* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Biology II Final Exam Practice
Epitranscriptome wikipedia , lookup
Site-specific recombinase technology wikipedia , lookup
X-inactivation wikipedia , lookup
Genetic code wikipedia , lookup
DNA damage theory of aging wikipedia , lookup
Genealogical DNA test wikipedia , lookup
Molecular cloning wikipedia , lookup
Therapeutic gene modulation wikipedia , lookup
Epigenomics wikipedia , lookup
Nucleic acid double helix wikipedia , lookup
Cell-free fetal DNA wikipedia , lookup
Non-coding DNA wikipedia , lookup
Cre-Lox recombination wikipedia , lookup
DNA supercoil wikipedia , lookup
Point mutation wikipedia , lookup
Designer baby wikipedia , lookup
Extrachromosomal DNA wikipedia , lookup
Vectors in gene therapy wikipedia , lookup
Artificial gene synthesis wikipedia , lookup
Nucleic acid analogue wikipedia , lookup
Primary transcript wikipedia , lookup
History of genetic engineering wikipedia , lookup
Biology II Final Exam Practice True/False Indicate whether the statement is true or false. ____ 1. A gamete has one-half the number of chromosomes of a regular body cell. ____ 2. Homologous chromosomes are two chromosomes with identical DNA sequences. ____ 3. Mendel’s work on garden pea plants resulted in the discovery that genetic traits of parents always blend together in subsequent generations. ____ 4. In humans, the ability to roll one’s tongue is a dominant trait. Therefore, a tongue roller can only have children who are also tongue rollers. ____ 5. The separation of genes during crossing over occurs more frequently between genes that are far apart on a chromosome than for genes that are close together. ____ 6. A woman with an X-linked dominant genetic disorder will have children who have a 50% chance to be affected by the trait also, regardless of their gender. ____ 7. The staff in the neonatal unit at a hospital are not sure which of two babies belong to which set of parents. The blood types of the babies are AB and O. One set of parents has blood types A and B. With this knowledge it would still be necessary to test the second couple. ____ 8. RNA polymerase has to bind to DNA for an enzyme to be synthesized. ____ 9. One strand of a double-stranded DNA helix is oriented in the 5' (carbon of the deoxyribose molecule) to 3' direction, while the complementary strand it is bonded to is oriented in the 3' to 5' direction. ____ 10. DNA nucleotides are always added to an existing strand at the 3-prime end. This means that during replication a leading and lagging strand are created. ____ 11. A vestigial structure in one organism can be defined as a reduced form of a functional structure in another organism. ____ 12. Natural selection is based on the concepts of excess reproduction, variation, inheritance, and the advantages of certain traits. ____ 13. Darwin developed his theory of evolution exclusively from his work on the Galapagos Islands. ____ 14. Fossils, although interesting, do not actually provide evidence of evolution. ____ 15. Homologous structures indicate a shared ancestry, while vestigial structures do not. ____ 16. Biochemical traits helped Darwin unravel his theory of evolution. ____ 17. In an incomplete dominance trait the traits are both dominant and both are shown phenotypically. ____ 18. Recombinant DNA is an example of genetic engineering. ____ 19. Blood type in humans is a multiple allele trait. ____ 20. In a polygenic trait the traits are blended (like white and black making gray). ____ 21. Females choosing males competing for territory is an example of artificial selection. ____ 22. In the sequence of hypothesized events leading to the evolution of eukaryotes, prokaryotes appeared first. ____ 23. The cell cycle is divided into interphase and mitosis. ____ 24. A pictures of an individual’s entire genome is called a karyotype. ____ 25. When two chromosomes fail to separate during meiosis the chromosomal mutation is called a frame-shift. Modified True/False Indicate whether the statement is true or false. If false, change the identified word or phrase to make the statement true. ____ 26. The following sequence could be a tRNA anticodon: GTG. _________________________ ____ 27. A mutation is a random change in genetic material. _________________________ ____ 28. Humans born with either below-normal or above-normal birth weights have a lower chance of survival than those born with average birth weights. Consequently, birth weights vary little in human populations. This form of natural selection is called directional selection. _________________________ ____ 29. Punctuated equilibrium and gradualism are two models that describe the cause of speciation. _________________________ ____ 30. Embryos of different organisms exhibit homologous structures during certain phases of development that become totally different structures in adult forms. This indicates that organisms evolved from different ancestors. _________________________ ____ 31. The structure on a chromosome where sister chromatids