Document 6533584

Key to Unit 6 Sample Problems (6-­‐1;T) 1. Which does NOT describe microbial catabolic reactions? a. endergonic (6-­‐2;In) 2. How do respiration and fermentation differ? c. fermentation transfers electrons to an organic molecule, respiration uses an inorganic molecule 3. If a cell performs the reaction below in the production of ATP, how would this cell be categorized? It is a(n) (6-­‐2;In) +
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C4H802 + NADH+H Acetoin a. C4H10O2 + NAD 2,3-­‐Butanediol fermenter (6-­‐3;In) 4. Observe Figure A-­‐1, page 426 in the Lab Manual. What would be the net ATP production if only one of the two Glyceraldehyde 3-­‐phosphates could be oxidized to pyruvic acid by a fermenting organism? a. 0 (6-­‐6;In) 5. A certain bacterium is inoculated into a PR lactose tube and a PR glucose tube. After 24 hours, the PR glucose is turbid and yellow with a bubble in the Durham tube. The PR lactose has not changed. Which is consistent with these observations? b. The organism ferments glucose but not lactose. (6-­‐6;An) 6. (6-­‐7;An) 7. Sucrose is a disaccharide made of a glucose bonded to a fructose. When a pure bacterial culture is incubated anaerobically in a validated (that is, base broth was run) PR sucrose, no color change or turbidity is observed. When a pure culture of the same bacterium is incubated anaerobically in a phenol red medium with glucose, the broth turns yellow. What conclusion is consistent with these observations? b. The bacteria can’t ferment sucrose because they lack an enzyme to digest it. What conclusion can you make about the bacteria in Question #6 concerning their ability to grow aerobically? e. There is not enough information given to make a conclusion. (6-­‐9;Ap) 8. Which of these has the potential to provide the greatest amount of energy to a cell? e. Oxidation of glucose to pyruvate. (6-­‐10;T) 9. When an organism converts glucose into carbon dioxide and ethanol to synthesize ATP, the synthesis of ATP occurs by: b. substrate phosphorylation. (6-­‐11;Ap) 10. In anaerobic respiration, the final electron acceptors receive the electrons at a higher energy level than oxygen does, so fewer molecules of ATP are produced per NADH and FADH2. One estimate suggests that with nitrate (NO3) as the electron receptor, NADH and FADH2 are each worth one ATP less than when cells respire aerobically. How many ATP are produced per glucose when certain bacterial cells respire anaerobically with nitrate as the final electron acceptor? c. 26 CALCULATION: 10 NADH X 2 ATP/NADH 2 FADH2 X 1 ATP/FADH2 SUBSTRATE PHOSPHORYLATION (6-­‐12;An) 11. TOTAL = 20 ATP = 2 ATP = 4 ATP 26 ATP Which experimental evidence DOES NOT support the chemiosmotic theory? c. demonstration that each pair of electrons from NADH is “worth” 3ATPs if O2 is the final electron acceptor. Sample Question Answer Keys -­‐ Page 1 (6-­‐14;Ap) 12. (6-­‐14;Ap) 13. Fats can be used as an energy source by removing two-­‐carbon fragments from the fatty acids and attaching them to Coenzyme A to form acetyl CoA. Assuming that a eukaryotic cell is respiring aerobically, how much ATP can be generated from each acetyl CoA produced from a fatty acid? a. 12 ATP CALCULATION: 3 NADH X 3 ATP/NADH 1 FADH2 X 2 ATP/FADH2 SUBSTRATE PHOSPHORYLATION TOTAL = 9ATP = 2 ATP = 1 ATP 12 ATP The amino acid cysteine can be hydrolyzed by some bacteria to H2S, NH3 and pyruvic acid. What is the potential maximum ATP yield per cysteine that is converted to pyruvic acid in an aerobically respiring bacterium? a. 15 CALCULATION: 4 NADH X 3 ATP/NADH = 12ATP 1 FADH2 X 2 ATP/FADH2 = 2 ATP SUBSTRATE PHOSPHORYLATION = 1 ATP TOTAL 15 ATP (6-­‐14;Ap) 14. Hydrolysis of triacylglycerols (mainly fats and oils) releases fatty acids and glycerol, so not only can the fatty acid components enter steps of aerobic respiration, but so can the glycerol. Glycerol is changed to dihydroxyacetone by the process show below. How many ATPs