C. | Reduction, because electrons would not be able TOUGHT the electron transport chain |
D. | Increase, because this would spontaneously establish a proton gradient |
E. | Increase, because this would make more oxygen available in the mitochondrial matrix |
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BLOOM’S LEVEL: 3. Apply LEARNING OUTCOME: 06.05.03 Explain the role of oxygen in respiration. SECTION: 06.05 TOPIC: Cellular Respiration |
54. | A cell produces approximately 30 ATP per glucose, however, if you calculated the total energy in a glucose molecule, 90 ATP should be generated. What is the best explanation for this difference?A. | Some of the energy is destroyed |
B. | Scientists have not figured out where the rest of the energy goes |
C. | ATP has more energy than originally thought |
D. | Some of the energy is used to do work in the cell |
E. | Some energy is lost as heat |
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BLOOM’S LEVEL: 3. Apply LEARNING OUTCOME: 06.06.01 Calculate the net ATP produced in aerobic metabolism. SECTION: 06.06 TOPIC: Cellular Respiration |
55. | The reactions in glycolysis are reversible, however, the conversion of pyruvate into acetyl CoA is irreversible in animal cells. Can fatty acids be converted back into carbohydrates?A. | No, because they enter metabolism as pyruvate |
B. | Yes, because they enter metabolism as acetyl CoA |
C. | No, because they enter metabolism as acetyl CoA |
D. | Yes, because they enter metabolism as pyruvate |
E. | Yes, because they enter metabolism in the Krebs cycle |
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BLOOM’S LEVEL: 4. Analyze LEARNING OUTCOME: 06.07.01 Compare and contrast where carbohydrate, fat and protein enter metabolism. SECTION: 06.07 TOPIC: Cellular Respiration |
56. | Which of the following could be electron acceptors in anaerobic respiration?
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BLOOM’S LEVEL: 2. Understand LEARNING OUTCOME: 06.08.01 Draw and explain the net reaction in anaerobic respiration. SECTION: 06.08 TOPIC: Cellular Respiration |
57. | During fermentation yeast produce ethanol, a toxic waste product. Why would yeast produce something that is toxic to themselves?A. | To produce NADH for energy |
B. | To regenerate NAD+ in order to continue glycolysis |
C. | To use up excess pyruvate in order to continue glycolysis |
D. | To conserve oxygen for other reactions |
E. | To avoid the production of lactic acid |
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BLOOM’S LEVEL: 2. Understand LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
58. | During fermentation ___ ATP can be produced per glucose, compared with ____ by aerobic respiration.
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BLOOM’S LEVEL: 2. Understand LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
59. | If you put a plant in a chamber and shine a light on it, the concentrations ______.A. | of O2 will rise and CO2 will fall |
B. | of O2 will fall and CO2 will rise |
C. | of both O2 and CO2 will rise |
D. | of both O2 and CO2 will fall |
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BLOOM’S LEVEL: 3. Apply LEARNING OUTCOME: 06.09.01 Compare and contrast respiration and photosynthesis. SECTION: 06.09 TOPIC: Cellular Respiration |
60. | If you put a plant in a chamber and place it in the dark, the concentrations ______.A. | of O2 will fall and CO2 will rise |
B. | of O2 will rise and CO2 will fall |
C. | of both O2 and CO2 will rise |
D. | of both O2 and CO2 will fall |
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BLOOM’S LEVEL: 3. Apply LEARNING OUTCOME: 06.09.01 Compare and contrast respiration and photosynthesis. SECTION: 06.09 TOPIC: Cellular Respiration |
61. | Most plants are:
C. | Heterotrophic ectotherms |
D. | Heterotrophic endotherms |
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BLOOM’S LEVEL: 2. Understand LEARNING OUTCOME: 06.09.01 Compare and contrast respiration and photosynthesis. LEARNING OUTCOME: 18.00.01 Describe the characteristics common to all plants. SECTION: 06.09 SECTION: 18.01 TOPIC: Cellular Respiration TOPIC: Investigating Life TOPIC: Plant Diversity |
62. | Philodendron flowers pass electrons through an electron transport chain to generate heat. What is the source of these electrons?
