl-CoA Fatty acids and triglyceridesUDP-glucose pyrophosphorylase catalyzes the removal of a pyrophosphate group from UTP as it synthesizes UDP-glucose. Why is this necessary, from a biochemical perspective? Both of the phosphate groups from UTP are needed to form UDP-glucose. There is no particular reason: wasteful reactions just happen sometimes. The pyrophosphate (after hydrolysis) is required to free up more phosphate for the synthesis of ATP by oxidative phosphorylation. The subsequent hydrolysis of pyrophosphate is a highly exergonic reaction, which pulls the equilibrium over towards UDP-glucose. Pyrophosphate is an allosteric activator of glycogen synthase, so this helps glycogen synthesis to proceed at a faster rate.
- Consider the complete oxidation of one mole of simple TAG containing behenic acid residues (22:0). I. For one mole of the fatty acid residue, determine the following: b. What is the number of ATP yield obtained from NADH coming from the complete β-oxidation of the fatty acid residueThe glucose/glucose-6-phosphate substrate cycle involves distinct reactions of glycolysis and gluconeogenesis that interconvert these two metabolites. Assume that under physiological conditions, [ATP] = [ADP]; [P;] = 1 mM. Consider the glycolytic reaction catalyzed by hexokinase: ATP + glucose ADP + glucose-6-phosphate AG = - 16.7 kJ/mol (a) Calculate the equilibrium constant (K) for this reaction at 298°K, and from that, calculate the maximum [glucose-6-phosphate]/ Iglucose] ratio that would exist under conditions where the reaction is still thermodynamically favorable. (b) Reversal of this interconversion in gluconeogenesis is catalyzed by glucose-6-phosphatase: glucose-6-phosphate + H20 = glucose + P AG" = -13.8 kJ/molIntramitochondrial 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]
- In working skeletal muscle under anaerobic conditions, glyceraldehyde 3-phosphate is converted to pyruvate (the payoff phase of glycolysis), and the pyruvate is reduced to lactate. Write balanced biochemical equations for all the reactions in this process, with the standard free-energy change for eachreaction. Then write the overall or net equation for the payoff phase of glycolysis (with lactate as the end product), including the net standard free-energy change.Decylic acid is a saturated fatty acid that occurs naturally in coconut oil and palm kernel oil. Calculate the net ATP yield when decylic acid undergoes complete B oxidation. The formula of decylic acid is shown below: (Given: The oxidation of one NADH yields 2.5 ATP; the oxidation of one FADH2 yields 1.5 ATP; and the oxidation of one acetyl CoA yields 10 ATP.) O 50 ATP O 52 ATP 66 ATP OH O 64 ATPConsider the complete oxidation of one mole of simple TAG containing behenic acid residues (22:0). I. For one mole of the fatty acid residue, determine the following: c. What is the number of ATP yield obtained from FADH2 coming from the complete β-oxidation of the fatty acid residue