attach is the centromere. _________________________ ____ 32. After telophase, the process of cell division is complete. _________________________ ____ 33. A person’s phenotype is the observable characteristics that can be observed and measured. _________________________ ____ 34. Genetic variation is measured in terms of the frequency of alleles in the gene pool of a population. _________________________ ____ 35. A species is generally defined as a group of natural populations that can interbreed. _________________________ ____ 36. When a group of individuals of the same species has become reproductively isolated from its parent population and has accumulated sufficient genetic differences to prevent interbreeding, convergence has occurred. _________________________ ____ 37. According to Darwin, individuals with traits better suited to their environment are more likely to survive. _________________________ ____ 38. Individuals can evolve. _________________________ ____ 39. The many different species of finches on the Galápagos Islands are an example of adaptive radiation. _________________________ Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 40. A white mouse whose parents are both white produces only brown offspring when mated with a brown mouse. The white mouse is most probably ____. a. homozygous recessive c. homozygous dominant b. heterozygous d. haploid ____ 41. In chickens, rose comb (R) is dominant to single comb (r). A homozygous rose-combed rooster is mated with a single-combed hen. All of the chicks in the F1 generation were kept together as a group for several years. They were allowed to mate only within their own group. What is the expected phenotype of the F2 chicks? a. 100% rose comb b. 75% rose comb and 25% single comb c. 100% single comb d. 50% rose comb and 50% single comb ____ 42. In mink, brown fur color is dominant to silver-blue fur color. If a homozygous brown mink is mated with a silver-blue mink and 8 offspring are produced, how many would be expected to be silver-blue? a. 0 c. 6 b. 3 d. 8 ____ 43. Using Figure 10-3, which process would result in the formation of chromosome C from chromosomes A and B? Figure 10-3 a. asexual reproduction b. independent assortment c. crossing over d. segregation Figure 10-5 ____ 44. What is the genotype of generation 1 in Figure 10-5? a. II c. ii b. Ii d. I ____ 45. Suppose an animal is heterozygous AaBb, and the traits are not linked. When meiosis occurs, what is the total number of possible combinations of gametes that can be made for these traits? a. 2 c. 6 b. 4 d. 8 ____ 46. A true-breeding tall pea plant is crossed with a true-breeding short pea plant, and all the offspring are tall. What is the most likely genotype of the offspring assuming a single-gene trait? a. tt c. TT b. Tt d. TT or tt ____ 47. In mice, black is dominant to white color and color is determined by a single gene. Two black mice are crossed. They produce 2 black offspring and one white offspring. If the white offspring is crossed with one of its parents, what percent of the offspring are expected to be white? a. 0 c. 50 b. 25 d. 75 ____ 48. Mendel crossed a true-breeding plant that produced green seeds with a true-breeding plant that produced yellow seeds to produce an F1 generation. The entire F1 generation produced yellow seeds. Then he crossed the F1 offspring with each other to produce the F2 generation. From the F2 generation, he counted 6022 yellow seeds.Which of these is the most likely estimate of the number of green seeds he collected from the F2 generation? a. 0 c. 6000 b. 2000 d. 18000 ____ 49. A heterozygous organism is best described as which of these? a. dominant c. hybrid b. genotype d. true-breeding Figure 11-1 ____ 50. Refer to Figure 11-1. If individual III-2 marries a person with the same genotype as individual I-1, what is the chance that one of their children will be afflicted with hemophilia? a. 0% c. 50% b. 25% d. 75% ____ 51. What type of inheritance pattern does the trait represented by the shaded symbols in Figure 11-1 illustrate? a. incomplete dominance c. codominance b. multiple alleles d. sex-linked ____ 52. For the trait being followed in the pedigree, individuals II-1 and II-4 in Figure 11-1 can be classified as ____. a. homozygous dominant c. homozygous recessive b. mutants d. carriers ____ 53. If a female fruit fly heterozygous for red eyes (XRXr) crossed with a white-eyed male (XrY), what percent of their offspring would have white eyes? a. 0% c. 50% b. 25% d. 75% ____ 54. A cross between a white rooster and a black hen results in 100% blue Andalusian offspring. When two of these blue offspring are mated, the probable phenotypic ratio seen in their offspring would be ____. a. 100% blue c. 75% blue, 25% white b. 75% black, 25% white d. 25% black, 50% blue, 25% white Figure 11-6 This pedigree shows the transmission of a rare disease that is dehabilitating but not lethal. Carriers are not shown. ____ 55. Which type of heredity does the pedigree in Figure 11-6 demonstrate? a. autosomal