Download Examination IV Key

MCMP 208 Exam IV Key - 1
Examination IV Key
MCMP 208 – Biochemistry for Pharmaceutical Sciences I
May 4, 2015
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
MCMP 208 Exam IV Key - 2
MATCHING. For problems 1 to 3, 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]









H3C
NH2

1. A product of the committed step in the de novo biosynthesis of purines. 
2. Made by salvage of guanine 
3. A product of the reaction that is intended to be inhibited indirectly as a result of directly inhibiting
dihydrofolate reductase using small molecule drugs. 
MCMP 208 Exam IV Key - 3
MULTIPLE CHOICE. For problems 4 to 26, 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]
4. Which of the following is not considered significant as a nutrient that is essential for good health and/or
development?







protein
fat
nucleic acids
carbohydrate
fiber
vitamins
minerals
5. In addition to its variable consumption of energy due to activity, muscle plays an important role in the
well-fed/post-prandial state by






oxidizing fatty acids from lipoproteins for energy production
releasing glucose into the blood
sparing blood glucose by using glucose from muscle glycogen
making triglycerides for later use during fasting
increasing its uptake and utilization of glucose, thereby lowering blood glucose
catabolizing amino acids for energy production
6. Metabolic syndrome








is a malabsorption syndrome
results from the nutritional deficiency of a specific nutrient
greatly increases the risk for developing type two diabetes mellitus
greatly increases the risk for developing type one diabetes mellitus
cannot be caused by obesity alone – some other causative agent is needed
is idiopathic (meaning that its cause is unknown)
is partially caused by an abnormally low BMI
always involves a negative nitrogen balance
7. The symptoms of gout are caused by






elevated levels of ribose-5-phosphate in the blood and tissues
precipitation of sodium urate crystals in joint and kidney tissues
increased purine catabolism
decreased purine salvage
increased activity of xanthine oxidase
decreased activity of xanthine oxidase
MCMP 208 Exam IV Key - 4
8. What kind of reaction is catalyzed by topoisomerases?









unwinding of DNA
stabilizing (lower the free energy of) negatively supercoiled DNA
destabilizing stem and loop structures in nucleic acids
forming and breaking down Holliday junctions
non-homologous end joining of DNA
changing the supercoiling of circular or anchored linear DNA
DNA slippage during replication
photo-reversal of UV-induced pyrimidine dimers in DNA
producing of micro RNAs
9. Six identical solutions containing a fully complementary pair of ssDNAs, each ssDNA being 17
nucleotides in length, are prepared and allowed to stand at room temperature (25ºC) for 24 hours. The
absorbance of 260 nm light is measured in the first solution. Then, immediately before measuring the
absorbance of 260 nm light by each of the other solutions, they are warmed to a specific temperature for
30 minutes. The resulting observations are shown in the table below:
Solution number Temperature 260 nm light absorbance
1
25ºC
0.420 absorbance units
2
30ºC
0.424 absorbance units
3
35ºC
0.442 absorbance units
4
40ºC
0.541 absorbance units
5
45ºC
0.754 absorbance units
6
50ºC
0.760 absorbance units
The Tm for this DNA in these solutions is










25ºC
between 25ºC and 30ºC
30ºC
between 30ºC and 35ºC
35ºC
between 35ºC and 40ºC
40ºC
between 40ºC and 45ºC
45ºC
more than 45ºC
The Tm is the 50% point between the minmum and maximum absorbance. 0.76-0.42= 0.34.
0.42+0.34/2=0.59AU. This is significantly above 40 degrees but below 45 degrees.
10. The region between and including the start codon and the termination codon in a mature mRNA is called








operon in prokaryotes and spliced mRNA in eukaryotes
untranslated sequence
leader sequence
trailer sequence
coding strand
non-coding strand
open reading frame
intronic sequence
MCMP 208 Exam IV Key - 5
11. The approximate size of the genome of homo sapiens (humans) is









