BIOCHEMISTRY I HOMEWORK III DUE 10/15/03 66 points total + 2 bonus points = 68 points possible Swiss-PDB Viewer Exercise Attached 1). 20 points total T or F (2 points each; if false, briefly state why it is false). Weak chemical interactions such as van der Waals interactions, hydrogen bonding and electrostatic interactions do not make a significant contribution to the overall protein folding free energy. Protein folding appears to be hierarchical in nature. Enzymes alter the amount of energy consumed (or liberated) in a reaction. Catalysts accelerate a reaction by increasing the G The steady-state assumption of the Michaels-Menton kinetic analysis assumes that at [S] >>>> [E], the rate in conversion of the ES intermediate to product is always faster than its formation from free E and S, hence product formation is favorable. In competitive inhibition, the inhibitor and substrate usually are chemically and/or structurally very similar to one another. Competitive inhibitors are effective in drug therapies for disease because they bind irreversibly to the active site. The end product of a catalytic reaction can often inhibit the enzyme. Enzymes are only regulated by the reversible non-covalent binding of small molecules called allosteric effectors. Histidine, because of its pka value, can act both as a general acid and general base in enzyme catalyzed reactions.
2). 10 points total Matching: Match the termis in the fist column with the correct descriptions in the second column. Column 1 Column 2 k in v = K[A] 1). K m d[p]/dt or -d[s]/dt 2). V max d[p]/dt = -d[s]/dt 3). First-order rate constant k in v=k[a] [B] 4). Michaelis constant k -1 /k 1 5). V max /E t (k -1 + k 2 )/k 1 6). The velocity or rate of a reaction v at [S] = infinity 7). Enzyme:substrate dissociation constant [S] when v = V max /2 8). 1/V max k cat 9).Equilibrium or steady-state 1/vwhen 1/[S] = 0 10. Second-orderrate constant 3). 3 points total Enzymes have three distinctive features that distinguish them from chemical catalysis. What are they?
Descriptive answers should be brief - no more than 3 (grammatically correct) sentences! 4). 6 points A researcher was studying the kinetic properties of β-galactosidase using an assay in which o-nitrophenol-β-galactoside (ONPG), a colorless substrate, is converted to galactose and o-nitrophenolate, a brightly-colored, yellow compound. Upon addition of substrate (at 0.25 mm) to a fixed amount of enzyme, o-nitrophenolate (ONP) production was monitored as a function of time by spectrophotometry at λ=410 nm. The following data were collected: Time (sec) Absorbance at 410 nm 0 0.000 15 0.158 30 0.273 45 0.360 60 0.429 75 0.484 90 0.529 150 0.652 210 0.724 270 0.771 330 0.805 390 0.830 450 0.849 510 0.864 A). Convert the absorbance at 410 nm to concentration of ONP using the following extinction coefficient. ε = 3.76 mm-1cm-1 B). Plot [ONP] vs. time. C). Determine the initial velocity of the reaction. D). Explain why the curve is nearly linear initially and later approaches a plateau. Explain why we are reaching a plateau at the concentration of ONP indicated by the curve.
4). 10 points total The enzyme aspartate transcarbamoylase catalyzes the condensation of carbamoyl phosphate and aspartic acid to form N-carbamoylaspartate. A plot of initial velocity vs. aspartate concentration is not a rectangular hyperbola as found for enzymes that obey Michaelis-Mentin kinetics. Rather, the plot is sigmoidal. Further, CTP shifts the curve to the right along the substrate axis and makes the sigmoidal shape more pronounced. ATP shifts the curve to the left and makes the sigmoidal shape less pronounced. A). What kind of enzyme is aspartate transcarbamoylase? B). What role does ATP play? (What would it be termed based on its role; explain how it affects the enzyme such that it has the described effect on the curve). C). What role does CTP play? (What would it be termed based on its role; explain how it affects the enzyme such that it has the described effect on the curve). 5). 2 points total What is a transition state analog?
6). 8 points total The steady state kinetics of an enzyme are studied in the absence and presence of an inhibitor the initial rate is given as a function of substrate concentration in the following table: v o (mm min -1 ) [S] mm No inhibitor With Inhibitor 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5 3.33 1.96 10 4.17 2.38 A). Why can we not determine V max and K m by directly plotting v o vs. [S]? B). Determine V max and K m in the presence and absence of inhibitor. C). What type of inhibition is represented? 7). 2 points Enzymes with k cat /Km values ~ 10 8 to 10 9 sec -1 are said to have reached "catalytic perfection". Why?
