Biology (MindTap Course List)
11th Edition
ISBN: 9781337392938
Author: Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg
Publisher: Cengage Learning
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Chapter 7, Problem 16TYU
Summary Introduction
To design: An experiment that will help to decide whether malonate is acting as a competitive inhibitor or a noncompetitive inhibitor.
Introduction: Enzymes are biological protein catalysts that alter the
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5. By using Excel or GoogleSheets. graph the Lineweaver-Burk plots for the behavior of
an enzyme for which the following experimental data are available. What are the Km and
Kwax values for the inhibited and uninhibited reactions? Is the inhibitor competitive or
noncompetitive?
[S]
(mM)
V, No Inhibitor
(mmol min-)
V, Inhibitor Present
(mmol min-')
1 × 10-4
5 × 10-4
1.5 x 10-3
2.5 x 10-3
5 x 10-3
0.026
0.010
0.092
0.136
0.040
0.086
0.150
0.120
0.165
0.142
1. Substrates and reactive groups in an enzyme’s active site must be precisely aligned in order for a productive reaction to occur. Why, then, is some conformational flexibility also a requirement for catalysis?
2. Some plants contain compounds that inhibit serine proteases. It has been hypothesized that these compounds protect the plant from proteolytic enzymes of insects and microorganisms that would damage the plant. Tofu, or bean curd, possesses these compounds. Manufacturers of tofu treat it to eliminate serine protease inhibitors. Why is this treatment necessary?
MAKE A GRAPH FOR ME ON GRAPH PAPER CALL IT ENZYMES VS RATE OF REACTION USING TABLE BELOW
GRAPH paper INSERTED BELOW
rules: data points must be an x or circled dot, must be on grid paper , the independant variable on the x axis and dependant variable on the y axis, must include titles
Regarding the data points:
- H2O2 + MnO2 Control #1: (Control #1, 5)- H2O2 + sand control #2: (Control #2, 0)- Plant versus Animal Liver Catalase: (Liver, 4)- Potato: Plant vs. Animal Catalase: (Potato, 3)- Substance Enzyme Concentration (Used Liver): (Liver Used, 4)- Substance Enzyme Concentration (Used H2O2): (Used H2O2, 1)
- Boiling Water Bath Temperature: (Boiling Water Bath, 5)- Ice Water Bath Temperature: (Ice Water Bath, 2)- HCl, or pH 3: (H 3, 4)- NaOH at pH 12: (pH 12, 2)- pH 7 (H2O): (assuming average of pH readings; pH 7, not specified)
The following explains how to display the graph:
Title: Factors versus Enzyme Activity Rate - Labels on X- and Y-axes: Factors and Rate of Enzyme…
Chapter 7 Solutions
Biology (MindTap Course List)
Ch. 7.1 - Define energy, emphasizing how it is related to...Ch. 7.1 - Use examples to contrast potential energy and...Ch. 7.1 - Prob. 1CCh. 7.2 - Prob. 3LOCh. 7.2 - Prob. 1CCh. 7.2 - Life is sometimes described as a constant struggle...Ch. 7.3 - Prob. 4LOCh. 7.3 - Prob. 5LOCh. 7.3 - Prob. 6LOCh. 7.3 - Prob. 1C
Ch. 7.3 - Prob. 2CCh. 7.4 - Explain how the chemical structure of ATP allows...Ch. 7.4 - Prob. 1CCh. 7.4 - Prob. 2CCh. 7.5 - Relate the transfer of electrons (or hydrogen...Ch. 7.5 - PREDICT Which has the most energy, the oxidized...Ch. 7.6 - Explain how an enzyme lowers the required energy...Ch. 7.6 - Describe specific ways enzymes are regulated.Ch. 7.6 - Prob. 1CCh. 7.6 - How does the function of the active site of an...Ch. 7.6 - How are temperature and pH optima of an enzyme...Ch. 7.6 - Prob. 4CCh. 7 - Which of the following can do work in a cell? (a)...Ch. 7 - Prob. 2TYUCh. 7 - Prob. 3TYUCh. 7 - Test Your Understanding 4. Diffusion is an (a)...Ch. 7 - Prob. 5TYUCh. 7 - Prob. 6TYUCh. 7 - Prob. 7TYUCh. 7 - Test Your Understanding 8. Induced fit means that...Ch. 7 - Prob. 9TYUCh. 7 - Prob. 10TYUCh. 7 - PREDICT In the following reaction series, which...Ch. 7 - Test Your Understanding 12. EVOLUTION link All...Ch. 7 - EVOLUTION LINK Some have argued that evolution is...Ch. 7 - Prob. 14TYUCh. 7 - Prob. 15TYUCh. 7 - Prob. 16TYU
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- In the following graph: A represents the product. B represents the energy of activation when enzymes are present. C is the free energy difference between A and D. C is the energy of activation without enzymes. E is the difference in free energy between the reactant and the products.arrow_forwardTest Your Understanding 8. Induced fit means that when a substrate binds to an enzymes active site, (a) it fits perfectly, like a key in a lock (b) the substrate and enzyme undergo conformational changes (c) a site other than the active site undergoes a conformational change (d) the substrate and the enzyme become irreversibly bound to each other (e) c and darrow_forwardC)|Myth: The specificity of an enzyme for its substrate is explained by the lock and key hypothesis. Fact: The lock and key hypothesis is outdated! What is our current model for understanding regarding how enzymes recognize and bind to substrates?arrow_forward
