1 BIOLOGY 112

PRACTICE (FINAL) EXAMINATION 2008 T1

Time allowed: 2.5 hr

There are 51 Multiple choice questions and 5 Short Answer questions on 16 pages (including this cover)

You are allowed to bring in one 8 x 11 inch review sheet – you can use both sides of the paper. You cannot

have additional papers taped or stapled to it. You can write or type your notes on the page.

You should try to do the practice exam as if it were a real exam.

Please read the information below – it will be on the cover of your real final exam.

READ THIS! FAILURE TO FOLLOW INSTRUCTIONS CAN DELAY THE GRADING OF YOUR EXAM A. MULTIPLE CHOICE QUESTIONS: On the computer answer sheet, in PENCIL, fill in the bubbles for your name (Last name FIRST, First name LAST) 8 digit student number. Other student information is NOT required. Turn the computer answer sheet 90o. IN INK, sign the computer answer sheet in the spaces provided. Record all multiple choice responses IN PENCIL on the computer answer sheet — NOT on the question paper. All single multiple choice responses have ONE BEST answer. Point values vary with the response. Students MUST record their multiple choice responses onto their computer sheet during the time allowed for the exam. NO writing will be permitted after the time allowed for the exam has expired. NO appeals will be considered. B. SHORT ANSWER QUESTIONS: Write your answer in the space provided. Answers written in alternate areas of the exam will not be marked.

2 B. RETURN OF THE QUESTION PAPER The question paper MUST be returned with the answer sheets. Answer sheets without a question paper will NOT be graded. In this regard, students are STRONGLY CAUTIONED to guard against theft of their question paper by other students. NO appeal relating to loss of a question paper by theft will be considered. C. STUDENT QUESTIONS The invigilators will NOT answer any clarifying questions during the exam. Students must interpret and answer each question on their own. This policy ensures that those students writing in a room not staffed by knowledgeable personal will not be disadvantaged relative to other students. FACULTY OF SCIENCE RULES GOVERNING FORMAL EXAMINATIONS #1. Each candidate must be prepared to produce, upon request, a Library/AMS card for identification. #2.Candidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities in examination questions. #3. No candidate shall be permitted to enter the examination room after the expiration of one-half hour from the scheduled starting time, or to leave during the first half hour of the examination. #4. Candidates suspected of any of the following, or similar, dishonest practices shall be immediately dismissed from the examination and shall be liable to disciplinary action:

• Having at the place of writing any books, papers or memoranda, calculators, computers, audio or video cassette players or other memory aid devices, other than those authorized by the examiners. • Speaking or communicating with other candidates • Purposely exposing written papers to the view of other candidates. The plea of accident or forgetfulness shall not be received.

#5. Candidates must not destroy or mutilate any examination material; must hand in all examination papers; and must not take any examination material from the examination room without permission of the invigilator.

3 1. A good rationale for why RNA is the primer used for DNA synthesis of the lagging strand is that:

A. Lagging strand synthesis must happen faster, so it needs more priming sites. B. Lagging strand synthesis takes place in a 5’ to 3’ direction. C. Lagging strand synthesis requires DNA strand separation. D. Because the initial helix is short, DNA polymerase might make mistakes when it starts replicating. E. Lagging strand synthesis takes place backwards so information transfer from RNA to DNA makes sense.

2. The properties that are important for forming the CsCl gradient used in the Meselson-Stahl experiment are:

1. the charge on the ions. 2. the density of CsCl. 3. the high diffusion of CsCl. 4. the centrifugal force acting on the CsCl. A. All 4 are important B. 1, 2, 3 are important C. 2, 3, 4 are important D. 1, 3, 4 are important E. 1, 2, 4 are important

3. In the regulation of the trp operon, production of mRNA used to synthesize the TrpEDCBA proteins is controlled at two different points. Which of the following contribute to this control?

1. Competition for the promoter between RNA polymerase and the repressor TrpR. 2. The affinity of the TrpR protein for tryptophan. 3. The frequency of transcription termination by RNA polymerase between the start of the mRNA

and the trpE gene. 4. The binding of nucleotide triphosphates to the RNA polymerase. 5. The affinity of the ribosome for the ribosome binding site of the leader peptide in the trp operon

mRNA.

