Name Period Mrs. Laux Take Home Test #9 on Chaps. 12 and 16 AP Biology DUE: MONDAY, DECEMBER 20, 2010 MULTIPLE CHOICE QUESTIONS

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Name Period Mrs. Laux Take Home Test #9 on Chaps. 12 and 16 AP Biology DUE: MONDAY, DECEMBER 20, 2010 MULTIPLE CHOICE QUESTIONS 1. The main reason scientists thought that proteins, rather than DNA, were
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Name Period Mrs. Laux Take Home Test #9 on Chaps. 12 and 16 AP Biology DUE: MONDAY, DECEMBER 20, 2010 MULTIPLE CHOICE QUESTIONS 1. The main reason scientists thought that proteins, rather than DNA, were the carriers of genetic material in the cell was: A. their presence within the nucleus. B. their abundance within the cell. C. the large number of possible amino acid combinations. D. their ability to self replicate within the cytoplasm. E. their ability to be exported from the cell. 2. In the experiments of Griffith, the conversion of nonlethal R-strain bacteria to lethal S- strain bacteria: A. was the result of genetic mutation. B. was an example of the genetic exchange known as transformation. C. supported the case for proteins as the genetic material. D. could not be reproduced by other researchers. E. was an example of conjugation. 3. Which of the following statements about DNA is false? A. DNA is capable of forming many different sequences. B. DNA contains thymine instead of uracil. C. DNA is double-stranded rather than single-stranded. D. DNA is only found in eukaryotic cells. E. DNA contains the sugar deoxyribose. 4. The bacteriophages used in Alfred Hershey s and Martha Chase s experiments showed that: A. DNA was injected into bacteria. B. DNA and protein were injected into bacteria. C. DNA remained on the outer coat of bacteria. D. proteins were injected into bacteria. E. proteins were responsible for the production of new viruses within the bacteria. 5. The two molecules that alternate to form the backbone of a polynucleotide chain are: A. adenine and thymine. B. cytosine and guanine. C. sugar and phosphate. D. base and sugar. E. base and phosphate. 6. Chargaff determined that DNA from any source contains about the same amount of guanine as. A. uracil B. thymine C. adenine D. cytosine E. guanine 7. used x-ray diffraction to provide images of DNA. A. Watson and Crick B. Crick and Wilkins C. Franklin D. Franklin and Crick E. Watson and Wilkins 12-1 Name Period Mrs. Laux Take Home Test #9 on Chaps. 12 and 16 AP Biology DUE: MONDAY, DECEMBER 20, X-ray diffraction studies are used to determine: A. the sequence of amino acids in protein molecules. B. the sequence of nucleic acids in nucleic acid molecules. C. the distances between atoms of molecules. D. the type of chemical under investigation. E. the wavelength of light emitted by chemicals. 9. X-ray crystallography showed that DNA: A. had the bases in the center of the molecule. B. had the sugars and phosphates on the outside of the molecule. C. was a very long molecule. D. was made of 2 strands. E. was a helix. 10. determined the structure of the molecule DNA. A. Crick and Wilkins B. Watson and Crick C. Franklin and Crick D. Franklin E. Watson, Crick, and Wilkins 11. The information carried by DNA is incorporated in a code specified by the: A. phosphodiester bonds of the DNA strand. B. number of separate strands of DNA. C. size of a particular chromosome. D. specific nucleotide sequence of the DNA molecule. E. number of bases in a DNA strand. 12. Why is DNA able to store large amounts of information? A. It contains a large number of different nucleotides. B. Its nucleotides can be arranged in a large number of possible sequences. C. It is capable of assuming a wide variety of shapes. D. The sugar and phosphates can be arranged in many different sequences. E. The nucleotides can be altered to form many different letters in the sequence Use the figure to answer the corresponding questions. 13. The portion of the molecule in box 5 of the accompanying figure is: A. a hydrogen bond. B. a phosphate. C. a nucleotide. D. a pyrimidine. E. a protein. 