- Fill the blank with the option below The substrate concentration at which an enzyme-catalyzed reaction proceeds at on-half its maximum velocity The rate enhancement of the enzyme catalyzed reaction over uncatalyzed reaction The rate of the reaction when the concentration of the products is zero and the reverse rate is negligible The rate at which the enzyme-substrate complex is formed The number of time an enzyme molecule transforms a substrate molecule per unit of time The proportionality constant that relates the velocity of a chemical reaction to the concentrations of the reactants The maximum velocity of an enzymatic reaction when the binding site is saturate with substrate A measure of the catalytic activity of an enzyme at low concentrations of substrate The rate constant for the equilibrium between the reactants and the enzyme-substrate complex 1. Vo 2. 4. 3. V₁ Acronym Parameter 6. 7. I 5. Km 8. 1 I 9. AAG 10. AG rate constant rate of disappearance I specificity constant…arrow_forwardInhibitor X exerts which of the following effects on the above enzyme (lactase)? (inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM) pure non-competitive inhibition uncompetitive inhibition competitive inhibition all of the above none of the abovearrow_forward4. Predict the reactivity of trypsin at pH 14. C 100 50 0° 20 40° 60°C Temperature 5. What is the optimal temperature for the enzyme in graph C? Where does the enzyme in graph C most likely function? 6. Using graph C: a. Explain what happens when hypothermia sets in (when enzymes get too cold!) Reaction Rate vs Substrate Concentration ege 3 of 4 635 words English (U.S.) Text Predictions: On DII F4 近 Percent maximum activityarrow_forward
- Inhibitor X exerts which of the following effects on the above enzyme (maltase)? (inhibitor X changes maltase activity to a Vo of 0.10 mM per minute when [S] = 0.125 mM, and a Vo of 0.25 mM per minute when [S] = 0.50 mM) competitive inhibition pure non-competitive inhibition uncompetitive inhibition all of the above none of the abovearrow_forwardDESIGN YOUR ENZYME AND SHOW THE REACTION! To trap glucose in a cell, the following reaction is catalyzed by the HEXOKINASE enzyme: GLUCOSE (G) + ATP -> GLUCOSE-6-Phosphate (G6P) + ADP Diagram and label the steps in the catalytic cycle of this reaction in the presence of the hexokinase enzyme. Show the steps of how the REACTANTS are converted to PRODUCTS (you can use shapes to represent the different molecules) From what you know about the structure and polarity of the reactants, predict the amino acid R groups that might be in the ACTIVE SITEarrow_forwardThe following statements refer to enzyme inhibition. Match the statement to the one of the following descriptors to which it is best associated. Descriptors: competitive inhibition; non-competitive inhibition; un-competitive; covalent inhibition. 9a. Inhibition is not reversed even after the inhibitor (1) is removed from solution by dialysis or drug metabolism/excretion. 9b. Inhibitor and substrate reversibly compete for occupancy of a common binding site 9c. The inhibitor binds reversibly only to the preformed E.S (enzyme-substrate) complex forming an inactive E.S.I. 9d. The inhibitor binds reversibly and independently of substrate to an allosteric site producing E.I or a ternary E.S.I complex which can't form product. 9f. The relative amount of inhibition decreases as [S] (the concentration of substrate) increases and S better competes for occupancy of the active site.arrow_forward
- Enzyme X follows Michaelis-Menten kinetics. You add an inhibitor to your enzyme and you notice that the Vmax has decreased while the Km for enzyme X has increased as a result of adding the inhibitor. What are you able to conclude from this information? The inhibitor must be competitive The amount of total enzyme available to catalyze the reaction in the presence of the inhibitor has likely decreased The enzyme has a higher affinity for its substrate in the presence of the inhibitor The substrate concentration required to reach 1/2 of the maximum velocity for this enzyme has increased as a result of the inhibitor More than one of the above are conclusions that can be drawn from this informationarrow_forwardA dichotomous key works by determining positive and negative reactions to different biochemicals. Why does this method allow us to identify one species of organism from another? – for this question, do not describe the dichotomous key procedure, think about why it works, what are biochemical reactions based on? Think enzymatic pathways, what are enzymes, what are they a reflection of? Keep the answer between 2-3 sentencesarrow_forward6X9 X What are the steps for an enzyme to create a product in order: 1. The energy of activation is lowered so the reaction can happen quicker. 2. Substrate attaches to the active site. 3. The product is created and released from the active site. 4. The energy of activation is lowered so the reaction can happen slower. 5. A specific substrate attaches to the active site. Selected Answer: Answers: 5 -- 4 --> 3 5--1--> 3 2 1 -> 3 5 -- 4 --> 3 2--> 4 --> 3arrow_forward
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