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Again, your weekly quizzes are a better indicator of the topics to be tested in the final exam.
1. It is true that:
(a) Fixation of carbon results in the net production of NAD+.
(b) Fixation of carbon results in the net production of ATP.
(c) RubisCO is an enzyme indicative of the reverse TCA cycle.
(d) Citrate lyase is an enzyme indicative of the hydroxyproprionate pathway.
(e) (a) and (b) only.
(a). Fixation of carbon involves reduction of a carbon atom from a carboxylic acid to an alcohol and the concurrent oxidation of NADH (or NADPH) to NAD+ or ( NADP+). See the class notes on the carbon cycle as well as Figures 16.19, 16.20 (pp. 588-590) and 19.10 (pp. 729-730).
2. Global warming is not increased by:
(a) The tricarboxylic acid (TCA) cycle.
(b) Industrial production of beer.
(c) Production of CO2 by methanotrophs.
(d) Breeding of agricultural ruminants.
(e) (c) and (d) only.
(c). The methanotrophs convert of CH4 to CO2. Since CO2 absorbs 5 times less infrared irradiation from the sun than CH4, this constitutes a net decrease in the greenhouse effect. See your class notes on the carbon cycle, in which the methanotrophs were explicitly discussed as a possibility for bioremediation.
3. Syntrophy is illustrated by the relationship(s) of:
(a) Bacteroids and legumes.
(b) Nitrosofying bacteria and nitrifying bacteria.
(c) Thiovolum and some giant tube worms in deep-sea thermal vents.
(d) (a) and (c) only.
(e) (a), (b), and (c).
(b). The relationship of nitrosofying and nitrifying bacteria is another example of syntrophy = "eating together" (pp. 738-740). You should have recognized the analogy to the reactions in the rumen discussed in class and pp. 650-652 of your text.
4. During the symbiosis of Rhizobia and legumes, the FixL protein regulates nitrogen fixation based on O2 and ATP/ADP ratio. In such a scheme, FixL would activate nitrogen fixation only if:
(a) Free O2 is low and the ATP/ADP ratio is low.
(b) Free O2 is high and the ATP/ADP ratio is high.
(c) Free O2 is high and the ATP/ADP ratio is low.
(d) Virtually all of the O2 is bound to leghemoglobin and the
ATP/ADP ratio is high.
(e) Virtually all of the O2 is bound to the FixL protein and
the ATP/ADP ratio is high.
(d). During the symbiosis of Rhizobia and legumes, the FixL protein activates nitrogen fixation only if virtually all of the O2 is bound to leghemoglobin and the ATP/ADP ratio is high.
5. Some cyanobacteria produce potent neurotoxins that, if ingested, will kill humans. These cyanobacteria are most likely to contaminate:
(a) Waters rich in organic carbon wastes but poor in phosphate.
(b) Waters that are anoxic.
(c) Waters rich in phosphate wastes but poor in organic carbon.
(d) Shellfish living in polluted waters.
(e) (c) and (d) only.
(e). Most cyanobacteria are obligate phototrophs, and many fix their own nitrogen (pp. 735-736). Therefore, their growth is not limited by carbon or nitrogen but by other substances, such as phosphate. Secondly, during discussions of mercury (pp. 665-666) and of water-borne pathogens (p. 564), we have also seen that filter feeders concentrate microorganisms in the waters where they live, along with their poisons.
6. The introduction of wastes rich in organic carbon into marine water causes:
(a) Rapid growth of microorganisms.
(b) Reduction in oxygen.
(c) Increase in hydrogen sulfide.
(d) (a) and (b) only.
(e) (a), (b), and (c).
(e). The introduction of wastes rich in organic carbon into the sea will cause rapid growth of microorganisms and consequent reduction in oxygen. Since marine water is rich in sulfate, those conditions are ideal for the growth of the obligately anaerobic sulfate reducing bacteria, most of which are chemoorganotrophs (pp. 749-752). The H2S gas that they produce accounts for the disgusting smell of some polluted coastal waters.
7. Consider the production of acetate and hydrogen from Syntrophomonas and the conversion of acetate to methane by some methanogens, as follows:
It is true that:
(a) The overall conversion of butyrate to CH4 and HCO3-
has an unfavorable delta Go = + 17.2 kJ.
(b) The Syntrophomonas reaction is unfavorable and is driven by
removal of the products.
(c) The overall reaction would only be favorable if the H2 and
HCO3- produced were also converted to methane.
(d) Both reactions occur in the rumen.
(e) (b) and (d) only.
(b). The overall reaction is favorable because the Synthophomonas reaction is driven by removal of the products (pp. 650-652). The second reaction does not occur in the rumen because acetoclastic methanogens grow too slowly to persist in the rumen.
