FINAL EXAM 2009 Solutions

FINAL EXAM 2009 Solutions - FINAL EXAM 2009 BME...

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Unformatted text preview: FINAL EXAM 2009 BME 339/CHE 339/BIO 335 SECTION 1: Short Answer questions 1. Bioreactors and animal cells (8 points) A. Explain using the appropriate questions how you would carry out an experiment to determine the Monod parameters mmax and Ks from experiments where you grow cells in shake flasks. Grow the bacteria in shake flasks with varying initial levels of substrate. Measure the biomass concentration at early time points and use the Monod model, dX/dt=μX, to get μ vs. S data. (The lag phase can be neglected if the bacteria are subcultured first). By taking the inverse of both sides of the other Monod equation, μ=μmaxS/(Ks+S), you get an eqation in y=mx+b form so you can plot 1/μ vs. 1/S  slope = Ks/μmax and y ­intercept = 1/μmax. B. Similarly explain how you can determine the Monod parameters µmax and Ks using a chemostat Use chemostat to grow bacteria at various inlet substrate levels (Sin) and dilution rates (D) Measure Sout (S) at exponential phase After obtaining the various data points [substrate levels (S), dilution rates (D)], plot their inverse values from the Monod Kinetics model to fit the Lineweaver Burk equation 1/D = Ks/umax * 1/S + 1/umax Y ­intercept gives 1/umax and Slope gives Ks/umax  ­ ­ ­> Ks Measure cell mass out (X) and Sout (S) during exponential phase and use X=YX/S(Sin ­S) to get YX/S C. Why are animal cells grown in incubators or reactors with 5% CO2 in the air? High levels of CO2 form carbonate/carbonic acid in the media which acts to buffer the media to a pH about 7.1 ­7.3. This is used in place of a high osmolarity of chemical buffers which the animal cells can’t tolerate. 2. Oxygen Transfer (7points total) A. What is the critical dissolved oxygen concentration? (1 point). The minimum O2 concentration needed to maintain ! ≥ 95%!!"# . B. How does the Oxygen Transfer Rate (OTR) depend on the interfacial area and the driving force for transfer? Write the equation and explain the meaning of ALL the terms (2 points). OTR = kL a (CO2*  ­ CO2) 1) a = interfacial area: larger area results in higher OTR 2) kL = driving force: mass transfer coefficient. 3) (CO2*  ­ CO2) = driving force (diff. btw. saturated O2 and actual O2 concentration): lower CO2 results in higher OTR C. At steady state OTR=OUR. What are three ways to make sure that the dissolved oxygen concentration remains constant as the biomass increases during the course of the fermentation? a. Increase !! ! b. Increase pressure of O2 c. Use fermenters with suitable design that gives rise to a higher !! ! d. Growth rate control (reduce ! so that oxygen demand is decreased) by slow addition D. Mammalian cells do not have cell walls and therefore they are particularly susceptible to damage in fermenters. How can cell damage be prevented when the cells are grown in high cell density bioreactors (1 point). Use a bubble/air ­lift reactor for mixing; immobilize the cells on polymer beads, membrane, axial impellers; addition of pluronic F ­68 surfactants into the medium in 3. Bioseparations (5 points) A. Many important proteins are isolated from human serum. The first step is usually salt fractionation (sometimes called fractionational precipitation). Explain in 2 ­3 lines what this is. (3 points) A method of separating proteins based on the principle that proteins are less soluble at high salt concentrations. The salt concentration needed for the protein to precipitate out of the solution differs from protein to protein. With knowledge of the solubility of each protein, separation via this technique is possible. B The next step is isoelectric point precipitation. Describe what this is in one sentence (1 point) The isoelectric point (pI) is the pH of a solution at which the net primary charge of a protein becomes zero, charges cancel, repulsive electrostatic forces are reduced, and dispersive forces predominate, resulting in aggregation & precipitation B. Immunoaffinity chromatography is then use to further separate serum proteins that are present in low concentrations such as the anti ­clotting protein Factor X. Explain in one sentence how this technique works (1 point). Solute containing of proteins pass by the immobilized antibodies on a stationary phase, which bind specifically to the anti ­clotting protein Factor X. 4. Cloning/Expression (14 points total): A. The human SIRT1 protein plays a critical role for aging. You know that the gene contains several introns and is about 5 kb in size. Fortunately the SIRT1 protein has been purified from cows in large amounts and you suspect that the cow and human proteins are very similar. Describe how you can clone the coding sequence of the SIRT1 gene (5 points). (1) Isolate cells from tissue of interest, lyse cells and purify mRNA, hybridize entire mRNA with oligo(dT) ­primer (2) Add reverse transcriptase and make DNA copy (complementary DNA strand) (3) Degrade the mRNA strand with alkali treatment, The 3’end of the single strand DNA forms a loop to insert and so these “double strandness” can be used as the primer. (4) DNA polymerase is added to elongate this complementary strand (5) Use Sl nuclease (which acts only on the single stranded loop) to cut this hairpin loop. Cut the double stranded cDNA with restriction enzymes, ligate into a vector and introduce recombinant DNA into host cells. B. You now try to express the protein in bacteria. You