Campbell Biology
12th Edition
ISBN: 9780135188743
Author: Urry
Publisher: PEARSON
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Textbook Question
Chapter 8.4, Problem 3CC
WHAT IF? Ø Malonate is an inhibitor of the enzyme succinate dehydrogenase. How would you determine whether malonate is a competitive or noncompetitive inhibitor?
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. Consider the enzymatic reaction scheme: Asparagine + H20 Aspartate + NH3:
a) calculate the specific activity of the enzyme, if in 30 seconds as a result of
a reaction involving 3 mg of the enzyme under optimal conditions (pH 8.0,
37 °C) 75 umol aspartate is obtained;
b) describe the reasons for the decrease in enzyme activity after incubation for
6.
10 minutes at 70 °C (provide an appropriate graph).
How does the allosteric enzyme kinetics different from M-M kinetics? Draw a graph to show allosteric enzyme kinetics (initial rate vs [S]. Show how the graph change in the presence of + effector and negative effector.
Suppose that the data below are obtained for an enzyme catalyzed reaction in the presence and absence of inhibitor Y.
[S] (mM)
V (mmol/mL*min)
Without Y
With Y
0.2
5.0
2.0
0.4
7.5
3.0
1.8
10.0
4.0
1.0
10.7
4.3
2.0
12.5
5.0
4.0
13.6
5.5
a.) Determine the type of inhibition that has occurred
b.) Does inhibitor Y combine with E, with ES or with both? Explain
c.) Calculate the inhibitor constant, Ki, for substance Y, assuming that the final concentration of Y in the reaction mixture was 0.3mM
Chapter 8 Solutions
Campbell Biology
Ch. 8.1 - Prob. 1CCCh. 8.1 - Describe the forms of energy found in an apple as...Ch. 8.1 - WHAT IF? If you place a teaspoon of sugar in the...Ch. 8.2 - Cellular respiration uses glucose and oxygen,...Ch. 8.2 - VISUAL SKILLS How would the processes of...Ch. 8.2 - Prob. 3CCCh. 8.3 - How does ATP typically transfer energy from an...Ch. 8.3 - Prob. 2CCCh. 8.3 - MAKE CONNECTIONS Does Figure 8.11a show passive...Ch. 8.4 - Many spontaneous reactions occur very slowly. Why...
Ch. 8.4 - Prob. 2CCCh. 8.4 - WHAT IF? Malonate is an inhibitor of the enzyme...Ch. 8.4 - Prob. 4CCCh. 8.5 - How do an activator and an inhibitor have...Ch. 8.5 - Prob. 2CCCh. 8 - Explain how the highly ordered structure of a cell...Ch. 8 - Explain the meaning of each component in the...Ch. 8 - Describe the ATP cycle: How is ATP used and...Ch. 8 - How do both activation energy barriers and enzymes...Ch. 8 - Prob. 8.5CRCh. 8 - Choose the pair of terms that correctly completes...Ch. 8 - Prob. 2TYUCh. 8 - Which of the following metabolic processes can...Ch. 8 - Prob. 4TYUCh. 8 - Some bacteria art metabolically active in hot...Ch. 8 - If an enzyme is added to a solution where its...Ch. 8 - Prob. 7TYUCh. 8 - EVOLUTION CONNECTION Some people argue that...Ch. 8 - Prob. 9TYUCh. 8 - WRITE ABOUT A THEME: ENERGY AND MATTER Life...Ch. 8 - Prob. 11TYU
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- -A hypothetical enzyme has a Km value of 10 mM and a kcat value of 100 s-1. A competitive inhibitor, present at its KI value of 7 mM, would result in the following new values: a) K'm =7mM,k'cat =50s-1 b) K'm = 10 mM, k'cat = 70 s-1 c) K'm = 10 mM, k'cat = 50 s-1 K'm = 5 mM, k'cat = 100 s-1 e) None of the above %3D %3D %3Darrow_forwardHello, For the image included: A) Which of the following statements are true of the malate dehydrogenase reaction in biochemical standard conditions? Choose all that apply. Group of answer choices: The reaction is spontaneous since ∆G°' is positive The reaction is spontaneous since ∆G°' is negative The reaction is not spontaneous since ∆G°' is positive The reaction is not spontaneous since ∆G°' is positive The equilibrium favors products since K is greater than 1 The equilibrium favors reactants since K is greater than 1 The equilibrium favors products since K is less than 1 The equilibrium favors reactants since K is less than 1 The reaction is always at equilibrium B) If the concentration of oxaloacetate is 107 times lower than the concentration of malate. Is the reaction spontaneous? Assume biochemical standard conditions. Group of answer choices No, because RTlnQ is very positive Yes, because RTlnQ is very positive No, because RTlnQ is very…arrow_forwardList 4 major types of inhibition modes and clearly indicate the effect on Vmax and KM for each mode?2. What is the effect of each of the 4 types of inhibitors on the initial rate of an enzyme catalyzed reaction?3. A potent inhibitor effectively inhibits an enzyme catalyzed reaction. What kind of a Ki value you would expect for a potent inhibitor?4. Considering PNPP → PNP reaction, would you expect to see more intense or pale color for the reaction that contain the inhibitor? Explain.arrow_forward
