This multiple choice test assessed knowledge of enzyme kinetics, glycolysis, and membrane structure. It included 15 multiple choice questions and 3 short answer questions. The short answer questions asked about: 1) drawing glycophospholipids and explaining bilayer formation, 2) describing an integral membrane protein, and 3) outlining steps to synthesize an antibody to break down oxycontin and why a carrier protein is used.

1. Multiple Choice
1. One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis.
Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be
referred to as the:
a. holoenzyme
b. prosthetic group
c. apoenzyme
d. coenzyme
e. substrate
2. Compare the two reaction coordinate diagrams below and select the answer that
correctly describes their relationship. In each case the single intermediate is the
ES complex.
a. The ES complex is given by #2 in (a) and #3 in (b).
b. The activation energy for the catalyzed reaction #5 in (a) and is #7 in (b).
c. (a) describes a strict “lock and key” model whereas (b) describes a transition-
state complementary model.
d. The activation energy for the uncatalyzed reaction is given by #5 + #6 in (a)
and by #7 + #4 in (b).
e. The contribution of binding energy is given by #5 in (a) and by #7 in (b).

2. 3. The concept of “induced fit” refers to the fact that:
a. when a substrate binds to an enzyme, the enzyme induces a loss of water
(desolvation) from the substrate.
b. substrate binding may induce a conformational change in the enzyme, which
then brings catalytic groups into proper orientation.
c. enzyme-substrate binding induces an increase in the reaction entropy, thereby
catalyzing the reaction.
d. enzyme specificity is induced by enzyme-substrate binding.
e. enzyme-substrate binding induces movement along the reaction coordinate to
the transition state.
4. Which of the following statements about a plot of V0 vs. [S] for an enzyme that
follows Michaelis-Menten kinetics is false?
a. Km is the [S] at which V0 = ½ Vmax.
b. The shape of the curve is a hyperbola.
c. The y-axis is a rate term with units of µm/min.
d. As [S] increases, the initial velocity of reaction, V0, also increases.
e. At very high [S], the velocity curve becomes a horizontal line that intersects
the y-axis at Km.
5. Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed
reaction could be written as
k1 k2
E+S ↔ ES → P
k-1
Using this reaction, the rate of breakdown of the enzyme-substrate complex can
be described by the expression:
a. k1 ([Et] - [ES])[S]
b. k-1 [ES] + k2 [ES]
c. k2 [ES]
d. k-1 [ES]
e. k1 ([Et] - [ES])
6. An enzyme-catalyzed reaction was carried out with the substrate concentration
initially 1,000 times greater than the Km for that substrate. After 9 minutes, 1% of
the substrate had been converted to product, and the amount of product formed in
the reaction mixture was 12 µmol. If, in a separate experiment, one-third as much
enzyme and twice as much substrate had been combined, how long would it take
for the same amount (12 µmol) of product to be formed?
a. 1.5 min
b. 3 min
c. 6 min
d. 13.5 min
e. 27 min

3. 7. In competitive inhibition, an inhibitor:
a. binds at several different sites on an enzyme.
b. binds reversibly at the active site.
c. binds only to the ES complex.
d. binds covalently to the enzyme.
e. lowers the characteristic Vmax of the enzyme.
8. A transition state analog:
a. resembles the transition-state structure of the normal enzyme-substrate
complex.
b. typically yields product more rapidly with an enzyme than the normal
substrate.
c. is less stable when binding to an enzyme than the normal substrate.
d. stabilizes the transition state for the normal enzyme-substrate complex.
e. resembles the active site of general acid-base enzymes.
9. Both water and glucose share an -OH that can serve as a substrate for a reaction
with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however,
is about a million times more reactive as a substrate than water. The best
explanation is that:
a. glucose has more -OH groups per molecule than does water.
b. the larger glucose binds better to the enzyme; it induces a conformational
change in hexokinase that brings active-site amino acids into position for
catalysis.
c. water normally will not reach the active site because it is hydrophobic.
d. water and the second substrate, ATP, compete for the active site, resulting in a
competitive inhibition of the enzyme.
e. the -OH group of water is attached to an inhibitory H atom while the glucose
-OH group is attached to C.
10. Which of these statements about the composition of membranes is generally true?
a. The lipid composition of all membranes of eukaryotic cells is essentially the
same.
b. All biological membranes contain cholesterol.
c. Free fatty acids are major components of all membranes.
d. The inner and outer membranes of mitochondria have different protein
compositions.
11. The fluidity of a lipid bilayer will be increased by:
a. decreasing the number of unsaturated positions.
b. increasing the length of the alkyl chains.
c. increasing the temperature.
d. decreasing the temperature.

