Biology Multiple Choice. 50 questions, 2 pt each. The following choices are for questions 1 5.

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1 Biology 3340 Spring 2007 Name Second Examination Version B Directions: Write your name in the correct space on the mark-sense sheet and the exam booklet. Both the exam booklet and the mark-sense sheet must be turned in at the end of the period. Indicate the version letter (A, B, C, or D) in the Key ID box in the upper left corner of the mark-sense sheet. Each question has only one correct answer. When a group of choices is used for more that one question, a choice may be used more than once. You may write in the exam booklet, but only the mark-sense sheet will be graded. No other paper, scratch paper, etc., may be used. Students must turn in the exam before leaving the room for any reason. A student may not continue working on the exam after having left the room. Multiple Choice. 50 questions, 2 pt each. The following choices are for questions 1 5. (a) thermotolerant (b) osmotolerant (c) obligate thermophile (d) obligate halophile (e) facultative thermophile 1. This refers to an organism capable of surviving exposure to elevated temperatures (for example, 70ºC) but not necessarily growing at the elevated temperature. 2. This refers to an organism that grows over a wide range of salt concentrations or ionic strengths; for example, Staphylococcus aureus. 3. This refers to an organism that is capable of growing at elevated temperatures (greater than 45ºC), but also capable of growing at lower temperatures as well. 4. This refers to an organism that requires elevated salt concentrations to grow; often require 0.2 M ionic strength or greater and may some may grow at 1 M or greater; example, Halobacterium. 5. This refers to an organism that requires elevated temperatures (greater than 45 C) for growth. The following choices are used for questions (a) propionic acid (b) homoserine lactone (c) formic acid (d) butanol (e) acetolactic acid 6. This refers to a transcriptional regulator, secreted by bacteria, that allows them to respond to population density. 7. This is one of the intermediates in mixed acid fermentation. 8. This is one of the intermediates in butanediol fermentation. 9. This is a fermentation end product of Clostridium acetobutylicum. 10. This is a fermentation end product of the bacteria used to ripen Swiss cheese. Biology 3340/01 & 02 Page 1 Second Exam, Version B

2 The information pertains to questions In lactic acid fermentation, pyruvic acid is converted to lactic acid. The following choices are used for questions (a) hydroxyl (alcohol) group (b) ester group (c) carboxylic acid group (d) carboxyl (ketone) group (e) amine group 11. Carbon #3 of lactic acid is a methyl group (-CH 3 ). Which of the above terms describes carbon #1 of lactic acid? 12. Carbon #3 of pyruvic acid is a methyl group (-CH 3 ). Which of the above terms describes carbon #1 of pyruvic acid? 13. Which term describes carbon #2 of lactic acid? 14. Which term describes carbon #2 of pyruvic acid? 15. What is the cofactor required for lactic acid fermentation? (a) NADH (d) Cofactor A (b) FADH (e) ATP (c) Cofactor Q The following choices are used for questions (a) It is simonized. (b) It is reduced. (c) It is phosphorylated. (d) It is oxidized. (e) It is hydrolyzed. 16. During lactic acid fermentation, what occurs to carbon #2 of pyruvic acid? 17. During lactic acid fermentation, what occurs to the cofactor? The following choices are used for questions (a) Phosphoketolase Pathway (b) Hexose Monophosphate Pathway (c) Entner-Douderoff Pathway (d) Embden-Meyerhoff-Parnas Pathway 18. In the first half of this pathway, a six-carbon glucose molecule is split into two glyceraldehyde 3-phosphate molecules. 19. In this pathway, a 5-C intermediate is split into glyceraldehyde 3-phosphate and acetyl CoA 20. This pathway is characteristic of Bifidobacterium and Leuconostoc. 21. This pathway is characteristic of Pseudomonas and related genera. 22. This pathway is characterized by interconversion of 5-C, 4-C, 7-C, 6-C, and 3-C intermediates, and it is the major pathway for pentose sugar synthesis. Biology 3340/01 & 02 Page 2 Second Exam, Version B

