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MCMP 208 Exam II Key - 1
Examination II Key
MCMP 208 – Biochemistry for Pharmaceutical Sciences I
March 7, 2017
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
The following accurate genetic code table may be useful to you for some parts of this exam.
U
C
A
G
UUU
Phe
UCU
Ser
UAU
Tyr
UGU
Cys
UUC
Phe
UCC
Ser
UAC
Tyr
UGC
Cys
U
UUA
Leu
UCA
Ser
UAA
Stop
UGA
Stop
UUG
Leu
UCG
Ser
UAG
Stop
UGG
Trp
CUU
Leu
CCU
Pro
CAU
His
CGU
Arg
CUC
Leu
CCC
Pro
CAC
His
CGC
Arg
C
CUA
Leu
CCA
Pro
CAA
Gln
CGA
Arg
CUG
Leu
CCG
Pro
CAG
Gln
CGG
Arg
AUU
Ile
ACU
Thr
AAU
Asn
AGU
Ser
AUC
Ile
ACC
Thr
AAC
Asn
AGC
Ser
A
AUA
Ile
ACA
Thr
AAA
Lys
AGA
Arg
AUG
Met
ACG
Thr
AAG
Lys
AGG
Arg
GUU
Val
GCU
Ala
GAU
Asp
GGU
Gly
GUC
Val
GCC
Ala
GAC
Asp
GGC
Gly
G
GUA
Val
GCA
Ala
GAA
Glu
GGA
Gly
GUG
Val
GCG
Ala
GAG
Glu
GGG
Gly
MULTIPLE CHOICE. For problems 1 to 17, 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]
1. Identify which set of proteins constitute the core protein components of nucleosomes.








H2A, H2B, H3, H4 These are the 4 core histone proteins.
leucine zipper proteins
DNA polymerases ( and ), PCNA (sliding clamp), MCM DNA helicase, Primase
Cdc6, Cdt1, MCM DNA helicase, ORC (origin replication complex)
Uvr A, B, and C
Mut S/L and MutH
TBP (TATA box binding protein), RNA polymerase, transcription factors (TFII A, B, D, E, G, H)
DNA polymerase I and DNA ligase
MCMP 208 Exam II Key - 2
2. The posttranslational modified residue shown to the right is best described
as






hydroxylation.
lipidation.
farnesylation.
acetylation.
N-linked glycosylation.
O-linked glycosylation. This residue displays a sugar modification
through the oxygen (O) of a serine side chain.
3. A common function of DNA helicases is








to introduce single or double strand breaks to relieve DNA supercoiling.
to bind to and stabilize the DNA double helix.
to bind to and stabilize single stranded DNA.
to form phosphodiester linkages between 5’ phosphomonoesters and 3’ hydroxyls on two DNA
strands.
to separate the annealed strands of duplex DNA. Helicase activity is critical to both DNA replication
and transcription as ssDNA templates are used in both processes.
to synthesize RNA primers for DNA polymerases in the process of replication.
to increase processivity of DNA polymerases.
to form Okazaki fragments.
4. Chargaff’s rules were critical information that helped Watson and Crick construct their structural model
of DNA. Which is the best description of Chargaff’s rules?







adenine pairs with thymine via 2 hydrogen bonds; cytosine pairs with guanine via 3 hydrogen bonds
phosphodiester bonds connect deoxyribose sugars
DNA does not contain sulfur atoms
bacteriophage always transfer DNA, not protein, to their host
DNA contains deoxyribose sugars, not ribose sugars
both proteins and DNA fold into helical conformations
the ratio of adenine to thymine and the ratio guanine to cytosine are both 1:1 in all organisms While
the relative ratio of A:G or A:C etc was variable among species, the A:T and G:T ratios were always
one.
 Van der Waals interactions stabilize the double helix
5. A difference between bacterial genomes and eukaryotic genomes is
 bacterial genomes can be made of either DNA or RNA, while eukaryotes only use DNA.
 the DNA of bacterial genomes is not supercoiled, while eukaryote genomic DNA is supercoiled.
 the majority of the DNA in bacterial genomes does not encode for gene products, while in eukaryotes
the major of the DNA does encode for gene products.
 prokaryotes can have a circular DNA molecule for a genome, while eukaryotes always have linear
molecules.
 bacterial genomes are packaged in nucleosomes, while eukaryotes do not package their genomes.
 bacterial genomes are typically much larger, containing more total base pairs, than eukaryotic
genomes.
MCMP 208 Exam II Key - 3
6. In the mechanism of mRNA splicing by the spliceosome, what is the initial nucleophile in the reaction at
the 5' splice site that forms the lariat intermediate.








