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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 ______________________________________________________________ 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. 1 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. 2 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. 3 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. 4 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 5 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 ______________________________________________ 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. 6