PHRM 836 Exam I key - 1
Examination I Key
PHRM 836 – Biochemistry for Pharmaceutical Sciences II
September 24, 2013
Correct answers in multiple choice questions are indicated in RED and underlined.
Correct answers to essay questions are indicated in RED in comic book font.
In some cases and explanation is provided in BLUE/BLUE
1. A relaxed circular 400 bp long dsDNA is compared to the same DNA with a single supercoil. Under
physiologic conditions but with no protein present, the supercoiled DNA has:
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more twisting distortion and more bending distortion than the relaxed circular DNA
more twisting distortion and the same bending distortion as the relaxed circular DNA
more twisting distortion and less bending distortion than the relaxed circular DNA
the same twisting distortion and more bending distortion than the relaxed circular DNA
the same twisting distortion and the same bending distortion as the relaxed circular DNA
the same twisting distortion and less bending distortion than the relaxed circular DNA
less twisting distortion and more bending distortion than the relaxed circular DNA
less twisting distortion and the same bending distortion as the relaxed circular DNA
less twisting distortion and less bending distortion than the relaxed circular DNA
In its lowest energy state the sc DNA would be in a plectonemic form (a figure 8), but would have the
normal number of helices per 400 bp. Thus the sc DNA would have no twisting distortion. The
plectonemic form actually has as much bending as a toroidal form, which would be two loops with
much tighter bending then the one larger circular DNA (when relaxed). Hence there is more bending
in the plectonemic form than the relaxed circular form.
2. The function of ATP hydrolysis by all type II topoisomerases is to
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hydrolyze the G-DNA phopsphodiester bonds
re-ligate the G-DNA
wind the DNA around the CTD
unwind the DNA from around the CTD
induce a large conformational change in the protein that is essential for the overall reaction
activate a tyrosine that is involved in breaking and re-ligating the G-DNA
3. Camptothecin is a therapeutic drug that directly inhibits
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telomerase
DNA polymerase
type I topoisomerases
type II topoisomerases
primase
DNA helicase
RNA polymerase II
cyclin-dependent protein kinases
nucleotide excision DNA repair
mRNA splicing
PHRM 836 Exam I key - 2
4. DNA helicases carry out their function by
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binding both bases in the next base pair and breaking the hydrogen bonds between them
binding to and twisting the deoxyriboses involved in the next base pair
moving along one of the DNA strands
moving along both the separated DNA strands
moving along the dsDNA, pulling a trailing plow-like region of the protein that splits the DNA
5. A licensing factor is a protein that
 is made during the G1 phase of the cell cycle and activates cdks
 is made during the G1 phase of the cell cycle and inhibits cdks
 is made during the G1 phase of the cell cycle, is essential for initiation of replication, then is
degraded immediately
 is made throughout the cell cycle but is degraded as soon as it is used
 is made in a cell cycle dependent manner, with levels peaking at a particular point in the cell cycle,
and activates cdks
 is made in a cell cycle dependent manner, with levels peaking at a particular point in the cell cycle,
and inhibits cdks
6. At the initiation of replication in mammals, the DNA is initially melted by the action of
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TFIIH
hexameric MCM
a dimer of hexameric MCMs
ORC
cdk
topoisomerase II
core histones
7. An example of an anti-cancer therapeutic agent that acts via reacting with DNA such that it creates
double strand DNA breaks is
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camptothecin
etoposide
5-flurouracil
methotrexate
cisplatin
gleevec
8. For a cellular gene to become an oncogene that causes cancer it must be mutated so that it is
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not expressed or is expressed at a level much below normal
either not expressed (or under-expressed) or is expressed in a non-functional form
expressed at a higher level than normal
expressed in a more active form than normal
expressed either at a higher level than normal or in a more active form than normal
