In: Chemistry

Suppose an enzyme has Km = 0.1mM and kcat = 10^3 s-1, and you perform an experiment in which [E]tot = 10-7M. Suppose that adding 0.1 µM of an inhibitor results in a Kmeff= 0.5 mM, and has no effect on kcat. Determine the substrate concentation that would be necessary to achieve a reaction velocity v of 5*10^-5Ms-1

Ans. Given, Kcat = 10^{3} s^{-1}

[E]_{tot} = 0.1 uM = 1.0 x10^{-7}
M
; [1 uM = 10^{-6} M]

# Now,

Kcat = Vmax / [E]_{tot}

Or, 10^{3} s^{-1} x (1.0 x 10^{-7} M) =
Vmax

Hence, **Vmax = 1.0 x 10 ^{-4} M
s^{-1}**

# In presence of Inhibitor-

Given, Kmeff = Km,app = 0.5 mM = 5.0 x 10^{-4}
M
; [1 mM = 10^{-3} M]

We have, **Vmax = 1.0 x 10 ^{-4} M
s^{-1} **
- remains unaffected of inhibitor

Vo = 5.0 x 10^{-5} M s^{-1}

Now, Using Michaelis- Menten equation- **Vo = Vmax [S] /
(Km +
[S]) **

(5.0 x 10^{-5} M s^{-1}) = (1.0 x 10^{-4} M
s^{-1}) x [S] / (5.0 x 10^{-4} M + [S])

Or, (5.0 x 10^{-5} M s^{-1}) / (1.0 x
10^{-4} M s^{-1}) =[S] / (5.0 x 10^{-4} M +
[S])

Or, 0.1 x (5.0 x 10^{-4} M + [S]) = [S]

Or, 5.0 x 10^{-5} M + 0.1[S] = [S]

Or, 5.0 x 10^{-5} M = [S] – 0.1[S]

Or, [S] = 5.0 x 10^{-5} M / 0.9

Hence, **[S] = 5.56 x 10 ^{-5} M**

Therefore, **required [S] = 5.56 x 10 ^{-5}
M**

the kcat of an enzyme is 1875 sec ^-1, catalytic efficeny of
same enzyme is 7.5*10^7, what is the km of enzyme?

Suppose you have an enzyme that has a
KM value of 10uM for its substrate and
a Vmax=
7umol-mg-1min-1. In the presence of a
reversible inhibitor, the KM value
increases to 40uM and the Vmax remains
the same. What type of inhibitor is this and what is the
KI value of this inhibitor?

The data for an Enzyme with a Kcat is 5,000m-1. This experiment
was performed by adding a given amount of substrate of 100
micrograms of enzyme in buffer solution. What is the molecular
weight of the enzyme?
Substrate Concentration mM
rate (Vo) micromol/min
1
167
2
250
4
334
6
376
100
498
1000
499

Q2. An enzyme has a Km of 2.8 * 10-5 M and
a Vmax of 53 μM/s.
(i) Calculate Vo if [S] = 3.7 * 10-4 M and
[I] = 4.8 *10-4 M for (a) a competitive inhibitor, (b) a
noncompetitive inhibitor and (c) an uncompetitive inhibitor.
(Assume Ki = 1.7 * 10-5 M.)
(ii) The degree of inhibition is given by i (%) = 100(1 -
Vi/Vo). Calculate the percent inhibition in
each of the three cases above.

An enzyme catalyzed reaction has a Km of 3.76 mM and a Vmax of
6.72 nM/s. What is the reaction velocity in nM/s,
when the substrate concentrations are:
A. 0.500 mM
B. 15.6 mM
C. 252 mM
D. Now assume you have added a competitive inhibitor that has a
concentration of 15.6 µM with a KI of 7.30 µM to the
enzyme in question 9. Calculate the velocity at the same substrate
concentrations as above:
E. 0.500 mM
F. 15.6...

Carbonic anhydrase (CA) has a 25,000-fold higher activity
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monophosphate decarboxylase (OMPD) (kcat = 40
s-1). However, OMPD provides more than a 1010 higher
“rate acceleration” than CA. Explain how this is possible.

In simple M-M kinetics, the units of kcat are s–1. Discuss how
the units are consistent with the name “turnover number.” If kcat
is large, what does that imply about the enzyme? Discuss the term
catalytic efficiency; what does it mean for an enzyme to be
efficient? How is it consistent with the term for catalytic
efficiency (kcat /Km)? Discuss the contribution of the terms kcat
and KM to the overall term of catalytic efficiency.
Discuss the meaning of enzyme...

A meteorite with mass 1.127E+3 kg has a speed of 123. m/s when
854. km above the Earth. It is falling vertically (ignore air
resistance) and strikes a bed of sand in which it is brought to
rest in 6.33 m. What is the average force (in N) exerted by the
sand on the meteorite?

How will you describe and contrast on enzyme activity
based on Km and Vmax from Michaelis menten equation? (low or high
Km/Vmax)
what does it mean when Km is High or low
what does it mean when Vmax is High or Low

1.) Suppose you will perform a test to determine whether
there is sufficient evidence to support a claim of a linear
correlation between two variables. Find the critical values of r
given the number of pairs of data n and the significance level
α.
n = 6, α = 0.05
Group of answer choices
r = ±0.811
r = 0.878
r = ±0.917
r = 0.811
2.) Suppose you will perform a test to determine whether
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