Lab 2- Peroxidase Extraction and Purification From Horseradish Roots, Summaries of Biology

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Lab 2: Protein Purification-PEROXIDASE Wednesday, September 7, 2022 PM LAB2 PEROXIDASE EXTRACTION AND PURIFICATION FROM HORSERADISH (Armoracia rusticana) ROOTS none dna To. Rot sre ote A\-22 Dear Student: If you're typing directly into this document, please use BLUE font color for anything you type so | can tell it apart from what | have written. INTRODUCTION Protein purification from biological tissue isa critically important skill that any lab biologist should be familiar with. In this lab exercise, we will earn about how biologists purify proteins for future use, Among the future uses could be chemical analysis, medical uses and others. We will try to purify a specific protein called peroxidase. The enzyme we use in this lab will be used by us in lab 3 and we will also investigate this enzyme in our IRPs, so this is a very important enzyme to learn about, We will purify this pratein from roots of horseradish. The structure of this enzyme is shown below as both a surface model and a ribbon model. As you can see, there are many alpha helix structures in this protein. In addition, the active site of the enzyme is dominated by an iron heme group. The protein is slighthy negative at cellular pH, 1< ay Active Site with Heme Group Plant peroxidases chemically reduce harmful H,O, and other similar reactive molecules to water by oxidizing some organic substrate, H.02 and other similar reactive molecules are very harmful to biological tissues because they oxidize important biological molecules (they cause oxidative stress). So peroxidase is an important stress enzyme that plants use to cope with the oxidative stress caused by H.0:, Without this enzyme, plants (and even animals) would have an abnormal buildup of H2O:, which could lead ta some damaging results for the cells. One way to see this reaction is to react H.O2 with a synthetic chemical called guaiacol. The product, called tetraguaiacol is an amber color and can easily be detected using a spectrophotometer Lab 2-1 2M on, ~OCHs — aC + any = _) + sno Cr 2 guaiacol peroxide ~| water tetraguaiacol Knowing the reaction above will be very useful toward understanding complete the procedures of this, lab. ‘The goal of this lab is to familiarize you with the techniques of protein purification and analysis. This lab will be conducted over a three-week period. First, we will start with a crude protein extraction and purification to separate prateins fram other macromolecules. In the second lab meeting, we will specifically purify LDH from other proteins. Finally, in the third lab, we'll analyze how well we did. 2A. CRUDE TOTAL PROTEIN EXTRACTION AND PARTIAL PURIFICATION BY SALTING OUT In 2A, you will extract and partially purify all proteins from root tissue of horseradish plant, The goal is to extract proteins fram the cells and tissues of the root and to separate proteins from other macromolecules, such as carbohydrates and nucleic acids. With most proteins, purifications must be done under cold conditions to prevent denaturation. The advantage of working with peroxidase is that it is heat-stable, so we can work at room temperature for lang periods, but should be stored in a frozen state. PROCEDURE Reagents and Materials: 1. Horseradish roots 11, Pipet tips 2. Homogenization buffer -> 0.1 M phosphate 12. Weigh boats buffer, pH 7.0. This buffer helps control the 13, 100 ml beaker (2/erp) pre chilled pH 14, 250 ml beaker (2/erp) pre chilled 3. 25 mM phosphate buffer, pH 7.5 45. 125 mi e flask (1/grp) pre chilled 4, Ammonium sulfate > salt used to 16. Small petri dish (1/grp) pre chilled precipitate proteins 17. Stir plates (1/erp) 5. Cheeseclath > used to filter homogenized 18, Stir bars (1/erp) chicken to remove large unhomogenized 19. Ice chunks and to remove lipids 20. Sharpies 6. 