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MCMP 208 Exam IV - 1
Examination IV Key
MCMP 208 – Biochemistry for Pharmaceutical Sciences I
May 2, 2012
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
MULTIPLE CHOICE. For problems 1 to 24, 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. A dsDNA that is 30bp long in 1M NaCl pH 7.4 buffer had a TM of 55°C. Which change in the sequence
of this dsDNA would increase the TM measured in the same conditions?
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Replace all As with Ts and all Ts with As
Replace all Cs with Gs and all Gs with Cs
Rearrange the base-pairs so all the purines in one strand, but keep the base-pairs the same
Rearrange the base-pairs so exactly half of the purines are in each strand but keep the base-pairs the
same
Replace some A-T base-pairs with G-C base pairs
Replace some G-C base-pairs with A-T base pairs
Put all the A-T base-pairs at the ends and the G-C base-pairs in the middle
Put all the G-C base-pairs at the ends and the A-T base-pairs in the middle
Rearranging the locations of base pairs (4, 7, and 8) or inverting the pairs between trands (1, 2, and 3)
without changing the fraction of base-pairs that are GC will not significantly alter the melting
temperature. Only increasing the GC content will increase the melting temperature.
2. One aspect of positively supercoiled DNA when compared to relaxed DNA that is always true is
 Positively supercoiled DNA is always bound to protein, while relaxed DNA may be free of protein.
 In positively supercoiled DNA, the DNA is a left-handed helix, while in relaxed DNA, the DNA
always has a right-handed helix
 Positively supercoiled DNA is always circular while relaxed DNA can be linear
 Positively supercoiled DNA is caused by under-winding the DNA, which results in fewer helices per
unit length than normal, while relaxed DNA always has a normal helical pitch of one turn per 10.5 bp
 Positively supercoiled DNA is caused by over-winding the DNA, which adds more helices per unit
length than normal, while relaxed DNA always has a normal helical pitch of one turn per 10.5 bp
B-DNA is always right handed, eliminating 3. Choice 1 cannot be correct because circular DNA can be
positively supercoiled without it being bound to proteins
MCMP 208 Exam IV - 2
3. During Southern blot DNA hybridization analysis, the unknown DNA is
 cut into pieces, separated on a gel and transferred (by blotting) onto nitrocellulose prior to
hybridization with labeled probe DNA
 labeled and hybridized with the immobilized probe DNA
 labeled and spotted onto a surface where it is immobilized, prior to hybridization with probe DNA
 cut into pieces, labeled, separated on a gel, and transferred by blotting onto nitrocellulose prior to
hybridization with probe DNA.
 hybridized with immobilized labeled probe DNA
What DNA is labeled is never immobilized, which eliminates all be choices 1 and 2. In a blot, the
unknown is immobilized, usually after gel separation. Choice 2 describes a DNA microarray.
4. In humans, double strand breaks in DNA are repaired via
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nucleotide excision repair
transcription-coupled repair
base-excision repair
restriction endonucleases
mismatch repair
non-homologous end joining
5. Not considering polymorphisms, how many different chromosomes do humans have?
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1
2
22
23
24 Two sex chromosomes and 22 autosomes
44
46
6. Approximately what percent of the human genome is involved in encoding protein?
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80% or more
40 to 60%
15 to 25%
5 to 10%
1 to 2%
less than 0.1%
MCMP 208 Exam IV - 3
7. An anti-codon is
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added onto the 3′ end of each tRNA
found in every amino-acyl tRNA synthetase
found once in every tRNA
found once in every intron
found once in every mRNA
found twice in every intron (one at each end)
found twice in every exon (one at each end)
found in the largest two rRNAs (one in each subunit)
found in mRNA once for every amino acid encoded by that mRNA
8. If PCR amplification using human genomic DNA as template is working properly, it produces a lot of
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the entire human genomic DNA
one of the human chromosomes
the two parts of human DNA that are complementary to the sequences found in the primers used
the human DNA that is in between the sequences that are complementary to the primers used
the human DNA that lies outside the sequences that are complementary to the primers used
9. In humans homologous recombination is primarily associated with DNA repair. The primary useful
outcome for migration (rolling) of Holliday junctions during homologous recombination is to
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exchange a short length of one strand of each of the DNAs
exchange a short length of double-stranded DNA
exchange large amounts of double-stranded DNA
make it possible to resolve the junction in the opposite way from which it was formed
extend the chromosomes at the telomeres
The only thing accomplished by rolling is exchange of a single strand. Resolution options are not
affected by rolling, and telomere extension has nothing to do with rolling Holliday junctions.
