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MCMP 208 Exam II Key - 1
Examination II Key
MCMP 208 – Biochemistry for Pharmaceutical Sciences I
March 7, 2016
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
MATCHING. For problems 1 to 4, a set of numbered answers is provided immediately below. For each
problem, select from the list of answers the single choice that best matches the item described in the
problem. Mark that answer on your answer sheet. An answer may be used more than once or not at all.
[3 points each]
1
O
2
NH2
NH
N
5
O
N
O
NH
NH
Uracil
O
N
Thymine
O
N
HN
NH
NH2
7-methyl guanine
7
S
4
NH
N
O
5-methyl cytosine
6
N
N
N
O
Dihydrouracil
O
3
O
N
Thiouracil
O
8
NH
O
Pseudouracil
O
N
N
NH
N
Hypoxanthine
(base present in
the nucleoside inosine)
1. The base present in DNA that serves as an epigenetic mark to promote heterochromatin formation. 
2. The base occasionally present in the wobble position of tRNAs that can base pair effectively with uracil,
adenine, and cytosine. 
3. The base that is one of the two standard pyrimidines present in DNA that is not a major component of
RNA. 
4. The base present in the nucleoside triphosphate that occurs as a post-transcriptional modification of the
5’-end of mRNAs in eukaryotes known as the 5’-cap. 
MCMP 208 Exam II Key - 2
MULTIPLE CHOICE. For problems 5 to 25, select from the list immediately following each question the
single most correct choice to complete the statement, solve the problem, or answer the question. Mark that
answer on your answer sheet. [3 points each]
5. Determine the mRNA sequence that is produced from the following DNA strand
5’- GCCATTTCCCGTTA-3’ (+) nontemplate or coding strand
3’- CGGTAAAGGGCAAT-5’ (-) template or noncoding strand








5’-CGGTAAAGGGCAAT-3’
5’-CGGUAAAGGGCAAU-3’
5’-TAACGGCAAATGGC-3’
5’-UAACGGCAAAUGGC-3’
5’-UAACGGGAAAUGGC-3’
5’-GCCAUUUCCCGUUA-3’
5’-GCCATTTCCCGTTA-3’
5’-AUUGCCCUUUACCG-3’
6. 5-bromouracil is a base analogue with a different tautomeric preference than the similar base thymine
due to the strong electron withdrawing properties of bromine. When 5-bromouracil is incorporated into
DNA in the place of thymine, what is the most likely outcome?
O
Br








OH
NH
N
H
a DNA double strand break
an indel mutation
a point mutation
a genome rearrangement
a Holliday junction
a thymine dimer
a cruciform structure
a newly incorporated RNA splice junction
O
Br
N
N
H
5-bromouracil
O
preferred
tautomer
The alternative tautomer causes base pairing with G instead of A, which can lead to a change in the base
as the DNA is replicated.
7. Which represents the correct assignment of genome size? (Mbp = 1 x 106 bases)






Human, 3 Mpb; E. Coli, 1 Mbp; lambda phage, 0.5 Mbp
Human, 3 Mpb; E. Coli, 0.5 Mbp; lambda phage, 1 Mbp
Human, 3200 Mpb; E. Coli, 4.6 Mbp; lambda phage, 0.05 Mbp
Human, 3200 Mpb; E. Coli, 0.05 Mbp; lambda phage, 4.6 Mbp
Human, 3200 Mpb; E. Coli, 4.6 Mbp; lambda phage, 5 Mbp
Human, 3200 Mpb; E. Coli, 1000 Mbp; lambda phage, 500 Mbp
MCMP 208 Exam II Key - 3
8. The Hershey-Chase experiment helped to establish that DNA was the genetic material. There were two
separate, critical experiments in this work. In one, the DNA of _______ was initially labeled with
________. In the other, the protein of _______ was initially labeled with ______.