recessive c. X-linked recessive b. autosomal dominant d. X-linked dominant ____ 56. Which series is arranged in order from largest to smallest in size? a. chromosome, nucleus, cell, DNA, nucleotide b. cell, nucleus, chromosome, DNA, nucleotide c. nucleotide, chromosome, cell, DNA, nucleus d. cell, nucleotide, nucleus, DNA, chromosome Figure 12-2 ____ 57. In which part of the cell does this process shown in Figure 12-2 take place? a. in the nucleus c. at the ribosomes b. in food vacuoles d. on the chromosome ____ 58. Structure III in Figure 12-2 represents a(n) ____. a. gene c. codon b. amino acid d. DNA molecule ____ 59. The process illustrated in Figure 12-2 is called ____. a. translation c. monoploidy b. replication d. transcription ____ 60. Which of the structures in Figure 12-2 are composed of RNA? a. II and IV c. I and V b. III and IV d. III and V Help Wanted Positions Available in the genetics industry. Hundreds of entry-level openings for tireless workers. No previous experience necessary. Must be able to transcribe code in a nuclear environment. Accuracy and Speed vital for this job in the field of translation. Applicants must demonstrate skills in transporting and positioning amino acids. Salary commensurate with experience. Executive Position available. Must be able to maintain genetic continuity through replication and control cellular activity by regulation of enzyme production. Limited number of openings. All benefits. Supervisor of production of proteins—all shifts. Must be able to follow exact directions from double-stranded template. Travel from nucleus to the cytoplasm is additional job benefit. Table 12-1 ____ 61. Applicants for the fourth job of the Help Wanted ad in Table 12-1, "Supervisor," could qualify if they were ____. a. DNA c. tRNA b. mRNA d. rRNA ____ 62. Applicants for the third job of the Help Wanted ad in Table 12-1, "Executive Position," could qualify if they were ____. a. DNA c. tRNA b. mRNA d. rRNA ____ 63. Which term best describes the structures shown in Figure 15-3? Figure 15-3 a. homologous c. analogous b. heterologous d. vestigial ____ 64. Why is the synthesis stage called this? a. because protein synthesis is taking place b. because DNA synthesis is taking place c. because it combines several smaller stages into one d. because the chromosomes come together ____ 65. Which of these has occurred by the end of prophase? a. Sister chromatids are separated. b. The spindle is beginning to form. c. The cell membrane has begun to pinch inward. d. The nuclear membrane has disappeared. Figure 9-4 ____ 66. Figure 9-4 illustrates which stage of mitosis? a. anaphase b. metaphase ____ 67. Which of the following occurs in telophase? a. chromosomes condense c. prophase d. telophase ____ 68. ____ 69. ____ 70. ____ 71. ____ 72. ____ 73. ____ 74. b. chromosomes line up c. chromosomes move to opposite poles d. chromosomes relax A cell has 12 chromosomes. How many chromosomes will each daughter cell have? a. 4 c. 12 b. 6 d. 24 What is cancer caused by? a. cell-membrane damage c. mutation b. metabolic poisoning d. immune-system damage Natural selection could not occur without a. artificial selection. c. competition for unlimited resources. b. gradual warming of Earth. d. genetic variation in populations. Darwin’s theory lacked an explanation for a. how evolution occurs. c. adaptations. b. inheritance of traits. d. competition. From the smallest functional units to the largest, the body is organized as follows: a. cells, systems, organs, tissues. c. cells, tissues, organs, systems. b. organs, cells, tissues, systems. d. systems, organs, tissues, cells. The heart and the blood vessels are separate organs that form the a. skeletal system. c. reproductive system. b. circulatory system. d. digestive system. The skin performs all of the following except a. protection. c. control of body temperature. b. sensation of heat. d. production of gametes. Completion Complete each statement. 75. An individual with the genotype Aa is ____________________ for the trait. 76. A geneticist refers to the appearance of an individual as his or her ____________________, while the genetic makeup is called the genotype. 77. Genes on separate chromosomes follow Mendel’s law of ____________________. 78. According to Mendel’s law of ____________________, two alleles for each trait separate during meiosis, and during fertilization, the two alleles unite. 79. A normal human karyotype displays ____________________ pairs of autosomes. 80. What are the 4 bases found in DNA? 81. What are the four bases found in RNA? 82. Describe the base pairing rule. 83. ______________________ are made of amino acids linked together by peptide bonds. 84. A codon is a sequence of _______ (number) bases. 85. Mutations are changes to an organism’s _________. 86. A mutation that involves a new bases inserted where another bases should have been is called a _________________ 87. A ________________ mutation causes all bases after the mutation to be out of order. 