Question 2 (15 points)

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2 Question 2 (15 points) Your undergraduate TAs were not very successful at performing the Protein Biochemistry experiments during Run-Through Week. Describe how the following mistakes would affect the purification or analysis of β-galactosidase activity. (A complete answer will say what the correct reagent does or why it is important--and how changing it will affect the experiment.) a) Anita ran out of APTG-agarose, so she decided to make her own affinity column using Xgal-agarose. Anita wanted to use APTG-agarose on her affinity column because APTG is an nonhydrolyzable substrate analog of lactose. Therefore, B-gal will stick to APTG on the column until it is eluted. Xgal can be cleaved by B-gal, and so a column containing Xgal would not retain B-gal. Instead, B-gal would flowthrough the column (and the flowthrough would be blue!). b) Jean forgot to wash her nitrocellulose membrane between the addition of primary and secondary antibodies during the Western blot. Primary antibody addition is followed by a wash to remove antibody that is bound nonspecifically to the membrane (i.e. not bound to B-gal). If Jean adds secondary antibody without washing, she ll likely have purple bands/blotches all over her blot, and not just in the location of B-gal. This will make it difficult to identify which band is Bgal. c) Matt used lactose instead of ONPG when performing his β-galactosidase assays. ONPG is a substrate of B-gal that is cleaved to form ONP and galactose. ONP production can be measured at A 420 because it is yellow, and this can be used to quantitate B-gal activity. Jed will be unable to quantify his Bgal activity, because neither product of lactose metabolism (glucose or galactose) is easy to assay. 1

3 Question 3 (30 points) a) (10 points) Label the following on the diagram of the IgG molecule shown below: A. Heavy chain B. Light chain C. C H1, C H2, C H3 (Constant Domains of Heavy Chain) D. V H and V L (Variable domains) E. Location of antigen binding sites (CDRs) CH1 VH CDRs light chain VL CH2 IgG CH3 heavy chain After 7.02 lab, you decide to do a UROP in a protein biochemistry lab. Your lab is interested in the AB complex, which is formed by two proteins ( A + B ). Your project is to see how mutations in A or B affect complex formation. You first decide to purify the wild type complex. After a number of purification steps, you have two bands left in your most pure sample (as judged by Coomassie staining of an SDS- PAGE gel). To identify the bands, you decide to perform a western blot. You probe your western blot with primary antibody (rabbit anti- A ), then a secondary antibody (goat anti-rabbit IgG) conjugated to alkaline phosphatase. Your NBT-BCIP reaction worked great and you see that one of the bands in your most pure sample is definitely protein A. Next, you want to check if the other band is protein B. Your postdoc tells you that you can strip the blot to remove the protein A-specific antibody (and, of course, the secondary antibody bound to it). Then you can reprobe the blot with a protein B--specific primary antibody. 2

4 Question 3 (continued) b) (9 points) You borrow the postdoc s protocol for Stripping Western Blots, and notice that it calls for treating the membrane with a solution containing β-mercaptoethanol and a detergent (Tween 20), and heating it to 50 C. Using your knowledge of antibody structure, explain how this treatment will strip the antibodies off the blot. In IgG, disulfide bonds connect the two heavy chains of IgG to each other, and connect each heavy chain to a light chain. Beta-mercaptoethanol (BME) (reducing agent) will reduce these disulfide bonds, causing the four antibody chains to separate. Detergent and heat will further denature the protein. Since the CDR (which recognizes the antigen protein A here) is composed of residues from both the heavy and light chains, the antibody will no longer have a CDR once the chains are separated. Thus, the antibody will no longer be able to bind protein A, and will be stripped off the blot. d) (5 points) Your postdoc had protein B injected into chickens to obtain protein B-- specific antibodies (IgG). Antibodies made in chickens are concentrated in the egg yolk. What is a simple way to purify protein B specific IgG molecules away from other proteins in the egg yolk? (Hint: this can be accomplished in ONE column). Use an affinity column with protein B bound to the matrix of the column. e) (6 points) After successfully stripping the blot, you probe it with your purified chicken anti B primary antibody. You then use the same secondary antibody as before (goat antirabbit IgG, alkaline phosphatase conjugate) and develop your blot. You see no bands on your blot. Explain what went wrong, and how you would change your protocol to fix the problem. (Assume that the second protein in your most pure sample is indeed protein B, and that both antibodies are functional i.e. not denatured.) The secondary antibody was a goat anti-rabbit IgG, while your protein B-specific primary antibody was raised in chickens. Thus, the secondary antibody will not recognize the constant domain of chicken IgG, and you d have no way to visualize your protein B on the blot. (Remember that alkaline phosphatase is conjugated to the secondary antibody.) To fix the problem, use a goat (or any animal but chicken) antichicken IgG conjugated to alkaline phosphatase to visualize the primary antibody, and thus protein B. 3