PHRM 836 Exam I key - 3
9. The initial DNA melting that occurs with the start of the synthesis of each mRNA transcript in
eukaryotes is produced by
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TBP binding to the TATA box
a DNA helicase that is part of TFIIH
RNA polymerase II
phosphorylation of DNA by a cdk
MCM
topoisomerase
10. The complexly interacting regulation of transcription by transcription regulatory factors is called
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a licensing system
a gene regulatory network
checkpoint control
epigenetics
DNA remodeling
enhancer element interactions
distal promoter element interactions
modular promoters
alternative promoters
RNA interference
11. Recent genome-wide studies, including ENCODE, show that 95% of the human genome is transcribed.
This has led to a proposal for a novel type of regulation of transcription that involves
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RNA splicing
RNA degradation at the exosome
chromatin remodeling
DNA methylation
transcription at one promoter affecting the transcription initiation at a nearby promoter
RNA interference
Licensing factors
12. If an exon in a gene is 305 bp long, and this exon includes an alternative 5´ splice site that divides this
exon into left and right parts, what length constraint commonly applies to the location of this alternative
splice site in this exon?
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The left part must be shorter than the right part
The right part must be shorter than the left part
The left part must be a multiple of 3 bp long
The right part must be a multiple of 3 bp long
The two parts must differ in length by a multiple of 3 bp
If the right part (which is optionally included or not) is not a multiple of 3 bp long, it will change the
reading frame when it is included versus not included in the mature mRNA. In rare cases, this
constraint might not apply: such as it is outside the ORF or it is paired with an alt 3´ splice site that
immediately follows it.
PHRM 836 Exam I key - 4
13. All of the following describe “protein recognition” except which phrase?
 contributes to the specificity of a protein-protein interaction
 arises from the chemical complementarity between an enzyme active site and substrate
 arises from the spatial complementarity between an enzyme active site and a heterotropic effector
 is involved with Hb binding O2.
 is involved with small-molecule antigens binding antibodies
 is involved with the association of subunits in an oligomeric protein
14. In chymotrypsin, serine transfers a proton to a histidine. Which of the following is correct?
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Serine is acting as a transition-state stabilization catalyst.
Chymotrypsin is relatively insensitive to pH changes.
Histidine is acting as a general acid.
Serine is acting as a general acid.
A covalent intermediate is formed between serine and histidine.
The above statement is false because serine transfers a proton to aspartic acid.
15. Which phrase best describes the structure/function relationship of serine proteases?
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all serine proteases have the same tertiary structure
the S1 residue is a serine
the P1 residue must be a Trp
the spatial orientation of the catalytic triad of Ser, His, Asp is conserved
P1 refers to one type of serine protease that recognizes the substrate S1
the active form of a serine protease is converted to the inactive zymogen form
serine proteases catalyze hydrolysis of polypeptides with Ser
none of the above
16. Which phrase best explains why a given antibody has very high affinity and specificity for a particular
antigen?
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hydrophobicity
disulfide bonds
hypervariable loops
IgG fold
C domains
-sandwich
carbohydrate linkages
PHRM 836 Exam I key - 5
17. The energy diagram below shows the energy as a function of reaction coordinate for an uncatalyzed and
enzyme-catalyzed reaction. Which phrase accurately describes an enzyme-catalyzed reaction?
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GO is the same for both uncatalyzed and catalyzed reactions.
At equilibrium, more product is formed for the catalyzed reaction than the uncatalyzed reaction.
At equilibrium, more product is formed for the uncatalyzed reaction than the catalyzed reaction.
The enzyme stabilizes the transition state and forms a stable transition-state complex.
The rate of the enzyme-catalyzed reaction increases as the energy of the ES complex decreases.
The transition state is not a stable complex and the enzyme does not stabilize the transition state.
E* is the energy difference between the catalyzed and uncatalyzed transition states of the reaction.
choices  and 
choices  and 
18. Which of the following phrases does NOT accurately describe the cooperativity of hemoglobin?
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interactions between the FG corner of one subunit and the C helix of an adjacent subunit
six-coordinated Fe2+ moves into the heme plane
rotation of one  dimer relative to the other  dimer
stabilizing oxyhemoglobin shifts the curve of saturation vs pO2 to the right.
His F8 interaction with heme iron contributes to the cooperativity mechanism
conversion of the deoxy form to oxyhemoglobin releases protons
the R state is stabilized by oxygen binding
none of the above (all statements are accurate)
19. Based on your knowledge about how 1,6 bisphosphoglycerate (BPG) alters the affinity for oxygen by
hemoglobin and about the structure of hemoglobin, what is the most probable site for the binding of
BPG?