an OH in the gamma phosphate of ATP.
the OH on the active site serine of the spliceosome ribonuclease.
a water molecule activated in the active site of the spliceosome ribonuclease.
the 2'-OH of an adenosine within the intron to be cleaved.
the 3'-OH of a guanosine in the 5' exon.
an OH of the 5' phosphate of a guanosine in the 3' exon
the SH on the active site cysteine of the spliceosome ribonuclease.
Mg2+ ions in the active site of the spliceosome ribonuclease.
7. Place the following steps of eukaryotic transcription in chronological order.
i. TBP (TATA box binding protein) within TFIID bind to the core promoter elements (CPEs) on DNA.
ii. A poly(A) signal is transcribed.
iii. RNA polymerase II dissociates from the mediator complex.
iv. 23 nucleotides of RNA are synthesized allowing the preinitiation complex to move away from the
core promoter.
v. General transcription factors (TFII A, B, E, G, H), RNA polymerase II, chromatin remodeler
proteins, and the mediator complex bind at core promoter elements (CPEs) on DNA forming the
preinitiation complex.






iii, v, i, iv, ii
v, iv, i, ii, iii
v, i, iv, iii, ii
i, v, iv, iii, ii
i, v, iii, iv, ii
i, v, iv, ii, iii
8. Which is not a mechanism involved in the regulation of transcription?






chromatin remodeling.
repressor proteins binding to gene operators.
gene expression level of transcription factors.
compartmentalization of transcription factors (sequestration in cytosol, e.g.).
regulation of transcription factor activity by phosphorylation.
DnaA proteins binding to oriC sites. This is part of the mechanism for DNA replication in
prokaryotes.
9. Which of step would result in a stabilization of the following duplexed (double stranded)
oligonucleotides?
5'-ATGGTATCAA-3'
3'-TACCATAGTT-5'
 exchanging a G:C base pair with an A:T base pair.
 exchanging an A:T base pair with a G:C base pair. G:C base pairs are more stable than A:T base
pairs, as they have an additional hydrogen bond and form more stable base stacking interactions.
 increasing the temperature.
 decreasing the ionic strength of the solution.
 adding a denaturing, water miscible solvent, such as formamide.
 shortening each oligonucleotide by two bases.
MCMP 208 Exam II Key - 4
10. Which of the following catalyzes the attachment of amino acids to tRNAs?








elongation factor EF-Tu
elongation factor EF Ts
the ribosome
aminoacyl tRNA synthetase
reverse transcriptase
RNA polymerase
ribonucleases
restriction endonucleases
11. This molecule directly provides the chemical energy used to translocate the ribosome down the mRNA as
translation proceeds.








ATP
TTP
GTP The GTPase eEF2 hydrolysis GTP causing a translocation of the ribosome along the mRNA.
deoxyATP
deoxyTTP
deoxyGTP
creatine phosphate
dolichol phosphate
12. In the general procedures of molecular cloning of DNA molecules, what is the best description of the role
of a vector?




