2. A 500ul reaction containing 0.5 pmol of an enzyme following Michaelis –-Menten kineties with saturating substrate concentration is studied. For this enzyme kl = 1 x 10° M-1 s', k-1 = 2 x 10's", and k2 = lx 10ʻ s". a. What is the Vmax for the enzyme considering that it has only one substrate binding site? b. What are the values of Km and Ks? Is this a rapid equilibrium enzyme or does it follow steady state kineties? Explain why. c. What is the substrate concentration needed to achieve half maximal velocity? Explain the basis for your answer. d. If the substrate concentration in part e is doubled, how many fold will the initial velocity increase?

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Chapter1: Biochemistry: An Evolving Science
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2. A 500µl reaction containing 0.5 pmol of an enzyme following Michaelis -Menten kinetics
with saturating substrate concentration is studied. For this enzyme kl = 1 x 10° M-1 s', k-1 =
2 x 10's', and k2 = 1 x 10° s'.
a. What is the Vmax for the enzyme considering that it has only one substrate binding
site?
b. What are the values of Km and Ks? Is this a rapid equilibrium enzyme or does it follow
steady state kinetics? Explain why.
c. What is the substrate concentration needed to achieve half maximal velocity? Explain the
basis for your answer.
d. If the substrate concentration in part e is doubled, how many fold will the initial
velocity increase?
e. What is the Km is the concentration of the enzyme is doubled?
f. What is the efficiency of this enzyme?
g. Would this enzyme be considered a perfect enzyme? Explain.
h. If 0.25 pmol of an irreversible or classic non-competitive inhibitor is added before starting
the reaction. What is the resulting Km for the enzyme?
i. A reversible inhibitor with Ki = 1 x 10* M is added to the reaction la. The observed
Vmax remained unchanged but Km was shifted to an increased value.
What type of inhibitor is this? Explain.
j. If the concentration of this inhibitor is 1 x 10“M, which will be the resulting apparent
final Km Value?
Transcribed Image Text:2. A 500µl reaction containing 0.5 pmol of an enzyme following Michaelis -Menten kinetics with saturating substrate concentration is studied. For this enzyme kl = 1 x 10° M-1 s', k-1 = 2 x 10's', and k2 = 1 x 10° s'. a. What is the Vmax for the enzyme considering that it has only one substrate binding site? b. What are the values of Km and Ks? Is this a rapid equilibrium enzyme or does it follow steady state kinetics? Explain why. c. What is the substrate concentration needed to achieve half maximal velocity? Explain the basis for your answer. d. If the substrate concentration in part e is doubled, how many fold will the initial velocity increase? e. What is the Km is the concentration of the enzyme is doubled? f. What is the efficiency of this enzyme? g. Would this enzyme be considered a perfect enzyme? Explain. h. If 0.25 pmol of an irreversible or classic non-competitive inhibitor is added before starting the reaction. What is the resulting Km for the enzyme? i. A reversible inhibitor with Ki = 1 x 10* M is added to the reaction la. The observed Vmax remained unchanged but Km was shifted to an increased value. What type of inhibitor is this? Explain. j. If the concentration of this inhibitor is 1 x 10“M, which will be the resulting apparent final Km Value?
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