Download Genetics 314 – Spring 2005

Name: __________________________________
Genetics 314 – Spring 2005
Make-up Exam 2 – 100 points
1. You have moved into a new laboratory and find a half full bottle of some
chemical that does not have a label on it. Campus safety office will not pick it up
unless you can prove it is a chemical they are required to handle, such as a
mutagen. You order special bacteria that have been designed to detect frameshift
mutations (Bacterial strain A) or base change mutations (Bacterial strain B). You
set up the following experiment to test your unknown by adding either water or
your chemical to the bacteria and get the following results.
Chemical or water
Water
Chemical
Bacteria strain
A
A
percent mutations
0.12
0.95
Water
B
0.13
Chemical
B
0.15
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a) Why did you test your bacterial strains with water and what would cause the
mutations that are observed with this treatment.
The bacterial strains were tested with water to determine the natural or
spontaneous rate of mutation. This needs to be known to determine if the
chemical in question actually causes an increase in the rate of mutations.
Mutations occur naturally in bacteria due to errors in proof-reading during
replication and tautomeric shifts changing how bases pair during replication.
b) Based on these results is the chemical a mutagen and what type of mutations
does it cause?
The chemical appears to cause a significant increase in mutations in bacterial
strain A. This would indicate that the chemical causes frame-shift mutations.
c) Does the type of mutation caused by this chemical have a greater potential to
produce non-functional protein products? Briefly explain your reasoning.
Frameshift mutations do have a greater potential to produce non-functional
protein products. The reason for this is that the addition or deletion of one
or two bases will not only change the codon where the mutation occurred but
it will also change the reading frame of the gene downstream of the mutation
changing all the codons in the gene after the site of the mutation. This level
of change in the resulting amino acid sequence would have a greater
potential to produce a non-functional protein product.
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Name: __________________________________
2. A friend of yours decides to play with your bacterial strains and takes them into
the genetic engineering lab and exposes them to the U.V. lamp in the gel imaging
room to produce mutations. To his surprise the mutation rate he observes is not
much higher (0.22) than what was observed with your water treatment.
a) Why did your friend expect to see mutations?
Ultra violet light is known to cause mutations by inducing the formation of
thymine dimmers in DNA. The presence of these thymine dimers can cause
problems during DNA replication.
b) Why did he not observe the number of mutations he expected?
The mutation rate was not as high as expected because bacteria contain two
repair systems for damage caused by ultra-violet light. The first is a light
activated system where the enzyme photolyase utilizes energy from light to
break the covalent bond between the adjacent thymines eliminating the
thymine dimers. There is a second excision – repair system that does not
require light energy that will cut out the area of the DNA strand that
contains the thymine dimers and then DNA polymerase will come in and use
the other strand of DNA as a template to replace the section of DNA that was
excised.
3. While researching the chemical reference books to identify your unknown
chemical mutagen you read about some chemicals that are base analogs of
adenine and guanine.
a) What is a base analog?
A base analog is a chemical that mimics the appearance of one of the bases
found in DNA that once inserted into DNA has the potential to pair
differently.
b) How does it cause a mutation?
Mutations occur due to the different pairing behavior of the base analog.
The change in pairing will lead to base change mutations in subsequent
cycles of replication.
4. You finally clear your laboratory of unknown chemicals and set up your lab to do
genetic engineering of bacteria. You are asked to take a gene created by another
lab and have the gene product mass produced. You would like to be able to turn
on and off production of the gene product in the bacteria fermentation tanks to
harvest the gene product.
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Name: __________________________________
a) What type of prokaryotic gene regulation system would you use to achieve
this type of control?
For having control to turn on and off gene expression the best system to use
would be an inducible regulatory system like the Lac operon.
b) Briefly describe how your system would work to turn on and turn off
expression of the gene.
An inducible system would have a regulatory protein called a repressor
protein that would bind to a DNA sequence called the operator site in the
promoter region of the operon preventing binding of RNA polymerase
resulting in no transcription or expression of the genes. The only way
expression could occur would be the addition of an inducer that could bind to
the repressor protein causing a conformational change making it impossible
for the repressor to bind to the operator site. With no repressor present,
RNA polymerase could bind and transcription could occur. To turn
expression off, one would just need to stop supplying the inducer.
5. A friend wants to help so decides to increase bacterial growth by adding glucose
to your fermentation tanks. While the additional glucose increases bacterial
growth, production of the gene product drops to almost zero. How could glucose
affect the regulatory system of your gene? Briefly describe how such a system
might work.
In some inducible systems there is a secondary form of regulation, an
example of this is catabolite repression in the Lac operon. Here the presence
of glucose will affect the level of a second product, cyclic-AMP, that is needed
to facilitate separation of the DNA strands for binding of the RNA
polymerase. So it is possible that glucose is inhibiting some form of activator
protein resulting in a decreased ability of RNA polymerase to bind. This in
turn leads to a decrease in transcription and gene expression.
A second possibility is glucose could act as a co-repressor, either by
competitive binding to the repressor, preventing binding of the inducer
allowing the repressor protein to stay bond to the operator region. Glucose
could also be a co-repressor that binds to an inactive repressor that in the
presence of glucose becomes active and binds to the DNA blocking
transcription and gene expression when glucose is available to the bacteria.