E. coli, radioactive phosphorous; E. coli, radioactive sulfur
E. coli, radioactive sulfur; E. coli, radioactive phosphorous
bacteriophage, radioactive phosphorous; bacteriophage, radioactive sulfur
bacteriophage, radioactive sulfur; bacteriophage, radioactive phosphorous
bacteriophage, radioactive phosphorous; E. coli, radioactive sulfur
E. coli, heavy nitrogen (15N); E. coli, light nitrogen (14N)
E. coli, light nitrogen (14N); E. coli, heavy nitrogen (15N)
bacteriophage, heavy nitrogen (15N); bacteriophage, light nitrogen (14N)
bacteriophage, light nitrogen (14N); bacteriophage, light nitrogen (15N)
bacteriophage, heavy nitrogen (15N); E. coli, light nitrogen (14N)
The experiment was to distinguish between proteins and DNA as the genetic material. Protein contains
sulfur (cysteine, methionine) and little or no phosphorous. DNA does not contain sulfur and has lots of
phosphorous in the phosphodiester backbone. They knew the bacteriophage must transfer the genetic
material to the E. coli. They used this selective radiolabeling to see which molecule was transferred to the
host and to the next generation of phage.
9. The packaging of the bacterial chromosome involves
 DNA complexed with histone proteins
 restriction enzymes cutting DNA at specific repeat sequences
 supercoiled DNA complexed with protein in a nucleoid Prokaryotes lack packaging of DNA into
nucleosomes with histone proteins.
 winding of DNA into the 30 nm fiber, which is wound into 200 nm filaments
 extensive cruciform DNA structures
 two replication forks which advance from the oriC initiation site
10. If a given double-stranded DNA is known to contain 40% cytosine bases, what is the expected
percentage of adenine?








5%
10%
15%
25%
30%
40%
60%
80%
By Chargaff’s ratios, 40% cytosine means the sequence must have 40% guanidine. That leaves 20% of
the bases for A and T. Again by Chargaff’s ratios, %A must equal %T. So A and T split the remaining
20%.
MCMP 208 Exam II Key - 4
11. Consider the following two 10-mer ssDNAs
Oligo #1: 5'-ATGGTATCAA-3'
Oligo #2: 5'-GTGGTCTCCT-3'
When hybridized to its perfect complement, which of these dsDNAs will have the lower Tm?
How will the Tm change in the presence of 1 M formamide (a water miscible organic solvent)?









Oligo #1; Tm increases
Oligo #1; Tm decreases
Oligo #1; Tm stays the same
Oligo #2; Tm increases
Oligo #2; Tm decreases
Oligo #2; Tm stays the same
Tm is the same; Tm increases
Tm is the same; Tm decreases
Tm is the same; Tm stays the same
The stability of dsDNA is related to its %GC content with higher GC content providing more stability.
Greater stabilization of the double helix gives a higher melting temperature. Oligo 1 has lower %GC than
oligo 2. Organic solvents destabilize the double helix structure by interfereing with the hydrophobic
interactions of the DNA bases. So the addition of formamide lowers the Tm.
12. In RNA, ______ codon(s) translate(s) to _______ amino acid(s).






1; 1
3; 1
1; 3
3; 3
1; 20
64; 1
A single codon composed of 3 bases in mRNA is translated to a single amino acid.
13. The HIV enzyme reverse transcriptase is critical to the viral life cycle because








it incorporates viral DNA into the host genome.
it reverses the order of RNA citrons through alternative splicing.
it binds to the CD4 cell surface proteins of helper T cells to initiate virus-cell fusion.
it reverses the flow of viral particles from into the cells (infection stage) to out of the cell (lytic
stage).
it converts the ssRNA genome into DNA using dNTPs from the host cell.
it reverses the viral DNA sequences through homologous recombination of repeat sequences.
it cleaves viral RNA polyprotein translation products to generate mature protein components of the
virion.
it replicates the host-incorporated viral DNA back into free DNA for incorporation into new viral
particles.
MCMP 208 Exam II Key - 5
14. In the replication of DNA, the dsDNA helix is unwound by an enzyme called _______ which generates
significant negative DNA supercoiling, which is relieved by an enzyme called ________.