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BLOOM’S LEVEL: 2. Understand LEARNING OUTCOME: 06.05.02 Diagram and explain the flow of electrons in the electron transport chain. SECTION: 06.05 TOPIC: Cellular Respiration TOPIC: Investigating Life |
63. | Philodendron flowers pass electrons through a mitochondrial electron transport chain to generate heat. What step is bypassed in generating heat?A. | Formation of a proton gradient |
B. | Transfer of electrons to oxygen |
C. | Donation of electrons to the electron transport chain |
D. | Excitation of electrons by sunlight |
E. | Release of electrons from water |
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BLOOM’S LEVEL: 3. Apply LEARNING OUTCOME: 06.05.02 Diagram and explain the flow of electrons in the electron transport chain. SECTION: 06.05 TOPIC: Cellular Respiration TOPIC: Investigating Life |
64. | What hypothesis were the scientists testing?A. | Does generating heat increase the metabolism of Philodendron? |
B. | Does generating heat decrease the metabolism of Philodendron? |
C. | Is heat being used by Philodendron to attract pollinators? |
D. | Is heat being used by Philodendron increase the rate of pollen production? |
E. | Does heat increase the rate of photosynthesis by Philodendron? |
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BLOOM’S LEVEL: 4. Analyze LEARNING OUTCOME: 06.09.01 Compare and contrast respiration and photosynthesis. LEARNING OUTCOME: 18.05.02 Explain the interaction of flowers and fruit with animals in angiosperm evolution. SECTION: 06.09 SECTION: 18.05 TOPIC: Cellular Respiration TOPIC: Investigating Life TOPIC: Plant Diversity |
65. | The rate of carbon dioxide production by Cyclocephala colasi beetles was used to measure their:A. | Ability to pollinate flowers |
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BLOOM’S LEVEL: 4. Analyze LEARNING OUTCOME: 06.01.01 Draw and explain the net reaction in aerobic respiration of glucose. SECTION: 06.01 TOPIC: Cellular Respiration TOPIC: Investigating Life |
66. | What is the dependent variable in figure 6.14?
B. | Rate of oxygen production |
C. | Amount of pollen collected |
D. | Rate of carbon dioxide production |
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BLOOM’S LEVEL: 4. Analyze LEARNING OUTCOME: 06.01.01 Draw and explain the net reaction in aerobic respiration of glucose. SECTION: 06.01 TOPIC: Cellular Respiration TOPIC: Investigating Life |
67. | What is the independent variable in figure 6.14?
B. | Rate of carbon dioxide production |
C. | Rate of oxygen production |
D. | Amount of pollen collected |
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BLOOM’S LEVEL: 4. Analyze LEARNING OUTCOME: 06.01.01 Draw and explain the net reaction in aerobic respiration of glucose. SECTION: 06.01 TOPIC: Cellular Respiration TOPIC: Investigating Life |
68. | What did the scientists conclude from figure 6.14?A. | That at higher temperatures resting beetles saved energy |
B. | That at higher temperatures active beetles saved energy |
C. | That at lower temperatures active beetles saved energy |
D. | That at lower temperatures resting beetles saved energy |
E. | That generating heat led to increased pollination of Philodendron flowers by beetles |
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BLOOM’S LEVEL: 4. Analyze LEARNING OUTCOME: 06.01.01 Draw and explain the net reaction in aerobic respiration of glucose. SECTION: 06.01 TOPIC: Cellular Respiration TOPIC: Investigating Life |
69. | Philodendron flowers have to use a lot of energy to attract beetles as pollinators. This is similar to the energy used to produce ____ by many other flowering plants to attract pollinators.
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BLOOM’S LEVEL: 4. Analyze LEARNING OUTCOME: 06.09.01 Compare and contrast respiration and photosynthesis. LEARNING OUTCOME: 18.05.02 Explain the interaction of flowers and fruit with animals in angiosperm evolution. SECTION: 06.09 SECTION: 18.05 TOPIC: Cellular Respiration TOPIC: Investigating Life TOPIC: Plant Diversity |
70. | Cells from a Philodendron flower have which of the following?
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BLOOM’S LEVEL: 2. Understand LEARNING OUTCOME: 06.09.01 Compare and contrast respiration and photosynthesis. LEARNING OUTCOME: 18.00.01 Describe the characteristics common to all plants. SECTION: 06.09 SECTION: 18.01 TOPIC: Cellular Respiration TOPIC: Investigating Life TOPIC: Plant Diversity |
True / False Questions 71. | Cellular respiration includes glycolysis, the Kreb’s cycle, and electron transport.
