Problem Set 4 KEY

 

 

8)                  Estimate Vmax and Km for the following data

[S] mM

v mM-min-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.       Lineweaver-Burke

Vmax = 1/0.0061 = 164; Km/Vmax = 0.1961, so Km = 164*.1961 = 32

 

b.      Eadie-Hofstee

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 kcat/Km for the enzyme

kcat is usually expressed in terms of sec-1, not min-1, co we need to convert (a):

                  kcat/Km is then:

                 

This is reasonably efficient, though not at the top end of enzyme efficiencies (>108).

 

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.                  kcat 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 Lineweaver-Burke plot (an Eadie-Hofstee 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)      Kmapp’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)      KI can be determined from the apparent Km’s and the relationship between Km and Kmapp for competitive inhibitors (see eq. 11.37a and fig. 11.20c).

 

      We plot Kmapp vs. [I].  The y-intercept is Km; the slope id Km/KI.

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 enzyme-catalyzed 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 Michaelis-Menten kinetics?

We can make a Lineweaver-Burke plot; the result is not linear, so Michaelis-Menten 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 y-intercept (corresponding to 1/Vmax) of about 0.2 sec/mM, giving Vmax of about 5 mM/sec.