8. Using a bacterial process, tetravalent uranium oxide (UO2) from low-grade ores can be converted to the leachable hexavalent form (UO2SO4) by ferric ions. This reaction is likely to involve:
(a) Neutral pH, because most bacteria cannot grow at acidic pH.
(b) Sulfur reducing bacteria.
(c) The precipitation of Fe(OH)3.
(d) A species of Thiobacillus.
(e) (c) and (d) only.
(d). Notice the analogy to the use of Thiobacillus (a sulfur oxidizing bacterium) for copper mining (pp. 663-665). At the acidic pHs thatThiobacillus requires for growth, precipitation of Fe(OH)3 would be insignificant.
9. We would be most likely to find an organism that is psychrophilic, anaerobic and obligately barophilic:
(a) In refrigerated vacuum-packed foods.
(b) Near thermal vents in the deep sea.
(c) In the deep sea but not near thermal vents.
(d) In the thermocline layer of a stratified lake.
(e) In a particle of surface soil.
(c). An organism that is psychrophilic (loves the cold), anaerobic (hates O2) and obligately barophilic (loves high pressure) would likely be in the deep sea but not near thermal vents.
10. Sake is a Japanese alcoholic drink produced from steamed rice (which is primarily starch) by the combined actions of Saccharomyces cerevisiae and one other organism. From this, we can deduce that:
(a) The other organism is one that excretes amylases, such as Aspergillus
oryzae.
(b) Sake is a wine rather than a beer.
(c) The other organism is a species of Acetobacter.
(d) The other organism must act on the rice before S. cerevisiae.
(e) (a) and (d) only.
(e). Notice the analogies to wine making and beer making (pp. 388-390). Sake is really a beer, though it is often called a wine. It is produced by a two-step fermentation. Aspergillus oryzae secretes amylases to convert starch in the steamed rice to sugars, then Saccharomyces cerevisiae ferments the sugars to ethanol.
11. Infants can be poisoned by honey or nitrate that is harmless to adults. This is because:
(a) Their gastro-intestinal tract is more alkaline than that of adults,
causing different microbial fauna to prevail.
(b) Their gastro-intestinal tract is more alkaline than that of adults,
leading to greater stability of nitrate and the toxin from Clostridium
botulinum.
(c) Their gastro-intestinal tract is aerobic.
(d) Sugar stimulates microbial growth in their gastro-intestinal tract.
(e) (a) and (c) only.
(a). Infants can be poisoned by honey (pp. 562-563) or nitrate (discussed in class) that is harmless to adults because their gastro-intestinal tract is more alkaline (i.e. less acidic) than that of adults, causing different microbial fauna to prevail.
12. In Brazil, the juice of sugar cane is fermented to produce large amounts of ethanol as an automotive fuel. This industry is not profitable and is heavily subsidized by the Brazilian government. A cheaper starting material would be the vast quantities of inedible plant materials (wood, corn husks, stalks, leaves, etc.) produced by agriculture. Such a process could be made practical by:
(a) Employing organisms with higher alcohol tolerance.
(b) Employing organisms that efficiently convert cellulose to sugars as
part of a two-stage fermentation.
(c) Employing organisms that excrete amylases as part of a two-stage fermentation.
(d) (a) and (b) only.
(e) (a) and (c) only.
(d). That unprofitable industry really does persist in Brazil. Ideally, we would want to employ (1) organisms that efficiently convert cellulose to sugars and (2) organisms with high alcohol tolerance that convert the sugars to ethanol.
13. The coliform test:
(a) Identifies water-borne pathogens.
(b) Is sufficient to prove that water is safe to drink.
(c) Reliably estimates the risk of pathogens in drinking water based on
its content of coliform bacteria.
(d) Reliably estimates the risk of pathogens in meat products based on
their content of coliform bacteria.
(e) All of the above.
(c). The coliform test reliably estimates the risk of pathogens in drinking water. It does not identify the pathogens, nor does it give information about contamination of the water with carcinogens, etc. It is not a reliable test of the quality raw meats, which usually contain large numbers of coliforms.
14. A hypothetical food-borne pathogen is an obligately aerobic organotroph that produces a heat-stable toxin. Which of these statements would be true of such an organism?
(a) It will establish itself in the gut and cause food infections.
(b) It will grow in canned foods that were prepared at neutral pH.
(c) It will grow in vacuum-packed meats that are not refrigerated.
(d) Reheating potentially contaminated foods will eliminate risks from
this organism.
(e) It will cause food poisoning.
(e). Our hypothetical food-borne pathogen will not cause food infection but will cause food poisoning, even if a contaminated food is reheated. It could not live in the gut, in canned foods, or vacuum-packed foods, all of which are anaerobic.