- Consider the enzyme-catalyzed conversion of a substrate S into a product in the presence of an inhibitor I with the following properties (mixed inhibition): (i) The inhibitor competes with S to bind the active site of the enzyme with dissoci- ation constant K¡ for the complex (EI). (ii) The inhibitor is also capable of binding the enzyme at a secondary site of the enzyme with dissociation constant K{ for the complex (IES). The complex (IES) is unable to form P. iii) The dissociation constants of the (IES) and (EI) complexes satisfy the relation K{ = K1/2. What is the value of the ratio vo/(k2 [E]o) when the concentration of the substrate is [S]o = KM/3 and the concentration of the inhibitor is [I]o = K{/4? Recall that KM is the Michaelis constant.arrow_forwardRecall that phosphonacetyl L - aspartate (PALA) is a potent inhibitor of ATCase because it mimics the two physiological substrates. However, in the presence of substrates, low concentrations of this unreactive bisubstrate analog increase the reaction velocity. On the addition of PALA, the reaction rate increases until an average of three molecules of PALA are bound per molecule of enzyme. This maximal velocity is 17-fold greater than it is in the absence of PALA. The reaction rate then decreases to nearly zero on the addition of three more molecules of PALA per molecule of enzyme. Why do low concentrations of PALA activate ATCase?arrow_forwardVmax [S] Vo = Km+ [S] Eadie-Hofstee plot Lineweaver-Burk (L-B) plot v=Vm-Km [S] Km 1 Vm Vm [S] The equations above apply for Michaelis-Menten enzyme kinetics but are presented in three different formats. For competitive inhibition The Michaelis-Menten equation becomes Vo=Vmax[S]/(aKm+[S]) Put the Michaelis-Menten equation for competitive inhibition in the Eadie-Hofstee format. The Y-intercept of this plot will bearrow_forward
- Select the incorrect statement. With regards to free energy ΔG of the reaction below E+S ⇌ ES Negative ΔG mean the reaction toward is facourable More negative value of ΔG indicates stronger binding to S to E It is possible to compute disassociation constant from the ΔG value alone It is possible to calculate the term ( ΔH – T ΔS) from the value of ΔGalone ΔG = 0 indicates (ES)/(E)(S) =1 None of the abovearrow_forwardCould you help me with a scheme following: Carboxylic acids do not readily undergo certain biochemical reactions; thus, they are often converted to thioesters or phosphonoesters for catalysis. What are two reactions that thioesters and phosphonoesters can undergo that carboxylic acids cannot? Show enzyme-catalyzed arrow pushing for both of these mechanisms.arrow_forwardTo investigate the effects of a newly discovered inhibitor and enzymology list obtained the results showed in the figure below. Which statement best describes the binding of its inhibitor on the enzyme? ~y=0.768r+4.7425 3-0.568x+46576 19 . . . . ● x X The inhibitor binds at the active site. The inhibitor binds away from the active site, but after substrate binding. The inhibitor binds away from the active site before or after substrate binding. The inhibitor binds close to the active site, but after substrate binding. The inhibitor binds close to the active site, but prior to substrate binding. red by a strainarrow_forward
- Given the active site and reaction mechanism below, what is the mechanism of irreversible inhibition of the inhibitor provided? NH* Active Site *H₂N. HN NH₂ +H₂N HN H₂N NH₂* Non-specific inhibition Uncompetitive Inhbitor Transition State Analog i Affinity-based inhibition Mechanism-Based Inhibition Reaction Mechanism *H₂N. NH HN- HN [*] H₂N NH₂ 2+Mn Mn²+ i +H₂N. H₂N i H₂N NH₂ HO Inhibitor *H₂N. B OH r View site informatiarrow_forward. The steady-state kinetics of an enzyme are studied in the absence and pres- ence of an inhibitor (inhibitor A). The initial rate is given as a function of substrate concentration in the following table: v[(mmol/L)min '] [S] (mmol/L.) No inhibitor Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38 (a) What kind of inhibition (competitive, uncompetitive, or mixed) is involved? (b) Determine Vmax and KM in the absence and presence of inhibitor.arrow_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_forward
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