TRUE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.02.01 Compare and contrast what occurs in glycolysis, the Krebs cycle, and electron transport chain. SECTION: 06.02 TOPIC: Cellular Respiration |
72. | Cellular respiration includes only the Kreb’s cycle and electron transport.
FALSE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.02.01 Compare and contrast what occurs in glycolysis, the Krebs cycle, and electron transport chain. SECTION: 06.02 TOPIC: Cellular Respiration |
73. | In Eukaryotic cells the electron transport system occurs in the mitochondria.
TRUE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.02.01 Compare and contrast what occurs in glycolysis, the Krebs cycle, and electron transport chain. SECTION: 06.02 TOPIC: Cellular Respiration |
74. | In prokaryotic cells the electron transport system occurs in the mitochondria.
FALSE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.02.01 Compare and contrast what occurs in glycolysis, the Krebs cycle, and electron transport chain. SECTION: 06.02 TOPIC: Cellular Respiration |
75. | Fermentation is a more efficient form of energy production than aerobic respiration.
FALSE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
76. | NADH is the only electron carrier produced during aerobic respiration.
FALSE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
77. | Oxidative phosphorylation occurs when protons pass through ATP synthase and combine with oxygen to make water.
TRUE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
78. | Substrate phosphorylation occurs when protons pass through ATP synthase and combine with oxygen to make water.
FALSE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
79. | Substrate phosphorylation occurs when a high energy “donor” molecule physically transfers a phosphate group to ADP.
TRUE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
80. | The electron transport chain extracts most of the potential energy in NADH by removing the energy from its electrons in a single step.
FALSE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
81. | According to the first law of thermodynamics some energy is lost as heat during energy transfers.
FALSE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
82. | Anaerobic respiration uses oxygen as its final electron acceptor at the end of the electron transport chain.
FALSE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
83. | Anaerobic respiration uses an inorganic molecule other than O2 as the electron acceptor at the end of the electron transport chain.
TRUE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
84. | Aerobic respiration uses oxygen as its final electron acceptor at the end of the electron transport chain.
TRUE |
BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. SECTION: 06.08 TOPIC: Cellular Respiration |
Chapter 006 – How Cells Release Energy Summary Category | # of Questions | BLOOM’S LEVEL: 1. Remember | 54 | BLOOM’S LEVEL: 2. Understand | 16 | BLOOM’S LEVEL: 3. Apply | 7 | BLOOM’S LEVEL: 4. Analyze | 7 | LEARNING OUTCOME: 06.01.01 Draw and explain the net reaction in aerobic respiration of glucose. | 11 | LEARNING OUTCOME: 06.02.01 Compare and contrast what occurs in glycolysis, the Krebs cycle, and electron transport chain. | 10 | LEARNING OUTCOME: 06.03.01 Describe where each step in respiration occurs in a mitochondria. | 3 | LEARNING OUTCOME: 06.04.01 Draw and explain the net reaction in glycolysis. | 10 | LEARNING OUTCOME: 06.05.01 Draw and explain the net reaction in the Krebs cycle. | 9 | LEARNING OUTCOME: 06.05.02 Diagram and explain the flow of electrons in the electron transport chain. | 7 | LEARNING OUTCOME: 06.05.03 Explain the role of oxygen in respiration. | 2 | LEARNING OUTCOME: 06.06.01 Calculate the net ATP produced in aerobic metabolism. | 4 | LEARNING OUTCOME: 06.07.01 Compare and contrast where carbohydrate, fat and protein enter metabolism. | 2 | LEARNING OUTCOME: 06.08.01 Draw and explain the net reaction in anaerobic respiration. | 2 | LEARNING OUTCOME: 06.08.02 Compare and contrast aerobic respiration, anaerobic respiration, and fermentation. | 16 | LEARNING OUTCOME: 06.09.01 Compare and contrast respiration and photosynthesis. | 8 | LEARNING OUTCOME: 18.00.01 Describe the characteristics common to all plants. | 2 | LEARNING OUTCOME: 18.05.02 Explain the interaction of flowers and fruit with animals in angiosperm evolution. | 2 | SECTION: 06.01 | 11 | SECTION: 06.02 | 10 | SECTION: 06.03 | 3 | SECTION: 06.04 | 10 | SECTION: 06.05 | 18 | SECTION: 06.06 | 4 | SECTION: 06.07 | 2 | SECTION: 06.08 | 18 | SECTION: 06.09 | 8 | SECTION: 18.01 | 2 | SECTION: 18.05 | 2 | TOPIC: Cellular Respiration | 84 | TOPIC: Investigating Life | 10 | TOPIC: Plant Diversity | 4 |
Chapter 007 – DNA Structure and Gene Function
Multiple Choice Questions 1. | The twisted ladder of DNA is composed of building blocks called:
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2. | The rungs of the DNA ladder:A. | Are formed by base pairs joined by covalent bonds |
B. | Are formed by base pairs joined by hydrogen bonds |
C. | Are formed by base pairs joined by phosphodiester bonds |
D. | Are formed by base pairs joined by hydrogen bonds |
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3. | In a molecule of DNA:A. | Cytosine pairs with adenine |
B. | Thymine pairs with uracil |
C. | Adenine pairs with uracil |
D. | Adenine pairs with thymine |
E. | Cytosine pairs with thymine |
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4. | The “Central Dogma” refers:A. | Genetic information passing from DNA to RNA by translation |
B. | Genetic information passing from RNA to protein by transcription |