50 mi centrifuge tubes pre chilled 21, 50 mi graduated cylinder (1/grp) 7. Blender > for homogenization 22. Ultracentrifuge at 5°C 8. Desalting column 23. Ring stand with clamps (1/erp) 9. Microfuge tubes 10. Disposable pipets Lab 2-2 dissolved proteins. To do this, we will add solid 1350. to our centrifuged supernatant slowly until we reach a salt saturation level of 80% (0.57 & (NH,).S0, per mi of filtrate) a. Knowing the volume of the supernatant, and knowing that you need 0.57 g of (NHa)2S0: per ml of filtrate, obtain the total amount of (NHs),SO. that you'll need and write down this amount: ‘Amt. of (NHa),SOs needed: \3.2 b. Slowly (over a period of 15 min) add 0.57 grams of ammonium sulfate per 1 ml of your centrifuged protein solution. It is best to perform this step in a chilled beaker on a magnetic stirrer. Place your sample into a beaker. Use a small magnetic stir bar to keep things mixed up. Avoid stirring too violently because this could shear (tear up) the proteins. If you see too many large bubbles forming then you are shearing your proteins. d. Stir for an additional 15 min after adding the ammonium sulfate to give the salt a chance to dissolve 11. Centrifugation — Centrifuge the sample as before in a pre-chilled balanced centrifuge. At the end, pour the supernatant into a separate 50-ml blue-capped sample tube labeled “SUPER” + GRP# and section..etc, Give this tube to your instructor for freezing. 12. Keep the pellet, consisting of precipitated proteins, in the centrifuge tube, QUESTIONS 1. Explain the purpose of why we salted out the proteins. Why was it done and how does it work? Sobing owt Je demokuring Hex prohiins , sensing, “Ho slung rn aay V sqending CE as ywrel Cen les Tnoreatiny 40 concintvation ef-s vnckis oaksa proteins lew soluble 2. Why do we need to remove the (NH),SO. alt before we go on to future steps? Briefly explain how the salt is removed from the protein. Ammonium ake binds Te Bre war ravlimles, Arce te frotine Yume in Witt mw sonbin inctvonls , decreasing, its Solubilthy, oma forwing, Percagitedion Ged Filtvation: a Arve Onwam oo - qplg wad tp parade dal foramt humiels bared om wT Desalting the Protein Sample Next, we must remove the ammonium sulfate salt fram the protein pellet. High concentrations of salt, such as ammonium sulfate, can interfere with subsequent protein purification steps so removing this ar Lab 2-4 salt is necessary. To remove the salt, we will use a chromatography technique called gel filtration chromatography. Gel filtration chromatography separates different chemicals by their tifferent sizes. The process of gel filtration chromatography employs a column (see figure below] that contains a buffer called the mobile phase and a semi solid (usually) material called the stationary phase. The sample to be desalted is placed in the column sample reservoir and gravity is used to pull the sample through the first frit (or filter] and into the column. Since the stationary phase consists of beads with small pores in them, the salt, which is a small molecule compared to the proteins, enters the pores and therefore spends more time in the beads. The larger proteins do not pass through the pores and simply go around the beads, Therefore, the larger proteins move faster through the column than the smaller salt ions. At the bottom of the column, the proteins come out (elute) first followed by the salt. This results in separation of the proteins from the salt. Once the sample has been desalted, the proteins can now be subjected to another kind of chromatography called ion exchange chromatography, which we'll begin on the second day of the procedure, Protein + Salt Solution >) oc [exe Gel Beads (e) (e) e) ie) ‘e) Column OOOO Oo ( 2 te) (ore) ‘ ie) OO =s SQ0 BSE Column bed Time 1 Resuspend ammonium sulfate pellet — add 2 ml of 25) BROsphate Buffel to the ammonium sulfate pellet. Gently mix the buffer and the solid material until the pellet dissolves. Keep on ice as much as possible during the procedure. Check the volume of the mixture to be sure that itis not more than 3 ml. Ifnecessary, add buffer so that the valume is 3 ml 2. Be sure that all the buffer used in storing the columns has eluted out of the column. Load the desalting column — load 3 ml of the mixture an the desalting column. Allow the liquid to drain to the frit (the plastic cover on the column resin). The