Essential Lab Safety and Biofuel Production Study Guide
School
University of California, Los Angeles**We aren't endorsed by this school
Course
CHEM 153L
Subject
Biology
Date
Dec 10, 2024
Pages
22
Uploaded by SargentLemur672
These are example questions from all the different topics we cover(ed) in the course. Alsoreview the learning outcomes as listed at the beginning of each lecture slide package. Considergenerating a mind/concept map of the course content for you to “see” how the various topicsin the course are related to each other.(google FREE mind mapping tools)Laboratory Safety (WILL BE ON THE EXAM)1.What do you need to consider about laboratory safety when working in a biochem lab (such asChem 153L lab)? How do you prepare yourself for lab experiments before you enter the laband what do you do inside the lab to stay safe and keep the environment safe?What do you do inside the lab, when an accident happens?Biofuels2.Describe the difference between first, second and third generation biofuels. Describe twoadvantages of biofuels as compared to fossil fuels.First= utilizes food crops and processed through things like fermentation- however it competes withthe food supply and can lead to higher food costsSecond= Utilizes food waste, does not compete with food supplyThird= Bioreactor with photosynthetic algae and lightTwo advantages of biofuels vs. fossil fuels: Biofuels are carbon Neutral (CO2 admitted is equal toCO2 absorbed by growth) & Biofuels are renewable and can be replenished quickly.3.How can modern biotechnology be applied to produce biofuel molecules?Control of the metabolic pathway to produce biofuels through large-scale protein overexpression ofeach metabolic step.4.While we discussed the advantage of using longer chain alcohol molecules for biofuels, whatcould be one reason for industry to still produce large amounts of ethanol as biofuel.Cost efficiency, ethanol is an established industry standard, and a lot of infrastructure alreadyexists.5.What experimental steps did you do in the laboratory to produce biofuels ?Inoculated fresh bacterial culture in antibiotics, induced isobutanol production pathway, enzymeassays to test amount of isobutanol produced per cell (per g glucose)6.How might you test whether a new microorganism suspected to produce biofuel is better orworse than existing microorganisms used for this purpose?COMPARE TO CONTROL STRAIN THAT WAS SHOWN TO PRODUCE ISOBUTANOL INLITERATURE.Run enzyme assays with controlled conditions and compare the enzymaticactivities.Protein overexpression (Group 2)7.Write down all experimental steps (like your flowchart) and understand the purpose of eachstep. Which result for this experiment would you expect, if one of the steps was omitted (one byone)?a.Add LB medium +AMPICILLIN +bacteria (overnight culture) to LB medium
b.Incubate for 30 minc.Remove two aliquots, label t=0.i.Aliquot #1: Microcentrifuge, save cell pellet for SDS-PAGE +immunoblottingii.Aliquot #2: Measure OD600.d.Add IPTG.e.Incubate for 1 hr, repeat step 3 but label t=1.f.Incubate for 2 hrs, repeat step 3 but label t=2.g.Incubate for 3 hr, repeat step 3 but label t=3.8.Remember the purpose of using relevant chemicals for this method (those that we covered inlecture). Which results would you expect if the incorrect molecule was used instead.a.IPTG (TO INDUCE OVEREXPRESSION OF POI, IF MISSING THIS STEP, POINOT OVEREXPRESSED), ampicillin (TO SELECT FOR CELLS THAT CARRY THECORRECT PLASMID, IF NOT CULTURE WILL BE OVERGROWN BY CELLS NOTCARRYING THE PLASMID AND NOT OVEREXPRESSING THE POI), LB medium(TO ESTABLISH FAVORABLE GROWTH CONDITIONS)9.Why is protein overexpression useful?a.The purpose is to produce the protein of interest in high amounts so it’s easier topurify and analyze.10.What is a trial overexpression? Why is a trial overexpression advised? Which parameters doyou adjust to find optimal overexpression conditions.a.Trial overexpression is finding optimal conditions for the POI overexpression. Thishelps us produce the most POI. The 3 parameters we adjust are: 1) temperature; 2)duration of induction; and 3) IPTG concentration11.Which components do you need on the plasmid for protein expression to work? List allcomponents with their respective function.a.Promoter: to initiate transcription of gene of interest by RNA polymerase (whichbinds tightly to a promoter region)b.Origin of replication: binding site for DNA polymerase to initiate replication andhence to make copies of the plasmids that can be passed on to next generationc.Operator: binding site for repressor, which will block RNA polymerase from bindingto the promoter and hence represses transcription of gene of interest (until we treatcells with inducer to “de-repress” transcription)d.lacI gene: encodes for the repressor protein, which binds to the operator andrepresses transcription of gene of interest until de-repressede.Ribosome binding sequence: included on DNA to generate a ribosome binding siteon mRNA, so that transcript can be bound by ribosomes for translationf.Selectable marker: coding sequence for an enzyme (ß-lactamase for ampicillinresistance). This enzyme cleaves ß-lactam ring in ampicillin and detoxifies theantibiotic. Cells with this marker will survive antibiotic treatment, while cells without itwill die.g.His6 tag: DNA sequence encoding for 6 histidine residues to be added to protein ofinterest, which allows the protein of interest to be purified by affinity chromatography
