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1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation...
An enzyme has a Km of 4.7X10^-5 M. If the Vmax of the preparation is (22 micromoles X liters^-1 Xmin^-1). What velocity would be observed in the presence of 2X10^-4 M substrate and 5X10^-4M of a. a competitive inhibitor b. a noncompetitive inhibitor c. an uncompetitive inhibiter Ki in all three cases is 3X10^-4M. What is the degree of inhibition in all three cases?
An enzyme binds to a competitive inhibitor with Kd = 1.2 × 10-6 M at pH 7.0 and 25°C.(a)At what inhibitor concentration will 75% of the enzyme be bound to the inhibitor if there is no substrate present? (b) This enzyme has a Km of 4.0 × 10-5 M and a Vmax of 50 μM/s. At a substrate concentration of 3.0 × 10-4 M, calculate (i) the velocity of reaction in the presence of the inhibitor at 4.8 x 10-5 M (ii) the degree of...
a. what are the values of Vmax and Km in the abscence if the
inhibitor what are the values of Vmax and Km in the presence of the
inhibitor?
b. what type of inhibition is it?
c. what is the dissociation constant (Ki) of the
inhibition?
***d. graph a linear scatter plot including equation.
Homework (CHE 407) The initial velocity for an enzyme-catalyzed reaction is measured at various initial substrate concentration [S]o, in the absence and in the presence of...
An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence of 4 M inhibitor, Vmax is 36 nmoles substrate converted per sec with no change in Km. Determine the Ki of the inhibitor. Identify the type of inhibitor with explanation.
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...
Q1: A marine microorganism contains an enzyme that hydrolyzes glucose-6- sulfate (S). The assay is based on the rate of glucose formation. The enzyme in a cell-free extract has kinetic constants of km = 6.7 x 10-4, Vmax = 300 nm/L/min in presence of 10-5M competitive inhibitor (Galactose-6-sulfate) and 2 x 10-5M substrate (Glucose-6-sulfate), velocity was 1.5 nmoles /L/min. a) Calculate Ki for Galactose-6-sulfate b) Calculate velocity in absence of the inhibitor
In a competitive inhibition, Km is increased while Vmax is unchanged. An enzyme is being assayed in the presence of a fixed amount of a competitive inhibitor. How could the rate be increased in this reaction?
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 uM/s. 5 4.5 4 a) What is the KM? KM: v. (mM/s) 3.5 3 2.5 2 1.5 1 0.5 b) If a pure non-competitive inhibitor was added to the assays, what would the resulting kinetics curve be like? Give a Km and Vmax in the presence of the inhibitor (write them below) and draw an appropriate curve...
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 uM/s. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 KM: 3mm V. (mM/s) 1.5 1 b) If a pure non-competitive inhibitor was added to the assays, what would the resulting kinetics curve be like? Give a Km and Vmax in the presence of the inhibitor (write them below) and draw an appropriate curve...
An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence of 4 microM inhibitor, Vmax is 36 nmoles substrate converted per sec with no change in Km. Determine the Ki of the inhibitor. What type of inhibitor is this?