 BPG binds the heme Fe and displaces oxygen
 BPG binds the heme on the opposite side of oxygen and causes the opposite direction of structural
changes as oxygen
 BPG binds at the 11 interface
 BPG binds at the interface between the two  dimers
 BPG binds on the surface near the mutation site the causes sickle cell Hb
 The binding site of BPG cannot be inferred from knowing its affect on O2 affinity and Hb structure
PHRM 836 Exam I key - 6
20. The phrase that accurately describes Hb structure or function is
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a homotetramer of all- chains
a dimer of dimers and two heme groups
four polypeptide chains with ionic interactions at the subunit interfaces
undergoes a monomer to tetramer equilibrium
four Mb chains associate to form Hb
multiple copies of the Ig fold
choices  and 
none of the above
21. Enzyme catalysis is regulated by a number of mechanisms. Also, the flux through metabolic pathways
must be regulated. Given the metabolic pathway below, which choice accurately describes the regulation
of compound G levels?
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G is regulated by chemical modification of the enzyme that catalyzes step 3
G serves as an inhibitory effector of the allosteric enzyme that catalyzes step 3
G inhibits the expression of the enzyme that catalyzes the formation of D
The reversible reaction of B to D is converted to an irreversible reaction.
G destabilizes the complex of D with the enzyme that catalyzes step 3
Both  and 
none of the above
22. One gene for alcohol dehydrogenase (ADH) has 3 types of alleles that differ in a single base. The gene
products, ADH 1, 2 and 3, have substitutions at position 47 and 369 (Table below). The crystal
structure of ADH 1 holoenzyme shows that Arg47 and Arg369 form ionic interactions with the
phosphate groups of NAD. Residues 47 and 369 are assumed to also be in close contact with the same
region of NAD in the 2 and 3 holoenzyme complexes. The pH dependence of catalysis differs for 1,
2 and 3 primarily as a result of ADH interaction with NAD. The following phrases are an accurate
statement of pH profiles of enzyme catalysis in general or for ADH excluding which phrase?
Position 47 Position 369 pH optimum
10
Arg
Arg
1
8.5
His
Arg
2
~7
Arg
Cys
3
The pH optimum near 10 for 1 reflects the pKa of Arg.
The KM for NAD is lower for 1 than 2 at pH 9.
A change in KM for substrate can change a pH profile.
A change in KM for cofactor can change a pH profile.
The pH dependence of ADH 3 catalysis indicates the KM for NAD decreases above pH 8.
Based on the crystal structure of the holoenzyme complex, 3 is likely to have lower affinity for
NAD than 1 or 2.
 none of the above (all are accurate statements)
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PHRM 836 Exam I key - 7
23. In the 3-dimensional structure of immunoglobulins,
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the number of C domains equals the number of V domains.
CL-VL associations form the complementary sites for binding antigens.
in each chain (H and L), the C and V domains fold onto one another, forming CV interactions.
high affinity binding of antigen arises from disulfide bonds formed with the antigen.
hinge regions connect globular domains.
-sheets align edge to edge.
24. [6 points] Explain why a cancer cell that no longer expresses functional p53 would be more easily killed
by a genotoxic agent than if that same cancer cell still expresses functional p53.
Functional p53 protein is needed for the delayed checkpoint control where cells stop
entry into S or M phases when there is DNA damage present. If p53 is absent, then this
delayed form of checkpoint control is absent. (While the immediate form of checkpoint
control will still be present when p53 is lost, it is not able to be sustained over a long
period of time.) When genotoxic anti-cancer agents are used to treat such cancer cells,
the ones with p53 will stop dividing for a long time while they repair the DNA damage.
Thus, cell growth will be slowed, but some, possibly most, cells will survive because they
repaired their DNA damage. But for cells lacking a functional p53, checkpoint control will
not be sustained and entry into S or M phases with significant amounts of DNA damage
present will result in cell death nearly all the time.
