Transcript Problem set answers

1. The crystal structure for the enzyme you are studying has recently been
published. The structure reveals a Glu45 reside in the active site and you
propose it represents a substrate binding residue. Sequence comparison
among 5 orthologs of the enzyme shows Glu45 is absolutely conserved.
Kinetic studies of an E45A point mutants results in a 103-fold lower Km for
substrate and a 104-fold lower Vmax relative to wild type enzyme. Interpret
these results with respect to the likely role of Glu45 in the catalytic cycle of the
enzyme.
The lower Vmax is consistent with the mutation resulting in a
decreased kcat, which is what one would expect if Glu45 is a catalytic
group in the active site. Therefore, the observed effect on Vmax is
consistent with the hypothesis.
The effect of the mutation on Km suggest one of two possibilities:
This mutation could result in an increased binding affinity as
evidenced by the decrease in Km. More likely, the enzyme follows
steady state kinetics for which Km=k2/k1, for which a decrease in
k2 (I.e., kcat) would result in a lower Km as observed. The data
presented do not allow one to distinguish between the two alternative
interpretations.
2. For the enzyme in question 1 you next prepare a E45D point mutant and
repeat the kinetic studies. These studies show a Km that is statistically
indistinguishable from wild type enzyme but a Vmax that is 10-fold lower.
Interpret these new results with respect to the effect of the E45D point
mutant.
The 10-fold lower Vmax suggests that mutation of Glu45 to Asp is not a
conservative mutation. Since the side chain of Asp is one methylene
shorter than Glu, the results suggests a decrease in Vmax due to a
correspondingly greater distance between the carboxlate side chain and
intermediate in the reaction cycle with which it must interact.
3. Contrast the effect on pKa for an aspartate versus a lysine residue within a
hydrophobic microenvironment of an enzyme active site.
A hydrophobic microenvironment will result in a localized lower dielectric
constant, which favors uncharged species. Therefore, the carboxylate side
chain of aspartate will appear to have a higher pKa (favoring the uncharged
protonated form over a broader pH range) and the lysine will exhibit a
lowered pKa (favoring an uncharged unprotonated form over a broader pH
range).
4. AlphaCP1 is a polyC/U binding protein that binds to polyC/U-containing 3' UTR
regions of certain mRNA species to stabilize the message and suppress translation.
Phosphorylation of aCP1 by S6 protein kinase is thought to result in reduced
binding to the mRNA and release of the protein from the polyC/U region, allowing
translation initiation. You have identified Ser395 as a putative phosphorylation site
for S6 protein kinase. To test this hypothesis, you generate an aCP1S395A point
mutant and find that it binds polyC/U but fails to undergo S6 protein kinasemediated phosphorylation. You are feeling pretty good about the data until your
advisor voices a concern that the result is not conclusive since the mutation may
have partially altered the structure of aCP1 in a way that allows mRNA binding but
prevents binding of S6 kinase. Propose an experiment in sufficient detail that
resolves this question.
The question is whether S6 kinase can distinguish wild type from
mutant αCP1. One way would be to express recombinant wild type and
mutant αCP1. (If done in parallel this would also obviate potential
differences due to differences in the preparation methods.) One would then
measure the initial rate of S6 kinase-catalyzed phosphorylation of wild
type αCP1 in order to determine Vmax and Km. One would then repeat the
experiment at different constant concentrations of point mutant (which
cannot be phosphorylated) and observe whether the mutant is a competitive
inhibitor. A secondary plot of Km,apparent versus [mutant]o would allow
determination of the KI, which should equal the Km for wild type protein if
the kinase is unable to distinguish between the two proteins.
1/vo
Increasing concentrations
of mutant
-Mutant
1/Km
1/Km,app
Km,app
1/[aCP1]o
Km
[Mutant]o
6. Defend the statement that general acid catalysis in an enzyme catalyzed reaction
is a type of proximity effect.
Proximity effects in enzyme catalysis arise by the precise orientation of
amino acid side chains within the active site. In the case of general acid
catalysis, this precise orientation results in the weak acid group being
positioned to optimally donate the proton within the catalytic cycle.
Therefore, general acid catalysis during an enzyme reaction is a special
case of a proximity effect.
7. In 25 words or less thoroughly but succinctly define entropy-enthalpy
compensation.
The unfavorable transition state entropy of activation (DS‡) is
compensated by the favorable enthalpy of binding (DH) of the
reactants in the active site.
8. A student in your lab is using a 125I-labeled peptide hormone to study its
binding to a membrane receptor. When the radiolabeled hormone was purchased
the product sheet listed a specific radioactivity of 53 milliCuries/mmole. If the half
life for 125I is 60 days, what is the specific radioactivity of the probe after 45 days?
