BCHS 3201 Final Exam Review Fall 2022 PDF

Final Exam Review need to What I BCHS 3201 go over/ask - - Practice Lab Math abt Jessica Stevens -...

Final Exam Review need to What I BCHS 3201 go over/ask - - Practice Lab Math abt Jessica Stevens - Practice Plasmid Digest size & calculating fragment Teaching Assistant - Ask /Go over Microarray Slide need to know how - Do we law formula Fall 2022 to use Brogg's concept of it or just the  Lab Math in 100 mL 1% (weight/volume or w/v) Ig = Making solutions from stock c1V1=c2V2 1 % (volume/volume v/V) or > ImL in 100 mL = - % solutions, w/v (1g  1mL) and v/v X dilutions (1X, 2X etc) Molarity: 3 7. ways concentration are expressed. calculating ame needed for desired concentration - mass 1 mole * = 6. 02x1023 particles Q : You want to make 150 mL of a 25 mM Soln of EDTA Given the MW. or EDTA is 292 glmol How much do. need ? A : 1 095g of EDTA you. Q : You need 240 mL of a Ix solution Given you have. a stock of 15X how would you make this ? A : 16m2 15X + 224 mL Practice Problems: Complete the practice problems below before beginning the lab exercise 1. Describe how you would prepare a 1- L aqueous solution of each of the following: a) 0.5 M glucose (M.W. 180.2) (180.2 g/mol)(0.5M glucose)(1L) = 90.1g glucose + H20 up to 1L c) 100 mM glucose (M.W. 180.2) (180.2 g/mol)(100mM)(M/1000mM)(1L) = 18.02g glucose + H20 up to 1L 2. Describe how you would prepare just 10 ml of each of the solutions below. a) 1 M glycine (M.W. 75.07) (75.07 g/mol)(1M)(10ml)(L/1000ml) = 0.75g + H20 up to 10ml b) 0.5 M glucose (M.W. 180.2) (180.2 g/mol)(0.5M glucose)(10ml)(L/1000ml) = 0.9g + H20 up to 10ml 3. If you mix 1 ml of the 1 M glycine solution in problem 2a with 9 ml of water, what is the final concentration of glycine in mM? (1M)(1ml) = (x)(10ml) x = 0.1M (1000mM/M) so x = 100mM 4. If you mix 1 ml of the 1 M glycine solution in problem 2a with 22 ml of water, what is the final concentration of glycine in mM? - ligman Q If I want to make 500mL : (1M)(1ml) = (x)(23ml) (Recipe 0 1% Coomassie Stain :. Coomassie Brilliant Blue , 40% x =.04M (1000mM/M) so x = 43.5mM Ethanol , 15% Acetic Acid) How much of each do you need given these reagents: Sidcoomassebrilliana A:0. 59 Coomassie + 200 mL Ethanol 5. Calculate the concentrations below in mM and mM. ap (a) 10 mg of glucose (M.W. 180.2) per 100 ml water - -- (180.2 g/mol)(x)(100ml)(L/1000ml) = (10mg)(g/1000mg) x = 0.000555M = 0.555mM = 555uM (b) 100 ml of a 2% alanine solution (M.W. 89.09) (89.09 g/mol)(x)(100ml)(L/1000ml) = 2g x = 0.22M = 220mM = 2.2x10^5 uM · 6. You have a 0.5 M stock solution of phosphate buffer. You need a working concentration of 20 mM. Describe how to make 200 ml of 20 mM phosphate buffer using the 0.5 M stock solution. - (0.5M)(x) = (200ml)(20mM)(M/1000mM) ↑ x = 8ml stock + 192ml H20 7. You need a 50% w/v glucose (M.W. 180.2) solution. Describe how you would make 1 L of this solution. 500 g of Glucose dissolved in 1L of H2O. 8. You bought a concentrated sports drink at 14X. How would you make 2L of 1X drink for your basketball team? Equation to use is C1V1=C2V2 (14x) x (V1) = (1X) x (2L) V1 = 0.1428 L = 142.8 ml 142.8 ml of sport drink should dissolve in 1857.2 ml of H2O. 9. You have 250 ml of a 10% NaCl solution (F.W. 58.4 ). What is the molarity (M) of the solution? Amount of NaCl available in 250 ml of a 10% NaCl solution = (10 g/100 ml) x (250 ml) = 25 g (250 ml) x (L/1000 ml) x ( M ) x ( 58.4 g/mol) = 25 g Solve the M then you get = 1.7 M [ M = (mol/L) ] 10. You have 250 ml of a 50 mM solution of NaCl (F.W. 58.4 ). Express the solution as a %. Solve the g amount first, (250 ml) x (L/1000 ml) x ( 50 mM ) x ( M/ 1000 mM) x (58.4 g/mol) = X g Solve X then you get = 0.73 g Now the percentage = (0.73 g/ 250 ml) x 100 ml = 0.29% 11. A packet of saccharine (F.W. 205.17) weighs 1.2 g. You like to add 1 packet per 100 ml of coffee. What is the molarity of the saccharine in the coffee you are drinking? (100 ml) x (L/1000 ml) x (X) x (205.17 g/mol) = 1.2 g X = 0.058 M  Pipetting Q : How is the minimum range determined? on a pipettor - 10 % of the