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KEY TO HOMEWORK QUESTIONS, LECTURE 8
CHAPTER 11: DNA REPLICATION
1.
In the Meselson-Stahl experiment, which of the 3 modes of replication could be ruled out after one round of replication? After two
rounds? Explain why.
Both the dispersive and semi-conservative models predicted that
after one round of replication, there would be one half-heavy band. In
the dispersive model, the original heavy band has been dispersed
evenly into the two daughter molecules, both of which also contain ½
light nitrogen. In the semi-conservative model, each new daughter
molecule contains one heavy strand (from the parent molecule) and one
light strand (newly synthesized). Since the DNA does not denature on
the gradient, one half-heavy band would be predicted by both models,
which is what was seen. The conservative model predicted that there
would be two bands: one heavy (containing both original heavy parent
strands) and one light band (both strands newly synthesized). Thus,
the first-round data ruled it out.
After two rounds of replication, the semi-conservative model predicts
two bands: one light band (daughter molecules with all new
nucleotides) and one half-heavy band (containing one of the parent
strands and one new, light strand) while the dispersive model predicts
only one ¼ heavy band. Since one-half heavy and one light band was
seen, the dispersive model was eliminated and the semi-conservative
model was strongly supported.
2.
Several temperature-sensitive mutant strains of E. coli display various
characteristics. Predict what enzyme or function is being affected by
each mutation.
a.
b.
Newly synthesized DNA contains many mismatched base pairs.
DNA polymerase has lost its proofreading function.
Okazaki fragments accumulate and DNA synthesis is never
completed.
Ligase activity has been eliminated.
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Spring 2006
c.
No initiation of replication occurs.
DnaA proteins may be missing or cis-elements at the ori are
mutated such that the DnaA proteins cannot bind them.
Alternatively, the A-T rich region may have been deleted in the
ori so that the region fails to denature when the DnaA proteins
bind their sites.
d.
Synthesis is very slow.
Processivity of the DNA polymerase is disrupted; it keeps
falling off the DNA and has to reattach, slowing down the
process.
Supercoiled strands are found to remain following replication,
which is never completed.
Topoisomerse (gyrase) activity is missing.
e.
3.
2
In 1994, telomerase activity was discovered in human cancer cells in
culture. Although telomerase is not active in human somatic tissue,
this discovery indicated that humans do contain the genes for
telomerase. Since inappropriate activation of telomerase can cause
cancer, why do you think that genes coding for this enzyme have been
maintained in the human genome throughout evolution? Are there any
types of human body cells where telomerase activation would be
advantageous or even necessary? Explain.
Telomerase must be expressed in germ-line tissue to maintain
telomere length from one generation to the next. Otherwise,
telomeres would shorten every generation until the human species
went extinct!
4.
The genome of Drosophila consists of approximately 1.6 x 108 base
pairs. DNA synthesis occurs at a rate of 30 base pairs per second. In
the early embryo, the entire genome is replicated in 5 minutes. How
many bidirectional origins of replication are required to accomplish
this feat?
If there was a single bi-directional origin, the process would take
1.6 x 108/2 x (1 sec/30 bp) = 2.7 x 106 sec
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Since it only takes 300 seconds (5 minutes), there must be
2.7 x 106 sec/ 300 sec = 8,889 origins
5.
DNA polymerases in all organisms only add the 5’ end of nucleotides to
the 3’ end of a growing DNA strand, never the other way around. One
possible reason for this is the fact that most DNA polymerases have
a proofreading function that would not be energetically possible if
DNA synthesis occurred in the 3’ to 5’ direction.
a.
Draw out the reaction that DNA polymerase would have to
catalyze if DNA synthesis occurred in the 5’ to 3’ direction.
BIO 184
b.
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Considering the information present in your drawing, speculate
as to why proofreading would be problematic, while it isn’t if
synthesis occurs in the 3’ to 5’ direction.
If synthesis occurred 3’ to 5’, the energy for the polymerization
reaction would have to come from the breaking of the
triphosphate of the last nucleotide added, not from the incoming
nucleotide. Thus, if DNA polymerase made a mistake and removed
the nucleotide it had just added, it would not have an energy
source for replacing it with a different nucleotide.
Therefore, proofreading is only possible if DNA polymerase
synthesizes the new strand 5’ to 3’. In this case, a mismatched
nucleotide can be removed and replaced because each incoming
nucleotide carries with it the energy source for its addition to
the growing chain.