4. 12. Which of the following statements about fatty acids is true?
a. Fatty acids with longer chain lengths have a higher melting point than fatty
acids with shorter chain lengths.
b. Saturated fatty acids have a lower melting point than unsaturated fatty acids.
c. Cis double bonds of unsaturated fatty acids cause tighter packing of
hydrophobic tails.
d. Double bonds in polyunsaturated fatty acids are almost always conjugated.
e. Fatty acids are found mostly as free acids.
13. Which of the following statements is not indicative of passive transport? Check
all that apply.
a. No energy is required.
b. ATP is required for transport.
c. Small organic molecules will diffuse based on molecular weight and solubility
in lipids.
d. Molecules are transported from low concentration to high concentration.
e. Molecules are transported from high concentration to low concentration.
14. Check all of the following that apply.
Enzyme catalysts:
a. shift the equilibrium of a reaction to favor products.
b. form asymmetric active sites due to their inherent chirality.
c. are sometimes permanently altered after releasing product.
d. often require prosthetic groups, which are transiently associated with a given
enzyme.
e. are often regulated by covalent modifications.
15. Lysozyme has a pH centered around pH 5.0. The active site of lysozyme contains
a glutamic acid residue (pKa = 5.5) and an aspartic acid residue (pKa = 4.0).
Which of the following statements is correct about the mechanism of lysozyme?
a. The glutamic acid residue is in a more polar environment than the aspartic
acid.
b. During the entire catalytic mechanism, the aspartic acid residue remains
unprotonated.
c. During the mechanism, an oxyanion transition state forms.
d. The glutamic acid residue acts as a general base catalyst.
16. The conversion of NAD+ to NADH is an example of reduction because
a. the pyridine ring loses electrons (and a hydrogen).
b. the pyridine ring gains electrons (and a hydrogen).
c. the adenine ring loses electrons.
d. the adenine ring gains electrons.

5. 17. The process of glycolysis
a. requires a pathway of chemically coupled phosphoryl-transfer reactions.
b. uses 2 ATP molecules and forms 2 ATP molecules and one NADH molecule.
c. occurs in the mitochondria.
d. converts glucose into two glycerate molecules.
18. Which of the following is not true of the reactions and enzymes in the glycolytic
pathway?
a. Triose phosphate isomerase is referred to as a “perfect enzyme” because
product is formed as soon as enzyme and substrate collide.
b. Phosphoglucose isomerase converts the glucose ring to a fructose ring.
c. Domains of phosphoglycerate kinase clamp down on the substrate so as to
exclude water from the reaction.
d. The reaction converting 2-phosphoglycerate to phosphoenolpyruvate is a
dehydration reaction.
e. Phosphofructokinase is stimulated by ATP and citrate.
19. Which statement about the gluconeogenesis pathway is false?
a. Most of the same glycolytic enzymes are utilized in gluconeogenesis.
b. The decarboxylation of oxaloacetate provides the free energy required for
synthesis of phosphoenolpyruvate.
c. Pyruvate carboxylase contains a biotin prosthetic group, which functions as a
CO2 carrier.
d. Oxaloacetate is generated in the mitochondria.
e. The hexokinase and phosphoglycerate kinase reactions of glycolysis are
reversed in gluconeogenesis via hydrolytic reactions that release Pi in
exorgonic processes.
20. Which of the following statements about glycogen metabolism is/are false?
I. Release of epinephrine leads to increased glycogen breakdown.
II. cAMP is a second messenger that triggers an increase in glycogen
synthesis.
III. Insulin results in dephosphorylation and subsequent activation of
glycogen synthase.
IV. Glucagon promotes glycogen breakdown.
V. Glucose-6-phosphate is an allosteric inhibitor of phosphorylase.
a. II only
b. V only
c. III and IV
d. II and V
e. III and V