maximum range Pipettor Theoretical Recommended Range (l) Range (l) P10 0-10 1-10 P20 0-20 2-20 P200 0-200 20-200 P1000 0-1000 100-1000  Plasmid Purification – Reaction Components Solution 1: Solution 2, lysis buffer: What is the purpose of EDTA? enzymeactivities SDS (detergent) – Chelate divalent metal cations so – Disrupt cell membrane (amphipatic detergent, interacts with and disrupts they are not available to DNAses as amphiphatic phospholipid bilayer of cofactors. Otherwise, they would membrane destroy the plasmid DNA NaOH Glucose – Denatures chromosomal & plasmid – Prevent premature instant cell lysis by DNA osmotic shock EDTA needs time to work * Solution 3: before cells are lysed Alkaline pH (Tris pH=8) Sodium Acetate (pH=5.5) reduces electrostatic interactions – Neutralize the lysis -n buffer between DNA and scaffolding proteins – Basepairing of the plasmid DNA is If solution I is missing EDTA what will happen to your plasmid? Q: restored the divalent cations– Precipitates the , A it risks : getting chopped by DWAse as are intact soDNAse is active, proteins/lipids/chromosomal DNA out (white , chunky stuff) chromosal DNA partially renatures; plasmid DNA - enatures  Q : If you have a linear band plasmid Plasmid Purification in your get after ↑ small DNA circular entered pieces into of bacteria A The prepwnetdoesthata b Why does plasmid DNA renature fully, Why TE/RNAse done bit too preparation was : but chromosomal does not? vigorously or a plasmid DWA ar... – DNA more stable in TE buffer (pipetting – Topologically intertwined, smaller 00 vortexiga – RNAse to remove soluable RNA and – RNA will also stay in solution chromosomal creaved avoid smears in gel DNA aneed ! the Why 90% Ethanol? Agarose Gel plasmich bromophenol blue – Precipitates DNA: it’s less polar than – Loading buffer: tracking dye, water and allows phosphate glycerol: higher density, makes DNA backbone to interact with sodium stay in bottom of the wells ions. Q Is chromosomal : DNA pelleted addition of after addition of 95 % ethanol ? So I'n III or – DNA is negatively charged (in pH=7), Why 70% Ethanol? A : ethanol After addition of SIII, the 95% precipates the plasmid. migrates towards positive pole – For washing: enough to prevent DNA – Separation by size, charge and shape from going back in solution, but allow – The higher the agarose %, the higher washing off excess salt the separation power (resolution) May sure its dried: Excessethanol will interefore I - with enzyme activity. =lineara -circular  Agarose Gel Electrophoresis Marker Which DNA fragments Linear run faster? I – Smaller - mutagen Supercoiled and DNAing & How does Ethidium - coiled Bromide/GelGreen - work? under bluelight Plasmid size: 2686 bp – Fluorescent dye, visible in UV, intercalates between base pairs DNA Why is GelGreen safer the EtBr? not mutagen ↳ a – Cannot pass through cell membrane inence RNA skin) not completely digested by RNAase  Restriction Enzymes · blunt out : enzymes Cleaves terminal residues from linear DNA makes straightwit thru – Exonuclease DNA strands , in same place & sticky cut enzymes · Cleaves internal residues makes : – Endonuclease overhang do staggered cuts not directly , pages unpar across from each other Recognizes the same sequence & cleaves at the same site – Isoschizomer Recognizes the same sequence & cleaves at different sites – Neoschizomer digest Off-site cleavage due to non-standard conditions – Star activity bendonucleases causes : glycerol too much enzyme ,lowionic Strength (low salt) , , , DMSOethanAnts ↑plt presenceof  Restriction Digest 0 Enzyme 1 – Cleaves at 500bp site Enzyme 2 – Cleaves at 1000bp site Plasmid X 2,000 bp - 500 Digest with Enzyme 1 – Fragment # ↑2 – Fragment size(s) , 000 pp Digest with Enzyme 1 + Enzyme 2 I – Fragment # 2 1,000 – Fragment size(s) 1 ,500 bp – Incomplete digest fragment # 500 &  ⑪ I , 000 5 bp A - 2118bp 3 2118 1820 1062 , , 2202 # 2 3150 , 1828 Plasmid Digest Practice Directions: Be sure to review the background materials and videos on plasmids, restrictions digests, and agarose gels before attempting the assignment below. Part I. Calculating band size from a digest and agarose gel. D Figure 1. pBLU plasmid from Carolina Biological Supply. This plasmid has an ampicillin resistance selectable marker and also codes for beta-galactosidase which is useful in blue/white screening of bacterial colonies. 