Download Student________________ Biochemistry I Homework III Due 10/13

Student________________
Biochemistry I
Homework III
Due 10/13/04
64 points total
(48 points based on text; 16 points for Swiss-PDB viewer exercise)
1). 20 points total
T or F; if false, provide a brief rationale as to why.
_ _ _ _ __ Only charged amino acids play a role in a proteins active site. false uncharged polar residues (e.g., ser, tyr) also participate; also
hydrophobic groups can play a role in substrate specificity
_ _ _ _ _ __ Both enzyme specificity and enzyme catalysis can be explained by the
"induced-fit model" for substrate binding. true
____Enzymes alter the amount of energy consumed (or liberated) in a reaction. F;
∆H remains unchanged in an enzyme catalyzed reaction.
____ Catalysts accelerate a reaction by increasing the ∆G≠ ∆ G≠ is the activation
energy, which they lower.
_____ The steady state assumption in Michaelis-Menten kinetics states that the ES
intermediate is consumed as rapidly as it is formed. T
_ _ _ _ _ A lower activation energy means the forward and reverse reactions both
proceed at faster rates. T
_ _ _ _ _ Biological reactions can have molecularities up to 5. F, never greater than
3
_ _ _ _ _ Vmax is reached when every enzyme is carrying out a chemical reaction. T
_ _ _ _ _ A large Km (10-1 M to 10-3 M) means that substrate binds tightly to enzyme. F;
a large Km indicates weak binding (look at a plot to convince yourself) .
_____ Affinity labels are an example of irreversible inhibitors. T
Student________________
2). (6 points) The curves below represent the effect of adding a positive allosteric
effector or a negative allosteric effector to an enzymatic reaction.
a). Identify on the plot which curve
represents the effect of a positive
allosteric effector and which curve
represents the effect of a negative
allosteric effector. right curve,
negative; left curve, positive.
2 points
b). How do positive and negative
allosteric effectors exert their effect
on an enzyme? 4 points.
In
general, they act by stabilizing different quaternary structures of the
molecule. Positive effectors stabilize quaternary R, negative effectors
stabilize quaternary T structures.
Positive effectors enhance affinity or
increase Km. Negative effectors decrease substrate affinity or decrease
Km.
Student________________
3). 9 points total
Properties of an enzyme involved in prostaglandin synthesis. Prostaglandins are fatty
acid derivatives that are responsible for producing fever and inflammation and its
associated pain. The are derived from the 20-carbon fatty acid arachidonic acid in a
reaction catalyzed by the enzyme prostaglandin endoperoxide synthase. This enzyme
uses oxygen to convert arachidonic acid to PGG2, an immediate precursor to many
different prostaglandins.
The kinetic data are given below in columns 1 & 2 for the reaction catalyzed by
prostaglandin endoperoxide synthase. Kinetic data were also collected in the
presence of ibuprofen, an inhibitor of prostaglandin endoperoxide synthase. Ibuprofen
inhibits the synthesis of protaglandins, thus aiding in reducing inflammation and pain.
The data collected in the presence of ibuprofen are given in column 1 & 3. Analyze
the data using a Lineweaver-Burke plot. What is the Vmax and Km of the
enzyme in the absence of ibuprofen? What kind of inhibitor is ibuprofen?
Why can't we use the direct plots of vo vs. [S] to determine Vmax and
Km? (Attach plots for full credit).
[Arachadonic Acid]
mM
Rate of formation of PGG2
(mM/min)
0.5
1
1.5
2.5
3.5
23.5
32.2
36.9
41.8
44.0
Rte of formation of PGG2
with 10 mg/mL ibuprofen
present
(mM/min)
16.67
25.25
30.49
37.04
38.91
+1 for use of
Lineweaver-Burke plot.
+1 for Vmax value
+1 for Vmax units
+1 for Km value
+1 for Km units.
+2 for type of inhibition
+2 for answer to final
question.
Vmax = 5 1 . 5 5
mM/min; Km =
0.598 mM;
ibuprofen is a
competitive inhibitor.
truly reach Vmax.
Student________________
Can't determine it directly because we can never
. 3 points
r a one-substrate, enzyme catalyzed reaction, double-reciprocal plots (Lineweaver-Burke
ts) were determined for three different enzyme concentrations. Which of the following
ree families of curves would expected to be obtained (circle the appropriate plot). Briefly
plain your choice in terms of the definitions of Vmax and Km.
