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BIOCHEMISTRY (CHEM 360)
EXAM 2
October 24, 2014
There are seven pages, ELEVEN questions, and a total of 114 points in this exam.
Please read each question carefully and possibly more than once. Good luck…
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1.
The active site of trypsin is illustrated below. Show the acyl enzyme formed after the hydrolysis of
the heptapeptide below (but before the hydrolysis of the acyl enzyme). You can use the template
for the active site, but you need to show the full structure (no “R” groups) of the acyl group.
1.1.
Circle the portion of the heptapeptide which fits into the specificity pocket.
(6)
2.
Physiological effects in the body usually compensate for environmental changes, as part of an
adaptation process. For example, a person at a high altitude would have increased levels of
2,3-BPG. Speculate possible reasons for this effect.
Oxygen concentration in the atmosphere decreases with increasing altitude.
Increased levels of 2,3-BPG increases the oxygen unloading of hemoglobin, thus preventing any
oxygen deficiency in the tissues.
(8)
3.
Thiamine pyrophosphate (TPP) shown below undergoes an acid-base reaction with an appropriate enzyme
to form a reactive species capable of forming a covalent bond with the proper substrate. Give the structure
of the reactive species. Use a generic enzyme and one-or-two curly arrows to illustrate the acid/base
reaction.
3.1.
The reaction below a substrate-TPP intermediate in the presence of a decarboxylase enzyme. Give
the structure of the previous intermediate, immediately before the reaction shown.
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4.
Complete the following isomerization reaction by providing structures of all the missing intermediates and
products. Show hydrogen ion transfers and brief details of mechanism.
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5.
Circle the best correct answer for the multiple choice questions given below:
5.1.
Carbon monoxide poisoning is due to the following interaction/change(s) in hemoglobin:
(1) Changing the oxidation state of iron.
(2) The tight CO bonding to the iron.
(3) The inability of CO to form a carbamoyl group on the N-terminus.
(4) The ability of CO to replace Fe2+ in the heme ring.
5.2.
Which of the following statements illustrates that hemoglobin is an allosteric
protein?
(1) 2,3-BPG replaces oxygen.
(2) Hemoglobin is a transporter.
(3) 2,3-BPG interacts with amino acid side chains in a different location than the
heme.
(4) Binding of oxygen causes conformational changes at locations other than the heme.
(5) Hemoglobin undergoes severe steric interactions.
5.3.
In a Michaelis-Menten plot of enzyme activity (reaction rate v.s. substrate concentration),
the reason that the curve reaches a plateau, and that it does not increase any further upon
adding more substrate, is the following:
(1)
(2)
(3)
(4)
(5)
5.4.
Which of the following reflects the mechanism of action for organophosphate nerve
agents such as Sarin?
(1)
(3)
(4)
(5)
5.5.
The active site is saturated with substrate
There is a competitive inhibitor present
There is a non-competitive inhibitor present
The enzyme is locked in an inactive conformation
All the substrate is converted into the product
Covalent modification of hemoglobin (2) Reversible modification of hemoglobin
Covalent inhibition of acetylcholine esterase
Reversible inhibition of acetylcholine esterase
Inhibition of chymotrypsin
The Michaelis-Menten constant, Km, is
(1)
(2)
(3)
(4)
The maximum velocity that any given enzyme reaction can achieve.
The maximal velocity, vmax, divided in half.
The measure of the catalytic activity of an enzyme.
The substrate concentration when the reaction is half-way toward the maximal
velocity.
(5) The substrate concentration which gives the best enzyme assay for an enzyme
reaction.
5.6.
Which of the following statements indicating the differences is between myoglobin and
hemoglobin is not correct?
(1) Hemoglobin consists of tetrameric units; whereas myoglobin is a monomeric protein.
(2) Partial pressure at half-saturation (P50) is lower for myoglobin.
(3) The oxygen binding curve of myoglobin is hyperbolic; whereas that of hemoglobin is
sigmoidal (S-shaped).
(4) Hemoglobin has higher affinity for oxygen compared to myoglobin.
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6.
The repeating unit of a bacterial cell wall polysaccharide is given below. However, the bacterial
cell wall is in fact constructed as a peptidoglycan “architecture”. Describe what is meant by a
peptidoglycan “architecture” by showing how and to what functionality the unit below is linked.
Note that you are not expected to give a whole cell wall structure here; it will be sufficient to link
a correct functional group to the correct position of the unit below.
(6)
7.
Structures of Vitamin A and Vitamin C are given below. Indicate which structure corresponds to
Vitamin A and which to Vitamin C.
Vitamin A
7.1.
It is popularly accepted that high doses of vitamin C prevent the common cold (although
not clinically/biochemically proven); whereas high doses of Vitamin A could be quite
hazardous. Explain why high doses of Vitamin C are tolerable, even if there are no
proven benefits; on the other hand excessive Vitamin A could pose a serious danger.
Vitamin C is water soluble; therefore can be excreted easily from the body. Vitamin A is
lipophilic and is likely to accumulate in the body particularly in the liver if ingested in
high doses.
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8.
NAD+ and lipoic acid, shown below, interact with same type of enzymes. How are these enzymes
classified?
Oxidase (or oxido-reductase)
NAD+ binds to enzymes non-covalently;
Whereas lipoic acid forms covalent bonds with free amino groups,
such as in lysine residues.
8.1.
Although the enzyme function is the same, there is usually a difference in the way each
of the two metabolites shown above interacts with the enzyme it is associated with.
Describe the difference (an illustration might be helpful).
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9.
One of the most significant steps in the cross-linking of polysaccharides in bacterial cell walls
(S. aureus) is the formation of an amide bond between the carboxyl group of D-alanine residue
and a pentaglycine bridge as shown below.
9.1. Give a narrative outline of the “linking”
process. Comment on the types and
functions of the enzyme(s) involved.
Try not to write any structures.
9.2. What activity would you target, if you
were working to design a drug which
would not allow the linking process to
take place?
Do not propose any specific drugs; just
summarize your strategy.
9.1.
Ironically, the amide bond formation is catalyzed by a “carboxypeptidase” enzyme,