Note that p53 also has a role in producing apoptosis in response to massive amounts of
DNA damage, particularly double strand DNA breaks. However, in this case, p53 makes
the cells more easily killed by genotoxic agents and radiation. This is the opposite of what
the question asks. Interestingly, it is likely that it is this response that provides a strong
selection for p53 mutations in cancers, and why cancers often convert from p53+ cells to
p53- after exposure to therapeutic doses of anti-cancer agents that are genotoxic.
25. [7 points] Topoisomerase IIA enzymes have three distinct structural regions that the T-DNA must pass
through during the reaction it catalyzes.
a. [6 points] In the three boxes below, indicate these three structural regions. You may identify them by
a commonly used name or a unique structural or functional aspect, so long as you distinguish the
three regions.
X The C-terminal Gate; which might also be referred to as the non-ATP-binding
gate
Y The G-DNA binding region; which might also be referred to as the catalytic
region (meaning region where the DNA-breaking catalysis occurs)
Z The N-terminal Gate; which might also be known as the ATP-binding region.
Note: The above three answers can be in any order.
PHRM 836 Exam I key - 8
b. [1 point] The descriptions you provided in part a have a letter, X, Y, or Z, in the upper left corner of
the box that contains the description of each structural region. Using these three letters, indicate the
order that the T-DNA pass through these regions during the reaction catalyzed by this enzyme.
N-terminal Gate, G-DNA binding region, then the C-terminal Gate.
Based on the answers given in the key above, this would be indicated as Z, Y, X (the
actual correct answer depends on the order of the answers given in part a.)
26. [4 points] The pH dependence of O2 binding to Hb is known as the Bohr effect; lower pH reduces O2
affinity. The figure shows one of the interactions in Hb that plays a role in the Bohr effect.
a. What is the physiological significance of the Bohr effect?
CO2, generated as a metabolic product in the tissues, leads to an increase in H+
and lower pH than in the lungs. Tissues: CO2  HCO3- + H+. Lower pH shifts
the equilibrium of Hb binding O2 to the oxy form through the law of mass action,
thereby promoting the delivery of O2 to the peripheral tissues from the lungs.
b. As stated in lecture, there are many sequence variations of Hb among the human population,
many of which do not affect Hb function while some do. If His 146 were substituted with
Val, what, if any, would be the affect on Hb function? Explain your answer.
The effect would be a significant reduction in the Bohr effect because His 146
is one source of the protons released upon conversion to deoxy Hb due to the
altered pKa of His 146. Also, deoxy Hb would be destabilized due to the loss of
the His146 salt bridge to Asp94 (shown in the figure).
27. [6 points] Histones have a single molecular mechanism involving DNA. By this molecular mechanism,
histones have an important or critical role in TWO distinct cellular functions. What are those two distinct
cellular functions?
1. Condensing the chromosomes so that they can segregate to the daughter cells at
mitosis.
2. Regulation of gene transcription
PHRM 836 Exam I key - 9
28. [8 points] CtXR is an allosteric enzyme regulated by ATP. CtXR converts the substrate xylose to the
product xylitol. The enzyme-substrate interactions are shown below in the scheme on the right. The plot
of initial velocity against [S] (below, left) shows CtXR activity in the absence of ATP, an allosteric
effector of CtXR. Addition of ATP leads to an increase in the KM value of xylose.
a. What kind of allosteric effector is ATP?
Inhibitory effector. The increase in Km corresponds to lower affinity for
xylose, and thus a decrease in activity.
b. Draw on the plot (above) a curve showing the effect of ATP in enzyme activity.
c. Predict the pH dependence of CtXR activity between pH 2 and pH 6, and explain your
reasoning.
The activity will be reduced at pH2 because Asp50 will be protonated and
substrate binding affinity will be lowered because of weaker/loss of Asp50
interactions with –OH groups of xylose.
d. Give one explanation for how ATP might alter the interactions shown in the figure.
ATP binding likely changes the CtXR structure so that the interactions in the
scheme are compromised. That is the hydrogen bonding of His113 or Tyr51 or
Asn309 or Asp50 could be lost due to the conformational change induced by
ATP. Another possibility is that potential hydrophobic interactions of the indole
ring of Trp23 with xylose are lost. Another possibility is that ATP changes
CtXR conformation in a way that alters the pKa of His113 or Asp50 in a direction
that perturbs interaction with xylose.