10. HIV particles include three virally encoded enzymes. Of these three viral enzymes, the one that acts first
during the infection process is
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gp120 env protein
p17 capsid protein
reverse transcriptase
DNA polymerase
integrase
RNA polymerase
protease
lipase
flipase
The other two virally encoded enzymes in the capsid are the integrase and protease. Gp1209 and p17 are
not enzymes. Flipase, lipase, RNA and DNA polymerases are cellular enzymes and are not encoded in
the viral genome.
MCMP 208 Exam IV - 4
11. A DNA helicase is an enzyme that
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relaxes supercoils in DNA
adds supercoils to DNA
makes a primer for DNA polymerase to use
dissociates the two strands of DNA
remodels chromatin by altering modifications of nucleosome proteins
remodels chromatin by methylating DNA
keeps single stranded DNA from re-annealing with its complementary strand
reversibly ligates nicks in single stranded DNA
12. During translation, the synthesis of the peptide bond occurs when
 the next amino acid is transferred off of its tRNA and onto the end of the growing polypeptide
 the polypeptide is transferred off of its tRNA and onto the incoming amino acid attached to its tRNA
 the polypeptide, temporarily covalently attached to the rRNA, is transferred onto the incoming amino
acid attached to its tRNA
 the next amino acid is transferred off of its tRNA to form a temporary covalent bond with the rRNA,
and then it is added to the end of the growing polypeptide
13. The activity of RNA polymerase does not require
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NTPs as magnesium chelates
a single stranded DNA template
a properly base-paired primer
none of the above (that is, all of the above are required by RNA polymerase)
14. Which one of these dsDNA sequences (only one strand is shown) has an inverted repeat that is 4 basepairs long and has no intervening sequence?
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GGGGTTTT
GAGACTCT
AGCTTCGA
AAGGAAGG
TGGCGCCA Draw all 6 choices out as a dsDNA and put arrows above the first 4 nt and look for
them in the next 4 nt on the other strand.
 TGCAGTAC
 none of the above
15. Complementary DNA is
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the parts of each gene which bind regulatory transcription factors
the part of each gene which binds general transcription factors, such as TATA box-binding protein
the strand of DNA that is complementary to the coding strand of DNA in genes
made from mRNA using an RNA-dependent DNA polymerase
made as a primer for Okazaki fragments during DNA replication
the template used by telomerase
any two dsDNAs that share the same sequence
MCMP 208 Exam IV - 5
16. An operon is a region of DNA where
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retroviral proviral DNA integrates into the chromosomal DNA
phage DNA integrates into the chromosomal DNA
transcription of a single protein-coding gene occurs, including the regulation of its expression
synthesis of a single polycistronic mRNA occurs, including the regulation of its expression
replication originates in bacteria
replication originates in eukaryotes
DNA-protein interactions result in DNA compaction
17. Which organ system, organ, or tissue has the most diverse set of roles in metabolism?
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the kidney
the liver
adipose tissue
muscle tissue
the vasculature
the gastrointestinal tract
the brain
the immune system
the skin
18. Deficiency of vitamin C intake causes scurvy because vitamin C is essential for
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bone mineralization
calcium absorption
carboxylation of glutamyl residues in proteins
hydroxylation of prolyl residues in proteins
phosphorylation of tyrosyl residues in proteins
heme synthesis
nucleotide synthesis
regeneration of methionine from homocystenine
producing thyroid hormone
N-linked protein glycosylation
19. If in the coding strand of DNA that encodes a protein, the sequence GCTAGTCTTGTG was mutated so
that it became GCTAGTCGTGTG, and the first nucleotide of this sequence is the first nucleotide of the
triplet of nucleotides that encodes an amino acid, what is the consequence of this mutation on the
protein’s amino acid sequence?
