BIOL:CONCEPT+INVEST.ETEXT
5th Edition
ISBN: 9781264154173
Author: Hoefnagels
Publisher: MCG
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Chapter 23.5, Problem 2MC
Summary Introduction
To describe:
The purpose of two control solutions.
Concept introduction:
Carnivorous plants have the mechanism to trap insects and are adapted to digest the protein matter contained in insects. They feature a deep cavity filled with liquid, insects such as flies are attracted to this cavity, often by color and by the nectar within.
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The following data, presented by G. Bowes and W. L. Ogre in J. Biol. Chem.
(1972) 247:2171-2176, describe the relative rates of incorporation of CO, by
Rubisco under N, and under pure Oz. Decide whether Oz is a competitive or
uncompetitive inhibitor.
[CO] (mM)
Under N2
Under O2
0.20
16.7
10
0.10
12.5
5.6
0.067
8.3
4.2
0.050
7.1
3.2
Why does a pure noncompetitive inhibitor not changethe observed KM?
Why does the apparent KM decrease in the presence ofan uncompetitive inhibitor?
Chapter 23 Solutions
BIOL:CONCEPT+INVEST.ETEXT
Ch. 23.1 - How do plants acquire C, H, O, N, and P?Ch. 23.1 - Prob. 2MCCh. 23.1 - Prob. 3MCCh. 23.2 - Prob. 1MCCh. 23.2 - Prob. 2MCCh. 23.2 - Prob. 3MCCh. 23.3 - Prob. 1MCCh. 23.3 - Prob. 2MCCh. 23.3 - Prob. 3MCCh. 23.4 - Prob. 1MC
Ch. 23.4 - Prob. 2MCCh. 23.5 - Prob. 1MCCh. 23.5 - Prob. 2MCCh. 23.5 - Prob. 3MCCh. 23 - Prob. 1MCQCh. 23 - Prob. 2MCQCh. 23 - Prob. 3MCQCh. 23 - Prob. 4MCQCh. 23 - Prob. 5MCQCh. 23 - Prob. 6MCQCh. 23 - Prob. 7MCQCh. 23 - Explain the relationship between transpiration and...Ch. 23 - Prob. 2WIOCh. 23 - Prob. 3WIOCh. 23 - Prob. 4WIOCh. 23 - Prob. 5WIOCh. 23 - Prob. 6WIOCh. 23 - Prob. 7WIOCh. 23 - Prob. 8WIOCh. 23 - Prob. 9WIOCh. 23 - Prob. 10WIOCh. 23 - Prob. 11WIOCh. 23 - Prob. 12WIOCh. 23 - Prob. 13WIOCh. 23 - Prob. 1PITCh. 23 - Prob. 2PITCh. 23 - Prob. 3PIT
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- Calculate, or make a best estimate of, the unknown factors in the situations listed in Table 1. Table 1. Data for problem 1. To find Data (i) Ek [K+]out = 4 mM; [K+]in = 130 mM (ii) [Cl-]in [Cl-]out = 570 mM; ECl = -65 mV (iii) ECl [Cl-]out = 150 mM; [Cl-]in = 8 mM; in a mammal (iv) [Na+]in Overshoot of action potential = +35 mV; saline [Na+] = 112 mM (v) [K+]in Blood [K+] = 3.2 mM; undershoot of action potential = -87 mV (vi) ECa [Ca2+]out = 5.6 mM; free [Ca2+]in = 0.8 mM (vii) [K+]out [K+]in = 350 mM; Ek = -82 mVarrow_forwardWhy is the induced-fit model a more likely model than the lock-and-key model?arrow_forwardThe following data, presented by G. Bowes and W. L. Ogre in J. Biol. Chem. (1972) 247:2171–2176, describe the relative rates of incorporation of CO2 by Rubisco under N2 and under pure O2. Decide whether O2 is a competitive or uncompetitive inhibitor.arrow_forward
- Compounds I, II, and III are in the following biochemical pathway: precursor compound I enzyme A enzyme B compound II ► compound III enzyme Mutation a inactivates enzyme A, mutation b inactivates enzyme B, and mutation c inactivates enzyme C. Mutants, each having one of these defects, were tested on minimal medium to which compound I, II, or III was added. Fill in the results expected of these tests by placing a plus sign (+) for growth or a minus sign (-) for no growth in the table below. Minimal medium to which is added Strain with mutation Compound I Compound II Compound IIIarrow_forwardWhat is the difference between pure and mixed noncompetitive inhibition?arrow_forwardIn mixed inhibition as shown below, please draw a lineweaver-burk plot when Kl is greater than KI'. Please draw plot with and without inhibitor and label axis and intercepts. Please explain what is happening to Km and Vmax. In mixed inhibition as shown below, please draw a lineweaver-burk plot when Kl is greater than KI'. Please draw plot with and without inhibitor and label axis and intercepts. Please explain what is happening to Km and Vmax. 1/v 1/[S1]arrow_forward
- How do we calculate Ki of an inhibitor? Also, is Kapp the same as Km or does Kapp involve more variables?arrow_forwardAn enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.arrow_forwardConsider the following set of data and answer the foarrow_forward
- Consider the following experimental data from another experiment: [S] 1.5 2.00 2.50 5.00 10.00 V (No inhibitor) mmol ml¹ min¹ 0.167 0.204 0.232 0.313 0.385 V (inhibitor) mmol ml¹¹ min¹¹ 0.115 0.143 0.167 0.250 0.333 Calculate Km and V max and determine whether this inhibitor is competitive, non-competitive or uncompetitive.arrow_forwardWhat is PEI and what is its purpose in this protocol?arrow_forwardFollowing is a Dixon plot for PNPP inhibition by inorganic phosphates. What answer choice shows the correct combination of the mode of inhibition and Ki of the EI complex?arrow_forward
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