Name:
Biochemistry 465
Hour Exam II
Spring 2004
All problems are worth 20 points, do any 5
1. a. A closed circular DNA molecule in its relaxed form has a linking number (Lk) of
500. About how many base pairs are in this piece of DNA?
How would the link (Lk) Twist (Tw) and writhe (Wr) of the DNA change under the
following conditions. (Possible answers include: increase, decrease, remain the same,
and undefined)
b. Nucleosomes are added to bind to the DNA.
c. One strand of the DNA is broken.
d. DNA gyrase and ATP are added to the solution.
e. The helix is denatured by heat.
f. Nucleosomes are added to bind to the DNA and a topoisomerase I is added as
well.
g. a Helicase is added to the DNA
Note that this is similar to a question from the text, but I added questions about Tw and
Wr that weren’t in the original problem
A. If CCC DNA is relaxed then Wr = 0, and Lk = Wr = 500 = number of times the 10.5
BP /turn of the DNA structure makes the two strands of DNA twist around each other.
If we have 500 turns and we know that we have 10.5 base pairs in each turn
then we have
500turns ×10.5 BP/turn = 5,250 base pairs
B. Since we haven’t changed the ccc nature of the DNA, the Ln remains unchanged.
Also since we haven’t changed the ionic strength of the medium, the Tw of the DNA will
also remain unchanged at 10.5 BP/turn. The only change we will see is in the Wr, and
this is not a net change, but is different in different parts of the molecule. The
nucleosome will wrap the DNA around itself and this induces a negative supercoil, or a
local change of Wr toWr-1. Since our DNA we have said above that the Lk and Tw of
the DNA is constant, we can’t have Wr change, so there must be a compensatory
change in the Wr to Wr+1 in the DNA that isn’t bound to the histone.
C. When the DNA gets nicked the Lk becomes undefined. Technically the DNA still has
its local twist so Tw should remain at 10.5 BP/turn, and the neck allows the DNA to
relax, so the Wr =0
D. Gyrase puts negative supercoils into the DNA so Lk9, Wr9, but again, since we didn’t
change the ionic strength of the medium, the local twist of the DNA Tw remains
constant
E. DNA is denatured, but it is still ccc so Lk must remain constant. As you untwist the
DNA to try to denature it, the Tw must 9, and since Lk is unchanged, Wr must 8to
compensate.
F. We start off like B above, but the addition of topo I allows the + Wr in the DNA that is
not bound to the nucleosome to relax back to zero. Since we still have - Wr in the DNA
wraped around the nucleosome, the net effect is that Lk9, Wr9 and Tw unchanged.
G. Helicase the enzyme that starts opening up DNA into single strands, but id doesn’t
change the ccc nature of the DNA. The net effect is like denaturation, Lk unchanged,
Tw9, and Wr8 to compensate.
2. Bacterial DNA is fairly simple; it usually one gene for each protein, plus a few control
sequences. Mammalian DNA, on the other hand is more complicated. Discuss the
more complicated structure of mammalian DNA. In your discussion focus on the
following terms: highly repetitive DNA, moderately repetitive DNA, satellite DNA,
centromeres, telomeres, and simple-sequence DNA.
Bacterial DNA consists of a single circular chromosome containing a few million
base pairs and a few hundred genes. Eukariotic DNA contains multiple, linear
chromosomes that are localized to nucleus, it contains roughly 600 million base pairs of
DNA and the sequence of 50-100 thousand genes. Eukariotic DNA has a much lower
ratio of genes to DNA because large amounts of the DNA contains DNA that does not
code for genes. Some of this non-coding DNA contains DNA sequences that are
relatively simple and are repeated over and over. We class this repetitive DNA into 2
major classes; Highly repetitive and moderately repetitive. Roughly 10% of mammalian
DNA is classed as highly repetitive and consists of sequences of only about 10 bp
repeated over and over. (This is also called satellite DNA). Roughly 20% of mammalian
DNA is classed moderately repetitive sequence and consists of sequences of up to
100 bp repeated 100's of times.
The complete function of the repetitive DNA is not entirely clear. Some of the
highly repetitive DNA is found in the centromeres, where is serves as the attachment
point for proteins that attach the DNA to the mitotic spindle. Repetitive DNA is also
used at the telomeres or the end of the chromosomes. Some of the moderately
repetitive DNA may simply be an artifact of transposons that have been carried in the
Eukariotic DNA, and serve no useful function
3. Compare and contrast the initiation of DNA replication and RNA transcription in E
coli. Include details like proteins and cofactors involved, DNA sequences required, how
often the activity takes place in the cell’s cycle.
