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What is the Turn Over Number(TON) for an enzyme reaction with a substrate having the Vmax...
An enzyme has a Km for substrate of 10 mM and Vmax of 5 mol L-1 sec-1 at a total enzyme concentration of 1 nM. At [S] = 10 mM, kcat is: A) 2500 per M per sec. B) 5000 per M per sec. C) 1250 per M per sec. D) 2500 per sec. E) 5000 per sec.
An enzyme catalyzes a reaction with a K of 6.50 mM and a Vmax of 2.75 mM s velocity, vo. for the following substrate concentrations a) 2.50 mM Calculate the reaction Number mM. sl b) 6.50 mM Number c) 14.0 mM Number mM.sl
An enzyme catalyzes a reaction with a Km of 5.00 mM and a Vmax of 1.85 mM s-1. Calculate the reaction velocity, vo. for the following substrate concentrations. a) 3.00 mM Number mM.s b) 5.00 mM Number mM.s c) 10.5 mM Number
What is the rate of an enzyme catalyzed reaction if the Vmax is 100µmol S→P/min and the Km is 7 mM and the substrate concentration is 11mM? Is the enzyme working at Vmax? What if the substrate concentration is raised to 25mM?
An enzyme catalyzes a reaction with a Km of 8.00 mM and a Vmax of 4.45 mM s-1. Calculate the reaction velocity, Vo, for the following substrate concentrations a) 1.00 mM Number mM.s-1 b) 8.00 mM Number mM s-1 c) 11.0 mM Number mM.s-1
Need help with number 13! I already asked about number 12. The inverse velocity and inverse substrate concentration relationship for an enzyme-catalyzed reaction is given below V Vmax Vmax S For the hydration of CO2 catalyzed by carbonic anhydrase, it was determined experimentally that (dm s mol 4023.9+ 39.934 at a total enzyme IS] concentration of 2.32 × 10-y mol-dm- What is the value of the Michaelis constant KM for this enzymatic reaction? (B). 9.92x103 mol dm3 (D). 100.8 mol...
An enzyme catalyzes a reaction with a Km of 9.00 mM and a Vmax of 3.95 mM·s–1. Calculate the reaction velocity, v0, for the following substrate concentrations: A. 1 mM B. 9 mM C. 11 mM
At high (saturating) substrate concentrations, the rate of an enzyme-catalyzed reaction approaches Vmax. How close does the reaction rate actually get to Vmax? Determine how high (i.e. how many times Km) the substrate concentration must be for the reaction rate to be: a. 98% Vmax (show your work) (2) b. 99% Vmax (answer only) (1) c. 99.9% Vmax (answer only) (1)
At high (saturating) substrate concentrations, the rate of an enzyme-catalyzed reaction approaches Vmax. How close does the reaction rate actually get to Vmax? Determine how high (i.e. how many times Km) the substrate concentration must be for the reaction rate to be: a. 98% Vmax (show your work) (2) b. 99% Vmax (answer only) (1) c. 99.9% Vmax (answer only) (1)
Vmax of an enzyme-catalyzed reaction is A. the rate observed when the enzyme active sites are saturated with substrate B. independent of the amount of enzyme present C. the rate observed at the highest substrate concentration that can be experimentally obtained D. the initial rate observed at very low substrate concentrations