88. Sex-linked traits are all found on the ______________ chromosome. 89. The process by which an individual is exactly copied is called _____________ 90. The snake pelvis is an example of a ____________________ structure. 91. Mitosis is the division of ____________________. 92. Interphase is divided into three stages: G1, ____________________, and G2. 93. Cytokinesis is division of the _____________________. Short Answer 94. Analyze the differences between Mendel's law of dominance and law of segregation. 95. Two parents with brown eyes have two children with blue eyes. Assume that only one gene for eye type is involved. How is this possible? Defend your answer by presenting possible genotypes for the individuals in the scenario. Table 11-1 96. Two couples, the Pages and the Bakers, had baby boys in the same hospital at the same time. There was a mixup in the hospital nursery. Use the information given in Table 11-1. Which baby belongs to which family? 97. Based on the Bakers' blood types, could Baby #1 be their child? Could Baby #2 be their child? Use Table 111 to explain your answer. 98. Describe the process of replication. 99. What is the difference between a codon and an anticodon? 100. Why is tRNA important in translation? Figure 12-4 101. Use the genetic code (Figure 12-4) to find the start codon in this mRNA sequence and write the sequence of amino acids this mRNA would translate into. CUGACGAUGCUCAACGGCAUAUAACGCGAG 102. Compare and contrast DNA and RNA 103. Put the following into correct sequence: translation transcription protein synthesis 104. List and describe the indirect evidence available to support the theory of evolution. 105. Summarize Darwin's theory of evolution by natural selection. 106. The idea of punctuated equilibrium in evolutionary history was developed by two paleontologists as a way to explain their observations of certain fossil records. Explain what pattern in the fossil record the punctuated equilibrium model fits. 107. Describe what Darwin found about the finches on the Galapagos Islands and why it provided evidence for evolution. 108. Why was artificial selection such an important form of evidence for the theory of evolution by natural selection proposed by Darwin? 109. How do scientists learn about organisms of the past when studying fossils? Figure 9-6 110. Cells A and F of Figure 9-6 show an early and a late stage of the same phase of mitosis. What phase is it? 111. Describe what happens in each stage of interphase. 112. Why is embryology useful in determining the evolutionary history of a species? Problem 113. In fruit flies, the allele for normal body (H) is dominant to the allele for hairy body (h), and the allele for red eye color (B) is dominant to the allele for brown (b). Use a Punnett square to determine the possible phenotypes and frequencies of the offspring from the cross Hhbb hhBb. 114. In Figure 12-5, use the letter P to label all of the phosphate groups. Use an S to label all the sugar molecules. For labeling the nitrogen bases, use a T for thymine and a C for cytosine. Guanine and adenine have been filled in for you. Circle and label a codon. Circle and label a nucleotide. Figure 12-5 Biology II Final Exam Practice Answer Section TRUE/FALSE 1. ANS: T During meiosis, the number of chromosomes is reduced to one-half the original number. PTS: 1 DIF: Bloom's Level B REF: 271 NAT: LS_2b STA: III.3.2 TOP: 10-1 2. ANS: F Homologous chromosomes contain the same genes as one another, but the DNA sequences vary. One homolog comes from an individual’s father, while the other comes from the mother. PTS: 1 DIF: Bloom's Level B REF: 270 NAT: LS_2a STA: III.3.1 TOP: 10-3 3. ANS: F He discovered that traits were discrete and could be “hidden” by other traits, but could reappear in another generation in their original form. PTS: 1 DIF: Bloom's Level B REF: 277–279 NAT: LS_2b STA: III.3.1 TOP: 10-4 4. ANS: F This statement would only be true if the tongue roller was homozygous (TT). If the tongue roller is heterozygous (Tt) and has children with someone else that carries the t allele, then it is possible for them to have a child who cannot roll his or her tongue. PTS: 1 DIF: Bloom's Level D REF: 279–281 NAT: LS_2b STA: III.3.1 TOP: 10-6 5. ANS: T The farther apart two loci are, the more likely it is that they will be separated by a crossover. PTS: 1 DIF: Bloom's Level B REF: 284 NAT: LS_2b STA: III.3.1 TOP: 10-8 6. ANS: T Only a dominant X-linked trait would appear equally in male and female offspring. PTS: 1 DIF: Bloom's Level D REF: 299–300 | 307–308 NAT: LS_2b STA: III.3.1 TOP: 11-5 7. ANS: T The parents with blood types A and B could have genotypes IaIi, and IbIi and so could have a child with either one of the blood types. If the second set of parents had the genotypes IiIi and IiIa, the answer would then be clear. PTS: 1 DIF: Bloom's Level D REF: 304 NAT: LS_2b STA: III.3.1 TOP: 11-4 8. ANS: T RNA polymerase controls the transcription of DNA to mRNA, which has to happen before an enzyme can be synthesized according