Initiation of DNA replication
One start sequence for the entire chromosome, called oriC
consists of 2 short repeats
One 13 bp sequence that is repeated 3 times
One 9 bp sequence that is repeated 4 times
will proceed in a bidirectional manner
About 20 DnaA molecules bind to the oriC sequence using ATP energy
HU then binds
In ths process the DNA double helix is opened up
DnaC then helps hexamers of DnaB bind at each end of the open loop of DNA
the DnaB is a helicase that now starts actively unwinding the DNA using ATP
energy and
DNA binding protein comes in to prevent the unwound DNA from
winding back up
The DnaB then serves as the start of the DNA polymerase complex that will
include DNA gyrase and primase a well as DNA polymerase, but that is
considered part of the elongation step
This process only occurs once in the cell’s life cycle.
Overall polymerase reaction includes 3'-5' proofreading so the overall error rate
is only about on error in 106 - 108 bases added, and the overall rate of the
reaction is about 250-1000 nucleotides/second.
Initiation of RNA transcription
Many start sequence spread out throughout the entire chromosome
Sequences are not all the same, but similar to each other
This consensus is between -10 and -35 nucleotide from the actual start
position
is called the promoter sequence in the -40 to -60 region called the UP
region to further enhance the start of transcription.
Rather than several proteins working in separate steps to open up the DNA,
stabilize it and start the polymerase reaction, a core protein complex
containing all needed enzymatic activities is formed with 5 proteins,
"2$$’T. A F factor is then added to this core, and it is the F facotor that
targets the polymerase to different promoter sequences. F is released at
end of initiation
Once this core complex is formed, it simply binds to the appropriate DNA
seqeunce and begins polymerization.
Overall rate of polymerization is only 50-90 nucleotides/sec
Complex contained no topoisomerase so the action of the complex leaves +
supercoils in front of it and negtive supercoils behind it
Complex only unwinds 17 bp of DNA in this process
There is no proofreading, so the error rate is 1 in 104-105
Overall process is unidirectional
4. Describe at least three different types of DNA damage that can occur and how that
can be repaired in a cell.
DNA is made incorrectly and contains bases that are no properly matched.
Mismatch repair
Identify original strand of DNA form methyl groups that are attached to the
DNA
Cut out relative long piece of DNA on the new strand between methyl
groups
repair with DNA pol III
C and A Bases of DNA have spontaneously deaminated
Base-excision repair
Cut out bad base at the sugar/base bond using a specific DNA
glycosylase
Nick with AP endonuclease that recognizes the DNA missing the bae
Pol I replaces bad base plus a few extras
Ligase reconnects strands
Certain dimers have formed between adjacent bases due to UV damage
Nucleotide excision repair
Multisubunit enzyme nicks 5th phosphate to 3' side 12-13 to 5' side
(Slightly larger in eukariotes)
DNA helicase separates strand- damaged strand is lost
DNA Pol I replaces missing bases
Ligase fills in the nicks
Other dimers have formed due between adjacent bases due to UV damage
Direct Repair
DNA photolyase
Uses light to direction repair T-T dimer
DNA has been methylated by an alkylating agent
Direct repair by methyl transferase
removes methyl and self destruct enzyme that did it
5. Compare and contrast the four different mechanisms used to splice out introns in
eukaryotic mRNA.
Group I - Needs only RNA ( no protein involved)
uses OH of GMP (GDP or GTP) to open a 5' end of intron
the 3' end of the Exon then attacks the 3' end of the intron and releases a
linear intron
Group II also needs only RNA
usually only for mitochondrial and chloroplast mRNA in fungae, algae or
plants
uses OH of a CAA sequence in intron to attack 5' end of intron
3' end of exon then attacks at 3' end of intron
releases a lariat of RNA
Group III needs RNA-protein complexes (snRNP’s)
Need 5 snRNP’s; U1, U2, U4, U5, and U6
U1 binds near 5' end of splice site
then with ATP binds U2, U4/U6 + U5 to make splicosome
1&4 released
OH of intron attacks 5' end of intron
3' end of exon attacks 3' end of intron
releases a U5,U6/U2 lariat
Group IV certain t-RNAs
endonuclease removes intron
leaves a 2'3' cyclic phosphoester at 3' end of intron
Changed to 2' OP with a phosphodiesterase
kinase puts phosphate on 5' end of exon
followed by ligase that puts on an A-P-P
Now 3'’ exon attacks 5' APP to make linkage and release AMP
2' PO4 removed
6. Besides splicing, eukaryotic mRNA requires other processing. What other
processing must be done, how is it done, and why is it done?
5' cap is added prevents attack by exonuclease at this end and will eventually
help in binding to eukariotic ribosome
some methylation also occurs on 2' OH’s of sugars near the 5' cap
Poly A tail added Poly A tail will help in binding to eukariotic ribosome
length of poly A tail seems to correlate with lifetime of mRNA
+ correlation in Eukariots
- correlation in E coli
Some RNA removed before tail is attached