Assume that an enzyme-catalyzed reaction follows the scheme shown:E+S SES→E + Pk₁ = 1 x 109/M-s k-1=2.5 x 10%/s k₂ = 3.4 x 107/sWhat is the dissociation constant for the enzyme-substrate, K,?What is the Michaelis constant, Km, for this enzyme?What is the turnover number, Keat, for this enzyme?What is the catalytic efficiency for the enzyme?If the initial Et concentration is 0.25mM, what is Vmax?

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Answered: Assume that an enzyme-catalyzed…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter25: Nitrogen Acquisition And Amino Acid Metabolism

Section: Chapter Questions

Problem 22P

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Indicate for the reactions below which type of enzyme and cofactor(s) (if any) wouldbe required to catalyze each reaction shown. 1) Fru-6-P + Ery-4-P <--> GAP + Sed-7-P2) Fru-6-P + Pi <--> Fru-1,6-BP + H2O3) GTP + ADP <--> GDP + ATP4) Sed-7-P + GAP <--> Rib-5-P + Xyl-5-P5) Oxaloacetate + GTP ---> PEP + GDP + CO 26) DHAP + Ery-4-P <--> Sed-1,7-BP + H 2O7) Pyruvate + ATP + HCO3- ---> Oxaloacetate + ADP + Pi
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.
Using the ActiveModel for aldose reductase, describe the structure of the TIM barrel motif and the structure and location of the active site.
Using the Nernst equation, calculate the equilibrium potential for Ca2 and for C1 from the following sets of data: a. Given [ Ca2+ ]0=1mM,[ Ca2+ ]i=100nM, find Eca2+ b. Given [ Cl- ]0=110mM,[ Cl- ]i=100mM, find Ecl
The Reactions and Meehanisms of the Leloir Pathway Write the reactions that permit galactose to be utilized in glycolysis. Write a suitable mechanism, tor one of these reactions.
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.
Explain how glucose is metabolized to yield ATP.
If phenylalanine was not an essential amino acid, would diet therapy (the elimination of phenylalanine from the diet) for PKU work?
MATHEMATICAL The enzyme -methylaspartase catalyzes the deamination of -methylaspartate [V. Williams and J. Selbin, J. Biol. Chem. 239, 1636 (1964)]. The rate of the reaction was determined by monitoring the absorbance of the product at 240 nm(A240). From the data in the following table, determine KM for the reaction. How does the method of calculation differ from that in Questions 26 and 27?
Do you think that the Ea for ATP hydrolysis is relatively low or high? Explain your reasoning.
Is aerobic respiration more or less efficient than glycolysis? Explain your answer.
b. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein E. To determine where the block in protein-E production occurred in each individual, the various intermediates were given to each individuals cel Is in culture. After a few weeks of growth with the intermediate, the cells were assayed for the production of protein E. The results for each individuals cells are given in the following table. A plus sign means that protein E was produced after the cells were given the intermediate listed at the top of the column. A minus sign means that the cells still could not produce protein E even after being exposed to the intermediate at the top of the column. Denote the point in the pathway in which each individual is blocked.

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