6. Written Answers
1.) a. Draw the basic chemical structure of all glycophospholipids, (2 points)
b. (4 points) (i) Explain why phosphoglycerides are capable of spontaneously
assembling into the bilayer structure found in biological membranes
but triacylglycerols are not.
(i) Triacylglycerols have three fatty acyl groups in ester linkage with
glycerol; they are very hydrophobic because the carboxyl groups, which
are involved in the ester linkages, cannot ionize. Phosphoglycerides have a
polar region at their head group, where a phosphate in a phosphodiester
linkage bears a full negative charge. The head group itself (serine,
ethanolamine, choline, etc.) may also be charged and is in any case polar.
Thus the phospholipids is amphipathic, having both polar and nonpolar
regions, and it forms lipid bilayers spontaneously in water.
(ii) What are the forces that drive bilayer formation?
(ii) These lipid bilayers are stabilized by the energy gained from burying
hydrophobic groups out of contact with water. A hydrophobic chain in
water forces the formation of a cage of immobilized water molecules
around it. When several hydrophobic regions cluster in a bilayer, the
surface area exposed to water decreases, and the water molecules in the
cage are released, accompanied by a gain in entropy that drives the
formation of the bilayer.

7. c. (4 points) A protein is found to extend all the way through the membrane of a
cell.
Describe this protein in terms of the location of particular types of amino acid
side chains in its structure and its ability to move within the membrane.
This integral membrane protein associates with the lipid bilayer through
hydrophobic interactions between domains containing many hydrophobic
amino acids and the fatty acyl chains of membrane lipids. Polar and
charged residues are located on portions of the protein that protrude out of
either face of the membrane. The protein is free to diffuse laterally in the
plane of the membrane, but cannot move across the lipid bilayer.
2.) a. For the reaction E + S ↔ ES → E + P, the Michaelis-Menten constant, Km, is
actually
a summary of three terms. What are they? (2 points) How is Km determined
graphically? (2 points)
Km = (k2 + k-1)/k1 where k-1 and k1 are the rate constants for the breakdown and
association, respectively, of the ES complex and k2 is the rate constant for the
breakdown of ES to form E + P. Km can be determined graphically on a plot of
V0 vs. [S] by finding the [S] at which V0 = ½ Vmax. More conveniently, on a
double plot, the x-axis intercept = -1/Km.
b. A biochemist obtains the following set of data for an enzyme that is known to
follow Michaelis-Menten kinetics. (4 points)
Substrate Initial
Concentration velocity
(µM) (µmol/min)
1 49
2 96
8 349
50 621
100 676
1000 698
5000 699

8. (i) Vmax for the enzyme is _____________________. Explain in one
sentence how you determine Vmax.
Vmax is about 700. In a plot of V vs. [S], the asymptote is Vmax. Simple
inspection of the data shows the approach to Vmax ---the rate increases by
only 1 unit when [S] increases fivefold.
(ii) Km for the enzyme is __________________________-. Explain in one
sentence how you determined Km.
Km is about 8 µM, the [S] at which the velocity is half-maximal. Because
Vmax is about 700, 12 Vmax is about 350. The [S] at that rate is about
8 µM.
c. Why is the Lineweaver-Burk (double-reciprocal) plot more useful than the
standard V vs. [S] plot in determining kinetic constants for an enzyme? (Your
answer should show typical plots with labeled axes). (3 points)
The plot of V vs. [S] is hyperbolic; maximum velocity is never achieved
experimentally, because it is impossible to do experiments at infinitely
high [S]. The Lineweaver-Burk transformation of the Michaelis-Menten
equation produces a linear plot that can be extrapolated to infinite [S]
(where 1/[S] becomes zero), allowing a determination of Vmax.