column will only drain if there is pressure by fluid above the frit. When you load your sample into the sample reservoir this creates pressure an the fluid (buffer) inside the column. The result is that the buffer elutes out of the column. Discard the flow thraugh, which is mostly buffer. At cellular pH, peroxidase is a weakly anionic protein (weakly -_ negative) and so we should use an anion exchange column : {positive}, The column we'll use is DEAE-sephacel: positive DEAE |X ou;cu, {diethylaminoethy!) covalently attached ta cellulose resin (see Iceitulose i figure to right) DEAE-Sephacel Bead The negative proteins will fiizeuce ne gear cet] (0-300 mM). As the buffers are passed through the column, we will collect 5 ions in test tubes. This ensures that we separate different proteins from each other. To know that an eluent (the stuff coming out of the column) is free of proteins, we will measure the absorbance of each Traction at 280 Hsnjx nm, a wavelength at which ALL proteins absorb light. Your group will divide itself into two sub-groups. Team A will perform all ion-exchange chromatography procedures and team B will assay all samples for total protein concentration. PROCEDURE Jon-Exchange Chromatography Reagents and Materi UV/Vis spectrophotometer Plastic cuvettes Microfuge tubes Blue capped plastic test tubes (20/erp} Microfuge racks Test tube racks for blue-capped tubes 50 ml falcon tubes (1/erp) Sharpies ee Ring stand with clamps (1/erp) 100 ml graduated cylinder (1/grp) Pipettors + tips lon-exchange (sephacel} chromatography columns (equilibrated) 1. You are now ready ta further purify your sample through ion-exchange chromatography. Obtain an ion exchange chromatography column and secure it to the ring stand. What you want to do is put the thawed and mixed desalted sample on the column and then subsequently remove: a, Allthe proteins that do not bind to the column b. And then all the proteins that bind weakly to the column And finally, all the proteins that bind strongly to the column 2. Label about 15-20 blue-capped plastic test tubes using label tape. Place the following information on the labels: sequential numbers {from 1 to 15), your group number, and your lab day 3. Let the storage buffer that is above the column frit flow out into a waste container and then discard it. Your instructor may have already done this for all the groups in the class in order to save time. 4, Place tube #1 in a plastic beaker containing ice below the column, Lab 2-7 10. a. L 13. 1 5 Next, Load the complete desalted protein solution onto the sephacel column. Begin to collect a5 ml fraction in tube #1, Its likely that tube #1 will not reach 5 ml with only your desalted sample. Therefore, once ALL of the desalted sample has flowed through past the top frit you can begin adding a 5 ml of 0 mM KC! buffer wash to the column. Once tube # 1 reaches the 5 ml mark, remove it from beneath the column and replace it with tube 2. Keep collecting 5 ml fractions in tube # 2 and all subsequent tubes. While collecting the fractions, ‘one member of the team can invert tube #1 to mix it and measure the absorbance of tube # 1 at 280 nm. Since proteins, absorb at 280 nm you can determine if proteins are present in the flow-through by measuring the absorbance of the fractions. Once the absorbance of tube #1 (F1} is measured, pour the contents of the cuvette back into the FL blue-capped tube and store this tube in a large plastic beaker filled with ice. Itis likely that the absorbance of Fi will be high because all un-bound proteins have flowed into it. The 0 mM KCI buffer will push out all these neutral or positively charged proteins. Now team B can assay the total protein concentration in fraction 1. SEE THE PROCEDURE STARTING ON PAGE 11. The protein assays can be an ongoing process as each fraction is collected (you don’t need to wait until all the fractions are collected to assay total protein concentration), Keep collecting 5 ml fractions and keep washing with 5 ml of buffer until the absorbance of the fraction reaches below 0.1, indicating that no more proteins are flowing out of the column. Once the