h.Gene of interest: encodes for protein of interest12.What is the function of the T7 promoter on the plasmid DNA?a)The T7 promoter is where T7 RNA polymerase binds for transcription13.What is the function of the operator sequence on the plasmid DNA?a)The operator sequence is where the repressor binds to PREVENT transcription14.What is cell density and how do you measure it?a)Cell density is the amount of cells in a unit volume of cultureb)You measure it using the OD 600 absorbance reading15.What does inducing protein expression really mean? And how does inducer contribute tothe control of the expression of the gene of interest?16.What happens if you induce a culture in the stationary phase?a)You will get lower yield because the cells are not actively dividing17.What happens if you induce a culture in the lag phase?a)Cells are still ADJUSTING TO THE CHANGE IN their environmentb)The cells will put their energy into protein overexpression rather than division and growthso they might not ever get to the log phase (START DIVIDING EXPONENTIALLY)c)There’s less protein overexpression OVERALL, BECAUSE THERE ARE LESS CELLSGROWING (INDIVIDUAL CELLS MIGHT PRODUCE A LOT OF POI)18.Why do you need to monitor the cell density of various samples?a)Normalization. To ensure THAT SAME NUMBERS OF CELL MATERIAL IS LOADED,SO THAT WE CAN CONCLUDE THAT VARIOUS amounts of protein DETECTED arebased on THE TREATMENT OF SAMPLES and not because more or less cell materialwas loaded.19.How is a mutation in the Lon protease (one particular protease) helpful for proteinoverexpression?a)Lon protease degrades proteins. A mutation IN THE LON PROTEASE THATDECREASES ITS ACTIVITY WILL LEAD TO INCREASED PROTEIN STABILITY (DUETO DECREASE IN DEGRADATION)20.What are the different phases to a microbial growth curve, and what would it look like as acurve?a)Lag phase: cells are adapting, little growth so low slopeb)Exponential (log phase): fast division so curve goes up fastc)Stationary phase: cells reach growing capacity of culture, line is flat and pretty stable(goes up and down a bit)d)Death phase: environment can’t support the cells anymore so downward slope of cellpopulation
SDS PAGE/Immunoblotting (Group 3)21.Write down all experimental steps (like your flowchart) and understand the purpose of eachstep. Which result for this experiment would you expect, if one of the steps was omitted (one byone)?22.Remember the purpose of using relevant chemicals for this method (those that we covered inlecture). Which results would you expect if the incorrect molecule was used instead.SDS - no SDS means no uniform negative charge and no uniform shape (think ofdifferent conformations of proteins when folded)Acrylamide - no acrylamide means no gel/PORES FOR SEPARATING PROTEINSBY SIZEBis-acrylamide - without, gel becomes FRAGILETEMED, APS - without, no radical reaction to initiate polymerization - NO GELFORMATIONREDUCING AGENTS (ß-MERCAPTOETHANOL OR DITHIOTREITOL) - without,size is not the only factor affecting movementRunning buffer - with wrong Tris pH, glycine might not enter gel, without glycine, nostacking effectTransfer buffer - wrong reagents can lead to poor transfer, no ethanol can makeproteins go through membraneBlocking - wrong reagents can lead to lots of background signal/nonspecific antibodybindingTTBS - wrong reagents can lead to background signal23.What is the purpose of running SDS-PAGE?a.SEPARATION OF PROTEINS BY MOLECULAR WEIGHT (CAN BE VISUALIZEDEITHER BY COOMASSIE STAINING TO SEE ALL PROTEINS IN SAMPLE) ORSEE Q2424.What is the purpose of immunoblotting?a.Visualize ONLY THE protein of interest25.What molecules are involved in gel formation? What are their respective roles?a.Acrylamide- monomer WILL POLYMERIZE TO FORM the gel matrix,Bisacrylamide- ALSO MONOMER THAT IS INVOLVED IN FORMING the gel matrix,crosslinks the linear strands of the polyacrylamide POLYMERTEMED and Ammonium persulfate- start the polymerization of the gel through aradical reaction
26.How do you treat the proteins before loading them onto the polyacrylamide gel? Why? (Thinkabout what would happen, if you missed any of the chemicals or steps necessary.)a.SDS - linearize protein (denature proteins) and coat evenly with negative chargeb.B-mercaptoethanol – reduce disulfide bridges to further denature proteinsc.Boiling27.Why is it advisable to use a stacking gel in SDS PAGE? What would we see if a stacking gelwas not used? Would proteins migrate?a.Prevents smearing in gel by allowing proteins to all start at the same position beforeentering the resolving gelb.If stacking gel not used, proteins would smearc.Proteins would still migrate according to size without stacking gel, but they wouldn’tform distinct bands28.Explain what is different in the separating gel (compared to the stacking gel) that allowsproteins of different sizes to migrate at different speeds.a.In the separating gel, the pore size is smaller versus the stacking gel, where we wanta larger pore size. This means that the only variable affecting the speed of theproteins is the molecular weight, wherein larger molecules will travel slower throughthe gel.29.How do you adjust the pore size in an acrylamide gel?a.Changing the ratio of acrylamide to bisacrylamide; the more bisacrylamide, thesmaller the pore because of increased frequency of crosslinkingb.But in general, the higher the concentration of acrylamide and bisacrylamide, thesmaller the pores30.What are the options of visualizing proteins following separation by SDS-PAGE?a.Coomassie, immunoblot31.How do you stain all proteins in the gel?a.Coomassie blue stain (incubate gel in staining solution, wash off solution, onlyproteins should be stained)32.Which pH conditions would you set for Coomassie staining?a.Should be at a pH high enough to which the dye molecule is completelydeprotonated/anionic to allow binding to all proteins on the PAGE; specifically (>2?)33.Why do all proteins migrate in the same direction in SDS PAGE, irrespective of their intrinsiccharge?a.Before loading the proteins in the gel, we treat them with Sodium-dodecyl-sulfate tocoat all the proteins evenly with a negative charge.34.Why can we ignore the various protein conformations when comparing their migration inSDS PAGE?a.Before loading the gel, all the proteins are treated with SDS which also is adetergent that disrupts the non-covalent bonds in the proteins. They are alsotreated with a reducing agent that breaks the disulfide bonds.35.Why does the molecular weight of proteins not affect their speed of migration in the stackinggel but in the resolving gel?
