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FST 123
Problem Set 4
Spring, 2009
Name
1. Discuss the following statement (which may be true or false):
"Enzymes catalyze chemical reactions by decreasing the activation
energy of the forward reaction only. The activation energy for the reverse
reaction remains the same."
Catalysts changes the "path" that the reactants follow to
become products - they change the energy surface that the reaction
must traverse. The energy surface is the same whether the reaction
is going forward or backwards. So, whatever the enzyme does to
catalyze the forward reaction, it must also do to catalyze the
reverse reaction. So, the activation energy for the reverse reaction
must also change.
Another way to look at is that catalysts do not alter the free
energy or equilibrium constant, but they alter the rate at which
equilibrium is attained. The equilibrium constant can be decomposed
into ratios of forward and reverse rate constants. At equilibrium,
Forward rate = reverse rate (so concentrations don't change). If
these reactions are both second order, you could write Forward rate
= kf[A][B] = kr[P][Q] = Reverse rate. The center pair in this
equality could be rewritten:
kf
[P][Q]
= [A][B]
which, you might notice, = Keq
kr
So, if Keq must remain constant when catalyst is added, then the
ratio of kf to kr must also remain constant. Therefore, the enzyme
must reduce the activation energies in both directions.
2. Based on a substantial amount of data obtained about an aldolase, a
biochemist postulated the following structures and interactions during
catalysis. This is the famous "Horecker" mechanism for muscle
aldolase. I did it on the board on Tuesday…
a) Identify the function of each of the following groups on the enzyme.
Lys-NH2
acts as a nucleophile to make the Schiff base, enzyme-bound
intermediate. The positively-charged nitrogen in the Schiff base
stabilizes the carbanion by providing the eneamine form shown on the
right side of the resonance structures.
-B
acts as a general base to abstract the proton from the CH2OH moiety
to create the carbanion that attacks the second substrate.
-Arg sidechain
provides a positively charged site to bind the phosphate group of
the substrate (and intermediate). NOte that I left off two protons
on the Arg sidechain.
-BH
acts as a general acid to protonate the carbonyl group of the second
substrate, making the carbonyl carbon more labile to the attack.
Lys
Arg
NH2
HN C HN
OH O
H
CH C CH2 O
PO3 -2
HN
B
Lys
OH NH
H C
B
Lys
OH NH
H C
C CH2 O
C CH2 O
PO3 -2
H
Arg
Arg
HN C HN
PO3 -2 HN
Lys
OH NH
H C
C CH2 O
BH O C H
CH2 OH
Substrate 2
HN C HN
PO3 -2 HN
b) Suggest how the Vo versus pH plots might look for [S]o >> KM and
[S]o << KM (i.e., draw them). Explain your answer. You might have to make
some assumptions about pKa values. If so, state them. Also, you may confine
your comments to ionizations that occur on the enzyme (or ES complex) and
ignore possible ionizations of the substrate or its intermediate forms.
Protonating the ε-amino group of the active-site lysine should
decrease Vmax drastically, but Lys has a high pKa. Protonating B
should abolish its basicity, so Vmax should also go down
drastically. If B has a lower pKa than Lys, its ionization will
appear in the Vo vs. pH curve at both high and low substrate
(because it decreases Vmax); it Lys has the lower pKa, then the Lys
protonation will be observed, instead.
KM would increase if the Arg proton were removed (or if the
phosphate were protonated - but I told you not to worry about that).
If KM goes up, Vmax/KM would go down, which would be observed at low
substrate concentration. However, the pKa of Arg is higher than that
of Lys, so Vmax would already be low and you probably wouldn't see
the effect of the Arg ionization on binding.(Bye the way - one
proton is missing from the guanidino group of the Arg - the form I
showed would not be cationic).
So, from the information you have, the Vo vs. pH curves at low
and high substrate should have the same shape - decreasing at higher
pH, with the pKa of either the Lys or B:.