10. 3.) You are faced with the task of synthesizing an effective catalytic antibody or
abzyme for the breakdown of the drug oxycontin which has the following
structure:
a. Briefly describe the steps that you would take to make such an antibody.
(i) Decide first in what manner the oxycontin might be broken down. For
example, one possibility you might think of is hydrolysis, i.e. the breaking of
bonds with the participation of water. Are there any regions of oxycontin that
might be susceptible to such a reaction? There are many other sorts of reactions
` you might imagine. The main thing is that you first envision the type of reaction
to be accomplished.
(ii) Given a particular type of reaction to be accomplished one then must design
and synthesize a stable transition state analog. Part of the design of such an analog
will be the incorporation of a group or region on the molecule that will enable it to
be attached to a carrier molecule (see part (b) below). This attachment site needs
to be designed in such a manner as to not interfere drastically with the availability
of the major transition state part of the molecule that will elicit antibodies.
(iii) collect blood & purify antibodies
(iv) test enzymatic activity of antibody preparation
(an alternative to the preparation of polyclonal antibodies described above is to prepare
monoclonal antibodies) (5 points)

11. b. Very often the antigen used to elicit a catalytic antibody is conjugated to a
carrier protein prior to injection into the animal that will be used to generate the
antibodies. Why is this?
Small antigens are typically combined with large carrier molecules such as
proteins because (i) the small molecules are very inefficient in generating
antibodies and the small molecules are often very rapidly removed from tissues in
the bloodstream and (ii) the carrier protein can often act as an adjuvant or
stimulant for antibody production (2 points)
(for grading part (i) is sufficient for 2 points)
c. You finally succeed in making a catabolic antibody of the desired
characteristics. How could you detect whether the antigen you used (minus the
carrier) was a competitive inhibitor of this enzyme?
Run a series of enzyme assays, varying the substrate concentration, in the
presence and absence of the antigen (the possible competitive inhibitor). Plot the
data using a Lineweaver-Burk plot (or equivalent). If the transition state analog is
a competitive inhibitor then the Lineweaver-Burk plot should look like Fig. 14-12
(below) with the lines with and without inhibitor intersecting on the Y axis.
(3 points)