C. | To inheritance of traits between generations |
D. | To production of energy in a cell |
E. | To the flow of genetic information in cells |
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5. | RNA differs from DNA in that:
D. | RNA is usually single stranded |
E. | RNA can catalyze chemical reactions |
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6. | DNA differs from RNA in that:
C. | DNA can catalyze chemical reactions |
D. | DNA is usually single stranded |
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7. | The type of RNA that carries the information that specifies a protein is:
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8. | The type of RNA that helps align the ribosome and mRNA is:
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9. | The type of RNA that carries each amino acid to the ribosome is:
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10. | Transcription and replication occur during ____________________ of the cell cycle.
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11. | The strand of DNA that encodes the RNA molecule during transcription is the:
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12. | A DNA sequence that signals a gene’s start is:
E. | An amino acid attachment site |
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13. | In eukaryotic cells sequences of mRNA that are removed from an mRNA molecule before being translated are:
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14. | The process used by cells to convert the mRNA “message” into a sequence of amino acids is:
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15. | Initiation, elongation, and termination are the three main steps in:A. | Replication and translation |
B. | Transcription and translation |
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16. | A three base sequence (loop) in tRNA that is complementary to a sequence of three bases in mRNA is:
E. | An amino acid attachment site |
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17. | A tRNA molecule is “bilingual” because it binds to:
C. | Promoters and amino acids |
D. | Codons of mRNA and amino acids |
E. | Promoters and terminators |
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18. | The step of translation in which an mRNA, a small ribosomal subunit, and the initiator tRNA are aligned together is:
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19. | The step of translation in which amino acids are added one at a time to the growing polypeptide is:
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20. | The step of translation in which release factors bind to a stop codon is:
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21. | Which of the following does not occur during translation’s termination step?A. | The initiator tRNA brings the amino acid methionine |
B. | Ribosomal subunits separate from each other |
C. | The polypeptide is released |
D. | A “stop” codon is reached by the ribosome |
E. | Release factors bind to the stop codon |
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22. | A group of prokaryotic genes and other segments of DNA that are controlled together is:
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23. | In the Lac operon, the attachment site for RNA polymerase is:
D. | The first lactose degrading gene of the operon |
E. | The last lactose degrading gene of the operon |
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24. | In the Lac operon, the protein that binds to the operator to prevent transcription is:
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25. | Proteins that initiate transcription in eukaryotes by recognizing sequences within the promoter region of a gene and attracting RNA polymerase are:
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26. | A change in a cell’s DNA sequence is:
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27. | In a “nonsense” mutation:A. | The codon that mutates does not cause a change in the amino acid specified |
B. | The codon that mutates causes a change in the amino acid specified |
C. | The codon that mutates causes a stop codon to occur instead of the placement of an amino acid |
D. | The mutation does not occur in a codon |
E. | The mutation is not in DNA |
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28. | In a “missense” mutation:A. | The codon that mutates does not cause a change in the amino acid specified |
B. | The codon that mutates causes a change in the amino acid specified |
C. | The codon that mutates causes a stop codon to occur instead of the placement of an amino acid |
D. | The mutation does not occur in a codon |
E. | The mutation is not in DNA |
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29. | In a “silent” mutation:A. | The codon that mutates causes a change in the amino acid specified |
B. | The codon that mutates does not cause a change in the amino acid specified |
C. | The codon that mutates causes a stop codon to occur instead of the placement of an amino acid |
D. | The mutation does not occur in a codon |
E. | The mutation is not in DNA |
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30. | In a “frameshift” mutation:A. | The codon that mutates causes a change in the amino acid specified |
B. | The codon that mutates causes a stop codon to occur instead of the placement of an amino acid |
C. | The mutation does not occur in a codon |
D. | May be caused by an addition or deletion of one or more nucleotides |
E. | The mutation is not in DNA |
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31. | Which of the following is not a common cause of mutation?