a.In the stacking gel, the pore size is large enough that the size of the proteinsdoes not affect their speed of migration.36.What would happen if the pores in the stacking gel were as small as those in the separatinggel? What would the gel look like?a.Instead of having the proteins line up, they would migrate based on size in thestacking gel37.Explain why we need a stacking gel and how the stacking effect works.a.Need a stacking gel so that all proteins start at the same position along gel frontbefore running through resolving gelb.Stacking effect:i.Glycine enters negatively charged at pH ~8.3-8.5ii.Then becomes a zwitterion in pH ~6.8 of stacking gel, which creates anarea of high local resistance and thus an area of high local voltage(Ohm’s law)iii.Chloride ions in buffer run faster and create an area of low resistance/lowvoltageiv.The voltage gradient pushes proteins that were higher in the gel downfaster so that all proteins line up along the gel frontv.Glycine then becomes negatively charged again, removing the voltagegradient, once they enter the resolving gel pH ~8.838.Is it possible to use SDS PAGE to determine if a protein is found as a monomer, dimer,trimer, etc in the cell? Explain your answer.a.Yes if the subunits are linked by covalent bonds, you can add and removedenaturants to separate the subunits. This might only work if the only covalentbonds are between subunits, as otherwise the shapes of the subunits will bedifferent. If the subunits are linked by non-covalent interactions, then SDS PAGEcan’t help.39.Describe the rationale of how immunoblotting works.a.Immunoblotting works by using antibodies to recognize a tag on a protein ofinterest and binding to it. To amplify detection of the protein, a secondaryantibody, which binds to the primary, is used. Through this binding, we canvisualize the presence of a protein of interest in a sample. This visualization isdue to an enzyme or dye molecule conjugated to the secondary antibody, andcomes in the form of colorimetric, chemiluminescence, and fluorescence.40.Remember why we needed to transfer proteins onto the membrane. How do you getproteins to move from gel to membrane without losing their position relative to each other?a.Transfer “horizontally” (relative to running the gel vertically)41.Why do you need a loading control?a.The loading control reaffirms our conclusion that the results we’re seeing in theimmunoblot (e.g. that one lane produced more protein than other lanes) is completelydue to the variability that we induced (e.g. time incubating) rather than due to randomvariability in number of cells.42.What are loading controls?
a.Either staining a part of the gel (all samples) with Coomassie to show that allother proteins are evenly represented and thus samples were loaded evenlyb.Or perform an immunoblot with antibody against a housekeeping protein, whichis typically expressed similarly under various different conditions. If thehousekeeping protein shows similar signal intensities between different samples,then those samples were evenly loaded43.Draw an IgG antibody molecule.(from lecture slides)44.What is the difference between primary and secondary antibody molecules?a.Primary binds directly to POI (can be monoclonal or polyclonal), secondary ispolyclonal and binds to primary45.Can a primary anditssecondary be from the same species?a.No, the secondary antibody has to be from a different species than the primary,otherwise it will not recognize it as foreign and won’t bind.46.What are the options of visualizing your protein of interest on the membrane?a.immunoblot (colorimetric, fluorescent, chemiluminescent visualization)47.Which detection method did you use in the course?a.Colorimetric enzyme-linked assay with alkaline phosphatase and NBT48.Recognize the structures of all molecules necessary here: acrylamide, bisacrylamide,TEMED, ammonium persulfate, sodium dodecyl sulfate, ß-mercaptoethanol,TRIS,bromophenol blue, Coomassie brilliant blue,.....Literature-based questions (thesekindof questions are likely to be on your exam)1. SDS-PAGE: While we use antibodies in the lab to label our protein of interest on thenitrocellulose membrane, scientists separated antibodymolecules themselves via SDS-PAGE here. Rememberwhat you learned about SDS PAGE and antibodymolecules in lecture and follow the questions below.
Guiding questions for your analysis:a.Why is Coomassie staining used here to visualize the proteins after SDS-PAGErather than immunoblotting?b.To visualize all proteins; purpose here is not to look for a specific protein (unlikely that bothchains will bind one antibody, and we want to visualize both chains)c.Thesameantibody was loaded in lanes 2 and 3, but the samples werenottreatedthe same prior loading. Which component of the protein sample buffer, when addedor left out, could affect the band pattern of antibody molecules in this way? Explainand specify the different treatment of samples in lanes 2 and 3.When the reducing agent is left out, it could affect the band pattern of theantibody molecules. The light and heavy chains are connected by a disulfidebridge, which is disrupted when treated by the reducing agent. This breaks up thetwo chains in the antibody and allows us to visualize them in lane 3.d.Remembering the structure of an IgG antibody, label the three bands you see inlanes 2 and 3.Which band(s) contain(s) the variable region of the antibody that binds to theepitope of the antigen molecule?All bands in this gel contain some portion of the variable region of the antibody because thevariable region is made up of heaving and light chainsBradford assay49.Write down all experimental steps (like your flowchart) and understand the purpose of eachstep. Which result for this experiment would you expect, if one of the steps was omitted (one byone)?50.Remember the purpose of using relevant chemicals for this method (those that we covered inlecture). Which results would you expect if the incorrect molecule was used instead.51.Describe the properties of Coomassie Brilliant Blue and why these properties allow us todetermine protein concentration. Recognize the Coomassie Brilliant Blue molecule.CBB dye has three protonation states that change the color of the dye in accordance with pH (Red:0, Green 1-2, Blue 2) For this experiment, we want the dye molecules to be red, so that there is nooverestimation of protein during the assay.Blue form is not stable unless we change protonation state52.What part of a protein binds to the Coomassie Brilliant Blue dye?Aromatic side chains and arginine side chains (positively charged side chains)53.How can you quantify protein concentrations in Bradford assay?Setting up a standard curve54.For which purpose did you use the Bradford assay in the lab course?Measuring unknown protein concentration in lab 1 and your purified YqhD in lab 555.Are proteins able to bind to the Coomassie Brilliant Blue dye at pH 4?yes56.Should you use the assay at pH 4?
