**Model 1****Idealized Graphs of Enzyme Inhibition**- **Competitive Inhibition:** - In the presence of an inhibitor (\[I\]), the slope (1/V\_max) increases, leading to increased K\_m,app, while V\_max,app remains unchanged.- **Uncompetitive Inhibition:** - The inhibitor reduces both V\_max and K\_m,app proportionally, leading to parallel lines.- **Mixed Inhibition:** - The presence of an inhibitor affects both V\_max,app and K\_m,app, altering both the slope and intercept.1. Using Model 1 as a reference, fill in the table (choices: increase, decrease, no change).| | A competitive inhibitor | An uncompetitive inhibitor | Mixed inhibitor ||-----------------------|-------------------------------------|------------------------------------|------------------------------------|| How does K\_m,app change? | Click or tap here to enter text. | Click or tap here to enter text. | Click or tap here to enter text. || How does V\_max,app change? | Click or tap here to enter text. | Click or tap here to enter text. | Click or tap here to enter text. |2. For each situation in Model 1, consider an inhibitor that is better than the one shown in the graph. Draw the graphs you see in Model 1, then draw a new line that would represent the new, better inhibitor. Paste a photo of your drawing below for: a. Competitive Inhibitor ![Graph Placeholder](#) b. Uncompetitive Inhibitor ![Graph Placeholder](#) **Module 5 Week 1 Assignment: Enzymes**This is a group assignment. You must work with a partner. Please be sure to have joined a group in Canvas and then upload one file for the group.The following kinetic data were obtained for an enzyme in the absence of inhibitor (1) or in the presence of two different inhibitors (2 and 3). The enzyme concentration was the same in all of the experiments. Graph these data as a Lineweaver-Burke plot.| [S] mM | v (umol/ml * sec) ||--------|--------------------|| | Enzyme (1) | Inhibitor (2) | Inhibitor (3) || 1 | 12 | 4.3 | 5.5 || 2 | 20 | 8 | 9 || 4 | 29 | 14 | 13 || 8 | 35 | 21 | 16 || 12 | 40 | 26 | 18 |a. Paste a photo of your Lineweaver-Burke plot here. Plot all three lines on one graph.b. Using the data, fill in the following table:| | Vmax | Km ||---------------|--------------------------|--------------------------|| Enzyme | Click or tap here to enter text. | Click or tap here to enter text. || | Apparent Vmax | Apparent Km || Inhibitor 2 | Click or tap here to enter text. | Click or tap here to enter text. || | Type of inhibition | || | Click or tap here to enter text. | || | Apparent Vmax | Apparent Km || Inhibitor 3 | Click or tap here to enter text. | Click or tap here to enter text. || | Type of inhibition | || | Click or tap here to enter text. | |

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Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Calculate KI' of the inhibitor from the information given. All information may not be needed to calculate. K'm = (29Ki+1.45x10^-10)/Ki Vmax = 11.7 µMs-1 Kcat = 130 s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µM Inhibitor Concentration = 5x10^-12
2. An enzyme-catalyzed reaction is carried out at several different substrate concentrations (listed below), either (1) with no inhibitor, or (2) in the presence of an inhibitor, I, at 0.63 g/L. Given the following data, determine: A. what type of inhibitor I is B. the value of Ki No Inhibitor: With Inhibitor: Substrate Concentration (g/L) 0.86 Reaction Velocity (g/L-s) Reaction Velocity (g/L-s) 0.179 0.161 2.35 3.80 4.90 6.10 18.4 0.398 0.567 0.666 0.726 0.950 0.322 0.407 0.443 0.495 0.639
2. Enzyme-catalyzed reactions. Answer the following with true or false. If false, explain why. (a) The initial rate of an enzyme-catalyzed reaction is independent of substrate concentration. (b) At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration. (c) The Michaelis constant Km equals the substrate concentration at which velocity (v) = Vmax/2. (d) The Km for a regulatory enzyme varies with enzyme concentration. (e) If enough substrate is added, the normal Vmax of an enzyme-catalyzed reaction can be attained even in the presence of a noncompetitive inhibitor. (f) The Km of some enzymes may be altered by the presence of metabolites structurally unrelated to the substrate. (g) The rate of an enzyme-catalyzed reaction in the presence of a rate-limiting concentration of substrate decreases with time. (h) The sigmoidal shape of the v versus [S] curve for some regulatory enzymes indicates that affinity of the enzyme for the…
Recall that vmax is achieved only at high substrate concentrations. Based on your answers toQuestion 5, do you predict that nmax will change when a competitive inhibitor is introduced?Why or why not?
2:31 The molecule shown below is? H,C -o HC -o H,C -o O glucose O cholesterol O none is correct O adenine O phospholipid
Part 1: Assess the following partial results section below by editing it for brevity by omitting any unnecessary parts (1 point), explain why you decided to remove certain sections (1 point): To evaluate inhibitory effects of the selected molecules, 10mM stock solutions of each molecule were prepared in DMSO. A reaction mixture (200μl) was prepared with the same formula optimized for the enzyme activity assay (0.1 M Tris-HCl ph 8, 0.1 M KCI, 25 mM NaCl, 0.25 mM ATP, and two units of inorganic yeast pyrophosphatase) with 10 µM of the sample molecule. The reaction mixture was incubated for 20 minutes at ambient temperature. Enzymatic reaction was triggered by addition of the substrate B (0.2 mM) and the absorbance of the product was monitored at 290 nm for 10 minutes. Six out of 15 sample molecules showed appreciable inhibition at 10 μM (Figure 5). Three of the molecules, A3, A6, and A7 exhibited more than 50% inhibition of the enzyme activity and were further diluted to find the minimal…

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