QUESTIONS RELATED TO THE SWISS-PDB VIEWER EXERCISE. 8). 15 points total + 2 bonus point Be brief - no more than 3 (grammatically correct) sentences per answer! You will have to use your text to answer some of these questions. Although we will cover some of this in class, it will be in one of the final lectures. A). What is the specificity of chymotrypsin? B). What is the specificity of trypsin? C). Chymotrypsinogen is a zymogen. Define zymogen. D). Describe the conversion of chymotrypsinogen to chymotrypsin. E). What holds the cleavage products together? CHYMOTRYPSIN VS. TRYPSIN COMPARISON F). Describe the structures of chymotrypsin and trypsin. G). List the hydrogen bonds and their lengths to histidine in the two structures. H). Bonus question: Why would we not see an expected hydrogen bond in a crystallographic structure? I). What is the oxyanion hole?
J). What are the roles of the N atoms of G193 and S195? K). Is the chemical characteristic of the oxyanion hole different between trypsin and chymotrypsin? L). How well (by eye) do the catalytic groups of the two active sites overlap? M). How well (by eye) does the residues enclosing the oxyanion hole overlap? CHYMOTRYPSIN VS. CHYMOTRYPSINOGEN COMPARISON N). Describe how (by eye) the overall polypeptide chain deviates between the two structures as compared to the chymotrypsin - trypsin overlap. O). How well (by eye) does the catalytic triad overlap, as compared to the chymotrypsin - trypsin overlap? P). How well does the oxyanion hole overlap, as compared to the chymotrypsin - trypsin overlap? Q). Bonus Question: Can your observations explain a decrease in enzymatic of chymotrypsinogen as compared to chymotrypsin?
Swiss-PDB Viewer Exercise SERINE PROTEASES Adapted from a Kinemage exercise by John H. Connor for Fundamentals of Biochemistry by Voet, Voet and Pratt. John Wiley and Sons, Inc, New York. 1999. In this exercise, we are looking at the structures and active sites of serine proteases. You will be opening up two structures at once, superimposing structures and looking at active site residues. Answer the appropriate homework questions as you go through this exercise. The serine proteases are enzymes that hydrolyze peptide bonds. They are characterized by an essential "catalytic triad" - three residues that are essential to their catalytic function. Of major importance to this catalytic triad is a highly reactive serine residue. You will need to download the following three pdb files bovine chymotrypsin: 4CHA bovine trypsin: 3PTN bovine chymotrypsinogen: 1CHG As a note, for some reason I had to go through a few extra steps to download the file. The following steps worked on my Mac (should be similar for PC). Click on download file (pdb format & no compression). This displayed the file to the screen. Under the FILE menu, choose download file. Enter the URL from the file window into the box and download the file. I want you to make the following comparisons Chymotrypsin to trypsin Chymotrypsin to trypsin.
CHYMOTRPYSIN: First open up chymotrypsin (4CHA) in Swiss-PDB Viewer. Note that there are two molecules of chymotrypsin showing in the view. This is because there are two molecules in the crystallographic asymmetric unit (a crystallography term for how many molecules are in a repeating unit). This is not because chymotrypsin is a dimer! We want to get rid of one of the molecules. In the control window, you will see in the very first column the letters A or B, denoting chain A or B. Scroll down to the first B. Click on the first B. All the residues in chain B will be highlighted in red. In the menus, go to BUILD! Remove selected residues. You should now only have one chain showing. View the structure as a ribbon structure. As a hint: In order to facilitate viewing of the structure and comparison of structures, set your sheets and helices to a very light color. CATALYTIC TRIAD: Highlight in GREEN the catalytic triad. These are residues His 57, Asp 102 and Ser 195. (I'm not really a control-freak when it comes to colors. However, we are comparing several structures, so following a specific color scheme will help you. You can choose your favorite colors). OXYANION HOLE: Highlight in a DARKER GREEN the residues of the oxyanion hole. These are 192-196. Rotate the molecule about the see the active site better. Leave the structure up and open up a second structure at the same time.
CHYMOTRYPSIN VS. TRYPSIN Without closing the chymotrypsin structure, open up the structure for trypsin (3PTN). You will see two structures on your screen. In the "control window", you will be able to select either 4CHA or 3PTN from the menu at the top of that little window. Highlight the catalytic triad in PINK and the oxyanion hole in DARKER PINK. View the structure as a ribbon structure. SUPERIMPOSE CHYMOTRYPSIN AND TRYPSIN Under the menu FIT! Best fit! Cα carbon atoms only. Rotate the structure. Pay special attention to the overall fit, the catalytic triad and the oxyanion hole. Answer the corresponding homework questions. Turn off the trypsin structure.
CHYMOTRYPSIN VS. CHYMOTRYPSINOGEN Without closing the chymotrypsin structure, open up the structure for chymotrypsinogen You will see two structures on your screen. View the structure as a ribbon structure. In the "control window", you will be able to select either 4CHA or 3PTN from the menu at the top of that little window. Highlight the catalytic triad in BLUE and the oxyanion hole in DARKER BLUE. Highlight the sites of cleavage in another color (red, perhaps). Ser 14 - Ile 16 Thr 147 - Asn 148 Superimpose the two structures as described above. Rotate the structure. Pay special attention to the overall fit, the catalytic triad and the oxyanion hole.