No, at pH 4 CBB will be in its blue form. We do not want this because it would result in anoverestimation of protein in the sample.57.Do all proteins bind equally well to Coomassie Brilliant Blue dye?No, proteins bind through certain side chains. The more of those amino acid side chainsthey contain, the better they bind the dye58.Why don’t we use Beer’s law to analyze the Bradford assay?We do not know the molar absorptivity constant of protein-dye complex.59.Under which circumstances can you claim that the protein concentration measured inBradford is of a single proteinRun on a gel stain with Coomassie to check if there is one band or several.60.What could be alternative ways to determine the concentration of a single protein?Immunoblotting, ELISA, enzyme assay (final exam relevant)61.Why isn’t the standard curve of Abs595vs. protein concentration linear over a long range?The Absorbance is limited by the amount of dye molecules loaded into the well. Once allthe dye molecules bind, no matter how much protein is in the sample, a difference inprotein concentration will not be detected. In addition, as the amount of proteinincreases, the red/green background will decrease because more dye molecules bind.62.How can you adjust your analysis to generate a longer linear correlation?Take ratio of 595/470 (blue/red) absorbance values and track how the absorbance ofred/green dye measured decreases with increasing protein. This ratio takes thisbackground absorbance decrease into account.63.What does a steeper slope in the Bradford assay calibration curve for BSA indicate aboutits interaction with the assay reagent compared to the wheat protein fractions?What does this indicate about the amino acid composition of BSA compared to the wheatprotein fractions?https://doi.org/10.1002/cche.10447-The steeper slope for BSA has a stronger positively charged amino acid compositionthan the wheat protein fractionsEnzyme assay - Group 2
64.Write down all experimental steps (like your flowchart) and understand the purpose of eachstep. Which result for this experiment would you expect, if one of the steps was omitted (one byone)?First, determined optimal pH conditions + enzyme activity (Km + Vmax, 340nm)Test at optimal conditions: 1) pH 2) temperature 3) substrate concentrationCombine mastermix (buffer, NAD+, ethanol/substrate) + enzyme dilutionsQuickly measure absorbance at 340 nm●Calculate mM/min using Beer’s Law → use specific enzyme activity (SEA) to calculateenzyme concentration per mL assay → multiply by d.f. to get original concentration65.Remember the purpose of using relevant chemicals for this method (those that we covered inlecture). Which results would you expect if the incorrect molecule was used instead.66.Write down substrates and products of an alcohol dehydrogenase reaction.a.Ethanol + NAD+ → Acetaldehyde + NADH + H+ (using ADH as the catalyst)67.When you set up this reaction in the lab, what else do you need to add? Why?a.Cofactor and buffer (KPhos at pH 7)68.Recognize the structure of these molecules (including cofactors)69.Indicate how you can measure the enzyme reaction (i.e. which molecule are you actuallymeasuring in the spectrophotometer). Keep in mind, if the absorbance is expected to rise or fallin the reaction you wrote above.a.As NAD+ is reduced to NADH, we measure the absorbance over time at 340 nm tosee the increase in NADH production (rise in absorbance).70.Differentiate what we mean by 0th order and 1st order reaction conditions? What would you doin the lab to set up one or the other?●0th order: velocity of rxn is independent of substrate concentration. Set up in lab withhigh substrate concentrations (plateau of MM graph)●1st order: velocity of rxn is dependent on substrate concentration. Set up in lab withlowsubstrate concentrations (beginning, steep portion of MM graph)71.Which graph would you pick to illustrate what 1storder or 0thorder reaction means? Label axes.Michaelis-Menten graph72.Draw a graph to illustrate what “initial rate” means. Label the axes. What is the issue withassays that are not in the initial rate?Absorbance vs time graph: initial rate, when the slope is linear; once it flattens out, backwardreaction started and rate no longer represents only the forward direction of the reaction73.What do the Kmand the Vmaxindicate about an enzyme reaction?
