Life Sciences 1a Practice Problems 5 1. Soluble proteins that are normally secreted from the cell into the extracellular environment must go through a series of steps referred to as the secretory pathway. Every step of the pathway is regulated, and there are many points where the journey of a secreted protein can go awry. a) Name two ways in which a protein could be mistargeted. You have isolated 4 mutant cell lines that mistarget the protein LSA. These mutant cell lines fail to properly secrete LSA into its normal location, the extracellular media. b) To study the mistargeting, you create a fusion of wild-type (normal) LSA to GFP and you express this LSA-GFP fusion protein in all four mutant cell lines. Explain how you could use these engineered cells to distinguish between whether the cell lines have a mutant endogenous LSA or whether they have a defective secretory pathway component? You perform the experiment described in b) with wild-type (normal cells) and your four mutant cell lines and get the following results: Strain Wild-type + LSA-GFP Mutant 1 + LSA-GFP Mutant 2 + LSA-GFP Mutant 3 + LSA-GFP Mutant 4 + LSA-GFP Location of LSA-GFP fluorescence extracellular media lysosome extracellular media extracellular media trans-golgi c) Which of these cell lines have secretory pathway defects and which have a defective endogenous LSA protein? d) Why is it important to perform the experiment with wild-type cells and LSA-GFP?
e) Next you wish to further determine the cause of the mistargeting of the endogenous LSA in all four cell lines. You use an antibody to fluorescently tag the endogenous LSA in each of these cell lines and obtain the results shown in the figure below. Give one possible explanation for mistargeting in each of the 4 mutant cell lines taking into account the data from your LSA-GFP experiment. Figure 1: Mistargeting of endogenous LSA in the secretory pathway. A) Schematic of cross section of a cell. ER=endoplasmic reticulum; Lys=lysosome; TG=trans Golgi. B) Wild-type cell expressing wild-type LSA protein. The location of the LSA is shown in green and was determined by tagging the endogenous LSA with a fluorescent antibody. C) LSA in mutant cell line 1. D) LSA in mutant cell line 2. E) LSA in mutant cell line 3. F) LSA in mutant cell line 4
2. Artificial liposomes are useful in vitro tools for studying the selective permeablility of cell membranes. You find that many macromolecules that easily pass from the external membrane into cells do not enter liposomes. What general feature of living cell membranes do the artificial liposomes lack that accounts for the above observations? 3. Label the structures of the cell indicated by the lines on the figure below: A. nucleus B. free ribosomes C. rough endoplasmic reticulum D. Golgi apparatus E. cytosol F. endosome G. plasma membrane H. lysosome I. mitochondrion J. peroxisome
4. If you were to remove the ER-retention signal from a protein that normally resides in the ER lumen, where do you expect the protein will ultimately end up? Briefly explain your reasoning. 5. You have created a GFP fusion to a protein that is normally secreted from yeast cells. You obtain a collection of three mutant yeast strains, each one defective in some aspect of the protein secretory process. Being a good scientist, you of course, also obtain a wild-type control strain. You decide to examine the fate of your GFP fusion protein in these various yeast strains and engineer the mutant strains to express your GFP fusion protein. However, in your excitement to do the experiment, you realize that you did not label any of the mutant yeast strains and no longer know which strain is defective in what process. You end up numbering your strains with the numbers 1 through 4, and then you carry out the experiment anyway, obtaining the following results (note that the black dots represent your GFP fusion protein): Name the process defective in each of these strains. Remember that one of these strains is your wild-type control. (Note: the ER, Golgi and secretory vesicles are shown above, but the nucleus has been omitted)
6. Before nuclear pore complexes were well understood it was unclear whether nuclear proteins diffused passively into the nucleus and accumulated there by binding to residents of the nucleus such as chromosomes, or were actively transported and accumulated regardless of their affinity for nuclear components. A classic experiment that addressed the problem used several forms of radioactive nucleoplasmin, which is a large pentameric protein involved in chromatin assembly. In this experiment, either the intact protein or the nucleoplasmin heads, tails, or heads with a single tail were injected into the cytoplasm or into the nucleus of frog oocytes (See Figure 1). All forms of nucleoplasmin, except heads, accumulated in the nucleus when injected in the cytoplasm, and all forms were retained in the nucleus when injected there. Figure 1: a) What portion of the nucleoplasmin molecule is responsible for localization in the nucleus? b) How do these experiments distinguish between active transport, in which a nuclear localization signal triggers transport by the nuclear pore complex, and passive diffusion, in which a binding site for a nuclear component allows accumulation in the nucleus?
7. Promoters are specific sequences of DNA that recruit transcription factors. a) Where is the position of the promoter relative to the transcriptional start site of a bacterial gene? b) Where is the position of the promoter relative to the transcriptional start site of a eukaryotic gene? c) Is the transcriptional start site also the start site for translation? Explain. 8. β-galactosidase is a bacterial protein that is often used to genetically label cells. When cells express β-galactosidase and are exposed to the proper substrate, a blue precipitate is deposited. In order to express this gene in eukaryotic cells, what portions of the gene structure (promoter, coding sequence, untranslated regions ) have to be modified, and which can be kept the same? Explain the significance of each component you mention. 9. A key step in the HIV lifecycle is the reverse transcription of the HIV RNA genome into DNA. This process is performed by reverse transcriptase (RT). a) Does the use of RNA as a genome violate the central dogma? Explain. b) Is RT normally found in eukaryotic cells? c) Why is AZT an effective drug in slowing down the replication of HIV?