Middle Graph: If Et is increased, Vmax will increase due to Vmax = kcat *
[E]t. However, Km is independent of enzyme concentration Km = (k -1 +
kcat) / k 1 and only reflects the rate constants for association and
dissociation of the ES complex. Km is unique for each substrate/enzyme
pair and can be viewed very loosely as reflecting the affinity of enzyme
for substrate. Affinity will not change if you have 1 µM or 2 µM or 5 µM…
enzyme concentration.
Student________________
5). 4points
Short answer (no more than 4 grammatically correct sentences)
What is the difference between amino acid sequence and amino acid composition?
What two techniques discriminate between amino acid sequence and amino acid
composition?
Amino acid sequence is the primary structure of a polypeptide (read
from N- to C- terminal) in which the amino acids are covalently linked by
the peptide bond. Protein sequencing by Edman Degradation provides
this information.
Amino acid composition refers to the total number of each amino acid in
a polypeptide chain. Polypeptide hydrolysis through acid conditions and
high temperature gives us this.
Student________________
6). 10 points
Predict the sequence of the peptide based on the given information. Partial credit
(up to 3 points) will be given for correct partial analysis, so justify your reasons.
Total credit ! only if correct sequence given.
a). Amino acid analysis of a heptapeptide gave the following results (in no particular
order
Asp
Glu
Leu
Lys
met
Tyr
Trp
NH4 +
The NH4+ product tells us that we can have Glx (Glu or Gln) or Asx (Asn or Asp)
b). Trypsin had no effect.
Because trypsin cleaves after lys and arg, it must be on the C-terminus
so, we write
x - x - x - x - x - x - LYS
c). The phenylthiohydantion released by Edman degradation was:
we can identify this derivative as tyrosine.
so we write:
TYR - x - x - x - x - x - LYS
d).
Chymotrypsin treatment yielded several products, including a dipeptide and a
tetrapeptide. The amino acid composition of the tetrapeptide was Glx, Leu, Lys and
met
Chymotrypsin cleaves after aromatic residues; so the only we can get a di
peptide and tetrapeptide is if trp is at position 3 .
This tells us the break-down must be:
TYR - X-TRP and
X-X-X-LYS
(or we can write the tetrapeptide
{glx,leu,met} -LYS
as
d). CNBr treatment yielded a tetrapeptide that had a net positive charge at pH 7 and
a tripeptide that a a zero net charge at pH 7.0
CNBr cleaves after Met.
*If CNBr leaves a tetrapeptide and a tripeptide,
TYR - x - TRP- MET - x - x - LYS
it must be at position 4!
This was the tricky part of this problem:
modified C-terminus - one without charge!
Student________________
CNBr treatment leaves a
So, NH3+-Tyr -X-TRP-Carboxy teminus where the C-terminal has no
charge. Based on the information, this must have a net + charge, so the
only way to get that is if asp is really asn!
So, Tyr-Asn-Trp-Met-X-X-Lys
e. Cleavage with Staph protease leaves a dipeptide and pentapeptide
The only way this would work is if glu is the third
so we have
TYR-ASN-TRP-MET-GLU-LEU-LYS
residue from the end,
Student________________
Homework III
Swiss PDB Viewer Exercise
16 points total
Download the structure 1IOZ from the Protein Data Bank Website.
a). This is the H isozyme of lactate dehydrogenase. What is an isozyme? Isozymes
have different quaternary forms that differ in their catalytic ability
b). What is the source of this LDH? human heart
c). What class of enzymes does LDH belong? oxidoreductase
d). What is the cofactor in LDH? NADH
e). What reaction does LDH catalyze?
lactate + NAD+ - ! pyruvate + NADH + H+
f). What is the oligomeric state of this structure? dimer
g). Describe the secondary structural content of LDH:
antiparallel β-sheets stabilized by α-helices (αβα fold)
h). Describe the location of the NADH:
buried in a cleft; stretched across each monomer
i). Oxamate is a competitive inhibitor. What is a competitive inhibitor? This would
affect which Michaelis-Menten parameter? A competitive inhibitor is a
substrate analog that binds at the active site. This would affect Km.
j). Is the oxamate in H-bonding distance from NADH? no
k). What is the distance between the reactive H's of the two molecules of NADH?
27.4 A
l). The active site is comprised of residues 99-110. What is the location of the active
site? What is its structure? surface, loop
m). What H-bond(s) are formed between oxamate and the active site loop?
Two of the Arg 106 side chain NHs with C=O of oxamate.