D. | Exposure to chemical substances |
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32. | A germline mutation occurs in cells that give rise to:
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33. | Transposable elements:A. | Are segments of RNA found in chromosomes |
B. | Are another name for translocations |
C. | Are DNA sequences that can “jump” within the genome |
D. | Result from damage to the chromosomes by things like radiation |
E. | Are formed during transcription |
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34. | An organism that receives recombinant DNA:
B. | Is a transgenic organism |
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35. | Treating a solution from S strain Streptococcus pneumoniae with an enzyme that destroys ____ would prevent it from transforming type R bacteria into a form that would cause pneumonia in mice.
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36. | If bacteria are infected with bacteriophage containing radioactive ____ the bacteria will now be radioactive.
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37. | If the DNA in a cell consists of 20% A, it will be ___ G.
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38. | If the DNA in a cell consists of 20% A, it will be ___ T.
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39. | The DNA sequence 5′ – ATGCATGC – 3′ will pair with which of the following DNA strands?
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40. | Why would it take more energy to separate the double-stranded region of DNA with the sequence GCGCGCGC than a region with the sequence ATATATAT?A. | GC base pairs form more phosphodiester bonds |
B. | The helix is wound more tightly in GC base pairs |
C. | GC base pairs form more hydrogen bonds |
D. | The sequence containing G’s and C’s is longer |
E. | GC base pairs form covalent bonds between DNA strands |
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41. | The DNA sequence 5′ – ATGCATGC – 3′ will pair with which of the following RNA strands?
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42. | A region of a chromosome contains two strands of DNA, yet only one is used to transcribe a gene because:A. | RNA polymerase binds to the promoter, initiating transcription on one strand of DNA |
B. | The start of transcription is random along the chromosome and doesn’t always occur on a gene |
C. | RNA polymerase binds to the terminator, stopping transcription on one strand of DNA |
D. | Both strands are transcribed, but only one strand of DNA contains a reading frame |
E. | All genes are found on the same strand of DNA in a chromosome |
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43. | If you compared the DNA sequence of a gene with the sequence of the mature mRNA that was transcribed from the gene you would find:A. | The mRNA is shorter because it does not contain exons |
B. | Both are the same length |
C. | The mRNA is shorter because it does not contain introns |
D. | The mRNA is shorter because each codon of three bases encodes only one amino acid |
E. | The mRNA is longer because each codon of one amino acid encodes three bases |
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44. | What would be the first codon translated in the mRNA sequence 5′ – GGAAUGAAACAGGAACCC – 3′?
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45. | Using the genetic code shown here, predict what type of mutation has occurred in the hemoglobin sickle cell anemia allele. Normal allele 5′ – GGAAUGAAACAGGAACCC – 3′ Mutant allele 5′ – GGAAUGAAACAGGUACCC – 3′A. | A point mutation, Val to Glu |
C. | Addition of a new stop codon |
D. | A point mutation, Glu to Val |
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46. | A diploid cell only has two copies of most genes, but can make hundreds of copies of a protein from those genes per second because during transcription:A. | Multiple codons can encode for the same amino acid |
B. | Multiple ribosomes can bind to the same mRNA simultaneously |
C. | The DNA can replicate making more copies of the gene |
D. | Once formed, proteins can feed back and help make more proteins |
E. | Multiple copies of the same mRNA can be produced from one gene |
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47. | A diploid cell only has two copies of most genes, but can make hundreds of copies of a protein from those genes per second because during translation:A. | Multiple codons can encode for the same amino acid |
B. | Multiple copies of the same mRNA can be produced from one gene |
C. | The DNA can replicate making more copies of the gene |
D. | Once formed, proteins can feed back and help make more proteins |
E. | Multiple ribosomes can bind to the same mRNA simultaneously |
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48. | Amanatin is a toxin found in the death cap mushroom, Amanita phalloides. It inhibits RNA polymerase, thus blocking:
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49. | If E. coli bacteria are grown in the presence of lactose:A. | The repressor will bind the operator allowing transcription of the lac operon genes |
B. | The repressor will not bind the operator preventing transcription of the lac operon genes |
C. | The repressor will not bind the operator allowing transcription of the lac operon genes |
D. | The repressor will bind the operator preventing transcription of the lac operon genes |
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50. | If E. coli bacteria are grown in the absence of lactose:A. | The repressor will not bind the operator allowing transcription of the lac operon genes |
B. | The repressor will bind the operator allowing transcription of the lac operon genes |
C. | The repressor will not bind the operator preventing transcription of the lac operon genes |
D. | The repressor will bind the operator preventing transcription of the lac operon genes |
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51. | In humans, typically only infants produce the enzyme lactase to break down lactose. Some adults have changes in _____ that allow transcription factors to bind and turn on expression of the lactase gene into adulthood.
