If an enzyme has a Km of 10 micromolar, what fraction of the enzyme is active at 100 micromolar substrate concentration?
Km means Michealis constant is concentration of substrate at which the half of the enzyme is active to it's maximum activity (Vmax). Means the concentration of substrate at which half of the enzyme sites are active. And Vmax is the double of the Km value.
So, sample enzyme has Km 10micromoles, that is half Vmax is 10 micromoles. Means Vmax is 20 micromoles. And we are provided 100 micromoles substrate that means saturated substrate solution for enzyme activity.
So, in this case all the active sites of enzyme are active. 100 percent fraction is active. And enzyme work with it's full efficiency.
If an enzyme has a Km of 10 micromolar, what fraction of the enzyme is active...
a. An enzyme has a Vmax of 100 umol/min and a Km of 40 uM. When substrate concentration is 40 uM what is the initial reaction rate? b. An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted with a irreversible active site specific inhibitor. After reaction with the inhibitor, the enzyme was assayed using a 2 mM concentration of substrate, and it gave a reaction rate of 20 umol/min. What percentage of the...
Part A An enzyme that follows Michaelis-Menten kinetics has a KM value of 10.0 uM and a kcat value of 201 s-1. At an initial enzyme concentration of 0.0100 uM, the initial reaction velocity was found to be 1.07 x 10- uM/s. What was the initial concentration of the substrate, [S], used in the reaction ? Express your answer in micromolar to three significant figures. ► View Available Hint(s) PO ALO O O ? [S]; = MM UM
Question2 An enzyme solution has a Vmax of 10 μΜ/sec and a Km of 10 μΜ. What is the velocity Vo of the enzyme at the following substrate concentrations? 1 uM 10 uM 100 HM Question3 For an enzyme, the following measurements have been made: Substrate concentration [S] Initial velocity Vo 10 20 40 90 120 180 300 500 10,000 50,000 0.12 0.20 0.30 0.42 0.45 0.39 0.53 0.56 0.60 0.60
What is the consequence of a high Km value of an enzyme? High concentration of the substrate is needed for the maximal enzyme activity. The enzyme is present in a large concentration in the cell. Moderate concentration of the substrate will be enough for maximal enzyme activity. This enzyme has high maximal velocity.
5. The Km of an enzyme of an enzyme-catalyzed reaction is 6.5 uM. What substrate concentration will be required to obtain 55% of Vmax for this enzyme? (10 pts)
Consider an enzyme with the following properties: – Km = 0.0050 M At what substrate conc. would the enzyme operate at one-fourth of its maximum rate? At a given substrate conc. (e.g., ½Km) what would the rate be as a fraction of Vmax?
2. (20 points) An enzyme catalyzes the reaction A | B. The initial rate of the reaction was measured as a function of the concentration of A. The following data were obtained: 0.08 |(A), micromolar V.,. nmoles/min 0.05 0.1 0.5 5 10 SO 100 500 1.000 5,000 10,000 20,000 2012 a) b) What is the of the enzyme for the substrate A? What is the K of the enzyme for the substrate A?
54. What is the catalyze reaction rate? Km=2 mM Kcat= 3 s-1 At the enzyme concentration=10 nM and substrate concentration= 3 mM
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.
At an enzyme concentration of 1microM the Vmax is 100microM/min and the Km is 20nm. At an enzyme concentration of 2microM, what would the V0 be at a substrate concentration of 40mM?