can be produced per molecule of glycerol by an aerobically respiring prokaryote when it catabolizes glycerol? e. 22 CALCULATION: 1 NADH FROM GLYCEROL X 3 ATP/NADH = 3 ATP 1 NADH FROM GLYCOLYSIS X 3 ATP/NADH = 3 ATP 1 NADH FROM TRANSITION X 3 ATP/NADH = 3 ATP 3 NADH FROM KREBS CYCLE X 3 ATP/NADH = 9 ATP 1 FADH2 FROM KREBS CYCLE X 2 ATP/FADH2 = 2 ATP SUBSTRATE PHOSPHORYLATION IN GLYCOLYSIS = 2 ATP SUBSTRATE PHOSPHORYLATION IN KREBS = 1 ATP TOTAL MADE 23 ATP ATP USED IN GLYCEROL CONVERIONS –1 ATP GRAND TOTAL 22 ATP (6-­‐15;K) 15. How many CO2 molecules are produced per glucose during respiration? d. six (6-­‐16; In) 16. What do chemoautotrophs and photoautotrophs have in common? b. Both fix CO2 in the Calvin-­‐Benson cycle. (6-­‐17;In) 17. How does cyanobacterial photosynthesis differ from purple sulfur bacterial photosynthesis? d. Cyanobacterial photosynthesis combines ATP production and NADP reduction, whereas purple sulfur bacteria separate ATP production and NADP reduction. (6-­‐18;K) 18. What is a similarity between cyclic and noncyclic photophosphorylation? a. Both require light. (6-­‐19;K) 19. In any photosynthetic organism, ______________ is a product of the light dependent reactions and __________________ is a reactant in the light independent reactions. e. ATP, CO2 (6-­‐20;K) 20. Electrons for reducing power in green sulfur bacteria come from c. H2S Sample Question Answer Keys -­‐ Page 2 Key to Unit 7 Sample Problems (7-­‐1; I) 1. How are uracil and thymine similar? d. “a” and “b” are similarities. (7-­‐2,3; T) 2. When bacteria are grown in a medium containing radioactive nitrogen (15N) all of the DNA nucleotides contain this heavy isotope. What will be the result after one generation if the bacteria are then transferred to a medium containing regular nitrogen (14N)? e. All the molecules will consist of one strand containing nothing but nucleotides with 14N and the other strand 15
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containing nothing but nucleotides with 15N. See Figure below. Blue lines are N and red lines are N . 15
Start with all N in nucleotides. 14
Move to N medium and after one generation: (7-­‐2,3; T) 3.What will be the result of continuing this experiment for another generation? c. Half the double stranded DNA molecules will contain nothing but nucleotides with 14N. The other half of the double stranded molecules will have one strand with nothing but nucleotides with 14N and the other strand will have nothing but nucleotides containing 15N. Start with DNA from Question 2: After another generation of replication (7-­‐12; Ap) 4.Write the complementary base sequence for the DNA strand shown below. ...AGCCGTCAGAGTCTA... ...TCGGCAGTCTCAGAT... (7-­‐12; Ap) 5.Write the complementary base sequence for the DNA strand shown below. ...CCGTACAGGATACGT... ...GGCATGTCCTATGCA... (7-­‐12; Ap) 6.The stretch of DNA shown below is in the middle of a gene. Use the codon map in your text (page 175) to determine the amino acid sequence encoded by it. The lower strand is the template (antisense) strand and is shown in the correct reading frame. ...ATGTATTCGGGGGACTAG... ...TACATAAGCCCCCTGATC... mRNA AUG UAU UCG GGG GAC UAG... Amino Acids Met-­‐Tyr-­‐Ser-­‐Gly-­‐Asp-­‐STOP Sample Question Answer Keys -­‐ Page 3 (7-­‐12; Ap) 7.This stretch of DNA includes the beginning of a gene. For what amino acid sequence does it code if the lower strand is the template (antisense) strand? ...GATGCATTTATTGAACAC... ...CTACGTAAATAACTTGTG... mRNA G AUG CAU UUA UUG AAC AC... Amino Acids Met-­‐His-­‐Leu-­‐Leu-­‐Asn-­‐Thr (7-­‐12; Ap) 8.This stretch of DNA includes the beginning of a gene. For what amino acid sequence does it code if the lower strand is the template (antisense) strand? ...GCCAATGTTCGCTATGGAGGC... ...CGGTTACAAGCGATACCTCCG... mRNA G CCA AUG UUC GCU AUG GAG GC... Amino Acids Met-­‐Phe-­‐Ala-­‐Met-­‐Glu-­‐Ala (7-­‐12; Ap) 9.What is the nucleotide sequence of the DNA template (antisense) strand