水处理微生物学课后答案

So, we need 4H2to use the last 8e-
（b）2 mol CO2and 1 molpropanal(CH3CH2CHO) uses 5C and 7+6+3=16 e-eq. We need one C and 8 e-. Then one mole CH4will be produced.
1.28Solution: for (a) and (b), please see the below figure,
CH3CH2CH2CH2OH
NAD+NADH+H+
CH3CH2CH2CHO
NAD+NADH+H+
+H2O
CH3CH2CH2COOH
HS-CoA
+ATP H2O+AMP+PPi
CH3CH2CH2COSCoA
b) Replication produces the double-strandeddeoxyribonucleotide.Transcription produces the string-strandedribonucleotide.
1.24mRNA is the transcribed genetic code used to direct protein synthesis.
1.18 Bacteria gain energy through the oxidation of an electron donor,and common electron donors are organic compounds.Algaecreatesorganic compounds-the algae themselves and their product. Thus in this context,becausethe primaryproduction generates their electron donor. (algaephotosynthesis also generates , anaccepterfor aerobicrespiration.)
1.31a, c
Generation of the proton motive force during aerobicrespiration,fromBrock biology of microorgani
1.371. NO 2.yes 3.perhaps4.yes-via mutation 5.yes
1.221、The feast and famine regime gains greater total polysaccharides and an increase in theα-glycosidicbonds, asα-glycosidicbonds are associated with storage materials, like glycogen .It appears that the feast and famine regime is causing the cells to accumulate more storage materials.
3FADH2=3*2ATP=6ATP
2GTP
-2ATP
Sum, 33ATP
1.29（a）C6H12O6has 6 carbons , 6×4=24 e-equivalent
2 mole CH3COOH take 4C, 2×8=16 e-equivalent ,
The 2 CO2take the last C
13、The membrane is a lipid bi-lager with a hydrophobic interior ionic species
17coenzymeis the best answer.
A coenzyme is non-protein molecule that is not a metal（a cofactor）. And not verytightly bond (the prostheticgroup ).Theapoenzymeis protein without the attached groups.
1.38The secondary, tertiary, or quaternary structure is established by hydrogen, disulfidebonds,and these can be destroyed by extreme in PH, which affect the change on protein functional groups ,like -SH, -OH, -COOH, and - , thus, an extreme PH can denature the enzymes nature and reduceituseorinefficientforcatalysis.
CH3CH2CH2CH2OH+3NAD++FAD+H2O+2HS-CoA+ATP= 2CH3COSCoA+3NADH+2H++ FADH2+AMP+PPi
CH3COSCoA +3NAD++FAD+GDP+Pi+3H2O=
2CO2+3NADH+3H++FADH2+GTP+HSCoA
Then, 9NADH+9H+= 9*3ATP=27 ATP
tRNAis the agent that matches the code（in mRNA）andbringsthe amino acidto the ribosome during protein production.
1.25
1、
01
1.27 Membrane-boundATPasescatalyze the reaction forming ATP fromADP+Pi. The reaction allows the re-entry of protons across the membrane in controlled amounts. Energy from the proton gradient (which required energy to form) is captured and used to drive the energy-consuming formation of ATP.
Solution:2、Polysaccharides glycogen are themainformof carbon and energy reserves. During feast period, the bacteria rapidly take up and sequester the substrate, by sequestering much more polysaccharides as an internal storage product, theydonotneed to have a rapid growth rate during the feast period, instead, they have a relatively steady growthrate that is based on the gradual utilization of internal storage products during the famine period.
1.23 a) Replication occurs only right before cell division. Transcription occurs whereas the cell needs the protein product, which could occurmore frequently.
FAD FADH2
CH3CHCHCOSCoA
H2O
CH3CHOHCH2COSCoA
NAD+
NADH+H+
CH3COCH2COSCoA
HS-CoA
CH3COSCoA+ CH3COSCoA
Then turn into the citric acid cycle
For (c), the overall reactionis ,
(2 +C6H12O6 2 CH3COOH+2 CO2+4H2C6H12O6 CH3CH2CHO+2CO2+CH4+ ）
1.30Cyanobacteriaare oxygenicphototrophs. isthe electron donor, and is product. Green and purple bacteria areanoxygenicphototrophs. They use or as the electron donor. They glow phototrophically only under anaerobicconditions ,astheir pigment is notsynthesized in the presence of oxygen .