1. What size band(s) would you expect if you digested the pBLU plasmid with BamHI? 5437BP 2. What size band(s) would you expect if you digested the pBLU plasmid with XmnI and Nde I? 3. Sketch in the bands you would see in the digests in questions 1 and 2 above in the appropriate lanes in the gel picture below. Uncut pBLU plasmid has been illustrated in the first gel lane. A molecular weight ladder has been provided in the lane on the far right.  pUC19 2,686 bp Figure 2. pUC19 plasmid. The polylinker cloning site (sometimes referred to as the multiple cloning site or MCS) has many restriction sites engineered in to allow researchers flexibility in selecting which enzymes to use to clone in their genes of interest. Researchers must select restriction enzymes that do not cut their gene of interest or at other sites in the plasmids. Enzymes recognition sites that are present in the polylinker site (MCS) are never found elsewhere in the plasmid. The polylinker site for puc19 is located at base pairs 396-454. The region is expanded below the plasmid illustration itself. 4. How many band(s) would you get on a gel if you cut pUC19 with PstI? 5. How many band(s) would you get on a gel if you cut pUC19 with PvuI? 6. How many band(s) would you get on a gel if you cut pUC19 with BamHI and PstI? 7. Based on your calculations above (questions 4-6), sketch in the bands you would see if you ran these digests out on an agarose gel. Assume all digests go to completion unless otherwise indicated in the lane labels. 8. What size bands would you get if you digested the Lambda phage DNA illustrated in Figure 3 with BamHI. The total length of the virus is 48,502 base pairs. Assume your reaction goes to completion. 9. What size bands would you get if you digested the Lambda phage DNA illustrated in Figure 3 with BamHI and the digest did not go to completion?  3 eciem e site indust multiple cloning · w/5369DP - multiple used restrictionwhen a W 2 fragments mid No 4117 & - stacked ismeanin McS times -  ① cutw/ only NDEL · linearize the plasmid & have 6538 bp BamHI (435 bp) & / Cut j BAMHI &NDEL 7 w/ · have 3 band fragments - ③ 01062 - 435 = 627pp - 3075-1062 = 2013 up full size of PDP2 : 6538 ③ A Sumupotherfragments: pDP - BamHI (1 , 002) Jo Ne · BamHI (435 bp) 23130- doesn't go to 898p j - completion 0e 9416- linear line we see may extra bands linear - - of extra bands- star activity 54 pDP2-BamHIl 4341 - inarris 2322 + 2027 -  resolving than PAGE ver Polyacrylamide gels - his ower the gel well made if you run marker as M. W. estimates of protein are a on for Used when resolving proteins or Coomassie Brilliant Blue: non- OG wool DNA fragments of nearly the same specifically stains all proteins size (in your case serum Proteins!) Just the M.W. does not proof the -uncolonized Made out of crosslinked Ineurotoxin! purified protein’s identity Especific polymerized Acrylamide and bis- - Western Blot: Proteins from gel are Acrylamide crosslinking forms get that proteins electrophoretically transferred to migrate Native (size, shape, charge, size thru membrane, membrane is blocked to cannot be determined) vs. SDS prevent unspecific binding, protein - PAGE (deantured, equally charged, of interest is identified through - only based on M.W.) antibody binding coupled to - Stacking vs. Resolving gelCdenser) colorimetric - (etc.) labeling film needed glycine trisglycine X-ray no so protins j - -creates wells separate by sample loaded can be moving to (t) end  Polyacrylamide Gel Electrophoresis Purpose of SDS(detergent – Coat proteins with uniform negative charge, disrupt native structure & makes proteins to Migrate based on size, not charge linear form go -added w/SDS Purpose of ß-Mercaptoethanol -Reduction (“break”) of disulfide bridges Q If : you're trying to see a desired