a.The Km is the concentration of substrate in which ½ of the enzyme’s max velocity ismaximized. Vmax indicates the maximize velocity of the enzyme in convertingsubstrate to product74.Differentiate between the different classes of enzyme inhibitors (binding site/effect on Vmax orKm). Sketch the binding. Start with the one you are most unfamiliar with and consider utilizing atable.a)competitive inhibition: inhibitor binds on the active site; the Vmax won’t change but Km goesupb)Uncompetitive: inhibitor binds ONLY to the ES complex; the Vmax decreases because someenzyme is being taken out, km decreases by the same degree,c)Pure noncompetitive: inhibitor binds to the free enzyme and to the ES complex WITH THESAME AFFINITY, Vmax decreases, Km stays the samed)Mixed: when Ki (dissociation constant) > Ki’ vmax decreases, km decreases, when Ki’> Ki,vmax decreases and Km increases75.What is the difference between KIand KI´?a)Ki is the dissociation constant of the inhibitor binding to the free enzymeb)Ki’ is the dissociation constant of the inhibitor binding to the enzyme substrate complex76.What is alpha with respect to competitive inhibition?𝑎 = 1 + [𝐼]/𝐾𝑖a)When alpha is 1 there’s no inhibitionb)alpha is greater than 1 for the competitive inhibitor -> it can increase by using a higherconcentration of inhibitor or by using an inhibitor with a higher affinity (and hence lowerKI)77.What kind of inhibitor is EDTA? Is it possible to recognize this mode of action using aLineweaver-Burk graph? If yes, how?a.EDTA APPEARS like a mixed inhibitor when analyzing a Lineweaver-Burk graph78.Is EDTA specifically inhibiting alcohol dehydrogenase OR will other enzymes be inhibited byEDTA? If yes, which ones?a)EDTA does not directly bind and inhibit the enzyme. Its inhibitory effect can be explained byEDTA sequestering metal cations so ALL enzymes requiring divalent metal cations will alsobe inhibited, just any enzyme that binds to metal cations to activate79.You are working in a biochemistry lab and are analyzing an enzyme. If you know themolarity of the enzyme, would you be able to determine its specific enzyme activity or itsturnover number? Describe the difference between the two.a.If you know the molarity of the enzyme, you would be able to determine itsturnover number (kcat). The specific enzyme activity and turnover number issimilar, they are only different in units, where specific enzyme activity measuresthe concentration of enzyme in mg and kcat measures in micromolar.80.Specific enzyme activity. You are measuring the maximum velocity of a dehydrogenaseenzyme by looking at the absorbance of 340 nm. During the initial rate phase, you measure
an absorbance change of negative 0.5/min. The molar absorptivity coefficient is 6.22 mM-1cm-1and the cuvette length is 1 cm. The kcatof this enzyme is 1300 min-1. Theoriginal undiluted enzyme stock concentration is unknown. You dilute the enzyme by thedilution factor of 15 before adding it to the assay. You then used 18 µL of this dilutedenzyme solution for the enzyme reaction in a total volume of 1.3 mL. What is the originalmolarity of the enzyme?Step 1: Find KcatΔA/t = ε?(Δc/t)0.5/min = ε?∆𝐶→ solve for change in concentration → ∆𝐶= 0.0804?𝑀/?𝑖??𝑖?Kcat= 1300 min-1 (given in the problem. If not given, calculate using Kcat= Vmax/ [E]total)Step 2: Find amount enzyme1 mM enzyme → 1300 mM of product/minX mM enzyme → 0.0804 mM of product/minX = 0.000062 mMStep 3: Account for dilution using C1V1 = C2V20.000062 mM * 1300 µL = X * 18 µLMultiply X by dilution factor (15)Original enzyme concentration = 0.067 mM81.Could you answer the above question, if you were given the specific enzyme activityinstead of the kcat(and no other change or additional info)?a)No, you need the molecular weight of the enzyme. Using specific enzyme activity allowsyou to solve enzyme concentration in mg/mL instead of molarity.82.What could be the issue if we used a first order enzyme assay for measuring enzymeconcentration?Underestimation of enzyme concentration83.When you plot the Michaelis-Menten graph, are all data points initial rates?a)Yes all the the data points are initial rates, they are just from different substrateconcentrations84.How can you determine KIexperimentally? (Hint: look at your Lab 6 worksheet)a)Competitive inhibition: Kmapp= αKmalpha = 1 + [I]/Ki→ solve for Ki85.How do you transfer the experimental data of product vs time to the Michaelis-Mentengraph?a.You will have to figure out the slope of the experimental data of product vs time.Then we can plot that slope to a Michaelis-Menten graph with its known substrateconcentration.
Affinity chromatography (Group 3)Potential figures for this topic:a.A microbial expression system for high-levelproduction of scFv HIV-neutralizing antibody fragments inEscherichia coli (Petrus et al., 2019)Explaina the difference between (a) and (b) and why the authorsperformed both analyses to evaluate the fractions….Gel A stained with Coomassie blue to visualize all proteins(evaluate the purity of our sample); immunoblot was done on gel B(specifically detect protein of interest)Which fraction do we consider for our protein of interest to becontained in?The eluate fractions (E9, E10, and E11)Why did the authors decide to run the flowthrough and thewashes?The flowthrough is the initial run of the sample through the column.In this fraction, all proteins with an affinity to the column will bind.The washes remove all the proteins, except the protein of interest,from the column. This ensures proper purification of the protein ofinterest.86.Write down all experimental steps (like your flowchart) and understand the purpose of eachstep. Which result for this experiment would you expect, if one of the steps was omitted (one byone)?87.Remember the purpose of using relevant chemicals for this method (those that we covered inlecture). Which results would you expect if the incorrect molecule was used instead?88.What is the purpose of affinity chromatography? Keep in mind the alternative methods(discussed during Week 9/10).a.The purpose is purification, to separate your protein of interest from as many otherproteins in the solution as possible.89.What part of the column allowed for binding of our protein of interest?a.Ni2+ ions form coordination with our protein’s imidazole tag.90.What is the nature of the interaction between the ligand and target in an affinity pair?a.The interaction is electrostatic; The positive charge of the Nickel and the negative charge ofthe imidazole create attraction.91.Besides the histidine tag and Ni2+, name two other affinity pairs (ligand/target) that can be usedfor affinity chromatography.a.Antibody/antigen, antigen/antibody, substrate or inhibitor/enzyme, enzyme/substrateor inhibitor92.Describe the various steps of an affinity chromatography and their respective purpose.