A. All of the items contribute to the control B. Only items 1, 2, 4 C. Only items 1, 3, 5 D. Only items 2, 3, 5 E. Only items 1, 2, 3

4. In the quorum sensing regulation there were several different proteins. One of them had a function of “detecting” the signal that the cell density is high. Which one was it?

A. LuxQ, the transmembrane protein. B. LuxO, the regulator of the luxR gene. C. LuxR, the regulator of the Lux DCABE operon. D. LuxC, one of the genes involved in light production.

4

5. A transposon inserts into region 3 of the leader sequence. What happens to the level of the enzyme tryptophan synthetase (made from TrpB and TrpA) when the tryptophan level is high?

A. The level is maximum. B. The level is intermediate. C. The level is minimum (basal). D. There is no tryptophan synthesis at all.

6. Reversion mutations are ones that change a strain with a mutation back to wild type. However, not all mutations can revert. Which if the following mutations cannot revert?

A. A missense mutation that changes a GC base pair to a CG base pair. B. An insertion mutation that adds two TTs to the DNA sequence GTTTTTCGTCGTCGT (at the bold

nucleotide) C. A deletion mutation that deletes the two Cs from the sequence AGGGGGCCGGGGA (the bold

ones) D. A missense mutation that changes a GC base pair to an AT base pair. E. Well, none of the above can revert.

7. DNA polymerase and RNA polymerase can make base pairing errors. Unlike DNA polymerase, however, RNA polymerase has no proof-reading function. If an error is made by RNA polymerase during the transcription of a gene encoding a ribosomal RNA molecule, what would be the LIKELY consequence to the cell in which the error occurred?

A. The genotype of the cell would be changed. B. The cell would die. C. The cell would live but its phenotype would be changed. D. There would be no significant consequence to the cell. E. ALL of the above are equally likely consequences.

8. Assume you are working in a lab. A graduate student in the lab has a result that suggests two DIFFERENT proteins are translated from the SAME mRNA molecule. When she asks you about it, which of the following would you include in your answer?

A. The coding regions for the two proteins must share a ribosome binding site. B. The two proteins must have the same function. C. The two genes for the proteins must be transcribed divergently. D. The coding regions for the two proteins must use the same reading frame. E. None of the above.

Leader peptide

Stop

codon

Trp

codons

1 2 3 4

rbs trpE

Leader peptide

Stop

codon

Trp

codons

1 2 3 4

rbs trpE

5 Consider the following nucleotide sequence. 5'-ATGTAAAATCAAAGAAAACCAACATCACATGCCGCCATGTAA-3' 3’-TACATTTTAGTTTCTTTTGGTTGTAGTGTACGGCGGTACATT-5' First, use a codon table (see Notes) to find the longest potential coding region. Use that region to answer Questions 9-11. Note: the sequences specifying the promoter, the transcription terminator and ribosome-binding site are not shown. 9. The 3rd amino acid encoded by the longest possible coding region is:

A. Gly B. Ala C. Arg D. Asp E. Ser

10. What would be the consequence of a mutation that changed the non-template strand sequence 5'-TTGAT-'3 to 5’-TTGTT-3’?

A. The mutation would result in a shorter protein. B. The mutation would result in a different amino acid being inserted into the protein. C. The mutation would result in a longer protein. D. The mutation would be silent, i.e., the amino acid sequence of the protein would not change. E. NONE of the above.

11. If the sequence above encoded a functional protein, what would be the likely effect on protein function of a mutation which changed the non-template strand sequence 5'-GGCA-3' to 5'-GGCT-3'?

A. No effect. B. Small effect. C. Large effect. D. Can’t say for sure.

12. A bacteriologist induces a mutation in a gene for a 100 amino acid protein. Upon further analysis it is found that when the mutant gene is expressed, a 50 amino acid mutant protein is made. Further, the mutant protein consists of the first 50 amino acids of the wild type protein. This mutant could be the result of:

A. a single base substitution. B. the deletion of 4 nucleotides. C. the deletion of 1 nucleotide. D. the insertion of 1 nucleotide. E. any of the above could explain the origin of the mutant protein.

6 13. The nif gene encodes the enzyme Nitrogenase catalyzes N2→ NH4

+ reactions. Azotobacter and Anabaena are two species of bacteria capable of N2 fixation to NH4

+. In the lab, when cells of a competent Nif– Azotobacter mutant are mixed with wild-type (Nif+) Azotobacter DNA, the Nif– Azotobacter mutants are transformed to the wild-type (Nif+) phenotype. However, using DNA from a Nif+ Anabaena strain, Nif+ Azotobacter transformants are not obtained. Which one of the following is a potential explanation for the latter result?

