Life Sciences 1A Midterm Exam 2. November 13, 2006

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1 Name: TF: Section Time Life Sciences 1A Midterm Exam 2 November 13, 2006 Please write legibly in the space provided below each question. You may not use calculators on this exam. We prefer that you use non-erasable pen when writing your answers. A significant number of completed exams will be photocopied by the teaching staff. Please write your name on each page of the exam. There are SIX multi-part questions in this exam. We recommend that you first read through all questions and begin with the questions that are easiest for you. Be sure to take a look at all questions before the end of the exam! Please write your TF s name and section time on the front page only. Question 1 /20 Question 2 /32 Question 3 /32 Question 4 /24 Question 5 /32 Question 6 /10 Total /150

2 1. The two graphs shown below depict membrane fluidity from two different membranes (a and b), neither of which have cholesterol. a. (4 points) Which graph above (a or b) corresponds to the more fluid membrane? Explain. A. The Tm is lower for organism A, meaning at any given temperature the membrane is more fluid. b. (6 points) List two changes in the structure of lipids that could lead to changes in membrane fluidity. In each case explain how that change would either increase or decrease membrane fluidity. 1. Change fatty acid chain length. Shorter chains form fewer Van der Waals interactions with neighboring chains, therefore making the membrane more fluid, conversely longer chains decrease fluidity because more VdW interactions would form. 2. Change the saturation of the fatty acids in the phospholipids. More unsaturated fatty acids (meaning more double bonds) do not pack well because each cis-double bond introduces a kink in the chain preventing the close apposition of adjacent phospholipid molecules. This decreases the Van der Waals interactions, causing the membrane to be more fluid. Conversely, more saturated fatty acid tails allow closer apposition of adjacent phospholipids, increased VdW interactions, and cause a decrease in fluidity. 1

3 c. (4 points) Draw the graph for fluidity if cholesterol was included in membrane a. Include the curve for without cholesterol on your graph for comparison. Without cholesterol With cholesterol d. (6 points) Explain how cholesterol affects the membrane fluidity at high and low temperatures. At high temperatures, cholesterol lowers membrane fluidity because the rigid 4-ring structure that interacts with the first few CH2 groups of the fatty acid tails helps keep the phospholipids in a more rigid alignment, causing less motion and therefore less fluidity. At low temperatures, cholesterol prevents the same phospholipids from packing as tightly as they normally could without cholesterol, making the membrane more fluid. 2

4 2. Erythropoietin (EPO) is a secreted human hormone that stimulates maturation of red blood cells. The human gene for EPO can be expressed in cells to produce large quantities of the protein for use as a drug to treat anemia. a. (4 points) Mature EPO protein is 165 amino acids long, however the EPO mrna (fully processed) is 1340 nucleotides. Provide two possible explanations for this discrepancy in length. Post-translational modification: Translated proteins can be modified after translation. One approach is proteolytic processing where a portion of the freshly translated nascent EPO protein is cleaved into a smaller protein. Only one of the peptide fragments is mature EPO, and the other fragment will be completely degraded. 5 and 3 untranslated regions in the mrna: the entire sequence of mature mrnas is rarely translated. They often possess 5 and 3 untranslated regions bracketing the coding region. These domains can be used by the cell to regulate the levels of translation and the stability of the mrna. b. (8 points) The gene for EPO can be expressed in cells to produce large quantities of protein for use as a drug to treat anemia. Scientists attempt to express the gene in E. coli (bacteria) by inserting the entire human EPO gene into bacteria. No EPO protein is produced at all. Provide two possible explanations. Incompatible promoters between eukaryotes and prokaryotes. Eukaryotic promoters and prokaryotic promoters are recognized by different transcription factors. When a eukaryotic gene is inserted into bacteria, the bacterial sigma factors are unable to recognize the eukaryotic promoter elements of EPO and thus cannot activate transcription. Presence of introns in eukaryotic genes that cannot be spliced out in prokaryotes. Even if EPO is transcribed, it will have introns in the RNA that is produced. Bacteria do not have the machinery necessary to remove the introns and thus a garbage EPO will be made and probably destroyed. Also plausible: insertion into the bacterial chromosome in a nontranscribable region. See answer to part d. 3