a.Equilibration: loading a small amount of imidazole to prevent too much non-specificbindingb.Loading: loading protein on columnc.Washing: using low concentration of imidazole to wash non-specifically boundproteins off the columnd.Elution: using high concentration of imidazole to knock off and collect purifiedHis-6-tagged protein93.What are the various ways to remove your protein of interest from the column (elution)?a.Add an extremely high concentration solution of imidazole to transfer attraction from proteinimidazole to free imidazoleb.INCREASE the salt concentrationc.Alter the pH94.Which elution method did you use in the lab? How/why did this work?a.We used elution method a; Various concentrations of imidazole95.Your roommate suggests you repeat the purification of alcohol dehydrogenase by affinitychromatography with ten histidine residues in the tag. Explain why this would or would notwork.a.It may work, but it might not work as well as using the His6 tag. This is because thebinding of a His10 tag to the column would be very strong, and we would need anextremely high concentration of imidazole to elute it out.96.Why is it useful to know the dissociation constant of the affinity pair utilized for affinitychromatography?a.Knowing the dissociation constant helps us figure out the concentration of elutingagent (in our case, imidazole) in washes to avoid washing out POI too early andelutions to ensure POI does get elutedb.The higher the dissociation constant, the weaker the interactions with the column,the lower the concentration needed to elute out; the lower the dissociationconstant, the higher the interactions with the column, the higher the concentrationneeded to elute out97.Troubleshooting: You are performing an affinity chromatography purification like inlaboratory 4. After collecting all the fractions, you take small aliquots of each to run them bySDS PAGE and check if you purified the protein efficiently by Coomassie staining. Explainshortly what might have happened in below scenarios.●Soluble lysate (after centrifugation to remove insoluble material) and flow throughcontain many protein bands (including a thick band in the size range of your proteinof interest), there are a couple of thin bands in wash step 1, all further fractionscontain no bands.a.Our protein likely completely PASSED THROUGH in the flow through step, whichis intended to only remove very poorly bound/non-bound components. We likelyadded a high concentration of imidazole too early in this wash step.●Soluble lysate (after centrifugation to remove insoluble material) and flow throughcontain many protein bands (including a thick band in the size range of your protein
of interest), there are a couple of thin bands in wash step 1, no bands in wash steps2 and 3, a thick band in eluate 1 and eluate 2.It seems like the affinity chromatography worked fine in this scenario, as youcollected pure protein in the elution. You could have captured more POI if morebinding sites in the column were available.●Soluble lysate (after centrifugation to remove insoluble material) and flow throughcontain many protein bands (without a thick band in the size range of your protein ofinterest), there are a couple of thin bands in wash step 1, all further fractions containno bands.a.If your soluble lysate didn’t contain any protein, it likely formed inclusion bodiesand showed up in the insoluble fraction OR WAS NOT EXPRESSED.Isobutanol production inE. coli- Group 198.Write down all experimental steps (like your flowchart) and understand the purpose of eachstep. Which result for this experiment would you expect, if one of the steps was omitted (one byone)?Look at flowchart99.Remember the purpose of using relevant chemicals for this method (those that we covered inlecture). Which results would you expect if the incorrect molecule was used instead.100.Whatare all the enzymes and substrates/products involved in the isobutanol pathway?(Imagine you are given two intermediates in the pathway asked to identify them, and whichspecific enzyme catalyzes the reaction between them)LECTURE SLIDES: remember the names of all enzymes (full names for full credit),intermediates and cofactors. Note which molecules leave during which reaction.101.How can we engineer bacterial strains that overexpress all five enzymes for production ofisobutanol?We can distribute all of the genes of the 5 enzymes onto a plasmid! But we’re going to need 2plasmids because there are a lot of genes.The first plasmid, pSA69, contains the genes for the first 3 enzymes in the pathway, alsS, ilvC,and ilvD. The plasmid also contains an origin of replication, kanamycin resistance gene, lacpromoter, and lac operator.The second plasmid different for each strain, but it contained the last two enzymes in thepathway, along with an ampicillin resistance gene, lac promoter and lac operator.102.If you plan to transform bacteria with two plasmids at the same time, what factors do you needto consider?They need to have two different selectable markers: so that we can make sure that only the cellswith both plasmids can surviveThey also need to have different origins of replication: to ensure that they are both being replicatedat the same frequency (because when two different plasmids with the same origin of replication arekept in the same cell, they will compete for replication factors and one might be replicated morefrequently than the other)
103.What was the reason we chose M9 minimal medium to grow the cells for isobutanolproduction?M9 minimal medium allows for accurate carbon tracing, which allows us to know that all of thecarbon sources for our cells come from glucose.104.Thinking about first and second generation biofuels, which carbon source should we considerto switch to?CELLULOSIC material (waste plant material) to not compete with food production105.Brieflydescribe the experimental workflow for isobutanol production in E. coli.106.Which controldid you use in our lab to determine how "active" the pathway was in makingisobutanol? Which alternative control could we use to test the same?The control strain we used is SA542; which is a parent E. coli gene that has a mutation in the YqhDgene. This was important to analyze how much isobutanol a strain without the overexpressedpathway could make, otherwise, it would be difficult to say how much isobutanol is really due to theoverexpression of the pathway.107.If you weretesting a new strain for production of propanol by overexpression of a differentpathway, which biological control would you consider to test the efficiency of theoverexpressed pathway in the cells?A positive control strain that was shown to produce propanol when engineered.108.Wanting to improve this isobutanol production pathway, which enzyme function would youtry to enhance (think of the Atsumi paper results)?Ketoacid decarboxylase was shown in Atsumi to be rate limitingIsobutanol detection (Group 2)109.Write down all experimental steps (like your flowchart) and understand the purpose of eachstep. Which result for this experiment would you expect, if one of the steps was omitted (one byone)?●After overnight incubation, measure OD600nm of each strain●Create mastermix containing Tris/Triton buffer, NAD+, MTTox (yellow), MPMSoxisobutanol, and ADH2●Prepare known isobutanol concentrations●Prepare samples of unknown isobutanol concentrations (dilutions for control strain, strain1, strain 2, strain 3)●Combine mastermix and unknown isobutanol concentrations●Incubate in dark for 1 hr●Measure absorbance at 570 nm (abs of MTTred purple)●Generate standard curve with absorbance values at 570 nm of known standards vsisobutanol concentration●Use standard curve to determine concentration of isobutanol produced●Determine isobutanol yield110.Remember the purpose of using relevant chemicals for this method (those that we covered inlecture). Which results would you expect if the incorrect molecule was used instead.111.Why is measuring the optical density (600 nm) essential here?