that codes for the amino acid sequence shown below? -­‐ Val -­‐ Ile -­‐ Asp -­‐ Gly -­‐ Pro -­‐ mRNA GU(U,C,A,G) -­‐ AU(U,C,A) -­‐ GA(U,C) -­‐ GG(U,C,A,G) -­‐ CC(U,C,A,G) antisense DNA CA(A,G,T,C) -­‐ TA(A,G,T) -­‐ CT(A,G) -­‐ CC(A,G,T,C) -­‐ GG(A,G,T,C) NOTE:This represents 384 possible combinations to produce the amino acid sequence!! (7-­‐12; Ap) 10. What is the nucleotide sequence of the DNA template (antisense) strand that codes for the amino acid sequence shown below? -­‐ Phe -­‐ Glu -­‐ Ala -­‐ His -­‐ Gln -­‐ mRNA UU(U,C) -­‐ GA(A,G) -­‐ GC(U,C,A,G) -­‐ CA(U,C) -­‐ CA(A,G) antisense DNA AA(A,G) -­‐ CT(T,C) -­‐ CG(A,G,C,T) -­‐ GT(A,G) -­‐ GT(T,C) NOTE:This represents 64 possible combinations to produce the amino acid sequence!! (7-­‐12,13; Ap) 11. What will be the amino acid sequence if a guanine is substituted for the first of the two adjacent adenines on the template (antisense) strand of the DNA molecule in Question #8. How do the amino acid sequences differ? antisense DNA ...CGGTTACGAGCGATACCTCCG (substitution is underlined) mRNA GCCAAUGCUCGCUAUGGAGGC amino acids ..Met -­‐ Leu -­‐ Ala -­‐ Met -­‐ Glu -­‐ Ala The substitution of G for A results in Leu being inserted into the peptide instead of Phe. The rest of the peptide is the same. Sample Question Answer Keys -­‐ Page 4 (7-­‐12,13; Ap) 12. What will be the amino acid sequence if a guanine is inserted between the two adjacent adenines on the template (antisense) strand of the DNA molecule in Question #8? How do the amino acid sequences differ? (Don’t include the mutation from Question #11.)
sense DNA ...CGGTTACAGAGCGATACCTCCG...(insertion is underlined) mRNA GCCAAUGUCUCGCUAUGGAGGC amino acids ...Met -­‐ Ser -­‐ Arg -­‐ Tyr -­‐ Gly -­‐ Gly All amino acids downstream from the insertion of G are different. (7-­‐7;T) 13. Fill in the table. Initiation = where the process begins; Elongation = the enzyme responsible; Termination = where the process ends. Process Initiation Elongation Termination DNA Replication Replication origin DNA Polymerase III Ter site (mostly) Transcription Promoter on DNA RNA Polymerase Termination site Translation Start Codon Peptidyl Tranferase Stop codon (AUG) (UAA, UGA, UAG) (7-­‐16;I) 14. Inducible genetic control is associated with b. genes of catabolic enzymes. (7-­‐16) 15. What would be the effect of a mutation in the lac operon repressor protein that makes it nonfunctional? The genes of the lac operon would a. be transcribed continuously, regardless of lactose’s presence or absence. Key to Unit 8 Sample Problems (8-­‐2; T) 1. Which is considered a recent criterion for categorizing bacteria? Comparison of a. antigens (8-­‐5; An) 2. Radioactive DNA from Species M, N, and O is allowed to hybridize with reference DNA from Species Z. The following data are recorded: % HYBRIDIZATION HYBRID M/Z 15% N/Z O/Z 89% 52% Which is true? b. N and Z are more closely related than M and Z. (8-­‐5; Ap) 3. Four types of bacteria have the following G+C percents: BACTERIUM A B C D G+C Percent 32 35 48 64 Which statement is best supported by the G+C percent data? e. Species B and D are not closely related. Sample Question Answer Keys -­‐ Page 5 (8-­‐5;Ap) 4. The rectangle below represents an electrophoresis gel used in a DNA sequencing experiment like the one discussed in class and in your text. The relevant nucleotides are labeled with a fluorescent chemical as follows: ddATP is red, ddTTP is blue, ddGTP is green, and ddCTP is yellow. Determine the nucleotide sequence of the strand in the gel. MIGRATION IS TO THE RIGHT. |
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The sequence of the strand in the gel is: 5’-­‐TCGAGGCAGTCCTATA-­‐3’ If you got 5’-­‐ATATCCTGACGGAGCT-­‐3’ then you probably forgot that nucleotides are added to the 3’ end of a growing polynucleotide chain. The smallest and fastest fragment will consist of primer plus the first nucleotide at the 5’ end. Sample Question Answer Keys -­‐ Page 6