DNA polymerase; helicase
DNA polymerase; DNA ligase
SSB; helicase
SSB; topoisomerase
helicase; topoisomerase
the 2-clamp; topoisomerase
topoisomerase; helicase
primase; helicase
15. Place the following steps of eukaryotic replication initiation in chronological order
i. Cdc6 and Cdr1 bind and recruit the MCM complex which contains a DNA helicase
ii. Polymerase  binds and adds RNA primers
iii. S phase cell cycle kinases Ddk and Cdk-cyclin E phosphorylate and activate preRC components
iv. DNA polymerases  and bind
v. In G1 cell cycle phase, a protein complex ORC (origin of replication complex) binds
vi. Clamp loader (RFC) and sliding clamp (PCNA) proteins bind increasing processivity
vii. MCM separates DNA strands






iii, v, i, vii, iv, ii, vi
iii, v, i, iv, vii, ii, vi
iii, v, i, vi, iv, vii, ii
v, i, iii, iv, vii, ii, vi
v, i, iii, vi, iv, vii, ii
v, i, iii, vii, iv, ii, vi
16. Which of the following best describes the function of telomerase at the telomere?






it synthesizes RNA primers onto DNA 5’ ends
it uses exonuclease activity to cleave 3’ overhangs
it synthesizes new DNA without the use of a template
it adds new DNA to both strands at the end of the chromosome
it adds new DNA to the shorter strand of an overhang until ends are matched
it adds new DNA to the longer strand of an overhang until there is sufficient length for addition of an
RNA primer
Telomerase has reverse transcriptase activity and uses an RNA template to extend the repeat sequences
on the long strand overhangs at chromosome ends until a RNA primer can be added and extended by the
normal replication machinery.
17. In the base excision repair mechanism of DNA repair, DNA glycosylases _________






catalyze a retro-Diels Alder reaction.
recognize DNA template strands via methylation marks
recognize DNA template strand status via nicks present in lagging strand
cleave the N-glycoside linkage of the damaged base and deoxyribose
remove 12-13 bases around the damaged base
promote branch migration of Holliday intermediates
MCMP 208 Exam II Key - 6
18. Homologous recombination is a mechanism used for






lengthen shortening telomeres
lagging strand DNA synthesis at replication forks
transcription termination
genome scanning for transcription start sites
DNA double strand break repair
RNA splicing
19. What best describes role of the eEF-1protein
 binds to stalled ribosomes at translation stop codons and promotes the release and dissociation of
ribosomes from mRNA
 a core component of the eukaryotic 48S initiation complex the binds the small 40S ribosome subunit
 binds the poly-A tails of mRNA to initiate translation
 a GTPase that binds aminoacyl-tRNAs and delivers them to the A site of the ribosome
 an ATP-dependent helicase that binds to mRNA hairpin (stem-loop) structures promoting
transcriptional termination
 promotes degradation of misfolded proteins in the unfolded protein response
20. In prokaryotes, the majority of the control of protein synthesis occurs at the level of _________.








chromatin modification (heterochromatin vs. euchromatin)
translation
mRNA localization
transcription
mRNA stability
replication
posttranslational modifications
activation and repression of ribosomal initiation and elongation factors
Protein translation is coupled with RNA transcription in prokaryotes, with ribosomes attaching to mRNA
as soon as the initiation sites are synthesized. So there is little opportunity for regulation at the translation
step.
21. D-seduheptulose is a ketoheptose, meaning it has 7 carbons and is a true carbohydrate. It can form a
pyranose structure. When in the pyranose form, how many chiral carbons does it have?