A. The Anabaena strain may have been cultivated in the presence of fixed nitrogen before its DNA was prepared.

B. Azotobacter DNA and Anabaena DNA may have very different ratios of A+T to G+C. C. The Anabaena strain may not have been competent when its DNA was prepared for the experiment. D. Preparation of the DNA may have failed to convert it from dsDNA to ssDNA. E. NONE of the above is a potential explanation for the result.

14. Which pair of words correctly fills in the blanks in the following sentence? In considering modes of horizontal gene transfer, the uptake of DNA released from a cell would most likely result in __________, while cell-to-cell contact would most likely result in __________.

A. transformation / conjugation. B. transduction / conjugation. C. conjugation / transformation. D. transformation / transduction. E. transduction / transformation.

15. When a transducing phage particle injects DNA into a bacterium:

A. The bacterium is usually killed because the phage initiates a lytic cycle and produces progeny phage. B. The DNA forms a plasmid and is passed to progeny. C. The DNA is from another bacterium. D. The DNA is phage DNA. E. The DNA will be degraded and the nucleotides will be used as nutrients .

16. The presence of restriction enzymes in the recipient that are not found in the donor would be MOST likely to prevent a successful gene transfer via which of the following mechanisms?

A. Transfer of a plasmid by conjugation. B. Transduction by a phage and recombination. C. Transformation of donor DNA by natural transformation. D. Restriction enzymes would prevent all of the above methods of gene transfer. E. Restriction enzymes would not prevent any of the above methods of gene transfer.

17. Which of the following is the best rationale for why conjugation is thought to be responsible for DNA transfer between unrelated bacteria.

A. Conjugative plasmids do not require recombination into the chromosome for survival in the recipient. B. The transfer of both strands of the DNA protects the DNA from restriction enzymes. C. The strand of DNA being transferred is nicked into pieces by enzymes in the donor. D. The recipient must have a surface receptor protein for the pilus. E. All cells use DNA as their genetic material.

7 18. Transformation and homologous recombination allow for the formation of heteroduplex DNA (i.e., areas where there may not be complementary base pairing between the two strands). Considering only the region of the DNA that contains the heteroduplex, which of the following would occur after errorless DNA replication of this molecule?

A. One daughter DNA would have two strands that are complementary to the donor DNA molecule while the other daughter DNA would have two strands complementary to the recipient DNA molecule.

B. Both daughter DNA molecules would have two strands complementary to the donor DNA molecule. C. Both daughter DNA molecules would have two strands complementary to the recipient DNA

molecule. D. Both daughters DNA would have one strand that is complementary to the donor DNA molecule and

one strand is complementary to the recipient DNA molecule E. None of the above is correct.

19. Assume a bacterium contains a conjugative plasmid (call it plasmid pC). The same bacterium contains a second plasmid (call it plasmid pM) that can be mobilized by pC. Which of the following statements would be TRUE about these two plasmids? (REVISED)

1. Plasmid pM contains a tra operon. 2. Proteins encoded on plasmid pC are required to transfer plasmid pM. 3. The nickase encoded on plasmid pC would also nick the oriT site on plasmid pM. 4. Plasmid pM has a gene for a nickase. 5. Plasmid pM could be transferred to a recipient that contains another plasmid. A. Statements 2, 3, 4, 5 are true. B. Statements 1, 2, 3, 4 are true. C. Only statements 1 and 4 are true. D. Only statements 3 and 4 are true. E. Statements 2, 3 and 5 are true.

20. It is common to use 37 C and 42 C as two growth temperatures to find temperature sensitive mutations. The phenotype of a strain of bacteria with a trpBts LOF mutation would be which of the following?

A. The cell would be wild type at 37 C but it wouldn’t grow under any conditions at 42 C. B. The cell would not grow at either 37 C or 42 C under any conditions. C. The cell would be wild type 37 C; at 42 C it would grow only if tryptophan is in the growth medium. D. The cell would require tryptophan at both 37 C and 42 C, but it would require more tryptophan at

42C. 21. Both Restriction endonucleases and transposases cut DNA. Which of the following describes a difference between these two enzymes?

