Question

An enzyme has a K_{M} of 47 μM. If the V_{max}
of the preparation is 22 μmol∙L^{-1}∙min^{-1}, what
velocity would be observed in the presence of 7 μM substrate and 4
μM of an uncompetitive inhibitor that has a K_{I} of 3
μM.

Answer #1

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?

a) The Km of an enzyme of an enzyme-catalyzed reaction is 7.5
µM. What substrate concentration will be required to obtain 65% of
Vmax for this enzyme? (same enzyme was used in part a and b)
b) Calculate the Ki for a competitive inhibitor whose
concentration is 7 x10-6 M. The Km in the presence of inhibitor was
found to be 1.2x10-5 M. ￼￼

You add an uncompetitive inhibitor to your enzyme assay.
What will happen to Km' and Vmax
', relative to Km and Vmax?
Choose all correct answers.
Group of answer choices
Vmax' is higher
Vmax' is the same
Km' is higher
Vmax' is lower
Km' is lower
Km' is the same

What is the value of Vi when Vmax is equal to 250 μmol/L/min and
KM is equal to four times the substrate concentration?

The kinetics of an enzyme were studied in the absence and
presence of two different concentrations of an inhibitor. Determine
the Km and Vmax of the uninhibited enzyme and
identify which class of inhibitor is present using the data
presented below. Use this data to determine the KI, the
binding constant of the inhibitor to the enzyme. Note that this
question will require you to use a computer-graphing program such
as EXCEL.
[S] mmol/L
vo (mmol/L)min-1
(no inhibitor)
vo (mmol/L)min-1...

THANK YOU
30- A species of Shewanella bacteria contains an enzyme
that catalyzes the dehalogenation of tetrachloroethene. The
KM is 120 μM and the Vmax
is 1.0 nmol · min−1 · mg−1. What is the
substrate concentration when the velocity is 0.75 nmol ·
min−1 · mL−1?

An experimenter performs two sets of enzyme kinetics
experiments. One set of experiments, when plotted as a
double-reciprocal plot, yielded a y-intercept of 0.043
μM-1·min and an x-intercept of -20.0 mM-1. A second set of
experiments, when similarly plotted, yielded a y-intercept
of 0.129 μM-1·min and an x-intercept of -60.0 mM-1. The
second set of experiments contained 75 nM of an uncompetitive
inhibitor. Given these data, what is the Ki' (in nM to the
nearest tenths) for this inhibitor? Hint:...

Vmax and Km of an enzyme are 25mM/s and 5mM, respectively. What
is initial velocity when substrate concentration is 10mM? Show work
and include proper unit.
10nM of enzyme was used in the above question. Based on the kinetic
parameters, has the enzyme achieved catalytic perfection? Show
work. Thank you!

Vmax and Km of an enzyme are 25mM/s and 5mM, respectively. What
is initial velocity when substrate concentration is 10mM? Show work
and include proper unit.
10nM of enzyme was used in the above question. Based on the kinetic
parameters, has the enzyme achieved catalytic perfection? Show
work. Thank you!

9.
A. Vmax is a poor metric for indicating
enzyme efficiency because:
1) It changes as a result of using competitive inhibitor
2) The total enzyme concentration affects the value of
Vmax
3) It requires measuring enzyme velocity
4) It is independent of the total enzyme concentration
5) It varies as a function of the KM value
of the substrate
B. What is the main disadvantage of using
kcat to compare the efficiencies of enzymes?
What is the main advantage of...

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