5 Question 4 (40 points) After 7.02, you take a UROP in a lab studying a commercially important enzyme, bluejeanase. Bluejeanase is produced by Bacillus levitilus, a bacterial species that thrives on college campuses. Your project involves purifying bluejeanase from naturally occurring wild B levitilus strains found around MIT and analyzing its kinetic properties. The postdoc you are working with has purified bluejeanase from the lab B. levitilus strain, and gives you her notes to help with your purification. She notes that after ammonium sulfate precipitation, bluejeanase is one of four major proteins in the pellet. These proteins have the following properties: Protein pi A 11.0 B 9.2 C 7.5 bluejeanase 3.5 a) (12 points) How would you purify bluejeanase away from contaminating proteins A, B, and C? (Hint: A complete answer will describe what type of column you d use, why you chose this column, under what conditions you d run the column, and how you d elute the column. Note that you do not have to separate A, B, and C from each other, just away from bluejeanase.) You could purify bluejeanase by running an anion exchange column at ph 6 (above 3.5, below 7.5). Under these conditions, bluejeanase is negatively charged (ph>pi), while proteins A, B, and C are all positively charged (ph<pi). Thus, bluejeanase will stick to the positively charged matrix of an anion exchange column, while ABC will flow through the column. Bluejeanase can be eluted by using buffers of increasing salt concentration (such as NaCl). The negatively charged chloride ions will compete with bluejeanase for the positively charged matrix, and bluejeanase will be eluted. (The key here was understanding what pi means (pi= ph at which a protein is at its IEP, and is thus neutral), describing what charge each protein would have under the conditions you chose, and knowing what type of columns allow you to purify proteins based on charge.) 4

6 (Question 4, continued) Scientists in your lab have discovered that bluejeanase acts on a compound called 501-Y. Cleavage of 501-Y produces a yellow pigment (Y) that can be quantitated in a spectrophotometer at 420 nm (A 420 ). (501-Y is to bluejeanase as ONPG is to β-gal.) After running your column from part a), you perform a quantitative assay on the fraction that appears to have the most bluejeanase activity and a Bradford assay to determine protein concentration. You collect the following data: sample name fraction from part a) total volume of sample protein concentration sample volume used in reaction 2 ml 0.6 mg/ml 5 µl of a 1:10 dilution reaction volume stop time of reaction 1 ml 5 minutes A 420 at stop 0.30 Note: the reaction was stopped by pipetting 0.5 ml of the reaction into a stop cuvette containing 0.5 ml of stop solution (just like in 7.02 lab). Important Conversion factors: 1 U of bluejeanase activity = 1 nmole Y released/min 1 nmole Y released/ml = A 420 b) (8 points) Using the data above, calculate the total activity in this sample (Show your work!) 1. A 420 in cuvette = 0.3 x 2 (diluted 1:2 from reaction tube) = 0.60 A 420 in reaction tube A 420 = 0.12 A 420 (reaction rate) 5 minutes 1 minute A 420 x 1 nmole Y/ml = 24 nmole Y/ml in reaction tube 1 minute A 420 minute nmole Y/ml in reaction tube x 1 ml (reaction volume) = 24 nmole Y = 24U minute minute 5. 24U in reaction tube = 48U 0.5 µl Bluejeanase (BJ) in reaction tube µl BJ 6. 48U x 2 ml (sample volume) x 1000 µl = 9.6 x 10 4 U µl BJ 1 ml 5

7 Note: If you got part b) wrong, you lost one point on parts c) and d) if your answer was incorrect only because of your error in part b) c) (5 points) What is the specific activity of this sample? (Show your work.) Specific Activity = total activity OR activity/ml total protein protein concentration (mg/ml) SA = 9.6 x 10 4 U (from part b) = 8 x 10 4 U 1.2 mg mg d) (5 points) If you began your purification with 10 ml of crude lysate containing 1.92x10 5 U/ml of bluejeanase activity and a total protein concentration of 144 mg, what fold purification did you achieve? (Show your work.) Fold purification = Specific Activity at any step (SA final) Specific Activity in Crude Lysate (SA initial) SA final = 8 x 10 4 U_ (from part c) mg SA initial = (10 ml)(1.92 x 10 5 U/ml) OR 1.92 x 10 5 U/ml = 1.33 x 10 4 U 144 mg 144 mg/10 ml mg SA final = 8 x 10 4 U/mg = 6X SA initial 1.33 x 10 4 U/mg 6

8 After all your work, you finally have purified bluejeanase and are ready to do enzyme kinetics. You start all your reactions, then realize with horror that your 501-Y solution is contaminated with 560-X (another substrate of bluejeanase that does not produce a yellow product)! e) (10 points) How will the presence of 560-X affect the development of yellow color (in comparison to reactions performed with just 501-Y)? What will happen to the apparent Km and Kcat of your purified bluejeanase? Justify your answer. In the reaction tube, you have both 501-Y and 560-X. When bluejeanase acts, it may act on 501-Y (which will produce a yellow color) or 560-X (which will not). Therefore, the presence of 560-X will cause the yellow color to develop more slowly than in reactions performed with just 501-Y (partial credit was given for less yellow, though ultimately the same yellow color will be reached). The apparent Km of the reaction will increase in the presence of 560-X, while the Kcat (which is Vmax/Etot) will remain the same. This can be most easily explained by looking at the reaction mechanism. Let s assume that this equation describes the formation of the yellow product upon cleavage of 501-Y (therefore E= bluejeanase, S = 501-Y, and ES = bluejeanase 501-Y complex): k1 k3 E + S > ES > E + P < k2 When there is a competition for enzyme between two substrates (501-Y and 560-G), the rate of formation of the ES complex will be reduced. (Another way to think of it is that you will need more 501-Y to form the same amount of ES complex.). If we assume that k2>>>k3, then the Km can be approximated at the Kd of the enzyme substrate complex (k2/k1). If k1 is lower in the presence of 560-X, then Km will increase. Once the ES complex (between 501-Y and bluejeanase) is formed, then the presence of 560-X will not affect further steps in the reaction. Therefore k3, which is Kcat, is unchanged. Another way to think of this is that neither Vmax nor Etot changes (because provided enough substrate, the maximal reaction velocity will be reached, and enzyme is not destroyed by reaction with 560-X.) Since Kcat = Vmax/Etot, then Kcat will remain the same. 7