a)Normalize the cell density so the number of cells do not contribute to the differences inproductivityb)We want to make sure we only have one variable, so the rest (including number of cells)needs to be controlled for112.Which reagents will you have to add to the assay for it to work? Be able to recognize thestructure of these molecules.a)Isobutanol (substrate), ADH2 (enzyme), all oxidized forms (NAD+, MPMSox, MTTox yellow),Tris buffer113.Which molecules should you NOT add to the assay? Why shouldn’t these molecules beadded to the assay? (remember worksheet lab 7)a)Reduced forms (NADH, MPMSred, MTTred purple). These would add additional electrons,causing false positive/overestimation of isobutanol.114.Which control did you use for this assay? Which of these is a positive/negative control?a)We did bothb)Positive controls: changing the concentrations of isobutanolc)negative control: no isobutanol115.What are possible reasons for a very low absorbance at 570 nm in the MTT assay? How canyou prove that the assay is working as intended?a)The cells did not produce a lot of isobutanol or your assay isn’t working as intendedb)You can use a positive and a negative control to show that both the positive and negativeresults are possible116.Draw the entire endpoint assay pathway (Isobutanol → MTT formazan).
117.Why did we use the coupled assay of MTT formazan production rather than measuringNADH directly at 340 nm?a)MTT does not have reverse reaction while the NADH doesb)If we were to measure NADH we would underestimate the isobutanol productionbecause we would also be measuring the reverse reaction118.What can you do to test if the assay is indeed measuring differences in isobutanolconcentration?a.Check with a standard curve with different concentrations of isobutanol to see ifyou get different signals119.How are you checking if it is this particular pathway that produces the isobutanol?a)Check with the control because the only thing being changed between the control andthe sample is the overexpression pathway120.Could you measure the concentration of propanol in this assay?a.Yes, since ADH can use other alcohols as substrate.Other uses of Antibodies in the Lab and in Medicine (will be covered in weeks 9 and 10)(Group 3)121.What are other experimental techniques which utilize antibodies?a.ELISA, immunoprecipitation, immuno-fluorescence microscopy, diagnosis andtreatment of cancer122.Briefly describe the role of antibody molecules for each technique.a.ELISA – detection of proteins in a sampleb.Immunoprecipitation – isolation of specific proteins and possible binding partnersc.Immuno-fluorescence microscopy – localization of specific proteins in cells(fluorescent dye conjugated)d.Diagnosis and treatment of cancer – antibodies conjugated to radionuclides to visualtumors via positron emission tomography scanning / can be targeted directly totumor for radiation therapy123.What are the different types of ELISA?a.Direct assay (enzyme attached to primary ab), indirect assay (enzyme attached tosecondary ab), capture assay/”sandwich” (polystyrene well coated with an additionalprimary ab unique from other primary ab, enzyme attached to secondary ab)124.What does “sensitivity” in the context of ELISA mean?a.Refers to the level of detection distinct from background noise the ELISA is capableof125.Which ELISA would you use if you are testing a patient’s immunity against measles? Why?a.INDIRECT, fill the well with antigen of interest then add patient samples. Anyantibody that binds to antigen in well will be captured and can then be detectedwith a secondary anti human IgG conjugated to an enzyme.126.How can antibodies be useful to diagnose and treat cancer?
