Microbiology: An Introduction (13th Edition)
13th Edition
ISBN: 9780134605180
Author: Gerard J. Tortora, Berdell R. Funke, Christine L. Case, Derek Weber, Warner Bair
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: An Introduction (13th Edition)
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 - 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 - 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 - 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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- In 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_forwardA 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_forward
- An enzyme that catalyzes the reaction X ⇌ Y is isolated from two bacterial species. The enzymes have the same Vmax but different Km values for the substrate X. Enzyme A has a Km of 2.0 μM, and enzyme B has a Km of 0.5 μM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X] = 1 μM. Which curve corresponds to which enzyme? Explain.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_forwardAn 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_forward
- Consider an enzyme-catalyzed reaction that follows Michaelis-Menten kinetics with KM =3.0 mmol·dm-3. What concentration of a competitive inhibitor with KI = 20 umol·dm-3 willreduce the rate of formation of product by 50% when the substrate concentration is held at 0.10mmol·dm-3.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_forwardDefine the following as they apply to an enzymatic reaction. Give an example of each. a. Enzyme-limited: b. Substrate-limited:arrow_forward
- The 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_forwardOne way of expressing the rate at which an enzyme can catalyze a reaction is to state its turnover number. The turnover number is the maximum number of substrate molecules that can be acted on by one molecule of enzyme per unit of time. The table gives the turnover number of four representative enzymes. Enzyme Substrate Turnover number (per second) Ribonuclease RNA 100 Fumarase fumarate 800 Lactate dehydrogenase lactate 1000 Urease urea 10,000 How many molecules of urea can one molecule of urease act on in 12.0 min ?arrow_forwardA mixed inhibitor of an enzyme (sometimes called a mixed non-competitive inhibitor) can decrease the rate of a reaction by any of the following EXCEPT by: a) binding to a site other than the active site of the enzyme. b) binding to the active site of the enzyme, preventing substrate binding. c) decreasing kcat. d) Increasing KM.arrow_forward
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