65 billion base pairs
32 billion base pairs
6.5 billion base pairs
3.2 billion base pairs
640 million base pairs
320 million base pairs
64 million base pairs
32 million base pairs
4 million base pairs
12. In typical gel electrophoresis separation of DNA




the DNA migrates to the positive electrode with the longest DNA moving the most rapidly
the DNA migrates to the positive electrode with the shortest DNA moving the most rapidly
the DNA migrates to the negative electrode with the longest DNA moving the most rapidly
the DNA migrates to the negative electrode with the shortest DNA moving the most rapidly
13. Sticky ends generated by cutting DNA with a restriction endonuclease are useful because those sticky
ends





make it possible to convert DNA into RNA
are a key aspect of dideoxynucleotide DNA sequencing
can direct the ligation of DNA pieces together in new arrangements
make it possible for the DNA to be transferred into bacteria for cloning
are essential for any kind of analysis by hybridization
14. The relative sizes of the sets (numbers of different molecular sequences for a given species) of the three
principal cellular RNAs are [This is asking about the size of the set of sequence, not the concentration]






tRNA > rRNA > mRNA
rRNA > tRNA > mRNA
mRNA > rRNA > tRNA
rRNA > mRNA > tRNA
tRNA > mRNA > rRNA
mRNA > tRNA > rRNA
15. Human immunodeficiency virus encodes in its genome three enzymes. This includes








a DNA-dependent RNA polymerase
a RNA-dependent RNA polymerase
a DNA-dependent DNA polymerase
a RNA-dependent DNA polymerase
Dicer
RISC
a topisomerase
a restriction endonuclease
MCMP 208 Exam IV Key - 6
16. Telomerase’s function is to






prevent any overhanging end at the telomeres
shorten each 5´end of the telomeric DNA so it is always shorter than the complementary 3´end
lengthen each 3´end of the telomeric DNA so it is always longer than the complementary 5´end
lengthen both the each 5´ends and 3´ends of the telomeric DNA
produce primers beyond the end of the telomeres for use during lagging strand DNA replication
make sure DNA is always negatively supercoiled
17. The function of a DNA helicase is to





create dsDNA from complementary ssDNAs
separate one strand of DNA from its complement without otherwise altering the DNA
keep complementary ssDNA strands from reannealing
prevent positive supercoils from forming ahead of the replication fork
change the topology of dsDNA
18. After a purine spontaneously is lost from DNA via hydrolysis of its glycosidic bond, this is repaired in
cells by ______________ DNA repair






mismatch
nucleotide excision
post-replication
transcription coupled
base excision
direct damage reversal mechanism
19. Recombination involving a Holliday junction in a large region of sequence homology is the predominant
mechanism used by human cells for






transposon insertion
retroviral integration
repair of double-strand breaks in DNA
post-replication repair of damage encountered during DNA replication
mismatch DNA repair
transcription coupled DNA repair
20. In general, how are all DNA microarrays different from all northern and Southern blots?





Microarrays do not involve hybridization
Microarrays do not use labeled nucleic acid
Microarrays use immobilized probes and labeled unknown nucleic acids
Microarrays use immobilized unknown nucleic acids and a labeled probe
Microarrays use proteins as probes
MCMP 208 Exam IV Key - 7
21. An experimental goal is to use human cell DNA as a template and to use two 30 nucleotide long primers
to PCR amplify a region of the X chromosome that includes base pairs 10,491,100 through 10,491,250
but no other DNA sequences from the X chromosome. The nucleotide numbering is from the p arm
telomere to the q arm telomere of this chromosome. Given the sequence of nucleotides (for the strand
going 5´ to 3´ from the p-arm telomere to the q arm telomere) in this region what sequences should be
used as primers to amplify just this region of the X chromosome?
 Same as nts 10,491,100 to 10,491,129 for one primer and the same as nts 10,491,221 to 10,491,250
for the other primer.
 Same as nts 10,491,100 to 10,491,129 for one primer and complementary to nts 10,491,221 to
10,491,250 for the other primer.
 Complementary to nts 10,491,100 to 10,491,129 for one primer and same as nts 10,491,221 to
10,491,250 for the other primer.
 Complementary to nts 10,491,100 to 10,491,129 for one primer and complementary to nts 10,491,221
to 10,491,250 for the other primer.
22. Consider a gene, transcribed from its only known promoter, involves 6 exons, which are 100, 200, 300,
400, 500, and 600 bp long, for a total exon length of 2100 bp. If the transcript is alternatively spliced such
that exons 2 and 5 are optionally retained or removed, and these alternative splicings occur independently
(i.e., the two alternative splicings are not correlated with each other), then the sizes of all the possible mature
mRNAs produced in cells will be
 1600 and 1900 bp
 1600 and 2100 bp
 2000 and 2100 bp
 1400 and 1600 bp
 1400 and 1900 bp
 1600 and 1900 bp
 1400, 1600, and 1900 bp
 1400, 1900, and 2100 bp
 1600, 1900, and 2100 bp
 1400, 1600, 1900, and 2100 bp
The exon numbers in these mRNAs are 1,3,4,6; 1,2,3,4,6; 1,3,4,5,6; and 1,2,3,4,5,6
23. One difference between transcription and DNA replication is