absorbance of a fraction goes below 0.1 and once there is no liquid above the top frit (everything has drain through the column}, you can add 10 ml of the first KCI salt solution, the SO mM KCI. Continue collecting 5 ml fractions below the column, You must indicate (in the table on the next page) which fraction was in place when you first started adding the 50 mM KCI. KCI will disladge weakly-binding proteins which will elute from the column. This salt concentration may be sufficient to dislodge the peroxidase fram the column, You should start to see the A280 values go up because proteins are now eluting from the column. If the A280 of the last fraction is still high, add another S ml of 50 mM KCI. Keep doing this until the A280 is less than 0.1 Repeat step 11 with all remaining KCI solutions (150 mM and 250 mM). Keep collection 5 ml fraction. Once the A280 of the last fraction from the 250 mM wash goes below 0.1, you are done with the ion-exchange procedure. Remember to indicate on the table which fraction first received ‘the 150 mM KCl and the 250 mM KCl. At the end, remove all the fraction tubes from the ice bath in which you are storing them, place them in blue wire rack and give this rack to your instructor to freeze. Record the results in the table below and also prepare a bar graph of your results with fraction number on the X-axis and absorbance at 280 nm on the Y-axis. Mark which fraction corresponds to which step in the procedure (50, 150, 250 mM KC}) Lab 2-8 Questions L._List the fractions (by fraction number) that showed up as absorbance peaks from the previous graph and carefully explain the causes of the high absorbance values for each peak (you must describe what class of protein makes up each peak and what elution buffer is responsible for eluting that class of protein}. con 2 VN I Kare Hee be pre of Lresiam Sans Loney eee NES, wee {wopor absevbamce =hi gh aman ¢ peer) by- by vile Wu, FLA * de aww, Fl\ Ran FES kan peetow which Ge-Samphe four SPAM sbmpl6s} ao you think will contain the highest overall protein Fully explain why you think these will contain the highest protein concentration? Fram oM the bectwus Evendinye, crude som Te hae HL bs probes Which sample do you think will have the highest peroxidase purity? Fully explain f\s per callecteA Atm, Fractions 2 and 3 Will have Oe Ka grest aw sunt t pers x AR pasty The olosov bance these samples gee ms te be He Bi phat iwhiceliny, ae Barbar Aine + protes shina Wi your errr. | ackivit. ‘ad 0 Protein Concentration of the Samples and Fractions In order to determine how well we purified peroxidase, we need to know something about the amount of peroxidase relative to all proteins in each sample. A very pure sample will contain a high concentration of peroxidase compared to other non-peroxidase proteins, To determine a sample’s “purity” we have to determine that sample’s specific activity of that protein. Specific activity is the activity of your target protein divided by the total protein concentration of a sample. Thus, itis clear we need to measure two things for each sample that we obtained: (1) the total protein cancentration and Lab 2-11 {2) the activity of LDH. In the next procedure, you will determine the total protein concentration of all the samples that you obtained. Next week we will determine the activity of the proteins in each sample. The method that we will use for determining protein concentrations is called the Bradford Assay, which you learned about in lab 1. Remember, the Bradford reagent binds to any and all proteins indiscriminately. When the Bradford reagent binds to a protein, the resulting product's color is blue and its absorbance has a maximum at 595 nm. The absorbance level depends an the pratein concentration. Protein + Bradford Reagent & PB Complex (clear) (brown) (blue) To measure the amount of protein in your unknown samples, you first need to calibrate the change in Bradford reagent absorbance induced by different amounts of protein. You already did this in lab 1 when you determined the extinction coefficient of BSA bound to the Bradford Reagent (remember?) Question In your own words, briefly explain