Problem Set 4 KEY
8) Estimate Vmax and Km for the following data
[S] mM 
v mMmin^{1} 
5 
22 
10 
39 
20 
65 
50 
102 
100 
120 
200 
135 
using:
a. Plot of v versus [S]
Eyeballing, can guess curve flattens out at
about Vmax = 150 mM/min, in which case Km is about 25 mM.
a. LineweaverBurke
Vmax = 1/0.0061 = 164;
Km/Vmax = 0.1961, so Km = 164*.1961 = 32
b. EadieHofstee
Km is about 30, Vmax
is about 158
9) Total enzyme concentration in (8) is 1 nM.
a. How many substrate molecules can an enzyme catalyze in 1 minute?
If we average the results from b and c above,
we get Vmax = 161mM/min and Km = 31 mM. Assuming those
values to be accurate, then:
_{}
b. Calculate k_{cat}/Km for the enzyme
k_{cat} is usually
expressed in terms of sec1, not min1, co we need to convert (a):
_{}
k_{cat}/Km is then:
_{}
This is reasonably efficient, though not at
the top end of enzyme efficiencies (>10^{8}).
15) Kinetics of an enzyme are studied with and without an inhibitor, giving the following results:
S (mM) 
V (mM/min) 
V+ (mM/min) 
1/S 
1/V 
1/V+ 
1.25 
1.72 
0.98 
0.8 
0.581395 
1.020408 
1.67 
2.04 
1.17 
0.598802 
0.490196 
0.854701 
2.5 
2.63 
1.47 
0.4 
0.380228 
0.680272 
5 
3.33 
1.96 
0.2 
0.3003 
0.510204 
10 
4.17 
2.38 
0.1 
0.239808 
0.420168 
a. What kind of inhibitor?
b. k_{cat} and Km with and without inhibitor
Okay – let’s plot the data:
Since Vmax changes
and Km is constant, this is noncompetitive inhibition.
Km = 2.5 mM
Vmax without inhibitor
is about 5 mM/min
Vmax with inhibitor is
about 3 mM/min
16) The same enzyme as in Problem 15 is analyzed in the presence of another inhibitor at two inhibitor concentrations (3 mM and 5 mM) with the following results. (Shaded numbers were calculated from the unshaded numbers as part of the solution.)
S 
V  I 
V + 3mM I 
V + 5mM I 
1/S 
1/V 
1/V+3mM 
1/V+5mM 
1.25 
1.72 
1.25 
1.01 
0.80 
0.58 
0.80 
0.99 
1.67 
2.04 
1.54 
1.26 
0.60 
0.49 
0.65 
0.79 
2.5 
2.63 
2 
1.72 
0.40 
0.38 
0.50 
0.58 
5 
3.33 
2.86 
2.56 
0.20 
0.30 
0.35 
0.39 
10 
4.17 
3.7 
3.49 
0.10 
0.24 
0.27 
0.29 
We can display the results on a LineweaverBurke plot (an EadieHofstee plot would work just as well).
Regression of the resulting lines gives the
following:
0mM inhib 
y =
0.4843x + 0.1951 

3mM inhib 
y =
0.755x + 0.1969 

5 mM inhib 
y =
1.0062x + 0.1861 

a)
Vmax is unchanged; Km varies with inhibitor
concentration. Therefore this is competitive inhibition.
b)
Km^{app}’s (apparent Km in
the presence of inhibitor) and Vmax are calculated
from the plot as:
Vmax 
5. 
Km 0 mM
I 
2.5 
Km 3 mM
I 
3.8 
Km 5 mM
I 
5.4 
c)
K_{I} can be determined from the apparent Km’s and the
relationship between Km and Km^{app} for
competitive inhibitors (see eq. 11.37a and fig.
11.20c).
_{}
We
plot Kmapp vs. [I]. The yintercept is Km; the slope id Km/K_{I}.
Km 
2.5 
Km/Ki 
0.5684 
Ki 
4.4 
Numbers may vary slightly depending on your
plot.
20) The following data are obtained for the steady state kinetics of an enzymecatalyzed reaction (shaded values are calculated):
[S] mM 
v mM/sec 
1/[S] 
1/v 
0.250 
0.260 
4.000 
3.846 
0.330 
0.450 
3.030 
2.222 
0.500 
0.920 
2.000 
1.087 
0.750 
1.800 
1.333 
0.556 
1.000 
2.500 
1.000 
0.400 
2.000 
4.100 
0.500 
0.244 
4.000 
4.800 
0.250 
0.208 
a) Does the enzyme exhibit MichaelisMenten kinetics?
We can make a LineweaverBurke
plot; the result is not linear, so MichaelisMenten
kinetics are violated.
(It is left as an exercise for the student to plot v versus [S] – the
resulting plot is slightly sigmoidal, implyimg cooperativity.)
b) What is Vmax?
My best eyeball guess from the above plot is
that – extrapolating the curve – we get a yintercept (corresponding to 1/Vmax)
of about 0.2 sec/mM, giving Vmax of about 5 mM/sec.