to the “template” carried in the mRNA. PTS: 1 DIF: Bloom's Level E REF: 337 NAT: LS_1c TOP: 12-7 9. ANS: T The two strands in a DNA molecule are oriented opposite to each other. PTS: 1 DIF: Bloom's Level C NAT: LS_2a TOP: 12-2 10. ANS: T The two strands of a DNA molecule run in opposite directions. REF: 331 PTS: 1 DIF: Bloom's Level C REF: 334 NAT: LS_2a TOP: 12-5 11. ANS: T PTS: 1 DIF: Bloom's Taxonomy B REF: 425 NAT: LS_3c TOP: 15-5 12. ANS: T PTS: 1 DIF: Bloom's Taxonomy C REF: 420 NAT: LS_3a STA: III.4.2 TOP: 15-2 13. ANS: F To develop his understanding of evolution, Darwin used his studies on H.M.S. Beagle and the work that he did later after his return to England. PTS: 1 DIF: Bloom's Taxonomy B STA: II.1.4 TOP: 15-1 14. ANS: F Fossils provide strong evidence of evolution. REF: 420–422 PTS: 1 DIF: Bloom's Taxonomy B REF: 423 NAT: LS_3c STA: III.4.1 TOP: 15-4 15. ANS: F Vestigial structures are homologous structures and both indicate a shared ancestry. PTS: 1 DIF: Bloom's Taxonomy C REF: 424–425 NAT: LS_3c STA: III.4.1 TOP: 15-5 16. ANS: F Darwin did not have any biochemical information, but later scientists used biochemical data to understand evolution. PTS: 1 DIF: Bloom's Taxonomy B NAT: LS_3c STA: II.1.4 TOP: 15-6 17. ANS: F False - this describes codominance PTS: 1 18. ANS: T PTS: 1 19. ANS: T There are three alleles - A, B and O PTS: 1 20. ANS: F This describes an incomplete dominance trait. REF: 427 PTS: 1 21. ANS: F This is sexual selection PTS: 1 22. ANS: T Review the endosymbiotic theory to see how eukaryotes are thought to have developed from prokaryotes. PTS: 1 DIF: Bloom's Level D REF: 405 STA: III.4.2 TOP: 14-5 23. ANS: F The cell cycle has three stages: interphase, mitosis, and cytokinesis. PTS: 1 DIF: Bloom's Level A NAT: LS_1d STA: III.1.1 TOP: 9-2 24. ANS: T PTS: 1 25. ANS: F This is called non-disjunction REF: 246 PTS: 1 MODIFIED TRUE/FALSE 26. ANS: F DNA codon The base thymine occurs in DNA, but not RNA. PTS: NAT: 27. ANS: REF: 28. ANS: 1 DIF: Bloom's Level E LS_1c TOP: 12-8 T PTS: 1 434 NAT: LS_2c STA: III.3.3 F, stabilizing selection REF: 336 | 338 DIF: Bloom's Taxonomy B TOP: 15-7 PTS: 1 NAT: LS_3a 29. ANS: F, rate DIF: Bloom's Taxonomy E STA: III.4.2 TOP: 15-8 REF: 434 PTS: 1 STA: III.4.2 30. ANS: F, common DIF: Bloom's Taxonomy C TOP: 15-9 REF: 440–441 PTS: 1 DIF: Bloom's Taxonomy C NAT: LS_3d STA: III.1.1 TOP: 15-5 31. ANS: T Sister chromatids are held together by the centromere. PTS: 1 NAT: LS_1d DIF: Bloom's Level A STA: III.3.1 TOP: 9-4 REF: 426 REF: 248 32. ANS: F cytokinesis Telophase is the end of mitosis, cytokinesis involves the division into two cells. 33. 34. 35. 36. PTS: NAT: ANS: OBJ: ANS: OBJ: ANS: OBJ: ANS: PTS: 37. ANS: OBJ: 38. ANS: 1 DIF: Bloom's Level B LS_1d STA: III.1.1 TOP: T PTS: 17.1.2 T PTS: 17.1.3 T PTS: 17.3.1 F, speciation 1 DIF: II T 16.2.1 F, Populations PTS: 1 39. ANS: T OBJ: 16.3.2 DIF: II REF: 251 9-4 1 DIF: I 1 DIF: I 1 DIF: I OBJ: 17.3.2 PTS: 1 DIF: II OBJ: 16.3.1 PTS: 1 DIF: III MULTIPLE CHOICE 40. ANS: A The most likely scenario is that the white mouse displays the recessive trait. If this is the case, then the white mouse must be homozygous. If the white mouse were either homozygous dominant or heterozygous, then it would likely produce white offspring when mated with a brown mouse. Feedback A B C D Correct. If the white mouse is heterozygous, then this would mean white is dominant and its brown mate is homozygous recessive. Such a cross would yield about half white and brown offspring. If the white mouse were homozygous dominant, then its offspring would have to be white. Sperm and eggs are haploid. Individual mice are not. PTS: 1 DIF: Bloom's Level D REF: 277–281 NAT: LS_2b STA: III.3.1 TOP: 10-6 41. ANS: B This is the classic situation in which the F1 generation (all heterozygous) are crossed to produce offspring in a 3 to 1 ratio of dominant to recessive. Feedback A B C Remember that the F1 individuals are heterogygous. Well done. Remember that the F1 individuals are heterogygous. D About 3/4 of the offspring will have rose comb. PTS: 1 DIF: Bloom's Level D REF: 277–281 NAT: LS_2b STA: III.3.1 TOP: 10-6 42. ANS: A The brown mink will always donate a dominant allele to its offspring, which means all the offspring will have the dominant phenotype. None of the offspring would be silver-blue, which is the recessive color. Feedback A B C D Well done. Remember, the brown mink is homozygous. Remember, the brown mink is homozygous. Brown is dominant to silver-blue. PTS: 1 DIF: Bloom's Level D REF: 277–281 NAT: LS_2b STA: III.3.1 TOP: 10-6 43. ANS: C Crossing over leads to new combinations of alleles on a given chromomosome. Feedback A B C D Asexual reproduction does not lead to new combinations of alleles. Review independent assortment. Correct. Check page 272. PTS: 1 DIF: Bloom's Level E NAT: LS_2b STA: III.3.2 TOP: 10-7 44. ANS: B The first generation are all heterozygotes, or Ii. REF: 272 Feedback A B C D The individuals in P1 are homozygous. Well done. Review pages 277–281. The