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BL4010
Problem Set 2
Fall 2008
1. Measurement of the rate constants for simple enzymatic reaction obeying MichaelisMenten kinetics gave the following results:
k1 = 2 x 108 M-1s-1, k2 = 1 x 103 s-1, k3 = 5 x 103 s-1
a) What is KD (the dissociation constant for the enzyme-substrate complex?
KD = k2/k1= 5x10-6M = 5M
b) What is Km (the Michaelis constant)?
Km =
k2 k3
-5
k1 =3x10 = 30M
c) What is kcat (the turnover number)? kcat = k3 = 5x103 s-1
d) What is kcat/Km? 1.67x108 M-1s-1
e) Does this enzyme approach catalytic perfection? yes..almost diffusion ~109
f) What is Vmax if 2 nmol/ml and saturating substrate are used?

Vmax = [E]total*kcat =2x10-9mol•ml-1*5x103 s-1 = 1x10-5 mol•ml-1•s-1
g) What are Vmax and Km using 4 nmol/ml enzyme and saturating substrate.
Vmax = 2x10-5 mol•ml-1•s-1
2. The turnover number of an enzyme is 40,000,000 per second. The Km is 0.11 M.
a) What is Vmax using 3 nmol/L?
Vmax = [E]total*kcat= 3x10-9 mol/103ml*4x107s-1=1.2x10-4 mol•ml-1•s-1 = 12 mmol•ml-1•s-1
Vmax =12 M•s-1
b) What is v when [S] = 0.75 M?
Vmax [S ]
v = K m [S ] =12Ms-1*0.75M/(0.11M+0.75M)= 9M2s-1/0.86M = 10.5 Ms-1
c) What is the catalytic efficiency? efficiency = kcat/Km = 4x107s-1/0.11M =3.6x108M-1s-1

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BL4010
Problem Set 2
Fall 2008
3. Fetal hemoglobin (HbF) contains serine in place of the cationic histidine at residue 143 in
HbA, between the  chains in the central cavity of the enzyme.
a) The P50 for HbF is 18 torr whereas the P50 for HbA is 26 torr. How do these values
explain the efficient transfer of oxygen from maternal blood to the fetus?
The affinity of fetal hemoglobin for oxygen is ~1.5 times that of matrernal
hemoglobin, thus it will be able to effectively compete for oxygen (effect transfer).
b) What is the likely explanation for the difference in the P50 values?
Negatively charged BPG (bisphosphoglycerate) is known to bind in the central
cavity. Replacement of a positive residue with a polar/neutral residue will likely
reduce binding, destabilizing the T state, allowing T to R transitions with an
accompanying increase in O2 binding.
c) Sketch a plot of fraction binding versus O2 for each variant.
1
Fraction
0.5
Bound
0
0
18
25
PO2 torr
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BL4010
Problem Set 2
Fall 2008
4. An antibody binds and antigen with a KD = 5 x 10-8 M. Calculate the antigen
concentration when the ratio bound is:
a) 0.2 b) 0.5 c) 0.6 d) 0.8
[AX ]
fraction bound = [A ][AX ] =0.5 when X = 5x10-8M
[A ][X ]
KD = [AX ] such that [AX] =
[ A][ X ]
KD
X]
[A ] [ A][
KD
fraction bound = 

[ A][ X ]
KD
[A ](1 [KX ]
D

)
[A ][ X ]
KD
[X ]
KD
(1 [KX ]
D
such that
[X]
KD
 ( KD  [X ] ) 
)
KD
KD
[X ]
K D [X ]
and [X] = fraction bound*(KD)/(1-fraction bound)
-8

so that a) [X] = 0.2*5x10-8M/(1-0.2)=1.25x10
M, b) 5x10-8M, c) 7.5x10-8M, d) 2x10-7M

