CCR Biology - Chapter 9 Practice Test - Summer 2012



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Name: Class: Date: CCR Biology - Chapter 9 Practice Test - Summer 2012 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Genetic engineering is possible because all organisms are based on the same a. proteins. b. DNA copies. c. plasmid loops. d. genetic code. 2. Scientists can study the effect of turning off a gene by using a. cloned animals. b. gene knockout organisms. c. DNA microarrays. d. DNA fingerprinting. 3. What is a clone? a. a genetically identical copy of a gene or organism b. a ring of bacterial DNA with foreign DNA inserted c. a mouse that has had some of its genes "turned off" d. a genetically engineered organism with new traits 4. Which of the following technologies uses electricity to sort DNA fragments based on their sizes? a. genetic screening b. gel electrophoresis c. DNA microarrays d. genetic engineering 5. In the restriction map shown in Figure 9.1, which line, or band, represents the smallest DNA fragments? a. A b. B c. C d. D 1

Name: 6. Which of the following processes could a scientist use to make an amount of DNA that is large enough to study? a. polymerase chain reaction b. genetic engineering c. bioinformatics d. gel electrophoresis 7. Which of the following is the study of entire genomes? a. gene sequencing b. proteomics c. genomics d. transgenics 8. In polymerase chain reaction, which of the following acts as the starting point for new DNA strands? a. cdna b. plasmids c. enzymes d. primers 9. The restriction enzymes used by scientists to cut DNA molecules at specific nucleotide sequences come from a. mice. b. yeasts. c. bacteria. d. viruses. 10. Which of the following experimental techniques involves treating a genetic disorder by replacing a defective gene with a functional gene? a. genetic screening b. bioinformatics c. genomics d. gene therapy 11. Suppose the restriction site for a particular restriction enzyme is: GATC CTAG. The enzyme cuts the DNA between the A and the T on the top strand, and between the T and the A on the bottom strand. How many restriction sites for this enzyme are there in the DNA sequence shown in Figure 9.2? a. 2 b. 3 c. 4 d. 5 12. What goal of the Human Genome Project has been achieved? a. learning how DNA sequences vary among all people b. sequencing all the base pairs of the human chromosomes c. understanding the function of each human gene d. discovering the proteins that play a part in cancer 2

Name: 13. Genetic screening can help determine a person's risk of passing on a genetic disorder through a combination of family history and a. recombinant DNA. b. gene knockouts. c. DNA microarrays. d. DNA testing. 14. The purpose of genetic engineering is to a. develop safer pesticides. b. study obesity and cancer. c. reduce Earth's biodiversity. d. give organisms new traits. 15. Proteomics is the study and comparison of a. proteins within and among species. b. genomes of various organisms. c. DNA sequence variations among people. d. transgenic organisms in different environments. 16. Transgenic organisms can express the foreign genes because all organisms are based on the same a. recombinant DNA. b. noncoding DNA areas. c. repeating nucleotides. d. genetic code. 17. Which of the following technologies can compare the expression of thousands of genes at once? a. DNA fingerprinting b. gel electrophoresis c. gene sequencing d. DNA microarrays 18. Using gene knockout organisms can reveal the function of a gene by a. rearranging the sequence of an organism's genes. b. showing the effect of turning off a gene. c. comparing the DNA fingerprint of two organisms. d. distinguishing between an original organism and its clone. 19. To quickly amplify the amount of DNA in a sample, which process would a scientist use? a. polymerase chain reaction b. bioinformatics c. DNA fingerprinting d. gel electrophoresis 3

Name: 20. Which of the following statements is true for the restriction map shown in Figure 9.1? a. fragments B and D smaller than A. b. fragments A and B smaller than D. c. fragments B and C same size as D. d. fragment B smaller than C. 21. How does gel electrophoresis separate DNA fragments? a. by genetic screening b. according to size c. DNA microarrays d. proteomic analysis 22. Which of the following components is required for PCR? a. cdna molecules b. restriction maps c. amino acids d. two primers 23. Which of the following terms means the study and comparison of proteins within and among organisms? a. genetic engineering b. proteomics c. genomics d. bioinformatics 24. Bacterial enzymes that cut DNA molecules at specific nucleotide sequences are called a. cdna enzymes. b. bacterial polymerases. c. restriction enzymes. d. DNA polymerases. 25. Scientists have inserted the gene for human insulin into plasmids. The new plasmids are an example of a. recombinant DNA. b. applied bioinformatics. c. therapeutic cloning. d. gene therapy. 4

