Between the following 4 Km values, select the one that indicates binding of the enzyme to its substrate with the highest affinity: Group of answer choices 10 nM 1000 uM 1 mM 10 pM
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32.
Between the following 4 Km values, select the one that indicates binding of the enzyme to its substrate with the highest affinity:
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- Data from enzyme inhibition are used to determine a Kmapp and Vmax PP. Comparison of these values with assays run without inhibitor are used to understand how the inhibition is occurring. This is useful for better understanding the active site as well as the practical aspect of pharmaceutical drugs. Below are idealized Line-Weaver Burke plots of different types of inhibitors. Comnetitive Uncomnetitive Mixed +Inh +Inh 4Inh Anh Inh Anh [S] [S] [S] a. How does the value of Vmax for the enzyme compare to the Vmax PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed b. How does the value of Km for the enzyme compare to the Km PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed c. For each situation in Model 1, consider an inhibitor that is better than the one shown on the graph. Answer the following questions for each type of inhibition: i. How would the KmPP change? ii. How would the Vmax PP change?An enzyme is present at a concentration of 1 nM and has a Vmax of 2 µM s-'. The Km for its primary substrate is 4 µM. Calculate kcat- kcat Calculate the apparent Vmax and apparent Km of this enzyme in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 2. Assume that the enzyme concentration remains at 1 nM. apparent Vmax µM s-1 apparent Km = µMAn enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.
- 1. A Lineweaver-Burk Plot is shown below. 30 25 Curve A y = 3.1207x + 2.4978 20 15 Curve B y = 1.0003x + 2.3602 10 5 -3 1 5 7 11 1/[Catechol] (mM1) With these curves, determine the following enzyme parameters. Show all pertinent solutions. a. Km of Curve A and Curve B b. Vmax of Curve A and Curve B c. Assuming that one of these curves corresponds to the kinetics of one enzyme and one substrate, which curve represents the effect of an inhibitor? Why do you say so? d. What type of inhibition is exhibited by your answer in question c? Why do you say so? 1/V, (units of activity 1)A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).You have performed protein purification on your new favorite enzyme using a protocol which involves the following steps/samples: crude extract, ammonium sulfate cut, ion exchange and gel filtration. You need to run 25ug of protein from your crude extract sample on an SDS-PAGE gel. You have determined that the protein concentration of your crude extract sample is 2.1mg/ml. Your total sample volume is 30ul. You have water for your diluent and 6x SDS-loading dye to prepare your sample. List the components of your prepared sample. Note: you will only have access to a P20 and a P200 to prepare this sample.
- A schematic representation of the enzyme IspD complexed to inhibitor 3, and a series of inhibitors 3-5 are shown below. Ala202 lle240 mwww NH NH Val263 ОН www HN N- lle177 HN 'N' CI 3 X = N 4 X = C-CN 5 X = C-COO IC50 274 µM IC50 140 nM IC50 35 nM NH2 HN Val266 N -N O-H---- N HN %3D Arg157 HN wwww lle265 Explain why structure 4 is a more potent inhibitor (lower IC50 value) than inhibitor 3 and why structure 5 is a much weaker inhibitor (higher IC50 value) than 3 and 4.A. Lineweaver-Burk plot of the enzyme with increasing amounts of substrate in the absence or the presence of the inhibitor is shown below. Graph A : x-intercept Graph B : x-intercept = - 0.012, y-intercept = 0.8 Graph C : x-intercept = - 0.027, y-intercept = 0.8 Graph D : x-intercept = - 0.039, y-intercept = 0.8 - 0.007, y-intercept = 0.8 Graph A 4 Graph B Graph C Graph D 1 -0,04 -0,02 0,00 0,02 0,04 1/[Substrate] (uM) (i) Which graph indicates an enzymatic reaction without inhibitor? (ii) Which type of inhibitor is it? Briefly explain. (iii) Which graph indicates the highest concentration of inhibitor? (iv) Calculate the Vmax and Km of the graph showing an enzymatic reaction with the lowest concentration of inhibitor. Show the steps of calculation and unit in your answers. Keep 2 decimal places in your answers. 1/Rate (umol/min)1 pt pt 9146 Bb 9146 Bb 1031 Class Etsy E Traps E Traps New Free Chat + ☆ 出口 keAssignment/takeCovalentActivity.do?locator-assignment-take [References] You do an enzyme kinetic experiment and calculate a Vmax of 118 μmol per minute. If each assay used 0.10 mL of an enzyme solution that had a concentration of 0.20 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol? (Enter your answer to two significant figures.) turnover number = sec-1 D 1 pt Submit Answer Try Another Version 2 item attempts remaining estion stion 5 on 6 7 1pt 1 pt 1 pt 1pt 1pt 1pt 1 pt 1 pt D is the substrate concentration multiplied by the catalytic constant. KM is equivalent to the substrate concentration multiplied by the ratio of rate constants for the formation and dissociation of the enzyme-substrate complex. KM is equivalent to the substrate concentration. KM is equivalent to the substrate concentration divided by 2 A: KM is equivalent to the substrate concentration…
- 5.50 1/V, min/umol 5.00 4.50 4.00 y = 0.9474x + 2.6649 y = 0.9997x + 2.032 0.00 1.00 2.00 2.50 3.00 1/[S], uM -1 Looking at the double reciprocal plot for an enzyme in the absence of inhibitor and in the presence of two concentrations of inhibitor, what would be the Vmax for the uninhibited enzyme? (bottom graph) Equation is given. Choose the one best answer. 3.50 3.00 2.50 2.00If the new higher KM value is 0.1 mM resulting in the new plot red curve is due to presence of enzyme inhibitor is the inhibitor reversible or irreversible?An enzyme has the following values: vmax's; 0.028 M s-' and 0.021 M s-' (I), and km's; 0.00198 M and 0.00197 M (I), when the total concentration of enzyme is 10-º M. The (I) values were determined with 1 microM inhibitor. Select the value closest to the k, for this enzyme. (hint 1 microM = 10-6 M).
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