5. You are studying a novel enzyme with site-directed mutagenesis and make the following

kinetic measurements:
Variant
kcat (s-1)
Km (M)
6
Wild-type
4 x 10
2 x 10-6
6
E99Q
4 x 10
2 x 10-3
A12K
4 x 102
2 x 10-3
1
H135A
2 x 10
2 x 10-6
a) Which residues are involved in substrate binding?
The residues that affect Km are E99 and A12.
b) Which residues are involved in catalysis?
The residues that affect kcat are A12 and H135.
c) Which variant is the least efficient?
efficiency is kcat/Km. A12K affects both, thus it is least efficient.
(wt 2x1012 M•s-1, E99Q 2x109 M•s-1, A12K 2x105 M•s-1, H135A 1x107 M•s-1)
6. Catalytic triad groupings of amino acid residues increase nucleophilic character of active
site serine, threonine or cysteine residues in enzymes that catalyze cleavage of amide and
ester bonds. Using chymotrypsin as a model, draw the expected arrangement of the catalytic
triads in the following enzymes.
a) Human cytomegaloviral protease: His, His, Ser
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BL4010
Problem Set 2
Fall 2008
b) -lactamase: Glu, Lys, Ser, c) Asparaginase: Asp, Lys, Thr
d) Hepatitis A protease: Asp, His, Cys
7. Estimate Vmax and Km using the following data:
[S] (M)
v (M/min)
-6
2.5 x 10
28
4.0 x 10-6
40
-5
1.0 x 10
70
2.0 x 10-5
95
4.0 x 10-5
112
-4
1.0 x 10
128
2.0 x 10-3
139
-2
1.0 x 10
140
Vmax  140 M/min, Km  1x10-5 M (at Vmax/2 or 70 µM/min)
8. Prostaglandins are a class of eicosanoids, fatty acid derivatives responsible for producing
fever and inflammation. The first enzyme in the conversion of arachidonic acid to
prostaglandin is inhibited by ibuprofen with the following kinetics:
Arachidonic acid (mM)
Rate( mM/min)
Rate with 10 mg/ml
ibuprofen (mM/min)
0.5
23.5
16.67
1.0
32.2
25.25
1.5
36.9
30.49
2.5
41.8
37.04
3.5
44.0
38.91
a) What are Km and Vmax for the enzyme?
Vmax = 1/0.0194 = 51.5 mM/min, Km = -(-0.0194)/(0.0116)=1.67 mM
b) What type of inhibitor is ibuprofen? Vmax no change, Km change = competitive
c) What is the KI of ibuprofen?
Km’ = -(-0.0194)/(0.0203) = 0.955 mM, Km’ = (1+[I]/KI) such that
KI = (1+10mg/ml(mmol/206.30mg)/1.67mM = 0.59 mM
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BL4010
Problem Set 2
Fall 2008
0.07
0.06
y = 0.0203x + 0.0194
0.05
1/rate
0.04
0.03
wt
+I
Linear (+I)
Linear (wt)
y = 0.0116x + 0.0194
0.02
0.01
0
-2
-1
0
1
2
3
-0.01
1/[S]
9. The enzyme urease enhances hydrolysis of urea at pH 8.0 by a factor of 1014. If a
quantity of urea can be hydrolyzed in 5 minutes by the enzyme, how long would it take to
hydrolyze this same quantity in the absence of the enzyme in a sterile solution?
If it takes the enzyme 5 minutes, and the enzyme increases the rate by 1014, the
normal rate would be 5x1014 min or
5x1014min•(hr/60min)•(day/24hr)•(year/365day)=9.5x108 yrs or 950 million years
10. A solution of hexokinase loses 50% of its activity in 12 minutes when heated at 45 C
but only 3% when in the presence of large concentrations of its substrate glucose. Describe
possible modes of interaction between the enzyme and substrate and explain how the
substrate delays loss of activity.
Glucose must be stabilizing the structure, most likely through hydrogen bonding
with the hydroxyl groups on the glucose with hydroxyl and/or amide groups in the
protein binding site.
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