Download Name: Chem 465 Biochemistry II - Test 3

Name:
Chem 465
Biochemistry II - Test 3
Multiple Choice 4 points each
1. The role of the Dam methylase is to:
A)
add a methyl group to uracil, converting it to thymine.
B)
modify the template strand for recognition by repair systems.
C)
remove a methyl group from thymine.
D)
remove a mismatched nucleotide from the template strand.
E)
replace a mismatched nucleotide with the correct one.
2. In base-excision repair, the first enzyme to act is:
A)
AP endonuclease.
B)
Dam methylase.
C)
DNA glycosylase.
D)
DNA ligase.
E)
DNA polymerase.
3. In homologous genetic recombination, RecA protein is involved in:
A)
formation of Holliday intermediates and branch migration.
B)
introduction of negative supercoils into the recombination products.
C)
nicking the two duplex DNA molecules to initiate the reaction.
D)
pairing a DNA strand from one duplex DNA molecule with sequences in
another duplex, regardless of complementarity.
E)
resolution of the Holliday intermediate.
4. RNA polymerase:
A)
binds tightly to a region of DNA thousands of base pairs away from the
DNA to be transcribed.
B)
can synthesize RNA chains de novo (without a primer).
C)
has a subunit called (lambda), which acts as a proofreading
ribonuclease.
D)
separates DNA strands throughout a long region of DNA (up to thousands
of base pairs), then copies one of them.
E)
synthesizes RNA chains in the 3' 6 5' direction.
5. Which of the following is not known to be involved in initiation by eukaryotic RNA
polymerase II?
A)
DNA helicase activity
B)
DNA polymerase activity
C)
Formation of an open complex
D)
Protein binding to specific DNA sequences
E)
Protein phosphorylation
6. Which one of the following is not true of the mRNA for ovalbumin?
A)
Exons are used for polypeptide synthesis.
B)
Introns are complementary to their adjacent exons and will form hybrids
with them.
C)
The mature mRNA is substantially shorter than the corresponding region
on the DNA.
D)
The mRNA is originally synthesized in the nucleus, but ends up in the
cytoplasm.
E)
The splicing that yields a mature mRNA occurs at very specific sites in the
RNA primary transcript.
7. Reverse transcriptase:
A)
can utilize only RNA templates.
B)
has a 3' 5' proofreading exonuclease but not a 5' 6 3' exonuclease.
C)
is activated by AZT.
D)
is encoded by retroviruses.
E)
synthesizes DNA with the same fidelity as a typical DNA polymerase.
Essay Questions (14 points each)
8. Compare and contrast base excision repair with nucleotide excision repair.
You can answer this but just outlining the two different pathways as shown below..
Base excision repair - Enzymes recognize common DNA lesions and remove just
the base from the DNA double helix to create an AP sites. The AP site is then
recognized by an AP endonuclease that nicks the damaged strand. DNA pol I then
does nick translation at this site, and removes enough of the DNA to replace the
missing base, and ligase then seals up the nick..
Nucleotide excision repair - Enzymes recognize common DNA lesions, that are
usually large and distort more of the helix, typically things like base dimers. In the E.
Coli system the enzyme exinuclease nicks the strand at 2 points, one 5 bases on the 3'
side and the other 12-13 based to the 5' side of the damage. DNA helicase then
unwinds the damaged DNA, and the small piece containing the damaged DNA is
released. DNA pol I then replaces the missing DNA, and ligase then seals up the nick
that remains in the strand.
Or once, outlined, you can put down similarities and differences
Difference
Base excision- recognizes a small defect on a single base
removes single base, but leaves DNA double stand intact
Uses AP endonuclease to put a single nick in DNA
Nucleotide excision- recognizes a larger defect like a base-base dimer
Uses Exiniclease to put 2 nicks on damaged DNA strand
Removes several bases around damaged area
Same
Both finish up repair with pol I and ligase
9. Give the mechanism of how site-specific recombination occurs (5 points), then use
this mechanism to explain how a piece of viral DNA can use a recombination event to
insert its viral DNA into a host chromosome (5 points). Now use this same model to
explain how recombination can be used to invert a sequence of DNA (4 points).
For the first 5 points I wanted a diagram like figure 25-38 (4th edition), and some
mention about the 4 different binding sites and how the DNA is covalently attached to
the enzyme through a tyrosine for part of the mechanism to preserve the energy of the
P-O band.
For the next 5 points you can use the same diagram, and make the upper blue
DNA the long chromosomal DNA and connect the two ends of the lower red DNA into a
loop to be the viral DNA inserting into the hose chromosome.