5 Shown below is the sequence of a gene from E. coli AGTCGTCCGTACGTTTGACAGAGTCATCCCATAAGCGTATATTATCGGCGGTGCAGTTACCC 3 -TCAGCAGGCATGCAAACTGTCTCAGTAGGGTATTCGCATATAATAGCCGCCACGTCAATGGG 4 CGTTCAACGCCGGTGGAATCGTGGCCGCCAGTCCATCTGGCGGCGTTTTCCAGATCGCATT-3 GCAAGTTGCGGCCACCTTAGCACCGGCGGTCAGGTAGACCGCCGCAAAAGGTCTAGCGTAA-5 c. (2 points) Which strand (top or bottom) is the coding (non-template) strand. top d. (6 points) At which location (1, 2, 3, or 4) would you insert the EPO protein coding sequence and expect EPO RNA and protein to be produced? Explain. Region 3. Downstream of promoters, upstream of transcriptional terminator locations. Bacterial sigma factor will bind at region 2 and based on the orientation of the sequence, will recruit RNA pol to transcribe the sequence further to the right (regions 3 and 4). Region 4 is after a transcriptional termination motif, so RNA pol is likely to cease transcribing prior to reaching region 4. Thus region 3 is the only region that will be transcribed. e. (6 points) When the EPO protein coding sequence is placed in the proper location and expressed in E. coli, EPO protein is translated, but not secreted. Moreover the purified protein is unable to stimulate red blood cell maturation. Based on your understanding of the differences in prokaryotic and eukaryotic biology, provide a rational explanation for these observations. EPO is normally secreted through the eukaryotic secretory pathway. Secreted proteins are often post-translationally modified (glycosylated, proteolytically cleaved, etc.) as they travel through the ER and Golgi, and these modifications can contribute to their ability to function. Such modifications couldn t occur to EPO expressed in bacteria because they lack a secretory pathway and would not necessarily be able to modify EPO in an identical fashion. 4

6 f. (6 points) Shown below is a portion of the EPO mrna. The start codon is located in this segment. 5 ggucgcugagggaccccggccaggcgcggagaugggggugcacga auguccugccuggcuguggcuucuccugucccugcugucgcuccc 3 Circle the start codon and translate the first five amino acids (a table containing the genetic code is provided at the end of the exam). First codon marked in red above. N MGVHEC C 5

7 3. The Reverse Transcriptase Assay is a method researchers use to quantify the amount of HIV. The enzymatic activity of Reverse Transcriptase (RT) is measured by using a reaction mixture containing the following components: Components 1) Virus sample (may or may not contain virus) 2) poly-riboadenosine (usually around nucleotides in length) 3) mild detergent 4) Buffer, ph 7 5) radioactive deoxythymidine triphosphate (dttp) a. (6 points) A mild detergent must be added to release the reverse transcriptase from the viral particles. Explain why detergent is necessary. HIV is surrounded by a lipid membrane, and the RT is enclosed within it. The detergent has to be added to disrupt the membrane and allow the other reagents to access the RT. b. (6 points) You find that the nucleic acid product of the reaction is single stranded. Is this surprising in light of the activities of RT and the reaction mixture? No. RT has two activities, the reverse transcription of RNA into DNA and the degradation of RNA. Based on the reagents available, polya template and dttp, the only DNA that can be synthesized is polydt. Concurrently the RNA template is degraded leaving the polydt behind. The complementary strand of DNA cannot be synthesized because no datp is present in the reaction mixture. c. (6 points) If a cellular polymerase was added as component 1 and components 2-5 are left unchanged, could this assay be used to measure its enzymatic activity? Explain. No. Cellular polymerases can only use DNA as templates. DNA pol synthesizes complementary DNA to DNA templates and RNA pol synthesizes complementary RNA to DNA templates. Component 2 is an RNA template, and thus it is impossible for any cellular polymerase to use it to make any additional polynucleotides. 6

8 d. (6 points) At the end of the experiment, the nucleic acid product is captured on a positively charged membrane. The amount of radioactivity on this membrane is proportional to the number of viral particles in the unknown sample. Draw in the expected result on the bar graph below if the following components are added as components 1: HIV, or HIV and AZT. e. (8 points) Draw in the expected results if components 2 and 5, polyriboadenosine and dttp, are replaced with poly-riboguanosine and dctp in the experiment, and the following components are used as component 1: HIV, or HIV and AZT. Explain your results. Without AZT same height as HIV column. With AZT, also same height. AZT is a T analog and will only be incorporated where T s are necessary. With the new template, C s are always inserted and AZT has no effect. 7

9 4. You decide to study the transport of glucose (structure drawn below) into red blood cells and intestinal cells. HO HO O OH OH OH You incubate the cells in a solution of 1M glucose in the presence and absence of an ATPase inhibitor. After 30 minutes, the concentration of glucose in the solution outside the cells is determined. Concentration of Glucose In Solution Outside Cells Red blood cells Intestinal cells No ATPase inhibitor 0.5M 0.1M ATPase inhibitor 0.5M 1M a. (6 points) What kind of transport do red blood cells use to import glucose? Are proteins necessary for this process? Explain. Passive transport presence or absence of ATP has no effect on transport. Proteins are necessary because glucose is large and polar and would not pass through the lipid bilayer without a carrier protein. b. (6 points) What kind of transport do intestinal cells use to import glucose? Are proteins necessary for this process? Explain. Active transport-the intestinal cells require energy in the form of ATP to remove glucose from the media as in the presence of ATPase inhibitors no glucose uptake is observed. Proteins are required to mediate the transport of the polar glucose molecule and to utilize the energy of ATP hydrolysis. 8

