Chapter 8 Problems
Page 1 of 6
11/1/2007
8.1 The decarboxylation of a beta-keto acid catalyzed by an enzyme can be measured by the
rate of formation of CO2. From the initial rates in the table, determine the MichaelisMenten constant for the enzyme and the maximal velocity by a graphical method.
First, the problem should really say “determine KM and Vmax for this enzyme-substrate
combination”. The values for these constants depend on what enzyme and what substrate
are being considered.
I chose to transform the data for an Eadie-Hofstee plot (v0/[S] vs. v0). For this plot, the
slope is -1/KM and the vertical axis intercept is Vmax/KM. The plot is below. From Excel,
I get the least-squares values for the slope (-2.26 M-1) and intercept (1.57 μmol/2min·M).
(I left the values in the units that were in the table.) From the slope, we find that KM = 1/(-2.26 M-1) = 0.442 M. From the intercept, Vmax = KM·intercept = (0.442 M)·(1.57
μmol/2min·M) = 0.695 μmol/2min = 0.347 μmol/min.
1.8
Eadie-Hofstee plot
1.6
1.4
1.2
v0/[S]
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
v0
0.5
0.6
0.7
Chapter 8 Problems
Page 2 of 6
11/1/2007
8.5. An enzymatic reaction has a simple Michaelis-Menten mechanism:
k1
k2
E + S ↔ ES → E + P
k-1
k1 and k-1 are very fast, k2 = 100 1/sec and KM = 10-4 M at 280 K while k2 = 200 1/sec
and KM = 1.5 × 10-4 M at 300 K. (a) For [S] = 0.1 M and [E]0 = 10-5 M, what is v0 at 280
K? (b) Calculate the activation energy for k2. (c) What is Keq at 280 K for E binding to
S? (d) What is ΔH0 for the formation of ES from E and S?
a) We are given KM, and need to calculate Vmax = k2·[E]0 = (100 1/sec)( 10-5 M) = 10-3 M/s.
Micaelis-Menten gives us
v0 =
Vmax [ S ] (10 −3 M / s )(0.1M )
=
= 10 − 3 M / s
−4
K M + [S ]
10 M + 0.1M
b) From the Arrhenius relationship k = Ae − E A / RT we can divide k2 at T = 280 K by k2’ at
300 K (I denote the different Ts by a prime on the second k) and then take the logarithm
of the ratio:
k
E ⎛1 1⎞
EA
1 ⎞
100 s −1
⎛ 1
−5
ln 2 = ln
= −.693 = − A ⎜ − ⎟ = −
−
⎟ = −(2.86 × 10 mol / J ) E A
⎜
'
−1
R ⎝T T'⎠
8.314 J / mol • K ⎝ 280 K 300 K ⎠
k2
200 s
so from this we have EA = (0.693)/(2.86 × 10-5 mol/J) = 24.2 kJ/mol.
c) Keq= k1/k-1 by definition. KM = (k-1 + k2)/k1 ≅ k-1/k1 since k-1 >> k2 from the problem
statement that the first reaction is very fast. Thus, Keq = 1/KM = 1/(10-4 M) = 104 M-1.
d) This problem is asking about how enthalpy of a reaction depends on the change of the
equilibrium constant with temperature. This is the van’t Hoff relationship:
ΔH 0 = − R
ln(10 −4 ) − ln(1.5 ×10 −4 )
(−9.21) − (−8.80)
Δ(ln K )
= −R
= −(8.314 J / K • mol )
= −114.2kJ / mol
−3
−3
1
1
1
(
)
Δ( )
−
(
3
.
57
10
3
.
