AVmaxReaction velocity (Vo)VmaxNoinhibitorKmWith noncompetitiveinhibitorBMaximal velocity, Vmax, isdecreased in the presenceof noncompetitive inhibitorKm[S]Michaelis constant, Km. is unchangedin the presence of a noncompetitive inhibitorVmaxVo1[S]NoncompetitiveinhibitorNoinhibitor224) Relation between Reaction Velocity and Substrate ConcentrationDetermine the fraction of Vmax that would be found at a substrate concentration of 12 Km, 2 Kmand 10 Km.

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Biochemistry

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Chapter1: Biochemistry: An Evolving Science

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The enzyme Y catalyzes the elementary reaction AB→A + B An enzyme concentration of 1.0 µM Y can convert a maximum of 0.5 µM AB to the products A and B per second. Note: Since the concentrations of species in the cytoplasm of cells is small, the concentration unit of micromolar (uM or 10-6 M) is used for consistency with biochemical systems.
What enzyme kinetic parameters are apparantly impacted by uncompetitive inhibitors? Vmax Km Both Km and Vmax Neither Km nor Vmax
Km value should be more than 10% apart, use non-competitive inhibitor
How would a competitive inhibitor change the Km and the Vmax for an enzyme? Explain why a competitive inhibitor would result in these changes.
1.With the data shown below, please answer the following: A. Which of the two inhibitors is more efficient at high substrate concentrations? At low substrate concentrations? Explain you reasoning
If the allosteric coefficient for ATCase increases from 250 to 1300, the rate of the N-carbamoylaspartate formation: O stays the same and the sigmoidal curve of reaction rate as a function of substrate shifts up. O decreases and the sigmoidal curve of reaction rate as a function of substrate shifts to the left. O increases and the sigmoidal curve of reaction rate as a function of substrate shifts to the left. O stays the same and the sigmoidal curve of reaction rate as a function of substrate shifts to the right. O decreases and the sigmoidal curve of reaction rate as a function of substrate shifts to the right. Question 9 Which of the following CANNOT be targeted for phosphorylation to modulate the activity of an enzyme? O Ser Ala O None of the amino acids listed can be targeted for phosphorylation to modulate enzyme activity The Tyr
Recall that vmax is achieved only at high substrate concentrations. Do you predict that vmax will change when an uncompetitive inhibitor is introduced? Why or why not?
Test Tube # Enzyme 1 (uL added) Product Formation (umol) 1 0 0 2 25 0.08 3 50 0.15 4 75 0.22 5 100 0.3 6 125 0.37 7 150 0.45 8 175 0.53 9 200 0.59 10 225 0.68 11 250 0.74 A. Make a graph is Excel/Google Sheets plotting Product formation (umol) on the y-axis and Enzyme volume added on the x-axis. Paste the graph below. Make sure to have axis labels and a title on the graph. B. What can you conclude from this graph, ie how is the amount of product formed changing as a function of the amount of enzyme present?
Do answer and explain plz.
How can competitive and pure noncompetitive inhibition be distinguished in terms of KM?
CHEM151/251 Biochemistry 1 a. Determine Km 1. The following data were obtained for a competitive inhibition study in which the [I] = 3 µM for each determination of vo in the presence of inhibitor. The Vmax = 200 μM P/min for both data sets. 200 Vo (UM P/min) 8 8 8 8 8 8 8 8 8 180 160 140 120 100 40 Name (Print)/ID #: the absence of inhibitor. Participation Question # 10 No Inhibitor 50 +Inhibitor 100 [Substrate] (UM) 150 b. Determine Km, app for the data obtained in presence of inhibitor. 200 c. Calculate the value for Ki. Note: a = 1 + [I]/Ki and a¹ = 1 + [I]/Ki*. for the data obtained in
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in the presence of a noncompetitive inhibitor 24) Relation between Reaction Velocity and Substrate Concentration Determine the fraction of Vmax that would be found at a substrate concentration of 12 Km, 2 Km and 10 Km.