The vector contains a DNA polymerase making several copies of a DNA sequence.
The vector amplifies DNA using short DNA sequences called primers.
The vector cuts DNA at specific sites to generate sticky ends.
The vector accepts the DNA to be cloned and is capable of being replicated in a host cell. The vector
is also a DNA molecule which is typically ligated with a DNA to be replicated. The vector will
contain the necessary sequences for the host cell to replicate.
 The vector is a host organism, such as E. coli, which is tasked with replicating a DNA sequence.
 The vector is a solid support, typically a nitrocellulose filter, that binds DNA prior to probing with
radiolabeled complementary DNA sequences.
13. The mechanism of peptidyl transferase catalyzed by the ribosome has been described as a concerted
proton shuttle mechanism. Where is the atom that acts as both a proton donor and acceptor in this
reaction?






The 2' oxygen of a rRNA uridine residue located at the P-site in the large ribosomal subunit.
The 2' oxygen of a rRNA uridine residue located at the A-site in the large ribosomal subunit.
The nitrogen of the incoming amino acid located on the aminoacyl tRNA within the A-site.
The 3' oxygen of the peptidyl-tRNA within the ribosomal P-site.
The 2' oxygen of the peptidyl-tRNA within the ribosomal P-site.
The oxygen of a water molecule activated by hydrogen bonding to a rRNA uridine residue located at
the P-site in the large ribosomal subunit.
MCMP 208 Exam II Key - 5
14. Why is the level of A1c useful diagnostic information?





It reflects the general health of an individual.
It reflects the current amount of liver glycolysis
It reflects the current amount of turnover of total body protein
It reflects the amount of turnover of total body protein during the past several weeks
It results from a chemical reaction and so indicates the average level of blood sugar for the past
several weeks.
 It reflects whether a patient is anemic or not
 It reflects whether or not the patient is unable to degrade glycosaminoglycans and therefore has a
form of mucopolysacchardosis
 It is the result of a reaction of carbohydrate with protein that reflects the average body temperature in
the past several weeks.
15. Glycosaminoglycans








are proteins with a lot of glycosylation
are large glycolipids
are linear polysaccharides containing acidic and amino sugars
are short branched oligosaccharides
are any oligosaccharide containing at least one sialic acid residue
are branched polysaccharides used for energy storage
are not able to be digested very well after being eaten by humans
are homopolysaccharides that are common in invertebrate exoskeletons
16. Short heteroglycans attached to lipids and proteins function primarily as






transporters of nutrients across the cell membrane
structures used in molecular recognition and sensing
storage of carbohydrates for use in energy generation by cells
an important part of the mechanism that targets those proteins and lipids to subcellular organelles
elements that stabilize proteins and lipids by protecting them from degradation
an important mechanism for stabilizing the high order structure of all proteins
17. A point mutation that converts the code for an amino acid into a premature stop signal is best described
as