d) In what way chemically is the use of AZT similar to the use of dideoxy nucleotides in DNA sequencing? 10. The virus that causes Hepatitis C (HCV) coopts human cellular machinery to replicate by first infecting cells of the liver. Like HIV, HCV has an RNA genome that needs to be replicated to produce new viral particles. HCV coopts the cellular machinery to replicate its genome by first translating its RNA into several different proteins. These proteins then generate an anti-sense copy of the original HCV genome, which is used as a template by additional viral proteins for the generation of additional sense copies of the HCV RNA genome. a) Based on your understanding of DNA and RNA synthesis, why does an antisense copy of the HCV RNA genome need to be made? b) What key cellular function does HCV coopt? c) If cells infected with HCV were treated with cycloheximide, a translation inhibitor, what would be the effect upon viral replication? Would this be a good treatment? d) If cells infected with HCV were treated with penicillin, a bacterial cell wall biosynthesis inhibitor, would this have any effect upon viral replication? Would this be a good treatment?
1. a) Mutated signal sequence Missing cellular protein machinery b) If the cell is mutated in the cellular secretory machinery than the LSA- GFP will be mistargeted. If the cell s own LSA protein is defective than the LSA-GFP should be secreted from the cell normally. c) Mutant 1 and 4 are defective in the secretory pathway Mutant 2 and 3 have a mutant LSA d) To make sure that the addition of the GFP tag did not interfere with the normal trafficking of LSA. e) Mutant 1- v-snare is mutated so now the vesicles containing LSA are targeted to the lysosome instead of the plasma membrane Mutant 2-lost its signal sequence so it no longer is targeted to the ER Mutant 3-mutated so now it is constantly retrieved to the ER, or lost its glycosylation and can no longer move to the Golgi Mutant 4-no longer interacts with coat protein or v-snare so it can t be packaged into vesicles for transport 2. Liposomes lack the protein components (i.e. channels) of living cells that permit the regulated exchange of macromolecules across the lipid bilayer. 3. 4. The protein would end up in the extracellular space. Normally, the protein would go from the ER to the Golgi apparatus, get captured because of its ER-retrieval signal, and return to the ER. However, without the ER-retrieval signal, the protein would evade capture, ultimately leave the Golgi via the default pathway, and become secreted into the extracellular space. The protein would not be retained anywhere else along the secretory pathway, as it presumably
has no signals to promote such localization since it normally resides in the ER lumen. 5. Strain A has protein accumulating in the ER, which means that this cell has a mutation that blocks transport from the ER to the Golgi apparatus. Strain B has secreted protein, and therefore is your wildtype control. Strain C has protein accumulating in the Golgi apparatus, and thus has a mutation that blocks exit of proteins from the Golgi apparatus. Strain D has protein accumulating in the cis-golgi network, and thus has a mutation that blocks the travel of proteins through the Golgi cisternae. 6. a) The portion of the nucleoplasmin responsible for localization in the nucleus must reside in the tail. The nucleoplasmin head does not localize to the nucleus when injected into the cytoplasm, and it is the only injected component that is missing a tail. b) These experiments suggest that the nucleoplasmin tail carries a nuclear localization signal and that the accumulation in the nucleus is not the result of passive diffusion. The observations involving complete nucleoplasmin or fragments that retain the tail do not distinguish between passive diffusion and active transport; they say only that the tail carries the important part of nucleoplasmin-be it a localization signal or a binding site. The key observations that argue against passive diffusion are the results with the nucleoplasmin heads. They do not diffuse into the cytoplasm when they are injected into the nucleus, nor into the nucleus when injected into the cytoplasm, suggesting that the heads are too large to pass through the nuclear pores. Since the more massive forms of nucleoplasmin with tails do pass through the nuclear pores, passive diffusion of nucleoplasmin is ruled out. 7. a) -10 and -35 sites (bases) upstream of the transcriptional start site. b) Most often upstream (more 5 ) from the transcriptional start site, however their position can be many 10-100 of kb away. c) No. The transcriptional start site is the start site for the transcription of RNA. The start site for translation is usually several bases 3 from the most 5 end of an RNA. 8. Leave the same: Coding sequence the genetic code between bacteria and eukaryotes has been conserved. Thus AUG in bacteria and eukaryotes bothe encode Methionine. Change: promoter different transcription factors are used in eukaryotes with different binding sites. Transcriptional termination different terminators are used polya signal required for proper eukaryotic targeting
9. HIV RT question a) No. Retroviruses may have caused a minor expansion in our understanding of the central dogma. However one still has to proceed from DNA to RNA to protein. You cannot skip any intermediate steps and move from protein directly to DNA or DNA directly to protein. Additionally proteins are still unable to selfreplicate and carry replicable information, that remains in the domain of the nucleic acids. b) No. c) AZT is a chain terminator and terminates the reverse transcription of DNA from HIV RNA. This is a function unique to retroviruses, and thus does not effect the rest of the host cells functions. d) Chain terminators are also used in DNA sequencing to generate a mix of DNA molecules that all have the same 5 start position, but are terminated at various positions with a nucleotide that can not have another nucleotide attached to it at the 3 position 10. HCV Question a) Proteins are usually only encoded on one strand of nucleic acid and has to be read in a specific direction. Thus if one were to replicate the HCV genome directly, one would produce a complementary copy of the genome that would encode for different proteins. The antisense RNA needs to be made so that the next generation of viruses is able to produce the proper proteins. b) Translation of the original RNA molecule into viral proteins c) Viral replication would be terminated at the first step. However this would also prevent normal translation and likely kill the cell. d) No. HCV infects eukaryotic cells without cell walls. Thus any cell wall inhibitor will have no effect upon HCV.