33
10
)
×
−
×
280 K
300 K
T
Chapter 8 Problems
Page 3 of 6
11/1/2007
8.10. Penicillinase inactivates penicillin. One form with MW = 30,000 has a single active
site, kcat = 2000 s-1, and KM = 5 × 10-5 M. In response to treatment with 5 μmol of
penicillin, a 1 mL suspension of bacteria release 0.04 μg of enzyme. (a) Assuming rapid
mixing and equilibration, what fraction of E will complex with penicillin early in the
reaction? (b) How long will it take for half of the initial amount of penicillin to be
inactivated? (c) What concentration of penicillin would cause the enzyme to react at half
maximal velocity? (d) A second suspension of bacteria releases 0.06 μg/mL of enzyme.
Will this affect the answers to (b) and (c) and, if so, by how much? (e) Modified
penicillin acts as a competitive inhibitor. If the affinity of E for penicillin and modified
penicillin is the same, what concentration of the inhibitor will reduce the rate of loss of
penicillin fivefold at low [penicillin]?
[ ES ]
[ ES ]
1
=
=
where
[ ES ] + [ E ] [ ES ] + K M [ ES ] 1 + K M
[S ]
[S ]
-5
the second step comes from KM = [E][S]/[ES]. Plugging in KM = 5 × 10 M and [S] = 5
μmol/1 mL = 5 × 10-3 M, we get f = 0.999.
a) The fraction of E complexed to S is f =
b) We can see that, from the M-M equation, when [S] = 2.5 × 10-3 M, v0 is still 0.98Vmax, so
the system will remain in the approximate zero-order domain. The velocity during this
time will be equal to Vmax to a good approximation. Vmax = kcat[E]0 = (2000 s-1)(0.04
μg/(30,000 g/mol)/(0.001 L) = 2.67 μM/s. The change in the concentration of S is -2.5 ×
10-3 M, so Δ[S]/Δt = -2.67 μM/s, or Δt = ( -2.5 × 10-3 M)/(-2.67 μM/s) = 937 sec = 15.6
min.
c) v0 = Vmax/2 when [S] = KM, so [penicillin] = 5 × 10-5 M.
d) Yes, this will affect part b) because [E]0 increases by 50%, so Vmax increases by 50% and
all velocities will thus increase by this factor. That means that Δt will decrease by a
factor of 1.5, or Δt = 625 sec. Part c) won’t change since the affinity doesn’t depend on
how much E is around.
e) If the inhibitor has the same affinity as penicillin, then KI = KM = 5 × 10-5 M. For a
competitive inhibitor we have
Vmax [ S ]
. So when v0(uninhibited) = 5v0(inhibited), we have
v0 =
⎛
[I ] ⎞
⎟⎟ + [ S ]
K M ⎜⎜1 +
K
I ⎠
⎝
Vmax [ S ]
5Vmax [ S ]
1
5
. That is,
=
or
=
K M + [S ]
⎛
⎛
⎛
0 ⎞
[I ] ⎞
[I ] ⎞
⎟⎟ + [ S ]
⎟⎟ + [ S ] K M ⎜⎜1 +
⎟⎟ + [ S ]
K M ⎜⎜1 +
K M ⎜⎜1 +
⎝ KI ⎠
⎝ KM ⎠
⎝ KI ⎠
K ⎛
[ I ] ⎞ [S ]
⎟+
so [ I ] = 4(K M + [ S ]) , thus it depend on how much S
K M + [ S ] = M ⎜⎜1 +
K M ⎟⎠ 5
5 ⎝
is present. If [S] << KM, then [I] = 4KM = 2 × 10-4 M.
