From the complete oxidation of glucose (glucose → 6CO2), how many total nucleotide triphosphates are yielded (be sure to deduct payback) as part of substrate level phosphorylation?
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From the complete oxidation of glucose (glucose → 6CO2), how many total
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- From the complete oxidation of glucose (glucose 6CO2), how many total nucleotide triphosphates are yielded (be sure to deduct payback) as part of substrate level phosphorylation?From the complete oxidation of glucose (glucose 6CO2), how many total nucleotide triphosphates are yielded (be sure to deduct payback) as part of substrate level phosphorylation? 灣During glycogen synthesis, glucose-1P is converted into a molecule called UDPG. This reaction also cleaves uridine triphosphate (UTP) forming uridine monophosphate and pyrophosphate (PPi). Provide four reasons why UTP can be used to power this reaction (no diagrams necessary).
- The reaction catalyzed by phosphorylase is readily reversible in vitro. At pH 6.8, the equilibrium ratio of orthophosphate to glucose 1-phosphate is 3.6. The value of ΔG°’ for this reaction is small because a glycosidic bond is replaced by a phosphoryl ester bond that has a nearly equal transfer potential. However, phosphorolysis proceeds far in the direction of glycogen breakdown in vivo. Suggest one means by which the reaction can be made irreversible in vivo.A solution of the enzyme hexokinase incubated at 45 °C lost 50% of its activity in 12 min, but when incubated at 45 °C in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 min. Suggest why thermal denaturation of hexokinase was retarded in the presence of one of its substrates. It is impossible for this result to be true. OA. Adding the substrate increases the weak forces that stabilize the enzyme. OB. The high concentration of substrate forms a barrier around the hexokinase. D. Adding the substrate results in protective covalent bonding.The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1
- 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]In what type of catalysis does Lys345 engage during the very first step of enolase's reaction cycle? Explain, in your own words, specifically why and how Lys345 falls into the category you chose.If the substrate for an enzyme catalyzed reaction contained a negative charge, which of the following amino acids would most likely be present in the active site to provide electrostatic destabilization of the ES complex? Leu Val O Asp O LYS QUESTION 15 What is the function of the histidine in the diagram of the mechanism at right with the substrate, a phenylalanine residue? (n) His Asp Nucleophilic affack by water stabilization of a metal ion covalent catalysis electrostatic attraction of the substrate O acid base catalysis 789 Mustly cloudy Click Save and Submit to save and submit. Click Save All Ansuwers to save all answers. here to search
- In bacteria, isocitrate dehydrogenase is regulated by phosphorylation of a specific Ser residue in the enzyme active site. X-ray structures of the phosphorylated and the nonphosphorylated enzyme show no significant conformational differences. How does phosphorylation regulate isocitrate dehydrogenase activity? O The phosphoryl group sterically hinders the substrate. O The phosphorylation bears a negative charge, which repels the substrate. O The phosphoryl group attracts positively charged Ca2* cations, which block the active site on the enzyme. None of the above.When enzyme solutions are heated, there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at 450C lost 50% of its activity in 12 minutes, but when incubated at 450C in the presence of a very large concentration of one of its substrates, it lost only 3% of its activity in 12 minutes. Suggest why thermal denaturation of hexokinase was retarded in the presence of one substratesA solution of [U 14C] glucose-1-phosphate (specific activity = 16,000 cpm/mmole) was incubated with glycogen and glycogen phosphorylase, an enzyme which adds glucose units on to glycogen. Radioactivity was incorporated into the glycogen primer at a rate = 2550 cpm/min. The rate of the enzymatic reaction in units of mmole glucose incorporated per minute is: (a) 0.016 mmol/min (b) 0.57 mmol/min (c) 0.16 mmol/min (d) 5.7 mmol/min