Name: 26. Suppose the restriction site for a particular restriction enzyme is: GATC CTAG. The enzyme cuts the DNA between the A and the T on the top strand, and between the T and the A on the bottom strand. How many fragments are produced from the DNA sequence shown in Figure 9.2? a. 2 b. 3 c. 4 d. 5 27. In the Human Genome Project, scientists have identified the sequence of base pairs of the human chromosomes, and are working to determine the a. sequence of opposite DNA strands. b. function of particular genes. c. first gene associated with a disease. d. code that DNA uses to make amino acids. 28. The process of DNA testing to determine a person's risk of having or carrying a genetic disorder is part of a. DNA fingerprinting. b. genetic engineering. c. the HapMap Project. d. genetic screening. 29. What is one of the benefits of genetic engineering of plants? a. reduced diversity among crop plants b. increased allergies to plants and pollen c. elimination of natural crop pollinators d. enhanced disease resistance in crops 30. One approach to gene therapy is to remove some of a patient's bone marrow stem cells and "infect" them with a virus that contains a functional gene. The stem cells are put back into the patient's bone marrow, so that they can a. divide and make more blood cells with the functional gene. b. kill the cells that contain the faulty gene. c. spread throughout the body to other types of cells. d. stimulate the immune system to attack harmful cells. 5

CCR Biology - Chapter 9 Practice Test - Summer 2012 Answer Section MULTIPLE CHOICE 1. ANS: D PTS: 1 DIF: Level A REF: act0976aaf18007e136_14 STA: KY 9-12.4.1.4 KY 9-12.4.2.3 KY 9-12.4.2.6 KY 9-12.4.2.11 KY 9-12.7.1.2 KY 9-12.7.2.8 KY 9-12.SC-HS-3.4.6 TOP: Ch 9 Test - A 2. ANS: B PTS: 1 DIF: Level A REF: act0976aaf18007e136_38 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 TOP: Ch 9 Test - A 3. ANS: A PTS: 1 DIF: Level A REF: act0976aaf18007e136_30 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 KY 9-12.4.2.11 KY 9-12.7.1.2 KY 9-12.7.2.8 TOP: Ch 9 Test - A 4. ANS: B PTS: 1 DIF: Level A REF: act0976aaf18007e136_46 TOP: Ch 9 Test - A 5. ANS: D PTS: 1 DIF: Level A REF: act0976aaf18007e136_22 STA: KY 9-12.4.1.4 KY 9-12.4.2.3 KY 9-12.4.2.6 KY 9-12.SC-HS-3.4.6 TOP: Ch 9 Test - A 6. ANS: A PTS: 1 DIF: Level A REF: act0976aaf18007e136_54 KY 9-12.7.2.8 TOP: Ch 9 Test - A 7. ANS: C PTS: 1 DIF: Level A REF: act0976aaf18007e136_62 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 TOP: Ch 9 Test - A 8. ANS: D PTS: 1 DIF: Level A REF: act0976aaf18007e136_70 KY 9-12.7.2.8 TOP: Ch 9 Test - A 9. ANS: C PTS: 1 DIF: Level A REF: act0976aaf18007e136_78 TOP: Ch 9 Test - A 10. ANS: D PTS: 1 DIF: Level A REF: act0976aaf18007e136_86 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 KY 9-12.4.2.11 KY 9-12.5.2.4 TOP: Ch 9 Test - A 11. ANS: B PTS: 1 DIF: Level A REF: act0976aaf18007e136_94 TOP: Ch 9 Test - A 12. ANS: B PTS: 1 DIF: Level A REF: act0976aaf18007e136_102 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 TOP: Ch 9 Test - A 13. ANS: D PTS: 1 DIF: Level A REF: act0976aaf18007e136_110 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 KY 9-12.4.2.11 KY 9-12.5.2.4 TOP: Ch 9 Test - A 14. ANS: D PTS: 1 DIF: Level A REF: act0976aaf18007e136_118 KY 9-12.7.1.2 KY 9-12.7.2.8 TOP: Ch 9 Test - A 1

