Final Exam 2008 Solutions

Final Exam 2008 Solutions - FINAL EXAM 2008 BME...

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Unformatted text preview: FINAL EXAM 2008 BME 339/CHE 339/BIO 335 1. Antibodies (15 points total) A. What is the hypervariable region of antibodies and what is its role in the recognition of antigens? (2 points) Part of the variable region that has a high ratio of different amino acids in a given position also called the complementarity determining region (CDR); the HV loops of the antibody create a surface that comes directly in contact with and is complementary chemically and geometrically to that of the antigen. B. Briefly describe the logic and the key steps in the generation of monoclonal antibodies via the hybridoma technology (6 points). i. Logic 1. Unfused myeloma cells cannot grow because they lack HGPRT 2. Unfused normal spleen cells cannot grow indefinitely because of their limited life span 3. Hybridoma cells are able to grow indefinitely b/c spleen cell partner supplies HGPRT and the myeloma partner is immortal ii. Key Steps 1. Ag immunization in mice 2. Isolate Ab producing B lymphocytes from spleen 3. Fuse B cells with myeloma cells to make B cells immortal 4. Culture in HAT medium; only hybrid cells survive 5. Test supernatants for those producing desired antibody 6. Isolate single cells from each antibody ­positive culture and subculture them 7. Test supernatants for desired antibodies; each positive subculture represents a clone and its antibodies are monoclonal (directed against a single determinant on the target antigen) 8. Scale up the size of the cultures of the successful clones C. What is the main problem with the use of murine monoclonal antibodies as a human therapeutics? What are chimeric antibodies (2 points)? They represent foreign proteins, so they elicit an immune response that results in the generation of human anti ­mouse antibodies, which can be detrimental Chimeric “humanized” antibody: mouse DNA encoding the binding portion of a monoclonal antibody merged with human antibody ­producing DNA in living cells, and the expression of this chimeric DNA through cell culture yielded half ­mouse, half ­human monoclonal antibody D. Describe how antibodies to a desired antigen can be isolated by phage display (5 points) 1. 2. 3. 2. Make combinatorial library Insert sequences into phagemid vector, which makes protein/antibody and displays it on outside Phagemid immobilizes on binding target sites that are complementary to the target protein/antibody 4. Wash away weak binders/antigens 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 Protein Engineering (6 points) A. Useful proteins with improved function can be engineered by fusing two genes encoding different proteins. How can two genes be fused using PCR? (2 points) Overlap extension PCR (describe): A PCR technique used to create one DNA fragment from two DNA fragments. It requires the 5’ region on one fragment to be complementary with the 3’ region of the second fragment. B. DNA shuffling or in vitro recombination followed by library screening is widely used to generate proteins with improved properties. What is it and what does it accomplish? A technique for generating genes that combine the mutations in a pool of parent genes 3. Vaccines (6 points) A. What is immunological memory and why is it important for protection against disease? First exposure to pathogen  few immune cells become programmed to recognize the pathogen and also persist in the body in a dormant state Re ­exposure to antigen  the memory cells become activated and proliferate rapidly to help mount a robust immune response quickly Allows for a quicker response to infection and eliminate the “adaption period” from adaptive immunity. B. What are the three mechanisms by which antibodies mediate protection from infectious agents (1.5 points) a. Binding and neutralization of key molecules produced by the pathogen b. Opsonization: antibodies that recognize pathogen surface molecules cover its surface and present it to phagocytic cells c. Complement activation: recruit innate immune proteins that kill pathogens by poking holes onto their membranes C. What is an adjuvant? (1 point) Material injected together with the antigen to stimulate a strong adaptive immune response 4. D. What is the main advantages and the two main disadvantages of subunit vaccines (1.5 points)? a. Advantages i. Better safety profile ii. Easier quality control b. Disadvantages i. Booster injections typically required ii. Injectable, oral delivery not an option iii. Elicit strong antibody response but typically not a CTL response Diagnostics (3 points) What is a molecular beacon and how does it work? Molecular beacons are hairpin DNAs that contain a dye (fluorophore & quencher) on each end. The two dyes are selected so that when they are close to each other the fluorescence energy emitted by the first (fluorophore) is absorbed/quenched by the second (quencher). When the hairpin hybridizes to the target DNA the molecular beacon stretches out such that the two dyes are no longer proximal to each other. This allows the fluorophore to fluoresce, meaning that the molecular beacon has found its complementary target DNA sequence. 