 No change in the amino acid sequence
 substitution of leucine to proline
 substitution of leucine to arginine The codons involved are CTT mutated to CGT. They are CUU and
CGU, respectively in the mRNA because the coding strand of DNA is the same sequence as the
mRNA. For this problem, the genetic code table on page 2 is very useful.
 substitution of histidine to arginine
 substitution of threonine to lysine
 substitution of threonine to arginine
 substitution of isoleucine to arginine
 substitution of isoleucine to lysine
MCMP 208 Exam IV - 6
20. There are many small molecules that selectively inhibit prokaryotic ribosome function, without having
significant inhibition of eukaryotic ribosome function. Given that ribosomes do the same function in all
organisms, how is this possible?
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The process of translation is completely different in prokaryotes compared to eukaryotes
The energetics of translation is very different in prokaryotes compared to eukaryotes
Prokaryotic and eukaryotic ribosomes have different numbers of subunts
The genetic code is completely different in prokaryotes compared to eukaryotes
Prokaryotic and eukaryotic ribosomes have similar large scale structural organization but very
different molecular structures
21. The role of signal recognition particle (SRP) is to recognize
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the sequence that signals the sites for N-linked protein glycosylation
a structure that regulates targeted protein degradation (proteosomal degradation)
an amino acid sequence in proteins being made that targets them to the secretory pathway
the amino acid sequence that is phosphorylated by protein kinases
that a viral particle has bound to the cell surface and initiates cellular defense processes
damage in DNA and activates DNA repair processes
splice junctions during RNA splicing
22. Which is not a component of metabolic syndrome?
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obesity
diabetes metabolic syndrome is a risk factor for diabetes and does not involve diabetes directly.
hypertension
dyslipidemia
insulin resistance
23. Why is methotrexate such an effective inhibitor of dTMP synthesis?
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Methotrexate is a substrate analog competitive inhibitor of thymidylate synthase
Methotrexate directly blocks the enzyme that makes formyl-tetrahydrofolate
Methotrexate directly blocks the enzyme that makes methylene-tetrahydrofolate
Thymidylate synthase oxidizes tetrahydrofoloate to dihydrofolate Thus converting THF to DHF and
thus requiring DHFR for every reaction that is catalyzed by thymidylate synthase.
 Methotrexate blocks utilization of the vitamin form of folic acid
24. The symptoms of gout are caused by
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xanthine
β-alanine
β-aminobutyrate
acetate
urate
cholate
β-hydroxybutyrate
MCMP 208 Exam IV - 7
ESSAY PROBLEMS. Write your answers to problems 25 to 30 in the space immediately below each
problem.
25. [6 points] Consider nucleosomes and chromatosomes
a. [2 points] What is the generic name for the proteins that are present in chromatosomes and
nucleosomes? What is their unusual characteristic that gives rise to the nucleosome structure?
Histones
They are unusually basic (rich in lysine), which is how they bind to DNA
(electrostatically)
b. [2 points] What is the other macromolecule involved in nucleosomes and chromatosomes? What
structural characteristic does it have when it is part of a nucleosome or chromatosome that it
generally does not have when it is not involved in one of these structures?
DNA
It is wound around the core histones in a negative (toroidal) supercoil
c. [2 points] Nucleosomes and chromatosomes have three distinct functions in cells. List two of these
functions.
Any 2 of these three are acceptable:
Compaction of DNA
Introduction of negative supercoils
Regulation of transcription or gene expression
26. [5 points] During DNA replication, explain why one strand of template DNA is not replicated at exactly
the same time as the other template strand is replicated. [Hint: This is why one strand is called the
“lagging strand.”]