3. The following data were obtained from initial rate measurements of an
enzyme-catalyzed reaction in the absence and in the presence of fixed levels
of a reversible inhibitor.
Double reciprocal plot:
from fits
1/V intercept = 8.33E-04
mM/min
so Vmax = 1200.00 mM/min
1/[S] intercept = -2 mM-1
so KM = 0.5 mM
[S] o
(mM)
0.2
0.4
0.8
1.6
3.2
0.2
0.4
0.8
1.6
3.2
0.2
0.4
0.8
1.6
3.2
[I]
Vo
(mM) (mM*min -1 )
0
342.9
0
533.3
0
738.5
0
914.3
0
1037.8
2
200.0
2
342.9
2
533.3
2
738.5
2
914.3
4
141.2
4
252.6
4
417.4
4
619.4
4
817.0
a) What is the KM of this enzyme for this substrate under these
conditions?
see values, above (i.e., with [I] = 0) KM = 0.5 mM
b) What is the Vmax for the reaction under these conditions?
see values, above (i.e., with [I] = 0) Vmax = 1200 mM/min
c) What kind of inhibition is observed?
lines intersecting in the 1/V axis indicate competitive inhibition
(see plot)
d) What is the KI for this inhibitor?
Work with the slope since the intercept is constant
K " [I] %
slope = M $1+ '
Vmax # K I &
Plug in values for one inhibitor concentration and solve for KI, or
(better) do a slope replot, then divide the intercept of the replot
by the slope of the replot.
!
slope of replot= 2.08E-04
intercept of replot = 4.17E-04
KI= 2.00
e) How would KI change if an alternate substrate had been used?
Assume that the alternate substrate bound better (i.e., had a lower KM)
but showed approximately the same Vmax.
Inhibitor binding is a property of the inhibitor, not the substrate.
This argument is definitely an approximation because substrate and
inhibitor bind at the same time and could influence one another's
binding. It's a very good argument in the case of competitive
inhibition.
4. Each of the following compounds has been found to inhibit several
enzymes. Suggest a mechanism for the inhibition, and indicate whether the
inhibitor is a reversible inhibitor.
a) p-chloromercuribenzoate - irreversible; binds strongly to
sulfhydryl groups (Cys sidechains)
b) N-ethylmaleimide- irreversible; bonds covalently. undergoes a
Michael addition with nucleophiles.
c) Methylmethanethiolsulfonate - irreversible; bonds covalently
to sulfhydryl groups (Cys sidechains). Look up the reaction in the
notes.
d) Ellman's reagent - irreversible; bonds covalently to
sulfhydryl groups (Cys sidechains)
e) O-Methyl isourea – is an electrophile that is easily attacked
by the sidechain amino group of Lysine, to form homoarginine. The
sidechain is therefore longer than that of either Arg or Lys, but
has the same charge. It is no longer nucleophilic. The reaction is
irreversible.
f) EDTA - from the early part of the inhibition lecture sequesters divalent and trivalent cations (i.e., metal ions); if a
metal ion is required for activity, the enzyme will be inactivated.
Inactivation is irreversible (by our definition), but activity may
be regained by addition of more metal ion.
g) diethyl pyrocarbonate - irreversible; an anhydride that
acylates His and Lys sidechains
5. From the sequence of the gene encoding a particular glycosidase, you know
that the protein
contains four Cys residues. Describe an experiment designed to determine
how many (if any)
disulfide bonds are present.
A simple titration experiment. Titrate the protein with Ellman’s
reagent, DTNB, and measure the absorbance of the leaving group,
whose concentration is equal to the concentration of Cys modified.
Calculate the moles of Cys modified per mole of enzyme. If no Cys
were modified, all Cys are in disulfide bonds.. If two are modified,
there is one disulfide bond. If all are modified, there are no
disulfide bonds. If you get an odd number, do the experiment again,
or check the molecular weight of the protein – you did something
wrong.