a) What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration...
a) What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration of 1.5x10-3 M if Km=2.9x10-4 and Ki=2x10-5?
b)To what concentration must the substrate be increased to re-establish the velocity at the original uninhibited value?
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a) What concentration of competitive inhibitor is required to
yield 75% inhibition at a substrate concentration...
6X 10+ 200 3. What concentration of competitive inhibitor is required to yield 75% inhibition at...
6X 10+ 200 3. What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration of 1.5 X 10M if Km-2.9 X 10+M and Ki-2. X 10M? The followina cat we winiaran Medan Termine whether
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...
Calculate the Ki for a competitive inhibitor whose concentration is 5 x10-4 M that raises the...
Calculate the Ki for a competitive inhibitor whose concentration is 5 x10-4 M that raises the Km of the substrate by 25%.
26. In a competitive inhibitor experiment, measured KM without inhibitor is 75 nM. After additing 3...
26. In a competitive inhibitor experiment, measured KM without inhibitor is 75 nM. After additing 3 M of inhibitor, the new Km is 225 nM. What is the closest value of a for the inhibition in this reaction? a. 1 b. 2 c. 3 d. 9 27. In the Question 26, what is the Ki for this inhibitor? a. 0.3 nM b. 0.75 nM c. 1.5 nM d. 3.0 nM 28. Which of the following about metabolism is INCORRECT?
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?
what will be the effect of increasing the substrate concentration in the presence of competitive inhibition?...
what will be the effect of increasing the substrate concentration in the presence of competitive inhibition? please use a drawing as well if you can. thank you
8. For the hypothetical data below, determine the type of inhibition and calculate KI Substrate concentration...
8. For the hypothetical data below, determine the type of inhibition and calculate KI Substrate concentration 4 mm 2 mM 1 mM 0.5 mm 0.2 mm Velocity without inhibitor 1.43 mm/min 1.11 mm/min 0.77 mm/min 0.48 mm/min 0.22 mm/min Velocity with inhibitor 1.08 mm/min 0.74 mm/min 0.45 mm/min 0.26 mm/min 0.11 mm/min
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...
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. T...
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. The rate of the reaction has been determined with and without an inhibitor. A total concentration of enzyme of 20 uM was used in the experiment. SHOW WORK AND UNITS!!! Without Inhibitor With Inhibitor [substrate] (mM)Rate of formation of te of formation of product product (mM/min) mM/min) 6.67 5.25 0.49 7.04 38.91 1.0 2.2 6.9 41.8 44.0 1.5 3.5 1 a) On the same graph, plot the data...
Enalaprilat is a competitive inhibitor of the angiotensin-converting enzyme (ACE), which cleaves the blood-pressure regulating peptide...
Enalaprilat is a competitive inhibitor of the angiotensin-converting enzyme (ACE), which cleaves the blood-pressure regulating peptide angiotensin I. ACE has a KM=52 μM for angiotensin I, which is present in plasma at a concentration of 79 μM. When enalaprilat is present at 2.8 nM, the activity of ACE in plasma is 11 % of its uninhibited activity. What is the value of KI for enalaprilat?
6X 10+ 200 3. What concentration of competitive inhibitor is required to yield 75% inhibition at a substrate concentration of 1.5 X 10M if Km-2.9 X 10+M and Ki-2. X 10M? The followina cat we winiaran Medan Termine whether
26. In a competitive inhibitor experiment, measured KM without inhibitor is 75 nM. After additing 3 M of inhibitor, the new Km is 225 nM. What is the closest value of a for the inhibition in this reaction? a. 1 b. 2 c. 3 d. 9 27. In the Question 26, what is the Ki for this inhibitor? a. 0.3 nM b. 0.75 nM c. 1.5 nM d. 3.0 nM 28. Which of the following about metabolism is INCORRECT?
8. For the hypothetical data below, determine the type of inhibition and calculate KI Substrate concentration 4 mm 2 mM 1 mM 0.5 mm 0.2 mm Velocity without inhibitor 1.43 mm/min 1.11 mm/min 0.77 mm/min 0.48 mm/min 0.22 mm/min Velocity with inhibitor 1.08 mm/min 0.74 mm/min 0.45 mm/min 0.26 mm/min 0.11 mm/min
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. The rate of the reaction has been determined with and without an inhibitor. A total concentration of enzyme of 20 uM was used in the experiment. SHOW WORK AND UNITS!!! Without Inhibitor With Inhibitor [substrate] (mM)Rate of formation of te of formation of product product (mM/min) mM/min) 6.67 5.25 0.49 7.04 38.91 1.0 2.2 6.9 41.8 44.0 1.5 3.5 1 a) On the same graph, plot the data...
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