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Problem Set #8 – Chem 391
Due in class on Thursday, November 10th
Name
1.
Please provide arrow-pushing mechanisms involving general acid/base catalysis for the following
reactions. You may find it useful to look up the structures of the compounds described below. In each case
include an “-AH” and “-B” in your scheme.
a.
The isomerization of D-glyceraldehyde to L-glyceraldehyde
b.
The hydrolysis of adenosine (the ribonucleoside) to α-ribose and adenine (Hint: adenine is released
in a tautomeric form different from the one I usually draw, so that N9 is not protonated).
2.
During fatty acid metabolism, an α,β-unsaturated thioester
is hydrated to create a β-hydroxy thioester as shown at right.
a.
Draw a simple arrow-pushing mechanism for this reaction
using general acid/base chemistry. There will be one intermediate in
the path (resonance stabilized).
b.
The reaction leads to S absolute configuration at the β carbon and with a proton being added to the
pro-R position α to the carbonyl. Discuss positioning of the general acid and base. Will they be on the same
or opposite faces of the alkene? Explain briefly (a diagram may help).
3.
Ketosteroid isomerase (KSI) catalyzes the transformation at
right via formation of an enol intermediate. The kinetic parameters
at 25˚C are kcat = 3.8 x 104 s-1 and Km = 1.3 x 10-4 M.
O
a.
O
The enzyme has two important active site residues, Tyr16 and Asp40, that contribute to general
acid/base catalysis. Write out a complete electron-pushing mechanism that invokes these two side
chains appropriately in the reaction (a PyMOL file, KSI.pse, shows the positions of Y16 and D40 as
green residues near a pink inhibitor). Take care to explicitly draw any intermediates in the reaction.
See above, purloined from PNAS (1997) 94, 11773
b.
Sketch the plot of kcat vs. pH that you would expect to see given your mechanism in “a”.
c.
Acetate can catalyze the reaction above with a rate constant of 1.8 M-1s-1 at pH 7. For the acetatecatalyzed reaction, ∆H≠ is 16 kcal/mol and ∆S≠ is –17 cal/mol•K. Compare the efficiency of KSI as
a catalyst to acetate ion. What is the rate enhancement achieved by using a protein catalyst?
Compare kcat/Km to kacetate: Ratio is (3.8 x 104 s-1/1.3 x 10-4 M)/(1.8 M-1s-1) = 1.6 x 108
d.
Suggest why KSI is a better catalyst invoking one entropic and one enthalpic rationale.
∆H≠: Arguably Tyr is a superior general acid to water or acetic acid because phenols are better
matched in pKa to the protonated intermediate.
∆S≠: Pre-organized relationship of GA & GB in active site removes entropic loss suffered by acetate
in solution.
.
4.
An enzyme contains two active site His residues at positions 20 and 40. It is proposed that they take
part in general acid/base catalysis. To investigate this hypothesis, the unnatural residue fHis (shown at right)
is used to replace the natural residues at each position. Based on the plot below, provide estimates of the
pKa of His20, fHis20, His40 and fHis40 and identify which residue is acting as general acid, and explain
how you know.
pKa of His20= 7.5
pKa of fHis20=6.8
pKa of His40= 5.4
pKa of fHis40=4.0
Which residue (His20 or His40) is the general acid and
what is the evidence?
His20 is general acid. When it gets deprotonated (above pH7.5) kcat decreases.
5.
Non-competitive inhibition has the following mechanism. Let’s assume that a non-competitive
inhibitor complexes with both the E•S complex and free enzyme with the same dissociation constant, Ki.
E + S
+I
E•S
E+P
+I
E•I
E•S•I
a.
Derive the above rate law, given the above
mechanism and assumptions.
v=
Vmax [S]
! [I] $
( K m +[S]) #1+ &
" Ki %
b.
Show algebraically that non-competitive
inhibition can be identified on the LineweaverBurk plot by a set of lines that intersect at the
(1/[S]) axis at -1/Km.
v
[E • S]
[E][S] / K m
[ E • S] =
=
=
Vmax [ E ]tot [E]+[E • S]+[E • I]+ [ E • S • I] [E]+[E][S] / K m +[E][I] / K i +[E][S][I] / K i K m
v
[S]
[S]
=
=
= ...
Vmax K m +[S]+ K m [I] / K i +[S][I] / K i K m (1+[I] / K i ) +[S](1+[I] / K i )
1 ( K m +[S]) α K m α 1
α
α = (1 + [I]/Ki)
=
=
⋅
+
v
Vmax [S]
Vmax [S] Vmax
when 1/v = 0 (x-intercept)…
K α 1
α
K mα 1
α
1
α Vmax
1
0= m ⋅
+
;
⋅
=−
;
=−
⋅
=−
Vmax [S] Vmax Vmax [S]
Vmax [S]
Vmax K m α
Km
6.
Aspartyl proteases catalyze hydrolysis of peptide bonds using only general acid/base catalysis.
a.
Draw an arrow pushing mechanism for amide bond hydrolysis that uses a general base to activate
the water nucleophile and a general acid to stabilize growing negative charge during nucleophilic attack. Use
these residues again to promote formation of products.
b
Conjugate acid of the General Base 1.3 and the General Acid 4.2
b.
It is “generally” accepted that two aspartyl residues in the active site provide the general base in
pepsin. The structural diagram, above right, represents the geometry of the two aspartyl residues with a
bound water, that acts as a nucleophile. The dotted lines represent H-bonds. Draw the structure of the two
aspartyl residues and water molecule, with all hydrogens, over the diagram.
How does the above structure explain the pKa of the conj. acid of the general base.
The low pKa is due to stabilization of the base via hydrogen bonding to the protonated Asp (possible
general acid) and H2O.