12-2 Name Period Mrs. Laux Take Home Test #9 on Chaps. 12 and 16 AP Biology DUE: MONDAY, DECEMBER 20, In the figure, the portion of the molecule in box is a pyrimidine. A. 1 B. 3 C. 4 D. 1 and 3 E. 3 and The portion of the molecule in box 3 of the associated figure is: A. a sugar. B. a protein. C. a pyrimidine. D. a purine. E. a nucleotide. 16. Hydrogen bonds can form between guanine and, and between adenine and. A. phosphate; sugar B. thymine; cytosine C. cytosine; thymine D. sugar; phosphate E. adenine; guanine 17. Two chains of DNA must run in direction(s) and must be if they are to bond with each other. A. the same; uncomplementary B. opposite; uncomplementary C. parallel; uncomplementary D. parallel; complementary E. antiparallel; complementary 18. Which of the following nucleotide sequences represents the complement to the DNA strand 5 AGATCCG- 3? A. 5 AGATCCG- 3 B. 3 AGATCCG- 5 C. 5 CTCGAAT- 3 D. 3 CTCGAAT- 5 E. 3 TCTAGGC Name Period Mrs. Laux Take Home Test #9 on Chaps. 12 and 16 AP Biology DUE: MONDAY, DECEMBER 20, How is a single strand of DNA able to serve as a template for the synthesis of another strand? A. Nucleotides pair with those of the original strand to form a new strand. B. Hydrogen bonds holding the two strands together are easy to break, allowing one strand to be a template. C. A single strand of DNA is not able to serve as a template. D. One strand of DNA directs the synthesis of a new strand on its partner. E. Both the pairing of nucleotides and the breaking of hydrogen bonds. 20. Which of the following best describes semiconservative replication? A. The translation of a DNA molecule into a complementary strand of RNA. B. A DNA molecule consists of one parental strand and one new strand. C. The number of DNA molecules is doubled with every other replication. D. The replication of DNA never takes place with 100% accuracy. E. The replication of DNA takes place at a defined period in the cell cycle. 21. The final product of DNA replication is: A. mrna, trna, and rrna molecules. B. a wide variety of proteins. C. DNA fragments. D. two DNA molecules, each of which contains one new and one old DNA strand. E. the enzymes needed for further processes, such as DNA polymerase. 22. Who first confirmed that the replication of DNA was semiconservative? A. Chargaff and Hershey B. Watson and Crick C. Avery and Griffith D. Meselson and Stahl E. Watson, Crick, and Wilkins 23. If DNA replication rejoined the 2 parental strands, it would be termed: A. dispersive. B. gradient. C. semiconservative. D. parental. E. conservative. 24. Meselson and Stahl separated DNA from different generations using: A. density gradient centrifugation. B. gel electrophoresis. C. an electron microscope. D. differential radioisotope labeling. E. None of these. 25. Which of the following cause the unwinding of the DNA double helix? A. DNA polymerase B. DNA helicase C. RNA primer D. primosome E. RNA polymerase 12-4 Name Period Mrs. Laux Take Home Test #9 on Chaps. 12 and 16 AP Biology DUE: MONDAY, DECEMBER 20, A replication fork: A. is only seen in prokaryotic chromosomes. B. is only seen in bacterial cells. C. is a Y-shaped structure where both DNA strands are replicated simultaneously. D. is a site where one DNA strand serves as a template, but the other strand is not replicated. E. is created by the action of the enzyme RNA polymerase. 27. In replication, once the DNA strands have been separated, reformation of the double helix is prevented by: A. DNA helicase enzyme. B. single-strand binding proteins. C. DNA polymerases. D. ATP. E. GTP. 28. Enzymes called form breaks in the DNA molecules to prevent the formation of knots in the DNA helix during replication. A. topoisomerases B. single-strand binding proteins C. DNA polymerases D. RNA polymerases E. DNA ligases 29. Which of the following adds new nucleotides to a growing DNA chain? A. DNA polymerase B. DNA helicase C. RNA primer D. primase E. RNA polymerase 30. Why does DNA synthesis only proceed in the 5 to 3 direction? A. Because DNA polymerases can only add nucleotides to the 3 end of a polynucleotide strand. B. Because the 3 end of the polynucleotide molecule is more electronegative than the 5 end. C. Because that is the direction in which the two strands of DNA unzip. D. Because that is the only direction that the polymerase can be oriented. E. Because the chromosomes are always aligned in the 5 to 3 direction in the nucleus. 