9.1.
The cross-linking process
involves an amide bond between
the

N-terminal glycine and the carbonyl
end of the 4th amino acid. This
would

require the activation of the
 carbonyl
group, which is possible with
 a serineprotease type enzyme cleaving
the

bond between the two D-ala
residues.

The resulting acyl enzyme is likely to

undergo attack by the N-terminal gly

forming the amide linkage shown
above.
9.2.
If the drug was designed to
inhibit the serine-protease type
enzyme, then carbonyl end of the 4th
amino acid would not get activated
and the cross-linking between gly and
D-ala would not take place.
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which is known to cata


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10.
The generic equation for an enzyme catalyzed reaction is given below:
10.1.
The Michaelis-Menten Equation is derived when the equation above is under
“steady-state” conditions. What is meant by “steady-state” conditions?
The rate of formation of the “enzyme-susbtrate complex” is equal to its rate of
dissociation.
10.2.
The enzyme catalyzed reaction may be influenced by different types of inhibition when
treated by inhibitors. Circle the region in the equation below, which illustrates
“non-competitive” inhibition.
10.3.
Under what conditions would the non-competitive inhibition would not be “purely”
non-competitive and have an uncompetitive edge?
Use some of the parameters above to answer the question.
When K2 > K1
10.4.
The magnitude of the rate constants shown above is a qualitative indication of how
tightly bound the enzyme substrate complex is.
The enzyme substrate complex is tightly bound if the relative magnitude(s) of the
following rate constant(s) is/are high “ k1” (fill in the blanks)
The enzyme substrate complex is loosely bound if the relative magnitude(s) of the
following rate constant(s) is/are high “k-1 and k2”
(fill in the blanks)
Please do not insert Km in the blank spaces to answer the questions above.
10.5.
Give an expression for the Michaelis-Menten constant, Km, using the appropriate
parameters in the equation.
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11.
A classical study carried out in a lab with simple non-biological reactants (as shown below) helped
to elucidate an important principle observed with reactions at enzyme active sites.
The results below show that product mixture II is formed much faster (107 fold) than product
mixture I.
Both products contain identical functional groups.
11.1.
The principal/effect mentioned above is referred to as the proximity effect
(fill in the blank)
11.2.
Identify two possible reasons (not necessarily biochemical) why product II is formed
faster than product I.
Intramolecular reaction
Restriction of free rotation
11.3.
Draw the structures for product mixture II.


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