a.Antibodies conjugated to radionuclides to visual tumors via positron emissiontomography scanning / can be targeted directly to tumor for radiation therapy127.How are antibody molecules generated?a.Immunize mouse (or other host animal) with antigen, allow cells to produceantibodies against the antigen, and collect serum for isolation (B-lymphocytes)b.Fuse B-lymphocytes with a mutant cell line derived from a tumor of B-lymphocytes(these mutant cells grow indefinitely in normal medium but die in selective medium)c.B-lymphocytes die after a few days in culture, and with the change of culture mediainto a more selective medium, of which only fusion cells/hybridomas growd.Supernatant tested for antibody – from an antibody-positive well, redistribute cells ~1cell per well so that after being allowed to multiply, positive clones will provide acontinuing source of antibody128.Explain the difference between polyclonal and monoclonal antibodies.a.Monoclonal - recognize a single (one) epitope on antigenb.Polyclonal – derived from multiple B-cells; recognize multiple epitopes on antigens129.When would you use a monoclonal versus a polyclonal antibody?Polyclonal always to amplify signal (for primary or secondary); monoclonal secondary notuseful; monoclonal primary useful when trying to differentiate between different versions of thesame protein (wildtype vs mutant form)130.When might a researcher use immunofluorescence microscopy as opposed toimmunoprecipitation?a.They might want to visualize where certain proteins localize in a cell rather thanpurify the protein sample – understanding localization of protein also gives insightto protein function
131.When would you suggest using immunoprecipitation?a.When you want to isolate the protein from a mix of proteins to further evaluate itsexpression, function, etc.Other protein separation methods (we will finish this during Week 10)132.State the overall goal of any of these protein separation methods.To separate the POI from as many proteins as possible.133.List the protein characteristic that is used for separation of proteins in respective methods.a.Protein solubility, Protein Size, Protein Charge134.How are you testing overall protein concentration and concentration of your protein of interestin the various protein separation techniques?Overall protein: absorbance at 280 nm or Bradford assay (or other assay that tests for overallprotein content); protein of interest: depending on the protein you can use a specific enzymeassay, or immunoblotting or ELISA to only detect your POI135.How do salts influence protein solubility?a.Salts disrupt electrostatic interactions which will increase solubility at low salt concentrationsor decrease solubility (precipitate formation may occur) at high salt concentrations136.What is the difference between salting in and salting out?a.Salting in is when a low concentration of salt is added so that the salt ions interact withamino acids and shield the charge which reduced the repulsion between protein molecules,making the molecules more solubleb.Salting out is when a high concentration of salt is added and this disrupts the interactionsbetween protein molecules and causes them to precipitate out of the solution137.How does the pH influence a protein's solubility?a.A change in pH will cause a different protonation stateb.When the pH is at PI of the protein, no net charge and higher likelihood to aggregate138.Why can differential and density gradient centrifugation help us to purify our protein of interest?a.These techniques both cause separation based on the protein size/density which will allowour POI with a specific size to be isolated from a mix of others139.This is mostly useful for proteins expressed in eukaryotic cells. Why?a.Eukaryotic cells have many organelles that have different sizes and densities which allowstheir separation into distinct layers with this technique140.If you want to isolate a protein naturally occuring in the cytoplasm, which centrifugation step(s)should you use for purification (differential centrifugation)?a.You can perform only the very long and fast centrifugation step to precipitate all organellesand polyribosomes in one pellet, leaving only the soluble proteins in the supernatant141.Describe why it is useful to use the relative elution volume for size exclusionchromatography.a.It helps normalize the elution volume to the total volume in the column so that you cancontrol for variables like different column sizes and flow rates142.Would an increase in salt concentration change the relative elution volume of your proteinof interest on a size exclusion column? Explain your answer!
a.Yes, since increasing the salt concentration will disrupt protein interactions and causingthe interaction between the matrix and the protein to become weaker, allowing fasterelution at a lower elution volume143.Which of the chromatography methods require a change in buffer composition for release ofthe POI?Affinity and ion exchange chromatography require us to change the buffer to release theprotein of interest from the column, whereas in size exclusion chromatography, no proteinswill bind to the matrix, so that it is more a question of how much buffer you need to get theprotein through (no bonds need to be broken)144.In size exclusion chromatography, would you choose a matrix material with pores that allowyour POI to enter the beads or exclude your POI from the beads? Explain!a.You want a matrix that has pores that allow your POI to enter the beads since it willmove through the column slower with the bead and therefore elute later145.Which other factor but size can affect the migration behavior of your protein through thesize exclusion matrix?a.The SHAPE of the protein146.Could you use size exclusion chromatography to figure out if your protein is a monomer,dimer, etc. in the cell?a.Run the column with native protein to determine MW → can use SDS treatment toseparate into subunits147.How are proteins separated in ion exchange chromatography?a.Proteins are separated by their net charge and they selectively bind to the148.Would you choose a positively or negatively charged matrix to separate a negativelycharged POI from other proteins?a.You would choose a positively charged matrix to separate a negatively chargedPOI149.In ion exchange chromatography, would you use an increase or decrease of saltconcentration to release your POI from the column?a.You would increase the salt concentration to release your POI150.What are the advantages of step and gradient elution, respectively?a.Gradient: The advantage is a higher resolution of proteins with different netcharges. The disadvantage is that it is much harder to set up, then simplywashing/eluting with buffers of different salt concentrations.b.Step: Easier to set upScientific controls:151.What is the purpose of using various control samples? For our lab experiments, be able todetermine which controls were used. Can you find control samples used in publications (Atsumipaper, other papers)?These questions are on the Atsumi paper that describes the project we are running in lab 7.152.What is the overall takeaway message of the Atsumi paper?
153.Which methods did the researchers use and what did they test/measure with the respectivemethods? Where in the paper do you find this information?154.Which biochemical pathway did the researchers modify to produce isobutanol?155.What class of enzymes are alcohol dehydrogenases (ADH)? What is the general reactionthey can catalyze? Do they require cofactors? If yes, specify which type?156.What are the sources (organisms) of the three ADH that were tested in the Atsumi et.al andout of these three which are the most suitable for isobutanol production in this system?157.How did Atsumi et al. prove the importance of YqhD in isobutanol production? Keep in mindthat E. coli has several genes encoding for alcohol dehydrogenases.158.What do the researchers show in figures 2 and 3? How do they use symbols to make iteasy for the reader to understand what is depicted?159.What do the researchers show in figure 4? Why is it important to test this?