MICROBIOLOGY-MASTERING MICRO.-ACCESS
11th Edition
ISBN: 9780321802705
Author: Tortora
Publisher: PEARSON
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Chapter 5, Problem 2A
The following graph shows the normal
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Please handraw this graph with all the necessary detailed information:
Imagine that I text enzyme rate for four different temperatures: 10 degrees celsius, 20 degrees celsisus, 30 degree celsius, and 40 degree celsius, in separate tubes. The enzyme appears to work faster as temperature increases, but completely ceases activity at 40 degrees celcius. Sketch a graph to show this outcome, but here you will graph product formation (nmoles/mL) vs. time (minutes). The graph should be 4 lines and HANDDRAWN. Include a legend if necessary. You do not need precise quantitivate values, but most show the correct trends on the graph.
In the scheme below which represents the mechanism of action for a large number of enzymes:
A+B⟺AB⟶C
The steady state approximation is reached when:
d[AB]/dt≈0
k2≫k1
k−1≫k1
k−1=k1
The fractional saturation of an enzyme is 80% at a substrate concentration of
20 mM. In the presence of 5 mM of a competitive inhibitor the fractional
saturation at the same substrate concentration is 40%. The Km for this
reaction is
mM, and the apparent Km is
mM. The value of alpha is
while the Ki
of the inhibitor is
mM.
Chapter 5 Solutions
MICROBIOLOGY-MASTERING MICRO.-ACCESS
Ch. 5 - Prob. 1RCh. 5 - DRAW ITUsing the diagrams below, show each of the...Ch. 5 - DRAW IT An enzyme and substrate are combined. The...Ch. 5 - Define oxidation-reduction, and differentiate the...Ch. 5 - There are three mechanisms for the phosphorylation...Ch. 5 - All of the energy-producing biochemical reactions...Ch. 5 - Fill in the following table with the carbon source...Ch. 5 - Write your own definition of the chemiosmotic...Ch. 5 - Why must NADH be reoxidized? How does this happen...Ch. 5 - NAME IT What nutritional type is a colorless...
Ch. 5 - Explain why, even under ideal conditions,...Ch. 5 - The following graph shows the normal rate of...Ch. 5 - Compare and contrast carbohydrate catabolism and...Ch. 5 - How much ATP could be obtained from the complete...Ch. 5 - The chemoautotroph Acidithiobacillus can obtain...Ch. 5 - Which substance in the following reaction is being...Ch. 5 - Which of the following reactions produces the most...Ch. 5 - Prob. 3MCQCh. 5 - Which of the following compounds has the greatest...Ch. 5 - Prob. 5MCQCh. 5 - Prob. 6MCQCh. 5 - Which culture produces the most lactic acid? Use...Ch. 5 - Which culture produces the most ATP? Use the...Ch. 5 - Which culture uses NAD+? Use the following choices...Ch. 5 - Which culture uses the most glucose? Use the...Ch. 5 - Haemophilus influenzae requires hemin (X factor)...Ch. 5 - The drug Hivid, also called ddC, inhibits DNA...Ch. 5 - The bacterial enzyme streptokinase is used to...
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- When studying the mechanism of the enzymatic reaction, functional groups were found that ensure the connection of the enzyme molecule with the substrate and take a direct part in the act of catalysis. What are these areas of the enzyme formed by these groups called? What functional structures form them and why?arrow_forwardIn an enzyme-catalyzed reversible reaction what happens when a) rate of change of enzyme-substrate complex concentration with time is positive b) rate of change of enzyme-substrate complex concentration with time is zero 9:0arrow_forwardThe graphs 3 and 4 representing 1/Vo = f(1/[S]o) have been done in the presence of a competitive (CI) and noncompetitive inhibitor (NCI). a- For each figure, determine from the relative position of the straight lines which one is obtained in presence of an inhibitor. b- Indicate which graph corresponds to the competitive inhibition and which one the noncompetitive inhibition. Justify your answer. c- Complete the graphs by indicating which values can be determined from the arrows. 3 1/No 1/[S]⁰ (4) 1/No 1/[S]oarrow_forward
- A researcher has measured the initial rate of an enzyme-catalyzed reaction as a function of substrate concentration in the presence and absence of 0.001μM inhibitor. She obtains the following data. What is the Vmax and Km for the "no inhibitor" and "+ inhibitor" experiments. Ans, What would be value of V0 for the "+inhibitor" at [S] = 0.4μM and the inhibitor is most likely a competitive, uncompetitive or mixed inhibitor.arrow_forwardKnowing that for a bacterial colony to be able to grow it must produce product "3" AND "4", use the information in the image to describe which enzyme(s) are that are Non-Functional in Colony C? Please note error in enzyme description at bottom of image. X converts 1 into 2; Y converts 1 into 3; and z converts 2 into 4.arrow_forwardAspartate transcarbamoylase, which is necessary for CTP production, is an essential enzyme for the human body. In the below graph, which line represents the rate of the reaction catalyzes by Aspartate transcarbamoylase? Explain.arrow_forward
- An experiment on enzyme-catalyzed reaction was conducted in the laboratory by a student. Results obtained are summarized in the table below. In all the experiments, the concentration of the enzyme is the same. Substrate Concentration Velocity (pmol/min) (pmol) 1.5 0.21 3 0.28 4 0.32 0.36 8 0.4 15 0.45 18 0.47 1. Plot or graph these results using the Lineweaver-Burk method. 2. Determine the KM and Vmax values. Show all equations and calculations.arrow_forwardOne of the hallmarks of competitive inhibition is that there is constant competition betweenthe substrate and the inhibitor for binding to the enzyme active site.a) If [inhibitor] >> [substrate], which compound “wins” (i.e., occupies the active site a greaterpercentage of the time)?b) If [substrate] >> [inhibitor], which compound “wins” (i.e., occupies the active site a greaterpercentage of the time)?arrow_forwardPlease note the reaction expression below. Which of the following rate constants describes the breakdown of the enzyme-substrate complex? There may be more than one answer. k₂ E+S OK-2 U k₂ 0 k₁ OK.1 k3 k.3 SES EPE+P K.3 K.₂ k.arrow_forward
- Define the following as they apply to an enzymatic reaction. Give an example of each. a. Enzyme-limited: b. Substrate-limited:arrow_forwardThe following reaction coordinate diagram charts the energy of a substrate molecule (S) as it passes through a transition state (X‡) on its way to becoming a stable product (P) alone or in the presence of one of two different enzymes (E1 and E2). How does the addition of either enzyme affect the change in Gibbs free energy (ΔG) for the reaction? Which of the two enzymes binds with greater affinity to the substrate? Which enzyme better stabilizes the transition state? Which enzyme functions as a better catalyst?arrow_forwardA graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (UO) at different substrate concentrations ([S]). First, move the line labeled Vmax to a position that represents the maximum velocity of the enzyme. Next, move the line labeled 1/2 Vmax to its correct position. Then, move the line labeled Km to its correct position. Estimate the values for V max and Km. 1/2 V max Michaelis-Menten curve 300 V max 275 250 225 200 v (μM/min) 175 150 K 2 125 m 100 75 50 25 0 0 10 20 20 50 58 30 40 50 60 60 [S] (μM) 70 10 60 80 90 100arrow_forward
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