Solutions to 7.02 Quiz II 10/27/05

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1 Solutions to 7.02 Quiz II 10/27/05 Class Average = 83 Standard Deviation = 9 Range Grade % A B C 17 > 54 D 1

2 Question 1 (56 points) While studying deep sea bacteria, you discover a new enzyme that catalyses the production of light. This enzyme is composed of two different polypeptides that interact as shown below where the backbone of polypeptide 1 is represented as a black line and the backbone of polypeptide 2 is represented as a gray line. Surface charges are shown as or -. Polypeptide 1 = 400 amino acids - - S S S S Polypeptide 2 = 600 amino acids a) An α-helix is an example what level of protein structure? α-helical structures represent secondary structure. (2 points) b) What type of bond(s) or force(s) stabilizes an α-helix or β-sheet? α-helical and β-sheet structures are stabilized by hydrogen bonds between the carboxyl and amino groups of the polypeptide backbone. (2 points) c) What effect might β-mercaptoethnanol have on this protein? Why? β-mercaptoethnanol breaks disulfide bonds, so will denature each of the two polypeptides of this protein. (2 points) d) What is the strongest possible interaction between the side chains of each of the following pairs of amino acids? Choose from covalent bonds, van der Waals forces, ionic bonds, and hydrogen bonds. amino acids strongest interaction (4 points, 2 points each) i) Ser and Gln Hydrogen bonds ii) Ala and Met van der Waals forces 2

3 Question 1, continued Your enzyme oxidizes compound W and in the process produces a faint yellow glow, so assaying activity of your enzyme is easy. You are ready to purify your enzyme. You begin with a crude lysate of bacteria, and after centrifugation, you find activity only in the supernatant. You first precipitate proteins out of the crude lysate supernatant at 15% ammonium sulfate, but there was no activity in the 15% ammonium sulfate pellet. You increase the amount of ammonium sulfate in the supernatant from 15% to 25% but there was no activity in the 25% ammonium sulfate pellet. You increase the amount of ammonium sulfate in the supernatant from 25% to 45%, and now all of the activity is in the pellet. e) In the lab, you precipitated β-galactosidase out of the crude lysate supernatant at 45% ammonium sulfate. Compare the precipitation above for enzyme X and the one in the lab for β-gal. Was one a better purification than the other or were they the same? Explain your answer. [6 points total] The precipitation of enzyme X was a better purification. (2 points) For enzyme X, you did sequential precipitations, each of which eliminated some of the other proteins. Thus in the pellet after your final 45% ammonium sulfate step you had proteins that precipitated from 25% - 45%. (2 points) In contrast, with the β-gal purification, you had all the proteins that precipitate between 0 and 45%. (2 points) You next use gel filtration chromatography on your resuspended 45% ammonium sulfate pellet. You collect 5 fractions (fraction 1 is collected first), and the only one that glows is fraction 4. f) If you were to assay β-gal in these same fractions and find β-gal activity in fraction 3, what comparison can you make between β-gal and your enzyme. In your answer, briefly explain how gel filtration chromatography separates molecules. [4 points total] Since find β-gal activity was found in an earlier fraction than enzyme X, the β-gal protein moved more quickly through the column. In size exclusion chromatography, smaller molecules have access to a greater % of the column volume and thus move more slowly through (2 points). This would indicate that enzyme X is smaller than β-gal (2 points). You prepare an affinity column by binding an analog of compound W to a gel matrix. You load fraction 4 (above) onto the column, wash, and elute. You obtain all the activity in a single 1 ml fraction. g) Why would you use an analog of compound W and not compound W itself? [3 points total] You want enzyme X to bind the column and stay bound. If you attach compound W to your affinity column, then when enzyme X binds, it can also catalyze the reaction and in doing so be released from the column. 3

4 Question 1, continued You ask a friend to run a SDS-PAGE gel on the samples that you collected during your purification. You return to find the gel and the following note. MW MW Here is your gel, unfortunately I can t remember which order I loaded the samples. Sorry! h) Assign each lane to the appropriate sample. (12 points, 2 each) Sample Lane Crude Lysate (CL) 6 Crude Lysate supernatant (CL-S) 1 45% ammonium sulfate pellet (AS-P) 3 Fraction 3 after gel filtration (F3) 4 Fraction 4 after gel filtration (F4) 5 Elution from affinity column (AF) 2 You do a second gel with the samples loaded in the following order: CL, CL-P, CL-S, AS-P, F3, F4, AF. You transfer this gel to nitrocellulose, and do a Western Blot. CL CL-P CL-S AS-P F3 F4 AF 4