a transversion mutation.
a silent mutation.
a missense mutation.
a misdirectional mutation.
a missing mutation.
a nonsense mutation.
MCMP 208 Exam II Key - 6
ESSAY PROBLEMS. Write your answers to problems 18 to 22 in the space immediately below each
problem.
18. A. [2 points] Electron micrographs of DNA in the process of general homologous recombination have
shown that chi structures generally have two pairs of equal length arms (as shown below). Explain why
chi structures exhibit this particular symmetry.
Homologous recombination is an exchange of DNA between similar or identical molecules
of DNA, such as homologous chromosomes. This symmetry arises because the similar
sequences occur at the same locations within the two molecules of DNA.
B. [2 points] The process of chi structure formation in general recombination involves assembly of two
DNA molecules, which we could call arbitrarily strand A and strand B. In the picture above, label
each arm of the chi structure as being from either the A or B strand. One arm is labeled for you.
A
B
B
B
B
A
Proc. Natl. Acad. Sci. USA. 73 (1976) 3001.
Each of the above two figures is a correct answer.
19. [6 points] Below are structures of the deoxyribonucleosides commonly present in DNA.
A. Indicate if each deoxyribonucleoside contains a purine or pyrimidine base.
B. The circle provided represents a view of the DNA double helix looking down the length of the helix from
above. Choose two bases that normally form base pairs and draw them within the circle taking care to place
the bases appropriately within the helix barrel. Draw just the bases with the glycosidic bond linking to the
circle perimeter and indicate the proper hydrogen bonding pattern.
C. Indicate the sides of the circle that represent the major groove and the minor groove in the double helix.
MCMP 208 Exam II Key - 7
Either drawing in the red box is a correct
answer for parts B and C. Note the also
correct are if the base pairs are drawn
inverted from right to left (i.e., C:G vs
G:C). Either A:T or G:C base pairs are
acceptable.
The narrower angle (on the minor groove
side) between glycosidic bonds should be
clearly less than 180 and greater than 90
degrees.
20. [6 points total] This problem concerns the dsDNA gene sequence from E. coli shown below.
1
2
3
5’- GCCATTTCCCGTTAAGGAGGTGCCATCTATGGAAGCCAGCTACTAA -3’ (+) coding strand
3’- CGGTAAAGGGCAATTCCTCCACGGTAGATACCTTCGGTCGATGATT-5’ (-) noncoding strand
A. [4 points] Site 1 is the transcriptional start site. Site 2 is a consensus sequence that when present in
the corresponding RNA transcript is called the Shine-Dalgarno sequence. Explain the role of this
sequence in transcripts in prokaryotes and explain where one would find the anti-Shine-Dalgarno
sequence.
The Shine-Dalgarno sequence is a ribosomal binding site found in bacteria. This site
functions to ensure the ribosome is able to find the proper AUG start for translation
initiation. The ribosome binds this site through Watson-Crick base pairing (RNA:RNA) to
the anti-Shine-Dalgarno sequence within the in the ribosomal RNA (rRNA) of the small
subunit.
B. [2 points] Using the provided genetic code table, determine the peptide/protein product of this gene.
Met-Glu-Ala-Ser-Tyr
MCMP 208 Exam II Key - 8
21. [6 points total] The Meselson-Stahl experiment established the mechanism of DNA replication as being a
semi-conservative.
A. [2 points] Briefly explain how DNA replication is semiconservative.
It is called semiconservative because upon replication two DNA copies are produced
and each contains one of the original strands and one new strand.
B. [4 points] The Meselson-Stahl experiment involved use of a heavy form of nitrogen (15N). Growing
bacteria on 15N containing media adds significant mass to the base pairs. Heavy (15N) and light
(14N) DNA can be distinguished on a density gradient with ultracentrifugation (shown below in the
control sample test tubes). In the experiment, bacteria grown overnight in heavy media were
transferred to light media and the DNA was analyzed by centrifugation after both one and two
doublings (1 and 2 generations). Use the test tube pictures below to indicate the location of the
DNA band(s) that Meselson and Stahl observed after 1 and 2 generations. Also, indicate the
location of the DNA that would have been observed had the mechanism of replication been
conservative.
The relative intensity of the DNA bands in the answer (in the test tubes in the red boxes)
won’t be required/graded, just the appropriate positions in the tube.
In the semiconservative mechanism (actual results), the DNA after just one replication has
one new strand (14N) and one old strand (15N). So, all observed DNA is at an intermediate
density. After an additional replication, now two of the four strands contain all new DNA
(14N) and two of the four contain one new (14N) and one old (15N) strand.
If the mechanism were conservative, this would mean that an entirely new dsDNA is
generated from a dsDNA template. Thus, the expected result with this mechanism yields the
original DNA template at the lower tube position due to the 15N and the new strand being
composed completely of 14N. With additional replications, all newly synthesized strands are
composed of 14N and the original 15N containing template will remain.
MCMP 208 Exam II Key - 9
22. [7 points total] This problem tests your knowledge of carbohydrate structure and nomenclature.
A. [4 points] D-isomaltose is the disaccharide that can be produced by digestion of amylopectin by an
enzyme that hydrolyzes all the linkages except the linkage involved uniquely at the branch points.
Draw the structure of α-D-isomaltose employing a standard Haworth projection.
B. [3 points] D-ribulose is the ketose with the same stereochemistry as D-ribose. Draw the structure of
the open chain form of D-ribulose employing a standard Fischer projection.