31. The 5 end of each Okazaki fragment begins with: A. the same RNA primer that began synthesis on the leading strand. B. a DNA primer binding to the template DNA. C. DNA polymerase binding to the template DNA. D. a separate RNA primer. E. a small DNA primer. 12-5 Name Period Mrs. Laux Take Home Test #9 on Chaps. 12 and 16 AP Biology DUE: MONDAY, DECEMBER 20, Primase is the enzyme responsible for: A. unwinding the DNA double strand to allow DNA polymerase access to the template DNA. B. introducing nicks into the DNA double strand in order to prevent the formation of knots. C. hydrolyzing ATP to facilitate DNA unwinding. D. making short strands of RNA at the site of replication initiation. E. forming a replication fork in the DNA double helix. 33. The DNA strand that is replicated smoothly and continuously is called the: A. primary strand. B. first strand. C. leading strand. D. alpha strand. E. lagging strand. 34. In DNA replication, the lagging strand: A. is synthesized as a series of Okazaki fragments. B. is synthesized as a complementary copy of the leading strand. C. pairs with the leading strand by complementary base pairing. D. is made up entirely of RNA primers. E. is not synthesized until the synthesis of the leading strand is completed. 35. Okazaki fragments are joined together by: A. RNA polymerase. B. DNA ligase. C. DNA polymerase. D. RNA ligase. E. primase. 36. How is the chromosome of a bacterial cell replicated? A. The linear DNA molecule is replicated from multiple origins of replication bidirectionally. B. The linear DNA molecule is replicated from one origin of replication bidirectionally. C. The circular DNA molecule is replicated from multiple origins of replication bidirectionally. D. The circular DNA molecule is replicated from one origin of replication bidirectionally. E. The circular DNA molecule is replicated from one origin of replication unidirectionally. 37. How are the chromosomes of a eukaryote cell replicated? A. The linear DNA molecules are replicated from multiple origins of replication bidirectionally. B. The linear DNA molecules are replicated from one origin of replication bidirectionally. C. The circular DNA molecules are replicated from multiple origins of replication bidirectionally. D. The circular DNA molecules are replicated from one origin of replication bidirectionally. E. The linear DNA molecules are replicated from one origin of replication unidirectionally. 12-6 Name Period Mrs. Laux Take Home Test #9 on Chaps. 12 and 16 AP Biology DUE: MONDAY, DECEMBER 20, Use the figure to answer the corresponding questions. 38. The correct designation for the DNA strand labeled C is: A. the leading strand. B. 3. C. Okazaki fragments. D. polymerase. E. None of these. 39. The segments labeled F are responsible for: A. linking short DNA segments. B. synthesizing the leading strand. C. forming the replication fork. D. initiating DNA synthesis. E. unwinding the DNA double helix. 40. The enzyme represented by the letter D is responsible for: A. linking short DNA segments. B. synthesizing the leading strand. C. forming the replication fork. D. forming nucleosomes. E. unwinding the DNA double helix. 41. The structures represented by the letter E are called: A. leading fragments B. Okazaki fragments. C. replication forks. D. nucleosomes. E. DNA polymerases. 42. Which of the following statements concerning nucleotide excision repair is FALSE? A. It is a type of mismatch repair. B. It involves a nuclease. C. It involves a DNA polymerase. D. It involves DNA ligase. E. It is implicated in xeroderma pigmentosum. 