Show your work.
k = 0.693/60 days = 0.0116 days-1
Specific activity after 45 days = 53 mCi/mmole•e-0.0116 days-1•45 days
= 53 mCi/mmole•0.59
= 32 mCi/mole
9. You are studying the role of mRNA binding proteins in translational regulation
of gene expression. You have found in a genetic screen that the ubiquitin
conjugating enzyme E2epf interacts with the RNA binding protein AUF1. To prove
this interaction, you express E2epf as an N-terminal GST fusion protein and
demonstrate in a glutathione-Sepharose pull down assay that 100 nM
concentration of the fusion is able to bind AUF1 (see Lane 2 of attached figure).
As a control, 100 nM GST alone fails to bind AUF1 (see Lane 1 of attached figure).
1 2
3 4
[In this variation on an IP assay, the glutathione binding
domain Glutathione S-Transferase (GST) is genetically
fused to the amino terminus of E2epf. The resulting
chimeric protein is incubated with a putative ligand, after
which glutathione-linked beads are added to precipitate
your GST fusion protein and any bound ligand. The
presence of bound ligand can be tested by Western blot,
as displayed in the figure which has been stained with
anti-AUF1 antibody.] The literature reports that AUF1 can
bind another RNA binding protein aCP1; to confirm this,
you incubate the same concentration of AUF1 with 100 nM
GST-aCP1 (Lane 3 of figure). Out of curiosity, the last
experiment is repeated but you also add 100 nM E2epf to
the mixture of GST-aCP1 and AUF1 (Lane 4 of figure).
Provide a succinct thermodynamic argument supporting the hypothesis that
E2epf binds aCP1. For extra credit, is there an alternative hypothesis also
consistent with the observations?
Because the density of the AUF1 band is greater in lane 4, the Western
blot suggests that a combination of aCP1 and E2epf bind AUF1 with
greater affinity than either alone. Because E2epf and aCP1 each bind
AUF1, one can construct the following model for the process:
AUF1 + E2epf
+
aCP1
K-1
K-2
K-2
aCP1•AUF1 + E2epf
AUF1•E2epf
+
aCP1
K-1
aCP1•AUF1•E2epf
In the scheme, the equilibrium constants represent Kdissociation constants.
Because the bands of bound AUF1 in lanes 2 and 3 are nearly equal, K-1 
K-2 ; however, because the combination of E2epf and aCP1 bind AUF1 with
greater affinity, K-1  K-1, requiring K-2  K-2.
The model requires E2epf and aCP1 to form a complex which binds AUF1
tighter than either alone. The tighter binding can be most easily
envisioned as involving additive binding surfaces, as shown
schematically:
E2epf AUF1
aCP1
There is one other model that is fits the available data that does not
require additive binding: AUF1 can bind E2epf and aCP1 at different
sites that result in a conformation change that makes the other
component bind tighter. This model would follow the same scheme as
above but not require interaction of E2epf and aCP1:
E2epf
AUF1
aCP1
10. The plasma membrane of human cells contains a ubiquitously distributed family
of tissue-specific receptor complexes involved in regulating cell adhesion,
growth, and differentiation.
These complexes contain a member of the
tetraspanin superfamily of transmembrane proteins that is non-covalently
associated with one or more membrane receptors or glycoproteins. The
tetraspanins also bind an additional 14 kDa subunit termed Leu-13. The complex
present on B lymphocytes (a type of immune cell circulating in the blood) is
reported to be composed of the tetraspanin CD81 (25 kDa) which is associated
with the transmembrane receptor CD19 (91 kDa) and the 110 kDa transmembrane
glycoprotein CD21 (see illustration). You have recently found that a protein
kinase, PI 3-kinase, binds to the cytoplasmic side of CD19.
CD21
Leu-13
outside
Cell membrane
cytoplasm
CD81
PI 3-kinase
CD19
In normal freshly isolated B lymphocytes and in cultured Daudi cells (an
established leukemic B lymphocyte cell line used in tissue culturing) Leu-13 is
absent from this complex or present at low levels. However, addition of pure
recombinant Leu-13 to either fresh B lymphocytes or Daudi cells reversibly
reconstitutes the complex. Reconstitution of the complex is accompanied by
complete inhibition of cell division. In this experiment, cell division is
quantitated by measuring the incorporation of the radioactive nucleotide 3Hthymidine into DNA during the 2 hours following addition of recombinant Leu13 to the cultures. You compare this data to an experiment conducted under
identical conditions but with freshly isolated normal human B lymphocytes.
Data for both experiments are summarized in Table 1 for experiments
conducted at 37 oC.
What is the quantitative difference, if any, in the binding of Leu-13 to CD81
present on normal B lymphocytes versus Daudi cells?