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52. | What type of mutation has occurred in the following? Normal allele 5′ – GGAAUGAAACAGGAACCC – 3′ Mutant allele 5′ – GGAAUGAAACAGGUACCC – 3′
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53. | What type of mutation has occurred in the following? Normal allele 5′ – GGAAUGAAACAGGAACCC – 3′ Mutant allele 5′ – GGAAUGAAAUCAGGAACCC – 3′
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54. | How can a single gene encode for more than one protein?A. | By mutating their genes |
B. | By removing some exons during splicing of mRNA |
C. | By transcribing multiple copies of mRNA |
D. | By using different codons |
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55. | In creating a transgenic organism, the ____ from one species is inserted into another species.
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56. | An antisense RNA is complementary to a mRNA and could:A. | Prevent transcription of the mRNA |
B. | Increase translation of the mRNA |
C. | Increase transcription of the mRNA |
D. | Prevent translation of the mRNA |
E. | Mutate the mRNA sequence |
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57. | How does this study support the study of a family with a language disorder support the role of genes in communication?A. | The identification of a single mutation in a single gene responsible for the disorder |
B. | The disorder is common in the family |
C. | The family members were all lacking vocal chords |
D. | The identification of multiple mutations in multiple genes responsible for the disorder |
E. | The identification of the same mutation in all patients with speech disorders |
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58. | Based on this study, FOXP2 must be expressed in which tissues?
E. | Muscle and nerves in the ear |
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59. | FOXP2 functions by:A. | Regulating cell division |
B. | Regulating mRNA splicing |
C. | Regulating the transcription of other genes |
D. | Acting as a neurotransmitter between neurons in the ear and brain |
E. | Acting as a neurotransmitter between neurons and muscles in the jaw |
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60. | FOXP2 sequences were the most different between mice and _____.
E. | All had the same number of changes from the mouse sequence |
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61. | Which mutation in DNA would lead to the change of a single amino acid in a protein?
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62. | The observation that all races of humans have the same FOXP2 allele is evidence that ____ favored this allele.
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63. | How did the researchers estimate that the original mutation happened 3-400,000 years ago?A. | By examining the fossil record |
B. | Using a molecular clock |
C. | They found the mutation in chimpanzee DNA |
D. | They found the mutation in Neandertal DNA |
E. | They found the mutation in 300,000 year old human remains |
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64. | Based just on what is known about FOXP2, could Neandertals speak?A. | No, they have the same FOXP2 mutation as humans |
B. | Yes, they have the same FOXP2 mutation as humans |
C. | Yes, they have the same FOXP2 sequence as chimpanzees |
D. | No, they have the same FOXP2 sequence as chimpanzees |
E. | Yes, they have the same FOXP2 sequence as both humans and chimpanzees |
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65. | How did the researchers initially estimate that the original mutation happened 200,000 years ago?A. | They found the mutation in Neandertal DNA |
B. | By examining the fossil record |
C. | They found the mutation in chimpanzee DNA |
D. | Using a molecular clock |
E. | They found the mutation in 300,000-year-old human remains |
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True / False Questions 66. | The process by which cells use the information of RNA molecules to make proteins is transcription.
True False |
67. | There are only two main types of RNA needed to make proteins. They are tRNA and rRNA.
