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- In a Lineweaver-Burk graph, the lines representing the uninhibited and inhibited enzyme catalyzed reaction meet each other on the x-axis. The type of inhibition which is occurring is: a) competitive b) noncompetitive c) uncompetitive d) allosteric CO2 exerts direct activity upon hemoglobin by: a) blocking oxygen from binding to the heme group b) displacing BPG from the central cavity c) oxidizing Fe+2 to Fe+3 which does not bind oxygen d) forming an N-terminal carbamate which favors the T-state The dominant motif found in hemoglobin and myoglobin is: a) helix-turn-helix b) twisted beta sheet c) beta barrel d) random coil Which of these is an ketohexose? a) fructose b) glucose c) ribose d) erythrose Which of these is a constitutional isomer of d-glucose? a) fructose b) galactose c) l-glucose d) ribose Which of these is an enantiomer of d-glucose? a) d-fructose b) d- galactose c) l-glucose d) d-ribose Which of these is a diastereomer of…You perform Michaelis-Menten kenetics on i) trypsin and ii) a mutant form of the same enzyme (a single amino acid has been changed). The specificity constant for the mutant was 10 times larger than for trypsin. a) Define the specificity constant and explain using this definition how a larger value might occur. b) Explain how, if at all, a non-competitive inhibitor of trypsin would affect the specificity constant.mTOR is a cytoplasmic kinase that regulates cell division. Its misregulation can lead to cancer. The canonical mTOR inhibitor, rapamycin, is FDA approved as a cancer treatment. Rapamycin is an allosteric inhibitor that does not bind anywhere near the substrate binding site of the enzyme. A) What is the name of this type of inhibition? B) Sketch a REPRESENTATIVE double-reciprocal/Lineweaver-Burke plot that includes the enzyme kinetics for BOTH the uninhibited and inhibited reaction. Be sure to: 1) Make ABSOLUTELY clear which curve is your uninhibited reaction and which is your inhibited reaction. 2) Label the X and Y axes. 3) Indicate what the X- and Y-intercepts represent with regards to Michaelis-Menten kinetics. C) Can increasing the substrate concentration overcome this type of inhibition? Explain in a sentence or two.
- An increase in PCO2 will result in which of the following? O a) An increased affinity between PO, and Hb, shifting the oxyhaemoglobin dissociation curve to the right b) An increased affinity between PO, and Hb, shifting the oxyhaemoglobin dissociation curve to the left O c) A decreased affinity between PO, and Hb, shifting the oxyhaemoglobin dissociation curve to the right O d) A decreased affinity between PO, and Hb, shifting the oxyhaemoglobin dissociation curve to the leftAn enzyme catalyzes the reaction M ßàN. The enzyme is present at a concentration of 1 nM, and the Vmax is 2 M s-1. The Km for substrate M is 4 μM. a)Calculate kcat. b)What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an α’ of 2.0?Consider the reaction catalyzed by PFK. In the presence of AMP, which of the following will be expected? Check all that apply: a)The Km for substrate would be decreased b) the initial velocity plot would show the curve shifted to the right c) the R state is stabilized d)the rate of the reaction is diminished
- Where do each of these 5 main themes occur in the chymotrypsin mechanism? 1) substrate specificity 2) induced fit 3) covalent catalysis 4) acid/base catalysis 5) transition state stabilizationFor a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.Which of the following is true under the following conditions: an enzyme displaying Michaelis-Menten kinetics where the enzyme concentration is 10 nM, the substrate concentration is 45 mM, and the Km is 50 µM? a) The enzyme has low catalytic efficiency for the substrate. b)The rate of catalysis is near half-maximal velocity. c)The enzymatic reaction is near maximal velocity. d)Halving the substrate concentration has little effect on the catalytic rate. e) There is not enough information provided.
- Drug D reversibly binds to enzyme E and inhibits its activity toward substrate S. D binds equally well whether or not S is bound. Sketch a graph of the expected 1/v vs. 1/[S] relationship for: A) The enzyme reacting with S in the absence of drug D, B) The enzyme reacting with S in the presence of a small amount of drug D, and C) The enzyme reacting with S in the presence of a large amount of drug D.13.) An unchanged apparent Km results from the action of an uncompetitive inhibitor because A) the presence of the inhibitor can shift the binding equilibrium toward the ES complex. B) the presence of the inhibitor can shift the binding equilibrium toward the E + S state. C) the presence of excess substrate shifts the binding equilibrium toward the ES complex. D) uncompetitive inhibitors do not alter the binding equilibrium between substrate and enzyme. E) Uncompetitive inhibitors always result in a lower apparent Km, not an unchanged apparent Km.Based on the kinetic constants below, which enzyme will most efficiently catalyze conversion of the substrate into product? A) Vmax = 10 uM s-1, KM = 10 µM B) Vmax = 10 uM s-1, KM = 0.01 µM C) Vmax = 1000 uM s-1, KM = 500 µM D) Vmax = 1 uM s-1, KM = 1 µM E) Vmax = 200 uM s-1, KM = 10 µM