the theory behind how we will determine the total protein concentration in our samples. Five Lwe mil ewan Ross obsor bance of each gample by setting Fea agectophetometin U4 S45 nen. Then, find Ahn anew nye of Hs opp corficest extinction. Nut wert Beon-Lambot wipalecte tte cs A : am Cgpation, ynanipalecte é Be ine pres Sawegle . Plage wn Ten dake £ Go te pol pretes concemty ak on. PROCEDURE DETERMINING THE PROTEIN CONCENTRATIONS OF YOUR SAMPLES 1. Once a § ml fraction is collected (from the previous IEC procedure), mix that fraction by inverting 3 times 2, Obtain 4-5 cuvettes and use these to perform the Bradford Assay. (A used cuvette can be reused by discarding the Bradford Reagent/protein mixture into the waste container in the hood, rinsing with isopropyl alcohol and then rinsing with water.) 3, Make a blank cuvette by adding 0.25 ml D.1, water to the cuvette and adding 1 ml of Bradford Reagent. Cover the cuvette with parafilm and invert. Wait 5 minutes and then invert the cuvette again invert to mix and then blank the spectraphotameter at S95 nm (the absorbance maximum of the Bradforc-Protein complex) with this sample. Lab 2-12 Total Protein Concentration (ug/ml) 7. Prepare a bar graph of your results with the sample identifier on the X-axis and the protein concentration (as wg/ml) on the Y-axis {It's OK to connect these data points with lines because there is na mathematical relationship between the X and ¥ variables). Protein Concentration in Sample ID(ug/ml) 250.0 200.0 150.0 100.0 50.0 0.0 EERE NTELEEELeAN As 4Henae Sgeeererecetr fee ik sake a SAMPLE ID Lab 2 F20 Questions 1. Which sample contained the highest protein cnannto DESMA 9 Saw ple SDs 2, Does this result agree with what you may have predicted back on page 11? Fully explain. No, Hew new deere “ye wi wy di chen From ny wnhnk nie tHe My Vesults ts het © - ee aa +o han fof ™ 7 a | sepeeted eneratteg du prsttins cameron tae eat tis dob tonal Lab 2-15 This measure of activity (increase in Once activity is determined, we can then determine a relative activity by dividing activity by the volume of enzyme used in the experiment (measured U/ml}. Aaiviy 7 Relative Activity — Activity +B ¥ Volume of Enzyme Used al Equation 5 Therefore (using the preceding definition of units): i we add 0.01 ml of an enzyme preparation, and find that the reaction proceeds at 10 pmol/min, the relative activity of the enzyme preparation is: 10j¢mol/ min _ 1000 jrmol/min _ L000U) 0.01 ml ml ml Equation 6 Itimately, the most important measure of purity is the specific ext. The ai and rele attr What we wantis a high activity ofthe enzyme without a lot teins around. To estimate this we use the concept of SPECIFIC ACTIVITY: ive Activ 7 Specific Activity — Relative Activi _ __Uiml U Tolal Protein Conecniration ag prowin# ml jag prouin Equation 7 which is the activity in units divided by the number of micrograms of total pratein in the sample. The number of milligrams of protein is a measure of tatal protein, active or inactive, enzyme or nan-enzyme, which is what you determined last week, As an example: Let's say you have three enzyme fractions, We take 200 ul of each and we assay for activity. We also take 0.5 ml of each and assay for total protein content. The results are as follows: Fraction Units (umol/min) | Protein Concentration (ug/ml) CH 100 3000 1 10 10 2 20 10 3 25 30 Lab 2-17 The specific activity of each fraction is calculated as follows: For fraction 1, we have 10 units that came from 200 pl or 0.2 ml, so we have 10U/0.2 ml = SOU/ml. Specific activity = U/ug = U.mi"/ug.ml* = 50/10 = 5 U/ua] For Fraction 2, the specific activity is 10 U/ug, for Fraction 3 it is 4.2 U/ug and for the CH. the specific activity is 0.17 U/ug. As you can see, even though the C.H. had the highest activity, it was not the purest sample. In fact, it was the least pure sample because it had the highest concentratian of total protein The purest fraction is Fraction 2. A useful thing to know is how pure your sampl: xd to the ung inthe CH the fold purification of Fraction 2 is oat = 60 fold purification Today you will measure the activity of the purification samples that