5. Protein phosphatase 1 (PP1) is a major eukaryotic serine/threonine protein phosphatase thatregulates many diverse cellular processes. One study has shown that phosphatidic acid (PA) is aninhibitor of PP1. The investigators measured the activity of PP1 in the presence and absence of theinhibitor PA. The concentration of PA was 300 nM.[Substrate](µM)Vo with PA(µM sec-1)0.00381Vo without PA(µM sec-1)0.0100.02090.0150.03350.006200.0250.04190.009310.0500.08380.0140a. Use Excel to make a Lineweaver-Burke plot showing both the noninhibited and the inhibitedreactions. Your Lineweaver-Burke plot should have labeled axes with units. The graph shoulddisplay both the y-intercept and the x-intercept and include linear trendline equations.b. Determine Vmax and Km for the enzyme in the uninhibited state. Be sure to include your units.c. Determine the type of inhibition for this inhibitor PA. Briefly explain.d. Calculate the K¡ for the inhibitor PA.

Using Lineweaver-Burk plot, we have to find Vmax, Km and Ki for the inhibitor.

5. Protein phosphatase 1 (PP1) is a major eukaryotic serine/threonine protein phosphatase that regulates many diverse cellular processes. One study has shown that phosphatidic acid (PA) is an inhibitor of PP1. The investigators measured the activity of PP1 in the presence and absence of the inhibitor PA. The concentration of PA was 300 nM. [Substrate] (µM) Vo with PA (иM sec-1) Vo without PA (иM sec-1) 0.0209 0.010 0.00381 0.015 0.0335 0.00620 0.025 0.0419 0.00931 0.050 0.0838 0.0140

5. Protein phosphatase 1 (PP1) is a major eukaryotic serine/threonine protein phosphatase that regulates many diverse cellular processes. One study has shown that phosphatidic acid (PA) is an inhibitor of PP1. The investigators measured the activity of PP1 in the presence and absence of the inhibitor PA. The concentration of PA was 300 nM. [Substrate] (µM) Vo with PA (иM sec-1) Vo without PA (иM sec-1) 0.0209 0.010 0.00381 0.015 0.0335 0.00620 0.025 0.0419 0.00931 0.050 0.0838 0.0140

Human Anatomy & Physiology (11th Edition)

11th Edition

ISBN:9780134580999

Author:Elaine N. Marieb, Katja N. Hoehn

Publisher:Elaine N. Marieb, Katja N. Hoehn

Chapter1: The Human Body: An Orientation

Section: Chapter Questions

Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...