2
3
4
5
6
7
Ketoses always have two primary alcohols and a ketone, meaning n-3 carbons have secondary alcohols,
all of which are chiral in monosaccharides. For seduheptulose, n=7, so there are 4 secondary alcohols.
There is a fifth chiral carbon, however, at carbon 2, the anomeric carbon, when in the cyclic hemiketal
(or ketal) state.
MCMP 208 Exam II Key - 7
22. Which disaccharide does not have a glycosidic linkage involving carbons 1 and 4?






fructose Fructose is not a disaccharide
sucrose Sucrose involves linkage between C1 of glucose and C2 of fructose
lactose
cellobiose
galactose Galactose is not a disaccharide
maltose
23. In adult humans, a condition where the small intestine fails to produce β-galactosidase








is known as Crohn’s syndrome
is an indicator of a pre-diabetes, also known as metabolic syndrome
is commonly the result of an autoimmune disease that attacks the mucosal cells of the small intestine
is known as fructose intolerance
is one type of mucopolysaccharidoses
is one type of sphingolipidoses
is known as lactose intolerance
results in the inability to digest insoluble cellulosic fiber
24. The carbohydrate found in most glycoproteins and in all glycolipids is
 always linear Only a very few are linear (GAG and extremely short glycoconjugates)
 always has a short chain length (as distinct from total number of sugars if there are branches) This is
true except for GAG in proteoglycans, but the question referred to most glycoproteins, which
generally excludes proteoglycans.
 always consists of the same sugar Almost all are heterosaccharides
 always contains acidic sugars This applies to proteoglycans and gangliosides but is not true for most
glycoproteins and other sphingolipids
 never contains acidic sugars This is not true for gangliosides, a common glycolipid, as well as GAG
 always contains branches This is not
25. Glycoconjugates function nearly always