A. Restriction endonucleases only cut chromosomal DNA but transposases do not. B. Restriction endonucleases may cut inside the DNA sequence that they bind to but transposases do

not. C. Restriction endonucleases may cut inside genes but transposases do not. D. Restriction endonucleases can ligate DNA ends together but transposases can not. E. Restriction endonucleases might cut DNA inside the gene that encodes them but transposases do not.

8 22. If we wanted to find a mutant E. coli with a mutation in either lacZ or lacY. What would we include in the medium?

1. lactose 2. glucose 3. X-gal 4. cAMP

A. All 4 B. 1, 2, 3 only C. 1, 2 only D. 1, 3 only E. Only 3

23. If we grow some cells on media with X-gal, lactose and glucose. Which of these mutants will we be able to find?

1. LOF mutants in lacI 2. LOF mutants in the gene for the enzyme that makes cAMP 3. GOF mutants in the lac ZYA promoter 4. LOF mutants in the gene for CAP

A. All of the 4 types B. Type 3 and 4 only C. Types 2 and 3 only D. Type 4 only E. Type 3, and maybe type 1

24. The conjugative plasmid pR100 has a copy number of 5. This means there are 5 copies of the plasmid in each cell. Consider the DNA transfer between a single donor E. coli containing pR100, and a single recipient cell and indicate which of the following describes what is going on. Immediately after the transfer is completed:

A. The donor will contain 4 copies of pR100 and the recipient will contain one copy. B. The donor will contain 5 copies of pR100 and the recipient will contain one copy. C. The donor will contain 4 copies of pR100 and the recipient will have a single strand of pR100 DNA. D. The donor will contain 5 copies of pR100 and the recipient will have a single strand of pR100 DNA. E. The donor will contain 4 double stranded copies of pR100, and one single strand of pR100 DNA and

the recipient will have a single copy of pR100 DNA. 25. In a transduction experiment your friend didn’t dilute the solution containing the phage sufficiently. As a result, in her experiment every cell will get more than one phage. Knowing all about this you tell her experiment it doomed because of which of the following:

A. There won’t be enough phage with bacterial DNA added to the cells. B. Cells that are infected with phage carrying bacterial DNA will also be infected with phage carrying

phage DNA. C. Cells that are infected with phage carrying bacterial DNA will also be infected with another phage

carrying bacterial DNA. D. The phage will compete for receptors on the cell surface and the infection will not work. E. There will be too many plaques to count when the phage are plated.

9 26. We talked about using restriction endonucleases to cut DNA into pieces for molecular cloning. Transposases also cut DNA. Which of the following are good rationales for why transposases would not be very useful for molecular cloning.

1.Transposases would probably cut the DNA too frequently because transposons can insert anywhere. 2. Since transposases can ligate DNA ends together, treating DNA with transposases might not actually cut the DNA. 3. Transposases don’t leave staggered ends when they cut the DNA. 4. Transposons would not cut the DNA unless it contained a transposon.

A. 1, 2, 3 B. 2, 3, 4 C. 1, 2, 4 D. 1, 3, 4 E. All 4 are good rationales

27. Suppose you were working in a lab and found that your strain of E. coli was suddenly a lac- mutant because it no longer is blue when grown on lactose plus X-gal. Which of the following would be a useful approach to finding out whether the mutation was caused by a transposon or not.

1. Check to see if glucose affects expression of the operon. 2. Check to see whether the cell has a plasmid or not. 3. Check to see if it’s possible to revert the lac- mutation to lac+

4. Use PCR to see if either the lacZ or lacY gene is larger in size than wildtype.

A. 1, 2, only would be useful B. 2, 3, only would be useful C. 3, 4 only would be useful D. 3 only would be useful E. 4 only would be useful

28. You have a bacterial strain that is resistant to the antibiotic tetracycline. You know it carries a conjugative plasmid (pRP1). However, you mixed your strain with a recipient and did not see transfer of tetracycline resistance. You continued to grow your strain for many generations and tested it again. This time there was transfer of the resistance phenotype by conjugation. Which of the following is a possible explanation for this observation?

A. A transposon carrying the resistance gene moved to pRP1. B. An IS element inserted upstream of the resistance gene. C. Your strain became diploid. D. Your strain was infected by a generalized transducing phage. E. The plasmid acquired a point mutation that results in tetracycline resistance.

10 29. Two closely related bacterial species, B. subtilis and B. cereus, are both competent and capable of being transformed by DNA from their own species. If DNA is extracted from wild-type organism B. subtilis, it cannot transform a B. cereus trp– to trp+. At what stage might the transformation process fail?