Chapter 8 Problems
Page 4 of 6
11/1/2007
8.16. RNA polymerase binds rapidly to DNA, then the complex rearranges slowly. This can
be diagrammed as
k1
A+B ↔ C
k2
k-1
C ↔
D
k
-2
fast
slow
At the start of the reaction 1 nM A and 1 nM B are present. At final equilibrium the
concentrations are [A] = [B] = 0.6 nM, [C] = 0.01 nM and [D] = 0.39 nM. (a) Calculate
K1 and K2 from the equilibrium concentrations. (b) Write the differential equations for
d[C]/dt and d[D]/dt. (c) Sketch curves for [A], [B] [C] and [D] vs. time. (d) The value of
k2 was found to be 0.05 sec-1. Calculate the value of k-2.
a) The equilibrium constants are gotten from the equilibrium concentrations: K1 =
[C]/[A][B] = (1 × 10-11 M)/(6 × 10-10 M)2 = 2.78 × 107 M-1. K2 = [D]/[C] = (3.9 × 10-10
M)/(1 × 10-11 M) = 39.0. (I kept the units for the first constant in the biochemistry style.)
b) d [C ] = k1[ A][ B ] − k −1[C ] − k 2 [C ] + k − 2 [ D ] = k1[ A][ B ] + k − 2 [ D ] − ( k −1 + k 2 )[C ] and
dt
d[ D]
= k 2 [C ] − k − 2 [ D ] .
dt
concentration (nM)
c) The first reaction is very fast, so it will come to equilibrium first. Then the second will
slowly come to equilibrium. We know the starting concentrations. A and B will rapidly
drop as the first reaction equilibrates and [C] increases. Then [D] will slowly increase to
its final value as [C] drops. The first reaction will react to the change in [C] with drops in
[A] and [B] to their final values. The
1
initial plateau when the second reaction
0.8
hasn’t yet kicked in will have [A] = [B] = 1
A
nM – x and [C] = x: [C]/[A][B] = x/(1 nM
and
0.6
B
2
7
-1
– x) = 2.78 × 10 M . If we expect x to
0.4
be small, then x = 2.78 × 10-11 M. Thus,
D
0.2
[A] = [B] = 9.72 × 10-10 M and [C] = 2.78
× 10-11 M at the intermediate point. A
C
0
0
0.2
0.4
0.6
0.8
1
sketch might look like
time (relative to equlib)
d) We know K2 which equals k2/k-2. So k-2 = k2/K2 = (0.05 s-1)/(39) = 1.28 × 10-3 s-1.
Chapter 8 Problems
Page 5 of 6
11/1/2007
8.21. A carboxypeptidase was found to have KM = 2 μM and kcat = 150 s-1 for substrate
A. (a) What is the initial rate of reaction for [A] = 5 μM and [E]0 = 0.01 μM? (b) The
presence of 5 mM of a competitive inhibitor decreased the initial rate by a factor of 2.
What is the value of KI? (c) A competing substrate B is added to part (a). Its KM = 10
μM and kcat = 100 s-1. Calculate vB/vA.
a) We need Vmax: Vmax = kcat[E]T = (150 s-1)(0.01 μM) = 1.5 μM/s. So plug into the M-M
equation:
V
[ S ] (1.5 × 10 −6 M / s )(5μM )
=
v0 = max
= 1.07 × 10 − 6 M / s
K M + [S ]
2 μM + 5μM
b) A competitive inhibitor makes the apparent KM change, so first calculate what the
apparent KM is:
v0 =
Vmax [ S ] (1.5 × 10 −6 M / s )(5μM ) 1.07 × 10 −6 M / s
or KM,app = 9.02 μM. The
=
=
K M + [S ]
K M , app + 5μM
2
⎛
[I ] ⎞
⎟ or, after a little algebra,
apparent KM in the presence of I is K M , app = K M ⎜⎜1 +
K I ⎟⎠
⎝
[ I ]K M
(5mM )(2 μM )
KI =
=
= 1.42mM .
K M , app − K M 9.02μM − 2μM
c) The problem as stated seems to ask for two different things: i) the relative rates if the
substrates are both present simultaneously at 5 μM and ii) the relative rates of hydrolysis
if A and B are considered separately in separate reaction. The second is easier to deal
with, so I do it first
1) We already have v0 for [A] = 5 μM from part a). Calculating for [B] = 5 μM gives
Vmax, B [ S ] (1.0 × 10 −6 M / s )(5μM )
v0, B =
=
= 0.333 × 10 − 6 M / s .
