Download Prescott`s Microbiology, 9th Edition Chapter 10 –Introduction to

Prescott’s Microbiology, 9th Edition
Chapter 10 –Introduction to Metabolism
GUIDELINES FOR ANSWERING THE MICRO INQUIRY QUESTIONS
Figure 10.2 Which reaction would release heat? Explain your answer.
Exergonic reactions have a negative delta G, and will have a positive change in enthalpy, and
thus release heat.
Figure 10.6 Why would energy be required to move electrons “up” the tower, from water to
nitrate, for example?
Looking at those pairs of redox reactions, taking an electron from water would be subtracting E’0
of 0.82, and giving the electron to nitrate NO3- to make nitrite NO2- would add E’0 of 0.421, so
the total reaction would have a change of E’0 = -0.399. Since delta G is inversely proportional
(this is a chance to show students how to “read” equations and understand what they mean. In
this case, because the equation G0’ = -nF . E’0 has a negative on the right side, the values must
be inversely related) to delta E, this reaction would have a positive delta G, and thus not be
spontaneous, and require energy input to drive forward.
Figure 10.7 Refer to figure 10.6 and determine theE0’ for NAD+/NADH and coenzyme
Q/CoQH2. Suggest a plausible E0’ value for FMN.
For NAD+/NADH it is -0.32, and for CoQ/CoQH2 it is 0.10. From figure 10.7, it can been seen
that FMN carries electrons between NADH and CoQ, thus is must be intermediate in E0’ value
as well, somewhere between -0.32 and 0.10. It is actually -0.219. As further thinking, ask
students to prediction the reaction, which would be FMN + 2 H+ + 2 e- -> FMNH2
Figure 10.17 Will an enzyme with a relatively high Km have a high or low affinity for its
substrate? Explain.
Since the Km is the substrate concentration at which the enzyme is operating at half of Vmax,
then a high Km would mean the enzyme needs a high concentration of substrate to bind to
operate. Therefore, the binding affinity must be low (affinity and Km are inversely related).
Figure 10.18 Based on this figure, explain the notion of a minimum inhibitory concentration of
sulfanilamide.
MIC is the lowest concentration of drug that will block growth of a microorganism. This is the
concentration of sulfanilamide applied outside the cell that will sufficiently permeate into the cell
and bind to the enzyme pictured (dihydropteroate synthetase) enough to outcompete PABA, thus
blocking synthesis of tetrahydrofolic acid, and thus in turn blocking synthesis of nucleotides, and
so the cells cannot grow, because they cannot run DNA replication. So, the binding affinity of
sulfanilamide for dihydropteroate synthetase is one factor in drug effectiveness. Have the
students predict other important factors (permeation into cell, half-life of stability, etc).
Figure 10.19 Is this allosteric effector a positive or negative effector?
Because the effector in this case allows binding of the substrate, it serves as a positive regulator.
Figure 10.21 What would happen to product formation if P and Q inhibited the formation of
intermediate E independently of one another?
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Prescott’s Microbiology, 9th Edition
If products inhibited an intermediate needed for both pathways independent of each other then
accumulation of one product would inhibit the production of both products. This is related to
Comp-Hyp-Inv question 3 in this chapter.
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© 2014 by McGraw-Hill Education. This is proprietary material solely for authorized instructor use. Not authorized for sale or distribution in
any manner. This document may not be copied, scanned, duplicated, forwarded, distributed, or posted on a website, in whole or part.