43., the ends of eukaryotic chromosomes, shorten with every cell replication event. A. Centromeres B. Telomeres C. Kinetochores D. Primosomes E. Nucleosomes 12-7 44. The ends of eukaryotic chromosomes can be lengthened by: A. apoptosis. B. reverse transcriptase. C. primase. D. telomerase. E. DNA polymerase. 45. Cancer cells differ from noncancerous cells in that: A. they have elevated levels of telomerase. B. they are virtually immortal. C. they have the ability to resist apoptosis. D. they can maintain telomere length as they divide. E. All of these. 46. Karyotyping is useful for determining: A. the number of chromosomes in an individual. B. if recessive traits are present. C. if a specific gene is missing from a chromosome. D. if gene mutations have occurred. E. the presence of sickle cell anemia. 47. The normal human karyotype contains autosomes. A. 22 B. 23 C. 44 D. 46 E The normal human karyotype contains chromosomes. A. 22 B. 23 C. 44 D. 46 E Colchicine is a drug that: A. arrests cells in mitotic telophase. B. stimulates meiosis. C. prevents DNA replication. D. arrests cells in mitotic metaphase. E. prevents cells from entering prophase. 16-8 Use the figure to answer the corresponding questions. 50. Based on the phenotypes of the third generation, the genotype of the father in the second generation must be: A. homozygous for albinism. B. heterozygous for hair color. C. heterozygous for albinism. D. X-linked. E. dominant. 51. The genotypes of the normal females in the second generation in the accompanying figure are: A. homozygous for albinism. B. heterozygous for hair color. C. heterozygous for albinism. D. X-linked. E. unknown. 52. The inheritance pattern demonstrated by the pedigree in the associated figure is: A. autosomal recessive. B. autosomal dominant. C. X-linked recessive. D. X-linked dominant. E. unknown from these data. 53. In genomic imprinting: A. a gene is imprinted with the phenotype of the parent, passing it to a progeny. B. the expression of a gene depends on the parental phenotype. C. a gene is imprinted with the age of the parent, passing it to the progeny. D. the expression of a gene depends on which parent it is inherited from. E. None of these. 54. The Human Genome Project: A. includes sequencing the entire human genome. B. includes identifying all of the genes in the human genome. C. will set the foundation for studying gene function in humans. D. will provide the basis for studying variations in gene sequences related to illness. E. All of these. 16-9 55. An individual who inherited the specific chromosome deletion from the would most likely have Prader-Willi syndrome rather than Angelman syndrome. A. 15; mother B. 15; father C. 21; mother D. 21; father E. None of these. 56. The Human Genome Project was completed in: A B C D E The human genome contains about genes. A. 25,000 B. 45,000 C. 60,000 D. 80,000 E. 120, About what percentage of the human DNA codes for polypeptides? A. 2% B. 5% C. 8% D. 50% E. 86% 59. Comparing the human and mouse genomes has revealed that: A. the mouse and human share few genes. B. the mouse and human genomes are identical. C. the mouse and human genomes have no genes in common. D. the mouse and human share most genes. E. None of these. 60. Polyploidy is: A. the presence of more than two of a certain chromosome. B. more than two of a certain chromosome. C. more than one chromosome. D. the presence of multiple sets of chromosomes. E. the general term for conditions such as Down or Turner syndrome. 61. Aneuploidies describe: A. a phenomenon that only occurs in plants. B. a condition in which an extra chromosome is present or one is absent. C. a defect that is always fatal in humans. D. an uncommon condition in humans. E. mutations that almost always have a beneficial effect on an individual 62. Autosomal aneuploidies arise by: A. chromosome breakage and rejoining. B. nondisjunction. C. errors in crossing-over. D. mistakes in chromosome replication. E. mutations. 63. The sperm in the figure below were most likely produced by: A. nondisjunction. B. mutation. C. X-linkage. D. translocation. E. normal meiosis. 