3 The following choices are used for questions (a) SO 4 2 H 2 S (b) NO 3 2 NO 2 (c) NO 2 NO 3 2 (d) NH 4 + NO 2 (e) H 2 H 2 O 23. This is the final step in the anaerobic respiration of Desulfovibrio and Desulfotomaculatum. 24. This is the final step in the anaerobic respiration of several gram-negative enteric species, such as E. coli. 25. This is the step in which a reduced electron donor is oxidized in the chemolithotrophy of Alcaligenes. 26. This is the step in which a reduced electron donor is oxidized in the chemolithotrophy of Nitrobacter. 27. This is the step in which a reduced electron donor is oxidized in the chemolithotrophy of Nitrosomonas. The following choices are used for questions (a) S O H 2 S (b) NO 2 N 2 (c) H 2 S S O (d) H 2 O O 2 (e) CO 2 CH This is the final step in the anaerobic respiration of Desulfuromonas. 29. This is the final step in the anaerobic respiration of Pseudomonas, Bacillus, and related species. 30. This represents the step in which a reduced electron donor is oxidized in the photosynthesis of Chlorobium. 31. This represents the step in which a reduced electron donor is oxidized in the photosynthesis of Cyanobacteria. 32. This represents the step in which an oxidized electron acceptor is reduced in the chemolithotrophy of methanogens. The following choices are used for questions (a) Nitrogen fixation (b) Nitrification (c) Mineralization (Ammonification) (d) Denitrification (e) Assimilatory nitrate reduction 33. In this process, ammonium is converted to nitrite, and nitrite is converted into nitrate. 34. In this process, nitrate and nitrite are converted into nitrogen (N 2 ). 35. In this process, nitrogen (N 2 ) is converted into ammonium. 36. The genera Nitrobacter and Nitrosomonas are notable for this process. Biology 3340/01 & 02 Page 3 Second Exam, Version B

4 The following choices are used for questions (a) Spontaneous mutation (b) Mutation by specific mispairing (c) Mutation by intercalating agents (d) Mutation by bypass of replication (e) Mutation by base analogs 37. Because 5 -bromouracil is very close in structure to thymine, it can substitute for thymine during DNA replication. However, it frequently pairs with a G instead of an A, causing this type of mutation. 38. In this type of mutation, a mutagenic agent chemically alters the structure of one of the bases, thereby changing its base-paring properties 39. In this type of mutation, flat planar aromatic mutagens, such as ethidium bromide, fit between the stacked bases in the center of the DNA helix. This distorts the geometry of the helix, thereby inducing mutations. 40. In this type of mutation, the one or more bases on the DNA template is damaged to the point that it can no longer base pair at all, requiring the cell to use error-prone mechanisms of repair such as the SOS repair system. 41. Thymine dimers, caused by exposure to ultraviolet light, cause this type of mutation. The following choices are used for questions (a) transformation (b) specialized transduction (c) random generalized transduction (d) conjugation 42. Gene distances given by this method are based on the time it takes for the genes to be transferred from donor to recipient cells. Therefore, the map distances on the chromosome maps are stated in units of minutes. 43. In order to be competent for this method, E. coli cells have to be treated with Ca In this method, DNA is transferred from a living donor bacterial cell into a living recipient bacterial cell. 45. In this method, the transfer of DNA is mediated by the presence of a sex pilus. 46. This method uses a bacteriophage intermediate that breaks the host chromosome into fragments, producing host DNA-capsid particles that can transfer any gene from the host to the recipient. 47. When a prophage is excised from a host chromosome, occasionally genes adjacent to the prophage attachment site are accidentally packaged into a phage capsid. This is how genes can be transferred from a donor to recipient in this method. 48. When an Hfr strain is crossed with an F - strain in an interrupted mating experiment, the DNA is transferred from the donor to the recipient by this mechanism. Biology 3340/01 & 02 Page 4 Second Exam, Version B

5 49. Out-of-focus images surrounded by a halo or color fringes are the result of this phenomenon. (a) spherical aberration (b) resolution (c) magnification (d) Koeller illumination (e) chromatic aberration 50. This statement describes chemolithotrophic metabolism. (a) Pyruvic acid is reduced to form a reduced organic acid or alcohol. During this process, NAD is regenerated, but no additional ATP is made. (b) Pyruvic acid is oxidized to form CO 2. Electrons are temporarily stored on NADH, then are transferred to an electron transport chain and a final inorganic electron acceptor. As a result, a lot of ATP is made. (c) Light energy is used to remove electrons from a reduced electron donor. The resulting electron transport produces ATP and NADH to be used for carbon fixation. (d) Glucose is partially oxidized to form pyruvic acid, with a small amount of ATP and NADH made. (e) A reduced inorganic electron donor is oxidized without the need for light energy. The resulting electron transport produces ATP and NADH to be used for carbon fixation. Biology 3340/01 & 02 Page 5 Second Exam, Version B