Download EXAM I (September 21, 2005) BIOCHEMISTRY 460 9:00 am section

Name______________________
EXAM I
(September 21, 2005)
BIOCHEMISTRY 460
9:00 am section
Be sure to put your name on each page of the exam
Name ______________________
Dr. DoGood in an effort to understand antibiotic resistance in pathogenic bacteria grows
both the wild-type (normal) strain and an antibiotic resistant strain. When she compares 2D gels
of the soluble fraction from both strains she finds a spot on the gel from the antibiotic resistant
strain that is not present in the wild-type strain.
1. To separate the soluble fraction from cell debris what technique would Dr. DoGood use? (5
pts).
Differential centrifugation, although centrifugation or some reasonable facsimile will do.
2. What are the properties of proteins, used in 2D gel electrophoresis to separate a mixture of
proteins. (10 pts).
Size or mass and isoelectric point
Dr. DoGood finds that the antibiotic resistance protein has a molecular weight of 34 kDa as
measured by gel filtration, and when chromatographed at pH 7.0, was retained on a DEAE
cellulose column.
3. Do you expect the isolelectric point of the antibiotic resistance protein to be less than or
greater than 7 and why. (5 pts)
Less than 7 since it is negatively charged at pH 7 (that is, sticks to a positively charged
resin).
4. In spite of having an apparent molecular weight of 34 kDa for the native protein, when run on
an SDS gel following treatment with mercaptoethanol and urea, two bands are observed, at 20
and 14 kDa. What does this suggest about the quaternary structure of the antibiotic resistance
protein and how could you explain this result. (10 pts).
It appears the native protein is a dimer (heterodimer to be precise). Since the dimer is
formed from treatment with mercaptoethanol and urea, the subunits must be held together
with disulfide bridges.
It turns out that the antibiotic resistance protein is a proteolytic enzyme which activates a 60 kDa
protein that is responsible for the break down of the antibiotic. The antibiotic resistance protein
activates the 60 kDa protein by catalyzing the cleavage of peptide bonds C-terminal to Leu, Val
and Ile.
5. What can you conclude about the specificity of the antibiotic resistance protein and the key
interactions that would stabilize the enzyme-substrate complex? Explain your answer (10 pts).
Clearly the specificity if for cleaving C-terminal to large aliphatic side chains. The side
chain binding must involve non-polar/hydrophobic interactions between the protein and
the substrate and must have are relatively large binding pocket or other feature to
accommodate the side chain.
6. Given that enzymes catalyze reactions, how would you explain the rate acceleration in context
of the transition state? (5 pts).
Enzymes, indeed catalysts in general lower the free energy of activation necessary to reach
the transition state.
7. Remembering that )G = )G0' + 1.36 log [products]/[substrates], hydrolysis reactions have a
large negative )G0'. In general would you expect these reactions to be spontaneous (yes or no),
and in general terms (use words not math) under what conditions would you expect the reaction
not to be spontaneous? (10 pts).
Yes, it would not be spontaneous when there are very high product concentrations relative
to substrate concentration.
After determining the sequence of the 34k kDa protein, it is clear that it is a structural homolog
of chymotrypsin.
8. What amino acids would you expect to find at the active site? (5pts)
A catalytic triad, Ser, His and Asp most likely.
9. If a side chain carboxyl group has a pKa of 4, what would you expect its charge to be two pH
units above the pKa? (5 pts).
Negative
10. In the context of forming an enzyme-substrate complex describe the lock and key hypothesis
and the induced fit hypothesis. (10 pts).
In a lock and key mechanism the substrate binding site has a structure that complements
the substrate, both structurally and chemically. In the induced fit mechanism after the
substrate associates with the enzyme a relatively large rearrangement of the enzyme
structure takes place which results in tight binding of the substrate. Note – you could have
used drawings to illustrate the key points.
It is found that the KM for the hydrolysis of the substrate is 7 :M and Vmax is 130
:moles/min/mg of enzyme.
11. What is the substrate concentration when the velocity of the reaction is 65 :moles/min/mg of
enzyme. (5pts)
7 :M
12. Draw a Lineweaver-Burke plot that you might expect for the antibiotic resistance protein and
indicate the features of the plot that reflect the KM and Vmax . (10 pts)
13. In the context of the active site of proteins briefly describe the role of the following features.
a. The protein tertiary structure. (5 pts)
Folds to position the amino acid site chains to bind the substrate and facilitate catalysis.
b. The type of interactions that result in binding the substrate. (5 pts)
Weak interactions-hydrogen bonding, non-polar and electrostatic