12. 4.) Here’s the BLOSUM50 amino acid substitution matrix used to align protein
sequences:
A R N D C Q E G H I L K M F P S T W Y V
A 5 -2 -1 -2 -1 -1 -1 0 -2 -1 -2 -1 -1 -3 -1 1 0 -3 -2 0
R -2 7 -1 -2 -4 1 0 -3 0 -4 -3 3 -2 -3 -3 -1 -1 -3 -1 –3
N -1 -1 7 2 -2 0 0 0 1 -3 -4 0 -2 -4 -2 1 0 -4 -2 -3
D -2 -2 2 8 -4 0 2 -1 -1 -4 -4 -1 -4 -5 -1 0 -1 -5 -3 -4
C -1 -4 -2 -4 13 -3 -3 -3 -3 -2 -2 -3 -2 -2 -4 -1 -1 -5 -3 -1
Q -1 1 0 0 -3 7 2 -2 1 -3 -2 2 0 -4 -1 0 -1 -1 -1 -3
E -1 0 0 2 -3 2 6 -3 0 -4 -3 1 -2 -3 -1 -1 -1 -3 -2 -3
G 0 -3 0 -1 -3 -2 -3 8 -2 -4 -4 -2 -3 -4 -2 0 -2 -3 -3 -4
H -2 0 1 -1 -3 1 0 -2 10 -4 -3 0 -1 -1 -2 -1 -2 -3 2 –4
I -1 -4 -3 -4 -2 -3 -4 -4 -4 5 2 -3 2 0 -3 -3 -1 -3 -1 4
L -2 -3 -4 -4 -2 -2 -3 -4 -3 2 5 -3 3 1 -4 -3 -1 -2 -1 1
K -1 3 0 -1 -3 2 1 -2 0 -3 -3 6 -2 -4 -1 0 -1 -3 -2 -3
M -1 -2 -2 -4 -2 0 -2 -3 -1 2 3 -2 7 0 -3 -2 -1 -1 0 1
F -3 -3 -4 -5 -2 -4 -3 -4 -1 0 1 -4 0 8 -4 -3 -2 1 4 -1
P -1 -3 -2 -1 -4 -1 -1 -2 -2 -3 -4 -1 -3 -4 10 -1 -1 -4 -3 -3
S 1 -1 1 0 -1 0 -1 0 -1 -3 -3 0 -2 -3 -1 5 2 -4 -2 -2
T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 2 5 -3 -2 0
W -3 -3 -4 -5 -5 -1 -3 -3 -3 -3 -2 -3 -1 1 -4 -4 -3 15 2 -3
Y -2 -1 -2 -3 -3 -1 -2 -3 2 -1 -1 -2 0 4 -3 -2 -2 2 8 -1
V 0 -3 -3 -4 -1 -3 -3 -4 -4 4 1 -3 1 -1 -3 -2 0 -3 -1 5
a. Some amino acids are “specialists” and play very specific roles in proteins that can be
difficult to fill using other amino acids. From the table, which single amino acid is most
likely not to be substituted for by another? How can you tell?
W (has the highest score on the diagonal, meaning that other residues substitute poorly
for it.) (2 points)
b. Some amino acids are more “generic” in their properties than others. From the table,
which 6 amino acids are most easily substituted for by others? How can you tell?
A, I, L, S, T, V (have lowest scores on the diagonal) (3 points)
c. What are the 3 most disfavored amino acid substitutions?
W for D, W for C, F for D (3 points)
d. Sequence alignment programs like BLAST work well for proteins and DNA, but very
poorly for aligning RNA molecules. Why is that the case?
BLAST assumes that each position in the alignment is independent of positions far away,
and ignores correlations between the sites. RNA molecules have many interactions in the
2D and 3D structure that require base-pairing between distant bases, so the positions are
strongly correlated, and accounted for poorly by this approximation. (2 points)

13. 5.) Consider the process of fermentation (13 metabolites / 12 enzymes) as
summarized below. Then answer the questions at right. Identify an enzyme by
its two metabolites.
1. Glucose (1)a) Identify compound Z=___acetaldehyde_________________
2. Compound “U” (2) b) F-2,6-BP (activates / inactivates) the
3. Fructose-6-P enzyme that converts metabolites _3__ to _4__.
4. Compound “V” (1) c) ATP is produced in glycolysis at step(s).
5. 3-P-Glycerate + DHAP _6_ to _7_, (and _10_ to _11_, __ to __, __ to
__)
+
6. Compound “W” (1) d) NAD is produced as _12___ goes to _13___.
7. 1,3 BisP-glycerate (1) e) The conversion of _6__ to _7__ involves the
8. Compound “X” a thiohemiacetal intermediate.
9. 2-Phosphoglycerate (1) f) The conversion of _8__ to _9__ involves the
10. Compound “Y” a phosphoryl-His group intermediate
11. Pyruvate (1) g) Oxidation during glycolysis takes place
12. Compound “Z” in converting _6__to _7__.).
13. Ethanol
Where (which carbon position) would 14C have to be located in glucose to ensure
that all the 14C activity was liberated as 14CO2 during alcoholic fermentation into
ethanol? __3 or 4___ (2 points)

14. +
6.) The standard reduction potential for the reduction of NAD to NADH is –0.315V
and that for the reduction of pyruvate to lactate is –0.185V. (1 F = 96,500 J / V)
a. Calculate the standard free energy (∆Go’) for the oxidation of lactate to
+
pyruvate by NAD . (4 points)
b. What is the equilibrium constant for this reaction? (2 points)
c. Draw the structure of NADH. (4 points)