protein & you see not Coomassie vs Western Blot corresponding band with Coomassie stain but Western blot , is it your protein ? – Which is more specific? Why? A : No Western blot – antibodies directed to a specific protein of interest Cantibody recognition site called a epitope)  Classic Glass slide Microarray * * ** * Affymetrix type Gene chip each poel has 40x10] To compare gene expression, RNA is extracted 5µm copies of a specific probe from sample and transcribed to cDNA.. 5µm Millions of identical probes / feature 5” taggedu) probestroy 1.28cm Up to I ~6,500,000 # probe e st ke emit features / chip lightthed ↓Pairssir. match perfectmismatch + 1.28cm probea Affymetrix Educator Resources gene Chips are synthesized onto silicate glass surface using photolithographic synthesis to create different probes for each individual feature simultaneously Photolithographic synthesis Scangene Each feature contains probes specific to one gene of the organism reatures of · each probe cDNA is transcribed to Biotin labeled RNA -> visualized with Biotin- binding Fluorophore Two compare 2 samples, each will be added to a separate chip and http://www.vetmed.iastate.edu/faculty_staff/users/phillips/Micro402/17-genomics/microarray.htm intensity of fluorescence is compared for each feature Uses of GeneChip Microstray food > testing - tenvironmental testing > agricultural biotech - livestock diagnostics or grading > - identity testing > - > individualized medicine - > - basic research - human diagnostics Q : What if someone forgets to use a photomask ? get will All side group caps - removed , not just the ones you want data, you notice agene Q : After analyzing your microarray a butnoexpression n thathasexpression inthecontrol was eliminated it as the gene's expression Yes, you keep - conditions. due to your experiment's  crystallized add the Q : What if someone forgets to we vacuum grease to the well when placing the coverslip ? A : the rate the in drop dry crystal of evaporation will increase, out and leading lattice. to loss of stability letting X-Ray Crystallography - ordered cysozyme y usedstructures to resolve need multiple Q which of the secrystal structures has % of , Izit-stain protein crystals : ? A A 8 you lattices : the best resolution blue (can enter protein crystal. - What is diffraction data larger protein Max Von Laue - lattice be not compact) converted into? – Showed X-rays were a form of – Electron density map electromagnetic radiation Process protein preparation & analysis > protein : crystallisation -- – Crystals are periodic arrangement - diffraction data measurement > - phase -> model building & estimation structure refinement of atoms Bragg’s Law -used a beam of X-rays containing – Angle of light scattering multiple wavelengths – William Henry & William Lawrence Role of PEG Bragg – Crowding agent to aid in the assembly of the protein into the lattice Hanging drop method · Role of Acetate Buffer : help maintain pla protein Ionic Since strength the reservoir of is &the buffer higher in thethan ionic strength droplet the drop, , Role of NaCl Stability from the drop will move to reservoir. isliquid higher or lower than the reservoir? – Keeps the protein soluble , preventinga -lower! resolving power (1 1A resolution). · For crustallization to happen : protein must good ! be pure soluble highly , conc monomeric &., or Stable , single species  Bonus Question A method to solve protein structures requiring a beam of x- rays resulting in multiple solutions to a set of equations was developed by…? – The Braggs!! i fe ci o range a blwatice ↑ 2dsint = order obtray no ↳ b/w incident very & lattice plane  Q During : seasonal changes plant's , leaves H2O & LO2 produce sugars /carbs) orange/yellow why ? turn shades of Photosynthesis (light E-> chemical E by Chloroplasts using to Molecular structure of Chlorophyll , · from A the plants pull back the Chlorophylls - : 7 Myabsorblightinbueresensmit the leaves for energy conservation leaving Chlorophylla methylgroup , · : > (CH2O) + n1 + H20 Carbon fixation chlorophyll b only the carotenoids. nHzD + nCO2 aldehyde group n · greenlight,explaining why plants look green. will lacking plastocyanin - : that is , Q If : I have a plant system reactions ? - NADPH and ATP be produced Carotenoids To - for NADPH A : Yes for ATP , no - water as occur in add DCIP, as - · round in all photosynthetic organisms the Hill rxn and forgot to an e-donor e-acceptor Q : 16 you were doing Stroma Chloroplast Cp coor thylakoid membrane whatwouldhappenbewudremingreen · located in the measured - 2 processes (inchloroplasts) - synthesis of sugar · light absorbed is Chlorophyll ; accessory protein transferred to ① light-dependent,thylakoid reactionstill reagent · NADPiS + · protection from light damage Hills An-chloroplasts [it20 + Ax Al20 + 02] light must be collected to be used > - - · redox rxn - Antenna complex group pigment mics that cooperate : a of anor. · acceptor transfer it to a rxn center complex. light energy & e-acceptor to absorb · H20 +PENADPHH* + 02 ~NADP + reduced to NADPH chlorophylls & carotenoids are part of the complex. · chioroplasts form ATP from ADP and Pilinorganic phosphate) in light's presence, in known as photo-phosphorylation. where in the Chloroplast are the Photosystems located ? - a process -generation of ATP in the chloroplast : protons are pumped through a channel · Cytochrome bef : evenly distributed throughout In our experimen t... in ATP synthase (Appase] from the lumen the stacked grana & the stroma lamellat to the stroma : 3 micIs ATP for each enzyme rotation located in · : diffusible e-carrier Plastocyanin - DCMU (Diuron) : an inhibitor the thylakoid lumen blw PSI that blocks the interaction produced · Plastoquinone : diffusible e-carrier located in NADPH and AtP provide the energy for the "dark reactions" in which & plastoquinone · the thylakoid membrane CO2 and H20 are converted to carbohydrates - DCIP/DPID (Hill reagent) : acts as an e-acceptor - blue when oxidized ② light-independent Calvin Cycle reduced , clear when In the lab - : H ↓ - enzymes : speed up hydroxylamine HO - N - H rateofan was theinhibitor. it binds to active site Kinetics - to measure "We use an colorimetric in heme-containing activity essay : assays allows you to follow enzymes. k1 E  S  ES  E  P changes based on color k2 measured by a spec. k -1 turnip Peroxidase neutral i a an ide D enzyme 𝐷𝑜𝑛𝑜𝑟 𝐻2 + 𝐻2𝑂2 → 𝐷𝑜𝑛𝑜𝑟 + 2𝐻2𝑂 ↳4 Guiacol + 2H2O2 - - ~ → 𝑇𝑒𝑡𝑟𝑎𝑔𝑢𝑖𝑎𝑐𝑜𝑙 - + 8 𝐻2𝑂 ↓ substrate brown product reducing agent in the conversion of hydrogen peroxide >water- It202 is reduced , while quiacol is oxidized , yielding a brown product called tetraguaico (X 470nm) =  Kinetics ↑ [s] ↑ Vmax k1 E  S  ES  E  P k2 k -1 Vmax S vo  K M  S The Michaelis - Menten equation T Vmax estimated v0 : Initial velocity (linear incline of reaction speed) measured for multiple Substrate concentrations. The Km is the substrate concentration where vo equals one- Umax : max velocity where the [S] is saturating. half Vmax or 50 % Umax All active sites are filled w/ substrated the in can't go any faster  The Double Reciprocal Plot or used to determine Lineweaver-Burk Plot exact values of Vmax 1  KM  1 1 1Vo max     Slope = vo  Vmax  S Vmax - Y intercept : 1/ Vmax Compare to y= mx+ b - Im xintercepto. special case : Umaxt-noncompetitive : There are three types of Kmax inhibitor binds to [E] & [ES] equal · inhibition kinetics: unchanged affinity. Intersect on graph ishappeningone St E P+ E E + > - ⑧ - [I]4 Vodkm2 Vmax unchanged Competitive % E+ SES > - P +E Umaxk ↑ Mixed km &/d E - E + S- ES = P+ E - ↑ X - - t (ESI) [EI] Uncompetitive [I]t Va↓ Kid Q : 1 you add an [I] to an enzyme & both Kmd & Vmax it is , uncompetitive inhibition Qiwatinn substrate (A) with your enzyme (X) Q : You mix 120 Ml of the date point you'd what 257 is get an Almin of 0 , motor. to that the the line-weaver-burke plot given use for M- cmt product is 32 9. coefficient of. absorption MW of (A) is 78g/mol (A) stock Path length is 1cm , , MW of (X) is 87 KDa , concentration was 8. 5 mg/d) 50mM , final volume , Cinal (x) Stock concentration was was 5mL? Answer 588M +; / = 0 0256 Mmol "mi 1/ (s) = :.