10 c. (6 points) The red blood cells and intestinal cells incubation in 1M glucose in the absence of ATPase inhibitors is continued. After 60 minutes, the concentration of glucose in the solution outside the cells is unchanged from 30 minutes (0.5M for red blood cells and 0.1M for intestinal cells). For each cell type, are the cells in equilibrium or steady state? Explain. Red blood cells are in equilibrium. The concentration of glucose inside and outside of the RBCs is constant and equal, and combined with the information from the table that this occurs in the presence or absence of ATPase inhibitors indicates that it is a process that does not require the input of energy. Intestinal cells are in steady state. The cellular concentration of glucose is maintained at a constant concentration, however this concentration is not equal to that of glucose outside of the cell. Moreover the table indicates that the establishment of this concentration requires the input of energy in the form of ATP. Both qualities that are consistent with steady state. d. (6 points) Would you expect the transport of O 2 to be affected by ATPase inhibitors in either red blood cells or intestinal cells? Explain. No, O 2 is a small, non-polar molecule, which can diffuse freely through the lipid bilayer. 9

11 5. The Vx protein is essential for the completion of the life cycle of a newly discovered retrovirus. To gain insight into its function, recombinant virus expressing the GFP fusion protein GFP-Vx instead of wildtype Vx is created. a. (8 points) A 902 bp segment of DNA encoding GFP is placed at the N- terminus of Vx to create a GFP-Vx fusion protein. Strong GFP fluorescence is observed in infected cells. The protein produced has the predicted molecular weight, however the amino acid sequence at the C- terminus does not match the sequence of wildtype Vx. What is the most likely explanation? Explain. Insertion of a 902 bp sequence into the open reading frame would result in a frameshift mutation, since 902 is not divisible by 3. Therefore, GFP and Vx in the fusion protein are not in the same reading frame, and so only functional GFP is produced. b. (8 points) Properly constructed GFP-Vx functions normally and results in GFP-Vx fluorescence primarily in the nucleus of infected cells. However, mutant GFP-Vx missing a stretch of 5 consecutive amino acids from the Vx portion of the protein is present only in the cytoplasm. What is the most likely role for this stretch of amino acids? Explain. This sequence most likely acts as a signal sequence which is recognized by some nuclear transport protein. This results in the translocation of GFP-Vx to the nucleus. Mutation or deletion of the sequence prevents nuclear translocation, so GFP-Vx remains in the cytoplasm. 10

12 c. (8 points) To test your hypothesis from part b, you attach the five amino acid sequence to GFP. When this form of GFP (5aa-GFP) is expressed in cells, GFP fluorescence is observed exclusively in the cytoplasm. When GFP alone is expressed it is also observed only in the cytoplasm. What is one possible explanation for this observation? Explain. This sequence is likely to be part of a signal patch sequence, perhaps with other regions of Vx. This patch is only made when Vx is allowed to assume its normal conformation. Also stating and explaining how the domain is necessary, but not sufficient is correct for this question. Or the domain is not necessary. d. (8 points) Based on the data provided, for nuclear localization is the 5 amino acid sequence necessary, sufficient, neither or both? Explain. This sequence is necessary, but not sufficient, for nuclear localization. If deleted, nuclear import is prevented (necessary), but addition of the sequence to another protein does not result in nuclear translocation (not sufficient). 11

13 6. You can successfully translate mature ovalbumin mrna using all the necessary components for translation in vitro. To further study translation you vary the conditions and make the following observations. a. (5 points) An ovalbumin RNA missing the 5 -cap is translated at a significantly lower level compared to the same ovalbumin RNA with a 5 - cap. Explain this observation. Since the 5 cap is recognized by translation initiation factors that recruit the small ribosomal subunit, the lack of the cap will prevent (or at least strongly impair) translation initiation. b. (5 points) When a chemically modified GTP that cannot be hydrolyzed is added, there is a significant decrease in the amount of ovalbumin protein translated. Explain this observation. A nonhydrolyzable GTP analog will prevent the eukaryotic equivalents of EF-Tu and EF-G from performing their roles in translation. In the case of the former, charged aminoacyl trnas will not be able to donate their amino acids to the end of the growing polypeptide chain since EF-TuGTP will not release them because they cannot hydrolyze to the EF-TuGDP form that is capable of releasing charged trnas. In the case of the latter, EF-GGTP will not contribute to translocation of the ribosome along the mrna, a process that also requires hydrolysis of the bound GTP. (full credit for either explanation) 12

14 The Genetic Code 13