MCMP 208 Exam II - 1
Examination II
MCMP 208 – Biochemistry for Pharmaceutical Sciences I
March 7, 2016
Name:______________________________________________
Instructions
1.
Check your exam to make certain that it has 9 pages including this cover page. Ask for a new copy
of the exam if you are missing any pages.
2.
Use a pencil for filling in the answer sheet for computerized grading.
3.
Write your name on the line above (this page) AND on the answer sheet for computerized grading.
Also write your student number on the answer sheet. Fill in the circles underneath where you write
your name and student number. Do NOT place any identification or signature anywhere on this
exam or on the answer sheet. Also, do NOT fill in anything for “section number” and do NOT write
your name on the back of the exam.
4.
Your answers to problems 1 through 25 will be graded by computer. All computer-graded questions
will be worth three points each. Put your answers for these problems on the answer sheet and fill in
the appropriate circle using a pencil. It is your responsibility to be certain that the answers for
these problems are correctly marked on the answer sheet at the time you hand in your
completed exam. You are strongly encouraged to double check this at the time you submit the
answer sheet, as no points will be given for having mistakenly marked your answer sheet. The
computer grading answer sheets will not be returned to the students, so you are encouraged to record
your answers to the computer-graded questions on this exam as well as on the answer sheet.
However, the computerized grading of your answers to these questions will be done solely on the
basis of the optically scanned answer sheet, not on what is recorded on your exam. The course policy
is that there will be no re-grading of the computer answer sheets.
5.
Examination problems 26 through 30 are to be answered with short answers, brief essays, or
drawings. Put your answer to each of these directly on this exam, in the space provided below each of
these questions.
6.
After completing your exam, both the exam and the answer sheet must be submitted for grading and
both must be identified only with your name.
7.
This is a closed-book, closed-notes exam. Calculators and electronic devices of any kind that can
store information or communicate (cell phones, smart watches, computers, tablets, headphones)
are not allowed to be used during this exam. Any electronic device that is not in a backpack,
bookbag, purse, pocket, etc, will be considered to be in use, so put away all electronic devices
including cell phones and calculators. Anyone observed to be using a calculator, any book,
handouts, notes, printed or handwritten material of any kind, or the exam or answer sheet of any other
student will be considered to have committed an act of academic dishonesty. Students that do so will
not be allowed to complete the exam and will be given a zero for this exam. Also remember that the
first instance of academic dishonesty also results in a grade of "F" in this course and reporting this
episode to the Pharmacy Dean and the Dean of Students office.
MCMP 208 Exam II - 2
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]
O
1
2
NH2
NH
N
O
5
N
O
NH
NH
Uracil
O
N
Thymine
N
HN
O
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.
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’
MCMP 208 Exam II - 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
O
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
Br
N
N
H
5-bromouracil
O
preferred
tautomer
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
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)
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
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
MCMP 208 Exam II - 4
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%
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
12. In RNA, ______ codon(s) translate(s) to _______ amino acid(s).






1; 1
3; 1
1; 3
3; 3
1; 20
64; 1
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 - 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
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 - 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
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
22. Which disaccharide does not have a glycosidic linkage involving carbons 1 and 4?






fructose
sucrose
lactose
cellobiose
galactose
maltose
MCMP 208 Exam II - 7
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
always has a short chain length (as distinct from total number of sugars if there are branches)
always consists of the same sugar
always contains acidic sugars
never contains acidic sugars
always contains branches
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
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.
MCMP 208 Exam II - 8
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?
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.
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.
b. [3 points] Provide 3 additional ways in which RNA transcription differs from DNA replication.
MCMP 208 Exam II - 9
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)?
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.
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?
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.