Biology (MindTap Course List)
10th Edition
ISBN: 9781285423586
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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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_forwardMAKE 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…arrow_forward
- C)|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_forward10arrow_forwardPlease help with both the partsarrow_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_forwardEnzyme 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_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_forward
- Please give the both answer.arrow_forwardFill in blanks with increased decreases or levels offarrow_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_forward
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Enzyme Kinetics; Author: MIT OpenCourseWare;https://www.youtube.com/watch?v=FXWZr3mscUo;License: Standard Youtube License