individuals have two alleles, not one. PTS: 1 DIF: Bloom's Level D NAT: LS_2b STA: III.3.1 TOP: 10-6 45. ANS: B There are four combinations: AB, Ab, aB, and ab. REF: 277–281 Feedback A B C D Remember the traits are not linked. Correct. Check page 275. Find all possible combinations of A with B and b and a with B and b. PTS: 1 NAT: LS_2b DIF: Bloom's Level D STA: III.3.1 TOP: 10-5 REF: 275 46. ANS: B Since the parents are true breeding, they are most likely homozygous (TT and tt). This means the offspring are most likely heterozygous, Tt. Feedback A B C D See page 280. Correct. But one parent was short. If all offspring are tall, then they cannot be TT and tt. PTS: 1 DIF: Bloom's Level D REF: 280 NAT: LS_2b STA: III.3.1 TOP: 10-6 47. ANS: C The original parents must be heterozygous (Bb) since they produced both black and white offspring. The white offspring must be homozygous recessive (bb). If the white offspring is crossed with its parent then half the offspring will be white. Feedback A B C D The black parent is heterozygous. The parents of this cross are Bb and bb. Correct. Review pages 277–281. PTS: 1 DIF: Bloom's Level D REF: 277–281 NAT: LS_2b STA: III.3.1 TOP: 10-6 48. ANS: B In the F2 generation, we expect a 3 to 1 ratio of yellow to green seeds. Thus, if there are approximately 6000 yellow seeds, then we expect about 2000 green seeds. Feedback A B C D Check page 278. Correct. Try again. There should be 3 times more yellow seeds than green seeds. PTS: 1 DIF: Bloom's Level D REF: 278 NAT: LS_2b STA: III.3.1 TOP: 10-4 49. ANS: C A heterozygous (Aa) individual is produced by a cross of two different true-breeding (AA and aa) individuals.Thus the heterozgous individual is a hybrid. Feedback A B C D This description refers to a trait or allele. This description refers to the genetic makeup of an individual. Correct. This refers to a homozygous individual. PTS: 1 NAT: LS_2b DIF: Bloom's Level B STA: III.3.1 TOP: 10-5 REF: 279 50. ANS: A Feedback A B C D You are right. You are on the right track. Please consider the pattern in the pedigree again. Please refer to page 299. PTS: 1 NAT: LS_2b 51. ANS: D DIF: Bloom's Level E STA: III.3.1 TOP: 11-1 REF: 299 Feedback A B C D You are on the right track. Please consider again all the factors. Please refer to pages 307-308. You are correct. PTS: 1 NAT: LS_2b 52. ANS: D DIF: Bloom's Level D STA: III.3.1 TOP: 11-5 REF: 299–300 | 307–308 Feedback A B C D Please consider again the reference to the individual, not the allele. Please consider the factors again. Please refer to pages 299-300. You are correct. PTS: 1 NAT: LS_2b 53. ANS: C DIF: Bloom's Level B STA: III.3.1 TOP: 11-3 REF: 299–300 Feedback A B C D Please consider all the factors. You are on the right track. You are correct. Please refer to pages 296-300. PTS: 1 NAT: LS_2b 54. ANS: D DIF: Bloom's Level C STA: III.3.1 TOP: 11-1 REF: 296–300 Feedback A B C D Please consider again the results of mating two from the second generation. You are on the right track. Please refer to pages 302-303. You are correct. PTS: 1 DIF: Bloom's Level D REF: 302–303 NAT: LS_2b STA: III.3.1 TOP: 11-4 55. ANS: C Because only the males demonstrate this disease, it shows that it is a recessive disease associated with the X chromosome. Feedback A B C D What is the pattern of those who exhibit the trait? Please refer to pages 307-308. You are correct. You are on the right track. PTS: 1 DIF: Bloom's Level E REF: 299–300 | 307–308 NAT: LS_2b STA: III.3.1 TOP: 11-1 56. ANS: B Nucleotides are the subunits of nucleic acids like DNA. DNA coils around histone proteins to form chromosomes, which are contained in the nucleus of a cell. Feedback A B C D Chromosomes are in the nucleus. That's correct! Check page 329. What is a chromosome made of? PTS: 1 DIF: Bloom's Level B NAT: LS_1c TOP: 12-3 57. ANS: C The process of translation takes place at the ribosomes. REF: 329 | 332 Feedback A B C D This process happens after messenger RNA leaves the nucleus. Where does messenger RNA go when it leaves the nucleus? That's correct! Chromosomes are made up of DNA and proteins. PTS: 1 DIF: Bloom's Level C REF: 338 NAT: LS_1c TOP: 12-6 58. ANS: B A tRNA molecule carries a specific amino acid corresponding to the anticodon of that tRNA molecule. Feedback A B C D What is the definition of a gene? That's correct! You're on the right track, but the anticodon is located somewhere else on this molecule. DNA does not have uracil as one of its bases. PTS: 1 DIF: Bloom's Level C REF: 338 NAT: LS_1c TOP: 12-6 59. ANS: A Translation is the process by which the mRNA “template” is used to form polypeptides. Feedback A B C D That's correct! Page 334 shows replication. Check the definition of monoploidy. Is any DNA involved in the pictured process? PTS: 1 DIF: Bloom's Level B REF: 338 NAT: LS_1c TOP: 12-6 60. ANS: A The mRNA “template” and the tRNA molecule are the only RNA structures. I is an anticodon composed of nucleotide bases, III is a polypeptide, and V is an amino acid. Feedback A B C D That's correct! Only