6. You are now asked to produce a product that requires a three enzyme biochemical
pathway. You are given the three genes that code for the enzymes but are told
you need to produce the product in an eukaryotic organism. Design a system that
would work in a eukaryotic organism that would turn on synthesis of the product
when you wanted it on. Remember, to get the product you would need all the
genes associated with the biochemical pathway on at the same time.
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Name: __________________________________
Operons, where one promoter region controls expression of several genes do
not exist in eukaryotes. The way to have three genes be transcribed
simultaneously in a eukaryotic organism would be to have the same enhancer
and up-stream activator sequences present upstream of each gene. In this
way when the activator proteins are present they will turn on expression of
all three genes at the same time
7. You discover your transformed eukaryotic cells have shut down expression of
your introduced genes. You know the gene is present but no translation is
occurring. Describe a pre or post-transcriptional regulatory system in eukaryotes
that could permanently shutdown expression of a gene.
The two most likely systems in eukaryotes that could permanently shutdown
expression of a gene would be methylation and siRNA. Methylation is a form
of pre-transcription regulation where the addition of a methyl group on
specific DNA bases serves as a signal to prevent transcription of the
methylated gene. Small interfering RNA is a form of post-transcriptional
control where specific endonucleases in a cell pick up short RNA sequences
of a given gene and use those sequences to identify mRNA transcripts for
elimination. Using base-complimentarity to recognize specific RNA
transcripts, siRNA can prevent gene expression by preventing the mRNA
from ever being translated. In both cases the end result is the permanent
suppression of expression of a gene.
8. You wander down the hall and listen in on a discussion in the virology lab. The
discussion was centered on how to reduce viral replication without hurting the
host bacterial cell.
a) One suggestion was to prevent rolling circle replication. Describe rolling
circle replication and explain why it would have a greater impact on virus
replication than bacterial replication.
Rolling circle replication is where one strand of circular DNA serves as a
template to produce multiple copies of either single-stranded or doublestranded linear DNA. DNA virus requires production of linear DNA for
packaging the viral genetic material into the protein capsules. The lack of
linear DNA would hinder this final step of virus replication. Bacteria only
use rolling circle replication for conjugation, a form of bacterial
recombination so bacterial replication would not be affected.
b) Another suggestion was to target enzymes specific for virus replication though
it would only work on certain types of virus. What type of virus would such a
control system work on and why would it not affect the host cell.
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Name: __________________________________
RNA virus requires non-host enzymes for replication. Targeting these
enzymes would severely limit replication of this type of virus but because the
enzymes are not required by the host for replication the host’s ability to
replicate would not be compromised. Examples of enzymes that could be
targeted are replicase for an RNA-RNA virus and reverse transcriptase for
retro viruses (RNA-DNA-RNA virus).
9. Recent news reports have indicated that the world may be facing a pandemic from
the avian (bird) flu. At present the virus can be only transferred between birds
and humans through direct contact. Of concern is the development of a
recombinant avian flu virus that can be more easily spread among humans.
a) What is required for virus recombination to occur?
Simultaneous infection by the two viruses is needed before recombination
can occur. Both viruses must be within the same cell for an exchange to
occur.
b) Describe one method of recombination that could lead to a more virulent avian
flu virus.
Once both viruses are within the same cell recombination can occur through
several mechanisms. The first would be a simple error in packaging where
RNA sequences from on virus get packaged with RNA sequences from
another virus. A second method would be direct recombination between two
pieces of genetic information facilitated by regions of complimentarity which
allow pairing and possible recombination to occur. A third method is by
copy-choice where RNA or DNA from the two viruses are in contact and
during replication the replicase or DNA polymerase starts on one piece and
moves over to an adjacent piece resulting in a final product that is a hybrid
between the two pieces of viral RNA or DNA.
10. The virology lab is all excited, they have isolated a gene for a protein that will
produce an immune response to the avian flu virus and they ask you to put the
gene into bacteria to mass produce it.
a) What is the easiest way to get a gene into bacteria?
Transformation
b) Briefly describe how it works.
Transformation is a form of bacterial recombination where bacteria will take
up free-floating DNA from the environment and some of it will become
incorporated into the bacterial chromosome. Transformation can be
facilitated by making the bacterial cells competent (more likely to take up the
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Name: __________________________________
DNA through both active and passive methods) and by having the DNA in
plasmid form so incorporation of the DNA into the bacterial chromosome is
not required for expression of the introduced gene in the bacteria.
11. The people in the bacteriology lab are using another form of bacterial
recombination to map the bacterial genome. Using four strains of bacteria they
come up with the following data:
Strain
first gene out
last gene out
A21
bio
gal
lac
mal
trp
his
B32
mal lac
gal
bio
cys
ala
C43
kan
bar
ala
cys
bio
gal
D54
lac
mal kan
bar
ala
cys
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a) What form of bacterial recombination would produce this type of data?
Conjugation (1 point given for just mentioning Hfr without mentioning
conjugation)
b) Why do the gene orders appear to differ among the different strains of
bacteria?
The F plasmid can insert in multiple sites in the bacterial chromosome and in
one of two orientations.
c) Draw a circular map of the bacterial genome based on this data indicating
insertion points and orientation for each bacterial strain.
There was a problem with the map in that the mal gene could have either
kan or trp adjacent to it on one side and lac on the other. This could not be
so there was an error in A21 or D54. Grading of this part of the question was
primarily based on insertion sites and orientation of the inserted F plasmid
due to the problem with the presence of two different genes on one side of the
mal gene.
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