BIOC301 Final Exam Prep Questions.docx

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Department
Biochemistry
Course Code
BIOC 301
Professor
Jason Read

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BIOC 301 Final Exam Prep Questions Protein Crystallography: 1 Why is determining protein structure important? a Structure → function (e.g. active sites, docking sites, signaling protein, how mutations affect function). b Used for drug discovery and development. 2 How can protein structure be determined? a X-ray crystallography (reconstructing shape of crystal from x-ray diffraction patterns. 3 What kind of bonds hold protein crystals together? How strong are these bonds? a non-covalent bonds that are quite weak. 4 How are protein crystals similar or different from mineral crystals? a Similar: lattice formation. b Different: contain solvent molecules (e.g. water), non-covalent bonds, more fragile, different crystals from the same protein. 5 What are the reagents and conditions necessary to make a crystal? a Solution of protein with sufficiently high concentration and purity b Appropriate solvent (aqueous buffer) c Choice of precipitant 6 What is vapour diffusion? a Solvent (water) slowly leaving a small sample (e.g. hanging drop) with low concentration and entering a reservoir with higher salt concentration. System must be sealed and isolated from external environment. Results in precipitation within the small sample (increase concentration). Basis for hanging drop method. 7 What are the biological roles of lysozyme? Where is it present? a enzyme hydrolysing beta-1,4-glycosidic linkage between N-acetylmuramic acid and N- acetylglucosamine in cell walls of bacteria. b Natural antibiotic and preservative. c Found in egg whites, mucus, tears, etc. 8 What is pI? a Isoelectric point = pH at which molecules carry no net charge. pH < pI = net positive charge. pH > pI = net negative charge. b pI = (pKa + pKb)/2 (for a 1 residue charged amino acid). DNA Techniques: 1 What are two ways to sequence DNA? What are they used for? Their pros and cons? a Maxam and Gilbert 1 DNA templates are radioactively labeled. 2 DNA is chemically cut at select locations, resulting in DNA strands of different length 3 Because we know what base was at the cut locations, it’s possible to work backwards and deduce the sequence. e.g. DNA footprinting: proteins (natural or added) on DNA strands protect the area of endonucleases. b Sanger 1 Double stranded DNA is put into 4 test tubes. 2 Each test tube contains regular 3 types of dNTP’s and 1 type of ddNTP (dideoxyNTP - no -OH on 2 or 3’). 3 ddNTP’s are randomly incorporated in DNA, halting replication. 4 Result in strands of various length. 5 Result can be run on a gel (or through a capillary tube). Because we know what base was the terminating one in the chain, it’s possible to work backwards to deduce the sequence. 2 How does Gel Electrophoresis separate DNA? a For 10 - 50,000 bp of double stranded linear DNA. b Agarose polymerize and form gel with holes DNA could migrate through. c Bigger DNA and higher [agarose], the longer it takes. Rate∝1/(log10(#bp)) d DNA migrate from cathode to anode. Protein Techniques: 1 How does a the Bradford Assay work? a Based on (concentration dependent) absorbance shift of dye to 595 nm. b Red/Green form = unbound, cationic. Blue form = bound, anionic. c Coomassie Blue binds readily to arginyl and lysyl residues of proteins, but not free amino acids. d First dye (red) donates free electrons to proteins’ ionisable groups → disrupt native protein structure → expose hydrophobic regions usually buried within the protein → non-covalent (Van der Waal) binding to non-polar regions of dye (positive amine group of amino acid close to negative charge of the dye) → stabilize blue form. e Maximum absorbance of the dye shifts due to new bonding to 595 nm. 2 What reagents/reactions could affect results from the Bradford Assay? a Sodium Dodecyl Suphate (SDS): i *Low+ → binds well with protein → inhibit protein binding to Coomassie → more red → Underestimate protein content. ii *High+ → Associate with dye → stabilize green form → more blue → overestimate protein content. b High buffer concentration → depletion of free protons → equivalent of too many electrons → overestimation of protein content. Vector Cloning: 1 Why is vector cloning used? a PCR not as accurate + linear DNA could not be expressed easily. b Lots of gene product could be produced. 2 What is recombinant DNA? Protein? a Recombinant DNA = DNA not usually found in nature (e.g. spliced together from 2 or more sources) b Recombinant DNA → Recombinant Protein. 3 What components are necessary for proper cloning? a Gene of interest: downstream of promoter. i.e. something to transcribe. b Expression Vector: plasmid, contains: i Ribosome binding sites (Shine-Dalgarno sequences). i.e. somewhere for ribosomes to bind and to start transcription. ii Promoter: controls level of transcription of a gene. i.e. when to transcribe. 4 What is induction? a Allowing bacteria to grow in numbers, then turning on the desired gene to produce large amounts of a desired protein. 5 What is the function of a normal lac operon? a Make beta-galactisidase to digest lactose when high [lactose] and/or low [glucose]. 6 What is the importance of T7 and T7 Polymerase? a T7 polymerase is an RNA polymerase originally from a bacteriophage. It is able to bind to the T7 promoter in the pET plasmid and initiate transcription. The gene for T7 is in the bacterial chromosome and could be transcribed by E. coli RNA polymerase. b T7 Promoter is the promoter sequence for T7 polymerase. It’s inserted within the pET plasmid, roughly in the same region as the LacI promoter. T7 polymerase can bind to it and initiate transcription. c When induced, T7 polymerase works in conjunction with E. coli polymerase to increase the transcription of the LacZ gene in the presence of lactose. 7 How does our “2 step” transcription control system work? a The lac operon in the E. coli genome has a T-7 RNAP coding region between the operator and Lac Z. The lac operon is inhibited by Lac Z (when lactose is not present). b When lactose is present in cells, it binds to LacI on the operator and causes Lac I to detach from the operator. This allows RNAP to bind to the promoter and initiate transcription. c Transcription of the chromosomal lac operon include the transcription of the T-7 polymerase gene. T-7 polymerase is made and acts as the RNAP for the plasmid lac Z gene. d When the concentration of glucose in the cell is low, cAMP levels increase. cAMP binds to CAP and causes it to attach to chromosomal lac operon and act as a transcription factor, increasing the action of E. coli polymerase and the level of transcription of T-7. This further upregulates T-7. e The vector may also be inserted with a gene for T-7 lysozyme, allowing better control of the cell (i.e. remove any effect of basal transcription of chromosal lac operon). 8 How could the his-tagged protein be removed from Ni-NTA beads? a Lower pH →His protonation → no binding to Ni. b Chelating agents: EDTA, EGTA → Remove Ni from NTA group → his-protein-metal complex (NTA resin needs Ni recharging). Both could denature protein or affect its activity. Therefore: c Imidazole → compete off His. Gentler. 9 What are the pros and cons of protein spectrophotometry and the Bradford Assay? Technique Pro Con Protein Specific extinction Assumes no contamination from other protein Spectrophotometry coefficient. Assumes appropriate blank (280 nm) Single data point Pure protein with known a.a. (to calculate
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