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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?
How can you find Kcat if you are only given Vmax, Km and [E] ? I tried using Kcat=Vmax/[E] but that didn't work. How do you know if [E] is the same as [E]total, and if it isn't how do you find it from this information: The Vmax for a particular enzyme is 10 nmols/L/s. The Km for its substrate is 5 microM. If the enzyme concentration is 10 nM, what is the kcat? a.80 nmoles/L/s b.8000 nmoles/L/s c.2 nmoles/L/s d.50 nmoles/L/s
The following table shows experimental results of the glucose transport rate, mM/sec, following facilitated diffusion by glucose carrier proteins. (Recall: the starting conc. L represents glucose added to one side of the membrane; distilled water, omM of glucose was added to the other side of the membrane). The rate of glucose transport was 0.0031 mm/sec with 8mM of glucose (run number 4, highlighted); the rate decreased to 0.0017 mM/sec with 10mM of glucose (run 5, highlighted). Why was the rate of glucose transport slower when the concentration gradient was increased? Experiment Results Run Number Solute 1 1 2 2 3 33 4 4 5 6 6 Na Ch Glucose Na Ch Glucose Na Ch Glucose Nat Ch Glucose Na Ch Glucose Nat Cl Glucose Start Conc. L Start Conc. R (MM) (mM) 0.00 0.00 2.00 0.00 0.00 0.00 8.00 0.00 0.00 0.00 2.00 0.00 0.00 0.00 8.00 0.00 0.00 0.00 10.00 0.00 2.00 0.00 2.00 0.00 Carriers 500 500 500 500 700 700 700 700 100 100 700 700 Rate (mm/sec) 0.0000 0.0008 0.0000 0.0023 0.0000 0.0010…
Two proteins, A and B, bind to the same ligand, L, with the binding curvesshown below. What is the dissociation constant, Kd , for each protein? Which protein (A or B) has a greater affinity for ligand L?
bo Consider a uniport system where a carrier protein transports an uncharged substance A across a cell membrane. Suppose that at a certain ratio of [A] inside to [A] outside, the AG for the transport of substance A from outside the cell to the inside, Macmillan Leaming A outside → Ainside, is 10.7 kJ/mol at 25°C. What is the ratio of the concentration of substance A inside the cell to the concentration outside? [A]inside [A]outside Choose the true statement about the transport of A under the conditions described. Because AG is positive, the ratio [A] inside/[A]outside must be less than one. Increasing [A]outside will cause AG for movement of Aoutside to Ainside to become a larger positive number. Movement of Ainside to Aoutside will be spontaneous. Decreasing the concentration of the uniport protein in the membrane will cause AG to become a smaller positive number.
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A transmembrane protein helps catalyze the transport of solute A from the INSIDE TO the OUTSIDE of a cell. The reaction can be represented by the reaction coordinate diagram below. It is possible to conclude that the concentration of A on the outside of the cell is than its concentration on the inside AND that it is_ to expect that the protein will at some point form a continuous path between th cytosol and the extracellular environment. A inside A outside reaction coordinate O lower; reasonable O higher; unreasonable O higher; reasonable free energy
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
Small molecules are used as inhibitors of protein action - as drugs. They most often do this by blocking the active site within the protein. Potential drugs can be screened computationally to determine if they are strongly bound to the protein. Figure 1 shows a possible conformation of a candidate drug molecule, 4-bromo-2- carboxymethylamide-pyrrole (abbreviation: BCMAP) at the active site of a protein (abbreviation: PR). Figure 2 shows the full protein structure whilst figure 3 shows a known inhibitor of the protein at the site, overlayed with another calculated conformer of BCMAP. (a) Explain what types of interactions, both intermolecular and intramolecular, that a molecular mechanics forcefield must be able to describe in order to be able to accurately determine the geometry of BCMAP in the protein. Identify which interactions will be the most important to describe accurately. Figure 1.4-bromo-2-carboxymethylamide-pyrrole (BCMAP) (C, N, O, and Br atoms in yellow, blue, red, and…
4. a. Calculate the KM (Michaelis constant) and the vmax (the maximum initial rate) for both substrates (sphingosine and ATP). Show your work, and be careful about units. b. threo-dihydrosphingosine, a stereoisomer of sphingosine, is an inhibitor of sphingosine kinase. What kind of inhibitor (competitive, uncompetitive, noncompetitive) is threo-dihydrosphingosine? Citing information from the Lineweaver-Burke plots, explain how you can tell.
1. Acid phosphatases are an important group of enzymes that can be detected in human bloodserum. Under slightly acidic conditions (pH 5.0), this group of enzymes can hydrolyzebiological phosphate esters as follows:R-O-P-O3-2 + H 2O R-OH + HO-P-O3-2.Acid phosphatases are produced and can be detected in erythrocytes, kidney, spleen, the liver,and prostrate gland. The enzyme from the prostrate gland is clinically important because anincreased activity in the blood is frequently an indication of cancer of the prostrate gland.Tartrate ion can strongly inhibit the phosphatase from the prostrate gland, but not acidphosphatases from other tissues. How can you use the information above to develop a specificprocedure for measuring the activity of the acid phosphatase of the prostrate gland in humanblood serum?
The KM of a Michaelis-Menten enzyme for a substrate is 1.0 x 104 M. At a substrate concentration of 0.20 M, v = 43 μmol/min. Calculate the rate of reaction when the substrate concentration is tenfold lower, 0.020M. 43 μmol min rate of reaction → Vmax Км substrate concentration →