6. Bioseparations (7 points). You are assigned to purify a 150 KDa intracellular protein produced in Bacillus subtilis, a Gram+ve bacterium with a thick wall. A. Describe two techniques you could use to break the cells without causing denaturation of the protein High speed ball mixer, high pressure homogenizer, enzymatic lysis, ultrasound/sonications B. Next you select to enrich the protein from the cell debris (DNA cell wall fragments etc) by fractional ammonium sulfate precipitation. How does it work? (3 points) Reduce the electrostatic repulsion between identical molecules by increasing the ionic strength and the forces holding the solvation shell around protein molecules, resulting in “salting out” or precipitation of different proteins based on their solubility at different AS concentration level C. After (b) you have elected to use ion exchange chromatography and gel filtration chromatography. What does each of these steps accomplish and what order would you use them in? (2 points) (d) anion exchange chromatography  ­ Protein purification technique using charge ­charge interaction between the protein in the sample and in the immobilized resin (1 point)  ­ Bound negative charged protein to the positive charged immobilized resin and selective elution using different salt concentration solution (1 point) (e) gel filtration column  ­ Separation based on the size (or hydrodynamic volume) of molecule (1 point)  ­ the smaller molecules get temporarily stuck in the porous resins, while the larger ones pass right by them (1 point) Ionic exchange chromatography to initially purify the protein based on charge gel filtration for final buffer exchange and polishing to purify the protein based on size 7. Growth Kinetics (from Exam I) (6 points) A. Your boss asked you to determine the µmax and Ks values for the organism Longhornious pestis which was recently discovered at the Royal Memorial stadium. You have in your disposal shake flasks, growth media (you wish to grow the bacteria with glucose as the limiting nutrient) and standard microbiological equipment (but no continuous stirred tank fermenetr. Describe what experiments you would need to do, the data you will have to obtain and how you will plot the data to calculate the parameters you need (4 points). 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. Briefly explain how you would calculate the mmax and Ks values if you had a set of X and S concentrations as a function of D from a chemostat experiment (2 points). Assuming Monod growth kinetics & chemostat with sterile feed: != C. !!"# ! !"#$%$&'$( !!"#$% ! !! !!  ! = !! !!"# ∗! + ! !!"# ! !!!"# !! ! ! = !! ! !!"# ! + ! !!"# Plot 1/D vs. 1/S ! − !"#$%&$'# = !!"#  !! !"#$% =  !!"# 1 ! − !"#$%&$'# = − !! BONUS QUESTION (2 points) How would you calculate the maintenance and true yield coefficient from the data in B above? Measure cell mass out (X) and Sout (S) during exponential phase and use X=YX/S(Sin ­S) to get YX/S ! ! ! =+ !!/! ! Then plot !!"#$ ! !!/! vs. 1/! 1 !!"#$ !"#$% = ! = !"#$%&$"$'& !"#$$%!%#&' ! − !"#$%&$'# = 8. Sterilization (3 points) Explain using an appropriate graph why sterilization at high temperatures for a short time can result in lower destruction of nutrients (relative to sterilization at a lower temperature for a longer time). As the temperature increases, the reaction rate, k, increases more rapidly for the reaction with the higher activation energy. Considering the difference between the two activation energies of spore destruction versus substrate destruction, an increase in temperature would accelerate spore destruction more than nutrient destruction. 9. Cloning/Expression (8 points) The gene fraternity is thought to be responsible for a disorder characterized by complete lapse of mental function in young adult males, particularly following moderate alcohol consumption. (a) The fraternity gene is expected to contain at least two introns. Describe how you would make a suitable library that will allow you to isolate the protein coding sequence of the gene (4 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 wish to clone the gene but the world’s supply of the FRATERNITY protein is less than 10 ng and therefore you cannot make antibodies for probing DNA libraries expressed in bacteria. Suggest a process for cloning the gene (no need to be verbose) (4 points). Need to determine the N ­terminal sequence, construct degenerate oligos and probe the library 1. Determine first few aa of N ­terminus (N ­terminal sequencing by Sanger + Edman degradation) using some of the 10 ng of protein. [1 pt] 2. Make degenerate oligos: determine all possible codons for the aa sequence from (1) and synthesize with radioactive label. [2 pt] 3. Make a cDNA library (since it is a HUMAN GENE). [2 pt] 4. Plate library, make a replica plate on nylon membrane or nitrocellulose filter and incubate with NaOH to denature DNA. [1 pt] 5. Add probe from step 2, wash, and identify colonies that hybridized (using autoradiography). Go back to master plate to find your colony. [2 pt] 6. Grow original colony and isolate plasmid. Use PCR to clone the fraternity gene from isolated plasmid. [2 pt] Note: in this problem you don’t have the gene. 10. Protein Secretion (3 points) A. Give three reasons why is secretion from the cell important for protein production?  ­ Easier purification  ­ Proper folding (disulfide bonds)  ­ Post ­translational modifications  ­ Correct N ­terminal residue B. What is a signal (or leader) peptide (1 point) A segment of about 15 ­30 amino acids at the N terminus of a protein that enables the protein to be secreted; prevents degradation C. What is disulfide bond isomerization? (0.5 points) Disulfide isomerase catalyzes the isomerization of incorrect disulfide bonds and to a lesser extent also oxidation of cysteines amino acids (that contain SH groups) 11. Miscl. Define in a few words (4 points, one point each) A. Crabtree effect Substrate inhibition on growth; occurs because cell cannot cope with excess nutrients and nutrient may shut off part of a metabolic process slowing the growth B. Polyhydrin promoter Strong promoter in insect cells C. Gene chip mRNA detection D. Dideoxynucleotide (think Sanger sequencing) DNA chain termination ...
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This note was uploaded on 12/13/2011 for the course BME 339 taught by Professor Georgiou during the Spring '11 term at University of Texas at Austin.

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