An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence of 4 M inhibitor, Vmax is 36...
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.
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An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence of 4 M inhibitor, Vmax is 36...
An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence...
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?
An enzyme has a Km of 4.7X10^-5 M. If the Vmax of the preparation is (22...
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?
1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation...
1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation is 224M/min, what velocity would be observed in the presence of 2 X 10^M substrate and 5 X10+M a competitive inhibitor. Ki is 3 X 10^M. What is the degree of inhibition? (10pts)
You have an inhibitor for an enzyme that you are studying. The concentration of inhibitor used is...
You have an inhibitor for an enzyme that you are studying. The concentration of inhibitor used is 5.50 µM. The following data was collected for the non-inhibited reaction as well as the reaction that was inhibited. mmol/(mL min) mmol/(mL min) mM Substrate Vo Substrate Vo + Inhibitor 0.200 5.000 3.751 0.400 7.500 4.998 0.800 10.000 5.995 1.000 10.700 6.173 2.000 12.500 6.807 4.000 13.600 7.143 a. Plot this data using Excel or a graphing program. Make sure you give your graph has a...
An enzyme binds to a competitive inhibitor with Kd = 1.2 × 10-6 M at pH 7.0 and 25°C.
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...
Q1: A marine microorganism contains an enzyme that hydrolyzes glucose-6- sulfate (S). The assay is based...
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
1. The kinetics of an enzyme was examined at various substrate concentrations in both the presence...
1. The kinetics of an enzyme was examined at various substrate concentrations in both the presence and absence of 3 mM inhibitor Z. The initial velocity data obtained are shown below: [S] (mmoles liter) v (mmoles"litermin) no inhibitor inhibitor Z 1.25 1.67 2.50 5.00 10.0 1.72 2.04 2.63 3.33 4.17 0.98 1.17 1.47 1.96 2.38 (4 pts) Estimat e Vmax and Kw in the presence and absence of inhibitor using the Michaelis Menton curve-fitting program on Kaleidagraph (see lab manual)....
An enzyme has a Km for substrate of 10 mM and Vmax of 5 mol L-1...
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.
a. what are the values of Vmax and Km in the abscence if the inhibitor what...
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...
Question2 An enzyme solution has a Vmax of 10 μΜ/sec and a Km of 10 μΜ....
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
1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation is 224M/min, what velocity would be observed in the presence of 2 X 10^M substrate and 5 X10+M a competitive inhibitor. Ki is 3 X 10^M. What is the degree of inhibition? (10pts)
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
1. The kinetics of an enzyme was examined at various substrate concentrations in both the presence and absence of 3 mM inhibitor Z. The initial velocity data obtained are shown below: [S] (mmoles liter) v (mmoles"litermin) no inhibitor inhibitor Z 1.25 1.67 2.50 5.00 10.0 1.72 2.04 2.63 3.33 4.17 0.98 1.17 1.47 1.96 2.38 (4 pts) Estimat e Vmax and Kw in the presence and absence of inhibitor using the Michaelis Menton curve-fitting program on Kaleidagraph (see lab manual)....
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...
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
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