One of the template strands can be replicated in the same direction that the fork is
opening, so its site of replication can be very near the fork.
The other strand must be synthesized in the opposite direction relative to the fork since
(a) DNA is anti-parallel (and thus the two template strands are in opposite directions) and
(b) DNA is always synthesized in the 5′ to 3′ direction.
The strand that is synthesized in the other direction is synthesized starting at the fork
and going back in the direction that the fork came from. This synthesis cannot start until
a primer is made very near the fork. For this reason, the region that is being replicated
(the Okazaki fragment) is always replicating DNA in a region where the other strand has
already been replicated.
MCMP 208 Exam IV - 8
27. [3 points] In Sanger DNA sequencing, DNA is synthesized by the typical primer extension reaction.
Other than this primer extension reaction and labeling of the DNA so it can be detected, what are the two
key methodological steps in Sanger DNA sequencing that make it possible to use this simple primer
extension reaction to determine the sequence of DNA?
1) Dideoxy nucleotides are used in the extension reaction to stop the extension reaction
at a particular nucleotide. These are done in 4 separate reactions, one for each base,
creating four sets of extension products that differ in which base is at their 3′ end.
2) Separation of the extension products by electrophoresis on a gel that can resolve
differences in length of just one nucleotide.
28. [3 points] Explain how it is known with certainty that nucleotide excision repair is essential in humans
for repairing pyrimidine dimmers in DNA.
People with xeroderma pigmentosum (XP) have the heritable triat of being sensitive to UV
light induced skin pathologies, including cancer. All the XP patients have a inherited a
defective allele of one of the enzymes involved in nucleotide excision repair.
29. [5 points] The initial transcript of a human gene encoding a protein is 22,920nt long. This initial
transcript has seven exon/intron segments (numbered 1 to 7, 5′ to 3′) with the following lengths:
Segment
1
2
3
4
5
6
7
Length (nt)
4500
5500
120
10100
160
2460
180
a. [3 points] Which segments are exons and how do you know this? How long is the mature mRNA?
[Show your work for partial credit]
Segments 1, 3, 5, and 7 are exons because the transcript must start and end with an
exon, because introns are removed by splicing together adjacent exons.
The mature mRNA is 4500+120+160+180 = 4960 nt long. (This is the sum of the
lengths of the exon segments)
b. [2 points] If the start codon is after 4300 of the first exon and the stop codon is located immediately
after 30 nt in the last exon, how long is the ORF (not inclucding the stop codon)? [Show your work
for partial credit]
510 nt
There are two ways to get to the answer:
MCMP 208 Exam IV - 9
One way: Sum the translated regions of the first and last exons with the lengths of
the two middle exons, which are completely translated: 200+120+160+30= 510 nt
Other way: Subtract the untranslated regions from the full length of the initial
transcript. The 5′ UTR is 4500-200 long = 4300nt. The 3′ UTR is 180-30 = 150 nt long.
The ORF is thus 4960 – 4300 – 150 = 510nt
30. [6 points] Not all of the purine and pyrimidine bases are synthesized (as nucleotides) de novo. Those not
made de novo are made by inter-conversion reactions. Indicate which nucleotide bases are made de novo
and then, for each of those made de novo, indicate which other nucleotide bases are made from them by
inter-conversion reactions.
Note: Either the base name or the nucleotide name/abbreviation can be used for the
answer. Both are not required. However, the correct nucleotide must be indicated.
(For purines:) hypoxanthine (as IMP) is made de novo then guanine (as GMP) and adenine
(as AMP) are made from hypoxanthine (as IMP) by inter-conversion reactions
(For pyrimidines:) uracil (as UMP) is made de novo then cytosine (as CTP) is made from
uracil (as UTP) and thymine (as dTMP) is made from uracil (as dUMP) by interconversion reactions