K M , B + [S ]
10 μM + 5μM
Therefore,
v0, B
v0, A
=
0.333 × 10 −6 M / s
1.07 × 10 − 6 M / s
= 0.311
Notice that this is not what you get if you apply equation 8.25 since that equation is valid
only if [A] and [B] are very small compared to their respective Km values (if the reactions
are in separate pots). However, as we will see, if both substrates are present together in
the same solution, then Eq. 8.25 does hold.
2) If both A and B are present together in the same solution, then simple MichaelisMenten analysis doesn’t hold. The reason is that, as far as A is concerned, some of the
enzyme that it can interact with is removed by the EB complex. We need to account for
this in the mathematics that leads to the expression for [EA], which gives us the velocity
of consumption of substrate A. This is done exactly the same way that it is done when a
competitive inhibitor I is present in the reaction mix. You can read about that on pages
Chapter 8 Problems
Page 6 of 6
11/1/2007
416-417 in the book. The whole “trick” is to note that the total amount of enzyme
present, [ET], equals the sum of [E], [EA] and [EB], so that
[EA] = [ET] – [E] – [EB]
Now, we know that [ EB] =
Menten, setting
1
[ E ][ B] (gotten in the same way as standard MichaelisKB
d [ EB]
= 0 = k1, B [ E ][ B] − k −1, B [ EB] − k 2, B [ EB] ). So
dt
[EA] = [ET] – [E] – [EB] = [ET] –[E] –
⎛
1
[ B] ⎞
⎟
[ E ][ B] = [ ET ] − [ E ]⎜⎜1 +
KB
K B ⎟⎠
⎝
If you put this into our derivation of the Michaelis-Menten equation, you will get
vA =
k cat , A [ ET ][ A]
⎛
[ B]
K M , A ⎜1 +
⎜ KM ,B
⎝
⎞
⎟ + [ A]
⎟
⎠
Following the same logic for substrate B, you get v B =
k cat , B [ ET ][ B]
⎛
[ A] ⎞⎟
K M , B ⎜1 +
+ [ B]
⎜ KM ,A ⎟
⎝
⎠
.
Taking the ratio of vB to vA gives
⎧⎪
⎛
k cat , B [ ET ][ B]⎨ K M , A ⎜1 +
⎜
⎪⎩
vB
⎝
=
vA
⎧⎪
⎛
k cat , A [ ET ][ A]⎨ K M , B ⎜1 +
⎜
⎪⎩
⎝
⎫ k cat , B
⎞
⎧⎪
⎟ + [ A]⎪⎬
[ B]⎨1 +
⎟
⎪⎩
⎪⎭ K M , B
⎠
=
⎧⎪
⎫⎪ k cat , A
[ A] ⎞⎟
[ A]⎨1 +
+ [ B]⎬
K M , A ⎟⎠
⎪⎩
⎪⎭ K M , A
[ B]
KM ,B
[ B]
[ A] ⎪⎫ k cat , B
+
[ B]
⎬
K M , B K M , A ⎪⎭ K M , B
=
[ A]
[ B] ⎪⎫ k cat , A [ A]
+
⎬
K M , A K M , B ⎪⎭ K M , A
This is exactly equation 8.25, and it is valid no matter how small or large are the
concentrations of A and B. The catalytic efficiencies for substrates A and B are,
respectively, kcat,A/KM,A = (150 s-1)/(2 μM) = 75 s-1μM-1 and , kcat,B/KM,B = (100 s-1)/(10
μM) = 10 s-1μM-1. So clearly the enzyme has a strong preference for A over B. When
the two substrates are at equal concentration, we have that the ratio of velocities vB/vA
equals the ratio of the catalytic efficiencies, so vB/vA = (10 s-1μM-1)/(75 s-1μM-1) = 0.133.
This holds when the two substrates are mixed together with the enzyme.