1. The binding of DNA at the cell surface. 2. The transfer of DNA through the cell membrane. 3. RecA dependent strand displacement. 4. Ligation of the single strand to the chromosome.

A. 1 only B. 1 and 2 C. 2 and 3 D. 1, 2 and 3 E. 1, 2, 3 and 4

30. Here are 4 statements concerning the lac operon. Which are correct?

1. The lacZ, lacY and lacA gene are transcribed from only 1 promoter preceding the lacZ gene. 2. The lacZ, lacY, lacA genes are transcribed from 3 promoter. 3. RNA polymerase is prevented from transcribing the operon when LacI, binds to the lac operator sequence. 4 lacI is expressed constitutively.

A. All 4 are correct. B. Only 1, 2, 3 are correct C. Only 1 and 3 are correct D. Only 1 and 4 are correct E. Only 1, 3 and 4 are correct

31. In the process of transposition in bacterial which of the following would be true?

A. The process overall depends on the recombination enzyme RecA. B. The overall process is not dependent on DNA synthesis. C. The overall process requires that the transposition donor site and the target site must have similar

DNA sequences. D. The overall process depends a plasmid DNA in the cell. E. The overall process depends on the endonuclease and ligation properties of the transposase.

32. Assume that you have a culture of lacZ– Streptococcus. As you know these cells can be transformed. You grow the cells so they are in the state of competence, and you add DNA from Streptococcus that were lacZ+ , but lacY–. Which of the choices correctly describes what type of media could you use to recover cells with a wildtype phenotype, and what would the colonies look like?

A. Media containing ampicillin and lactose; wild-type cells will grow into colonies. B. Media containing glucose and X-gal; look for blue colonies. C. Media containing glucose and X-gal; look for white colonies. D. Media containing glucose and lactose; wild-type cells will grow into colonies. E. Media containing lactose; wild-type cells will grow into colonies.

11 33. In old textbooks on microbiology, you might read that in conjugation the DNA goes through the pilus. This is not correct. It can be shown by finding which one of the mutants that fails to transfer DNA?

A. A mutant that doesn’t make a pilus. B. A mutant that makes a pilus but it can’t attach to a recipient. C. A mutant that makes a pilus that attaches to the recipient, but doesn’t form close contact mating pairs. D. A mutant that makes mating pairs but doesn’t transfer DNA. E. A mutant that transfers single strand DNA but doesn’t ligate it into a circle in the recipient.

34. Which one of the following is not necessary in the process of conjugation?

A. Synthesis of a pilus extending from the donor. B. Synthesis of a receptor molecule on the surface of the recipient. C. Formation of close contact between the the donor and recipient. D. Formation of a channel between the donor and recipient. E. Expression of plasmid genes in the recipient.

35. Under what conditions can the recipient in a bacterial conjugation become a donor?

A. The DNA being transferred must include a conjugative plasmid. B. The plasmid DNA must be completely transferred. C. The DNA being transferred must include a transposon. D. The DNA being transferred must include a mobilizable plasmid. E. The DNA being transferred must be nearly the same in sequence to the recipient genome.

36. Which of the following are essential characteristics of a plasmid in Bacteria?

1. It must contain transposons. 2. It must contain genes that encode antibiotic resistances. 3. It must contain a gene for a nickase. 4. It must contain sequences that permit the cell to replicate it. 5. It must contain sequences that encode a DNA polymerase.

A. All five statements apply B. Statements 2, 3, 4 apply C. Statements 1, 3, 5 apply D. Statement 3 only applies E. Statements 4 only applies

37. A trpA- E. coli cell is treated so that it can be used in a electroporation experiment. The DNA that is electroporated into the cell is a plasmid DNA that contains a gene that provide resistance to the antibiotic ampicillin and it includes a copy of the trp operon that has a mutation in the trpB gene (trpB-) and it also has a deletion of the trpR gene. The likely phenotype of cells that are ampicillin resistant will be which of the following?

A. Able to grow without any tryptophan in the medium. B. Able to grow when there is a small amount of tryptophan in the medium. C. Able to grow only when there is a large amount intermediate amount of tryptophan in the medium. D. Unable to grow in any condition.

12 38. For heterotrophs, what role does a sugar such as glucose play?