you got in lab 2B. Next, you wil calculate the specifi les as a way of Question Based on your understanding of the procedure, predict which sample(s) you think will have the highest peroxidase activity. Which samples will have the highest peroxidase specific oa will aie be the same? Explain your ray os pa nae ete ee fod poe have meee a Avighusk sche advily No key arent some, Slee ryalen Se tee Staci te Ree ee You wll frst addveaction mixture to spectrophotometry lve subsample of your purification samples to these and measure the absorbances over time in your spectrophotometer. Details are below: 1. Prepare a reaction mixture by mixing the following into a 25 ml glass flask: 4.5 ml guaiacol + 5.5 ml H:02 + 10 ml sodium phosphate buffer. Cover the flask with parafilm and invert to mix. 2. Obtain as many small-volume spectrophotametry cuvettes as the number of samples you have. The samples you will analyze are the same samples for which you determined the total protein concentration in lab 2B. Label these tubes with tape using the appropriate name for the sample to be analyzed, Be careful not to place the tape in the path of the light. [DO NOT WRITE DIRECTLY ON ‘THE CUVETTES!) 3, Precut many small square pieces of parafilm. 4, Into each cuvette add 990 pl of the reaction mix. Lab 2-18 7. Next prepare a scatter plot of all the samples that seem to show positive slopes (that show there is a reaction taking place). Do not plot samples that show no reaction (the ones that have a negative slope or that do not have a consistent upward trend in absorbance aver time}. The graph should have time (in minutes) on the X-axis and absorbance on the Y-axis. Place the data for all samples on the same graph (one graph with multiple plots). Timefin mins) affecting Absorbance(280nm) we ago em per 5 G01 O57 Tine (i) Lab 2-20 8 Next, determine the slope of each plot from the graph you prepared on the previous page. Again, ONLY do this for samples that had a positive slope. This slope represents the initial reaction rate for each sample that you analyzed. Place this slope in the second column in the table below. 9. Next, determine the activity of peroxidase in units (U) for each sample you plotted by using the formula initial ren slope (abs min~>) [0.0266 mele Units U) x 10-2 = umole min’ ore} sem The value 0.0266 (222) 1 cm tis the molar extinction coefficient of tetraguaiacol at 480 nm and it allows us to convert changes in absorbance of tetraguaiacol to changes in amount of. ‘tetraguaiacol at that wavelength. Write the result in the table above. 10. Next, determine the relative activity (U/ml) of each sample by dividing by the volume of sample used: Aerivity (W) Relative Acttotey ~ Fo of Sample Und (OTH) And write the results in the table above. Be sure to write down all your calculations on an extra sheet of paper. 11. Next, determine the specific activity of the samples by dividing the relative activity by the concentration of protein in the sample that you determined in lab 2B (see table on page 12) Relative Activity (U/ml) u Total Corrected Prowwin Concentration (agiml) jag protein \Specitie Activity — 12, Finally, determine the fold purification of each sample, which is calculated as the specific activity of ‘a sample divided by the specific activity of the crude homogenate (so by definition, the fald purification of the CH will be 1). Write all the results inthe table on the next page. Lab 2-21 ‘Sample Slope of the line = Specific Initial Reaction Rate Relative Activity Fold 0. ZB¢SarsOpancans | AAW) haciary U/m) —(U/ug__| Purification A435) MMgsyay be way: 339 protein) Crude * 7s ~ i Cte SZ 1 Homegered 726 0.0017 O.WNE4! 0. dor! O2e ace Supernaténkes* ONTO (TAOTA 0-09 CTA WV Coz a 0-0 - . Desaltd Q2q04 9.00102 b.tetcy 0.05218 9.¢4887 F109 .0O47L 9 .D0018 [6.01774 0.40335 osiz3 720-0024 0.003% o.ca4es —O-fozS! 0. 46ers 0-0-9236 —0:0000174-D-© OB 7—0/0- 0246-043 0.01567 0.00057 0.059% 0.00231 0. 4zns F4O 0021S 0.000028 p. 09 %*T o.010027 0. O1g17 5 0-00107 0.000776 pd? 402 0.podhr 0.67795 VO UVSY— OSS Poo z OP ONT P-osHET 0.03712 9.00 p.1sasr 0.40109 0.36987 F7 9 OS068 Donte biesrh wales cn 0.0296 Fe