15. ANS: A PTS: 1 DIF: Level A REF: act0976aaf18007e136_126 STA: KY 9-12.4.1.2 KY 9-12.4.1.3 KY 9-12.4.2.1 KY 9-12.4.2.3 KY 9-12.4.2.6 KY 9-12.SC-HS-3.4.1 KY 9-12.SC-HS-3.4.3.a KY 9-12.SC-HS-3.4.3.b TOP: Ch 9 Test - A 16. ANS: D PTS: 1 DIF: Level B REF: act0976aaf18007e137_202 STA: KY 9-12.4.1.2 KY 9-12.4.1.3 KY 9-12.4.1.4 KY 9-12.4.1.7 KY 9-12.4.2.1 KY 9-12.4.2.3 KY 9-12.4.2.6 KY 9-12.SC-HS-3.4.6 TOP: Ch 9 Test - B 17. ANS: D PTS: 1 DIF: Level B REF: act0976aaf18007e137_219 STA: KY 9-12.4.1.2 KY 9-12.4.1.3 KY 9-12.4.1.7 KY 9-12.4.2.1 KY 9-12.4.2.6 KY 9-12.SC-HS-3.4.6 TOP: Ch 9 Test - B 18. ANS: B PTS: 1 DIF: Level B REF: act0976aaf18007e137_227 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 TOP: Ch 9 Test - B 19. ANS: A PTS: 1 DIF: Level B REF: act0976aaf18007e137_243 STA: KY 9-12.4.1.2 KY 9-12.4.1.3 KY 9-12.4.1.7 KY 9-12.4.2.1 KY 9-12.4.2.6 KY 9-12.SC-HS-3.4.6 TOP: Ch 9 Test - B 20. ANS: A PTS: 1 DIF: Level B REF: act0976aaf18007e137_210 STA: KY 9-12.4.1.4 KY 9-12.4.2.3 KY 9-12.4.2.6 KY 9-12.SC-HS-3.4.6 TOP: Ch 9 Test - B 21. ANS: B PTS: 1 DIF: Level B REF: act0976aaf18007e137_235 TOP: Ch 9 Test - B 22. ANS: D PTS: 1 DIF: Level B REF: act0976aaf18007e137_259 STA: KY 9-12.4.1.2 KY 9-12.4.1.3 KY 9-12.4.1.7 KY 9-12.4.2.1 KY 9-12.4.2.6 KY 9-12.SC-HS-3.4.6 TOP: Ch 9 Test - B 23. ANS: B PTS: 1 DIF: Level B REF: act0976aaf18007e137_251 STA: KY 9-12.4.1.2 KY 9-12.4.1.3 KY 9-12.4.1.7 KY 9-12.4.2.1 KY 9-12.4.2.3 KY 9-12.4.2.6 KY 9-12.SC-HS-3.4.1 KY 9-12.SC-HS-3.4.3.a KY 9-12.SC-HS-3.4.3.b KY 9-12.SC-HS-3.4.6 TOP: Ch 9 Test - B 24. ANS: C PTS: 1 DIF: Level B REF: act0976aaf18007e137_267 TOP: Ch 9 Test - B 25. ANS: A PTS: 1 DIF: Level B REF: act0976aaf18007e137_275 KY 9-12.7.2.8 TOP: Ch 9 Test - B 26. ANS: C PTS: 1 DIF: Level B REF: act0976aaf18007e137_283 TOP: Ch 9 Test - B 27. ANS: B PTS: 1 DIF: Level B REF: act0976aaf18007e137_291 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 TOP: Ch 9 Test - B 28. ANS: D PTS: 1 DIF: Level B REF: act0976aaf18007e137_299 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 KY 9-12.4.2.11 KY 9-12.5.2.4 TOP: Ch 9 Test - B 29. ANS: D PTS: 1 DIF: Level B REF: act0976aaf18007e137_307 KY 9-12.7.1.2 KY 9-12.7.2.8 TOP: Ch 9 Test - B 2

30. ANS: A PTS: 1 DIF: Level B REF: act0976aaf18007e137_315 STA: KY 9-12.4.1.7 KY 9-12.4.2.3 KY 9-12.4.2.11 KY 9-12.5.2.4 TOP: Ch 9 Test - B 3

CCR Biology - Chapter 9 Practice Test - Summer 2012 [Answer Strip] _ A 6. _ D 13. _ A 20. _ C 26. _ D 1. _ C 7. _ D 14. _ B 2. _ D 8. _ A 15. _ B 27. _ A 3. _ C 9. _ D 16. _ D 28. _ B 21. _ B 4. _ D 10. _ D 17. _ D 29. _ D 22. _ D 5. _ B 11. _ B 18. _ A 30. _ B 23. _ A 19. _ C 24. _ B 12. _ A 25.

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