one of these is made of RNA. What is an anticodon? These are related to each other, but they are not made of RNA. PTS: 1 DIF: Bloom's Level C REF: 336 | 338 NAT: LS_1c TOP: 12-6 61. ANS: B mRNA is formed from the template strand of DNA, and it carries the “code” from the nucleus to the ribosomes. Feedback A B C D Does DNA leave the nucleus? That's correct! tRNA carries amino acids. Look at page 336. PTS: 1 DIF: Bloom's Level B REF: 336 NAT: LS_1c TOP: 12-6 62. ANS: A DNA is the genetic material that replicates and is passed along when a cell divides. DNA controls the production of enzymes and other proteins. Feedback A B C D That's correct! Does mRNA replicate? tRNA carries amino acids. Check page 333 for clues. PTS: 1 DIF: Bloom's Level D REF: 326–333 NAT: LS_1c STA: III.3.2 TOP: 12-1 63. ANS: C Not all anatomically similar structures are evidence of evolution. Analogous structures are not inherited from a common ancestor. Feedback A B C D Good try! Read page 426 of your text again. Try again! Read page 426 of your text again. Correct! You are on the right track. Read page 426 of your text again. PTS: 1 DIF: Bloom's Level B NAT: LS_3d STA: III.5.1 TOP: 15-5 64. ANS: B “Synthesis” refers to the synthesis of DNA during this phase. REF: 426 Feedback A B C D This is not what is being made. That's correct! It refers to something being made. Review the material on page 247. PTS: 1 DIF: Bloom's Level B NAT: LS_1d STA: III.1.1 TOP: 9-3 65. ANS: D By the end of prophase the nuclear membrane has disintegrated. REF: 247 Feedback A B C D This occurs during anaphase. This occurs at the beginning of prophase. This occurs in telophase. That's right! PTS: 1 DIF: Bloom's Level B REF: 249–250 NAT: LS_1d STA: III.1.1 TOP: 9-4 66. ANS: B During metaphase, the spindle apparatus aligns the sister chromatids in the center, or equator, of the cell. Feedback A B C D Try again. That's correct! See page 250 for help. Look at what the chromosomes are doing. PTS: 1 DIF: Bloom's Level B REF: 250 NAT: LS_1d STA: III.1.1 TOP: 9-4 67. ANS: D In telophase, the chromosomes arrive at the poles and begin to relax, or decondense. Feedback A B C D That's prophase. That's metaphase. That's anaphase. That's correct! PTS: 1 DIF: Bloom's Level A REF: 251 NAT: LS_1d STA: III.1.1 TOP: 9-4 68. ANS: C During mitosis, the cell’s replicated genetic material separates and the cell prepares to divide into two cells. Feedback A B C D Did you consider all the factors? Too low. This would occur in meiosis. That's correct! That's too many. PTS: 1 DIF: Bloom's Level C REF: 248 NAT: LS_1d STA: III.1.1 TOP: 9-4 69. ANS: C Cancer can have diverse causes, all of which result in mutation in a cell’s DNA. Feedback A B C D 70. 71. 72. 73. 74. This will just kill a cell. This doesn't cause cancer. That's right! Have a look at page 254. PTS: NAT: ANS: STA: ANS: STA: ANS: STA: ANS: STA: ANS: STA: 1 LS_1d D III.4.H.2 B III.4.H.2 C III.1.H.1 B III.2.H.4 D III.2.H.4 DIF: Bloom's Level C TOP: 9-7 PTS: 1 DIF: II REF: 254 PTS: 1 DIF: II OBJ: 16.2.3 PTS: 1 DIF: III OBJ: 34.1.3 PTS: 1 DIF: I OBJ: 34.1.3 PTS: 1 DIF: III OBJ: 34.4.2 OBJ: 16.2.1 COMPLETION 75. ANS: heterozygous PTS: 1 NAT: LS_2b 76. ANS: phenotype DIF: Bloom's Level A STA: III.3.1 TOP: 10-5 REF: 279 PTS: 1 DIF: Bloom's Level A NAT: LS_2a STA: III.3.1 TOP: 10-5 77. ANS: independent assortment REF: 279 PTS: 1 DIF: Bloom's Level B REF: 280 NAT: LS_2b 78. ANS: segregation STA: III.3.1 TOP: 10-5 PTS: 1 NAT: LS_2b 79. ANS: 22 DIF: Bloom's Level B STA: III.3.1 TOP: 10-5 PTS: 1 DIF: Bloom's Level B NAT: LS_2b TOP: 11-7 80. ANS: adenine, thymine, guanine, cytosine REF: 279 REF: 311 PTS: 1 81. ANS: adenine, uracil, guanine, cytosine PTS: 1 82. ANS: A pairs with T and C pairs with G PTS: 1 83. ANS: proteins PTS: 1 84. ANS: 3 PTS: 1 85. ANS: DNA PTS: 1 86. ANS: substitution PTS: 1 87. ANS: frame shift. PTS: 1 88. ANS: x PTS: 1 89. ANS: cloning. PTS: 1 90. ANS: vestigial PTS: 1 TOP: 15-5 91. ANS: a nucleus DIF: Bloom's Taxonomy D REF: 425 PTS: 1 NAT: LS_1d 92. ANS: synthesis DIF: Bloom's Level B STA: III.1.1 TOP: 9-2 REF: 246 PTS: 1 NAT: LS_1d 93. ANS: cytoplasm PTS: 1 NAT: LS_1d DIF: Bloom's Level A STA: III.1.1 TOP: 9-3 REF: 247 DIF: Bloom's Level A STA: III.1.1 TOP: 9-5 REF: 252 SHORT ANSWER 94. ANS: Answers may include: The law of dominance deals with individual genes and their influence. The law of segregation explains how these genes are separated during meiosis. PTS: 1 DIF: Bloom's Level E REF: 279 NAT: LS_2b STA: III.3.1 TOP: 10-5 95. ANS: The parents are both heterozygous, Bb. Though the odds of having a blue-eyed child are 1 in 4, it is certainly possible that two out of two children could be blue eyed (bb). The odds that both children will have blue eyes is 1/4 1/4 = 1/16 or 6.25% PTS: 1 DIF: Bloom's Level F REF: 280 NAT: LS_2b STA: III.3.1 TOP: 10-6 96. ANS: Baby #2 must belong to the Bakers because only Baby #1 can belong to the Pages. PTS: 1 DIF: Bloom's Level E REF: 299–300 NAT: LS_2b STA: III.3.1 TOP: 11-3 97. ANS: Either Baby #1 or Baby #2 