transcription requires primers while replication does not
only one strand is transcribed while both strands are replicated
transcription can occur in either direction while replication can only occur in one direction
transcription involves polymerization 3´ to 5´ while replication involves 5´ to 3´ polymerization
DNA must be single stranded for replication but transcription does not require DNA to be melted
24. The TATA box in eukaryotic DNA determines





which genes are transcribed
where transcription starts and its direction
where transcription ends
where replication originates
where exon junction boundaries are located
MCMP 208 Exam IV Key - 8
25. Given a peptide sequence of threonine-tyrosine-serine-tryptophan-isoleucine, how many possible mRNA
sequences might encode this?









36
48
64
96
128
144 The degeneracy by aa for this sequence is 4, 2, 6, 1, 3. The product of these is 144.
256
288
384
26. During translation in eukaryotes, how is the 3´ end of the mRNA involved?




It provides the stop codon
It stops translation if there is no stop codon in the mRNA
It forms a circular RNA molecule by being ligated to the 5´ end of the mRNA
It assists in the formation of the initiation complex involving the ribosome large subunit, mRNA and
stating amino-acyl tRNA
 It prevents the formation of the initiation complex by increasing the hydrolysis of the amino acid
from the initial amino acyl tRNA.
 It is not involved at all in eukaryotic translation
ESSAY PROBLEMS. Write your answers to problems 27 to 30 in the space immediately below each
problem.
27. [5 points] Diagram the pathways of interconversion of purine nucleotides without showing intermediates
or other molecules, but showing each of the following:
a. The nucleotides being interconverted, using names or standard abbreviations (no structures)
b. The directions of metabolite flow in the pathways in your diagram
c. Which of the purine nucleotide(s) is/are made by de novo biosynthesis by using an arrow labeled “de
novo biosynthesis” pointing at the purine nucleotide(s) which is/are made by de novo biosynthesis
GMP
de novo biosynthesis
IMP
AMP
Note that the three may be arranged in any manner and the arrows arranged in any manner
as long as the 5 pathways are indicated connecting the NMPs is above.
MCMP 208 Exam IV Key - 9
28. [6 points] Concerning the stability of dsDNA:
a. How does increasing the temperature affect the stability of dsDNA, and what is the molecular
explanation for this?
This decreases the stability of dsDNA. Heat decreases the hydrophobic bonding between
bases, lessening base-staking energy a force that stabilizes dsDNA.
b. How does increasing the concentration of formamide (a denaturant) affect the stability of dsDNA,
and what is the molecular explanation for this?
This decreases the stability of dsDNA. Denaturants like formamide decrease the
hydrophobic bonding between bases, lessening base-staking energy a force that stabilizes
dsDNA.
c. How does increasing the concentration of sodium chloride affect the stability of dsDNA, and what is
the molecular explanation for this?
This increases the stability of dsDNA. Monovalent cations such as sodium are attracted
to the negatively charged phosphates and this reduces the electrostatic repulsion of the
nearby phosphates, thus reducing this destabilizing force.
29. [6 points total] This question is about nucleosomes.
a. [1 point] What is the structural state of DNA that is in a nucleosome compared to DNA that is not in
a nucleosome?
The DNA in nucleosomes is negatively supercoiled while DNA not involved in nucleosomes
is not supercoiled.
b. [1 point] What else is in a nucleosome besides DNA?
Histones
c. [2 points] Nucleosomes have two primary functions. For one of these functions, describe the function
and then describe how the nucleosomes carry out that function.
They compact DNA by making the DNA have less length than if the DNA was not in a
supercoil.
d. [2 points] For the other function of nucleosomes, describe the function and then describe how the
nucleosomes carry out that function.
They provide the negative supercoiling of DNA that is needed in advance of the DNA
replication fork.
30. [5 points] This question is about the proofreading activity of replicative DNA polymerases.
a. [3 points] Describe the reaction catalyzed during proofreading.
Hydrolysis of the phosphodiester bond linking the nucleotide at the 3´ end of the primer.
or
3´to 5´exonuclease activity [Both of these are equally correct answers]
b. [2 points] Describe when the proofreading reaction is very likely to occur.
MCMP 208 Exam IV Key - 10
The proofreading reaction is most likely when the nucleotide at the end of the 3´ end
of the primer IS NOT properly base-paired causing the polymerase to stall.