Download Genetics 314 – Spring 2007

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Genetics 314 – Spring 2007
Exam 1 – Key
1. It took several experiments to convince the scientific world that DNA carried the
genetic information in a cell. Describe one of these experiments and how the
results demonstrated that DNA carried genetic information.
The two experiments were: 1) The Avery, Macleod and McCarty experiment where
they repeated the Griffith’s experiment with heat-killed Pneumoccocus but used
enzymes to eliminate proteins, RNA or DNA prior to transformation of the rough
non-virulent bacteria demonstrating that transformation could only be halted with
the DNAase treated heat-killed Pneumoccocus; 2) The Hershey-Chase experiment
where they labeled the DNA and protein of a bacteria phage with different isotopes,
allowed the phage to infect bacteria and then determined where the radioactivity
was (within the bacteria or in the supernatant) after putting the bacteria into a
blender after infection demonstrating the labeled DNA entered the cell while the
labeled protein stayed in the supernatant. Half credit was given it Griffith’s original
experiment was described involving smooth and rough Pneumoccocus because this
experiment did not prove or demonstrate that DNA carried the genetic information
but did give the first indication that DNA could be the carrier.
2. Much can be learned about DNA by simply heating and cooling it.
a) What information can be gained about DNA from different sources by
determining the temperature the DNA denatures, briefly explain your answer.
The temperature needed to denature DNA will give an indication of the G-C to A-T
ratio because the temperature needed to denature a given piece of DNA is
dependent on the number of hydrogen bonds holding the strands of DNA together.
The more G-C base pairs in the DNA translates to more hydrogen bonds the higher
the temperature for denaturing since G-C pairs are held together by 3 hydrogen
bonds while A-T pairs are held together by 2 hydrogen bonds.
b) Cooling the DNA and determining the rate of renaturation can also give you
information concerning the nature of the DNA. What causes differences in the
rate of renaturation of DNA as it cools?
Several aspects of a given piece of DNA will cause differences in the rate of
renaturation. The first is genome size which relates to the number of unique DNA
sequences. The larger the genome the longer the DNA will take to renature. The
second aspect of DNA that can be determined is the presence of highly repetitive
and moderately repetitive DNA sequences in a DNA sample. The presence of highly
repetitive sequences will cause an initial rapid rate of renaturing such as that
observed with DNA from eukaryotic organisms
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c) You are given the following Cot curves A and B. Based on the curves what can
you say about the composition of the DNA of each sample and which would be
from a prokaryote and which would be from a eukaryote?
A.
B.
A. Eukaryotic DNA – initial rapid renaturing due to the presence of highly
repetitive DNA followed by renaturing of moderately repetitive DNA and
finally renaturing of the unique DNA sequences of which there are a
relatively high number.
B. Prokaryotic DNA – curve is based on number of unique sequences / genome
size.
Once DNA was determined to be a double helix the next question was how did it
replicate.
a) What were the three possible ways DNA could replicate and which one was
proven to be correct?
Conservative, Semi-conservative, and Dispersive
DNA was found to replicate in a semi-conservative manner
b) The manner that DNA replicated was proven using an isotope of nitrogen. Using
N15 it indicate original DNA strands and N14 for newly synthesized strands
diagram what would be expected after one replication and after two replications.
1st replication
2nd replication
N15
N14
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3. You are given the following strand of DNA
3’ TAC CCC GCT TTT ATA GGG CGA ACC AAC CCG GGC ATT 5’
a)
What is the sequence of the complimentary strand of DNA? In your answer label the
3’ and 5’ ends.
5’ ATG GGG CGT AAA TAT CCC GCT TGG TTC GGC CCG TAA 3’
b)
What would be the mRNA sequence if you used the original DNA strand as a
template? In your answer label the 3’ and 5’ ends.
5’ AUG GGG CGU AAA UAU CCC GCU UGG UUC GGC CCG UAA
3’
c)
Using the codon table at the end of the exam, give the amino acid sequence that is
coded for by the mRNA strand that you produced in your answer to ‘b’.
Met
d)
Did any amino acids repeat in you polypeptide? How is this possible if none of the
codons repeated? Briefly explain your answer.
Yes, several amino acids repeated even though no codons repeated. This is because of the
redundancy of the genetic code where more than one codon can code for the same amino
acid.
4. While the DNA sequence in question 4 codes for the amino acid sequence of a protein
it will not allow for the DNA sequence to be transcribed and translated. What other
DNA sequences need to be added to insure the gene is transcribed and translated in a
cell. In your answer briefly describe the role of each sequence.
Promoter sequence – needed to provide a binding site for sigma factor/RNA polymerase
for the start of transcription.
Leader sequence – needed to provide a binding site for the small ribosomal sub-unit for
the start of translation.
Termination sequence – needed to signal the release of the RNA polymerase for the end
of transcription.
5. When adding these additional sequences of DNA why is it important to know if the
gene is going to be put into a prokaryotic cell or eukaryotic cell for transcription and
translation.
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Because prokaryotes and eukaryotes have slightly different enzymes that recognize
different sequences for initiation and termination of transcription and translation. To
make sure you get expression of a gene you need to have the proper promoter, leader and
termination sequences to match the organisms enzymes involved in transcription and
translation.