A. A source of protons source B. A carbon source C. A source of energy D. An electron source E. All of the above

39. You are comparing a group of molecules. Which of the following would describe the best energy source on a per mole basis (i.e. yield the greatest amount of ATP per mole utilized) among the group?

A. The molecule that has the highest electronegativity. B. The molecule that has the lowest electronegativity. C. The molecule that is in its most oxidized state. D. The molecule that is most chemically inert. E. The molecule with the most carbon atoms.

40. A bacterium is carrying out the following reaction: (REVISED) NO2

- + H2O + NADP+ → NO3- + NADPH + H+

Which of the following are correct?

1. In the reaction, the bacterium is using NO2– as an energy source.

2. The information identifies the bacterium as a chemolithototroph. 3. In the reaction, the bacterium is using NO2

– as a source of electrons. 4. In the reaction, the bacterium is transferring electrons to NADP+. 5. The reaction could be used to establish a H+ gradient across the cytoplasmic membrane.

A. All 5 are correct. B. Only 1, 2, 3, 5 are correct C. Only 2, 3, 4 are correct D. Only 3, 4 are correct E. Only 4, 5 are correct

41. Consider a bacterium carrying out the following metabolic reaction:

Which of the following statements is correct?

A. The organism could be a chemolithotroph or a phototroph. B. NADPH is being used to reduce the carbon atoms in the C-source. C. ATP is being used to push/drive the reaction past its equilibrium position so more C3H6O3

product is made. D. The purpose of the process is to build cellular organic molecules from an inorganic C source. E. ALL of the above are CORRECT.

13 42. Consider the reaction that is carried out by some Bacteria: 4 NADPH + 6 H+ + NO3

- → 4 NADP+ + NH4+ + 3 H2O

Which of the following are correct about this process? 1. NADPH accepts electrons from NO3

- becoming reduced to NADP+. 2. The N in NO3

- is being reduced in this reaction. 3. The reaction is likely to be used by organisms to produce a proton gradient. 4. The reaction could be used as part of the assimilation of nitrogen.

A. All 4 are correct. B. Only 1, 2, 3 are correct C. Only 1, 3, 4 are correct D. Only 2, 3 are correct E. Only 2, 4 are correct

43. Molecular hydrogen can be reacted with molecular oxygen in the laboratory: (REVISED)

A similar reaction is carried out by some bacteria to synthesize ATP.

In reaction (2) less energy appears as heat because the energy released: 1. has been used to transport H+ into the periplasm. 2. has been “trapped” (or conserved) in ATP. 3. has been used to transport H+ into the cytoplasm. 4. has been used to keep the bacterium warm.

A. All 4 are correct. B. Only 1, 2, 3 are correct C. Only 1, 2, 4 are correct D. Only 1, 2 are correct E. Only 2, 3 are correct

44. Which of the following completes the following sentence so that it is true. In oxygenic photophosphorylation, the electrons that move through the electron transport chain end up on:

A. a light harvesting complex. B. a reduced inorganic molecules. C. oxygen reducing it to water. D. reduced NADPH. E. the terminal phosphate in ATP.

14 45. E. coli can metabolize glucose (C6H12O6) as an energy source via aerobic or anaerobic respiration (O2 versus NO3

- as the terminal electron acceptor). If grown on identical amounts of glucose, a culture of E. coli will reach a higher cell density when grown aerobically compared to grown using NO3

- as the terminal electron acceptor. The best explanations for this include: 1. O2 diffuses faster than NO3

- 2. The use of O2 rather than NO3

- allows for the synthesis of more ATP per electron extracted 3. The use of O2 rather than NO3

- allows for more complete oxidation of the C in glucose 4. The use of O2 rather than NO3

- makes the cytoplasmic membrane less permeable to H+ ions.

A. All 4 are good explanations B. Only 1, 2, 3 are good explanations C. Only 1 is good explanations D. Only 2 is good explanations E. Only 3, and 4 are good explanations

46. Consider the metabolism of an organism whose synthesis ATP is linked to the following reaction: CO2 + 4 H2 → CH4 + 2 H2O Which of the following statements about metabolism is true?

A. It shows that the organism uses CO2 as a carbon source. B. It is a form of fermentation. C. It is a form of aerobic respiration. D. It uses CO2 as a TEA. E. It shows that the organism is a member of the domain Bacteria.

47. Which of the following reactions would result in a net input of material into the organic reservoirs of the elements in the global cycles?