could be theirs. If Mrs. Baker is /Bi and Mr. Baker is /Ai or /A/A, then their baby could be /Ai, making it Baby #1. If Mrs. Baker is /Bi and Mr. Baker is /Ai, then their baby could be ii, making it Baby #2. PTS: 1 DIF: Bloom's Level F REF: 304 NAT: LS_2b STA: III.3.1 TOP: 11-4 98. ANS: First, DNA helicase unwinds and separates the double-stranded DNA molecule like a zipper unzipping. The weak hydrogen bonds between the complementary nucleotides break and the two DNA strands separate. Then free nucleotides attach to the exposed nucleotides of the DNA strands, catalyzed by DNA polymerase. From one DNA molecule there are now two DNA molecules. Each one of the DNA molecules has a strand from the original DNA and one new strand. PTS: 1 DIF: Bloom's Level B REF: 333–335 NAT: LS_2a STA: III.3.1 TOP: 12-4 99. ANS: A codon is a three-base code for a specific amino acid. An anticodon is a tRNA triplet of nitrogen bases that bonds to a complementary codon on the messenger RNA. PTS: 1 NAT: LS_1c DIF: Bloom's Level B TOP: 12-6 REF: 338 100. ANS: Transfer RNA brings an amino acid to the ribosome in the process of translating the mRNA code into a protein. PTS: 1 DIF: Bloom's Level A NAT: LS_1c TOP: 12-6 101. ANS: CUG ACG AUG CUC AAC GGC AUA UAA CGC GAG Met Leu Asn Gly Ile STOP REF: 338 PTS: 1 DIF: Bloom's Level C REF: 328 NAT: LS_1c TOP: 12-8 102. ANS: DNA has the sugar deoxyribose while RNA has the sugar ribose. DNA is double stranded while RNA is single stranded. DNA is only found in the nucleus of euklaryotes while RNA can be found elsewhere in the cell. DNA and RNA both carry genetic information. DNA and RNA both contain A, C and G PTS: 1 103. ANS: transcription translation protein synthesis PTS: 1 104. ANS: Fossils can show evolutionary changes over time. Anatomical studies indicate evolutionary relationships. Vestigial structures indicate evolutionary pathways. Embryological development and genetic comparisons show evolution from a common ancestor. PTS: 1 DIF: Bloom's Level C NAT: LS_3c STA: III.4.1 TOP: 15-4 | 15-5 | 15-6 105. ANS: Darwin's theory of evolution by natural selection can be summarized by the following four statements. First, variations exist within populations. Second, some variations are more advantageous for survival and reproduction than others. Organisms produce more offspring than can survive. Finally, over time, offspring of survivors will make up a larger proportion of the population. PTS: 1 NAT: LS_3a 106. ANS: DIF: Bloom's Level C STA: III.4.2 TOP: 15-2 REF: 418–421 Punctuated equilibrium is an explanation for fossil records in which an organism appears almost without change over many geological strata, or many years. Then there is a sudden change in the anatomy of the organism with no evidence of in-between stages. PTS: 1 STA: II.1.1 107. ANS: DIF: Bloom's Taxonomy C TOP: 15-9 REF: 440–441 Darwin found several species of finch on the islands that were found nowhere else on Earth. Yet the finches on the Galapagos resembled finches from mainland South America. Darwin reasoned that a South American finch had come to the Galapagos Islands and over time had changed and given rise to new species on the islands that were different from each other and from the South American species. PTS: 1 DIF: Bloom's Taxonomy D REF: 418–419 NAT: LS_3d STA: II.1.4 TOP: 15-1 108. ANS: Artificial selection works, just like natural selection, by selecting certain inherited traits for greater representation in the next generation. Different traits are selected under artificial selection and natural selection, but if artificial selection works, which it does, then natural selection should also work. PTS: 1 DIF: Bloom's Taxonomy E REF: 420 NAT: LS_3a STA: II.1.4 TOP: 15-1 109. ANS: Scientists identify different kinds of fossils found in sedimentary rock layers. From the way the rock is formed, scientists can determine approximately when the organism lived and what the climate was like. By observing many different layers, conclusions can be drawn about the evolution of certain organisms from their fossils. PTS: 1 NAT: LS_3c 110. ANS: prophase DIF: Bloom's Level B TOP: 14-1 REF: 393–395 PTS: 1 DIF: Bloom's Level D REF: 248–250 NAT: LS_1d STA: III.1.1 TOP: 9-4 111. ANS: During G1, a cell is growing, carrying out normal functions, and preparing to replicate its DNA. During synthesis, the cell copies its DNA in preparation for mitosis. Finally, G2 is the stage in which the cell prepares for division of the nucleus. PTS: 1 DIF: Bloom's Level B REF: 247 NAT: LS_1d STA: III.1.1 TOP: 9-3 112. ANS: because it reveals similar patterns and structures that may indicate a shared evolutionary history PTS: 1 DIF: III OBJ: 16.2.2 PROBLEM 113. ANS: The phenotypic ratio is 1 normal body hair red eye: 1 normal body brown eye: 1 hairy body red eye: 1 hairy body brown eye. See Solution 10-2. Solution 10-2 PTS: 1 NAT: LS_2b 114. ANS: See Solution 12-6. DIF: Bloom's Level C STA: III.3.1 TOP: 10-6 PTS: 1 DIF: Bloom's Level C NAT: LS_1c | LS_2a TOP: 12-2 REF: 282 REF: 329–330 | 338