Concept explainers
(a)
Interpretation:
The hypothesis for the different results of ibuprofen and indomethacin inhibitors should be determined.
Concept introduction:
The substances that are added in a reaction in order to slow down it or to prevent the formation of unwanted products are called inhibitors. These are just the opposite of catalysts. There are three types of inhibitors:
- Competitive inhibitors: These occupy the region where the substrate is to be placed and therefore it blocks the active site of the enzyme. Methotrexate is an example of a competitive inhibitor.
- Uncompetitive inhibitors: These bind to enzyme-substrate complex and therefore the enzyme becomes unable to react with the substrate. Lansoprazole is an example of an uncompetitive inhibitor.
- Non-competitive inhibitors: These bind to some region of the enzyme other than the active site and alter its shape. Phenethyl isothiocyanate is an example of a non-competitive inhibitor.
(b)
Interpretation:
The results with the use of aspirin should be determined.
Concept introduction:
The substances that are added in a reaction in order to slow it down or to prevent the formation of unwanted products are called inhibitors. These are just the opposite of catalysts. There are three types of inhibitors:
- Competitive inhibitors: These occupy the region where the substrate is to be placed and therefore it blocks the active site of the enzyme. Methotrexate is an example of a competitive inhibitor.
- Uncompetitive inhibitors: These bind to enzyme-substrate complex and therefore the enzyme becomes unable to react with the substrate. Lansoprazole is an example of an uncompetitive inhibitor.
- Non-competitive inhibitors: These bind to some region of the enzyme other than the active site and alter its shape. Phenethyl isothiocyanate is an example of a noncompetitive inhibitor.
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Chapter 12 Solutions
BIOCHEM-ACHIEVE(FIRST DAY DISCOUNTED)
- 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)arrow_forwardenzyme-inhibitor complex requires 450 kJ.mol-1 to dissociate and that it displays kinetics somehow similar to non-competitive inhibition, this enzyme is good to use to inhibit toxanthine oxidase in the case of hyperuricemia and gout?arrow_forwardName: Hendric Nduwuba DNP and Cellular Respiration Summary https://www.youtube.com/watch?v=eLtRan2qFx0 Your assignment today is to summarize in two paragraphs (8 sentences total) the effects of dinitrophenol (DNP) on the human body. Make sure you explain what DNP is, the stages of cellular respiration, and how DNP alters the metabolism of the person using it (side effects). The following vocabulary terms need to be used: cellular respiration, adenosine triphosphate, and dinitrophenol (DNP). Your objective is to describe the role of ATP in cellular respiration in order to explain how DNP affects the rate of metabolism. The first paragraph is an introduction to the summary and the second paragraph is an explanation of the main idea. You will be graded on this criteria: main idea, explanation of cellular respiration, explanation of DNP, and paragraph structure.arrow_forward
- enzyme-inhibitor complex requires 450 kJ.mol-1 to dissociate and that it displays kinetics somehow similar to non-competitive inhibition, does this is inhibitor good to use to inhibit toxanthine oxidase in the case of hyperuricemia and gout?arrow_forwardIn 2-page worth of words (around 500), discuss CYCLAMATE's regulation, allowable levels, and what group of people are at high risk for the side effects of CYCLAMATE in the body.arrow_forwardPlease help me with this one. Cause and Effect: 1. A wet pipette was used to transfer 10.00 mL of 1M HNO3 in the determination of ΔHrxn. [magnitude of ΔHrxn] a. Increase b. Decrease c. No effect 2.The test tube used in the determination of the ΔHrxn for nitric acid was still wet with water. [ magnitude of ΔHrxn] a. Increase b. Decrease c. No effect 3.The actual concentration of the sodium hydroxide used in the calibration part of the experiment was lower than the stated value. [Ccal] a. Increase b. Decrease c. No effect 4.The actual concentration of the sodium hydroxide used in the calibration part of the experiment was lower than the stated value. However, the NaOH solution used in the determination of ΔHrxn was of the correct concentration. [magnitude of ΔHrxn] a. Increase b. Decrease c. No effect 5.The thermometer and stopper were inserted to the test tube 5 minutes after adding the base in the calibration part. [Ccal] a. Increase b. Decrease c. No effectarrow_forward