need to make as much protein as possible and moreover it is of course important that your produce active protein. You suspect translation of the SIRT1 gene might be a problem. What could be the reason(s) for that and how could this problem be fixed (3 points)  ­ mRNA stem ­loop structure; silent mutation of gene to prevent stem ­loop  ­ protein degradation; use protease ­deficient cell strains  ­ insoluble proteins/misfolding; use fusion proteins or chaperones C. ALSO you have information that the protein might be toxic and as soon it is produced the cell can no longer divide although it can still make protein. What do you need to do to get a high level of protein production (1 point). Use inducible promoter, allowing the cell to divide and not constantly produce toxic recombinant proteins D. Unfortunately this is really a difficult protein to work with. You managed to solve the problems in B and C above but now you find that the protein aggregates. Why would that occur? Suggest two approaches for producing soluble SIRT1 protein (3 points). Because partially folded intermediates form during the process of folding. Such intermediates are often sticky, have a propensity to aggregate and don’t have enough time to reach the correct conformation. 1. Lyse the cells 2. Isolate inclusion bodies (centrifugation, microfiltration) 3. Add urea 8M or GuHCl to carry out unfolding 4. Allow refolding under renaturing conditions E. Why are insect cells are used for protein expression and what is the significance of the Baculovirus polyhydrin gene in that regard (2 points)  ­ The promoter for the polyhedrin gene is exceptionally strong and it is not required for virus production. Replacement of the polyhedrin gene with a coding sequence for your heterologous protein, followed by infection of cultured insect cells, would result in large amounts of the heterologous protein.  ­ Also, because of the similarity of posttranslational modification systems between insects and mammals, the recombinant protein may mimic closely, if not precisely, the authentic form of the original protein. 5. DNA Sequencing and Diagnostics (10): A. Describe succinctly how 454 sequencing works. For clarity you can just list the key steps and what is accomplished in each. Please do not exceed the space allocated (8 points).  ­ Emulsion PCR  ­ Sequencing ­by ­synthesis  ­ Chemiluminescent enzymes: sulfurylase and luciferase  ­ Detect light signals for sequence readouts  ­ Single stranded template DNA immobilized beads are deposited onto the PicoTiterPlate™ device.  ­ Addition of nucleotides by nucleotide flow results in a reaction that generates a light signal that is recorded by the CCD camera. A. The easiest way to detect whether an individual has the mutant gene X is to use PCR/oligonucleotide ligation (PCR ­OLA). How does this work (2 points)? 6. Vaccines (6 points) A. What is the difference between innate and adaptive immunity? What is the role of adjuvants in vaccines? (3 points). Innate immunity – recognize patterns that occur only in pathogens (foreign, non ­human cells) causing inflammation and production of molecules that kill bacterial pathogens and virally infected cells Adaptive immunity – production of specialized antibodies and cytotoxic T lymphocytes which recognize pathogens with high specificity and mediate more potent cell killing Adjuvants  ­ material injected with antigen to stimulate a strong adaptive immune response B. What are attenuated live vaccines and what is their major drawback (2 points). Live pathogenic agents that do not cause disease b/c it’s been mutated to a non ­pathogenic form Major drawback is that possibility of recombination/reversion mutation can result in active pathogen C. What are Cytotoxic T lymphocytes (CTLs) and why are they important in protecting us from disease (1 point). Immune cells programmed to recognize surface markers on diseased human cells and to then destroy them They are important for destruction of infected human cells and prevention of spreading of infection 7. Protein Engineering/Antibodies (15 points) A. What is DNA shuffling? BRIEFLY (in 2 ­3 sentences) describe how it works (4 points) A technique for generating genes that combine the mutations in a pool of parent genes 1. DNAse I fragments a set of parent genes 2. PCR w/o primers (instead overlapping fragments anneal to each other and are then extended by DNA polymerase) until extended to the size of parental genes 3. PCR w/ primers to complement ends of strands B. Proteins that bind a desired target molecule are routinely isolated from large libraries of protein variants (for example libraries generated by random mutagenesis) and are then screened by phage display. Explain in the SPACE below how that works (5 points) 1. Make combinatorial library 2. Insert sequences into phagemid vector, which makes protein and displays it on outside 3. Phagemid immobilizes on binding target sites that are complementary to the target protein 4. Wash away weak binders 5. Recover tight binders to infect E. coli amplification (to enrich library with tight binders) 6. Enrich tight binding library with another round of washing D. Draw a diagram describing how ELISA assays work (2 points). E. What is the difference between chimeric, humanized and fully human antibodies and why these differences are significant for therapeutic purposes? (2 points). Chimeric – entire VH and VL domains from the mouse antibody are transplanted into a human antibody; more immunogenic than humanized