64. Turner syndrome is an example of a condition. A. monosomic B. disomic C. trisomic D. polyploid E. fragile site 65. Autosomal monosomy is not seen in live births because: A. its effects are so small as to be overlooked. B. its effects do not set in until adulthood. C. its effects are so lethal as to cause spontaneous abortion early in pregnancy. D. it only occurs in sex chromosomes and therefore does affect nonreproductive function. E. None of these. 66. Nearly half of the pregnancies that end in miscarriage have: A. trisomy of chromosome 21. B. homozygosity for a recessive lethal allele. C. major chromosomal abnormalities. D. Klinefelter syndrome. E. Down syndrome. 67. Persons having an XO karyotype are sterile females. They have syndrome. A. Turner B. Klinefelter C. Down (trisomy form) D. Phenylketonuria E. Down (translocation form) 68. Down syndrome is an example of a condition. A. monosomic B. disomic C. trisomic D. polyploid E. transgenic 16-11 69. Which of the following statements concerning Down syndrome is FALSE? A. Down syndrome involves nondisjunction. B. Down syndrome individuals have 47 chromosomes per somatic cell. C. Down syndrome incidence increases with increasing material age. D. Down syndrome individuals have an extra chromosome 22. E. Down syndrome is variable in expression. 70. Aneuploidies for the sex chromosomes are more viable because: A. they carry less information that most chromosomes. B. of a lack crossing-over between the X and the Y. C. they are smaller than most chromosomes. D. their information is carried on other chromosomes. E. of dosage compensation. 71. Persons having an XXY karyotype are nearly normal males but produce few or no sperm. They have syndrome. A. Turner B. Klinefelter C. Down (trisomy form) D. Phenylketonuria E. Down (translocation form) 72. A karyotype reveals that an individual is XYY. Based on your knowledge of human genetics you correctly conclude that this individual is phenotypically and. A. female; with criminal tendencies B. female; sterile C. female; fertile D. male; sterile E. male; fertile 73. Translocation occurs when: A. part of a chromosome breaks off and attaches to a nonhomologous chromosome. B. part of a chromosome breaks off and attaches to a homologous chromosome. C. crossing-over events occur. D. genes move from one area on a chromosome to another area on the same chromosome. E. a Y chromosome replaces an X chromosome in a female cell. 74. In a chromosomal inversion a segment of a chromosome is: A. reversed. B. duplicated. C. lost. D. attached to a nonhomologous chromosome. E. None of these. 75. Cri du chat syndrome arises from: A. trisomy for chromosome 5. B. deletion of part of chromosome 5. C. a 14/21reciprocal translocation. D. nondisjunction. E. a duplication of part of chromosome 76. Which of the following statements about fragile X syndrome is not correct? A. It results in learning and attention disabilities in males. B. It results in hyperactivity in males. C. The location of the fragile X site is exactly the same in all of an individual s cells. D. It results in the formation of shorter-than-normal sequences of CGG. E. Females are usually heterozygous and are therefore more likely to have a normal phenotype. 77. Which of the following are inborn errors of metabolism? A. PKU and alkaptonuria B. sickle cell anemia and hemophilia C. cystic fibrosis and Tay-Sachs disease D. Huntington s disease and fragile X syndrome E. cri-du-chat syndrome and Down syndrome 78. Why are most infants with phenylketonuria usually healthy at birth? A. The defect causing the condition has usually been repaired by that time. B. The mother breaks down phenylalanine for both herself and her fetus. C. The infant can produce sufficient quantities of the enzymes needed by birth. D. Phenylalanine can be completely excreted in infants. E. Phenylalanine is completely metabolized in infants. 79. Maternal PKU can result in serious effects in fetuses. This problem occurs because: A. the fetus is homozygous for PKU and therefore cannot metabolize phenylalanine. B. the fetus is usually homozygous dominant and therefore has PKU. C. the mother
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