6. If a chemical was discovered that prevented complimentary pairing of RNA nucleotides
with other RNA nucleotides, what impact would this have on transcription and would
the impact be the same in prokaryotes and eukaryotes?
This would have the potential to inhibit the formation of hairpin loops in the mRNA
while it is immediately after it has been transcribed. The formation of these sections of
double-stranded RNA form a structure that is part of the mechanism to stop
transcription in prokaryotes. In the case of rho independent termination the lack of these
hairpin structures would result in RNA transcription not being terminated. The lack of
formation of double stranded RNA structures in eukaryotes may not interfere with
termination of transcription in eukaryotes because little is known about the mechanism
of termination of transcription in eukaryotes.
7. You are cleaning out a lab refrigerator and find a tube labeled wonder protein DNA.
You decide you want to know what this wonder protein is and what type of organism it
originated from. You transform it into a bacteria (prokaryote) and a yeast (eukaryote).
You recover a protein from each type of cell but the protein from the yeast cell is
functional and the protein from the bacteria is non-functional. Why the difference in
functionality and how does this help you determine the source organism for the gene?
The difference in functionality is due to the presence of introns in the gene that
remain in the transcript produced in a bacteria compared to the transcript
produced in the yeast. This is because the yeast will have the splicesome complex
needed to process out the intorns prior to translation. If the introns are not
removed additional amino acids could be added to the polypeptide rendering it nonfunctional. Based on the information of functionality of the protein produce the
‘wonder’ gene is a eukaryotic gene.
8. You want to replicate a piece of DNA in vitro so you buy several DNA
replication kits off of E-Bay for half the price it normally would cost. You add
your DNA to the kit and do not get complete synthesis.
a) Using the first kit you discover no replication has occurred. You read the fine
print and see that it say ‘RNA nucleotides sold separately’ could this be your
problem? Briefly explain your answer.
Yes, for DNA polymerase to synthesize DNA it needs an open 3’ end to start
producing a new DNA strand. This open 3’ end is produced by primase utilizing
RNA nucleotides to produce a RNA primer. Without RNA nucleotides no primer
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could be produced resulting in no open 3’ end to be available for the start of DNA
systhesis
b) You check the other kit and find it does include RNA nucleotides. Feeling
confident you run the reaction again and discover that the DNA does replicate but
now there are RNA nucleotides in the DNA and one of the strands is not a
continuous piece of DNA but is a series of short pieces. What is wrong now?,
why do you have pieces?, and how would you correct the problem?
If DNA still has RNA nucleotides present within the strand it means that DNA pol I
is missing or non-functional. The role of DNA pol I is to remove the RNA primer
and replace the RNA nucleotides with DNA nucleotides. To solve the problem, add
DNA pol I. If gaps still remain in the lagging or discontinuous strand of DNA it
may be necessary to also add ligase to connect the final phosphor-diester bond.
9. You switch to working on transcription of a gene in vitro so again you go on-line
and bid on a transcription kit from E-bay. You notice it says for prokaryotic
genes only.
a) Why would it be necessary to limit an in vitro kit for transcription to be only for
prokaryotic or eukaryotic organisms?
The reason is that the enzymes provided in the kit would be specific for promoter,
leader and termination sequences unique to either prokaryotes or eukaryotes. If the
kit is only for prokaryotic genes then prokaryotic promoter, leader and termination
sequences will be needed to insure proper transcription and translation of the gene.
b) You determine your gene is from a prokaryotic organism so you proceed with
transcription. You discover all your transcripts are longer than expected? What
could be happening and how would you fix it.
There could be a problem with the termination sequence of the gene or the
termination sequence being used was rho dependent and rho was not present in the
kit. Since you contributed the DNA and it should have the correct sequence for
termination it most likely was a problem with the kit lacking rho. The way to solve
this is to add rho.
10. You decide to attempt to translate your transcript and even though it is too long
you still get the correct sized polypeptide, how is this possible?
The reason for this is that termination of translation has no relationship with the
termination of transcription. As long as the stop codon is in the correct location on
the mRNA transcript, translation will halt at the proper codon and the polypeptide
will be the correct size.
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11. Being wary of ordering kits off of E-Bay you decide to read the labels on the kits
before you buy them. You find that some of the translation kits contained 45
different tRNAs, others 54 different tRNAs while a third had 61 tRNAs. Even
though they differed for number of tRNAs they all claimed they would
successfully produce the polypeptide produce.
a) How is this possible?
Technically, because of the redundancy of the genetic code and the fact that some
tRNAs have a modified base in the third base spot in the anti-codon allowing that
tRNA to match several different codons, it is possible to have fewer tRNAs than
there are codons. The exceptions are the tRNAs that correspond to the codons for
methionine and tryptophan since these two amino acids are only coded for by one
codon.
b) If you used the 45 different tRNA kit and you discover translation never started
what would you suspect is missing from your kit and why is it important?
Most likely the kit did not contain F-met which is a modified methionine that is
needed for the initiation of translation. It is also possible that the lone tRNA needed
for methionine was missing which would also prevent translation from occurring.
This is because before the ribosome can assemble an activated tRNA carrying F-met
must be positioned on the small ribosomal subunit. Without this modified amino
acid or tRNA assembly of the ribosome would not occur resulting in no translation.
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