while in the cell nucleus
while associated with the cell membrane
while outside of the cell
while in the endoplasmic reticulum
while in the lysosome
after secretion by cells
MCMP 208 Exam II Key - 8
ESSAY PROBLEMS. Write your answers to problems 26 to 30 in the space immediately below each
problem.
26. [3 points] Inadequate dietary intake of vitamin C (ascorbate) can result in scurvy, which is caused by
collagen instability. The posttranslational modification of proline residues in collagen by an
oxidoreductase enzyme is required for proper collagen function. Name this posttranslational modification
(which also occurs on lysines) and draw the structure of a modified proline residue within a peptide
backbone.
Hydroxylation (1 point)
(1 point for proline structure) (1 point for hydroxyl group)
27. [5 points] The unidirectional nature of DNA synthesis results in the formation of Okazaki fragments in
the process of DNA replication.
a. [2 points] What is the direction of DNA synthesis by DNA polymerases and how does this result in
Okazaki fragments?
DNA is synthesized only in the 5’ to 3’ direction. (1 point)
This creates a problem at the replication fork where DNA is synthesized as the fork is
opened. Due to the antiparallel nature of DNA, synthesis of one strand can proceed
continuously (the leading strand) while the other must be synthesized discontinuously
(the lagging strand) in segments called Okazaki fragments. (1 point)
b. [3 points] Draw a replication fork which includes 3 Okazaki fragments. Appropriately label the 5’
and 3’ of all DNA strands, the template strands, leading strand, lagging strand, and the Okazaki
fragments.
Labeling of 5’ and 3’ ends (1 point)
Labeling of leading strand, lagging strand,
and template (1 point)
Labeling of the Okazaki fragments (1 point)
MCMP 208 Exam II Key - 9
28. [5 points] Unlike the fork structure that occurs during DNA replication, the structure formed in RNA
transcription has been described as a bubble.
a. [2 points] Provide an explanation why RNA transcription produces a bubble-like structure rather than
a fork.
In transcription, only one strand of oligonucleotides is synthesized instead of two. The
DNA duplex is opened to allow one strand to serve as a template, which is base paired
to RNA as it is synthesized. This involves 12-14 bp of RNA:DNA duplex within the
transcription machinery while the other non-template strand is ssDNA creating the
bubble structure. As RNA is synthesized, the DNA duplex is reformed as the nontemplate stand reanneals with the template. If two new strands were synthesized and
both DNA strands served as templates, you would not expect the templates to
reanneal but remain as RNA:DNA duplexes to give a fork structure.
b. [3 points] Provide 3 additional ways in which RNA transcription differs from DNA replication.
Any of the following acceptable (3 are required for full credit)
 No primer is required
 No sliding clamp
 Ribo-NTPs not dNTPs
 UTP instead of dTTP
 No proofreading activity
 either strand can serve as a template
 Transcription must be terminated by specific sequences
29. [7 points] Concerning mRNA translation and protein synthesis:
a. [2 points] Provide a distinct role for both the small and the large ribosomal subunits in protein
synthesis (i.e. a role distinct to the particular subunit)?
Small subunit: any one of the following (1 point)
 Binds mRNA first to begin initiation
 Uses anti-Shine Dalgarno sequence to find AUG start site (prokaryotes)
 Uses cap binding and RNA scanning to find AUG start site (eukaryotes)
 Matches mRNA codon to a tRNA anticodon in the decoding center within the
ribosome A site
Large subunit: any one of the following (1 point)
 Binds to mRNA complex second to complete initiation complex
 Contains the exit tunnel for the nascent polypeptide chain
 Contains the residues that catalyze peptide bond formation in the peptidyl
transferase center (PTC)
 Contains the GTPase associated region (GAR) that drives GTP hydrolysis when
associated with certain translation factors
MCMP 208 Exam II Key - 10
b. [3 points] Eukaryotic translation control mechanisms are particularly complex compared to
prokaryotes. List and briefly describe 3 mechanisms of translational control present in eukaryotes not
present in prokaryotes.
Any 3 of the following
 mTOR-Mediated Translational Control – mTOR mediated activation of initiation
and elongation factors
 mRNA export-control over mRNA release from the nucleus, translation occurs in
the cytoplasm
 mRNA stability-rate of translation often related to abundance, half-lives of
mRNA can very from 20 min-over 24 hours, certain sequences and RNA binding
proteins can effect degradation rates
 Negative Translation Control - he translation of certain specific mRNAs is known
to be blocked by binding repressor proteins
c. [2 points] Eukaryotes lack a particular recognition element present that is present in prokaryotic
mRNAs that facilitates positioning of the ribosome at the starting AUG codon. What is this purinerich RNA sequence called in prokaryotes and how do eukaryotic ribosomes successfully find their
proper start sites?
This sequence is called the Shine-Dalgarno sequence. (1 point)
In eukaryotes, several initiation proteins bind to the 5’ cap of an mRNA and recruit the
small ribosome subunit (40S) to the 5’ end. The complete 48S preinitiation complex
scans the mRNA in an ATP dependent process until the AUG start site is found. (1
point)
30. [5 points] A number of unrelated plant species generate cardiac glycosides as secondary metabolites. The
aglycone parts of these agents have a similar steroid structure. Cardiac glycosides have interesting
pharmacology and have been used as poisons on arrow tips in many areas of eastern Africa. Digitoxin is
one of these cardiac glycosides made by the foxglove plant. The aglycone of digitoxin is called
digitoxigenin and the 3-hydroxyl group of digitoxigenin is linked to a homo-trisaccharide. The sugar
involved is called digitoxose, which is 2,6-dideoxyl-D-allose in the β-pyranose form. D-allose is the 3epimer of D-glucose. The sugars are linked β1,4 and the linkage to the aglycone is via C1. Using “DIG”
to represent digitoxigenin, draw the structure of digitoxin. In the structure you draw, explicitly
indicate the location of all the bonds and atoms, including all of the all hydrogens, by writing the atomic
symbol of each of the atoms.