because more murine parts Humanized – loop sequences from a mouse antibody can be transplanted into a human antibody without loss of affinity for the antigen; slightly immunogenic Fully human – no part of the antibody is murine (entirely non ­immunogenic) F. What is the Fc domain of an antibody and why is it important? (2 points)  ­tail region of an antibody that interacts with cell surface receptors called Fc receptors and some proteins of the complement system  ­ binds to various cell receptors and complement proteins and mediates different physiological effects of antibodies  ­ allows antibodies to activate the immune system Name: ___________________ EID:___________________ SECTION II Multiple choice (15 points, 1 point per question) 1. Sometimes proteins are separated by liquid ­liquid extraction. This means that: A. First you add one organic solvent to separate the protein and then a second solvent. B. You mix the protein with water miscible organic solvents such as butanol. C. You generate an aqueous 2 ­phase solution by adding a polymer like PEG and a second component which could be either a salt or another polymer. D. You simply add starch to the protein solution and the starch separates from the proteins. 2. Which method cannot be used to break yeast. A. A blender usually containing steel beads for more impact. B. Lipase enzymes that hydrolyze the cell membrane. C. A sonicator that generates ultrasound. D. High pressure homogenizer E. None of the above; all these methods will work just fine. 3. The Crabtree effect is: A. Formation of products such as antibiotics in stationary phase. B. A way for the cell to get rid of excess NADH by producing ethanol or acetate. C. The consumption of substrate to needed to maintain the cell viable. D. Describes the effect of salts on protein solubility. 4. A higher T during sterilization is preferable because: A. Contaminants are destroyed faster. B. The activation energy for the destruction of nutrients is higher than that of cells. C. There is less destruction of nutrients. D. All of the above. 5. Which of the following methods can be used to determine the amount of a specific mRNA (e.g. interleukin receptor 2) in cells: A. Northern blotting. B. Gene chips (DNA microarrays). C. High throughput sequencing of cDNA D. All of the above (A, B and C). E. Southern blotting 6. To recover correctly folded proteins from inclusion bodies it is necessary to: A. Denature the protein using urea and then remove the denaturant by dilution to allow it to fold into its native conformation. B. Denature the protein and then separate the folded protein by chromatography. C. Incubate the inclusion bodies with chaperones and ATP . D. Use a detergent to dissolve the aggregate E. All of the above. 7. What are the essential parts of a plasmid cloning vector? A. An origin of replication, a gene that allows selection of cells transformed with the plasmid and convenient restriction sites. B. An origin of replication, a gene that allows selection of cells transformed with the plasmid and a promoter followed by a ribosome binding site. C. A means for packaging the DNA into a virus. D. An origin of replication for E.coli, a gene that allows selection of cells transformed with the plasmid and second origin of replication for a different host cell such as yeast. 8. Which part of plant biomass is most difficult to degrade biologically? A. Cellulose B. Hemicellulose C. Lignin D. Starch E. Cell wall glucans 10. “Humanization” is: A. A process for isolating human stem cells B. The engineering of proteins and, in particular, antibodies to remove sequences that can elicit an immune response when injected into a human C. The generation of hybrid cells (cell fusions) between human B cells and mouse myeloma cells D. None of the above 11. In the hybridoma technology: A. B cells are grown in bioreactors to produce antibodies B. B cells are immortalized by infecting with viruses C. B cells are immortalized i.e. are made to grow for many generations by fusion to cancer (myeloma) cells D. T cells are used to stimulate antibody production in the test tube 12. A molecular beacon is: A. An antibody that is conjugated to an enzyme and used for protein detection. B. A DNA molecule that is used for the fluorescent detection of a target DNA C. A DNA molecule that is used in the OLA assay to detect single nucleotide mutations in DNA. D. The process by which light is emitted during the incorporation of bases in high throughput DNA sequencing. E. An antibody labeled with a radionucleotide that is used to detect tumors in the body. 13. Subunit vaccines are: A. Safer than attenuated vaccines B. Consist of immunogenic components from a pathogen. C. Have a long self ­life D. Usually have to be used with an adjuvant E. All of the above. F. A and D only. 14. Very large genes (> 15 Kb) for genome engineering can be constructed by: A. Gene synthesis machines B. Cloning and ligating smaller fragments C. Overlap PCR of 0.5 to 1 kb fragments D. By C above and by taking advantage of recombination in yeast. E. All of the above. F. A, C and D 15. Yeast artificial chromosomes (YACs): A. Are used routinely for cloning in yeast B. Contain the 2 mm origin of replication and a selectable marker C. Contain a yeast kinetochore region. D. Are used for cloning very large pieces of DNA E. Are not stably maintained in yeast cells and therefore progeny cells not containing YACs arise at a high frequency. ...
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