- Decoupling agents such as 2,4-DNP can result in altered metabolic activity. Explain what 2,4-DNP is, describe how it alters metabolic activity, and why this could be dangerous.arrow_forwardLong explanations are not needed. Direct answers would suffice. a. High concentration of glucose 6-phosphate is inhibitory to the enzyme hexokinase. I. True II. False b. Which compound is formed when NAH reduces pyruvate (anaerobic condition)? I. lactate II. alpha-ketoglutarate III. glucose IV. oxaloacetatearrow_forwardTABLE 3-LACTATE PRODUCTION IN FORTIFIED HEMOLYSATES OF HUMAN ERYTHROCYTES* Substrate Glucose Glucose Lactate production† No. of experiments pH 6 7.1 2.03 ± 0.91 6 7.8 4.76 ± 1.09 7-1 10-73 +1-88 5 7.8 12.34 ±2.92 5 7.0 7-15±0.73 5 7-7 (b)( ) In mature erythrocytes (red blood cells) the end product of glycolysis is lactate because of the absence of mitochondria. On the right is a table comparing the rate of lac- tate production in hemolysates (lysed cells) of human RBCs as a function of pH with dif- ferent substrates introduced into the glyco- lytic pathway. The hemolysate was fortified with 30 μmoles substrate, 7.5 μmoles MgCl2, 10 μmoles disodium phosphate, 1.5 μmoles NAD and 5 μmoles ATP in a volume of 5 mL. The rate of lactate production is given as μmoles of lactate/g Hb/hr at 37° C, buffered to either pH 7.1 or 7.8, as indicated. According to the results in the table which glycolytic enzyme is rate-limiting? Explain. Glucose-6-phosphate Glucose-6-phosphate Fructose-1,6-diphosphate…arrow_forward
- Each plot has its own limitations. Match the plot with the correct statement on limitations. 1.Hanes-Woolf 2.Michaelis Menten 3.Lineweaver-Burk limitations: A. This plot appears to allow for more accurate calculation of these parameters, but is prone to error, as the y-axis takes the reciprocal of the rate of reaction –this increases any small errors in measurement at low substrate concentration, where they are most likely to occur. B.The advantage of this plot is that it does not not overemphasizing the data obtained at low [S]. The disadvantage is that any inaccurate measures of substrate concentration will be exaggerated as [S] is used on both axes. C.It is difficult to obtain a meaningful value of Km from this plot because of the high substrate concentrations needed to reach Vmax. Vmax has to be estimated by extrapolation- which is partly subjective and introduces error. D.The advantage of this plot is that it overcomes uneven spacing of points, and undue weight of a…arrow_forwardDirections: Solve the following problem: The enzyme ẞ-methylaspartase catalyzes the deamination of ẞ-methylaspartate: CH,NH, CH OOC-CH-CH-COOOOC-CH-CH-COO+NH mesaconate Williams and Selbin The effects of hydroxymethylaspartate as an inhibitor for this enzyme was studied. The following data wer Substrate Concentration obtained: Reaction Rate without inhibitor (mM) 1 × 10-4 5 x 10-4 1.5 x 10-3 2.5 x 10-3 5 x 10-3 (mM/s) 0.026 0.092 0.136 0.150 0.165 Reaction Rate with inhibitor (mM/s) 0.010 0.040 0.086 0.120 0.142 Use Lineweaver-Burk plot to determine the KM and Vmax of the enzyme in the absence of inhibitor. Moreover, determine as well whether the inhibitor is competitive or noncompetitive. Show the graphs and calculations below.arrow_forwardGlucose can be isomerized to fructose to glucose isomerase. The enzyme kinetics of this enzyme was studied in the presence of an inhibitor X. Using the data provided, complete the table and determine what type of inhibitor is X. 1/ [S] , 1/M 1/V, w/o inhibitor Min/M 0.22 1/V, w/ inhibitor 0.33 0.27 0.2 0.16 0.20 0.14 0.13 0.16 0.11 0.11 0.14 0.09 0.11 0.13 Km/Vmax 1/Vmax Km Vmaxarrow_forward
- BiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage Learning