1. Nitrogen fixation 2. H2S oxidation 3. Phosphate assimilation 4. CO2 fixation 5. Methane formation

A. All five input elements into organic molecules globally. B. 1 - 4 input elements into organic molecules globally. C. 3, 4, 5 input elements into organic molecules globally. D. 1, 2 only input elements into organic molecules globally. E. 1, 3, 4 input elements into organic molecules globally.

48. Which of the following would describe the role of chemolithotrophs in the global elemental cycles. (REVISED)

A. They convert reduced organic forms of the elements to reduced inorganic forms. B. They convert reduced inorganic forms of the elements to oxidized forms. C. They convert oxidized forms of the elements to oxidized organic forms. D. They convert oxidized forms of the elements to reduced inorganic forms. E. They are capable of all of the above reactions.

15 49. Fish tanks have a metabolic problem. The nutrient input into a fish tank is the fish food which is protein rich (fish are carnivores). The result is that fish excrete reduce nitrogen (think of this as NH4

+). This changes the pH in the tank. This problem can be cured if certain types of bacteria grow in the tank. Which types would they be?

a) A chemoheterotroph that uses aerobic oxidative phosphorylation. b) A chemoheterotroph that uses anaerobic oxidative phosphorylation. c) A chemolithotroph that uses aerobic oxidative phosphorylation. d) A chemolithotroph that uses anaerobic oxidative phosphorylation. e) A phototroph that uses anaerobic photophosphorylation.

50. Which of the following findings supports the endosymbiont theory for mitochondria? (REVISED)

1. Mitochrondria have DNA 2. Mitochrondria have ribosomes that resemble those in bacteria 3. Mitochrondria are missing in eukaryotes that are fermenting, such as yeast 4. Mitochrondria can divide outside cells 5. Mitochrondria contain ATP synthase enzymes in their membrane.

a) All five support the theory b) Only 1, 2, 3, 4 c) Only 1, 3, 5 d) Only 2, 4, 5 e) Only 1, 2, 5

51. In cells that are fermenting lactose, the electrons removed from lactose are:

A. placed on NAD+ to make NADH + H+; NADH + H+ delivers the electrons to anabolic reactions. B. placed on an inorganic molecule when the cells are growing in anaerobic conditions. C. returned to an organic molecule derived from the breakdown of lactose. D. used in the electron transport chain to establish a proton gradient and then returned to a derivative of

lactose. E. are used to reduce O2, producing H2O as the by-product.

52. Here is a diagram showing the fermentation of the amino acid threonine to the end product propanol. The carbons in threonine are numbered and where each C ends up is indicated. In this process which of the following statements is true.

A. Carbon 1, 2, 3, 4 are all reduced. B. Carbon 4 is oxidized and the rest are reduced. C. Carbon 4 is oxidized, carbon 1 is unchanged, carbon 2, 3 are reduced. D. Carbon 3 ad 4 are oxidized and carbons 1 and 2 are reduced. E. Carbon 3 and 4 are oxidized, carbon 2 is reduced and carbon 1 is unchanged.

16 Short Answer Questions

1. Topic = Mutations

In the lab, you are working with wildtype E. coli cells. The boss tells you to grow 20 small cultures (10 ml each). You make up the 20 flasks with media so the cells to grow. You then add 0.01 ml of a 'starter culture" that had 108 cells per ml to each flask. The next day, you plate some of the cells from each flask on media with X-gal and lactose. In 19 flasks you find that 1/108 cells make white colonies (so the cells are mutant) rather than blue (as made by the wild type). In one flask, 1/103 cells make white colonies. What would you conclude and how would you test your conclusion? 2. Topic = Gene Transfer

Explain why a gene would spread faster in a population if on a conjugative plasmid rather than a chromosome. 3. Topic = Replication. In class we said that DNA polymerase has some odd properties. One if these is that it requires a “3’OH” to start synthesis. Describe what effect this odd property this has on DNA replication. 4. Topic = Energy We said in lecture that there are two ways to make ATP in cells. Describe the difference between these two. They share one feature, what is it? 5. Topic = Environment and Microbes

Describe where each of the classes of various organisms live in the soil and why. Aerobic chemoheterotroph Anaerobic chemoheterotroph Aerobic lithoautotroph Anaerobic lithoautotroph Anaerobic photoautotroph (or anoxygenic?) Aerobic photoautotroph (oxygenic)