For the last 4 points, use the same diagram. Our chromosomal DNA will start
from the upper left hand corner. Label this ‘1' where it comes out on the upper right
hand corner, label it ‘2'. Now loop this DNA around so it comes in on the lower left, and
label this ‘3', and where the DNA come out on the lower right ‘4'. Next go through the
recombination as written. When recombination is complete the DNA sequence reads
1,3,2,4 si the DNA between 2 and 3 has been inverted!
10. Explain the complete process of DNA dependent RNA synthesis in eukariots using
RNA polymerase II.
Start with a diagram like figure 26-9 (4th edition)
TBP (TATA-binding protein) binds to DNA at the TATA box
Transcription factor TFIIB binds on both sides of TBP to further stabilize DNAProtein complex. (Factor TFIIA can also bind here to make stronger)
TFIIF and Pol II form a separate complex that now binds to TFIIB part of DNA
complex
TFIIE and TFIIH now bind to make closed complex
Using ATP energy, TFIIH acts as a helicase to unwind DNA to open up the DNA
and make the ‘open’ complex.
Above process actually take 30 different peptides
TFIIH uses ATP to phosphorylates Pol II many places on the CTD
(carboxyterminal domain). Once phosphorylated, the Pol II begins to make RNA from
the DNA template. After about 60-70 nucleotides are synthesized, TFIIE and TFIIH fall
off the complex and several elongation factors bind so elongation of mRNA continues.
Book does not really cover termination except to say that when termination is
complete the mRNA is released, and the Pol II is dephosphorylated.
11. Continue on from question 10 to explain all the m-RNA processing events that occur
on the RNA polymerase II complex as the m-RNA is synthesized.
Processing of the mRNA begins about in the middle of the above process, after
only 20-30 nucleotides have been made. At this point the m-RNA reachs the CAP
complex taht is associated with the CTD of Pol II. This CAP complex consists of
phosphohydrolase to remove the 5' terminal phosphate, guanylyltransferase to add the
guanosine cap, and guanine-7-methyltransferase to methylate the guanine cap at the 7
position. There may also be a 2'-O-methyltransferases to methylate the initial 1 or 2
ribose sugars at the 2' position. After the cap is complete the CAP complex is released
from the CTD, but a Cap Binding protein (CBP) hangs on to the cap and keeps it
associated with the CTD of the Pol II.
Splicing of the mRNA occurs as the mRNA continues to be synthesized on the
Pol II because the CTD of the Pol II also binds splicesomes. The spliceosome consists
of U1, U2, U4, U5 and U6 snRNP’s (small nuclear ribonucleoproteins) and this complex
splices out each intron almost as quickly as it is synthesized.
The final piece of processing, removing the last several bases, and adding the
poly A tail also takes place on a large enzyme complex associated with the CTD of Pol
II. The complex consists of an endonuclease that clips the mRNA at a site between a
(5')AAUAAA(3') 10-30 residues before the clip and a GU rich region 20-40 residues
after the clip site. Polyandenylate cyclase then adds 80-250 A’s to the tails of the RNA.
12. In Chapter 24 you learned that much of the human genetic material consists of
transposons. In Chapter 25 you learned that most transposons integrate using a
recombination event. In Chapter 26 we learn that most eukariots transposons are
retrotransposons. Put these three chapters together; what is a retrotransposon and
what is the life cycle of a retrotransposon? How does a piece of RNA in integrated over
and over into the DNA of a eukariotic cell.
A retrotransposon is a transposon that is replicated through an RNA intermediate.
Retrotransposons are most common in eukariotic cells. A typical retrotransposon life
cycle with start with a sequence of DNA that has already been integrated into the host
chromosome. This piece of DNA resembles the sequence of on RNA virus in that is
usually contains a pol gene that would contain the coding for a protease needed for
processing the synthesized protein in the protease, a reverse transcriptase and an
integrase (or recombinase or endonuclease) The DNA noes not, however, contain the
env gene that would normally code for the virus coat protein.
When this piece of DNA is transcribed, the virus portion of the gene is actually spliced
out of the mRNA. This spliced mRNA is translated separately into an endonuclease
and a reverse transcriptase. The endonuclease introduces a double strand break into a
random site on the DNA, and the RNA is incorporated into the DNA at this site in a
‘reverse splicing’ reaction. The reverse transcriptase then makes a DNA copy of the
mRNA and incorporates into the chromosomal DNA while destroying the original RNA.
1B
2C
3A
4B
5 B
6. B
7. D