Question

In the Michaelis-Menten mechanism, under what conditions will the rate law be at a maximum?

A.k_{2}[E]_{0} <<
K_{m} |

B.k_{2}[E]_{0} >>
K_{m} |

C.[S]_{0} >> K_{m} |

D.[S]_{0} << K_{m} |

Answer #1

For any enzyme that follows simple Michaelis-Menten kinetics,
when the initial velocity of the reaction is 1/5 of Vmax what is
the Substrate concentration?
A) KM << [S]
B) KM = 1/5[S]
C) KM = 4[S]
D) KM = 5[S]
E) KM = [S]

The Michaelis-Menten equation is an expression of the
relationship between the initial velocity,V0, of an enzymatic
reaction and substrate concentration, [S]. There are three
conditions that are useful for simplifying the Michaelis-Menten
equation to an expression from which the effect of [S] on the rate
can be more readily determined. Match the condition (e.g. [S] = Km)
with the statement(s) that describe it: 1. Doubling [S] will almost
double the rate. 2. Half of the active sites are occupied by...

1) For an enzyme that displays Michaelis-Menten kinetics, what
is the initial velocity as a function of Vmax when:
i) [S] = Km
ii) [S] = 0.1 Km
iii) [S] = 50Km
2) An enzyme (follows Michaelis-Mentin Kinetics) has Km = 0.5
M. The initial velocity is 0.2 Mmin-1 at substrate concentration
of 50 M. What is the Vmax? What is the initial velocity when a)
[S] = 2 M and b) [S] = 0.5 M?
3) What will be...

Enzyme E, which follows Michaelis- Menten kinetics, catalyzes
the same reaction upon three different substates that are
strucurally related (S1,S2, and S3) When the kinetic data for the
three reactions are determined under the same reaction conditions,
the data indicated below is otabined. The Km (Km1) for E and S1 is
1uM, that for E and S2 (Km2) is 80nM, and that for E and S3 (Km3)
is 20nM. The Vmax for E and S1 is 115uMol/min (vmax1), Vmax2 is...

Which of the following in not true for
an enzymatic reaction that shows Michaelis-Menten kinetics?
Group of answer choices
the rate steadily decreases over the course of the reaction
Vmax depends on the concentration of the enzyme
in the presence of an inhibitor the v vs. [S] plot is a
hyperbola
the equilibrium constant is lower as compared to the uncatalyzed
reaction
KM is determined by finding the [S] at which v =
Vmax/2

solve the michaelis-menten equation for km when vo=vmax/2. what
does this reveal about the relationship between the [substrate] and
the enzyme? Generally, what does it mean when the Km is high?
low?

Michaelis-Menten kinetics define the way in which enzymes behave
at various substrate concentrations. Unlike the linear reaction
rates discussed in organic chemistry, enzymes can only turn over
substrate at a rate that is equal to or less than the amount of
time it takes to bind to the enzyme, therefore, at high enough
substrate concentrations, there will be no effect on the reaction
rate. This is termed reaching Vmax, or the maximum velocity the
enzyme can turn over product. From...

Enzyme Kinetics Please explain reasoning for answers.
How does the [S] affect the rate of an enzymatic reaction? Why
don’t enzyme catalyzed rates increase linearly with [S]?
How will Km (Michaelis Constant) it affect the rate of an
enzymatic reaction? What will it change on the hyperbolic curve?
Why is the kcat/Km ratio a superior way to compare one enzyme to
another as compared to either of the two constants alone?
What happens to the Michaelis-Menten equation when [S] <<...

Based on the three step mechanism below, what is the rate law
for the reaction 2A + 2B →E + G?
A + B ⇌ D (Fast equilibrium)
D + B →E + F ( slow)
A + F → G (fast

What is the rate law for the following mechanism?
CH3COOC2H5 + H2O →
CH3COOC2H6+ +
OH-
(Slow)
CH3COOC2H6+
→ CH3COOH +
C2H5+
(Fast)
C2H5+ +
OH-
→
C2H5OH
(Fast)
A) Rate =
k[CH3COOC2H5][H2O]2
B) Rate =
k[C2H5OH]
C) Rate = k[CH3COOH]
D) Rate =
k[CH3COOC2H5]
E) Rate =
k[CH3COOC2H5][H2O]
The answer is E, but could you please
give a detailed response on how to get to this answer? Thank
you.

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