Consider the following reactions of the biogeochemical cycles:
A. H2 + HCO3- ----------> CH4
B. NH3 ----------> R-NH2
C. NH3 ----------> NH2OH
D. NH2OH ----------> NO2-
E. NO2- ----------> NO3-
F. NO3- ----------> N2
G. Fe2+ ----------> Fe3+
H. H2S ----------> So
I. So ----------> SO42-
J. SO42-----------> H2S
K. SO42-----------> R-SH
L. CH4 ----------> CH3OH
M. CH3OH ----------> CO2
N. CO2 ----------> (CH2O)6
Give the genus of one organism that catalyzes:
15 to 18: The genus is only the first part of the formal binomial name of an organism.
15. Reaction A: Any genus in Table 20.5 (p. 827) that consumes H2 and CO2
16. Reaction G: See pp. 592-594.
17. Reaction J: The genus of any sulfate reducer in Table 19.10, p. 750.
18. Reaction N: The genus of any carbon fixing organism, e.g. Anabaena.
19. Of the biogeochemical cycle reactions listed above, which one(s) would be used only under anoxic conditions:
(a) N only.
(b) A and G.
(c) A, F, and J.
(d) B, K, and N.
(e) E, F, J, K, and N.
(c). As I've reminded you many times, the use of alternative electron acceptors to O2 is restricted to anoxic conditions.
20. Of the biogeochemical cycle reactions listed above, the one(s) not limited to microorganisms is (are):
(a) N only.
(b) A and G.
(c) A, F, and J.
(d) B, K, and N.
(e) none.
(d). Green plants fix carbon, and they assimilate ammonia or sulfate into amino acids.
21. Of the biogeochemical cycle reactions listed above, a monooxygenase is involved in:
(a) L only.
(b) C and L.
(c) L and M.
(d) I, L, and M.
(e) C, D, E, and M.
(b). Reaction L involves methane monooxygenase (pp. 746-747); reaction C involves ammonia monooxygenase (pp. 595-596).
22. Consider the following organic compounds: Which one(s) would resist biodegradation?
(a) Compound A only.
(b) Compound B only.
(c) Compounds A and D only.
(d) Compounds B and D only.
(e) All of the above.
(d). It is not possible to oxidize compound B aerobically, with subsequent beta-oxidation. Beta-oxidation is also impossible for compound D after ring cleavage between the two OHs following their oxidation (pp. 615-616). (This type of question is a favorite on the GREs.)
23. Oxygen-saturated water has ~ 10 mg O2/L. Before secondary waste treatment, 2 mg O2/Lwas measured in a water sample that had been saturated with oxygen and left at 20oC for 5 days. After secondary treatment, 8 mg O2/Lwas measured in the water after saturation with oxygen and 5 days at 20oC. We have removed:
(a) 80% of BOD.
(b) 75% of BOD.
(c) 60% of BOD.
(d) 20% of BOD.
(e) None of the above.
(b). The fraction of BOD remaining after a treatment (compared to the BOD before) is a measure of the efficiency of that liquid waste treatment. I gave an example of an analogous calculation in class. With secondary treatment, we have removed: 100 x [(10 - 2 mg/L) - (10 - 8 mg/L)]/ (10 - 2 mg/L) = 75% of BOD.
24. Fifty patrons who ate from the same restaurant buffet have developed symptoms consistent with contamination of their food by Staphylococcus aureus. The suspected food is a hollandaise sauce, which you have been asked to test. Briefly describe four ways in which you would confirm or dismiss S. aureus as the contaminating agent (Limit yourself to methods discussed in the lectures. Be specific, but use no more than two sentences for each approach.).
Here, it is important to give a molecular biological or microbiological approach and also to indicate (1) a method of detection and (2) what you would expect if S. aureus is or is not the contaminant. Examples:
Approach 1. A fast and inexpensive approach would be to treat a dilution of the sauce directly with fluorescently labeled antibodies specific to surface protein(s) of S. aureus and then examine the sample by fluorescence microscopy. Fluorescently labeled cells would indicate the presence of S. aureus.
Approach 2. Another fast and inexpensive approach would be to coat a plate with a dilution of the sauce and use an enzyme-linked antibody specific to Enterotoxin A in an ELISA . Development of color would demonstrate the presence of S. aureus.
Approach 3. Alternatively, we could try to PCR the enterotoxin gene sequence from a sample of the sauce. If this were possible, then we would conclude that S. aureus was present.
Approach 4. Alternatively, we could PCR 16S rRNA sequences from a sample of the sauce and probe them with a radiolabeled oligonucleotide that hybridizes specifically at a signature S. aureus 16S rRNA sequence. A positive signal, e.g. radioactive band(s) on a gel after hybridization to the PCR products, would indicate the presence of S. aureus.