This a reaction going on in your muscle cells right this very minute: Exhibit 6AThis is a reaction going on in your muscle cells right this very minute:450s-1dihydroxyacetone phosphate500 s-1glyceraldehyde-3-phosphateThe enzyme triose phosphate isomerase catalyzes this reaction in the forward direction as part of glycolytic pathway. It follows simple Michaelis-Menten kinetics:E+Sk₁k₁k₂ESE+PTypical cellular concentrations: triose phosphate isomerase = 0.1 nM=dihydroxyacetone phosphate 5 μM glyceraldehyde-3-phosphate = 2 μMRefer to Exhibit 6A. "Hindrate" is an inhibitor of triose phosphate isomerase. When it is added to cells at a concentration of 0.1 nM, the enzyme's KM for the substrate is unchanged, but theapparent Vmax is altered to 50 nM/sec.a. This is a noncompetitive inhibitor.b. This is an irreversible inhibitor.c. This is an uncompetitive inhibitor.d. This is a competitive inhibitor.

● Enzyme Inhibition is of two types: A) Reversible and B) Irreversible.● Reversible enzyme…

Answered: Exhibit 6A This is a reaction going on…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter18: Glycolysis

Section: Chapter Questions

Problem 11P

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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.
Many of the mechanisms involved in the regulation of glycogen metabolism involve covalent modification of enzymes. In particular, reversible phosphorylation/ dephosphorylation plays a central role.(a) Estimate the ΔG°′ for the dephosphorylation of phosphorylated Ser-14of glycogen phosphorylase (catalyzed by phosphoprotein phosphatase 1).Explain your reasoning.(b) Estimate the ΔG°′ for the phosphorylation of Ser-14 of glycogen phosphorylase by ATP (catalyzed by phosphorylase kinase). Explain yourreasoning.(c) Would you expect both reactions to be favorable under physiologicalconditions (i.e., to possess a negative ΔG)?
ATP is a (+) allosteric effector, and CTP is a (-) allosteric effector of theenzyme ATCase. Both of these heterotropic effectors bind to the regulatorysubunits on ATCase. The substrates of ATCase, aspartate and carbamoylphosphate, bind the enzyme active site with positive cooperativity (i.e.,they exert a “+” homotropic effect on activity). As the concentrations ofthe substrates change from values where [S] ≪ KM to values where [S] issaturating ([S]≫ KM), how will the binding constants for each of the twoallosteric effectors change? In other words, does ATP bind ATCase withhigher affinity when [S] is low or high? Does CTP bind ATCase with higheraffinity when [S] is low or high?
The oxidation of glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate by glyceraldehyde 3-phosphate dehydrogenase has an unfavourable equilibrium constant (K'eq= 0.08; G′° = 6.3 kJ/mol), yet the flow at this point in the glycolytic pathway is smooth. How does the cell get out of the unfavourable equilibrium?
The oxidation of glyceraldehyde 3-phosphate to 1,3-bisphosphoglycerate, catalyzed by glyceraldehyde 3- phosphate dehydrogenase, proceeds with an unfavorable equilibrium constant ( K'eq= 0.08; ΔG′° = 6.3 kJ/mol), yet the flow through this point in the glycolytic pathway proceeds smoothly. How does the cell overcome the unfavorable equilibrium?
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.
is this stement false? Intracellular concentrations in resting muscle are as follows: Fructose-6-phosphate (1.0 mM)Fructose-(1-6)-bisphosphate (10.0 mM)AMP (0.1 mM)ADP (0.5 mM)ATP (5.0 mM)Pi (10.0 mM)Under the above conditions the Phosphofructokinase reaction in muscle is more exergonic than under standard conditions.
Although phosphoesters are generally not energetic enough to donate the phosphate to make a high energy bond (e.g., a phosphoanhydride), the phosphate in PEP is used to convert ADP to ATP in the pyruvate kinase reaction. Carefully explain the ‘quirk’ of this reaction that makes it energetically feasible.
Could you help me with a scheme following: Carboxylic acids do not readily undergo certain biochemical reactions; thus, they are often converted to thioesters or phosphonoesters for catalysis. What are two reactions that thioesters and phosphonoesters can undergo that carboxylic acids cannot? Show enzyme-catalyzed arrow pushing for both of these mechanisms.
n your own words and with your own diagrams, describe the catalytic reaction mechanism of the enzyme acetylcholinesterase. Discuss the functional roles of each member of the catalytic triad plus whatever other residues that you find important. Describe the actual movements of protons, and the making and breaking of bonds. What types of chemistries is the enzyme using? (what is the enzyme doing, in term of reaction
Pyruvate carboxylase catalyzes the first step of gluconeogenesis. ATP + HCO3─ + pyruvate → oxaloacetate + ADP + Pi ∆G0’ = ─2.1 kJ mol-1 a) Calculate ΔG for this reaction under the following physiological conditions: 370C, pH 7 [pyruvate] = [HCO3─] = 4.0 mM [oxaloacetate]= 2.0 mM [ATP] = 3.5 mM [Pi] = 5.0 mM [ADP] = 1.8 mM
Does the complete oxidation of tridecanoic acid (C13:0) make more ATP than the complete oxidation of 9-tetradecenoic acid (C14:1, Δ9) inside human body? Show your calculation.

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