5 Nitrocellulose membrane Question 1, continued i) When you assayed your crude lysate pellet (CL-P) and supernatant (CL-S), all the enzyme activity was in the CL-S. When you look at the western, there is clearly a signal in the CL-P lane. How can you explain this discrepancy? [4 points total] Your assay measures activity, not protein (2 points). For an enzyme to be active, it must be properly folded maintain the correct 3-D shape. If you had denatured enzyme X protein in a sample, it would never be measured by the assay (2 points). j) Explain why you might see two bands. [4 points total] Enzyme X is composed of two distinct polypeptides (3 points) that are of different sizes. (1 point) k) Look at the western data above and given your answer to (j), which of the following detection systems (Schemes A D) would give the results seen on the nitrocellulose membrane shown above. Circle all that apply. [13 points total] Scheme A Scheme B Scheme C Scheme D 1. Rabbit polyclonal antibody against your enzyme 2. Goat antibody against rabbit IgG conjugated to alkaline phosphatase 3. Treat membrane with a substrate for β-gal and watch for color development. 1. Rabbit polyclonal antibody against your enzyme 2. Rabbit antibody against goat IgG conjugated to alkaline phosphatase. 3. Treat membrane with a substrate for alkaline phosphatase and watch for color development. 1. Rabbit monoclonal antibody against your enzyme 2. Goat antibody against rabbit IgG conjugated to alkaline phosphatase 3. Treat membrane with a substrate for alkaline phosphatase and watch for color development. 1. Mouse polyclonal antibody against your enzyme 2. Goat antibody against mouse IgG conjugated to alkaline phosphatase. 3. Treat membrane with a substrate for alkaline phosphatase and watch for color development. l) For the one(s) that you did not choose, explain why they would not give you the Western blot results seen above. D: Should be circled. (4 points) A: In this case, we have used a substrate for β-gal to develop the color, but using antibodies, have localized alkaline phosphatase to the enzyme X protein. alkaline phosphatase will not cleave the β-gal substrate. (3 points) B: The secondary antibody is anti-goat, and will not recognize the rabbit primary antibody. (3 points) C: A monoclonal antibody would only detect one of the two bands. (3 points) 5

6 Question 2 (44 points) You are studying a transcription factor, tf1, that is related to the transcription factors encoded by the hox genes. You have identified the DNA sequence to which tf1 binds. To assay for tf1, you have developed a way to monitor binding to this DNA sequence. The sequence of the tf1 protein is shown below. Met Lys Lys Leu Asp Ala Arg Lys Gly His Gly Val Ala Phe Ser Ile Arg Ala Gly Trp Val Gly Lys Thr His Arg a) Circle a contiguous stretch of 8 amino acids that you would expect to find internal in the folded protein at ph 7. Why did you choose these amino acids? Any of the above boxes are equal (8 points). They are hydrophobic (2 points). b) You begin with 10ml crude lysate, centrifuge it, and determine that the supernatant (9 ml) contains tf1. As your first step in the purification, you choose sucrose gradient centrifugation. A sucrose gradient separates molecules based upon what two properties? Size, shape and density (2 points, 1 each) c) After determining which fraction contains the transcription factor, you plan to use column chromatography to further purify tf1. What type of column would you use to best purify tf1, an affinity column or an ion exchange column? Why did you choose this type of column as opposed to a different type? [4 points total] An affinity column (2 points) would allow purification of tf1 only as opposed to an ion exchange column that would also allow purification of other proteins that have similar charge (2 points). OR: An ion exchange column because we had better results with our DEAE column than with our affinity column. (half credit) Your chosen column separates proteins based on what property? An affinity column would separate tf1 based upon the ability to bind specific DNA sequence. This binding is determined by the 3-D shape of the enzyme. (1 point). OR: An ion exchange separates based on charge (full credit if consistent) Explain what the matrix of your column is and how this matrix allows purification of tf1 specifically. Your affinity column is composed of a water-insoluble matrix to which you have attached the specific DNA sequence to which tf1 binds (3 points). Of all the proteins remaining in the sample, only tg1 will bind to the column, all the other will pass through. (1 points). OR: The ion exchange column will have a negatively charge water-insoluble matrix. Because tf1 has an overall positive charge, it will preferentially bind to the column. (full credit if consistent) How would you elute tf1 from your chosen column? How does this work? Tf1 can be eluted from the affinity column by changing the ph or adding an excess of the binding DNA. ph change would alter the shape of the protein enough that is no longer binds to the column. Excess DNA would out-compete the column. (3 points) OR: Tf1 can be eluted from the ion exchange column by increasing the salt concentration in the washes. (full credit if consistent) What is one possible disadvantage to using your chosen column for purification of tf1? 6

7 If tf1 was eluted from the affinity column by changing the ph, the 3D shape of the protein can be altered, and activity can be lost. (3 points) OR: Other proteins of similar charge can be purified along with tf1 on an ion exchange column. (full credit if consistent) 7

8 Question 2, continued d) To measure purification, you need to know how much total protein remains in your samples. In 7.02, you used two techniques to examine the amount of total protein in your samples. i) Name one of these techniques. Bradford Assay (2 point), ii) Name the other technique. SDS-PAGE (2 point), e) In a few sentences describe the quantitative technique. Include the key reagents. [ 7 points total] The Bradford assay is the quantitative technique. It uses a dye, Coomassie Brillant Blue that binds to all protein, and when bound to protein it changes color. That color change can be measured by a spectrophotometer as a change in the absorbance at 595 nm. To use this assay to measure amount of total protein, you need to first generate a standard curve using known amounts of a protein, such as BSA, f) If you had successfully purified tf1 in your experiment, what would the data from this technique look like? Assume you used the samples: CL, CL-S, Post sucrose fraction with activity and post column fraction with activity. You would see a progressive decrease in the absorbance at 595 nm where crude lysate has the highest absorbance. (3 points) g) Assume that you had successfully purified tf1 in your experiment, and used the data from your quantitative technique to calculate specific activity for these same four samples (CL, CL-S, Post sucrose fraction with activity and post column fraction with activity). Which sample would you expect to have the highest specificity activity, and which sample would you expect to have the lowest specific activity? Because you are enriching for tf1, and decreasing total protein, You would see a progressive increase in the specific activity where crude lysate has the lowest and the post column fraction has the highest. (3 points) 8