Solution Manual Environmental Biotechnology Principles and Applications 2nd edition by Rittmann McCarty – Updated 2024
Complete Solution Manual With Answers
Sample Chapter Is Below
Environmental Biotechnology – Principles and Applications
Edition 2 – Problem Solutions
Chapter 3 – Biochemistry, Metabolism, Genetics, and Information Flow
Problems According to Major Subjects Addressed
Enzymes
1, 2, 17, 26
Genetics
4, 5, 6, 7, 10, 11, 15, 16
Metabolism
3, 8, 9, 14, 18, 19, 20, 21, 22, 23, 24, 25
Morphology and Taxonomy
12, 13, 173.1 3.2 3.3 3.4
3.5
3.6
1. Competitive
2. Non-Competitive
3. Uncompetitive
1. Constitutive
2. Inducible
Case Electron Donor Electron Acceptor Feasible ∆G°’, kJ/e–
eq
a acetate Acetate
(methanogenesis)
Yes 23.5 – 27.4 = -3.9
b acetate Fe3+ (reduction to
Fe2+)
Yes -74.3 – 27.4 = -101.7
c acetate H+ (reduction to
H2)
No 39.9 – 27.4 = +12.3
d glucose H+ (reduction to
H2)
Yes 39.9 – 41.35 = – 1.45
e H2 CO2
(methanogenesis)
Yes -39.9 + 23.5 = – 16.4
r H2 NO3
–
(denitrification to
N2)
Yes -72.2 – 39.9 = -112.1
g S0 NO3
–
(denitrification to
N2)
Yes -72.2 – 23.5 = – 98.7
h CH4 NO3
–
(denitrification to
N2)
Yes -7272 – 23.5 = -95.7
i NH4
+ (oxidation
to NO2
–)
SO4
2- (reduction to
H2S+HS–)
No 32.9 + 20.85 = 53.8
RNA – usually single-stranded; uses AUCG
DNA – usually double stranded; uses ATCG3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15 DNA
Information storage and replication
RNA
Formation of proteins and the information-processing apparatuses (ribosome, mRNA, and tRNA)
Fe2+ Both potentially, although donor is observed
O2 Acceptor
CO2 Acceptor
NH4
+ Donor
SO4
2- Acceptor
NO2
– Both: acceptor to N2, donor to NO3
–
NO3
– Acceptor
HS– Donor
CH4 Donor
Fe3+ Acceptor
Couple E°’ = (Eacc – Edon)°’, volts Rank
H2/Fe3+ 0.76 – (-0.42) = 1.18 2
H2S/O2 CH4/NO3
– 0.74 – (-0.24) = 0.99 4
0.82 – (-0.22) = 1.04 3
0.82 – (-0.41) = 1.23 1
H2S/NO3
– 0.75 – (-0.25) = 0.97 5
H2/O2 Fe2+/O2 0.82 – (0.77) = 0.05 6
The DNA polymerase requires a high-energy tri-phosphate bond on the 5’ end of the incoming
base. It also requires an available 3’C-OH site to bond to the growing strand. By having the old
DNA strand starting 5’ to 3’, this guarantees an available 3’ site for the DNA polymerase.
In most cases, starting the RNA polymerase one base upstream would have no impact. This is true
because the ribosome always begins translation at the start codon (AUG) within the sequence on
the mRNA. The only time an effect would occur is if the new start created an new AUG code that
was not to be present as a start signal. In that case, the protein formed would be entirely different.
The membrane is a lipid bi-layer with a hydrophobic interior. Ionic species, being hydrophilic, do
not pass through the membrane directly. Instead, membrane transport proteins embedded in the bi-
layer span the membrane and act as a conduit for hydrophilic solutes. The transport is selective
and my require an energy expenditure.
Similarities
Both are storage sites for genetic information
Both use the same bases: C, T, C, G
Differences
Prokaryote DNA is circular
Eukaryote DNA is super-coiled
Coenzyme is the best answer. A coenzyme is a non-protein molecule that is not a metal (a
cofactor) and not tightly bound (a prosthetic group). The apoenzyme is a protein without the
attached groups. -39.9 + 23.5 = – 16.4
(a) Replication occurs only right before cell division. Transcription occurs whenever the cell
needs to make a protein product, which could be continuously to never.3.16 3.17 3.18 3.19 (b) Replication produces the double-stranded deoxyribonucleotide. Transcription produces the
single-stranded ribonucleotide.
mRNA is the transcribed genetic code used to direct protein synthesis.
tRNA is the agent that recognizes the code (in the mRNA) and brings the correct amino acid to the
ribosome during translation to make the protein product.
1. Regulates transcription (common)
2. Is a structural member of flagella (common)
3. Is a structural member of the ribosome (common)
4. Is part of the cell wall of Archaea
5. Is part of the glycocalyx (extracellular capsule) of some prokaryotes.
6. Is a structural member of pili in some bacteria
7. Is a porin in the outer membrane of gram-negative bacteria
8. Forms histones in eukaryotes
9. Is part of the capsid coat of viruses
10. Forms part of the gas vacuoles of some buoyant prokaryotes.
ATPase catalyzes a reversible reaction that forms ATP from ADP + Pi. The reaction allows the re-
entry of protons (H+) across the membrane in controlled amounts. Energy from movement of
protons across a proton gradient (which required energy to form) is captured and used to drive the
energy-consuming (and conserving) formation of ATP. The gradient is called the Proton Motive
Force (PMF).
The full mineralization of butanol is carried out with a series of dehydrogenations, hydroxylation,
and activations characteristic of -oxidation, as follows.3.20 C6H12O6 has 6 carbon and 24 e– equivalents.
(a) 2 mol CH3COOH take up 4 C and 2×8 = 16 e– eq. This means that we need to find a place
for 2 C and 8 e– eq. The 2 C eq are in 2CO2, which had no e– eq. The 8 e– eq. are in 4H2.
(b) 2 mol CO2 and 1 mol propanal (CH3CH2CHO) take up 5 C and 7 = 6 + 3 = 16 e–
eq.
Thus, we need to find a home for 1 C and 8 e– eq. It is CH4.3.21 3.22 3.23 3.24 Cyanobacteria are oxygenic phototrophs. H2O is the electron donor, and O2 is the oxidized
product.
Green and purple bacteria or anoxygenic phototrophs. They use H2 or H2S as the electron donor.
The grow phototrophically only in anaerobic conditions, as their pigments are not synthesized in
the presence of O2.
Correct are numbers 1 & 3
Correct is number 3
Only number 2 is false3.25
3.26 The secondary, tertiary, or quaternary structure is established by hydrogen, sulfide, and
electrostatic bonds. These bonds can be disrupted by extremes in pH, which affects the charge on
protein functional groups such as -SH, -OH, -COOH, and -NH2. Thus, an extreme pH can denature
the protein’s structure and render it useless or inefficient for catalysis.
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