True False |
68. | The type of RNA that delivers amino acids to the ribosome during protein synthesis is tRNA.
True False |
69. | In eukaryotes transcription copies just one gene from one DNA strand, but replication copies both strands of an entire chromosome.
True False |
70. | A gene is any DNA sequence that is transcribed to mRNA only.
True False |
71. | A gene is any DNA sequence that is transcribed to any type of RNA.
True False |
72. | A poly A tail helps attach the ribosome to the mRNA molecule and may also determine how long the mRNA is retained within the cell.
True False |
73. | Cystic fibrosis may be caused by a protein that does not fold correctly into its final form.
True False |
74. | Because cells are very efficient in the processes that they carry out, very little energy is needed for the cell to make proteins.
True False |
75. | In eukaryotes an mRNA molecule can only be translated by one ribosome at a time.
True False |
76. | In the bacterium E. coli, in the absence of lactose, a repressor protein binds to the operator region of the DNA, preventing the transcription of the genes of the lactose operon.
True False |
77. | In eukaryotic cells gene expression may be controlled by not allowing mRNA to leave the nucleus.
True False |
78. | Translocation during meiosis is not a cause of mutation.
True False |
79. | A somatic mutation is a mutation that occurs in cells that give rise to gametes.
True False |
80. | A transgenic organism is an organism that receives all of its genes from only a single parent.
True False |
81. | Gene therapy might someday be used to treat many genetic disorders by removing faulty genes from somatic cells and replacing them with functional gene copies.
True False |
82. | A proteome is all of the proteins an organism expresses throughout its life.
True False |
Chapter 007 – DNA Structure and Gene Function Key Multiple Choice Questions 1. | The twisted ladder of DNA is composed of building blocks called:
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.02.01 Identify the components of double-stranded DNA. SECTION: 07.02 TOPIC: Gene Regulation |
2. | The rungs of the DNA ladder:A. | Are formed by base pairs joined by covalent bonds |
B. | Are formed by base pairs joined by hydrogen bonds |
C. | Are formed by base pairs joined by phosphodiester bonds |
D. | Are formed by base pairs joined by hydrogen bonds |
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.02.01 Identify the components of double-stranded DNA. SECTION: 07.02 TOPIC: Gene Regulation |
3. | In a molecule of DNA:A. | Cytosine pairs with adenine |
B. | Thymine pairs with uracil |
C. | Adenine pairs with uracil |
D. | Adenine pairs with thymine |
E. | Cytosine pairs with thymine |
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.02.01 Identify the components of double-stranded DNA. SECTION: 07.02 TOPIC: Gene Regulation |
4. | The “Central Dogma” refers:A. | Genetic information passing from DNA to RNA by translation |
B. | Genetic information passing from RNA to protein by transcription |
C. | To inheritance of traits between generations |
D. | To production of energy in a cell |
E. | To the flow of genetic information in cells |
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.03.01 Explain the roles of DNA, RNA, and protein in the central dogma. SECTION: 07.03 TOPIC: Gene Regulation |
5. | RNA differs from DNA in that:
D. | RNA is usually single stranded |
E. | RNA can catalyze chemical reactions |
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.03.01 Explain the roles of DNA, RNA, and protein in the central dogma. SECTION: 07.03 TOPIC: Gene Regulation |
6. | DNA differs from RNA in that:
C. | DNA can catalyze chemical reactions |
D. | DNA is usually single stranded |
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.03.01 Explain the roles of DNA, RNA, and protein in the central dogma. SECTION: 07.03 TOPIC: Gene Regulation |
7. | The type of RNA that carries the information that specifies a protein is:
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.03.01 Explain the roles of DNA, RNA, and protein in the central dogma. SECTION: 07.03 TOPIC: Gene Regulation |
8. | The type of RNA that helps align the ribosome and mRNA is:
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.03.01 Explain the roles of DNA, RNA, and protein in the central dogma. SECTION: 07.03 TOPIC: Gene Regulation |
9. | The type of RNA that carries each amino acid to the ribosome is:
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.03.01 Explain the roles of DNA, RNA, and protein in the central dogma. SECTION: 07.03 TOPIC: Gene Regulation |
10. | Transcription and replication occur during ____________________ of the cell cycle.
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BLOOM’S LEVEL: 1. Remember LEARNING OUTCOME: 07.04.01 Describe the events in transcription. SECTION: 07.04 TOPIC: Gene Regulation |
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